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d362a02ea6
| Author | SHA1 | Date | |
|---|---|---|---|
 Krishna Ayyalasomayajula
|
d362a02ea6 |
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@@ -1,41 +1 @@
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# LaTeX build artifacts
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*.aux
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*.log
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*.out
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*.toc
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*.lof
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*.lot
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*.fls
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*.fdb_latexmk
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*.synctex.gz
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*.ent
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*.ptc
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*.nlo
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*.nls
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*.dvi
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*.ps
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*.bbl
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*.blg
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*.run.xml
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*.sav
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*.ilg
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# PDFs except main.pdf (tracked)
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*.pdf
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!main.pdf
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# Editor swap/backup files
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*~
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.*.swp
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.*.swo
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#*#
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# macOS
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.DS_Store
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# Build directories
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build/
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_build/
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# Opencode
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.opencode/**
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@@ -1,37 +0,0 @@
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# =============================================================================
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# Makefile — Differential Equations Handbook
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# =============================================================================
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TARGET = main
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TEX = pdflatex
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BIBTEX = bibtex
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.PHONY: all quick clean view distclean
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## all: Full build with nomenclature indexing and 3-pass pdflatex
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all:
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$(TEX) -interaction=nonstopmode $(TARGET).tex
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@if [ -f $(TARGET).nlo ]; then \
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makeindex -s nomencl.ist -o $(TARGET).nls $(TARGET).nlo; \
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fi
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$(TEX) -interaction=nonstopmode $(TARGET).tex
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$(TEX) -interaction=nonstopmode $(TARGET).tex
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## quick: Single pdflatex pass for rapid iteration
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quick:
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$(TEX) -interaction=nonstopmode $(TARGET).tex
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## clean: Remove all auxiliary files
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clean:
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rm -f $(TARGET).aux $(TARGET).log $(TARGET).out $(TARGET).toc \
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$(TARGET).lof $(TARGET).lot $(TARGET).nls $(TARGET).nlo \
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$(TARGET).vrb $(TARGET).synctex.gz $(TARGET).fls $(TARGET).out \
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$(TARGET).blg $(TARGET).bbl $(TARGET).run.xml
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## view: Open the resulting PDF
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view:
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xdg-open $(TARGET).pdf
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## distclean: Clean + remove PDF
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distclean: clean
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rm -f $(TARGET).pdf
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@@ -1,408 +0,0 @@
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\appendix
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\section{Solution Method Summary}
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\label{app:solution_summary}
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This appendix provides comprehensive quick-reference tables for every solution method covered in the handbook. Use these tables to identify the correct method for a given equation type and to recall key formulas.
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\subsection{First-Order ODE Methods}
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\label{sec:appA_first_order}
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\begin{table}[htbp]
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\centering
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\caption{First-Order Solution Methods (see \cref{ch:first_order})}
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\label{tab:appA_first_order_methods}
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\begin{tabular}{p{2.5cm} p{4.5cm} p{4.5cm} p{3.5cm}}
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\toprule
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\textbf{Method} & \textbf{Equation Form} & \textbf{Solution / Procedure} & \textbf{When to Use} \\
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\midrule
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Separable &
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$\displaystyle \frac{\diff y}{\diff x} = g(x)\,h(y)$ &
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$\displaystyle \int \frac{\diff y}{h(y)} = \int g(x)\,\diff x + C$ &
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RHS factors as $g(x)\cdot h(y)$ \\[8pt]
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Linear &
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$\displaystyle y' + p(x)y = g(x)$ &
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$y = \dfrac{\displaystyle \int \mu(x)g(x)\,\diff x + C}{\mu(x)}$, \qquad $\mu = \exp\!\bigl(\int p(x)\,\diff x\bigr)$ &
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Can be written in standard linear form \\[8pt]
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Exact &
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$M(x,y)\,\diff x + N(x,y)\,\diff y = 0$, \qquad $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ &
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Find $\psi(x,y)$ such that $\dfrac{\partial \psi}{\partial x} = M$, $\dfrac{\partial \psi}{\partial y} = N$; solution is $\psi(x,y) = C$ &
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Exactness test $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ passes \\[8pt]
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Bernoulli &
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$\displaystyle y' + p(x)y = g(x)\,y^n$ &
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Substitute $v = y^{1-n}$; equation becomes linear in $v$ &
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RHS is a power of $y$ times a function of $x$ \\[8pt]
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Homogeneous &
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$\displaystyle \frac{\diff y}{\diff x} = F\!\left(\frac{y}{x}\right)$ &
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Substitute $y = vx$; equation becomes separable in $v$ &
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RHS is a function of the ratio $y/x$ \\[8pt]
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\bottomrule
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\end{tabular}
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\end{table}
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\subsection{Second-Order Linear Homogeneous (Constant Coefficients)}
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\label{sec:appA_second_order_hom}
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Consider the equation $a\,y'' + b\,y' + c\,y = 0$ with $a,b,c \in \R$, $a \neq 0$.
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\begin{table}[htbp]
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\centering
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\caption{Characteristic Equation Cases (see \cref{ch:second_order_homogeneous})}
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\label{tab:appA_homogeneous_cases}
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\begin{tabular}{p{3cm} p{3cm} p{6cm} p{3cm}}
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\toprule
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\textbf{Discriminant} & \textbf{Roots of } $ar^2+br+c=0$ & \textbf{General Solution} & \textbf{Phase / Behavior} \\
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\midrule
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$b^2 - 4ac > 0$ \small (distinct real) &
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$r_1 = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$, \ $r_2 = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$ &
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$y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ &
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Exponential growth/decay; overdamped \\[12pt]
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$b^2 - 4ac = 0$ \small (repeated) &
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$r = -\dfrac{b}{2a}$ &
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$y = c_1 e^{rx} + c_2 x\,e^{rx}$ &
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Critically damped \\[12pt]
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$b^2 - 4ac < 0$ \small (complex conjugate) &
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$r = \alpha \pm i\beta$, \qquad $\alpha = -\dfrac{b}{2a}$, \ $\beta = \dfrac{\sqrt{4ac-b^2}}{2a}$ &
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$y = e^{\alpha x}\bigl[c_1\cos(\beta x) + c_2\sin(\beta x)\bigr]$ &
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Damped oscillation; underdamped \\[8pt]
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\bottomrule
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\end{tabular}
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\end{table}
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\begin{keyresult}
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\textbf{Superposition principle.} For $L[y]=0$ linear homogeneous, any linear combination of solutions is again a solution. A \emph{fundamental set} $\{y_1,y_2\}$ has $W(y_1,y_2)\neq 0$, and the general solution is $y = c_1 y_1 + c_2 y_2$.
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\end{keyresult}
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\subsection{Second-Order Linear Nonhomogeneous}
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\label{sec:appA_second_order_nonhom}
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Consider $a\,y'' + b\,y' + c\,y = g(x)$. The general solution is $y = y_h + y_p$, where $y_h$ is the homogeneous solution and $y_p$ is a particular solution.
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\begin{table}[htbp]
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\centering
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\caption{Nonhomogeneous Solution Methods (see \cref{ch:second_order_nonhomogeneous})}
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\label{tab:appA_nonhomogeneous_methods}
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\begin{tabular}{p{3cm} p{6cm} p{4cm}}
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\toprule
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\textbf{Method} & \textbf{Formula / Procedure} & \textbf{When to Use} \\
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\midrule
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Undetermined Coefficients &
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See \cref{tab:appA_undetermined_guess} for guess table. Plug $y_p$ into ODE, solve for unknown coefficients. &
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$g(x)$ is a polynomial, exponential, sine/cosine, or finite sums/products thereof \\[12pt]
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Variation of Parameters &
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$\textstyle y_p(x) = -y_1(x)\!\int \frac{y_2(x)\,g(x)}{a\,W(x)}\,\diff x + y_2(x)\!\int \frac{y_1(x)\,g(x)}{a\,W(x)}\,\diff x$
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where $\textstyle W = y_1 y_2' - y_2 y_1'$. &
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Any $g(x)$ for which the integrals can be evaluated; requires the fundamental set $\{y_1,y_2\}$ \\[8pt]
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\bottomrule
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\end{tabular}
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\end{table}
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\begin{table}[htbp]
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\centering
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\caption{Guess Table for Undetermined Coefficients (see \cref{ch:second_order_nonhomogeneous})}
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\label{tab:appA_undetermined_guess}
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\begin{tabular}{p{4.5cm} p{5.5cm}}
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\toprule
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\textbf{Form of } $g(x)$ & \textbf{Guess for } $y_p(x)$ \\
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\midrule
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$P_n(x)$ (polynomial of degree $n$) &
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$A_n x^n + A_{n-1}x^{n-1} + \cdots + A_1 x + A_0$ \\[8pt]
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$A\,e^{kx}$ &
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$C\,e^{kx}$ \\[6pt]
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$A\sin(\omega x)$ or $A\cos(\omega x)$ &
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$C_1\cos(\omega x) + C_2\sin(\omega x)$ \\[6pt]
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$A\,e^{kx}\sin(\omega x)$ or $A\,e^{kx}\cos(\omega x)$ &
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$e^{kx}\bigl[C_1\cos(\omega x) + C_2\sin(\omega x)\bigr]$ \\[8pt]
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Sums of the above &
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Sum of the corresponding guesses \\[8pt]
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\bottomrule
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\end{tabular}
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\end{table}
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\begin{keyresult}
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\textbf{Modification rule.} If any term of the guess $y_p$ is already a solution of the homogeneous equation (i.e.\ duplicates a term in $y_h$), multiply the \emph{entire} corresponding group by $x$. If $x$-multiplication still duplicates, multiply by $x^2$, and so on.
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\end{keyresult}
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\subsection{Laplace Transform Summary}
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\label{sec:appA_laplace}
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\begin{table}[htbp]
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\centering
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\caption{Common Laplace Transforms (see \cref{ch:laplace_transforms})}
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\label{tab:appA_laplace_transforms}
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\begin{tabular}{l l}
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\toprule
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\textbf{$f(t)$} & \textbf{$\mathcal{L}\{f(t)\} = F(s)$} \\
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\midrule
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$1$ & $\dfrac{1}{s}$ \\[8pt]
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$t^n$ \quad ($n \in \N$) & $\dfrac{n!}{s^{n+1}}$ \\[8pt]
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$e^{at}$ & $\dfrac{1}{s-a}$ \\[8pt]
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$\sin(\omega t)$ & $\dfrac{\omega}{s^2+\omega^2}$ \\[8pt]
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$\cos(\omega t)$ & $\dfrac{s}{s^2+\omega^2}$ \\[8pt]
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$\sinh(at)$ & $\dfrac{a}{s^2-a^2}$ \\[8pt]
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$\cosh(at)$ & $\dfrac{s}{s^2-a^2}$ \\[8pt]
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$e^{at}\sin(\omega t)$ & $\dfrac{\omega}{(s-a)^2+\omega^2}$ \\[8pt]
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$e^{at}\cos(\omega t)$ & $\dfrac{s-a}{(s-a)^2+\omega^2}$ \\[8pt]
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$t\,\sin(\omega t)$ & $\dfrac{2\omega s}{(s^2+\omega^2)^2}$ \\[8pt]
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$t\,\cos(\omega t)$ & $\dfrac{s^2-\omega^2}{(s^2+\omega^2)^2}$ \\[8pt]
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$u(t-a)$ \small (unit step) & $\dfrac{e^{-as}}{s}$ \\[8pt]
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$\delta(t-a)$ \small (Dirac delta) & $e^{-as}$ \\[8pt]
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\bottomrule
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\end{tabular}
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\end{table}
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\begin{table}[htbp]
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\centering
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\caption{Laplace Transform Properties (see \cref{ch:laplace_transforms})}
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\label{tab:appA_laplace_properties}
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\begin{tabular}{p{3.2cm} p{7.5cm}}
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\toprule
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\textbf{Property} & \textbf{Formula} \\
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\midrule
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Linearity &
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$\mathcal{L}\{a\,f(t) + b\,g(t)\} = a\,F(s) + b\,G(s)$ \\[10pt]
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First Shift (exponential) &
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$\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ \\[10pt]
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Second Shift (time delay) &
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$\mathcal{L}\{u(t-a)\,f(t-a)\} = e^{-as}F(s)$ \\[10pt]
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Derivatives &
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$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$ \\[4pt]
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& $\mathcal{L}\{f''(t)\} = s^2 F(s) - s\,f(0) - f'(0)$ \\[4pt]
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& $\mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1}f(0) - \cdots - f^{(n-1)}(0)$ \\[10pt]
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Multiplication by $t^n$ &
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$\mathcal{L}\{t^n f(t)\} = (-1)^n \dfrac{\mathrm{d}^n}{\mathrm{d}s^n}F(s)$ \\[10pt]
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Convolution &
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$\mathcal{L}\{f * g\} = F(s)\,G(s)$, \quad $(f*g)(t) = \displaystyle\int_0^t f(\tau)\,g(t-\tau)\,\diff\tau$ \\[10pt]
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Integral &
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$\displaystyle \mathcal{L}\!\left\{\int_0^t f(\tau)\,\diff\tau\right\} = \frac{F(s)}{s}$ \\[10pt]
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\bottomrule
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\end{tabular}
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\end{table}
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\begin{keyresult}
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\textbf{Solving IVPs with Laplace.} (1) Take the Laplace transform of the ODE (using derivative formulas to incorporate initial conditions). (2) Solve the resulting algebraic equation for $Y(s)$. (3) Find $y(t) = \mathcal{L}^{-1}\{Y(s)\}$ via partial fractions and the transform table.
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\end{keyresult}
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\subsection{Systems of Linear ODEs}
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\label{sec:appA_systems}
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For $\mathbf{x}' = A\,\mathbf{x}$ with $A$ a constant $n \times n$ matrix (see \cref{ch:systems}):
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\begin{table}[htbp]
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\centering
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\caption{Eigenvalue Cases for 2$\times$2 Systems (see \cref{ch:systems})}
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\label{tab:appA_systems_cases}
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\begin{tabular}{p{2.8cm} p{3.5cm} p{6.5cm}}
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\toprule
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\textbf{Eigenvalue Case} & \textbf{Eigenvalues / Vectors} & \textbf{General Solution} \\
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\midrule
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Real distinct &
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$\lambda_1 \neq \lambda_2 \in \R$; eigenvectors $\mathbf{v}_1, \mathbf{v}_2$ &
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$\mathbf{x}(t) = c_1 e^{\lambda_1 t}\,\mathbf{v}_1 + c_2 e^{\lambda_2 t}\,\mathbf{v}_2$ \\[12pt]
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Complex conjugate &
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$\lambda = \alpha \pm i\beta$; eigenvector $\mathbf{a} \pm i\mathbf{b}$ &
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$\mathbf{x}(t) = c_1\bigl[e^{\alpha t}\cos(\beta t)\,\mathbf{a} - e^{\alpha t}\sin(\beta t)\,\mathbf{b}\bigr]$
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$+ c_2\bigl[e^{\alpha t}\sin(\beta t)\,\mathbf{a} + e^{\alpha t}\cos(\beta t)\,\mathbf{b}\bigr]$ \\[12pt]
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Repeated (diagonalizable) &
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$\lambda_1=\lambda_2=\lambda$; two independent eigenvectors $\mathbf{v}_1,\mathbf{v}_2$ &
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$\mathbf{x}(t) = c_1 e^{\lambda t}\,\mathbf{v}_1 + c_2 e^{\lambda t}\,\mathbf{v}_2$ \\[10pt]
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Repeated (defective) &
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$\lambda$ (single eigenvector $\mathbf{v}_1$; generalized eigenvector $\mathbf{v}_2$) &
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$\mathbf{x}(t) = c_1 e^{\lambda t}\,\mathbf{v}_1 + c_2 e^{\lambda t}\bigl(t\,\mathbf{v}_1 + \mathbf{v}_2\bigr)$ \\[8pt]
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\bottomrule
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\end{tabular}
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\end{table}
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\begin{keyresult}
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\textbf{Matrix exponential.} The solution to $\mathbf{x}' = A\mathbf{x},\; \mathbf{x}(0)=\mathbf{x}_0$ is $\mathbf{x}(t) = e^{At}\,\mathbf{x}_0$, where $e^{At} = \displaystyle\sum_{k=0}^{\infty} \frac{(At)^k}{k!}$. When $A$ is diagonalizable, $e^{At} = P\,e^{\Lambda t}\,P^{-1}$ with $\Lambda$ diagonal.
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\end{keyresult}
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\subsection{Series Solutions}
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\label{sec:appA_series}
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\begin{table}[htbp]
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\centering
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||||
\caption{Series Solution Methods (see \cref{ch:series_solutions})}
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\label{tab:appA_series_methods}
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\begin{tabular}{p{2.5cm} p{3.8cm} p{5.5cm} p{3.5cm}}
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||||
\toprule
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||||
\textbf{Method} & \textbf{Equation Type} & \textbf{Solution Form} & \textbf{Key Details} \\
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\midrule
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||||
Power series &
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$p(x)$ and $q(x)$ analytic at ordinary point $x_0$ in \ $y''+p(x)y'+q(x)y=0$ &
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$y = \sum_{n=0}^{\infty} a_n (x-x_0)^n$ &
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Substitute into ODE, equate coefficients to find recurrence for $a_n$ \\[12pt]
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Euler--Cauchy &
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$x^2 y'' + \alpha x y' + \beta y = 0$, \ $x > 0$ &
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Try $y = x^r$; indicial equation $r(r-1)+\alpha r+\beta = 0$ &
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Distinct roots $r_1\neq r_2$: $y=c_1 x^{r_1}+c_2 x^{r_2}$. Repeated: $y=x^r(c_1+c_2\ln x)$ \\[12pt]
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Frobenius &
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Regular singular point at $x_0$ in $y''+p(x)y'+q(x)y=0$ &
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$y = x^r\sum_{n=0}^{\infty} a_n x^n$, \quad $a_0\neq 0$ &
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Indicial equation from lowest power. Cases depend on $r_1-r_2$: non-integer, zero, or positive integer \\[8pt]
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\bottomrule
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||||
\end{tabular}
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||||
\end{table}
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\subsection{Nonlinear Systems}
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\label{sec:appA_nonlinear}
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||||
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\begin{table}[htbp]
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||||
\centering
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||||
\caption{Linearized Classification via Jacobian (see \cref{ch:nonlinear_systems})}
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\label{tab:appA_nonlinear_classification}
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||||
\begin{tabular}{p{3.2cm} p{3cm} p{3cm} p{3.8cm}}
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||||
\toprule
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||||
\textbf{Type} & \textbf{Trace } $T=\mathrm{tr}(J)$ & \textbf{Determinant } $D=\det(J)$ & \textbf{Eigenvalue Pattern} \\
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||||
\midrule
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||||
Stable node & $T<0$ & $D>0$, $T^2-4D>0$ & $\lambda_1,\lambda_2 < 0$, real \\[6pt]
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||||
Unstable node & $T>0$ & $D>0$, $T^2-4D>0$ & $\lambda_1,\lambda_2 > 0$, real \\[6pt]
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||||
Saddle point & $-$ & $D<0$ & $\lambda_1 < 0 < \lambda_2$, real \\[6pt]
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||||
Stable spiral & $T<0$ & $D>0$, $T^2-4D<0$ & $\alpha \pm i\beta$, $\alpha < 0$ \\[6pt]
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||||
Unstable spiral & $T>0$ & $D>0$, $T^2-4D<0$ & $\alpha \pm i\beta$, $\alpha > 0$ \\[6pt]
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||||
Center & $T=0$ & $D>0$ & $\pm i\beta$ (pure imaginary) \\[6pt]
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||||
\bottomrule
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||||
\end{tabular}
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||||
\end{table}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Hartman--Grobman theorem.} Near a hyperbolic equilibrium point ($\Re(\lambda_i)\neq 0$), the nonlinear system is topologically conjugate to its linearization. The phase portrait of the linearized system accurately describes the local behavior.
|
||||
\end{keyresult}
|
||||
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||||
\subsection{PDE Solution Methods}
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||||
\label{sec:appA_pde}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{PDE Methods and Solution Forms (see \cref{ch:heat_equation}, \cref{ch:wave_and_laplace})}
|
||||
\label{tab:appA_pde_methods}
|
||||
\begin{tabular}{p{3.2cm} p{2.5cm} p{2.5cm} p{6cm}}
|
||||
\toprule
|
||||
\textbf{Equation} & \textbf{Domain} & \textbf{Method} & \textbf{Solution Form} \\
|
||||
\midrule
|
||||
Heat equation \ $\displaystyle u_t = \alpha\,u_{xx}$ & $0<x<L$ &
|
||||
Separation of variables + Fourier series &
|
||||
$\displaystyle u(x,t) = \sum_{n=1}^{\infty} b_n \sin\!\Bigl(\frac{n\pi x}{L}\Bigr)
|
||||
\exp\!\Bigl(-\alpha\,\frac{n^2\pi^2}{L^2}\,t\Bigr)$ \\[12pt]
|
||||
Wave equation \ $\displaystyle u_{tt} = c^2\,u_{xx}$ & $x \in \R$ (infinite) &
|
||||
d'Alembert's formula &
|
||||
$\displaystyle u(x,t) = \tfrac{1}{2}\bigl[f(x-ct)+f(x+ct)\bigr]
|
||||
+ \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,\diff s$ \\[12pt]
|
||||
Wave equation \ $\displaystyle u_{tt} = c^2\,u_{xx}$ & $0<x<L$ (finite, Dirichlet) &
|
||||
Separation + Fourier sine series &
|
||||
$\displaystyle u(x,t) = \sum_{n=1}^{\infty} \sin\!\Bigl(\frac{n\pi x}{L}\Bigr)
|
||||
\bigl[A_n\cos(\tfrac{n\pi c t}{L}) + B_n\sin(\tfrac{n\pi c t}{L})\bigr]$ \\[12pt]
|
||||
Laplace \ $\displaystyle u_{xx}+u_{yy}=0$ & Rectangle $0<x<a$, $0<y<b$ &
|
||||
Separation of variables &
|
||||
$\displaystyle u(x,y) = \sum_{n=1}^{\infty} A_n \sin\!\Bigl(\frac{n\pi x}{a}\Bigr)
|
||||
\sinh\!\Bigl(\frac{n\pi y}{a}\Bigr)$ \\[10pt]
|
||||
Laplace \ $\displaystyle u_{rr}+\tfrac{1}{r}u_r+\tfrac{1}{r^2}u_{\theta\theta}=0$ & Disk $r<a$ &
|
||||
Polar separation &
|
||||
$\displaystyle u(r,\theta) = A_0 + \sum_{n=1}^{\infty} r^n\bigl[A_n\cos(n\theta)+B_n\sin(n\theta)\bigr]$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Separation of variables (general procedure).} For a PDE $F(x,t,u,u_x,u_t,\dots)=0$, assume $u(x,t)=X(x)T(t)$. Substitute and separate variables to obtain two ODEs, one for $X$ and one for $T$, linked by a separation constant. Solve the spatial ODE as a Sturm--Liouville eigenvalue problem (\cref{ch:boundary_value_problems}), then expand the initial/boundary data in the resulting eigenfunction basis.
|
||||
\end{keyresult}
|
||||
|
||||
\subsection{Mechanical and Electrical Applications}
|
||||
\label{sec:appA_applications}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Damping Cases for Spring-Mass Systems (see \cref{ch:mechanical_applications})}
|
||||
\label{tab:appA_damping}
|
||||
\begin{tabular}{p{3.2cm} p{2.8cm} p{4.5cm} p{4cm}}
|
||||
\toprule
|
||||
\textbf{Case} & \textbf{Condition} & \textbf{Solution $y(t)$} & \textbf{Behavior} \\
|
||||
\midrule
|
||||
Undamped & $c=0$ &
|
||||
$y(t) = C_1\cos(\omega_0 t) + C_2\sin(\omega_0 t)$, \ $\omega_0 = \sqrt{k/m}$ &
|
||||
Persistent oscillation \\[10pt]
|
||||
Underdamped & $0 < c < 2\sqrt{km}$ &
|
||||
$y(t) = e^{-ct/(2m)}\bigl[C_1\cos(\omega_d t) + C_2\sin(\omega_d t)\bigr]$, \ $\omega_d = \sqrt{\omega_0^2-(c/2m)^2}$ &
|
||||
Decaying oscillation \\[10pt]
|
||||
Critically damped & $c = 2\sqrt{km}$ &
|
||||
$y(t) = C_1 e^{-\omega_0 t} + C_2 t\,e^{-\omega_0 t}$ &
|
||||
Fastest decay without oscillation \\[10pt]
|
||||
Overdamped & $c > 2\sqrt{km}$ &
|
||||
$y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$, \ $r_{1,2} = \dfrac{-c \pm \sqrt{c^2-4km}}{2m}$ &
|
||||
Slow non-oscillatory decay \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Mechanical--Electrical Analogy (see \cref{ch:mechanical_applications})}
|
||||
\label{tab:appA_mech_elec}
|
||||
\begin{tabular}{l l l l}
|
||||
\toprule
|
||||
\textbf{Mechanical} & \textbf{Electrical (charge)} & \textbf{Mechanical} & \textbf{Electrical (flux)} \\
|
||||
\midrule
|
||||
Mass $m$ & Inductance $L$ & Mass $m$ & Capacitance $C$ \\
|
||||
Damping $c$ & Resistance $R$ & Damping $c$ & Conductance $G$ \\
|
||||
Spring constant $k$ & Inverse capacitance $1/C$ & Spring constant $k$ & Inverse inductance $1/L$ \\
|
||||
Force $F(t)$ & Voltage $E(t)$ & Force $F(t)$ & Current $I(t)$ \\
|
||||
Displacement $y(t)$ & Charge $q(t)$ & Velocity $v(t)$ & Flux $\phi(t)$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Qualitative Analysis and Numerical Methods}
|
||||
\label{sec:appA_qualitative}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Equilibrium Classification for $y' = f(y)$ (see \cref{ch:qualitative})}
|
||||
\label{tab:appA_equilibrium}
|
||||
\begin{tabular}{p{3.5cm} p{4cm} p{4.5cm}}
|
||||
\toprule
|
||||
\textbf{Condition at } $y^*$ (where $f(y^*)=0$) & \textbf{Stability} & \textbf{Phase Line Behavior} \\
|
||||
\midrule
|
||||
$f'(y^*) < 0$ & Asymptotically stable (sink) & Solutions near $y^*$ converge to $y^*$ \\[6pt]
|
||||
$f'(y^*) > 0$ & Unstable (source) & Solutions near $y^*$ diverge away from $y^*$ \\[6pt]
|
||||
$f'(y^*) = 0$ & Inconclusive (semi-stable or higher-order) & Requires higher-order analysis; may be a node or saddle \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Euler's Method for Numerical Approximation (see \cref{ch:qualitative})}
|
||||
\label{tab:appA_euler}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Algorithm} & \\
|
||||
\midrule
|
||||
Given $y' = f(t,y)$, $y(t_0) = y_0$, step size $h$: \\[6pt]
|
||||
\quad $t_{n+1} = t_n + h$ \\[4pt]
|
||||
\quad $y_{n+1} = y_n + h\,f(t_n, y_n)$ \\[10pt]
|
||||
Local truncation error: $\mathcal{O}(h^2)$ \\
|
||||
Global error: $\mathcal{O}(h)$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Fourier Series Summary}
|
||||
\label{sec:appA_fourier}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Fourier Series Coefficients (see \cref{ch:fourier_series})}
|
||||
\label{tab:appA_fourier}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Series} & \textbf{Coefficients} \\
|
||||
\midrule
|
||||
Full range $[-L,L]$: \ $f(x) \sim \dfrac{a_0}{2} + \displaystyle\sum_{n=1}^{\infty} \Bigl[a_n\cos(\tfrac{n\pi x}{L}) + b_n\sin(\tfrac{n\pi x}{L})\Bigr]$ &
|
||||
$\displaystyle a_0 = \frac{1}{L}\int_{-L}^{L} f(x)\,\diff x$ \\[8pt]
|
||||
& $\displaystyle a_n = \frac{1}{L}\int_{-L}^{L} f(x)\cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[8pt]
|
||||
& $\displaystyle b_n = \frac{1}{L}\int_{-L}^{L} f(x)\sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[10pt]
|
||||
Even $f$ (cosine series): & $b_n = 0$, \ $\displaystyle a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[10pt]
|
||||
Odd $f$ (sine series): & $a_n = 0$, \ $\displaystyle b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
@@ -1,154 +0,0 @@
|
||||
% =============================================================================
|
||||
% appB_transform_tables.tex
|
||||
% Appendix B: Transform and Integral Tables
|
||||
% =============================================================================
|
||||
|
||||
\section{Transform and Integral Tables}
|
||||
\label{app:transform_tables}
|
||||
|
||||
This appendix provides comprehensive reference tables for Laplace transforms (\cref{ch:laplace_transforms}), Fourier series formulas (\cref{ch:fourier_series}), and common integrals used throughout the handbook. All entries are presented without derivation; see the referenced chapters for proofs and worked examples.
|
||||
|
||||
\subsection{Laplace Transform Table}
|
||||
\label{sec:appB_laplace_table}
|
||||
|
||||
The Laplace transform is defined in \cref{eq:laplace_definition} as
|
||||
\[
|
||||
\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\,\diff t,
|
||||
\]
|
||||
provided the integral converges. The following table lists the most frequently encountered transforms.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Common Laplace Transforms}
|
||||
\label{tab:laplace_transforms}
|
||||
\begin{tabular}{l l p{3cm}}
|
||||
\toprule
|
||||
\textbf{$f(t)$} & \textbf{$\mathcal{L}\{f(t)\} = F(s)$} & \textbf{Conditions} \\
|
||||
\midrule
|
||||
$1$ & $\dfrac{1}{s}$ & $s > 0$ \\[8pt]
|
||||
$t$ & $\dfrac{1}{s^{2}}$ & $s > 0$ \\[8pt]
|
||||
$t^{n}$ & $\dfrac{n!}{s^{n+1}}$ & $s > 0,\; n \in \N$ \\[8pt]
|
||||
$e^{at}$ & $\dfrac{1}{s-a}$ & $s > a$ \\[8pt]
|
||||
$t\,e^{at}$ & $\dfrac{1}{(s-a)^{2}}$ & $s > a$ \\[8pt]
|
||||
$\sin(bt)$ & $\dfrac{b}{s^{2}+b^{2}}$ & $s > 0$ \\[8pt]
|
||||
$\cos(bt)$ & $\dfrac{s}{s^{2}+b^{2}}$ & $s > 0$ \\[8pt]
|
||||
$\sinh(bt)$ & $\dfrac{b}{s^{2}-b^{2}}$ & $s > |b|$ \\[8pt]
|
||||
$\cosh(bt)$ & $\dfrac{s}{s^{2}-b^{2}}$ & $s > |b|$ \\[8pt]
|
||||
$t\,\sin(bt)$ & $\dfrac{2bs}{(s^{2}+b^{2})^{2}}$ & $s > 0$ \\[8pt]
|
||||
$t\,\cos(bt)$ & $\dfrac{s^{2}-b^{2}}{(s^{2}+b^{2})^{2}}$ & $s > 0$ \\[8pt]
|
||||
$e^{at}\sin(bt)$ & $\dfrac{b}{(s-a)^{2}+b^{2}}$ & $s > a$ \\[8pt]
|
||||
$e^{at}\cos(bt)$ & $\dfrac{s-a}{(s-a)^{2}+b^{2}}$ & $s > a$ \\[8pt]
|
||||
$u_{c}(t) \text{ (unit step)}$ & $\dfrac{e^{-cs}}{s}$ & $s > 0$ \\[8pt]
|
||||
$\delta(t-a) \text{ (Dirac)}$ & $e^{-as}$ & $a \ge 0$ \\[8pt]
|
||||
$t^{n}e^{at}$ & $\dfrac{n!}{(s-a)^{n+1}}$ & $s > a,\; n \in \N$ \\[8pt]
|
||||
$\sin^{2}(bt)$ & $\dfrac{2b^{2}}{s(s^{2}+4b^{2})}$ & $s > 0$ \\[8pt]
|
||||
$\cos^{2}(bt)$ & $\dfrac{s^{2}+2b^{2}}{s(s^{2}+4b^{2})}$ & $s > 0$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Laplace Transform Properties}
|
||||
\label{sec:appB_laplace_properties}
|
||||
|
||||
The following algebraic and operational properties make the Laplace transform a powerful tool for solving linear differential equations.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Laplace Transform Properties}
|
||||
\label{tab:laplace_properties}
|
||||
\begin{tabular}{@{}l p{12cm}@{}}
|
||||
\toprule
|
||||
\textbf{Property} & \textbf{Formula} \\
|
||||
\midrule
|
||||
Linearity & $\mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s)$ \\[6pt]
|
||||
First shift & $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ \\[6pt]
|
||||
Second shift & $\mathcal{L}\{u_{c}(t)\,f(t-c)\} = e^{-cs}F(s)$ \\[6pt]
|
||||
Derivative & $\mathcal{L}\{f'(t)\} = s F(s) - f(0)$ \\[6pt]
|
||||
Second derivative & $\mathcal{L}\{f''(t)\} = s^{2}F(s) - s f(0) - f'(0)$ \\[6pt]
|
||||
$n$th derivative & $\mathcal{L}\{f^{(n)}(t)\} = s^{n}F(s) - s^{n-1}f(0) - \cdots - f^{(n-1)}(0)$ \\[6pt]
|
||||
Integral & $\displaystyle \mathcal{L}\!\left\{\int_{0}^{t} f(\tau)\,\diff\tau\right\} = \frac{F(s)}{s}$ \\[8pt]
|
||||
Convolution & $\mathcal{L}\{(f*g)(t)\} = F(s)\,G(s)$ \\[6pt]
|
||||
$t$-multiplication & $\mathcal{L}\{t^{n}f(t)\} = (-1)^{n}\dfrac{\mathrm{d}^{n}F}{\mathrm{d}s^{n}}$ \\[8pt]
|
||||
Frequency differentiation & $\mathcal{L}\{t\,f(t)\} = -\dfrac{\mathrm{d}F}{\mathrm{d}s}$ \\[8pt]
|
||||
Final value theorem & $\displaystyle \lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s)$ \\[8pt]
|
||||
Initial value theorem & $\displaystyle \lim_{t\to 0^{+}} f(t) = \lim_{s\to\infty} sF(s)$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
The convolution integral $(f*g)(t)$ appearing in the table above is defined by
|
||||
\[
|
||||
(f*g)(t) = \int_0^t f(t-\tau)\,g(\tau)\,\diff\tau.
|
||||
\]
|
||||
|
||||
\subsection{Fourier Series Formulas}
|
||||
\label{sec:appB_fourier_series}
|
||||
|
||||
Fourier series decompose a periodic function $f(x)$ of period $2L$ into sine and cosine harmonics, as developed in \cref{ch:fourier_series}. The following tables collect the most essential formulas.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Fourier Series on $[-L, L]$}
|
||||
\label{tab:fourier_series}
|
||||
\renewcommand{\arraystretch}{1.3}
|
||||
\resizebox{\textwidth}{!}{%
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Formula} & \textbf{Expression} \\
|
||||
\midrule
|
||||
Full series & $\displaystyle f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \Bigl[a_n \cos\!\Bigl(\frac{n\pi x}{L}\Bigr) + b_n \sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\Bigr]$ \\[10pt]
|
||||
$a_0$ & $\displaystyle a_0 = \frac{1}{L}\int_{-L}^{L} f(x)\,\diff x$ \\[8pt]
|
||||
$a_n$ & $\displaystyle a_n = \frac{1}{L}\int_{-L}^{L} f(x)\cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[8pt]
|
||||
$b_n$ & $\displaystyle b_n = \frac{1}{L}\int_{-L}^{L} f(x)\sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[8pt]
|
||||
Half-range cosine & $\displaystyle a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[8pt]
|
||||
Half-range sine & $\displaystyle b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x$ \\[8pt]
|
||||
Complex coefficients & $\displaystyle c_n = \frac{1}{2L}\int_{-L}^{L} f(x)\,e^{-i n\pi x/L}\,\diff x$ \\[8pt]
|
||||
Complex series & $\displaystyle f(x) = \sum_{n=-\infty}^{\infty} c_n\,e^{i n\pi x/L}$ \\[8pt]
|
||||
Parseval's identity & $\displaystyle \frac{1}{L}\int_{-L}^{L} |f(x)|^{2}\,\diff x = \frac{a_0^{2}}{2} + \sum_{n=1}^{\infty} \bigl(a_n^{2} + b_n^{2}\bigr)$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}%
|
||||
}
|
||||
\end{table}
|
||||
|
||||
\subsection{Common Integral Table}
|
||||
\label{sec:appB_common_integrals}
|
||||
|
||||
The following integrals are used throughout the handbook, particularly in separation of variables, integrating factor methods, and Fourier coefficient calculations.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Common Indefinite Integrals}
|
||||
\label{tab:common_integrals}
|
||||
\begin{tabular}{@{}l p{11.5cm}@{}}
|
||||
\toprule
|
||||
\textbf{Integrand} & \textbf{Result} \\
|
||||
\midrule
|
||||
$\displaystyle \int x^{n}\,\diff x$ & $\displaystyle \frac{x^{n+1}}{n+1} + C \qquad (n \neq -1)$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{x}\,\diff x$ & $\displaystyle \ln|x| + C$ \\[8pt]
|
||||
$\displaystyle \int e^{ax}\,\diff x$ & $\displaystyle \frac{1}{a}e^{ax} + C$ \\[8pt]
|
||||
$\displaystyle \int \sin(ax)\,\diff x$ & $\displaystyle -\frac{1}{a}\cos(ax) + C$ \\[8pt]
|
||||
$\displaystyle \int \cos(ax)\,\diff x$ & $\displaystyle \frac{1}{a}\sin(ax) + C$ \\[8pt]
|
||||
$\displaystyle \int \sec^{2}(ax)\,\diff x$ & $\displaystyle \frac{1}{a}\tan(ax) + C$ \\[8pt]
|
||||
$\displaystyle \int \csc^{2}(ax)\,\diff x$ & $\displaystyle -\frac{1}{a}\cot(ax) + C$ \\[8pt]
|
||||
$\displaystyle \int \tan(ax)\,\diff x$ & $\displaystyle -\frac{1}{a}\ln|\cos(ax)| + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{\sqrt{a^{2}-x^{2}}}\,\diff x$ & $\displaystyle \arcsin\!\Bigl(\frac{x}{a}\Bigr) + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{a^{2}+x^{2}}\,\diff x$ & $\displaystyle \frac{1}{a}\arctan\!\Bigl(\frac{x}{a}\Bigr) + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{x\sqrt{x^{2}-a^{2}}}\,\diff x$ & $\displaystyle \frac{1}{a}\,\mathrm{arcsec}\!\Bigl(\frac{|x|}{a}\Bigr) + C$ \\[8pt]
|
||||
$\displaystyle \int \sqrt{a^{2}-x^{2}}\,\diff x$ & $\displaystyle \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\arcsin\!\Bigl(\frac{x}{a}\Bigr) + C$ \\[10pt]
|
||||
$\displaystyle \int e^{ax}\sin(bx)\,\diff x$ & $\displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\bigl(a\sin(bx)-b\cos(bx)\bigr) + C$ \\[10pt]
|
||||
$\displaystyle \int e^{ax}\cos(bx)\,\diff x$ & $\displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\bigl(a\cos(bx)+b\sin(bx)\bigr) + C$ \\[10pt]
|
||||
$\displaystyle \int x e^{ax}\,\diff x$ & $\displaystyle \frac{e^{ax}}{a^{2}}\bigl(ax-1\bigr) + C$ \\[8pt]
|
||||
$\displaystyle \int x \sin(ax)\,\diff x$ & $\displaystyle \frac{\sin(ax)}{a^{2}} - \frac{x\cos(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int x \cos(ax)\,\diff x$ & $\displaystyle \frac{\cos(ax)}{a^{2}} + \frac{x\sin(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \ln(x)\,\diff x$ & $\displaystyle x\ln(x) - x + C$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Integration by parts reminder.} For products of functions that appear in Fourier coefficient computations:
|
||||
\[
|
||||
\int u\,\mathrm{d}v = uv - \int v\,\mathrm{d}u.
|
||||
\]
|
||||
This formula underlies the $x e^{ax}$, $x \sin(ax)$, and $x \cos(ax)$ entries in \cref{tab:common_integrals} and is essential for evaluating Fourier coefficients of piecewise-linear functions in \cref{ch:fourier_series}.
|
||||
\end{keyresult}
|
||||
@@ -1,139 +0,0 @@
|
||||
\section{Integral Tables}
|
||||
\label{app:integral_tables}
|
||||
|
||||
This appendix provides commonly used integral formulas encountered throughout the handbook. See \cref{ch:first_order} for separable equations, \cref{ch:second_order_homogeneous} and \cref{ch:second_order_nonhomogeneous} for second-order ODE techniques, \cref{ch:laplace_transforms} for Laplace-domain integration, and \cref{ch:fourier_series} for trigonometric integrals in series expansions.
|
||||
|
||||
\subsection{Basic Integrals}
|
||||
\label{sec:appC_basic}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Basic Integral Formulas}
|
||||
\label{tab:integrals_basic}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Integrand} & \textbf{Result} \\
|
||||
\midrule
|
||||
$\displaystyle \int x^n \, dx$ & $\displaystyle \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{x} \, dx$ & $\displaystyle \ln|x| + C$ \\[8pt]
|
||||
$\displaystyle \int e^{ax} \, dx$ & $\displaystyle \frac{e^{ax}}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int a^x \, dx$ & $\displaystyle \frac{a^x}{\ln(a)} + C$ \\[8pt]
|
||||
$\displaystyle \int \sin(ax) \, dx$ & $\displaystyle -\frac{\cos(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \cos(ax) \, dx$ & $\displaystyle \frac{\sin(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \tan(ax) \, dx$ & $\displaystyle -\frac{\ln|\cos(ax)|}{a} + C = \frac{\ln|\sec(ax)|}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \sec^2(ax) \, dx$ & $\displaystyle \frac{\tan(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \csc^2(ax) \, dx$ & $\displaystyle -\frac{\cot(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \sec(ax)\tan(ax) \, dx$ & $\displaystyle \frac{\sec(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \csc(ax)\cot(ax) \, dx$ & $\displaystyle -\frac{\csc(ax)}{a} + C$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Integrals Involving Exponential and Trigonometric Functions}
|
||||
\label{sec:appC_exptig}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Exponential--Trigonometric Integrals}
|
||||
\label{tab:integrals_exptig}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Integrand} & \textbf{Result} \\
|
||||
\midrule
|
||||
$\displaystyle \int e^{ax}\sin(bx)\,dx$ & $\displaystyle \frac{e^{ax}\bigl(a\sin(bx) - b\cos(bx)\bigr)}{a^2 + b^2} + C$ \\[8pt]
|
||||
$\displaystyle \int e^{ax}\cos(bx)\,dx$ & $\displaystyle \frac{e^{ax}\bigl(a\cos(bx) + b\sin(bx)\bigr)}{a^2 + b^2} + C$ \\[8pt]
|
||||
$\displaystyle \int x e^{ax}\,dx$ & $\displaystyle e^{ax}\!\left(\frac{x}{a} - \frac{1}{a^2}\right) + C$ \\[8pt]
|
||||
$\displaystyle \int x \sin(ax)\,dx$ & $\displaystyle \frac{\sin(ax)}{a^2} - \frac{x\cos(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int x \cos(ax)\,dx$ & $\displaystyle \frac{\cos(ax)}{a^2} + \frac{x\sin(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \ln(x)\,dx$ & $\displaystyle x\ln(x) - x + C$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Integrals Involving Inverse Trigonometric Functions}
|
||||
\label{sec:appC_inverse_trig}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Inverse Trigonometric Integrals}
|
||||
\label{tab:integrals_inverse_trig}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Integrand} & \textbf{Result} \\
|
||||
\midrule
|
||||
$\displaystyle \int \frac{1}{\sqrt{a^2 - x^2}}\,dx$ & $\displaystyle \arcsin\!\left(\frac{x}{a}\right) + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{a^2 + x^2}\,dx$ & $\displaystyle \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{x\sqrt{x^2 - a^2}}\,dx$ & $\displaystyle \frac{1}{a}\,\text{arcsec}\!\left(\frac{|x|}{a}\right) + C$ \\[8pt]
|
||||
$\displaystyle \int \sqrt{a^2 - x^2}\,dx$ & $\displaystyle \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\!\left(\frac{x}{a}\right) + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{\sqrt{x^2 + a^2}}\,dx$ & $\displaystyle \ln\!\bigl|x + \sqrt{x^2 + a^2}\,\bigr| + C = \text{arcsinh}\!\left(\frac{x}{a}\right) + C$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Hyperbolic Integrals}
|
||||
\label{sec:appC_hyperbolic}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Hyperbolic Integral Formulas}
|
||||
\label{tab:integrals_hyperbolic}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Integrand} & \textbf{Result} \\
|
||||
\midrule
|
||||
$\displaystyle \int \sinh(ax)\,dx$ & $\displaystyle \frac{\cosh(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \cosh(ax)\,dx$ & $\displaystyle \frac{\sinh(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \tanh(ax)\,dx$ & $\displaystyle \frac{\ln|\cosh(ax)|}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \text{sech}^2(ax)\,dx$ & $\displaystyle \frac{\tanh(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \text{csch}^2(ax)\,dx$ & $\displaystyle -\frac{\coth(ax)}{a} + C$ \\[8pt]
|
||||
$\displaystyle \int \text{sech}(ax)\tanh(ax)\,dx$ & $\displaystyle -\frac{\text{sech}(ax)}{a} + C$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Additional Common Integrals}
|
||||
\label{sec:appC_additional}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Additional Common Integral Formulas}
|
||||
\label{tab:integrals_additional}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Integrand} & \textbf{Result} \\
|
||||
\midrule
|
||||
$\displaystyle \int \frac{1}{x^2 - a^2}\,dx$ & $\displaystyle \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{1}{a^2 - x^2}\,dx$ & $\displaystyle \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{x}{\sqrt{a^2 + x^2}}\,dx$ & $\displaystyle \sqrt{a^2 + x^2} + C$ \\[8pt]
|
||||
$\displaystyle \int \frac{dx}{x\sqrt{a^2 + x^2}}$ & $\displaystyle -\frac{1}{a}\ln\left|\frac{a + \sqrt{a^2 + x^2}}{|x|}\right| + C$ \\[8pt]
|
||||
$\displaystyle \int \sqrt{x^2 + a^2}\,dx$ & $\displaystyle \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\!\bigl|x + \sqrt{x^2 + a^2}\,\bigr| + C$ \\[8pt]
|
||||
$\displaystyle \int \sqrt{x^2 - a^2}\,dx$ & $\displaystyle \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\!\bigl|x + \sqrt{x^2 - a^2}\,\bigr| + C$ \\[8pt]
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\subsection{Integration by Parts Formula}
|
||||
\label{sec:appC_parts}
|
||||
|
||||
The method of integration by parts, frequently used in solving ODEs (\cref{ch:first_order}), is based on the product rule:
|
||||
\begin{equation}
|
||||
\label{eq:integration_by_parts}
|
||||
\int u \, dv = uv - \int v \, du.
|
||||
\end{equation}
|
||||
Choose $u$ and $dv$ so that $\int v \, du$ is simpler than the original integral.
|
||||
|
||||
\subsection{Reduction Formulas}
|
||||
\label{sec:appC_reduction}
|
||||
|
||||
Reduction formulas allow integrals of higher powers to be expressed in terms of lower powers:
|
||||
|
||||
\begin{align}
|
||||
\label{eq:reduction_sin}
|
||||
\int \sin^n(x) \, dx &= -\frac{1}{n}\sin^{n-1}(x)\cos(x) + \frac{n-1}{n}\int \sin^{n-2}(x) \, dx, \\
|
||||
\label{eq:reduction_cos}
|
||||
\int \cos^n(x) \, dx &= \frac{1}{n}\cos^{n-1}(x)\sin(x) + \frac{n-1}{n}\int \cos^{n-2}(x) \, dx, \\
|
||||
\label{eq:reduction_xe}
|
||||
\int x^n e^{ax} \, dx &= \frac{x^n e^{ax}}{a} - \frac{n}{a}\int x^{n-1}e^{ax} \, dx.
|
||||
\end{align}
|
||||
|
||||
These formulas are especially useful when solving higher-order ODEs (\cref{ch:second_order_homogeneous,ch:second_order_nonhomogeneous}) and evaluating Fourier coefficients (\cref{ch:fourier_series}).
|
||||
@@ -1,37 +0,0 @@
|
||||
\section{Notation Glossary}
|
||||
\label{app:notation}
|
||||
|
||||
This appendix provides a glossary of notation used throughout the handbook.
|
||||
|
||||
\nomenclature{$y$}{Dependent variable; unknown function}
|
||||
\nomenclature{$y(t)$}{Dependent variable as a function of time $t$}
|
||||
\nomenclature{$y(x)$}{Dependent variable as a function of spatial variable $x$}
|
||||
\nomenclature{$x$}{Independent variable (spatial)}
|
||||
\nomenclature{$t$}{Independent variable (time)}
|
||||
\nomenclature{$\lambda$}{Eigenvalue; parameter in characteristic equation}
|
||||
\nomenclature{$\omega$}{Angular frequency}
|
||||
\nomenclature{$\omega_0$}{Natural angular frequency}
|
||||
\nomenclature{$\alpha, \beta, \gamma$}{General constants; damping ratio}
|
||||
\nomenclature{$\delta$}{Dirac delta function}
|
||||
\nomenclature{$\mu$}{Separation constant; parameter}
|
||||
\nomenclature{$\theta$}{Angle; phase shift}
|
||||
\nomenclature{$\phi$}{Eigenfunction; angle}
|
||||
\nomenclature{$A, B, C$}{General constants of integration}
|
||||
\nomenclature{$a, b, c$}{Coefficients in differential equations}
|
||||
\nomenclature{$n, m, k$}{Integer indices}
|
||||
\nomenclature{$f(t)$}{Input/forcing function}
|
||||
\nomenclature{$F(s)$}{Laplace transform of $f(t)$}
|
||||
\nomenclature{$\mathcal{L}$}{Laplace transform operator}
|
||||
\nomenclature{$\mathcal{L}^{-1}$}{Inverse Laplace transform}
|
||||
\nomenclature{$\mathcal{F}$}{Fourier transform operator}
|
||||
\nomenclature{$u(t)$}{Unit step (Heaviside) function}
|
||||
\nomenclature{$y_h$}{Homogeneous solution}
|
||||
\nomenclature{$y_p$}{Particular solution}
|
||||
\nomenclature{$W(y_1, y_2)$}{Wronskian of $y_1$ and $y_2$}
|
||||
\nomenclature{$\mathbf{A}$}{Coefficient matrix in systems}
|
||||
\nomenclature{$\mathbf{x}(t)$}{State vector}
|
||||
\nomenclature{$I_n$}{Identity matrix of size $n$}
|
||||
\nomenclature{$e^{\mathbf{A}t}$}{Matrix exponential}
|
||||
\nomenclature{$\mathbf{u}, \mathbf{v}$}{Vectors in phase plane}
|
||||
\nomenclature{$u(t-a)$}{Shifted unit step function}
|
||||
\nomenclature{$*$}{Convolution operator}
|
||||
@@ -1,433 +0,0 @@
|
||||
\section{Classification of Differential Equations}
|
||||
\label{sec:ch01_classification}
|
||||
\label{ch:introduction}
|
||||
|
||||
A \textbf{differential equation} (DE) is a relation between an unknown function and its derivatives. Differential equations are the language of dynamical systems, appearing in physics, engineering, biology, economics, and throughout the applied sciences.
|
||||
|
||||
\subsection{Order and Degree}
|
||||
|
||||
\begin{definition}[Order]
|
||||
The \textbf{order} of a differential equation is the order of the highest derivative appearing in the equation.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Degree]
|
||||
The \textbf{degree} of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial in its derivatives.
|
||||
\end{definition}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Determine the order and degree of the following equations:}
|
||||
|
||||
\begin{enumerate}
|
||||
\item $\displaystyle \frac{\diff^3 y}{\diff x^3} + 2\frac{\diff y}{\diff x} = x$
|
||||
|
||||
The highest derivative is $y'''$, so this is a \textbf{third-order} equation. The highest derivative appears to the first power, so the \textbf{degree is 1}.
|
||||
|
||||
\item $\displaystyle \left(\frac{\diff^2 y}{\diff x^2}\right)^{\!2} + \frac{\diff y}{\diff x} = 0$
|
||||
|
||||
The highest derivative is $y''$, so this is \textbf{second-order}. The highest derivative is squared, so the \textbf{degree is 2}.
|
||||
|
||||
\item $\displaystyle \frac{\diff y}{\diff x} = e^{y} + \sin x$
|
||||
|
||||
Highest derivative is $y'$, so \textbf{first-order}. It appears to the first power, so \textbf{degree 1}.
|
||||
\end{enumerate}
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Linearity}
|
||||
|
||||
\begin{definition}[Linearity]
|
||||
An $n$-th order differential equation is \textbf{linear} if the unknown function $y$ and all of its derivatives appear to the first power and are not multiplied together. A linear $n$-th order DE has the general form
|
||||
\[
|
||||
a_n(x)\,y^{(n)} + a_{n-1}(x)\,y^{(n-1)} + \cdots + a_1(x)\,y' + a_0(x)\,y = g(x),
|
||||
\]
|
||||
where $a_0, a_1, \dots, a_n$ and $g$ are functions of $x$ only.
|
||||
\end{definition}
|
||||
|
||||
If any term violates these conditions, the equation is \textbf{nonlinear}.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Classify as linear or nonlinear:}
|
||||
|
||||
\begin{enumerate}
|
||||
\item $y'' + 3y' + 2y = \cos x$
|
||||
|
||||
All derivatives appear to first power; coefficients depend only on $x$. \textbf{Linear.}
|
||||
|
||||
\item $y'' + y\,y' + y = 0$
|
||||
|
||||
The term $y\,y'$ multiplies $y$ by its derivative. \textbf{Nonlinear.}
|
||||
|
||||
\item $\sin(y) + y' = x$
|
||||
|
||||
The function $y$ appears inside a transcendental function. \textbf{Nonlinear.}
|
||||
\end{enumerate}
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Homogeneity}
|
||||
|
||||
\begin{definition}[Homogeneity (linear DEs)]
|
||||
A linear differential equation is \textbf{homogeneous} if $g(x) = 0$; i.e., every term contains $y$ or a derivative of $y$. It is \textbf{non-homogeneous} (or inhomogeneous) if $g(x) \neq 0$.
|
||||
\end{definition}
|
||||
|
||||
\begin{workedexample}
|
||||
\begin{enumerate}
|
||||
\item $y'' - 4y = 0$ \quad $\longrightarrow$ \textbf{Homogeneous linear, second-order.}
|
||||
\item $y'' - 4y = e^{2x}$ \quad $\longrightarrow$ \textbf{Non-homogeneous linear, second-order.}
|
||||
\item $y' + y^2 = 0$ \quad $\longrightarrow$ \textbf{Nonlinear} (not classified as homogeneous/heterogeneous in the linear sense).
|
||||
\end{enumerate}
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Classification Summary}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{ODE Classification Checklist:}
|
||||
\begin{enumerate}
|
||||
\item \textbf{Order:} What is the highest derivative? (e.g., $y^{(3)}$ $\Rightarrow$ third order)
|
||||
\item \textbf{Degree:} What power is the highest derivative raised to?
|
||||
\item \textbf{Linearity:} Does $y$ and every derivative appear only to the first power, never multiplied together?
|
||||
\item \textbf{Homogeneity (linear only):} Is $g(x) = 0$? If yes, homogeneous.
|
||||
\end{enumerate}
|
||||
\end{keyresult}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Classification of example differential equations}
|
||||
\label{tab:ch01_classification_table}
|
||||
\begin{tabular}{@{} l l c c c @{}}
|
||||
\toprule
|
||||
\textbf{Equation} & \textbf{Order} & \textbf{Degree} & \textbf{Linear?} & \textbf{Homogeneous?} \\
|
||||
\midrule
|
||||
$y' + 2y = e^x$ & 1 & 1 & Yes & No \\
|
||||
$y'' - 3y' + 2y = 0$ & 2 & 1 & Yes & Yes \\
|
||||
$(y'')^2 + y' = 0$ & 2 & 2 & No & --- \\
|
||||
$y' = xy + y^2$ & 1 & 1 & No & --- \\
|
||||
$y^{(4)} + y'' = 0$ & 4 & 1 & Yes & Yes \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
The remainder of this handbook develops solution methods organized by these classifications; see \cref{ch:first_order} for first-order techniques and \cref{ch:qualitative} for qualitative methods.
|
||||
|
||||
% =============================================================================
|
||||
% INITIAL AND BOUNDARY VALUE PROBLEMS
|
||||
% =============================================================================
|
||||
|
||||
\section{Initial and Boundary Value Problems}
|
||||
\label{sec:ch01_ivp_bvp}
|
||||
|
||||
A differential equation by itself typically admits a family of solutions. Additional conditions—initial or boundary values—restrict the family to a specific solution relevant to the physical or mathematical problem.
|
||||
|
||||
\subsection{Initial Value Problems (IVP)}
|
||||
|
||||
\begin{definition}[Initial Value Problem]
|
||||
An \textbf{initial value problem} (IVP) consists of a differential equation together with conditions specifying the value of the unknown function and/or its derivatives at a single point $x_0$:
|
||||
\[
|
||||
y^{(n)} = f(x, y, y', \dots, y^{(n-1)}), \qquad
|
||||
y(x_0) = y_0,\; y'(x_0) = y_0',\; \dots,\; y^{(n-1)}(x_0) = y_0^{(n-1)}.
|
||||
\]
|
||||
An $n$-th order IVP requires $n$ initial conditions.
|
||||
\end{definition}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{IVP examples:}
|
||||
|
||||
\begin{enumerate}
|
||||
\item First-order IVP:
|
||||
\[
|
||||
\frac{\diff y}{\diff x} = 2x, \qquad y(0) = 3.
|
||||
\]
|
||||
Integrating: $y(x) = x^2 + C$. The condition $y(0) = 3$ gives $C = 3$, so $y(x) = x^2 + 3$.
|
||||
|
||||
\item Second-order IVP:
|
||||
\[
|
||||
\frac{\diff^2 y}{\diff x^2} = 6x, \qquad y(0) = 1,\; y'(0) = -2.
|
||||
\]
|
||||
Integrating twice: $y'(x) = 3x^2 + C_1$, $y(x) = x^3 + C_1 x + C_2$. The conditions give $C_1 = -2$ and $C_2 = 1$, so $y(x) = x^3 - 2x + 1$.
|
||||
\end{enumerate}
|
||||
\end{workedexample}
|
||||
|
||||
IVPs model \textbf{evolution} problems: given the state of a system at an initial time, predict its future behavior. Typical applications include population growth, radioactive decay, projectile motion, and electrical circuits driven by initial charge/charge rate.
|
||||
|
||||
\subsection{Boundary Value Problems (BVP)}
|
||||
|
||||
\begin{definition}[Boundary Value Problem]
|
||||
A \textbf{boundary value problem} (BVP) consists of a differential equation together with conditions at two or more distinct points:
|
||||
\[
|
||||
y^{(n)} = f(x, y, y', \dots, y^{(n-1)}), \qquad
|
||||
y(a) = \alpha,\; y(b) = \beta, \; \dots
|
||||
\]
|
||||
\end{definition}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Second-order BVP:}
|
||||
\[
|
||||
\frac{\diff^2 y}{\diff x^2} = 0, \qquad y(0) = 0,\; y(1) = 1.
|
||||
\]
|
||||
Integrating: $y(x) = C_1 x + C_2$. The conditions give $C_2 = 0$ and $C_1 = 1$, so $y(x) = x$.
|
||||
\end{workedexample}
|
||||
|
||||
BVPs typically arise in \textbf{steady-state} or \textbf{spatial} problems: heat distribution along a rod, deflection of a beam, or electrostatic potential between electrodes. Unlike IVPs, BVPs may have no solution, a unique solution, or infinitely many solutions.
|
||||
|
||||
\subsection{Comparison}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{IVP vs.\ BVP comparison}
|
||||
\label{tab:ch01_ivp_bvp}
|
||||
\begin{tabular}{@{} l l l l @{}}
|
||||
\toprule
|
||||
& \textbf{IVP} & \textbf{BVP} & \\
|
||||
\midrule
|
||||
\textbf{Conditions at:} & Single point $x_0$ & Two or more distinct points \\
|
||||
\textbf{Typical problem:} & Time evolution, dynamics & Steady state, spatial distribution \\
|
||||
\textbf{Existence:} & Usually guaranteed (Picard--Lindel\"of) & Not guaranteed \\
|
||||
\textbf{Uniqueness:} & Usually guaranteed & May fail or have multiple solutions \\
|
||||
\textbf{Example:} & $y' = y,\; y(0) = 1$ & $y'' = 0,\; y(0)=0,\; y(1)=1$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
% =============================================================================
|
||||
% DIRECTION FIELDS
|
||||
% =============================================================================
|
||||
|
||||
\section{Direction Fields}
|
||||
\label{sec:ch01_direction_fields}
|
||||
|
||||
A direction field provides a geometric picture of the solution family of a first-order ODE without solving it analytically.
|
||||
|
||||
\subsection{Construction}
|
||||
|
||||
For a first-order ODE $\displaystyle \frac{\diff y}{\diff x} = f(x, y)$, the direction field is constructed as follows:
|
||||
|
||||
\begin{enumerate}
|
||||
\item Choose a rectangular grid of points $(x_i, y_j)$ in the $xy$-plane.
|
||||
\item At each grid point, compute the slope $m_{ij} = f(x_i, y_j)$.
|
||||
\item Draw a short line segment centered at $(x_i, y_j)$ with slope $m_{ij}$.
|
||||
\end{enumerate}
|
||||
|
||||
Solution curves are trajectories that are everywhere tangent to the direction field. Reading the field visually reveals qualitative features: equilibrium solutions (horizontal segments), asymptotic behavior, and regions of rapid growth or decay.
|
||||
|
||||
\begin{hintbox}
|
||||
Look for \textbf{isoslopes}: curves along which $f(x, y)$ is constant. For $\frac{\diff y}{\diff x} = x - y$, isoslopes are the lines $y = x - c$, along which every segment has the same slope $c$.
|
||||
\end{hintbox}
|
||||
|
||||
\subsection{Direction Field Diagram}
|
||||
|
||||
The following TikZ figure shows the direction field for $\displaystyle \frac{\diff y}{\diff x} = x - y$ over the region $[-2, 2] \times [-2, 2]$, together with two representative solution curves.
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=1.3]
|
||||
% Axes
|
||||
\draw[->] (-2.3, 0) -- (2.3, 0) node[below right] {$x$};
|
||||
\draw[->] (0, -2.3) -- (0, 2.3) node[above left] {$y$};
|
||||
|
||||
% Direction field: dy/dx = x - y
|
||||
\foreach \x in {-2,-1.6,-1.2,-0.8,-0.4,0,0.4,0.8,1.2,1.6,2} {
|
||||
\foreach \y in {-2,-1.6,-1.2,-0.8,-0.4,0,0.4,0.8,1.2,1.6,2} {
|
||||
\pgfmathsetmacro{\slope}{\x - \y}
|
||||
\pgfmathsetmacro{\dx}{0.15}
|
||||
\pgfmathsetmacro{\dy}{\slope * 0.15}
|
||||
% Clamp to avoid excessively long segments
|
||||
\pgfmathsetmacro{\len}{sqrt(\dx*\dx + \dy*\dy)}
|
||||
\pgfmathsetmacro{\scale}{min(0.35 / max(\len,0.001), 1)}
|
||||
\pgfmathsetmacro{\sdx}{\dx * \scale}
|
||||
\pgfmathsetmacro{\sdy}{\dy * \scale}
|
||||
\draw[gray!70, very thin] (\x-\sdx,\y-\sdy) -- (\x+\sdx,\y+\sdy);
|
||||
}
|
||||
}
|
||||
|
||||
% Solution curve 1: y = x - 1 + 2*e^{-x}, passing through (0,1)
|
||||
\draw[SteelBlue, thick, domain=-2:2, samples=80, smooth]
|
||||
plot (\x, {\x - 1 + 2*exp(-\x)});
|
||||
|
||||
% Solution curve 2: y = x - 1 - 2*e^{-x}, passing through (0,-3)
|
||||
\draw[SteelBlue!60, thick, domain=-1:2, samples=80, smooth]
|
||||
plot (\x, {\x - 1 - 2*exp(-\x)});
|
||||
|
||||
% Equilibrium line y = x (slope = 0 everywhere)
|
||||
\draw[ForestGreen!70, dashed, domain=-2:2, samples=40]
|
||||
plot (\x, {\x});
|
||||
\node[ForestGreen!80, font=\footnotesize] at (1.8, 1.2) {$y=x$};
|
||||
|
||||
% Tick marks
|
||||
\foreach \t in {-2,-1,1,2} {
|
||||
\draw (\t, 0.05) -- (\t, -0.05) node[below, font=\tiny] {$\t$};
|
||||
\draw (0.05, \t) -- (-0.05, \t) node[left, font=\tiny] {$\t$};
|
||||
}
|
||||
\end{tikzpicture}
|
||||
\caption{Direction field for $\frac{\diff y}{\diff x} = x - y$. Blue curves are solution trajectories. The dashed green line $y = x$ is the locus of zero slope (equilibrium for the shifted system).}
|
||||
\label{fig:ch01_direction_field}
|
||||
\end{figure}
|
||||
|
||||
\subsection{Sketching Solution Curves}
|
||||
|
||||
Given a direction field, individual solution curves can be sketched by hand:
|
||||
|
||||
\begin{enumerate}
|
||||
\item \textbf{Trace the flow:} Start at a point and follow the segments, drawing a smooth curve tangent to the local arrows.
|
||||
\item \textbf{Respect the grid:} The curve should never cross a segment at an angle significantly different from the segment's slope.
|
||||
\item \textbf{Identify special features:} Equilibrium points (zero slope), asymptotes, and inflection points often emerge from the field pattern.
|
||||
\end{enumerate}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Sketch the solution of $\frac{\diff y}{\diff x} = x - y$ passing through $(0, 1)$.}
|
||||
|
||||
\begin{enumerate}
|
||||
\item At $(0, 1)$, the slope is $f(0, 1) = 0 - 1 = -1$. Draw a segment with slope $-1$.
|
||||
\item Moving right: at $(1, 0.5)$ (approximately), the slope is $1 - 0.5 = 0.5$. The curve bends upward.
|
||||
\item The line $y = x$ is where the slope is zero. For $y > x$, the slope is negative (curves bend down toward $y = x$). For $y < x$, the slope is positive (curves bend up toward $y = x$).
|
||||
\item Thus the line $y = x$ acts as an attractor. The solution starting at $(0, 1)$ approaches the line $y = x$ as $x \to \infty$.
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
\textbf{Verification:} The analytic solution is $y(x) = x - 1 + 2e^{-x}$. Indeed, $\lim_{x \to \infty} (y(x) - x) = -1$, and the transient $2e^{-x}$ decays, consistent with the field.
|
||||
\end{workedexample}
|
||||
|
||||
Direction fields also underpin numerical methods such as Euler's method (see \cref{ch:qualitative}).
|
||||
|
||||
% =============================================================================
|
||||
% EXISTENCE AND UNIQUENESS
|
||||
% =============================================================================
|
||||
|
||||
\section{Existence and Uniqueness}
|
||||
\label{sec:ch01_existence_uniqueness}
|
||||
|
||||
A fundamental question before attempting to solve any differential equation is: \emph{does a solution exist?} and \emph{is it unique?} The Picard--Lindel\"of theorem provides the foundational answer for first-order IVPs.
|
||||
|
||||
\subsection{The Picard--Lindel\"of Theorem}
|
||||
|
||||
\begin{theorem}[Picard--Lindel\"of]
|
||||
\label{thm:picard_lindelof}
|
||||
Consider the first-order IVP
|
||||
\[
|
||||
\frac{\diff y}{\diff x} = f(x, y), \qquad y(x_0) = y_0.
|
||||
\]
|
||||
Let $R = \{(x, y) : |x - x_0| \leq a,\; |y - y_0| \leq b\}$ be a closed rectangle centered at $(x_0, y_0)$ with $a, b > 0$.
|
||||
|
||||
If the following two conditions hold:
|
||||
\begin{enumerate}
|
||||
\item \textbf{Continuity:} $f(x, y)$ is continuous on $R$;
|
||||
\item \textbf{Lipschitz condition:} There exists a constant $L > 0$ such that for all $(x, y_1), (x, y_2) \in R$,
|
||||
\[
|
||||
|f(x, y_1) - f(x, y_2)| \leq L\,|y_1 - y_2|.
|
||||
\]
|
||||
\end{enumerate}
|
||||
Then there exists an interval $|x - x_0| \leq h > 0$ (with $h \leq a$) on which a \textbf{unique} solution $y = \varphi(x)$ exists.
|
||||
\end{theorem}
|
||||
|
||||
\begin{remark}
|
||||
A \textbf{sufficient} (but not necessary) condition for the Lipschitz condition is that $\pd{f}{y}$ exists and is bounded on $R$. In that case, one may take $L = \sup\limits_{(x, y) \in R} \bigl|\pd{f}{y}\bigr|$.
|
||||
\end{remark}
|
||||
|
||||
\subsection{The Lipschitz Condition}
|
||||
|
||||
The Lipschitz condition essentially requires that $f$ does not change too rapidly with respect to $y$. Geometrically, it prevents solution curves from diverging too quickly from one another.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Verify the Lipschitz condition for $f(x, y) = 2xy$ on the rectangle $R = [-1, 1] \times [-1, 1]$.}
|
||||
|
||||
Compute the partial derivative:
|
||||
\[
|
||||
\pd{f}{y} = 2x.
|
||||
\]
|
||||
On $R$, $|x| \leq 1$, so $\bigl|\pd{f}{y}\bigr| \leq 2$. Thus $f$ satisfies a Lipschitz condition with $L = 2$ on $R$. By \cref{thm:picard_lindelof}, the IVP $y' = 2xy,\; y(0) = 1$ has a unique solution on some interval around $x = 0$.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Non-uniqueness when the Lipschitz condition fails.}
|
||||
|
||||
Consider the IVP
|
||||
\[
|
||||
\frac{\diff y}{\diff x} = y^{2/3}, \qquad y(0) = 0.
|
||||
\]
|
||||
Here $f(y) = y^{2/3}$. The partial derivative with respect to $y$ is
|
||||
\[
|
||||
\pd{f}{y} = \tfrac{2}{3}\,y^{-1/3},
|
||||
\]
|
||||
which is \textbf{unbounded} as $y \to 0$. The Lipschitz condition fails in any rectangle containing $y = 0$.
|
||||
|
||||
Indeed, there are infinitely many solutions:
|
||||
\[
|
||||
y(x) = 0 \quad \text{(trivial solution)},
|
||||
\]
|
||||
and for any $c \geq 0$,
|
||||
\[
|
||||
y(x) =
|
||||
\begin{cases}
|
||||
0, & 0 \leq x \leq c, \\
|
||||
\dfrac{(x - c)^3}{27}, & x > c,
|
||||
\end{cases}
|
||||
\]
|
||||
all satisfy the IVP. The solution branches off from the trivial solution at any chosen point $c$.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{hintbox}
|
||||
A quick test for potential non-uniqueness: check whether $\pd{f}{y}$ blows up at or near the initial condition point. If it does, the Lipschitz condition likely fails.
|
||||
\end{hintbox}
|
||||
|
||||
\subsection{Interval of Validity}
|
||||
|
||||
\begin{definition}[Interval of Validity]
|
||||
The \textbf{interval of validity} (or interval of existence) of a solution $y = \varphi(x)$ to an IVP is the \textbf{largest} open interval $I$ containing $x_0$ on which the solution exists and is differentiable.
|
||||
\end{definition}
|
||||
|
||||
The interval of validity may be restricted by:
|
||||
\begin{itemize}
|
||||
\item Singularities in the coefficients of the equation;
|
||||
\item Points where the Lipschitz condition breaks down;
|
||||
\item Points where the solution itself becomes unbounded (blow-up).
|
||||
\end{itemize}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Find the interval of validity for $y' = y^2,\; y(0) = 1$.}
|
||||
|
||||
This is separable: $\displaystyle \frac{\diff y}{y^2} = \diff x \;\Rightarrow\; -\frac{1}{y} = x + C$. With $y(0) = 1$, we get $C = -1$, so $y(x) = \frac{1}{1 - x}$.
|
||||
|
||||
The solution has a vertical asymptote at $x = 1$. Since the initial point is $x_0 = 0$, the interval of validity is
|
||||
\[
|
||||
(-\infty, 1).
|
||||
\]
|
||||
The solution blows up at $x = 1$, so it cannot be extended beyond this point.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Interval of validity for $y' = \sqrt{y},\; y(0) = 1$.}
|
||||
|
||||
Separating: $\displaystyle \frac{\diff y}{\sqrt{y}} = \diff x \;\Rightarrow\; 2\sqrt{y} = x + C$. With $y(0) = 1$, we get $C = 2$, so $y(x) = \left(\frac{x + 2}{2}\right)^2 = \frac{(x+2)^2}{4}$.
|
||||
|
||||
To find the interval of validity, check that the solution satisfies the original ODE: $y' = \frac{x+2}{2}$ while $\sqrt{y} = \frac{|x+2|}{2}$. These are equal only when $x + 2 \geq 0$, i.e., $x \geq -2$. For $x < -2$, we have $y' < 0$ while $\sqrt{y} \geq 0$, so the formula fails the ODE. At $x = -2$, the solution reaches $y = 0$ (the equilibrium). Therefore, the interval of validity is
|
||||
\[
|
||||
(-2, \infty).
|
||||
\]
|
||||
Note that although the Lipschitz condition fails at $y = 0$ (since $\pd{f}{y} = \frac{1}{2\sqrt{y}}$ is unbounded there), the particular solution passing through $y(0) = 1$ is valid on $(-2, \infty)$, and uniqueness is preserved on this interval.
|
||||
\end{workedexample}
|
||||
|
||||
\section{Summary}
|
||||
\label{sec:ch01_summary}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Chapter 1 summary: key concepts in differential equations}
|
||||
\label{tab:ch01_summary}
|
||||
\begin{tabular}{@{} p{3.8cm} p{5.5cm} @{}}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key Definition / Formula} \\
|
||||
\midrule
|
||||
\textbf{Order} & Highest derivative present (e.g., $y'''$ $\Rightarrow$ 3rd order) \\
|
||||
\textbf{Degree} & Power of the highest-order derivative when polynomial in derivatives \\
|
||||
\textbf{Linear DE} & $a_n(x)y^{(n)} + \dots + a_1(x)y' + a_0(x)y = g(x)$ \\
|
||||
\textbf{Nonlinear DE} & $y$ or derivatives appear to non-integer powers, multiplied together, or inside nonlinear functions \\
|
||||
\textbf{Homogeneous (linear)} & $g(x) = 0$ in the linear form \\
|
||||
\textbf{IVP} & DE + conditions at a single point: $y(x_0) = y_0, \dots$ \\
|
||||
\textbf{BVP} & DE + conditions at two or more distinct points \\
|
||||
\textbf{Direction field} & Plot of short line segments with slope $f(x_i, y_j)$ at grid points \\
|
||||
\textbf{Picard--Lindel\"of} & Continuity + Lipschitz $\Rightarrow$ local existence and uniqueness \\
|
||||
\textbf{Lipschitz condition} & $|f(x, y_1) - f(x, y_2)| \leq L|y_1 - y_2|$; sufficient: $\bigl|\pd{f}{y}\bigr| \leq L$ \\
|
||||
\textbf{Interval of validity} & Largest interval containing $x_0$ where the solution exists and is unique \\
|
||||
\textbf{Non-uniqueness example} & $y' = y^{2/3},\; y(0) = 0$ admits infinitely many solutions \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
The chapters that follow build directly on these foundations. Solution methods for first-order equations are developed in \cref{ch:first_order}; qualitative analysis and numerical methods appear in \cref{ch:qualitative}; and linear second-order theory begins in \cref{ch:second_order_homogeneous}.
|
||||
@@ -1,505 +0,0 @@
|
||||
\section{First-Order Methods}
|
||||
\label{ch:first_order}
|
||||
|
||||
\usetikzlibrary{positioning}
|
||||
|
||||
\subsection{Separable Equations}
|
||||
\label{sec:ch02_separable}
|
||||
|
||||
A first-order ODE is \textbf{separable} when it can be written as
|
||||
\begin{equation}
|
||||
\frac{\diff y}{\diff x} = g(x)\,h(y)
|
||||
\end{equation}
|
||||
i.e.\ the right-hand side factors into a function of $x$ times a function of $y$.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Separable equations.} Rewrite as $\dfrac{\diff y}{h(y)} = g(x)\,\diff x$ and integrate both sides:
|
||||
\[
|
||||
\int \frac{\diff y}{h(y)} = \int g(x)\,\diff x + C.
|
||||
\]
|
||||
Solve for $y$ explicitly when possible; otherwise leave the solution in implicit form.
|
||||
\end{keyresult}
|
||||
|
||||
\begin{hintbox}
|
||||
If $h(y_0)=0$, the constant function $y(x)=y_0$ is always a solution (an \emph{equilibrium}). These may be lost when dividing by $h(y)$, so check separately.
|
||||
\end{hintbox}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $\dfrac{\diff y}{\diff x} = x\,y$.
|
||||
|
||||
\textbf{Solution.} Separate variables:
|
||||
\[
|
||||
\frac{\diff y}{y} = x\,\diff x.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
\ln|y| = \frac{x^2}{2} + C.
|
||||
\]
|
||||
Exponentiate:
|
||||
\[
|
||||
|y| = e^{C}\,e^{x^2/2} \quad\Longrightarrow\quad y = C_1\,e^{x^2/2},
|
||||
\]
|
||||
where $C_1 = \pm e^{C}$ is an arbitrary nonzero constant. Including the equilibrium $y=0$ (lost when dividing by $y$), the general solution is
|
||||
\[
|
||||
y(x) = C\,e^{x^2/2}, \qquad C \in \R.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $\dfrac{\diff y}{\diff x} = \dfrac{2x}{3y^2}$.
|
||||
|
||||
\textbf{Solution.} Separate:
|
||||
\[
|
||||
3y^2\,\diff y = 2x\,\diff x.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
y^3 = x^2 + C.
|
||||
\]
|
||||
This implicit form is perfectly acceptable. Solving explicitly gives $y = \sqrt[3]{x^2 + C}$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Linear First-Order Equations}
|
||||
\label{sec:ch02_linear_first_order}
|
||||
|
||||
A first-order ODE is \textbf{linear} if it can be written in the \emph{standard form}
|
||||
\begin{equation}
|
||||
\frac{\diff y}{\diff x} + p(x)\,y = g(x).
|
||||
\label{eq:linear_std}
|
||||
\end{equation}
|
||||
The coefficient of $\diff y/\diff x$ \textbf{must be $1$}; if the equation arrives with a leading coefficient $a(x)$, divide through first.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Integrating factor.} For \cref{eq:linear_std}, the integrating factor is
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int p(x)\,\diff x\right).
|
||||
\]
|
||||
Multiplying the equation by $\mu(x)$ produces
|
||||
\[
|
||||
\frac{\diff}{\diff x}\bigl[\mu(x)\,y\bigr] = \mu(x)\,g(x),
|
||||
\]
|
||||
so the general solution is
|
||||
\[
|
||||
y(x) = \frac{1}{\mu(x)}\left(\int \mu(x)\,g(x)\,\diff x + C\right).
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\textbf{Derivation.} With $\mu(x) = \exp\!\bigl(\int p(x)\,\diff x\bigr)$, the chain rule gives
|
||||
$\mu'(x) = \mu(x)\,p(x)$. Multiplying \cref{eq:linear_std} by $\mu(x)$:
|
||||
\[
|
||||
\mu y' + \mu p\,y = \mu g \quad\Longrightarrow\quad \mu y' + \mu' y = \mu g.
|
||||
\]
|
||||
The left side is exactly the product-rule derivative $\dfrac{\diff}{\diff x}[\mu y]$, yielding the result.
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Pitfall.} If the equation is $a(x)y' + b(x)y = f(x)$ with $a(x)\neq 1$, you \emph{must} divide by $a(x)$ to obtain standard form before computing $\mu(x)$. Forgetting this is the most common error.
|
||||
\end{hintbox}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y' + \dfrac{2}{x}\,y = x^3$.
|
||||
|
||||
\textbf{Solution.} The equation is already in standard form with $p(x)=\dfrac{2}{x}$ and $g(x)=x^3$.
|
||||
Compute the integrating factor:
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int \frac{2}{x}\,\diff x\right)
|
||||
= \exp\!\bigl(2\ln|x|\bigr)
|
||||
= |x|^2 = x^2.
|
||||
\]
|
||||
(We drop the absolute value since $x^2 \ge 0$.)
|
||||
|
||||
Multiply the entire equation by $x^2$:
|
||||
\[
|
||||
x^2 y' + 2x\,y = x^5
|
||||
\quad\Longrightarrow\quad
|
||||
\frac{\diff}{\diff x}\bigl[x^2 y\bigr] = x^5.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
x^2 y = \frac{x^6}{6} + C.
|
||||
\]
|
||||
Divide by $x^2$:
|
||||
\[
|
||||
y(x) = \frac{x^4}{6} + \frac{C}{x^2}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $2x\,y' + 3y = 6x$.
|
||||
|
||||
\textbf{Solution.} The leading coefficient is $2x \neq 1$. Divide through:
|
||||
\[
|
||||
y' + \frac{3}{2x}\,y = 3.
|
||||
\]
|
||||
Now $p(x) = \dfrac{3}{2x}$ and $g(x) = 3$. Integrating factor:
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int \frac{3}{2x}\,\diff x\right)
|
||||
= \exp\!\left(\tfrac{3}{2}\ln|x|\right)
|
||||
= |x|^{3/2}.
|
||||
\]
|
||||
Work with $x>0$ for simplicity, so $\mu(x) = x^{3/2}$. Multiply:
|
||||
\[
|
||||
x^{3/2}y' + \tfrac{3}{2}x^{1/2}y = 3x^{3/2}
|
||||
\quad\Longrightarrow\quad
|
||||
\frac{\diff}{\diff x}\!\left[x^{3/2}y\right] = 3x^{3/2}.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
x^{3/2}y = \int 3x^{3/2}\,\diff x = 3\cdot\frac{2}{5}x^{5/2} + C = \frac{6}{5}x^{5/2} + C.
|
||||
\]
|
||||
Solve for $y$:
|
||||
\[
|
||||
y(x) = \frac{6}{5}x + \frac{C}{x^{3/2}}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Exact Equations}
|
||||
\label{sec:ch02_exact}
|
||||
|
||||
An ODE written in differential form
|
||||
\begin{equation}
|
||||
M(x,y)\,\diff x + N(x,y)\,\diff y = 0
|
||||
\end{equation}
|
||||
is \textbf{exact} if there exists a scalar function $\psi(x,y)$ such that
|
||||
$\pd{\psi}{x} = M$ and $\pd{\psi}{y} = N$. The solution is then given implicitly by
|
||||
$\psi(x,y) = C$.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Exactness test.} The equation is exact \emph{if and only if}
|
||||
\[
|
||||
\pd{M}{y} = \pd{N}{x}.
|
||||
\]
|
||||
When exact, the potential function is
|
||||
\[
|
||||
\psi(x,y) = \int M\,\diff x \;+\; \int\!\left[N - \pd{}{y}\!\left(\int M\,\diff x\right)\right]\!\diff y.
|
||||
\]
|
||||
The solution is $\psi(x,y) = C$.
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $(2xy + y^3)\,\diff x + (x^2 + 3xy^2)\,\diff y = 0$.
|
||||
|
||||
\textbf{Solution.} Here $M = 2xy + y^3$ and $N = x^2 + 3xy^2$.
|
||||
Check exactness:
|
||||
\[
|
||||
\pd{M}{y} = 2x + 3y^2, \qquad \pd{N}{x} = 2x + 3y^2.
|
||||
\]
|
||||
They match, so the equation is exact.
|
||||
|
||||
Integrate $M$ with respect to $x$:
|
||||
\[
|
||||
\int (2xy + y^3)\,\diff x = x^2y + xy^3 + h(y).
|
||||
\]
|
||||
Differentiate with respect to $y$ and equate to $N$:
|
||||
\[
|
||||
\pd{}{y}\bigl[x^2y + xy^3 + h(y)\bigr] = x^2 + 3xy^2 + h'(y) \stackrel{!}{=} x^2 + 3xy^2.
|
||||
\]
|
||||
Thus $h'(y) = 0$, so $h(y)$ is constant. The potential function is
|
||||
\[
|
||||
\psi(x,y) = x^2y + xy^3.
|
||||
\]
|
||||
The implicit solution is
|
||||
\[
|
||||
x^2y + xy^3 = C.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Integrating factors for non-exact equations.}
|
||||
If $\pd{M}{y} \neq \pd{N}{x}$, the equation can sometimes be made exact by multiplying by an integrating factor $\mu$.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Integrating factors.}
|
||||
\begin{itemize}
|
||||
\item If $\displaystyle \frac{\pd{M}{y} - \pd{N}{x}}{N}$ depends \emph{only on $x$}, then
|
||||
$\mu(x) = \exp\!\left(\int \frac{\pd{M}{y} - \pd{N}{x}}{N}\,\diff x\right)$.
|
||||
\item If $\displaystyle \frac{\pd{N}{x} - \pd{M}{y}}{M}$ depends \emph{only on $y$}, then
|
||||
$\mu(y) = \exp\!\left(\int \frac{\pd{N}{x} - \pd{M}{y}}{M}\,\diff y\right)$.
|
||||
\end{itemize}
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $(3xy + y^2)\,\diff x + (x^2 + xy)\,\diff y = 0$.
|
||||
|
||||
\textbf{Solution.} $M = 3xy + y^2$, $N = x^2 + xy$. Check:
|
||||
\[
|
||||
\pd{M}{y} = 3x + 2y, \qquad \pd{N}{x} = 2x + y.
|
||||
\]
|
||||
Not exact. Test for an integrating factor depending on $x$ alone:
|
||||
\[
|
||||
\frac{\pd{M}{y} - \pd{N}{x}}{N}
|
||||
= \frac{(3x+2y) - (2x+y)}{x^2 + xy}
|
||||
= \frac{x + y}{x(x+y)}
|
||||
= \frac{1}{x}.
|
||||
\]
|
||||
This depends only on $x$, so
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int \frac{1}{x}\,\diff x\right) = \exp(\ln|x|) = x.
|
||||
\]
|
||||
Multiply the original equation by $\mu = x$:
|
||||
\[
|
||||
(3x^2y + xy^2)\,\diff x + (x^3 + x^2y)\,\diff y = 0.
|
||||
\]
|
||||
Verify: $\pd{\tilde{M}}{y} = 3x^2 + 2xy = \pd{\tilde{N}}{x}$. Exact.
|
||||
|
||||
Integrate $\tilde{M}$ with respect to $x$:
|
||||
\[
|
||||
\int (3x^2y + xy^2)\,\diff x = x^3y + \tfrac{1}{2}x^2y^2 + h(y).
|
||||
\]
|
||||
Differentiate with respect to $y$:
|
||||
\[
|
||||
x^3 + x^2y + h'(y) \stackrel{!}{=} x^3 + x^2y \quad\Longrightarrow\quad h'(y)=0.
|
||||
\]
|
||||
The potential is $\psi(x,y) = x^3y + \dfrac{1}{2}x^2y^2$, and the solution is
|
||||
\[
|
||||
x^3y + \tfrac{1}{2}x^2y^2 = C.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Bernoulli Equations}
|
||||
\label{sec:ch02_bernoulli}
|
||||
|
||||
A \textbf{Bernoulli equation} has the form
|
||||
\begin{equation}
|
||||
\frac{\diff y}{\diff x} + p(x)\,y = g(x)\,y^n, \qquad n \neq 0,1.
|
||||
\end{equation}
|
||||
For $n=0$ the equation is linear; for $n=1$ it is also linear. The substitution $v = y^{1-n}$ transforms it into a linear equation.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Bernoulli substitution.} Let $v = y^{1-n}$. Then
|
||||
\[
|
||||
v' = (1-n)\,y^{-n}\,y',
|
||||
\]
|
||||
and substituting into the Bernoulli equation yields the \emph{linear} equation in $v$:
|
||||
\[
|
||||
v' + (1-n)\,p(x)\,v = (1-n)\,g(x).
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y' + \dfrac{1}{x}\,y = x\,y^3$.
|
||||
|
||||
\textbf{Solution.} This is Bernoulli with $p(x)=\dfrac{1}{x}$, $g(x)=x$, and $n=3$.
|
||||
Substitute $v = y^{1-3} = y^{-2}$. Then $v' = -2y^{-3}y'$.
|
||||
|
||||
Multiply the original equation by $-2y^{-3}$:
|
||||
\[
|
||||
-2y^{-3}y' - 2y^{-3}\cdot\frac{1}{x}\,y = -2y^{-3}\cdot x\,y^3.
|
||||
\]
|
||||
Recognizing $v = y^{-2}$ and $v' = -2y^{-3}y'$:
|
||||
\[
|
||||
v' - \frac{2}{x}\,v = -2x.
|
||||
\]
|
||||
This is a linear equation in $v$. Integrating factor:
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int -\frac{2}{x}\,\diff x\right) = \exp(-2\ln|x|) = \frac{1}{x^2}.
|
||||
\]
|
||||
Multiply by $\mu$:
|
||||
\[
|
||||
\frac{1}{x^2}v' - \frac{2}{x^3}v = -\frac{2}{x}
|
||||
\quad\Longrightarrow\quad
|
||||
\frac{\diff}{\diff x}\!\left[\frac{v}{x^2}\right] = -\frac{2}{x}.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
\frac{v}{x^2} = -2\ln|x| + C
|
||||
\quad\Longrightarrow\quad
|
||||
v = x^2\bigl(C - 2\ln|x|\bigr).
|
||||
\]
|
||||
Substitute back $v = y^{-2}$:
|
||||
\[
|
||||
\frac{1}{y^2} = x^2\bigl(C - 2\ln|x|\bigr)
|
||||
\quad\Longrightarrow\quad
|
||||
y(x) = \pm\frac{1}{x\sqrt{\,C - 2\ln|x|\,}}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y' - \dfrac{2}{x}\,y = x^2\,y^2$.
|
||||
|
||||
\textbf{Solution.} Bernoulli with $n=2$, $p(x)=-\dfrac{2}{x}$, $g(x)=x^2$.
|
||||
Substitute $v = y^{1-2} = y^{-1} = \dfrac{1}{y}$. Then $v' = -y^{-2}y'$.
|
||||
|
||||
Multiply the original equation by $-y^{-2}$:
|
||||
\[
|
||||
-y^{-2}y' + \frac{2}{x}\,y^{-1} = -x^2.
|
||||
\]
|
||||
In terms of $v$:
|
||||
\[
|
||||
v' + \frac{2}{x}\,v = -x^2.
|
||||
\]
|
||||
Integrating factor:
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int \frac{2}{x}\,\diff x\right) = x^2.
|
||||
\]
|
||||
Multiply:
|
||||
\[
|
||||
x^2v' + 2xv = -x^4
|
||||
\quad\Longrightarrow\quad
|
||||
\frac{\diff}{\diff x}\!\left[x^2v\right] = -x^4.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
x^2v = -\frac{x^5}{5} + C
|
||||
\quad\Longrightarrow\quad
|
||||
v = -\frac{x^3}{5} + \frac{C}{x^2}.
|
||||
\]
|
||||
Since $v = 1/y$:
|
||||
\[
|
||||
y(x) = \frac{1}{-\dfrac{x^3}{5} + \dfrac{C}{x^2}}
|
||||
= \frac{5x^2}{C - x^5}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Homogeneous Substitutions}
|
||||
\label{sec:ch02_homogeneous_substitutions}
|
||||
|
||||
A first-order ODE is \textbf{homogeneous} (of degree zero) if
|
||||
\begin{equation}
|
||||
\frac{\diff y}{\diff x} = F\!\left(\frac{y}{x}\right).
|
||||
\end{equation}
|
||||
That is, the right-hand side is a function of the ratio $y/x$ alone.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Homogeneous substitution.} Let $y = vx$. Then
|
||||
\[
|
||||
\frac{\diff y}{\diff x} = v + x\frac{\diff v}{\diff x}.
|
||||
\]
|
||||
The equation becomes
|
||||
\[
|
||||
v + x\frac{\diff v}{\diff x} = F(v)
|
||||
\quad\Longrightarrow\quad
|
||||
\frac{\diff v}{\diff x} = \frac{F(v) - v}{x},
|
||||
\]
|
||||
which is separable:
|
||||
\[
|
||||
\int \frac{\diff v}{F(v) - v} = \int \frac{\diff x}{x} = \ln|x| + C.
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y' = \dfrac{x+y}{x-y}$.
|
||||
|
||||
\textbf{Solution.} Rewrite the right side:
|
||||
\[
|
||||
\frac{x+y}{x-y} = \frac{1 + y/x}{1 - y/x} = F\!\left(\frac{y}{x}\right).
|
||||
\]
|
||||
The equation is homogeneous. Set $y = vx$, so $y' = v + xv'$:
|
||||
\[
|
||||
v + x\frac{\diff v}{\diff x} = \frac{1+v}{1-v}.
|
||||
\]
|
||||
Isolate the derivative:
|
||||
\[
|
||||
x\frac{\diff v}{\diff x} = \frac{1+v}{1-v} - v = \frac{1+v - v(1-v)}{1-v} = \frac{1+v^2}{1-v}.
|
||||
\]
|
||||
Separate:
|
||||
\[
|
||||
\frac{1-v}{1+v^2}\,\diff v = \frac{\diff x}{x}.
|
||||
\]
|
||||
Split the left integrand:
|
||||
\[
|
||||
\int \frac{1}{1+v^2}\,\diff v - \int \frac{v}{1+v^2}\,\diff v = \int \frac{\diff x}{x}.
|
||||
\]
|
||||
These are standard integrals:
|
||||
\[
|
||||
\arctan(v) - \tfrac{1}{2}\ln(1+v^2) = \ln|x| + C.
|
||||
\]
|
||||
Substitute back $v = y/x$:
|
||||
\[
|
||||
\arctan\!\left(\frac{y}{x}\right) - \tfrac{1}{2}\ln\!\left(1 + \frac{y^2}{x^2}\right) = \ln|x| + C.
|
||||
\]
|
||||
Using $\ln(1 + y^2/x^2) = \ln\!\bigl((x^2+y^2)/x^2\bigr) = \ln(x^2+y^2) - 2\ln|x|$, the solution simplifies to
|
||||
\[
|
||||
\arctan\!\left(\frac{y}{x}\right) - \tfrac{1}{2}\ln(x^2+y^2) = C.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y' = \dfrac{xy + y^2}{x^2}$.
|
||||
|
||||
\textbf{Solution.} Rewrite:
|
||||
\[
|
||||
y' = \frac{y}{x} + \frac{y^2}{x^2} = F\!\left(\frac{y}{x}\right).
|
||||
\]
|
||||
Homogeneous with $F(v) = v + v^2$. Set $y = vx$:
|
||||
\[
|
||||
v + x\frac{\diff v}{\diff x} = v + v^2
|
||||
\quad\Longrightarrow\quad
|
||||
x\frac{\diff v}{\diff x} = v^2.
|
||||
\]
|
||||
Separate:
|
||||
\[
|
||||
\frac{\diff v}{v^2} = \frac{\diff x}{x}.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
-\frac{1}{v} = \ln|x| + C
|
||||
\quad\Longrightarrow\quad
|
||||
v = -\frac{1}{\ln|x| + C}.
|
||||
\]
|
||||
Substitute back $v = y/x$:
|
||||
\[
|
||||
y(x) = -\frac{x}{\ln|x| + C}.
|
||||
\]
|
||||
The equilibrium solution $y=0$ (corresponding to $v=0$) is also valid.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch02_summary}
|
||||
|
||||
The following flowchart guides method selection. Test methods in the order shown; the first applicable method solves the equation.
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[node distance=1.3cm and 1.2cm,
|
||||
>=Stealth,
|
||||
box/.style={draw, rounded corners=3pt, fill=LightSteelBlue!20, minimum width=3cm, align=center, font=\small, text width=3.2cm},
|
||||
sol/.style={draw, rounded corners=3pt, fill=ForestGreen!15, minimum width=3cm, align=center, font=\small, text width=3.0cm},
|
||||
arrow/.style={->, >=Stealth, thick, shorten <=1pt, shorten >=1pt}]
|
||||
|
||||
\node[box, fill=LemonChiffon!40] (start) {\textbf{Given:} $y' = f(x,y)$};
|
||||
\node[box, below=of start] (t1) {\textbf{Can $f$ be written}\\as $g(x)\,h(y)$?};
|
||||
\node[box, below=of t1] (t2) {\textbf{Is it linear}\\$y'+p(x)y=g(x)$?};
|
||||
\node[box, below=of t2] (t3) {\textbf{Is it Bernoulli}\\$y'+py=gy^n$?};
|
||||
\node[box, below=of t3] (t4) {\textbf{Is $f(x,y)=F(y/x)$?}\\(Homogeneous)};
|
||||
\node[box, below=of t4] (t5) {\textbf{Write as $M\,dx+N\,dy=0$.}\\Exact?};
|
||||
\node[box, below=of t5, fill=Salmon!20] (gen) {\textbf{Non-exact:}\\Try integrating\\factor or other methods};
|
||||
|
||||
% Solution boxes
|
||||
\node[sol, right=2cm of t1] (s1) {Separable:\\Integrate};
|
||||
\node[sol, right=2cm of t2] (s2) {Integrating factor:\\$\mu=e^{\int p\,dx}$};
|
||||
\node[sol, right=2cm of t3] (s3) {Substitute\\$v=y^{1-n}$};
|
||||
\node[sol, right=2cm of t4] (s4) {Substitute\\$y=vx$};
|
||||
\node[sol, right=2cm of t5] (s5) {Potential\\$\psi(x,y)=C$};
|
||||
|
||||
% Decision arrows
|
||||
\draw[arrow] (start) -- (t1);
|
||||
\draw[arrow] (t1) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t2);
|
||||
\draw[arrow] (t2) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t3);
|
||||
\draw[arrow] (t3) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t4);
|
||||
\draw[arrow] (t4) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t5);
|
||||
\draw[arrow] (t5) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (gen);
|
||||
|
||||
% Yes arrows
|
||||
\draw[arrow] (t1) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s1);
|
||||
\draw[arrow] (t2) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s2);
|
||||
\draw[arrow] (t3) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s3);
|
||||
\draw[arrow] (t4) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s4);
|
||||
\draw[arrow] (t5) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s5);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{First-order solution methods}
|
||||
\label{tab:ch02_summary}
|
||||
\begin{tabular}{l l p{6cm}}
|
||||
\toprule
|
||||
\textbf{Type} & \textbf{Form} & \textbf{Method} \\
|
||||
\midrule
|
||||
Separable & $y' = g(x)h(y)$ & Rewrite as $\dfrac{\diff y}{h(y)} = g(x)\,\diff x$; integrate both sides \\
|
||||
Linear & $y' + p(x)y = g(x)$ & Integrating factor $\mu(x) = \exp\!\left(\int p(x)\,\diff x\right)$ \\
|
||||
Exact & $M\,\diff x + N\,\diff y = 0,\;\pd{M}{y}=\pd{N}{x}$ & Find potential $\psi$ with $\pd{\psi}{x}=M,\;\pd{\psi}{y}=N$; $\psi=C$ \\
|
||||
Non-exact & $M\,\diff x + N\,\diff y = 0,\;\pd{M}{y}\neq\pd{N}{x}$ & Multiply by integrating factor $\mu(x)$ or $\mu(y)$ \\
|
||||
Bernoulli & $y' + p(x)y = g(x)y^n$ & Substitute $v=y^{1-n}$; solve resulting linear equation \\
|
||||
Homogeneous & $y' = F(y/x)$ & Substitute $y=vx$; solve resulting separable equation \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
@@ -1,644 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch03_qualitative.tex -- Chapter 3: Qualitative Analysis and Numerical Methods
|
||||
% =============================================================================
|
||||
|
||||
\section{Qualitative Analysis and Numerical Methods}
|
||||
\label{ch:qualitative}
|
||||
|
||||
\subsection{Autonomous Equations and Phase Lines}
|
||||
\label{sec:ch03_autonomous}
|
||||
|
||||
An \textbf{autonomous differential equation} is a first-order ODE in which the right-hand side depends only on the dependent variable and \emph{not} explicitly on the independent variable~$t$:
|
||||
\begin{equation}
|
||||
\label{eq:autonomous}
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = f(y).
|
||||
\end{equation}
|
||||
Because $f$ has no explicit $t$-dependence, the direction field is invariant under horizontal translation. Solutions simply shift left or right along the $t$-axis without changing shape.
|
||||
|
||||
\paragraph{Equilibrium solutions.}
|
||||
An \textbf{equilibrium solution} (or \textbf{critical point}) is a constant solution $y(t) \equiv y^*$ that satisfies
|
||||
\begin{equation}
|
||||
f(y^*) = 0.
|
||||
\end{equation}
|
||||
At such a point the derivative vanishes and the solution is stationary. Every root of $f(y) = 0$ gives one equilibrium.
|
||||
|
||||
\paragraph{Phase line construction.}
|
||||
The \textbf{phase line} is a one-dimensional sketch of the dynamics of \cref{eq:autonomous}. It is constructed in three steps:
|
||||
\begin{enumerate}
|
||||
\item \textbf{Identify equilibria:} Solve $f(y) = 0$.
|
||||
\item \textbf{Determine the sign of $f(y)$} on each interval between consecutive equilibria.
|
||||
If $f(y) > 0$ on an interval, solutions increase ($y$ moves upward on the phase line).
|
||||
If $f(y) < 0$, solutions decrease.
|
||||
\item \textbf{Draw the phase line:} Mark each equilibrium on a vertical line and place arrows on each interval indicating the direction of motion.
|
||||
\end{enumerate}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Phase line for $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}t} = y(y-1)(y-2)$.}
|
||||
|
||||
\textit{Step 1: Equilibria.} Set $f(y) = y(y-1)(y-2) = 0$. The roots are
|
||||
\[
|
||||
y_1^* = 0,\qquad y_2^* = 1,\qquad y_3^* = 2.
|
||||
\]
|
||||
|
||||
\textit{Step 2: Sign of $f(y)$ on each interval.} Test a sample point in each interval:
|
||||
\[
|
||||
\begin{array}{c|c|c}
|
||||
\text{Interval} & \text{Sample } y & f(y) \\ \hline
|
||||
(-\infty, 0) & y = -1 & (-)(-)(-) = - \;\; \text{(decreasing)} \\
|
||||
(0, 1) & y = 0.5 & (+)(-)(-) = + \;\; \text{(increasing)} \\
|
||||
(1, 2) & y = 1.5 & (+)(+)(-) = - \;\; \text{(decreasing)} \\
|
||||
(2, \infty) & y = 3 & (+)(+)(+) = + \;\; \text{(increasing)}
|
||||
\end{array}
|
||||
\]
|
||||
|
||||
\textit{Step 3: Phase line.}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
% Vertical phase line
|
||||
\draw[thick] (0,0) -- (0,5);
|
||||
\draw[->] (0,0) -- (0,-0.3);
|
||||
\draw[->] (0,5) -- (0,5.3);
|
||||
|
||||
% Equilibrium points
|
||||
\filldraw[black] (0,0.8) circle (2.5pt) node[right=4pt] {$y^*=0$};
|
||||
\filldraw[black] (0,2.5) circle (2.5pt) node[right=4pt] {$y^*=1$};
|
||||
\filldraw[black] (0,4.2) circle (2.5pt) node[right=4pt] {$y^*=2$};
|
||||
|
||||
% Arrows on intervals
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,0.35) -- (-0.15,0.0);
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,1.6) -- (-0.15,1.25);
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,3.3) -- (-0.15,2.95);
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,4.8) -- (-0.15,4.55);
|
||||
|
||||
% Labels
|
||||
\node[below] at (0,-0.5) {$y$};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Solutions starting between $0$ and $1$ approach $y=0$ from above or $y=1$ from below as $t$ increases.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Phase line for $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}t} = y^2 - 4$.}
|
||||
|
||||
\textit{Step 1:} $y^2 - 4 = 0 \implies y^* = -2,\; +2$.
|
||||
|
||||
\textit{Step 2:} Test intervals:
|
||||
\[
|
||||
\begin{array}{c|c|c}
|
||||
\text{Interval} & \text{Sample } y & f(y) \\ \hline
|
||||
(-\infty, -2) & y = -3 & 9 - 4 = +5 \;\; \text{(increasing)} \\
|
||||
(-2, 2) & y = 0 & 0 - 4 = -4 \;\; \text{(decreasing)} \\
|
||||
(2, \infty) & y = 3 & 9 - 4 = +5 \;\; \text{(increasing)}
|
||||
\end{array}
|
||||
\]
|
||||
|
||||
\textit{Step 3:}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\draw[thick] (0,0) -- (0,5);
|
||||
\draw[->] (0,0) -- (0,-0.3);
|
||||
\draw[->] (0,5) -- (0,5.3);
|
||||
|
||||
\filldraw[black] (0,1.25) circle (2.5pt) node[right=4pt] {$y^*=-2$};
|
||||
\filldraw[black] (0,3.75) circle (2.5pt) node[right=4pt] {$y^*=2$};
|
||||
|
||||
% Arrow: below -2, going up
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,0.5) -- (-0.15,0.85);
|
||||
% Arrow: between -2 and 2, going down
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,3.0) -- (-0.15,2.2);
|
||||
% Arrow: above 2, going up
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,4.5) -- (-0.15,4.15);
|
||||
|
||||
\node[below] at (0,-0.5) {$y$};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
The interval $(-2,2)$ is a ``funnel'' where all solutions decrease toward $y=-2$ as $t \to \infty$.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Phase line for $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}t} = y^2$.}
|
||||
|
||||
\textit{Step 1:} $y^2 = 0 \implies y^* = 0$ (the only equilibrium).
|
||||
|
||||
\textit{Step 2:} For $y \neq 0$, $y^2 > 0$, so $f(y) > 0$ on both $(-\infty,0)$ and $(0,\infty)$. Solutions increase on both sides of the equilibrium.
|
||||
|
||||
\textit{Step 3:}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\draw[thick] (0,0) -- (0,5);
|
||||
\draw[->] (0,0) -- (0,-0.3);
|
||||
\draw[->] (0,5) -- (0,5.3);
|
||||
|
||||
\filldraw[black] (0,2.5) circle (2.5pt) node[right=4pt] {$y^*=0$};
|
||||
|
||||
% Arrow below 0: going up (toward 0)
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,1.5) -- (-0.15,1.15);
|
||||
% Arrow above 0: going up (away from 0)
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,3.8) -- (-0.15,3.45);
|
||||
|
||||
\node[below] at (0,-0.5) {$y$};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Solutions approach $y^*=0$ from below but depart from above. This is the hallmark of a \textbf{semi-stable} equilibrium (discussed in \cref{sec:ch03_stability}).
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Stability Analysis}
|
||||
\label{sec:ch03_stability}
|
||||
|
||||
\paragraph{Linear stability test.}
|
||||
Suppose $y^*$ is an equilibrium of $\mathrm{d}y/\mathrm{d}t = f(y)$. Expand $f$ in a Taylor series about $y^*$:
|
||||
\begin{equation}
|
||||
f(y) = f(y^*) + f'(y^*)(y - y^*) + \frac{f''(y^*)}{2}(y - y^*)^2 + \cdots.
|
||||
\end{equation}
|
||||
Since $f(y^*) = 0$, the linearized equation near $y^*$ is
|
||||
\begin{equation}
|
||||
\label{eq:linearized}
|
||||
\frac{\mathrm{d}u}{\mathrm{d}t} = f'(y^*)\,u, \qquad u = y - y^*.
|
||||
\end{equation}
|
||||
The solution is $u(t) = u(0)\,e^{f'(y^*) t}$, which immediately yields three cases.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Stability classification for autonomous equations $\mathrm{d}y/\mathrm{d}t = f(y)$.}
|
||||
|
||||
Let $y^*$ be an equilibrium with $f(y^*) = 0$.
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{l l p{5.5cm}}
|
||||
\toprule
|
||||
\textbf{Type} & \textbf{Condition} & \textbf{Behavior} \\
|
||||
\midrule
|
||||
Asymptotically stable (sink)
|
||||
& $f'(y^*) < 0$
|
||||
& Nearby solutions converge to $y^*$ as $t \to \infty$. \\[4pt]
|
||||
Unstable (source)
|
||||
& $f'(y^*) > 0$
|
||||
& Nearby solutions depart from $y^*$ as $t$ increases. \\[4pt]
|
||||
Semi-stable (node)
|
||||
& $f'(y^*) = 0$
|
||||
& Solutions approach $y^*$ from one side and depart on the other. \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Classify the equilibria of $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}t} = y(y-1)(y-2)$.}
|
||||
|
||||
We have $f(y) = y(y-1)(y-2) = y^3 - 3y^2 + 2y$, so
|
||||
\[
|
||||
f'(y) = 3y^2 - 6y + 2.
|
||||
\]
|
||||
|
||||
Evaluate at each equilibrium:
|
||||
\begin{align*}
|
||||
f'(0) &= 3(0)^2 - 6(0) + 2 = 2 > 0
|
||||
&&\implies y^* = 0 \text{ is \textbf{unstable (source)}}. \\
|
||||
f'(1) &= 3(1)^2 - 6(1) + 2 = -1 < 0
|
||||
&&\implies y^* = 1 \text{ is \textbf{asymptotically stable (sink)}}. \\
|
||||
f'(2) &= 3(4) - 12 + 2 = 2 > 0
|
||||
&&\implies y^* = 2 \text{ is \textbf{unstable (source)}}.
|
||||
\end{align*}
|
||||
|
||||
These classifications are consistent with the phase line drawn in the previous section: $y=1$ attracts solutions from both sides, while $y=0$ and $y=2$ repel.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Semi-stable equilibrium: $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}t} = y^2$.}
|
||||
|
||||
Here $f(y) = y^2$ and $f'(y) = 2y$. At the equilibrium $y^* = 0$:
|
||||
\[
|
||||
f'(0) = 0.
|
||||
\]
|
||||
The linear test is inconclusive (derivative is zero), so we examine the sign of $f(y)$ directly:
|
||||
$f(y) = y^2 > 0$ for all $y \neq 0$. Solutions approach $y^*=0$ from below but depart to the right from above. This confirms the \textbf{semi-stable (node)} classification.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{All three types in one equation: $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}t} = y(y-1)^2$.}
|
||||
|
||||
Equilibria: $y^* = 0$ and $y^* = 1$.
|
||||
|
||||
Compute $f'(y) = (y-1)^2 + 2y(y-1) = (y-1)(3y-1)$.
|
||||
|
||||
\begin{align*}
|
||||
f'(0) &= (-1)(-1) = 1 > 0
|
||||
&&\implies y^*=0 \text{ is \textbf{unstable (source)}}. \\
|
||||
f'(1) &= (0)(2) = 0
|
||||
&&\implies \text{linear test inconclusive.}
|
||||
\end{align*}
|
||||
|
||||
For $y^*=1$, inspect the sign of $f(y) = y(y-1)^2$:
|
||||
Since $(y-1)^2 \geq 0$ always and $y > 0$ near $y=1$, we have $f(y) > 0$ on both sides of $y=1$.
|
||||
Solutions increase on both sides, so $y^*=1$ is \textbf{semi-stable} (approached from below, departed from above).
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{The Logistic Equation}
|
||||
\label{sec:ch03_logistic}
|
||||
|
||||
\paragraph{Biological motivation.}
|
||||
The simplest population model is exponential growth, $\mathrm{d}P/\mathrm{d}t = rP$, which predicts unbounded growth. In reality, resources are finite. The \textbf{logistic equation} introduces a carrying capacity $K > 0$ that caps the population:
|
||||
\begin{equation}
|
||||
\label{eq:logistic}
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = r\,y\left(1 - \frac{y}{K}\right), \qquad r > 0,\; K > 0.
|
||||
\end{equation}
|
||||
The factor $(1 - y/K)$ reduces the per-capita growth rate as $y \to K$. When $y \ll K$ the dynamics are approximately exponential; when $y \to K$ the growth rate vanishes.
|
||||
|
||||
\paragraph{Phase line analysis.}
|
||||
Set $f(y) = ry(1 - y/K)$. The equilibria are:
|
||||
\begin{align*}
|
||||
f(y) &= 0 \implies y\left(1 - \frac{y}{K}\right) = 0 \implies y^*_1 = 0,\quad y^*_2 = K.
|
||||
\end{align*}
|
||||
|
||||
Compute the derivative:
|
||||
\[
|
||||
f'(y) = r - \frac{2r}{K}y.
|
||||
\]
|
||||
|
||||
Evaluate at each equilibrium:
|
||||
\[
|
||||
f'(0) = r > 0 \;\;\text{(unstable source)}, \qquad
|
||||
f'(K) = r - 2r = -r < 0 \;\;\text{(asymptotically stable sink)}.
|
||||
\]
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\draw[thick] (0,0) -- (0,5);
|
||||
\draw[->] (0,0) -- (0,-0.3);
|
||||
\draw[->] (0,5) -- (0,5.3);
|
||||
|
||||
\filldraw[black] (0,0.6) circle (2.5pt) node[right=4pt] {$y^*=0$ \small (unstable)};
|
||||
\filldraw[black] (0,4.4) circle (2.5pt) node[right=4pt] {$y^*=K$ \small (stable)};
|
||||
|
||||
% Arrow between 0 and K: going up toward K
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,2.5) -- (-0.15,2.15);
|
||||
% Arrow below 0: going down (y < 0 is non-physical but mathematically valid)
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,0.2) -- (-0.15,-0.05);
|
||||
% Arrow above K: going down toward K
|
||||
\draw[-{Latex[length=4mm]}, red!70!black, very thick] (-0.15,4.9) -- (-0.15,4.65);
|
||||
|
||||
\node[below] at (0,-0.5) {$y$};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
All solutions with $y(0) > 0$ approach the carrying capacity $K$ as $t \to \infty$.
|
||||
|
||||
\paragraph{Exact solution by separation of variables.}
|
||||
We solve \cref{eq:logistic} with initial condition $y(0) = y_0$:
|
||||
\begin{align*}
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t}
|
||||
&= r\,y\left(1 - \frac{y}{K}\right) \\[6pt]
|
||||
\frac{\mathrm{d}y}{y\left(1 - \dfrac{y}{K}\right)}
|
||||
&= r\,\mathrm{d}t.
|
||||
\end{align*}
|
||||
|
||||
Decompose the left-hand side using partial fractions:
|
||||
\[
|
||||
\frac{1}{y\left(1 - \frac{y}{K}\right)}
|
||||
= \frac{K}{K y - y^2}
|
||||
= \frac{1}{y} + \frac{1}{K - y}.
|
||||
\]
|
||||
\textit{Verification:} $\displaystyle\frac{1}{y} + \frac{1}{K-y} = \frac{K-y+y}{y(K-y)} = \frac{K}{y(K-y)}$.
|
||||
|
||||
Integrate both sides:
|
||||
\begin{align*}
|
||||
\int \left(\frac{1}{y} + \frac{1}{K-y}\right) \mathrm{d}y
|
||||
&= \int r\,\mathrm{d}t \\
|
||||
\ln|y| - \ln|K-y| &= rt + C \\
|
||||
\ln\left|\frac{y}{K-y}\right| &= rt + C \\
|
||||
\frac{y}{K-y} &= A\,e^{rt}, \qquad A = \pm e^C.
|
||||
\end{align*}
|
||||
|
||||
Solve for $y$:
|
||||
\begin{align*}
|
||||
y &= (K-y)\,A\,e^{rt} \\
|
||||
y(1 + A e^{rt}) &= K A e^{rt} \\
|
||||
y(t) &= \frac{K A e^{rt}}{1 + A e^{rt}}
|
||||
= \frac{K}{1 + \dfrac{1}{A}e^{-rt}}.
|
||||
\end{align*}
|
||||
|
||||
Apply the initial condition $y(0) = y_0$:
|
||||
\[
|
||||
y_0 = \frac{K}{1 + \tfrac{1}{A}}
|
||||
\implies 1 + \tfrac{1}{A} = \frac{K}{y_0}
|
||||
\implies \frac{1}{A} = \frac{K - y_0}{y_0}.
|
||||
\]
|
||||
Denoting $\displaystyle A' = \frac{K - y_0}{y_0}$, the final form is
|
||||
\begin{equation}
|
||||
\label{eq:logistic_solution}
|
||||
y(t) = \frac{K}{1 + A' e^{-rt}},
|
||||
\qquad A' = \frac{K - y_0}{y_0}.
|
||||
\end{equation}
|
||||
As $t \to \infty$, $e^{-rt} \to 0$ and $y(t) \to K$, confirming the phase line prediction.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Logistic growth with $r = 0.5$, $K = 100$, $y_0 = 10$.}
|
||||
|
||||
The equation is $\mathrm{d}y/\mathrm{d}t = 0.5\,y(1 - y/100)$.
|
||||
|
||||
The constant $A'$ is
|
||||
\[
|
||||
A' = \frac{100 - 10}{10} = 9.
|
||||
\]
|
||||
The solution is
|
||||
\[
|
||||
y(t) = \frac{100}{1 + 9\,e^{-0.5t}}.
|
||||
\]
|
||||
|
||||
At $t = 5$:
|
||||
\[
|
||||
y(5) = \frac{100}{1 + 9\,e^{-2.5}}
|
||||
= \frac{100}{1 + 9 \cdot 0.0821}
|
||||
= \frac{100}{1.739} \approx 57.5.
|
||||
\]
|
||||
The population reaches about 58\% of its carrying capacity after 5 time units.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Euler's Method}
|
||||
\label{sec:ch03_eulers_method}
|
||||
|
||||
Many differential equations do not admit closed-form solutions. \textbf{Euler's method} provides a simple numerical procedure to approximate the solution.
|
||||
|
||||
\paragraph{Derivation from Taylor expansion.}
|
||||
Let $y(t)$ be the exact solution of $y' = f(t,y)$. Expand $y(t_{n+1})$ about $t_n$ using Taylor's theorem:
|
||||
\begin{equation}
|
||||
y(t_{n+1}) = y(t_n) + h\,y'(t_n) + \frac{h^2}{2}\,y''(\xi_n),
|
||||
\qquad \xi_n \in (t_n, t_{n+1}),
|
||||
\end{equation}
|
||||
where $h = t_{n+1} - t_n$ is the step size. Since $y'(t_n) = f(t_n, y(t_n))$,
|
||||
\[
|
||||
y(t_{n+1}) = y(t_n) + h\,f(t_n, y(t_n)) + \frac{h^2}{2}\,y''(\xi_n).
|
||||
\]
|
||||
Euler's method retains only the first two terms, discarding the $O(h^2)$ remainder.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Euler's method.}
|
||||
Given $y' = f(t,y)$ with $y(t_0) = y_0$ and step size $h$:
|
||||
\begin{align}
|
||||
t_{n+1} &= t_n + h, \label{eq:euler_t} \\
|
||||
y_{n+1} &= y_n + h\,f(t_n, y_n). \label{eq:euler_y}
|
||||
\end{align}
|
||||
\end{keyresult}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Error analysis.}
|
||||
|
||||
\begin{itemize}
|
||||
\item \textbf{Local truncation error} (error per step):
|
||||
\[ \tau_{n+1} = \frac{h^2}{2}\,y''(\xi_n) = O(h^2). \]
|
||||
\item \textbf{Global truncation error} (accumulated error at a fixed $t$ after $N = (t-t_0)/h$ steps):
|
||||
\[ \varepsilon_N = O(h). \]
|
||||
\end{itemize}
|
||||
|
||||
Halving the step size roughly halves the global error, but doubles the number of steps.
|
||||
\end{hintbox}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Approximate $y' = y$, $y(0) = 1$ using Euler's method with $h = 0.1$.}
|
||||
|
||||
The exact solution is $y(t) = e^t$. Here $f(t,y) = y$, so the iteration is
|
||||
\[
|
||||
y_{n+1} = y_n + 0.1\,y_n = 1.1\,y_n.
|
||||
\]
|
||||
|
||||
Starting from $y_0 = 1$:
|
||||
\begin{center}
|
||||
\begin{tabular}{c c c c c c}
|
||||
\toprule
|
||||
$n$ & $t_n$ & $y_n$ (Euler) & $y(t_n) = e^{t_n}$ (exact) & Local error $\tau$ & Cumulative error \\
|
||||
\midrule
|
||||
0 & 0.0 & 1.0000 & 1.00000 & $-$ & 0.0000 \\
|
||||
1 & 0.1 & 1.1000 & 1.10517 & 0.00059 & 0.0052 \\
|
||||
2 & 0.2 & 1.2100 & 1.22140 & 0.00061 & 0.0114 \\
|
||||
3 & 0.3 & 1.3310 & 1.34986 & 0.00063 & 0.0189 \\
|
||||
4 & 0.4 & 1.4641 & 1.49182 & 0.00066 & 0.0277 \\
|
||||
5 & 0.5 & 1.6105 & 1.64872 & 0.00069 & 0.0382 \\
|
||||
6 & 0.6 & 1.7716 & 1.82212 & 0.00072 & 0.0505 \\
|
||||
7 & 0.7 & 1.9487 & 2.01375 & 0.00075 & 0.0650 \\
|
||||
8 & 0.8 & 2.1436 & 2.22554 & 0.00079 & 0.0819 \\
|
||||
9 & 0.9 & 2.3579 & 2.45960 & 0.00083 & 0.1017 \\
|
||||
10 & 1.0 & 2.5937 & 2.71828 & 0.00087 & 0.1246 \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
At $t = 1.0$, Euler's method gives $y_{10} = 2.5937$ while the exact value is $e \approx 2.7183$. The relative error is
|
||||
\[
|
||||
\frac{|2.71828 - 2.59374|}{2.71828} \approx 4.58\%.
|
||||
\]
|
||||
|
||||
The error grows because each step's $O(h^2)$ local error compounds. With the same $h = 0.1$, the error at $t = 1$ is proportional to $h \cdot t \approx 0.1$. To reduce the error by a factor of 10, use $h = 0.01$.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Euler's method on $y' = -2y$, $y(0) = 3$, with $h = 0.5$.}
|
||||
|
||||
Here $f(t,y) = -2y$, so $y_{n+1} = y_n - h(2y_n) = y_n(1 - 2h)$.
|
||||
|
||||
With $h = 0.5$: $y_{n+1} = y_n(1 - 1) = 0$ for $n \geq 1$. This is an \textbf{unstable} outcome: any numerical noise would blow up because $|1 - 2h| = 0$ is at the boundary.
|
||||
|
||||
The exact solution is $y(t) = 3e^{-2t}$. At $t = 1$: $y(1) = 3e^{-2} \approx 0.406$.
|
||||
|
||||
Euler with $h = 0.5$ gives $y_2 = 0$ at $t = 1$, which is far off.
|
||||
|
||||
With a smaller step $h = 0.1$: $y_{n+1} = y_n(1 - 0.2) = 0.8\,y_n$.
|
||||
\[
|
||||
y_{10} = 3 \cdot (0.8)^{10} = 3 \cdot 0.1074 \approx 0.322.
|
||||
\]
|
||||
This is much closer to the exact $0.406$.
|
||||
|
||||
\textit{Lesson:} For stiff equations or large negative eigenvalues, small step sizes are essential for Euler's method.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Applications}
|
||||
\label{sec:ch03_applications}
|
||||
|
||||
\subsubsection{Newton's Law of Cooling}
|
||||
\label{sec:ch03_cooling}
|
||||
|
||||
\textbf{Physical setup.} A hot object placed in a cooler environment loses heat at a rate proportional to the temperature difference between the object and the ambient medium.
|
||||
|
||||
\textbf{Model.} Let $T(t)$ be the object's temperature and $T_a$ the constant ambient temperature. Newton's law states:
|
||||
\begin{equation}
|
||||
\label{eq:newton_cooling}
|
||||
\frac{\mathrm{d}T}{\mathrm{d}t} = k\,(T - T_a), \qquad k < 0.
|
||||
\end{equation}
|
||||
The equilibrium is $T^* = T_a$ (the object eventually reaches ambient temperature). Since $f'(T_a) = k < 0$, the equilibrium is asymptotically stable.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{A cup of coffee cools from $90^\circ$C to $70^\circ$C in 10 minutes in a $20^\circ$C room. How long until it reaches $40^\circ$C?}
|
||||
|
||||
\textit{Step 1: Solve the ODE.} Separate variables:
|
||||
\[
|
||||
\frac{\mathrm{d}T}{T - T_a} = k\,\mathrm{d}t
|
||||
\implies \ln|T - T_a| = kt + C
|
||||
\implies T(t) = T_a + (T_0 - T_a)e^{kt}.
|
||||
\]
|
||||
|
||||
With $T_a = 20$ and $T_0 = 90$:
|
||||
\[
|
||||
T(t) = 20 + 70\,e^{kt}.
|
||||
\]
|
||||
|
||||
\textit{Step 2: Determine $k$.} At $t = 10$:
|
||||
\[
|
||||
70 = 20 + 70\,e^{10k}
|
||||
\implies e^{10k} = \frac{50}{70} = \frac{5}{7}
|
||||
\implies k = \frac{1}{10}\ln\!\left(\frac{5}{7}\right) \approx -0.0336.
|
||||
\]
|
||||
|
||||
\textit{Step 3: Find the time to reach $40^\circ$C.}
|
||||
\[
|
||||
40 = 20 + 70\,e^{-0.0336t}
|
||||
\implies e^{-0.0336t} = \frac{20}{70} = \frac{2}{7}.
|
||||
\]
|
||||
\[
|
||||
-0.0336t = \ln\!\left(\frac{2}{7}\right)
|
||||
\implies t = \frac{\ln(2/7)}{-0.0336}
|
||||
= \frac{-1.253}{-0.0336} \approx 37.3 \text{ minutes}.
|
||||
\]
|
||||
|
||||
The coffee reaches $40^\circ$C after approximately \textbf{37.3 minutes}.
|
||||
\end{workedexample}
|
||||
|
||||
\subsubsection{Mixing Problems}
|
||||
\label{sec:ch03_mixing}
|
||||
|
||||
\textbf{Physical setup.} A tank contains a solution of salt and water. Brine flows in at a known concentration and rate, and the well-stirred mixture flows out at a known rate. We wish to track the amount of salt $Q(t)$ in the tank.
|
||||
|
||||
\textbf{Model.} Let $V(t)$ be the volume, $c_1$ the concentration of incoming brine, and $r_1$, $r_2$ the inflow and outflow rates. The rate of change of salt is
|
||||
\begin{equation}
|
||||
\label{eq:mixing}
|
||||
\frac{\mathrm{d}Q}{\mathrm{d}t} = \underbrace{c_1 r_1}_{\text{rate in}}
|
||||
- \underbrace{\frac{Q(t)}{V(t)}\,r_2}_{\text{rate out}}.
|
||||
\end{equation}
|
||||
If $r_1 = r_2 = r$ then $V(t) = V_0$ is constant, and the equation becomes linear:
|
||||
\begin{equation}
|
||||
\frac{\mathrm{d}Q}{\mathrm{d}t} + \frac{r}{V_0}\,Q = c_1 r.
|
||||
\end{equation}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{A tank holds 100~L of brine with 20~kg of dissolved salt. Brine containing 0.5~kg/L flows in at 3~L/min. The well-mixed solution flows out at 3~L/min. Find $Q(t)$ and the long-term salt content.}
|
||||
|
||||
\textit{Parameters:} $V_0 = 100$~L, $Q(0) = 20$~kg, $c_1 = 0.5$~kg/L, $r = 3$~L/min.
|
||||
|
||||
\textit{ODE:} Since inflow = outflow, $V(t) = 100$ is constant.
|
||||
\[
|
||||
\frac{\mathrm{d}Q}{\mathrm{d}t} = (0.5)(3) - \frac{Q}{100}(3)
|
||||
= 1.5 - \frac{3}{100}Q
|
||||
= 1.5 - 0.03\,Q.
|
||||
\]
|
||||
|
||||
\textit{Solve:} This is a linear first-order equation. The integrating factor is
|
||||
\[
|
||||
\mu(t) = e^{\int 0.03\,\mathrm{d}t} = e^{0.03t}.
|
||||
\]
|
||||
Multiply through:
|
||||
\[
|
||||
\frac{\mathrm{d}}{\mathrm{d}t}\!\left(Q e^{0.03t}\right)
|
||||
= 1.5\,e^{0.03t}.
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
Q(t)\,e^{0.03t} = \frac{1.5}{0.03}\,e^{0.03t} + C
|
||||
= 50\,e^{0.03t} + C,
|
||||
\]
|
||||
\[
|
||||
Q(t) = 50 + C\,e^{-0.03t}.
|
||||
\]
|
||||
|
||||
Apply $Q(0) = 20$: $20 = 50 + C \implies C = -30$.
|
||||
\[
|
||||
\boxed{Q(t) = 50 - 30\,e^{-0.03t}}.
|
||||
\]
|
||||
|
||||
\textit{Long-term behavior:} As $t \to \infty$, $Q(t) \to 50$~kg. This makes physical sense: the equilibrium concentration matches the inflow concentration, so the tank eventually holds $0.5 \times 100 = 50$~kg of salt.
|
||||
|
||||
\textit{Phase line interpretation:} The autonomous equation $Q' = 1.5 - 0.03Q$ has one equilibrium at $Q^* = 50$. Since $f'(Q^*) = -0.03 < 0$, it is asymptotically stable.
|
||||
\end{workedexample}
|
||||
|
||||
\subsubsection{Falling Body with Air Resistance}
|
||||
\label{sec:ch03_falling_body}
|
||||
|
||||
\textbf{Physical setup.} A body of mass $m$ falls under gravity. Air resistance opposes the motion. We model the speed $v(t)$ (positive downward).
|
||||
|
||||
\textbf{Model with linear drag.} The drag force is proportional to speed: $F_{\text{drag}} = -kv$. Newton's second law gives
|
||||
\begin{equation}
|
||||
\label{eq:falling_linear}
|
||||
m\,\frac{\mathrm{d}v}{\mathrm{d}t} = mg - kv.
|
||||
\end{equation}
|
||||
The equilibrium (terminal velocity) is found by setting $v' = 0$:
|
||||
\[
|
||||
mg - kv^* = 0 \implies v^* = \frac{mg}{k}.
|
||||
\]
|
||||
Since $f'(v^*) = -k/m < 0$, the terminal velocity is asymptotically stable: any initial speed eventually converges to $v^*$.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{A 10~kg object falls under gravity ($g = 9.8$~m/s\textsuperscript{2}) with linear air resistance ($k = 2$~N$\cdot$s/m). Find the terminal velocity and $v(t)$ if the object is dropped from rest.}
|
||||
|
||||
\textit{Terminal velocity:}
|
||||
\[
|
||||
v^* = \frac{mg}{k} = \frac{10 \cdot 9.8}{2} = 49 \text{ m/s}.
|
||||
\]
|
||||
|
||||
\textit{Solve the ODE:} Divide by $m = 10$:
|
||||
\[
|
||||
\frac{\mathrm{d}v}{\mathrm{d}t} = 9.8 - \frac{2}{10}v = 9.8 - 0.2\,v.
|
||||
\]
|
||||
Separate:
|
||||
\[
|
||||
\frac{\mathrm{d}v}{9.8 - 0.2v} = \mathrm{d}t
|
||||
\implies -\frac{1}{0.2}\ln|9.8 - 0.2v| = t + C.
|
||||
\]
|
||||
With $v(0) = 0$: $-\frac{1}{0.2}\ln(9.8) = C$.
|
||||
\[
|
||||
\ln|9.8 - 0.2v| = -0.2t + \ln(9.8)
|
||||
\implies 9.8 - 0.2v = 9.8\,e^{-0.2t},
|
||||
\]
|
||||
\[
|
||||
\boxed{v(t) = 49\bigl(1 - e^{-0.2t}\bigr)}.
|
||||
\]
|
||||
|
||||
\textit{Check:} As $t \to \infty$, $v(t) \to 49$~m/s (terminal velocity). At $t = 5$~s:
|
||||
\[
|
||||
v(5) = 49(1 - e^{-1}) = 49(1 - 0.368) \approx 31.1 \text{ m/s}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\textbf{Model with quadratic drag.} For high-speed motion (e.g., skydiving), the drag force is proportional to $v^2$:
|
||||
\begin{equation}
|
||||
\label{eq:falling_quadratic}
|
||||
m\,\frac{\mathrm{d}v}{\mathrm{d}t} = mg - kv^2.
|
||||
\end{equation}
|
||||
The terminal velocity is $v^* = \sqrt{mg/k}$. The qualitative behavior is the same: solutions converge to $v^*$ asymptotically. The approach, however, is different from the linear case.
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch03_summary}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Chapter 3 Summary: Qualitative Analysis and Numerical Methods}
|
||||
\label{tab:ch03_summary}
|
||||
\begin{tabular}{l p{5.5cm}}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key Formula/Method} \\
|
||||
\midrule
|
||||
Autonomous equation & $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t} = f(y)$, no explicit $t$-dependence \\
|
||||
Equilibrium & $y^*$ such that $f(y^*) = 0$ \\
|
||||
Phase line & Vertical line with equilibria marked and arrows showing flow direction \\
|
||||
Stability (sink) & $f'(y^*) < 0 \;\Rightarrow$ asymptotically stable \\
|
||||
Stability (source) & $f'(y^*) > 0 \;\Rightarrow$ unstable \\
|
||||
Semi-stable (node) & $f'(y^*) = 0$; sign of $f$ determines approach/departure \\
|
||||
Logistic equation & $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t} = ry\!\left(1 - \frac{y}{K}\right)$ \\
|
||||
Logistic solution & $\displaystyle y(t) = \frac{K}{1 + A e^{-rt}}, \;\; A = \frac{K-y_0}{y_0}$ \\
|
||||
Euler's method & $y_{n+1} = y_n + h\,f(t_n, y_n)$ \\
|
||||
Euler global error & $O(h)$ \\
|
||||
Newton's cooling & $\displaystyle T(t) = T_a + (T_0 - T_a)e^{kt}, \; k < 0$ \\
|
||||
Mixing (constant $V$) & $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t} + \frac{r}{V_0}Q = c_1 r$ \\
|
||||
Linear drag terminal velocity & $v^* = mg/k$ \\
|
||||
Quadratic drag terminal velocity & $v^* = \sqrt{mg/k}$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
@@ -1,751 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch04_second_order_homogeneous.tex
|
||||
% Chapter 4: Second-Order Linear Homogeneous Equations
|
||||
% =============================================================================
|
||||
|
||||
\section{Second-Order Homogeneous}
|
||||
\label{ch:second_order_homogeneous}
|
||||
|
||||
\subsection{The Characteristic Equation}
|
||||
\label{sec:ch04_characteristic_equation}
|
||||
|
||||
We consider the \textbf{second-order linear homogeneous differential equation with constant coefficients}:
|
||||
\begin{equation}
|
||||
\label{eq:second_order_homogeneous}
|
||||
a\,y'' + b\,y' + c\,y = 0,
|
||||
\qquad a, b, c \in \R,\;\; a \neq 0.
|
||||
\end{equation}
|
||||
This is the simplest class of second-order ODEs that is both tractable analytically and broadly applicable to physical systems (mechanical vibrations, electrical circuits, etc.).
|
||||
|
||||
\paragraph{Derivation.} The key insight is that the exponential function $y = e^{rx}$ has derivatives proportional to itself, making it a natural candidate for solutions of linear equations with constant coefficients. Substituting the \textbf{ansatz}
|
||||
\[
|
||||
y = e^{rx}, \qquad y' = r\,e^{rx}, \qquad y'' = r^2\,e^{rx}
|
||||
\]
|
||||
into \cref{eq:second_order_homogeneous} gives
|
||||
\[
|
||||
a\,r^2 e^{rx} + b\,r\,e^{rx} + c\,e^{rx} = 0.
|
||||
\]
|
||||
Since $e^{rx} \neq 0$ for all $x$, we divide through and obtain the \textbf{characteristic equation} (also called the \textbf{auxiliary equation}):
|
||||
\begin{equation}
|
||||
\label{eq:characteristic}
|
||||
a\,r^2 + b\,r + c = 0.
|
||||
\end{equation}
|
||||
This is an ordinary quadratic equation. Its roots are given by the quadratic formula:
|
||||
\begin{equation}
|
||||
\label{eq:quadratic_formula}
|
||||
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
|
||||
\end{equation}
|
||||
|
||||
The nature of the roots depends on the \textbf{discriminant}
|
||||
\begin{equation}
|
||||
\Delta = b^2 - 4ac.
|
||||
\end{equation}
|
||||
There are three cases.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Three cases for second-order homogeneous equations.}
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{l l l}
|
||||
\toprule
|
||||
\textbf{Discriminant} & \textbf{Roots} & \textbf{Solution form} \\
|
||||
\midrule
|
||||
$\Delta > 0$ & Two distinct real roots $r_1 \neq r_2$ & $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ \\[6pt]
|
||||
$\Delta = 0$ & One repeated real root $r = -b/(2a)$ & $y = c_1 e^{rx} + c_2 x\,e^{rx}$ \\[6pt]
|
||||
$\Delta < 0$ & Complex conjugates $r = \alpha \pm i\beta$ & $y = e^{\alpha x}\bigl[c_1 \cos(\beta x) + c_2 \sin(\beta x)\bigr]$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{keyresult}
|
||||
|
||||
We treat each case in detail in the following subsections. First, two quick examples to illustrate the method.
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - 5y' + 6y = 0$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 - 5r + 6 = 0 \quad\Longrightarrow\quad (r - 2)(r - 3) = 0.
|
||||
\]
|
||||
The roots are $r_1 = 2$ and $r_2 = 3$ (two distinct real roots, $\Delta = 25 - 24 = 1 > 0$). The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{2x} + c_2 e^{3x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + 4y = 0$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + 4 = 0 \quad\Longrightarrow\quad r^2 = -4 \quad\Longrightarrow\quad r = \pm 2i.
|
||||
\]
|
||||
Here $\alpha = 0$ and $\beta = 2$, so the general solution is
|
||||
\[
|
||||
y(x) = c_1 \cos(2x) + c_2 \sin(2x).
|
||||
\]
|
||||
This represents undamped oscillations (we will see the physical interpretation in \cref{sec:ch04_complex_roots}).
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Case 1: Distinct Real Roots}
|
||||
\label{sec:ch04_distinct_real_roots}
|
||||
|
||||
When $\Delta = b^2 - 4ac > 0$, the characteristic equation \cref{eq:characteristic} has two distinct real roots
|
||||
\[
|
||||
r_1 = \frac{-b + \sqrt{\Delta}}{2a}, \qquad r_2 = \frac{-b - \sqrt{\Delta}}{2a}.
|
||||
\]
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Distinct real roots.} If $r_1 \neq r_2$ are real, then $\{e^{r_1 x}, e^{r_2 x}\}$ is a fundamental set of solutions and the general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}.
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Verification.} Each exponential satisfies the ODE by construction (that is how the characteristic equation was derived). The general solution follows from the superposition principle (\cref{thm:superposition}): any linear combination of solutions is again a solution.
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - y' - 2y = 0$ with initial conditions $y(0) = 3$ and $y'(0) = 1$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 - r - 2 = 0 \quad\Longrightarrow\quad (r - 2)(r + 1) = 0.
|
||||
\]
|
||||
Roots: $r_1 = 2$, $r_2 = -1$. The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{2x} + c_2 e^{-x}.
|
||||
\]
|
||||
|
||||
Differentiate:
|
||||
\[
|
||||
y'(x) = 2c_1 e^{2x} - c_2 e^{-x}.
|
||||
\]
|
||||
|
||||
Apply initial conditions:
|
||||
\[
|
||||
\begin{cases}
|
||||
y(0) = c_1 + c_2 = 3, \\
|
||||
y'(0) = 2c_1 - c_2 = 1.
|
||||
\end{cases}
|
||||
\]
|
||||
Adding the two equations: $3c_1 = 4$, so $c_1 = \dfrac{4}{3}$. From the first equation: $c_2 = 3 - \dfrac{4}{3} = \dfrac{5}{3}$.
|
||||
|
||||
The solution is
|
||||
\[
|
||||
y(x) = \frac{4}{3}\,e^{2x} + \frac{5}{3}\,e^{-x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $2y'' + 7y' + 3y = 0$ with $y(0) = 4$ and $y'(0) = -5$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
2r^2 + 7r + 3 = 0.
|
||||
\]
|
||||
Using the quadratic formula:
|
||||
\[
|
||||
r = \frac{-7 \pm \sqrt{49 - 24}}{4} = \frac{-7 \pm 5}{4}.
|
||||
\]
|
||||
So $r_1 = \dfrac{-7 + 5}{4} = -\dfrac{1}{2}$ and $r_2 = \dfrac{-7 - 5}{4} = -3$.
|
||||
|
||||
The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{-x/2} + c_2 e^{-3x}.
|
||||
\]
|
||||
|
||||
Differentiate:
|
||||
\[
|
||||
y'(x) = -\tfrac{1}{2}\,c_1\,e^{-x/2} - 3c_2\,e^{-3x}.
|
||||
\]
|
||||
|
||||
Apply initial conditions:
|
||||
\[
|
||||
\begin{cases}
|
||||
y(0) = c_1 + c_2 = 4, \\
|
||||
y'(0) = -\tfrac{1}{2}\,c_1 - 3c_2 = -5.
|
||||
\end{cases}
|
||||
\]
|
||||
From the first equation: $c_1 = 4 - c_2$. Substitute into the second:
|
||||
\[
|
||||
-\tfrac{1}{2}(4 - c_2) - 3c_2 = -5 \quad\Longrightarrow\quad -2 + \tfrac{1}{2}c_2 - 3c_2 = -5.
|
||||
\]
|
||||
\[
|
||||
-\tfrac{5}{2}c_2 = -3 \quad\Longrightarrow\quad c_2 = \frac{6}{5}, \qquad c_1 = 4 - \frac{6}{5} = \frac{14}{5}.
|
||||
\]
|
||||
|
||||
The solution is
|
||||
\[
|
||||
y(x) = \frac{14}{5}\,e^{-x/2} + \frac{6}{5}\,e^{-3x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - 4y = 0$ with $y(0) = 0$ and $y(1) = e^2 - e^{-2}$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is $r^2 - 4 = 0$, so $r = \pm 2$. The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{2x} + c_2 e^{-2x}.
|
||||
\]
|
||||
|
||||
Apply the first condition:
|
||||
\[
|
||||
y(0) = c_1 + c_2 = 0 \quad\Longrightarrow\quad c_2 = -c_1.
|
||||
\]
|
||||
The solution simplifies to $y(x) = c_1(e^{2x} - e^{-2x})$.
|
||||
|
||||
Apply the second condition:
|
||||
\[
|
||||
y(1) = c_1(e^2 - e^{-2}) = e^2 - e^{-2} \quad\Longrightarrow\quad c_1 = 1.
|
||||
\]
|
||||
|
||||
The solution is
|
||||
\[
|
||||
y(x) = e^{2x} - e^{-2x} = 2\sinh(2x).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Case 2: Repeated Real Roots}
|
||||
\label{sec:ch04_repeated_roots}
|
||||
|
||||
When $\Delta = 0$, the characteristic equation has a single (double) root
|
||||
\[
|
||||
r = -\frac{b}{2a}.
|
||||
\]
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Repeated real root.} If $r$ is a double root, the general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{rx} + c_2\,x\,e^{rx}.
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Why the $x$ factor?} With only one root $r$, the function $e^{rx}$ gives us just one solution. We need a second linearly independent solution to form the general solution of a second-order equation. The method of \textbf{reduction of order} provides this second solution.
|
||||
|
||||
\subparagraph{Derivation via reduction of order.}
|
||||
Suppose $y_1(x) = e^{rx}$ is one solution of \cref{eq:second_order_homogeneous} (with $\Delta = 0$). We seek a second solution of the form
|
||||
\[
|
||||
y_2(x) = v(x)\,y_1(x) = v(x)\,e^{rx},
|
||||
\]
|
||||
where $v(x)$ is an unknown function to be determined. Compute the derivatives:
|
||||
\[
|
||||
y_2' = (v' + rv)\,e^{rx}, \qquad y_2'' = (v'' + 2rv' + r^2 v)\,e^{rx}.
|
||||
\]
|
||||
Substitute $y_2$ into the ODE $y'' + \frac{b}{a}y' + \frac{c}{a}y = 0$ (dividing \cref{eq:second_order_homogeneous} by $a$):
|
||||
\[
|
||||
\bigl(v'' + 2rv' + r^2 v\bigr)e^{rx} + \frac{b}{a}\bigl(v' + rv\bigr)e^{rx} + \frac{c}{a}\,v\,e^{rx} = 0.
|
||||
\]
|
||||
Dividing by $e^{rx}$ and collecting terms:
|
||||
\[
|
||||
v'' + \left(2r + \frac{b}{a}\right)v' + \left(r^2 + \frac{b}{a}r + \frac{c}{a}\right)v = 0.
|
||||
\]
|
||||
The $v$-coefficient is exactly the characteristic polynomial evaluated at $r$, which vanishes because $r$ is a root:
|
||||
\[
|
||||
r^2 + \frac{b}{a}r + \frac{c}{a} = 0.
|
||||
\]
|
||||
For a repeated root, $r = -b/(2a)$, so the $v'$-coefficient is also zero:
|
||||
\[
|
||||
2r + \frac{b}{a} = 2\!\left(-\frac{b}{2a}\right) + \frac{b}{a} = 0.
|
||||
\]
|
||||
We are left with the simple equation
|
||||
\[
|
||||
v''(x) = 0 \quad\Longrightarrow\quad v(x) = Ax + B.
|
||||
\]
|
||||
Choosing $A = 1$ and $B = 0$ gives $y_2(x) = x\,e^{rx}$, a second linearly independent solution. The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{rx} + c_2 x\,e^{rx}.
|
||||
\]
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Quick check.} You can verify that $y = x\,e^{rx}$ satisfies $ay'' + by' + cy = 0$ when $r = -b/(2a)$ and $b^2 = 4ac$ by direct substitution. This is always a safe verification step.
|
||||
\end{hintbox}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - 6y' + 9y = 0$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 - 6r + 9 = 0 \quad\Longrightarrow\quad (r - 3)^2 = 0.
|
||||
\]
|
||||
Double root: $r = 3$. The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{3x} + c_2 x\,e^{3x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + 4y' + 4y = 0$ with $y(0) = 2$ and $y'(0) = 0$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + 4r + 4 = 0 \quad\Longrightarrow\quad (r + 2)^2 = 0.
|
||||
\]
|
||||
Double root: $r = -2$. The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{-2x} + c_2 x\,e^{-2x}.
|
||||
\]
|
||||
|
||||
Differentiate:
|
||||
\[
|
||||
y'(x) = -2c_1 e^{-2x} + c_2 e^{-2x} - 2c_2 x\,e^{-2x}
|
||||
= \bigl(-2c_1 + c_2 - 2c_2 x\bigr)e^{-2x}.
|
||||
\]
|
||||
|
||||
Apply initial conditions:
|
||||
\[
|
||||
\begin{cases}
|
||||
y(0) = c_1 = 2, \\
|
||||
y'(0) = -2c_1 + c_2 = 0 \quad\Longrightarrow\quad c_2 = 2c_1 = 4.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
The solution is
|
||||
\[
|
||||
y(x) = 2\,e^{-2x} + 4x\,e^{-2x} = 2(1 + 2x)\,e^{-2x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Case 3: Complex Conjugate Roots}
|
||||
\label{sec:ch04_complex_roots}
|
||||
|
||||
When $\Delta = b^2 - 4ac < 0$, the characteristic equation has two complex conjugate roots:
|
||||
\[
|
||||
r = \alpha \pm i\beta, \qquad
|
||||
\alpha = -\frac{b}{2a}, \quad
|
||||
\beta = \frac{\sqrt{4ac - b^2}}{2a}.
|
||||
\]
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Complex conjugate roots.} If $r = \alpha \pm i\beta$ with $\beta > 0$, the general real-valued solution is
|
||||
\[
|
||||
y(x) = e^{\alpha x}\Bigl[c_1 \cos(\beta x) + c_2 \sin(\beta x)\Bigr].
|
||||
\]
|
||||
Equivalently, using amplitude–phase form:
|
||||
\[
|
||||
y(x) = A\,e^{\alpha x}\cos(\beta x - \phi),
|
||||
\]
|
||||
where $A = \sqrt{c_1^2 + c_2^2}$ and $\phi = \arctan(c_2/c_1)$.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Derivation from Euler's formula.}
|
||||
The complex-valued solutions are $e^{(\alpha + i\beta)x}$ and $e^{(\alpha - i\beta)x}$. By Euler's formula,
|
||||
\[
|
||||
e^{(\alpha + i\beta)x} = e^{\alpha x}\,e^{i\beta x}
|
||||
= e^{\alpha x}\bigl[\cos(\beta x) + i\,\sin(\beta x)\bigr].
|
||||
\]
|
||||
Because the ODE has real coefficients, both the real and imaginary parts are themselves real solutions:
|
||||
\begin{align*}
|
||||
\Re\!\left[e^{(\alpha+i\beta)x}\right] &= e^{\alpha x}\cos(\beta x), \\
|
||||
\Im\!\left[e^{(\alpha+i\beta)x}\right] &= e^{\alpha x}\sin(\beta x).
|
||||
\end{align*}
|
||||
These two functions are linearly independent (one is not a constant multiple of the other), so they form a fundamental set. The general solution is their linear combination, giving the formula above.
|
||||
|
||||
\paragraph{Physical interpretation: damped oscillations.}
|
||||
The parameter $\alpha$ governs the exponential growth ($\alpha > 0$) or decay ($\alpha < 0$) of the amplitude, while $\beta$ determines the angular frequency of the oscillation. In mechanical systems, this corresponds to \textbf{underdamped motion}: the mass oscillates about equilibrium while the amplitude decays exponentially (if $\alpha < 0$). When $\alpha = 0$ (i.e.\ $b = 0$), we have pure undamped oscillations with constant amplitude.
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + 2y' + 5y = 0$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + 2r + 5 = 0.
|
||||
\]
|
||||
Discriminant: $\Delta = 4 - 20 = -16 < 0$. The roots are
|
||||
\[
|
||||
r = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i.
|
||||
\]
|
||||
Here $\alpha = -1$ and $\beta = 2$. The general solution is
|
||||
\[
|
||||
y(x) = e^{-x}\Bigl[c_1 \cos(2x) + c_2 \sin(2x)\Bigr].
|
||||
\]
|
||||
This represents damped oscillations with amplitude decaying like $e^{-x}$.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + 4y' + 13y = 0$ with $y(0) = 2$ and $y'(0) = 3$.
|
||||
|
||||
\textbf{Solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + 4r + 13 = 0.
|
||||
\]
|
||||
Discriminant: $\Delta = 16 - 52 = -36 < 0$. The roots are
|
||||
\[
|
||||
r = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i.
|
||||
\]
|
||||
So $\alpha = -2$ and $\beta = 3$. The general solution is
|
||||
\[
|
||||
y(x) = e^{-2x}\Bigl[c_1 \cos(3x) + c_2 \sin(3x)\Bigr].
|
||||
\]
|
||||
|
||||
Differentiate (product rule):
|
||||
\[
|
||||
y'(x) = -2\,e^{-2x}\bigl[c_1 \cos(3x) + c_2 \sin(3x)\bigr]
|
||||
+ e^{-2x}\bigl[-3c_1 \sin(3x) + 3c_2 \cos(3x)\bigr].
|
||||
\]
|
||||
|
||||
Apply initial conditions:
|
||||
\[
|
||||
\begin{cases}
|
||||
y(0) = c_1 = 2, \\[6pt]
|
||||
y'(0) = -2c_1 + 3c_2 = 3 \quad\Longrightarrow\quad -4 + 3c_2 = 3 \quad\Longrightarrow\quad c_2 = \dfrac{7}{3}.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
The solution is
|
||||
\[
|
||||
y(x) = e^{-2x}\left(2\cos(3x) + \frac{7}{3}\sin(3x)\right).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Superposition Principle}
|
||||
\label{sec:ch04_superposition}
|
||||
|
||||
The entire theory rests on the linearity of the differential operator.
|
||||
|
||||
\begin{theorem}[Superposition Principle]
|
||||
\label{thm:superposition}
|
||||
Let $L[y] = ay'' + by' + cy$ with constant $a, b, c$ and $a \neq 0$. If $y_1(x)$ and $y_2(x)$ are solutions of $L[y] = 0$, then any linear combination
|
||||
\[
|
||||
y(x) = c_1 y_1(x) + c_2 y_2(x)
|
||||
\]
|
||||
is also a solution, for arbitrary constants $c_1, c_2 \in \R$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{proof}
|
||||
Compute:
|
||||
\[
|
||||
L[c_1 y_1 + c_2 y_2]
|
||||
= a(c_1 y_1 + c_2 y_2)'' + b(c_1 y_1 + c_2 y_2)' + c(c_1 y_1 + c_2 y_2).
|
||||
\]
|
||||
By linearity of differentiation:
|
||||
\[
|
||||
= c_1\bigl(ay_1'' + by_1' + cy_1\bigr) + c_2\bigl(ay_2'' + by_2' + cy_2\bigr)
|
||||
= c_1 \cdot 0 + c_2 \cdot 0 = 0.
|
||||
\]
|
||||
Therefore $c_1 y_1 + c_2 y_2$ satisfies the ODE.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Fundamental Set of Solutions]
|
||||
\label{def:fundamental_set}
|
||||
Two solutions $\{y_1, y_2\}$ of \cref{eq:second_order_homogeneous} form a \textbf{fundamental set of solutions} on an interval $I$ if they are linearly independent on $I$. The general solution is then
|
||||
\[
|
||||
y(x) = c_1 y_1(x) + c_2 y_2(x),
|
||||
\]
|
||||
where $c_1, c_2$ are arbitrary constants.
|
||||
\end{definition}
|
||||
|
||||
\begin{workedexample}
|
||||
Verify that $y_1(x) = e^{3x}$ and $y_2(x) = e^{-2x}$ form a fundamental set for $y'' - y' - 6y = 0$, and write the general solution.
|
||||
|
||||
\textbf{Solution.} Check each function:
|
||||
\[
|
||||
y_1 = e^{3x} \;\Rightarrow\; y_1' = 3e^{3x}, \; y_1'' = 9e^{3x}.
|
||||
\]
|
||||
\[
|
||||
L[y_1] = 9e^{3x} - 3e^{3x} - 6e^{3x} = 0. \quad \checkmark
|
||||
\]
|
||||
|
||||
\[
|
||||
y_2 = e^{-2x} \;\Rightarrow\; y_2' = -2e^{-2x}, \; y_2'' = 4e^{-2x}.
|
||||
\]
|
||||
\[
|
||||
L[y_2] = 4e^{-2x} + 2e^{-2x} - 6e^{-2x} = 0. \quad \checkmark
|
||||
\]
|
||||
|
||||
Linear independence: the ratio $y_1/y_2 = e^{5x}$ is not constant, so $y_1$ and $y_2$ are linearly independent. They form a fundamental set.
|
||||
|
||||
The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{3x} + c_2 e^{-2x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Show that $\{y_1, y_2\} = \{e^{2x}, 2e^{2x}\}$ is \emph{not} a fundamental set for $y'' - 4y = 0$, even though both are solutions.
|
||||
|
||||
\textbf{Solution.} Clearly $y_2 = 2y_1$, so the two functions are linearly dependent. Any linear combination gives
|
||||
\[
|
||||
c_1 e^{2x} + c_2 (2e^{2x}) = (c_1 + 2c_2)e^{2x},
|
||||
\]
|
||||
which spans only a one-dimensional solution space. We cannot represent all solutions (we are missing the $e^{-2x}$ component). A proper fundamental set is $\{e^{2x}, e^{-2x}\}$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Wronskian and Linear Independence}
|
||||
\label{sec:ch04_wronskian}
|
||||
|
||||
To systematically determine whether two solutions form a fundamental set, we use the \textbf{Wronskian}.
|
||||
|
||||
\begin{definition}[Wronskian]
|
||||
\label{def:wronskian}
|
||||
For two differentiable functions $y_1(x)$ and $y_2(x)$, the \textbf{Wronskian} is
|
||||
\[
|
||||
W(y_1, y_2)(x) = \det\begin{pmatrix} y_1(x) & y_2(x) \\[4pt] y_1'(x) & y_2'(x) \end{pmatrix}
|
||||
= y_1(x)\,y_2'(x) - y_1'(x)\,y_2(x).
|
||||
\]
|
||||
\end{definition}
|
||||
|
||||
\begin{theorem}[Wronskian Test]
|
||||
\label{thm:wronskian_test}
|
||||
Let $y_1$ and $y_2$ be two solutions of \cref{eq:second_order_homogeneous} on an interval $I$. Then:
|
||||
\begin{enumerate}
|
||||
\item If $W(y_1, y_2)(x_0) \neq 0$ for some $x_0 \in I$, then $\{y_1, y_2\}$ is a fundamental set on $I$.
|
||||
\item If $W(y_1, y_2)(x) = 0$ for all $x \in I$, then $y_1$ and $y_2$ are linearly dependent on $I$.
|
||||
\end{enumerate}
|
||||
In particular, $W(y_1, y_2)(x)$ is either identically zero on $I$ or never zero on $I$.
|
||||
\end{theorem}
|
||||
|
||||
The Wronskian gives a practical computational test: evaluate $W$ at any single point; if it is nonzero, the solutions are linearly independent.
|
||||
|
||||
\begin{workedexample}
|
||||
Compute the Wronskian of $y_1(x) = e^{3x}$ and $y_2(x) = e^{-2x}$.
|
||||
|
||||
\textbf{Solution.} We have $y_1' = 3e^{3x}$ and $y_2' = -2e^{-2x}$.
|
||||
\[
|
||||
W(y_1, y_2)(x) = e^{3x} \cdot (-2e^{-2x}) - 3e^{3x} \cdot e^{-2x}
|
||||
= -2e^{x} - 3e^{x} = -5e^{x}.
|
||||
\]
|
||||
Since $W(x) = -5e^x \neq 0$ for all $x$, the two functions are linearly independent. They form a fundamental set.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Compute the Wronskian of $y_1(x) = e^{rx}$ and $y_2(x) = x\,e^{rx}$ (the repeated-root case).
|
||||
|
||||
\textbf{Solution.} We have $y_1' = r\,e^{rx}$ and $y_2' = (1 + rx)\,e^{rx}$.
|
||||
\[
|
||||
W(y_1, y_2)(x) = e^{rx} \cdot (1 + rx)\,e^{rx} - r\,e^{rx} \cdot x\,e^{rx}
|
||||
= (1 + rx)\,e^{2rx} - rx\,e^{2rx} = e^{2rx}.
|
||||
\]
|
||||
Since $W(x) = e^{2rx} \neq 0$ for all $x$, these two functions are linearly independent. This confirms that $\{e^{rx}, x\,e^{rx}\}$ is a valid fundamental set for the repeated-root case.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{remark}
|
||||
The connection between the Wronskian and fundamental sets is fundamental (pardon the pun): a pair of solutions is a fundamental set \emph{if and only if} their Wronskian is nonzero on the interval. This is the standard criterion used in practice.
|
||||
\end{remark}
|
||||
|
||||
\subsection{Abel's Identity}
|
||||
\label{sec:ch04_abels_identity}
|
||||
|
||||
Abel's identity provides a formula for the Wronskian without computing derivatives, by relating it directly to the coefficients of the ODE.
|
||||
|
||||
\begin{theorem}[Abel's Identity]
|
||||
\label{thm:abels_identity}
|
||||
Let $y_1$ and $y_2$ be two solutions of the standard-form equation
|
||||
\begin{equation}
|
||||
\label{eq:standard_form}
|
||||
y'' + p(x)\,y' + q(x)\,y = 0
|
||||
\end{equation}
|
||||
on an interval $I$, where $p(x)$ and $q(x)$ are continuous. Then the Wronskian satisfies
|
||||
\[
|
||||
W(x) = W(x_0)\,\exp\!\left(-\int_{x_0}^{x} p(t)\,\diff t\right)
|
||||
\]
|
||||
for any $x_0 \in I$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{proof}
|
||||
Let $W(x) = y_1 y_2' - y_1' y_2$. Differentiate:
|
||||
\[
|
||||
W' = y_1' y_2' + y_1 y_2'' - y_1'' y_2 - y_1' y_2'
|
||||
= y_1 y_2'' - y_1'' y_2.
|
||||
\]
|
||||
Since $y_1$ and $y_2$ both satisfy $y'' + p(x)y' + q(x)y = 0$, we have
|
||||
\[
|
||||
y_1'' = -p(x)y_1' - q(x)y_1, \qquad
|
||||
y_2'' = -p(x)y_2' - q(x)y_2.
|
||||
\]
|
||||
Substitute:
|
||||
\begin{align*}
|
||||
W' &= y_1\bigl[-p y_2' - q y_2\bigr] - \bigl[-p y_1' - q y_1\bigr]y_2 \\
|
||||
&= -p\,y_1 y_2' - q\,y_1 y_2 + p\,y_1' y_2 + q\,y_1 y_2 \\
|
||||
&= -p\,(y_1 y_2' - y_1' y_2) = -p(x)\,W(x).
|
||||
\end{align*}
|
||||
Thus $W' + p(x)W = 0$, a first-order separable equation. The solution is
|
||||
\[
|
||||
W(x) = W(x_0)\,\exp\!\left(-\int_{x_0}^{x} p(t)\,\diff t\right).
|
||||
\]
|
||||
\end{proof}
|
||||
|
||||
\begin{remark}
|
||||
For the constant-coefficient equation \cref{eq:second_order_homogeneous}, dividing by $a$ puts it in standard form with $p(x) = b/a$ (constant). Abel's identity then gives
|
||||
\[
|
||||
W(x) = W(x_0)\,\exp\!\left(-\frac{b}{a}(x - x_0)\right).
|
||||
\]
|
||||
Since $W(x_0) \neq 0$ for a fundamental set, this confirms that $W(x)$ is never zero (exponential of a real number is always positive).
|
||||
\end{remark}
|
||||
|
||||
\begin{workedexample}
|
||||
Use Abel's identity to find the Wronskian of two solutions of $y'' + 3y' + 2y = 0$, given that $W(0) = 5$.
|
||||
|
||||
\textbf{Solution.} In standard form, $p(x) = 3$. By Abel's identity:
|
||||
\[
|
||||
W(x) = W(0)\,\exp\!\left(-\int_0^x 3\,\diff t\right)
|
||||
= 5\,e^{-3x}.
|
||||
\]
|
||||
The Wronskian is $W(x) = 5e^{-3x}$, which is always positive (confirming the solutions are linearly independent).
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Verify Abel's identity for $y'' - 4y = 0$ using the fundamental set $\{e^{2x}, e^{-2x}\}$.
|
||||
|
||||
\textbf{Solution.} In standard form, $p(x) = 0$. Abel's identity predicts
|
||||
\[
|
||||
W(x) = W(x_0)\,\exp\!\left(-\int_{x_0}^{x} 0\,\diff t\right) = W(x_0).
|
||||
\]
|
||||
The Wronskian should be constant. Compute it directly:
|
||||
\[
|
||||
W(e^{2x}, e^{-2x}) = e^{2x}(-2e^{-2x}) - (2e^{2x})e^{-2x} = -2 - 2 = -4.
|
||||
\]
|
||||
Indeed, $W(x) = -4$ is constant, in agreement with Abel's identity.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Practical tip.} Abel's identity is especially useful when you know one solution and need to find another. It tells you the Wronskian \emph{without} computing derivatives, and combined with the formula $v' = W/y_1^2$ from reduction of order (\cref{sec:ch04_reduction_of_order}), it directly yields the second solution.
|
||||
\end{hintbox}
|
||||
|
||||
\subsection{Reduction of Order}
|
||||
\label{sec:ch04_reduction_of_order}
|
||||
|
||||
Reduction of order is a systematic method for finding a second linearly independent solution when one solution $y_1(x)$ is already known. We saw this technique in \cref{sec:ch04_repeated_roots} to derive the $x\,e^{rx}$ factor; here we state it in full generality.
|
||||
|
||||
\paragraph{Derivation.}
|
||||
Consider the standard-form equation $y'' + p(x)y' + q(x)y = 0$. Suppose $y_1(x)$ is a known solution. We seek a second solution of the form
|
||||
\begin{equation}
|
||||
\label{eq:reduction_ansatz}
|
||||
y_2(x) = v(x)\,y_1(x),
|
||||
\end{equation}
|
||||
where $v(x)$ is to be determined. Differentiate:
|
||||
\begin{align*}
|
||||
y_2' &= v'y_1 + vy_1', \\
|
||||
y_2'' &= v''y_1 + 2v'y_1' + vy_1''.
|
||||
\end{align*}
|
||||
Substitute into the ODE:
|
||||
\[
|
||||
\bigl(v''y_1 + 2v'y_1' + vy_1''\bigr) + p(x)\bigl(v'y_1 + vy_1'\bigr) + q(x)\,v\,y_1 = 0.
|
||||
\]
|
||||
Collect terms:
|
||||
\[
|
||||
v''y_1 + v'\bigl(2y_1' + p(x)y_1\bigr) + v\bigl(y_1'' + p(x)y_1' + q(x)y_1\bigr) = 0.
|
||||
\]
|
||||
The $v$-term vanishes because $y_1$ is a solution. We obtain a first-order equation in $u = v'$:
|
||||
\[
|
||||
y_1\,u' + \bigl(2y_1' + p(x)y_1\bigr)u = 0.
|
||||
\]
|
||||
This is separable:
|
||||
\[
|
||||
\frac{u'}{u} = -\frac{2y_1'}{y_1} - p(x) = -2\frac{d}{dx}\bigl(\ln|y_1|\bigr) - p(x).
|
||||
\]
|
||||
Integrating:
|
||||
\[
|
||||
\ln|u| = -2\ln|y_1| - \int p(x)\,\diff x + C,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
u = \frac{v'}{y_1^2} \cdot y_1^2 = \frac{C}{y_1^2}\,\exp\!\left(-\int p(x)\,\diff x\right).
|
||||
\]
|
||||
Noting that $W(x) = W(x_0)\exp\!\left(-\int p\,\diff x\right)$ by Abel's identity, we can write this as
|
||||
\begin{equation}
|
||||
\label{eq:reduction_formula}
|
||||
v'(x) = \frac{W(x)}{y_1(x)^2}.
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Reduction of order formula.}
|
||||
Given one solution $y_1(x)$ of $y'' + p(x)y' + q(x)y = 0$, a second linearly independent solution is
|
||||
\[
|
||||
y_2(x) = y_1(x)\int \frac{1}{y_1(x)^2}\,\exp\!\left(-\int p(x)\,\diff x\right)\diff x.
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
Given that $y_1(x) = e^{3x}$ is a solution of $y'' - 6y' + 9y = 0$, find a second linearly independent solution using reduction of order.
|
||||
|
||||
\textbf{Solution.} The equation in standard form has $p(x) = -6$. By Abel's identity (or direct computation):
|
||||
\[
|
||||
\exp\!\left(-\int p(x)\,\diff x\right) = \exp\!\left(-\int (-6)\,\diff x\right) = e^{6x}.
|
||||
\]
|
||||
Using \cref{eq:reduction_formula}:
|
||||
\[
|
||||
v'(x) = \frac{e^{6x}}{(e^{3x})^2} = \frac{e^{6x}}{e^{6x}} = 1.
|
||||
\]
|
||||
Integrate: $v(x) = x$ (choosing the simplest antiderivative with $C=0$).
|
||||
Therefore
|
||||
\[
|
||||
y_2(x) = v(x)\,y_1(x) = x\,e^{3x}.
|
||||
\]
|
||||
This recovers the repeated-root solution we derived earlier. The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{3x} + c_2 x\,e^{3x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Given that $y_1(x) = x$ is a solution of $x^2 y'' - x y' + y = 0$, find a second solution.
|
||||
|
||||
\textbf{Solution.} Write the equation in standard form by dividing by $x^2$:
|
||||
\[
|
||||
y'' - \frac{1}{x}y' + \frac{1}{x^2}y = 0.
|
||||
\]
|
||||
Here $p(x) = -\dfrac{1}{x}$. Compute the integrating factor:
|
||||
\[
|
||||
\exp\!\left(-\int p(x)\,\diff x\right) = \exp\!\left(-\int \!\left(-\frac{1}{x}\right)\!\diff x\right)
|
||||
= \exp\!\bigl(\ln|x|\bigr) = |x|.
|
||||
\]
|
||||
For $x > 0$, this is simply $x$. Now apply the reduction of order formula:
|
||||
\[
|
||||
v'(x) = \frac{x}{y_1(x)^2} = \frac{x}{x^2} = \frac{1}{x}.
|
||||
\]
|
||||
Integrate: $v(x) = \ln|x|$. The second solution is
|
||||
\[
|
||||
y_2(x) = y_1(x)\,v(x) = x\,\ln|x|.
|
||||
\]
|
||||
The general solution (for $x > 0$) is
|
||||
\[
|
||||
y(x) = c_1\,x + c_2\,x\,\ln x.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch04_summary}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Second-order linear homogeneous equations with constant coefficients}
|
||||
\label{tab:ch04_cases}
|
||||
\begin{tabular}{l l l p{5cm}}
|
||||
\toprule
|
||||
\textbf{Discriminant} & \textbf{Roots} & \textbf{Solution} & \textbf{Key idea} \\
|
||||
\midrule
|
||||
$\Delta > 0$ & $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ & $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ & Two distinct real exponentials \\[8pt]
|
||||
$\Delta = 0$ & $r = -\dfrac{b}{2a}$ (double) & $y = c_1 e^{rx} + c_2 x e^{rx}$ & Extra $x$ from reduction of order \\[8pt]
|
||||
$\Delta < 0$ & $\alpha \pm i\beta$ & $e^{\alpha x}[c_1 \cos(\beta x) + c_2 \sin(\beta x)]$ & Damped oscillation via Euler's formula \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Theory results: Wronskian, Abel's identity, reduction of order}
|
||||
\label{tab:ch04_theory}
|
||||
\begin{tabular}{l p{6.5cm}}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key formula} \\
|
||||
\midrule
|
||||
Wronskian & $W(y_1, y_2)(x) = y_1 y_2' - y_1' y_2$ \\
|
||||
Wronskian test & $W(x_0) \neq 0 \;\Rightarrow\;$ fundamental set on $I$ \\
|
||||
Abel's identity & $W(x) = W(x_0)\,\exp\!\left(-\displaystyle\int_{x_0}^{x} p(t)\,\diff t\right)$ \\
|
||||
Abel (constant coeff.) & $W(x) = W(x_0)\,\exp\!\left(-\dfrac{b}{a}(x - x_0)\right)$ \\
|
||||
Reduction of order & $y_2(x) = y_1(x)\displaystyle\int \frac{e^{-\int p(x)\,\diff x}}{y_1(x)^2}\,\diff x$ \\
|
||||
Superposition & $L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2] = 0$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Problem-solving workflow.}
|
||||
\begin{enumerate}
|
||||
\item Write the ODE in standard form (divide by $a$ if needed).
|
||||
\item Form the characteristic equation $ar^2 + br + c = 0$.
|
||||
\item Compute the discriminant $\Delta = b^2 - 4ac$.
|
||||
\item Apply the appropriate case from \cref{tab:ch04_cases}.
|
||||
\item If initial/boundary conditions are given, determine $c_1$ and $c_2$.
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
@@ -1,797 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch05_second_order_nonhomogeneous.tex
|
||||
% Chapter 5: Second-Order Linear Nonhomogeneous Equations
|
||||
% =============================================================================
|
||||
|
||||
\section{Second-Order Nonhomogeneous}
|
||||
\label{ch:second_order_nonhomogeneous}
|
||||
|
||||
We now extend the theory from \cref{ch:second_order_homogeneous} to the nonhomogeneous case. The general equation is
|
||||
\begin{equation}
|
||||
\label{eq:nonhomogeneous_general}
|
||||
a\,y'' + b\,y' + c\,y = g(x),
|
||||
\qquad a, b, c \in \R,\;\; a \neq 0,
|
||||
\end{equation}
|
||||
where $g(x)$ is a given \textbf{forcing function} (also called the \textbf{source term} or \textbf{nonhomogeneous term}). When $g(x) \equiv 0$, we recover the homogeneous equation \cref{eq:second_order_homogeneous}. The presence of $g(x)$ models external driving forces in physical systems --- a periodically driven pendulum, a forced electrical circuit, or a building shaken by an earthquake.
|
||||
|
||||
The general strategy in all cases is the same: first solve the associated homogeneous equation, then find one particular solution of the nonhomogeneous equation. The two parts combine to give the complete solution.
|
||||
|
||||
\subsection{General Solution Structure}
|
||||
\label{sec:ch05_general_solution}
|
||||
|
||||
\begin{theorem}[Structure of the General Solution]
|
||||
\label{thm:nonhomogeneous_structure}
|
||||
Let $L[y] = ay'' + by' + cy$ with constant $a, b, c$ and $a \neq 0$. If $y_h(x)$ is the general solution of the associated homogeneous equation $L[y] = 0$, and $y_p(x)$ is \emph{any} particular solution of the nonhomogeneous equation $L[y] = g(x)$, then the general solution of $L[y] = g(x)$ is
|
||||
\[
|
||||
y(x) = y_h(x) + y_p(x).
|
||||
\]
|
||||
\end{theorem}
|
||||
|
||||
\begin{proof}
|
||||
Define the linear operator $L[y] = ay'' + by' + cy$. We need to show that $y = y_h + y_p$ satisfies $L[y] = g(x)$ for \emph{every} choice of the constants in $y_h$.
|
||||
|
||||
By linearity of differentiation (the \textbf{superposition principle}, \cref{thm:superposition}):
|
||||
\[
|
||||
L[y_h + y_p] = L[y_h] + L[y_p].
|
||||
\]
|
||||
Since $y_h$ solves the homogeneous equation, $L[y_h] = 0$. Since $y_p$ is a particular solution, $L[y_p] = g(x)$. Therefore:
|
||||
\[
|
||||
L[y_h + y_p] = 0 + g(x) = g(x).
|
||||
\]
|
||||
Thus $y_h + y_p$ is a solution of the nonhomogeneous equation. Conversely, if $y$ is any solution of $L[y] = g(x)$, then $L[y - y_p] = L[y] - L[y_p] = g(x) - g(x) = 0$, so $y - y_p = y_h$ for some homogeneous solution $y_h$, meaning $y = y_h + y_p$.
|
||||
\end{proof}
|
||||
|
||||
The homogeneous part $y_h(x)$ retains two arbitrary constants $c_1, c_2$ from the general solution found in \cref{ch:second_order_homogeneous}. The particular solution $y_p(x)$ contains no free parameters --- we only need \emph{one} specific function that works.
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Two-step workflow for every nonhomogeneous problem.}
|
||||
\begin{enumerate}
|
||||
\item \textbf{Solve the homogeneous equation} $ay'' + by' + cy = 0$ using the characteristic equation method from \cref{ch:second_order_homogeneous}. Obtain $y_h(x) = c_1 y_1(x) + c_2 y_2(x)$.
|
||||
\item \textbf{Find one particular solution} $y_p(x)$ using either the method of undetermined coefficients or variation of parameters.
|
||||
\end{enumerate}
|
||||
Combine: $y(x) = y_h(x) + y_p(x)$.
|
||||
\end{hintbox}
|
||||
|
||||
We now develop the two main methods for finding $y_p(x)$.
|
||||
|
||||
\subsection{Method of Undetermined Coefficients}
|
||||
\label{sec:ch05_undetermined_coefficients}
|
||||
|
||||
The method of \textbf{undetermined coefficients} (UC) is an algebraic technique that works when the forcing function $g(x)$ has a \textbf{simple form}. The idea is straightforward: guess the \emph{shape} of $y_p$ based on the form of $g(x)$, but leave the coefficients unknown. Then substitute the guess into the ODE and solve for those coefficients.
|
||||
|
||||
\paragraph{When to use.} Undetermined coefficients applies only when $g(x)$ is one of the following (or a sum/product of them):
|
||||
\begin{itemize}
|
||||
\item A polynomial $P_n(x)$ of degree $n$.
|
||||
\item An exponential $e^{\alpha x}$.
|
||||
\item A sine or cosine function $\sin(\beta x)$ or $\cos(\beta x)$.
|
||||
\item Products of the above, such as $e^{\alpha x} P_n(x)$ or $e^{\alpha x}\sin(\beta x)$.
|
||||
\end{itemize}
|
||||
If $g(x) = \ln(x)$, $1/x$, $\tan(x)$, or any other function outside this family, undetermined coefficients \textbf{will not work}. Use variation of parameters instead (\cref{sec:ch05_variation_of_parameters}).
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Limitation warning.} Undetermined coefficients is an \emph{educated guessing} method. It exploits the fact that polynomials, exponentials, and trigonometric functions are closed under differentiation. Functions like $\ln x$ and $\tan x$ do not share this property, so the method breaks down.
|
||||
\end{hintbox}
|
||||
|
||||
\subsubsection{Guess Table}
|
||||
\label{sec:ch05_guess_table}
|
||||
|
||||
The following table gives the initial guess for $y_p$ based on the form of $g(x)$. The constants $A_k, B_k$ are the ``undetermined coefficients'' to be found by substitution.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Guess table for the method of undetermined coefficients.}
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Form of $g(x)$} & \textbf{Initial guess for $y_p(x)$} \\
|
||||
\midrule
|
||||
$P_n(x) = a_n x^n + \cdots + a_0$ & $A_n x^n + A_{n-1} x^{n-1} + \cdots + A_0$ \\[8pt]
|
||||
$e^{\alpha x}$ & $A\,e^{\alpha x}$ \\[8pt]
|
||||
$\sin(\beta x)$ \textbf{or} $\cos(\beta x)$ & $A\sin(\beta x) + B\cos(\beta x)$ \\[8pt]
|
||||
$e^{\alpha x} P_n(x)$ & $e^{\alpha x}(A_n x^n + \cdots + A_0)$ \\[8pt]
|
||||
$e^{\alpha x}\sin(\beta x)$ \textbf{or} $e^{\alpha x}\cos(\beta x)$ &
|
||||
\begin{tabular}[t]{@{}l@{}}
|
||||
$e^{\alpha x}\bigl[(A_n x^n + \cdots + A_0)\cos(\beta x)$ \\
|
||||
\quad $+ (B_n x^n + \cdots + B_0)\sin(\beta x)\bigr]$
|
||||
\end{tabular} \\[12pt]
|
||||
Product of two forms & Multiply the individual guesses \\[8pt]
|
||||
Sum of two forms $g_1 + g_2$ & Sum the individual guesses \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{keyresult}
|
||||
|
||||
\subsubsection{Modification Rule}
|
||||
\label{sec:ch05_modification_rule}
|
||||
|
||||
There is one important caveat: if any term of your initial guess \textbf{already appears in the homogeneous solution} $y_h$, the guess will fail (substitution leads to $0 = g(x)$, a contradiction). In this case, you must \textbf{modify} the guess.
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Modification rule.} If a term in the initial guess for $y_p$ is also a solution of the homogeneous equation:
|
||||
\begin{enumerate}
|
||||
\item Multiply that term by $x$.
|
||||
\item If the modified term \emph{still} appears in $y_h$, multiply by $x$ again (i.e.\ by $x^2$).
|
||||
\end{enumerate}
|
||||
For a second-order equation, $x^2$ is always sufficient, because the homogeneous solution space has dimension~2.
|
||||
\end{hintbox}
|
||||
|
||||
\paragraph{Why this works.} The modification multiplies the guess by $x$ (or $x^2$), which introduces new functional forms (like $x e^{rx}$ or $x^2 e^{rx}$) that are \emph{not} solutions of the homogeneous equation, while preserving the algebraic structure needed for substitution.
|
||||
|
||||
\subsubsection{Worked Examples}
|
||||
\label{sec:ch05_uc_examples}
|
||||
|
||||
\paragraph{Example 1: Polynomial forcing.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - 3y' + 2y = 4x + 1$.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 - 3r + 2 = 0 \quad\Longrightarrow\quad (r-1)(r-2) = 0.
|
||||
\]
|
||||
Roots: $r_1 = 1$, $r_2 = 2$. The homogeneous solution is
|
||||
\[
|
||||
y_h(x) = c_1 e^x + c_2 e^{2x}.
|
||||
\]
|
||||
|
||||
\textit{Step 2: Particular solution.} Here $g(x) = 4x + 1$, a first-degree polynomial. From the guess table, we try
|
||||
\[
|
||||
y_p(x) = Ax + B.
|
||||
\]
|
||||
No term overlaps with $y_h$ (exponentials vs.\ polynomials), so no modification is needed. Differentiate:
|
||||
\[
|
||||
y_p' = A, \qquad y_p'' = 0.
|
||||
\]
|
||||
Substitute into the ODE:
|
||||
\[
|
||||
0 - 3A + 2(Ax + B) = 4x + 1.
|
||||
\]
|
||||
Collect terms:
|
||||
\[
|
||||
2A\,x + (2B - 3A) = 4x + 1.
|
||||
\]
|
||||
Equate coefficients of like powers of $x$:
|
||||
\[
|
||||
\begin{cases}
|
||||
2A = 4 \quad\Longrightarrow\quad A = 2, \\
|
||||
2B - 3A = 1 \quad\Longrightarrow\quad 2B - 6 = 1 \quad\Longrightarrow\quad B = \dfrac{7}{2}.
|
||||
\end{cases}
|
||||
\]
|
||||
The particular solution is $y_p(x) = 2x + \dfrac{7}{2}$.
|
||||
|
||||
\textit{Step 3: General solution.}
|
||||
\[
|
||||
y(x) = c_1 e^x + c_2 e^{2x} + 2x + \frac{7}{2}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Example 2: Exponential forcing.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + y' - 6y = 12e^{3x}$.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + r - 6 = 0 \quad\Longrightarrow\quad (r-2)(r+3) = 0.
|
||||
\]
|
||||
Roots: $r_1 = 2$, $r_2 = -3$. Thus
|
||||
\[
|
||||
y_h(x) = c_1 e^{2x} + c_2 e^{-3x}.
|
||||
\]
|
||||
|
||||
\textit{Step 2: Particular solution.} Here $g(x) = 12e^{3x}$. From the guess table, we try
|
||||
\[
|
||||
y_p(x) = A\,e^{3x}.
|
||||
\]
|
||||
Check for overlap: $e^{3x}$ is not in $y_h$ (which contains $e^{2x}$ and $e^{-3x}$), so no modification needed. Differentiate:
|
||||
\[
|
||||
y_p' = 3A\,e^{3x}, \qquad y_p'' = 9A\,e^{3x}.
|
||||
\]
|
||||
Substitute:
|
||||
\[
|
||||
9A\,e^{3x} + 3A\,e^{3x} - 6A\,e^{3x} = 12e^{3x}.
|
||||
\]
|
||||
Factor out $e^{3x}$ (never zero):
|
||||
\[
|
||||
(9 + 3 - 6)A = 12 \quad\Longrightarrow\quad 6A = 12 \quad\Longrightarrow\quad A = 2.
|
||||
\]
|
||||
So $y_p(x) = 2e^{3x}$.
|
||||
|
||||
\textit{Step 3: General solution.}
|
||||
\[
|
||||
y(x) = c_1 e^{2x} + c_2 e^{-3x} + 2e^{3x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Example 3: Trigonometric forcing.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + 4y = 8\cos(2x)$.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + 4 = 0 \quad\Longrightarrow\quad r = \pm 2i.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
y_h(x) = c_1 \cos(2x) + c_2 \sin(2x).
|
||||
\]
|
||||
|
||||
\textit{Step 2: Particular solution.} Here $g(x) = 8\cos(2x)$. From the guess table, we try
|
||||
\[
|
||||
y_p(x) = A\cos(2x) + B\sin(2x).
|
||||
\]
|
||||
\textbf{Overlap detected!} Both $\cos(2x)$ and $\sin(2x)$ appear in $y_h$. We must apply the modification rule: multiply the entire guess by $x$:
|
||||
\[
|
||||
y_p(x) = x\bigl[A\cos(2x) + B\sin(2x)\bigr] = Ax\cos(2x) + Bx\sin(2x).
|
||||
\]
|
||||
Check again: $x\cos(2x)$ and $x\sin(2x)$ are \emph{not} in $y_h$, so we are good. Differentiate (product rule):
|
||||
\[
|
||||
\begin{aligned}
|
||||
y_p' &= A\cos(2x) - 2Ax\sin(2x) + B\sin(2x) + 2Bx\cos(2x), \\
|
||||
y_p'' &= -2A\sin(2x) - 2A\sin(2x) - 4Ax\cos(2x) + 2B\cos(2x) + 2B\cos(2x) - 4Bx\sin(2x) \\
|
||||
&= -4A\sin(2x) - 4Ax\cos(2x) + 4B\cos(2x) - 4Bx\sin(2x).
|
||||
\end{aligned}
|
||||
\]
|
||||
Substitute into $y'' + 4y = 8\cos(2x)$:
|
||||
\[
|
||||
\bigl[-4A\sin(2x) - 4Ax\cos(2x) + 4B\cos(2x) - 4Bx\sin(2x)\bigr]
|
||||
+ 4\bigl[Ax\cos(2x) + Bx\sin(2x)\bigr] = 8\cos(2x).
|
||||
\]
|
||||
The $x$-terms cancel:
|
||||
\[
|
||||
-4A\sin(2x) + 4B\cos(2x) = 8\cos(2x).
|
||||
\]
|
||||
Equate coefficients:
|
||||
\[
|
||||
\begin{cases}
|
||||
-4A = 0 \quad\Longrightarrow\quad A = 0, \\
|
||||
4B = 8 \quad\Longrightarrow\quad B = 2.
|
||||
\end{cases}
|
||||
\]
|
||||
So $y_p(x) = 2x\sin(2x)$.
|
||||
|
||||
\textit{Step 3: General solution.}
|
||||
\[
|
||||
y(x) = c_1 \cos(2x) + c_2 \sin(2x) + 2x\sin(2x).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Example 4: Product of exponential and trigonometric.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - 2y' + 5y = 10e^x\cos(2x)$.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 - 2r + 5 = 0.
|
||||
\]
|
||||
Discriminant: $\Delta = 4 - 20 = -16 < 0$. Roots:
|
||||
\[
|
||||
r = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
y_h(x) = e^x\bigl[c_1 \cos(2x) + c_2 \sin(2x)\bigr].
|
||||
\]
|
||||
|
||||
\textit{Step 2: Particular solution.} Here $g(x) = 10e^x\cos(2x)$. From the guess table for $e^{\alpha x}\cos(\beta x)$ with $\alpha = 1, \beta = 2$, we try
|
||||
\[
|
||||
y_p(x) = e^x\bigl[A\cos(2x) + B\sin(2x)\bigr].
|
||||
\]
|
||||
\textbf{Overlap detected!} Both $e^x\cos(2x)$ and $e^x\sin(2x)$ appear in $y_h$. Apply the modification rule: multiply by $x$:
|
||||
\[
|
||||
y_p(x) = x\,e^x\bigl[A\cos(2x) + B\sin(2x)\bigr].
|
||||
\]
|
||||
Now differentiate. Let $y_p = x\,e^x\,(A\cos(2x) + B\sin(2x))$. Set $u = x\,e^x$ and $v = A\cos(2x) + B\sin(2x)$ for clarity. Then $u' = e^x(1 + x)$ and $v' = -2A\sin(2x) + 2B\cos(2x)$.
|
||||
\[
|
||||
\begin{aligned}
|
||||
y_p' &= u'v + uv' = e^x(1+x)\bigl[A\cos(2x) + B\sin(2x)\bigr] + x\,e^x\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr], \\
|
||||
y_p'' &= e^x(1+x)\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr] + e^x\bigl[A\cos(2x) + B\sin(2x)\bigr] \\
|
||||
&\quad + e^x(1+x)\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr] + x\,e^x\bigl[-4A\cos(2x) - 4B\sin(2x)\bigr] \\
|
||||
&= 2e^x(1+x)\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr] + e^x\bigl[A\cos(2x) + B\sin(2x)\bigr] \\
|
||||
&\quad - 4x\,e^x\bigl[A\cos(2x) + B\sin(2x)\bigr].
|
||||
\end{aligned}
|
||||
\]
|
||||
Substitute $y_p$, $y_p'$, and $y_p''$ into the ODE $y'' - 2y' + 5y$. Collect the coefficient of $e^x\cos(2x)$ and $e^x\sin(2x)$ (the $x$-terms will cancel due to the modification rule):
|
||||
|
||||
Coefficient of $e^x\cos(2x)$: $4B + A - 2A + 5A = 4B + 4A$.
|
||||
|
||||
Coefficient of $e^x\sin(2x)$: $-4A + B - 2B + 5B = -4A + 4B$.
|
||||
|
||||
Set equal to the RHS $10e^x\cos(2x)$:
|
||||
\[
|
||||
\begin{cases}
|
||||
4A + 4B = 10, \\
|
||||
-4A + 4B = 0 \quad\Longrightarrow\quad A = B.
|
||||
\end{cases}
|
||||
\]
|
||||
From the second equation, $A = B$. Substituting into the first: $8A = 10$, so $A = \dfrac{5}{4}$ and $B = \dfrac{5}{4}$.
|
||||
|
||||
The particular solution is
|
||||
\[
|
||||
y_p(x) = \frac{5}{4}\,x\,e^x\bigl[\cos(2x) + \sin(2x)\bigr].
|
||||
\]
|
||||
|
||||
\textit{Step 3: General solution.}
|
||||
\[
|
||||
y(x) = e^x\Bigl[c_1\cos(2x) + c_2\sin(2x)\Bigr] + \frac{5}{4}\,x\,e^x\bigl[\cos(2x) + \sin(2x)\bigr].
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Example 5: Modification rule (repeated-root overlap).}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - 4y' + 4y = 6e^{2x}$.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 - 4r + 4 = 0 \quad\Longrightarrow\quad (r-2)^2 = 0.
|
||||
\]
|
||||
Double root: $r = 2$. Thus
|
||||
\[
|
||||
y_h(x) = c_1 e^{2x} + c_2 x\,e^{2x}.
|
||||
\]
|
||||
|
||||
\textit{Step 2: Particular solution.} Here $g(x) = 6e^{2x}$. The initial guess is $y_p = A\,e^{2x}$.
|
||||
|
||||
\textbf{Overlap check:} $e^{2x}$ is in $y_h$. Modify: multiply by $x \Rightarrow y_p = Ax\,e^{2x}$.
|
||||
|
||||
\textbf{Overlap check again:} $x\,e^{2x}$ is \emph{also} in $y_h$. Modify again: multiply by $x$ once more $\Rightarrow y_p = Ax^2 e^{2x}$.
|
||||
|
||||
Now $x^2 e^{2x}$ is \emph{not} in $y_h$, so this is our final guess. Differentiate (product rule):
|
||||
\[
|
||||
\begin{aligned}
|
||||
y_p &= A x^2 e^{2x}, \\
|
||||
y_p' &= A(2x e^{2x} + 2x^2 e^{2x}) = 2A x\,e^{2x} + 2A x^2 e^{2x}, \\
|
||||
y_p'' &= 2A e^{2x} + 4A x\,e^{2x} + 4A x\,e^{2x} + 4A x^2 e^{2x} \\
|
||||
&= 2A e^{2x} + 8A x\,e^{2x} + 4A x^2 e^{2x}.
|
||||
\end{aligned}
|
||||
\]
|
||||
Substitute into $y'' - 4y' + 4y = 6e^{2x}$:
|
||||
\[
|
||||
\begin{aligned}
|
||||
&\bigl[2A e^{2x} + 8A x\,e^{2x} + 4A x^2 e^{2x}\bigr]
|
||||
- 4\bigl[2A x\,e^{2x} + 2A x^2 e^{2x}\bigr]
|
||||
+ 4\bigl[A x^2 e^{2x}\bigr] \\
|
||||
&= 2A e^{2x} + (8A - 8A)x\,e^{2x} + (4A - 8A + 4A)x^2 e^{2x} \\
|
||||
&= 2A e^{2x}.
|
||||
\end{aligned}
|
||||
\]
|
||||
Set equal to $6e^{2x}$:
|
||||
\[
|
||||
2A = 6 \quad\Longrightarrow\quad A = 3.
|
||||
\]
|
||||
The particular solution is $y_p(x) = 3x^2 e^{2x}$.
|
||||
|
||||
\textit{Step 3: General solution.}
|
||||
\[
|
||||
y(x) = c_1 e^{2x} + c_2 x\,e^{2x} + 3x^2 e^{2x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Example 6: Polynomial times exponential.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + 2y' + y = 3x\,e^{-x}$.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + 2r + 1 = 0 \quad\Longrightarrow\quad (r+1)^2 = 0.
|
||||
\]
|
||||
Double root: $r = -1$. Thus
|
||||
\[
|
||||
y_h(x) = c_1 e^{-x} + c_2 x\,e^{-x}.
|
||||
\]
|
||||
|
||||
\textit{Step 2: Particular solution.} Here $g(x) = 3x\,e^{-x} = e^{-x} P_1(x)$ where $P_1(x) = 3x$ is a first-degree polynomial. From the guess table, the initial guess would be
|
||||
\[
|
||||
y_p(x) = e^{-x}(Ax + B).
|
||||
\]
|
||||
\textbf{Overlap check:} $e^{-x}$ and $x\,e^{-x}$ are both in $y_h$. We must multiply the entire guess by $x^2$:
|
||||
\[
|
||||
y_p(x) = x^2 e^{-x}(Ax + B) = A x^3 e^{-x} + B x^2 e^{-x}.
|
||||
\]
|
||||
Differentiate:
|
||||
\[
|
||||
\begin{aligned}
|
||||
y_p &= (Ax^3 + Bx^2)e^{-x}, \\
|
||||
y_p' &= (3Ax^2 + 2Bx)e^{-x} - (Ax^3 + Bx^2)e^{-x}
|
||||
= (-Ax^3 + (3A-B)x^2 + 2Bx)\,e^{-x}, \\
|
||||
y_p'' &= (-3Ax^2 + 2(3A-B)x + 2B)e^{-x} - (-Ax^3 + (3A-B)x^2 + 2Bx)e^{-x} \\
|
||||
&= \bigl[Ax^3 + (-3A + 2(3A-B) - (3A-B))x^2 + \dots\bigr]e^{-x}.
|
||||
\end{aligned}
|
||||
\]
|
||||
Computing $y_p''$ more carefully with the product rule:
|
||||
\[
|
||||
\begin{aligned}
|
||||
y_p' &= e^{-x}\bigl(-Ax^3 + (3A-B)x^2 + 2Bx\bigr), \\
|
||||
y_p'' &= e^{-x}\Bigl(Ax^3 - (6A-B)x^2 + (6A-4B)x + 2B\Bigr).
|
||||
\end{aligned}
|
||||
\]
|
||||
Now substitute into $y'' + 2y' + y$:
|
||||
\[
|
||||
\begin{aligned}
|
||||
y_p'' + 2y_p' + y_p &= e^{-x}\Bigl[
|
||||
\bigl(Ax^3 - (6A-B)x^2 + (6A-4B)x + 2B\bigr) \\
|
||||
&\quad + 2\bigl(-Ax^3 + (3A-B)x^2 + 2Bx\bigr)
|
||||
+ (Ax^3 + Bx^2)
|
||||
\Bigr] \\
|
||||
&= e^{-x}\Bigl[
|
||||
(A - 2A + A)x^3
|
||||
+ \bigl(-(6A-B) + 2(3A-B) + B\bigr)x^2 \\
|
||||
&\quad + \bigl((6A-4B) + 4B\bigr)x + 2B
|
||||
\Bigr] \\
|
||||
&= e^{-x}\Bigl[ 0\cdot x^3 + 0\cdot x^2 + 6Ax + 2B \Bigr].
|
||||
\end{aligned}
|
||||
\]
|
||||
Set equal to $g(x) = 3x\,e^{-x}$:
|
||||
\[
|
||||
\begin{cases}
|
||||
6A = 3 \quad\Longrightarrow\quad A = \dfrac{1}{2}, \\
|
||||
2B = 0 \quad\Longrightarrow\quad B = 0.
|
||||
\end{cases}
|
||||
\]
|
||||
So $y_p(x) = \dfrac{1}{2}x^3 e^{-x}$.
|
||||
|
||||
\textit{Step 3: General solution.}
|
||||
\[
|
||||
y(x) = c_1 e^{-x} + c_2 x\,e^{-x} + \frac{1}{2}x^3 e^{-x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Variation of Parameters}
|
||||
\label{sec:ch05_variation_of_parameters}
|
||||
|
||||
Variation of parameters (VoP) is a \textbf{universal} method for finding a particular solution. Unlike undetermined coefficients, it works for \emph{any} forcing function $g(x)$, provided we can evaluate the resulting integrals. The trade-off is that VoP typically involves more computation.
|
||||
|
||||
\subsubsection{Full Derivation}
|
||||
\label{sec:ch05_vop_derivation}
|
||||
|
||||
\textbf{Setup.} Consider the nonhomogeneous equation in \textbf{normalized form}:
|
||||
\begin{equation}
|
||||
\label{eq:normalized_nonhomogeneous}
|
||||
y'' + p(x)\,y' + q(x)\,y = g(x).
|
||||
\end{equation}
|
||||
Let $\{y_1(x), y_2(x)\}$ be a fundamental set of solutions for the associated homogeneous equation $y'' + p(x)y' + q(x)y = 0$. Their Wronskian is
|
||||
\[
|
||||
W(x) = W(y_1, y_2)(x) = y_1(x)\,y_2'(x) - y_1'(x)\,y_2(x) \neq 0.
|
||||
\]
|
||||
|
||||
\textbf{Ansatz.} We look for a particular solution of the form
|
||||
\begin{equation}
|
||||
\label{eq:vop_ansatz}
|
||||
y_p(x) = u_1(x)\,y_1(x) + u_2(x)\,y_2(x),
|
||||
\end{equation}
|
||||
where $u_1(x)$ and $u_2(x)$ are unknown functions to be determined. This is called ``variation of parameters'' because we have replaced the constant coefficients $c_1, c_2$ of the homogeneous solution with variable functions $u_1(x), u_2(x)$.
|
||||
|
||||
Differentiate $y_p$:
|
||||
\begin{equation}
|
||||
\label{eq:vop_first_derivative}
|
||||
y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2'.
|
||||
\end{equation}
|
||||
We have two unknown functions but only one equation (the ODE). To resolve this underdetermination, we impose an \textbf{auxiliary condition}:
|
||||
\begin{equation}
|
||||
\label{eq:vop_constraint}
|
||||
u_1' y_1 + u_2' y_2 = 0.
|
||||
\end{equation}
|
||||
This is not a restriction --- it is a convenient choice that simplifies the algebra. With this constraint, the first derivative simplifies to
|
||||
\[
|
||||
y_p' = u_1 y_1' + u_2 y_2'.
|
||||
\]
|
||||
Differentiate once more:
|
||||
\begin{equation}
|
||||
\label{eq:vop_second_derivative}
|
||||
y_p'' = u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''.
|
||||
\end{equation}
|
||||
|
||||
\textbf{Substitution.} Substitute $y_p$, $y_p'$, and $y_p''$ into the ODE \cref{eq:normalized_nonhomogeneous}:
|
||||
\[
|
||||
\bigl[u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''\bigr]
|
||||
+ p(x)\bigl[u_1 y_1' + u_2 y_2'\bigr]
|
||||
+ q(x)\bigl[u_1 y_1 + u_2 y_2\bigr] = g(x).
|
||||
\]
|
||||
Group terms by $u_1$ and $u_2$:
|
||||
\[
|
||||
u_1'\,y_1' + u_2'\,y_2'
|
||||
+ u_1\bigl[y_1'' + p(x)y_1' + q(x)y_1\bigr]
|
||||
+ u_2\bigl[y_2'' + p(x)y_2' + q(x)y_2\bigr] = g(x).
|
||||
\]
|
||||
Since $y_1$ and $y_2$ are solutions of the homogeneous equation, the bracketed terms vanish. We are left with
|
||||
\[
|
||||
u_1'\,y_1' + u_2'\,y_2' = g(x).
|
||||
\]
|
||||
|
||||
\textbf{System of equations.} Combining this with the constraint \cref{eq:vop_constraint}, we have a $2 \times 2$ linear system for $u_1'$ and $u_2'$:
|
||||
\begin{equation}
|
||||
\label{eq:vop_system}
|
||||
\begin{cases}
|
||||
u_1' y_1 + u_2' y_2 = 0, \\[4pt]
|
||||
u_1' y_1' + u_2' y_2' = g(x).
|
||||
\end{cases}
|
||||
\end{equation}
|
||||
|
||||
\textbf{Solution via Cramer's rule.} Write the system in matrix form:
|
||||
\[
|
||||
\begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix}
|
||||
\begin{pmatrix} u_1' \\ u_2' \end{pmatrix}
|
||||
=
|
||||
\begin{pmatrix} 0 \\ g(x) \end{pmatrix}.
|
||||
\]
|
||||
The determinant of the coefficient matrix is the Wronskian:
|
||||
\[
|
||||
\det = y_1 y_2' - y_1' y_2 = W(x).
|
||||
\]
|
||||
Since $\{y_1, y_2\}$ is a fundamental set, $W(x) \neq 0$ and the system is solvable. By Cramer's rule:
|
||||
\begin{align}
|
||||
u_1'(x) &= \frac{\det\begin{pmatrix} 0 & y_2 \\ g(x) & y_2' \end{pmatrix}}{W(x)}
|
||||
= \frac{-y_2(x)\,g(x)}{W(x)}, \label{eq:vop_u1prime} \\
|
||||
u_2'(x) &= \frac{\det\begin{pmatrix} y_1 & 0 \\ y_1' & g(x) \end{pmatrix}}{W(x)}
|
||||
= \frac{y_1(x)\,g(x)}{W(x)}. \label{eq:vop_u2prime}
|
||||
\end{align}
|
||||
|
||||
Integrate to find $u_1$ and $u_2$ (we may choose any antiderivatives; we set the constants of integration to zero since we need only \emph{one} particular solution):
|
||||
\begin{align}
|
||||
u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x, \\
|
||||
u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x.
|
||||
\end{align}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Variation of parameters formula.}
|
||||
|
||||
Given a fundamental set $\{y_1, y_2\}$ of the homogeneous equation associated with \cref{eq:normalized_nonhomogeneous}, a particular solution is
|
||||
\begin{equation}
|
||||
\label{eq:vop_formula}
|
||||
y_p(x) = -y_1(x)\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x
|
||||
+ y_2(x)\int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x,
|
||||
\end{equation}
|
||||
where $W(x) = y_1 y_2' - y_1' y_2$.
|
||||
|
||||
\textbf{CRITICAL:} The equation \emph{must} be in normalized form (coefficient of $y''$ equals~1) before applying this formula. For $a\,y'' + b\,y' + c\,y = \tilde{g}(x)$, first divide by $a$ so that the right-hand side becomes $g(x) = \tilde{g}(x)/a$.
|
||||
\end{keyresult}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Normalized form warning.} A very common mistake is to apply the VoP formula to $a\,y'' + b\,y' + c\,y = \tilde{g}(x)$ directly, using $\tilde{g}(x)$ as $g(x)$. This is \textbf{incorrect}. The derivation assumed the coefficient of $y''$ is~1, so you \emph{must} divide the entire equation by $a$ first, making $g(x) = \tilde{g}(x)/a$.
|
||||
\end{hintbox}
|
||||
|
||||
\subsubsection{Worked Examples}
|
||||
\label{sec:ch05_vop_examples}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' + y = \sec(x)$ on the interval $(-\pi/2, \pi/2)$.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is $r^2 + 1 = 0$, so $r = \pm i$. Thus
|
||||
\[
|
||||
y_h(x) = c_1 \cos x + c_2 \sin x.
|
||||
\]
|
||||
We take $y_1(x) = \cos x$ and $y_2(x) = \sin x$.
|
||||
|
||||
\textit{Step 2: Wronskian.}
|
||||
\[
|
||||
W(x) = \cos x \cdot \cos x - (-\sin x) \cdot \sin x = \cos^2 x + \sin^2 x = 1.
|
||||
\]
|
||||
The Wronskian is constant, which simplifies the integrals.
|
||||
|
||||
\textit{Step 3: Apply VoP.} The equation is already in normalized form ($g(x) = \sec x$). Compute:
|
||||
\[
|
||||
\begin{aligned}
|
||||
u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x
|
||||
= -\int \frac{\sin x \cdot \sec x}{1}\,\diff x
|
||||
= -\int \tan x\,\diff x
|
||||
= \ln|\cos x|, \\[8pt]
|
||||
u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x
|
||||
= \int \frac{\cos x \cdot \sec x}{1}\,\diff x
|
||||
= \int 1\,\diff x
|
||||
= x.
|
||||
\end{aligned}
|
||||
\]
|
||||
|
||||
\textit{Step 4: Particular solution.}
|
||||
\[
|
||||
y_p(x) = u_1(x)\,y_1(x) + u_2(x)\,y_2(x)
|
||||
= \ln|\cos x|\,\cos x + x\,\sin x.
|
||||
\]
|
||||
|
||||
\textit{Step 5: General solution.}
|
||||
\[
|
||||
y(x) = c_1 \cos x + c_2 \sin x + \cos x\,\ln|\cos x| + x\,\sin x.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - 4y' + 3y = \dfrac{e^{2x}}{1 + e^x}$ using variation of parameters.
|
||||
|
||||
\textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 - 4r + 3 = 0 \quad\Longrightarrow\quad (r-1)(r-3) = 0.
|
||||
\]
|
||||
Roots: $r_1 = 1$, $r_2 = 3$. Thus
|
||||
\[
|
||||
y_h(x) = c_1 e^x + c_2 e^{3x}.
|
||||
\]
|
||||
Take $y_1(x) = e^x$ and $y_2(x) = e^{3x}$.
|
||||
|
||||
\textit{Step 2: Wronskian.}
|
||||
\[
|
||||
W(x) = e^x \cdot 3e^{3x} - e^x \cdot e^{3x} = 3e^{4x} - e^{4x} = 2e^{4x}.
|
||||
\]
|
||||
|
||||
\textit{Step 3: Apply VoP.} The equation is in normalized form with $g(x) = \dfrac{e^{2x}}{1 + e^x}$. Compute:
|
||||
\[
|
||||
\begin{aligned}
|
||||
u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x
|
||||
= -\int \frac{e^{3x} \cdot \dfrac{e^{2x}}{1 + e^x}}{2e^{4x}}\,\diff x \\
|
||||
&= -\frac{1}{2}\int \frac{e^{5x}}{e^{4x}(1 + e^x)}\,\diff x
|
||||
= -\frac{1}{2}\int \frac{e^x}{1 + e^x}\,\diff x.
|
||||
\end{aligned}
|
||||
\]
|
||||
Substituting $u = 1 + e^x$, $\diff u = e^x\,\diff x$:
|
||||
\[
|
||||
u_1(x) = -\frac{1}{2}\int \frac{1}{u}\,\diff u = -\frac{1}{2}\ln|u|
|
||||
= -\frac{1}{2}\ln(1 + e^x).
|
||||
\]
|
||||
|
||||
Next:
|
||||
\[
|
||||
\begin{aligned}
|
||||
u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x
|
||||
= \int \frac{e^x \cdot \dfrac{e^{2x}}{1 + e^x}}{2e^{4x}}\,\diff x \\
|
||||
&= \frac{1}{2}\int \frac{e^{3x}}{e^{4x}(1 + e^x)}\,\diff x
|
||||
= \frac{1}{2}\int \frac{1}{e^x(1 + e^x)}\,\diff x.
|
||||
\end{aligned}
|
||||
\]
|
||||
Use partial fractions: $\dfrac{1}{e^x(1+e^x)} = \dfrac{1}{e^x} - \dfrac{1}{1+e^x} = e^{-x} - \dfrac{1}{1+e^x}$.
|
||||
\[
|
||||
\begin{aligned}
|
||||
u_2(x) &= \frac{1}{2}\int \left(e^{-x} - \frac{1}{1+e^x}\right)\diff x \\
|
||||
&= \frac{1}{2}\Bigl(-e^{-x} - x + \ln(1+e^x)\Bigr).
|
||||
\end{aligned}
|
||||
\]
|
||||
(The second integral: $\int \frac{1}{1+e^x}\diff x = x - \ln(1+e^x)$, obtained by writing $\frac{1}{1+e^x} = 1 - \frac{e^x}{1+e^x}$.)
|
||||
|
||||
\textit{Step 4: Particular solution.}
|
||||
\[
|
||||
\begin{aligned}
|
||||
y_p(x) &= u_1(x)\,y_1(x) + u_2(x)\,y_2(x) \\
|
||||
&= -\frac{1}{2}\ln(1+e^x)\,e^x + \frac{1}{2}\Bigl(-e^{-x} - x + \ln(1+e^x)\Bigr)e^{3x} \\
|
||||
&= -\frac{1}{2}e^x\ln(1+e^x) - \frac{1}{2}e^{2x} + \frac{1}{2}e^{3x}\ln(1+e^x) - \frac{1}{2}x\,e^{3x}.
|
||||
\end{aligned}
|
||||
\]
|
||||
|
||||
\textit{Step 5: General solution.}
|
||||
\[
|
||||
y(x) = c_1 e^x + c_2 e^{3x} - \frac{1}{2}e^x\ln(1+e^x)
|
||||
+ \frac{1}{2}e^{3x}\ln(1+e^x) - \frac{1}{2}e^{2x} - \frac{1}{2}x\,e^{3x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $2y'' + 4y' + 2y = e^{-x}$ using variation of parameters. Note that this equation is \textbf{not} in normalized form.
|
||||
|
||||
\textbf{Solution.} \textit{Step 0: Normalize the equation.} Divide the entire equation by $2$:
|
||||
\[
|
||||
y'' + 2y' + y = \frac{1}{2}\,e^{-x}.
|
||||
\]
|
||||
Now $g(x) = \dfrac{1}{2}e^{-x}$.
|
||||
|
||||
\textit{Step 1: Homogeneous solution.} The characteristic equation is
|
||||
\[
|
||||
r^2 + 2r + 1 = 0 \quad\Longrightarrow\quad (r+1)^2 = 0.
|
||||
\]
|
||||
Double root $r = -1$. Take $y_1(x) = e^{-x}$ and $y_2(x) = x\,e^{-x}$.
|
||||
|
||||
\textit{Step 2: Wronskian.}
|
||||
\[
|
||||
W(x) = e^{-x}\bigl(e^{-x} - x\,e^{-x}\bigr) - (-e^{-x})(x\,e^{-x})
|
||||
= e^{-2x}(1 - x) + x\,e^{-2x} = e^{-2x}.
|
||||
\]
|
||||
|
||||
\textit{Step 3: Apply VoP.}
|
||||
\[
|
||||
\begin{aligned}
|
||||
u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x
|
||||
= -\int \frac{x\,e^{-x} \cdot \tfrac{1}{2}e^{-x}}{e^{-2x}}\,\diff x
|
||||
= -\frac{1}{2}\int x\,\diff x
|
||||
= -\frac{1}{4}x^2, \\[8pt]
|
||||
u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x
|
||||
= \int \frac{e^{-x} \cdot \tfrac{1}{2}e^{-x}}{e^{-2x}}\,\diff x
|
||||
= \frac{1}{2}\int 1\,\diff x
|
||||
= \frac{1}{2}x.
|
||||
\end{aligned}
|
||||
\]
|
||||
|
||||
\textit{Step 4: Particular solution.}
|
||||
\[
|
||||
y_p(x) = u_1(x)\,y_1(x) + u_2(x)\,y_2(x)
|
||||
= -\frac{1}{4}x^2\,e^{-x} + \frac{1}{2}x^2\,e^{-x}
|
||||
= \frac{1}{4}x^2\,e^{-x}.
|
||||
\]
|
||||
This is consistent with the undetermined coefficients result from Example~5 (the modification rule also gave $x^2 e^{-x}$ for this case).
|
||||
|
||||
\textit{Step 5: General solution.}
|
||||
\[
|
||||
y(x) = c_1 e^{-x} + c_2 x\,e^{-x} + \frac{1}{4}x^2\,e^{-x}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Method Comparison}
|
||||
\label{sec:ch05_method_comparison}
|
||||
|
||||
Both undetermined coefficients (UC) and variation of parameters (VoP) find particular solutions, but they have different strengths and limitations.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Undetermined coefficients vs.\ variation of parameters}
|
||||
\label{tab:ch05_method_comparison}
|
||||
\begin{tabular}{p{2.8cm} p{4.8cm} p{6.2cm}}
|
||||
\toprule
|
||||
\textbf{Aspect} & \textbf{Undetermined Coefficients} & \textbf{Variation of Parameters} \\
|
||||
\midrule
|
||||
Applicable $g(x)$ & Polynomials, $e^{\alpha x}$, $\sin/\cos(\beta x)$, sums/products & \emph{Any} continuous $g(x)$ (in principle) \\
|
||||
Requires $y_h$? & Only for overlap check & Yes --- need a full fundamental set \\
|
||||
Requires $W(x)$? & No & Yes \\
|
||||
Normalized form? & Not required & \textbf{Required} (coefficient of $y''$ must be~1) \\
|
||||
Computation & Algebra (solve linear system) & Integration (may be difficult or non-elementary) \\
|
||||
Speed (simple $g(x)$) & Fast --- algebraic, no integrals & Slower --- integrals needed \\
|
||||
Speed (complex $g(x)$) & \textbf{N/A} --- method fails & Always applicable (if integrals exist) \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Decision guide.}
|
||||
\begin{enumerate}
|
||||
\item Is $g(x)$ a polynomial, exponential, sine/cosine, or a product of these? \textbf{Yes} $\Rightarrow$ use undetermined coefficients (simpler algebra).
|
||||
\item Is $g(x)$ anything else ($\ln x$, $\tan x$, rational functions, etc.)? \textbf{Yes} $\Rightarrow$ use variation of parameters.
|
||||
\item If you are unsure, variation of parameters is always a safe fallback (provided you can compute the integrals).
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
|
||||
\paragraph{Example comparison.} Consider $y'' + y' - 2y = 4x$.
|
||||
|
||||
Using \textbf{undetermined coefficients}: guess $y_p = Ax + B$, substitute, solve the linear system. Takes about 5 lines of algebra.
|
||||
|
||||
Using \textbf{variation of parameters}: find $y_h$, compute $W(x)$, evaluate two integrals involving $\int x e^x\,\diff x$ and $\int x e^{-2x}\,\diff x$ (by parts). Takes about 15 lines of computation.
|
||||
|
||||
Both give the same answer, but UC is clearly faster here. However, if $g(x) = \ln x$, only VoP can be applied.
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch05_summary}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Nonhomogeneous second-order equations: complete reference}
|
||||
\label{tab:ch05_summary_table}
|
||||
\begin{tabular}{l p{8cm}}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key formula / rule} \\
|
||||
\midrule
|
||||
General solution & $y(x) = y_h(x) + y_p(x)$ (\cref{thm:nonhomogeneous_structure}) \\
|
||||
Homogeneous part & From \cref{ch:second_order_homogeneous}: $y_h = c_1 y_1 + c_2 y_2$ \\
|
||||
\midrule
|
||||
\textbf{UC: polynomial $P_n(x)$} & $y_p = A_n x^n + A_{n-1} x^{n-1} + \cdots + A_0$ \\
|
||||
\textbf{UC: $e^{\alpha x}$} & $y_p = A\,e^{\alpha x}$ \\
|
||||
\textbf{UC: $\sin(\beta x)$ or $\cos(\beta x)$} & $y_p = A\sin(\beta x) + B\cos(\beta x)$ \\
|
||||
\textbf{UC: $e^{\alpha x} P_n(x)$} & $y_p = e^{\alpha x}(A_n x^n + \cdots + A_0)$ \\
|
||||
\textbf{UC: $e^{\alpha x}\sin(\beta x)$ or $e^{\alpha x}\cos(\beta x)$} & $y_p = e^{\alpha x}\bigl[(A_n x^n + \cdots + A_0)\cos(\beta x) + (B_n x^n + \cdots + B_0)\sin(\beta x)\bigr]$ \\
|
||||
\midrule
|
||||
\textbf{UC modification rule} & If guess term $\in y_h$, multiply by $x$; if still $\in y_h$, multiply by $x^2$ \\
|
||||
\midrule
|
||||
\textbf{VoP formula} & $y_p = -y_1\displaystyle\int\frac{y_2 g}{W}\,\diff x + y_2\displaystyle\int\frac{y_1 g}{W}\,\diff x$ \\
|
||||
\textbf{VoP constraint} & Equation must be normalized: $y'' + py' + qy = g$ (coeff.\ of $y''$ = 1) \\
|
||||
\textbf{VoP system} & $u_1' y_1 + u_2' y_2 = 0$, $\; u_1' y_1' + u_2' y_2' = g(x)$ \\
|
||||
\textbf{VoP Cramer's rule} & $u_1' = -\dfrac{y_2 g}{W}$, $\; u_2' = \dfrac{y_1 g}{W}$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Problem-solving workflow for nonhomogeneous equations.}
|
||||
\begin{enumerate}
|
||||
\item \textbf{Identify the form of $g(x)$.} If it is a polynomial, exponential, trig function, or their products/sums, prefer undetermined coefficients. Otherwise, use variation of parameters.
|
||||
\item \textbf{Solve the homogeneous equation} $ay'' + by' + cy = 0$ and obtain $y_h(x)$.
|
||||
\item \textbf{If using UC:} Write the guess based on the table, check for overlap with $y_h$, modify if needed, substitute, and solve for coefficients.
|
||||
\item \textbf{If using VoP:} \emph{First} normalize the equation (divide by $a$ if needed). Compute $W(x)$, then evaluate the two integrals for $u_1$ and $u_2$.
|
||||
\item \textbf{Combine:} $y(x) = y_h(x) + y_p(x)$. Apply initial conditions if given.
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
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|
||||
% =============================================================================
|
||||
% ch09_series_solutions.tex
|
||||
% Chapter 9: Series Solutions of Differential Equations
|
||||
% =============================================================================
|
||||
|
||||
\section{Series Solutions}
|
||||
\label{ch:series_solutions}
|
||||
|
||||
When the coefficients of a linear differential equation are not constant, the methods of \cref{ch:second_order_homogeneous} and \cref{ch:second_order_nonhomogeneous} (characteristic equations, undetermined coefficients, variation of parameters) generally fail. In such cases, \textbf{series solutions} provide a powerful alternative: we represent the unknown solution as an infinite series and determine the coefficients by substituting into the differential equation.
|
||||
|
||||
This chapter develops three closely related methods:
|
||||
\begin{enumerate}
|
||||
\item The \textbf{power series method} for equations with analytic coefficients (\cref{sec:ch09_power_series}).
|
||||
\item \textbf{Euler--Cauchy (equidimensional) equations}, a special class admitting closed-form solutions (\cref{sec:ch09_euler_cauchy}).
|
||||
\item The \textbf{Frobenius method}, which extends the power series idea to equations with regular singular points (\cref{sec:ch09_frobenius}).
|
||||
\end{enumerate}
|
||||
|
||||
All three methods share a common theme: substitute a series ansatz, reindex, collect coefficients, and solve the resulting recurrence relation.
|
||||
|
||||
\subsection{Power Series Method}
|
||||
\label{sec:ch09_power_series}
|
||||
|
||||
\paragraph{When to use.} The power series method applies to the second-order linear homogeneous equation
|
||||
\begin{equation}
|
||||
\label{eq:power_series_ode}
|
||||
y'' + p(x)\,y' + q(x)\,y = 0
|
||||
\end{equation}
|
||||
when both $p(x)$ and $q(x)$ are \textbf{analytic} at a point $x_0$ (typically $x_0 = 0$). A function is analytic at $x_0$ if it equals its Taylor series in a neighborhood of $x_0$. This includes polynomials, $e^x$, $\sin x$, $\cos x$, and rational functions away from their poles.
|
||||
|
||||
If $p(x)$ or $q(x)$ has a singularity at $x_0$, the power series method in its basic form fails --- we must instead use the Frobenius method (\cref{sec:ch09_frobenius}).
|
||||
|
||||
\paragraph{Step-by-step algorithm.}
|
||||
|
||||
\begin{enumerate}
|
||||
\item \textbf{Ansatz.} Assume a power series solution about $x_0 = 0$:
|
||||
\[
|
||||
y(x) = \sum_{n=0}^{\infty} a_n\,x^n,
|
||||
\qquad a_n \text{ to be determined.}
|
||||
\]
|
||||
\item \textbf{Differentiate term-by-term:}
|
||||
\[
|
||||
y'(x) = \sum_{n=1}^{\infty} n\,a_n\,x^{n-1},
|
||||
\qquad
|
||||
y''(x) = \sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}.
|
||||
\]
|
||||
\item \textbf{Substitute} $y$, $y'$, and $y''$ into the ODE \cref{eq:power_series_ode}.
|
||||
\item \textbf{Reindex} all sums so they involve the same power $x^k$. This typically means shifting the index: set $k = n - m$ in each sum so every sum starts at the same $k$ and involves $x^k$.
|
||||
\item \textbf{Collect coefficients} of each power $x^k$. The resulting coefficient expression must vanish for every $k$, producing a \textbf{recurrence relation} for the coefficients $a_n$.
|
||||
\item \textbf{Solve the recurrence} to find the coefficients in terms of $a_0$ and $a_1$ (which are arbitrary --- they correspond to the two degrees of freedom in a second-order ODE).
|
||||
\item \textbf{Write the general solution} as a linear combination of the two independent series.
|
||||
\end{enumerate}
|
||||
|
||||
\paragraph{Radius of convergence.}
|
||||
|
||||
\begin{theorem}[Radius of Convergence]
|
||||
\label{thm:radius_of_convergence}
|
||||
Consider the equation \cref{eq:power_series_ode} with $p(x)$ and $q(x)$ analytic at $x_0 = 0$. If $R_p$ and $R_q$ denote the radii of convergence of the Taylor series of $p(x)$ and $q(x)$ at $x_0 = 0$, then the power series solution $y(x) = \sum_{n=0}^{\infty} a_n x^n$ converges at least on the interval $(-R, R)$ where
|
||||
\[
|
||||
R = \min(R_p,\; R_q).
|
||||
\]
|
||||
Equivalently, $R$ is at least as large as the distance from $x_0$ to the nearest singularity of $p(x)$ or $q(x)$ in the complex plane.
|
||||
\end{theorem}
|
||||
|
||||
In practice, most textbook problems have polynomial coefficients (so $R_p = R_q = \infty$), and the series converges for all $x \in \R$. When $p(x)$ or $q(x)$ is a rational function, $R$ is the distance from $x_0$ to the nearest pole.
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Quick radius check.} If the ODE has polynomial coefficients, the power series converges for all $x$. If coefficients are rational, compute the distance from the expansion point to the nearest pole. For example, $y'' + \frac{1}{1-x^2} y = 0$ expanded about $x_0 = 0$ has poles at $x = \pm 1$, so $R = 1$.
|
||||
\end{hintbox}
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $y'' - y = 0$ using the power series method and show that the solution recovers the exponential functions.
|
||||
|
||||
\textbf{Solution.} Here $p(x) = 0$ and $q(x) = -1$, both analytic everywhere. Assume
|
||||
\[
|
||||
y(x) = \sum_{n=0}^{\infty} a_n\,x^n.
|
||||
\]
|
||||
Differentiate:
|
||||
\[
|
||||
y' = \sum_{n=1}^{\infty} n\,a_n\,x^{n-1},
|
||||
\qquad
|
||||
y'' = \sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}.
|
||||
\]
|
||||
Substitute into $y'' - y = 0$:
|
||||
\[
|
||||
\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}
|
||||
- \sum_{n=0}^{\infty} a_n\,x^n = 0.
|
||||
\]
|
||||
Reindex the first sum by setting $k = n - 2$ (so $n = k + 2$):
|
||||
\[
|
||||
\sum_{k=0}^{\infty} (k+2)(k+1)\,a_{k+2}\,x^k
|
||||
- \sum_{k=0}^{\infty} a_k\,x^k = 0.
|
||||
\]
|
||||
Both sums now involve $x^k$ starting at $k = 0$. Collect coefficients of $x^k$:
|
||||
\[
|
||||
(k+2)(k+1)\,a_{k+2} - a_k = 0
|
||||
\quad\Longrightarrow\quad
|
||||
a_{k+2} = \frac{a_k}{(k+2)(k+1)}.
|
||||
\]
|
||||
This is the \textbf{recurrence relation}. It connects every even coefficient to $a_0$ and every odd coefficient to $a_1$.
|
||||
|
||||
\textit{Even coefficients:}
|
||||
\[
|
||||
\begin{aligned}
|
||||
a_2 &= \frac{a_0}{2 \cdot 1} = \frac{a_0}{2!}, \\
|
||||
a_4 &= \frac{a_2}{4 \cdot 3} = \frac{a_0}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{a_0}{4!}, \\
|
||||
a_6 &= \frac{a_4}{6 \cdot 5} = \frac{a_0}{6!}, \\
|
||||
&\;\;\vdots \\
|
||||
a_{2m} &= \frac{a_0}{(2m)!}.
|
||||
\end{aligned}
|
||||
\]
|
||||
|
||||
\textit{Odd coefficients:}
|
||||
\[
|
||||
\begin{aligned}
|
||||
a_3 &= \frac{a_1}{3 \cdot 2} = \frac{a_1}{3!}, \\
|
||||
a_5 &= \frac{a_3}{5 \cdot 4} = \frac{a_1}{5!}, \\
|
||||
a_7 &= \frac{a_5}{7 \cdot 6} = \frac{a_1}{7!}, \\
|
||||
&\;\;\vdots \\
|
||||
a_{2m+1} &= \frac{a_1}{(2m+1)!}.
|
||||
\end{aligned}
|
||||
\]
|
||||
|
||||
The general solution is
|
||||
\[
|
||||
y(x) = a_0 \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m)!}
|
||||
+ a_1 \sum_{m=0}^{\infty} \frac{x^{2m+1}}{(2m+1)!}.
|
||||
\]
|
||||
These are the Taylor series for $\cosh x$ and $\sinh x$:
|
||||
\[
|
||||
y(x) = a_0 \cosh x + a_1 \sinh x.
|
||||
\]
|
||||
Equivalently, writing $c_1 = \frac{a_0 + a_1}{2}$ and $c_2 = \frac{a_0 - a_1}{2}$:
|
||||
\[
|
||||
y(x) = c_1 e^x + c_2 e^{-x}.
|
||||
\]
|
||||
This recovers the exponential solutions found via the characteristic equation in \cref{ch:second_order_homogeneous}. The power series method works, but for constant-coefficient equations it is overkill --- the characteristic equation is much faster. The power series method truly shines when coefficients are \emph{not} constant.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the Legendre equation $(1-x^2)y'' - 2xy' + 2y = 0$ using the power series method. Find a polynomial solution.
|
||||
|
||||
\textbf{Solution.} This is a special case of the \textbf{Legendre equation}
|
||||
$(1-x^2)y'' - 2xy' + \ell(\ell+1)y = 0$ with $\ell = 1$.
|
||||
The coefficients $p(x) = \dfrac{-2x}{1-x^2}$ and $q(x) = \dfrac{2}{1-x^2}$ have poles at $x = \pm 1$, so by \cref{thm:radius_of_convergence} the series converges for $|x| < 1$.
|
||||
|
||||
Assume $y = \sum_{n=0}^{\infty} a_n x^n$. Then
|
||||
\[
|
||||
y' = \sum_{n=1}^{\infty} n\,a_n\,x^{n-1},
|
||||
\qquad
|
||||
y'' = \sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}.
|
||||
\]
|
||||
Substitute into $(1-x^2)y'' - 2xy' + 2y = 0$:
|
||||
\[
|
||||
(1-x^2)\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}
|
||||
- 2x\sum_{n=1}^{\infty} n\,a_n\,x^{n-1}
|
||||
+ 2\sum_{n=0}^{\infty} a_n\,x^n = 0.
|
||||
\]
|
||||
Expand $(1-x^2)y''$:
|
||||
\[
|
||||
\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}
|
||||
- \sum_{n=2}^{\infty} n(n-1)\,a_n\,x^n
|
||||
- \sum_{n=1}^{\infty} 2n\,a_n\,x^n
|
||||
+ \sum_{n=0}^{\infty} 2a_n\,x^n = 0.
|
||||
\]
|
||||
Reindex the first sum by $k = n-2$ (so $n = k+2$):
|
||||
\[
|
||||
\sum_{k=0}^{\infty} (k+2)(k+1)\,a_{k+2}\,x^k
|
||||
- \sum_{n=2}^{\infty} n(n-1)\,a_n\,x^n
|
||||
- \sum_{n=1}^{\infty} 2n\,a_n\,x^n
|
||||
+ \sum_{n=0}^{\infty} 2a_n\,x^n = 0.
|
||||
\]
|
||||
All sums now involve powers of $x$ starting from $x^0$. Renaming the dummy index to $k$ everywhere and collecting coefficients of $x^k$:
|
||||
\[
|
||||
(k+2)(k+1)\,a_{k+2}
|
||||
- k(k-1)\,a_k
|
||||
- 2k\,a_k
|
||||
+ 2a_k = 0,
|
||||
\]
|
||||
where we understand $a_k = 0$ for $k < 0$. Simplify the $a_k$ coefficient:
|
||||
\[
|
||||
-k(k-1) - 2k + 2 = -k^2 + k - 2k + 2 = -k^2 - k + 2 = -(k-1)(k+2).
|
||||
\]
|
||||
The recurrence relation is
|
||||
\[
|
||||
(k+2)(k+1)\,a_{k+2} - (k-1)(k+2)\,a_k = 0.
|
||||
\]
|
||||
For $k \neq -2$ (always true since $k \geq 0$), divide by $(k+2)$:
|
||||
\[
|
||||
(k+1)\,a_{k+2} - (k-1)\,a_k = 0
|
||||
\quad\Longrightarrow\quad
|
||||
a_{k+2} = \frac{k-1}{k+1}\,a_k.
|
||||
\]
|
||||
|
||||
Now compute coefficients:
|
||||
\[
|
||||
\begin{aligned}
|
||||
a_2 &= \frac{0-1}{0+1}\,a_0 = -a_0, \\
|
||||
a_4 &= \frac{2-1}{2+1}\,a_2 = \frac{1}{3}(-a_0) = -\frac{a_0}{3}, \\
|
||||
a_6 &= \frac{4-1}{4+1}\,a_4 = \frac{3}{5}\!\left(-\frac{a_0}{3}\right) = -\frac{a_0}{5}.
|
||||
\end{aligned}
|
||||
\]
|
||||
|
||||
For the odd series:
|
||||
\[
|
||||
\begin{aligned}
|
||||
a_3 &= \frac{1-1}{1+1}\,a_1 = 0 \cdot a_1 = 0.
|
||||
\end{aligned}
|
||||
\]
|
||||
Since $a_3 = 0$, every subsequent odd coefficient vanishes ($a_5 = \frac{3-1}{3+1} a_3 = 0$, etc.). The odd series terminates immediately.
|
||||
|
||||
If we choose $a_1 = 1$ and $a_0 = 0$, the solution is simply
|
||||
\[
|
||||
y(x) = a_1 x = x.
|
||||
\]
|
||||
Let us verify: $y = x \Rightarrow y' = 1, y'' = 0$. Substituting into the ODE:
|
||||
\[
|
||||
(1-x^2) \cdot 0 - 2x \cdot 1 + 2 \cdot x = -2x + 2x = 0. \quad \checkmark
|
||||
\]
|
||||
|
||||
Thus $y_1(x) = x$ is a \textbf{polynomial solution} (a Legendre polynomial of degree~1, denoted $P_1(x)$). The second solution involves the infinite even series, which is related to the Legendre function of the second kind $Q_1(x)$.
|
||||
|
||||
\textbf{Note.} Legendre polynomials $P_n(x)$ are polynomial solutions of the Legendre equation for integer $\ell = n$. They arise in physics (e.g.\ spherical harmonics in quantum mechanics, gravitational potential theory) and form an orthogonal basis on $[-1, 1]$.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{remark}
|
||||
The power series method often produces solutions that are recognized as classical \textbf{special functions}. Bessel functions (Frobenius method, \cref{sec:ch09_frobenius}), Legendre polynomials (as in Example~2 above), Hermite polynomials, and Laguerre polynomials all arise as power series solutions of specific ODEs. These special functions have their own properties, tables, and computational implementations, making the power series approach not just theoretical but deeply practical.
|
||||
\end{remark}
|
||||
|
||||
\subsection{Euler--Cauchy Equations}
|
||||
\label{sec:ch09_euler_cauchy}
|
||||
|
||||
The Euler--Cauchy equation (also called the \textbf{equidimensional equation}) is a special second-order ODE with variable coefficients that nevertheless admits closed-form solutions through a power ansatz.
|
||||
|
||||
\paragraph{Form and substitution.} The equation has the form
|
||||
\begin{equation}
|
||||
\label{eq:euler_cauchy}
|
||||
x^2\,y'' + \alpha\,x\,y' + \beta\,y = 0,
|
||||
\qquad x > 0,
|
||||
\end{equation}
|
||||
where $\alpha, \beta \in \R$. The key observation is that every term is \textbf{equidimensional}: the $k$-th derivative is multiplied by $x^k$. This suggests the ansatz
|
||||
\[
|
||||
y(x) = x^r,
|
||||
\]
|
||||
where $r$ is a constant to be determined.
|
||||
|
||||
\paragraph{Derivation of the indicial equation.}
|
||||
Compute the derivatives of $y = x^r$:
|
||||
\[
|
||||
y' = r\,x^{r-1}, \qquad y'' = r(r-1)\,x^{r-2}.
|
||||
\]
|
||||
Substitute into \cref{eq:euler_cauchy}:
|
||||
\begin{align*}
|
||||
x^2 \cdot r(r-1)\,x^{r-2} + \alpha\,x \cdot r\,x^{r-1} + \beta \cdot x^r &= 0, \\
|
||||
r(r-1)\,x^r + \alpha\,r\,x^r + \beta\,x^r &= 0, \\
|
||||
\bigl[r(r-1) + \alpha\,r + \beta\bigr]\,x^r &= 0.
|
||||
\end{align*}
|
||||
Since $x^r \neq 0$ for $x > 0$, we must have
|
||||
\begin{equation}
|
||||
\label{eq:euler_indicial}
|
||||
r(r-1) + \alpha\,r + \beta = 0.
|
||||
\end{equation}
|
||||
This is the \textbf{indicial equation} (also called the \textbf{characteristic equation} of the Euler--Cauchy equation). It is a quadratic in $r$:
|
||||
\[
|
||||
r^2 + (\alpha - 1)\,r + \beta = 0,
|
||||
\]
|
||||
with solutions
|
||||
\[
|
||||
r = \frac{-(\alpha-1) \pm \sqrt{(\alpha-1)^2 - 4\beta}}{2}.
|
||||
\]
|
||||
|
||||
\paragraph{Three cases.}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Euler--Cauchy: solution forms based on the indicial equation.}
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Roots of \cref{eq:euler_indicial}} & \textbf{General solution} \\
|
||||
\midrule
|
||||
Distinct real $r_1 \neq r_2$ &
|
||||
$y(x) = c_1\,x^{r_1} + c_2\,x^{r_2}$ \\[10pt]
|
||||
Repeated $r_1 = r_2 = r$ &
|
||||
$y(x) = x^{r}\bigl[c_1 + c_2\,\ln(x)\bigr]$ \\[10pt]
|
||||
Complex $\lambda \pm i\omega$ &
|
||||
$y(x) = x^{\lambda}\bigl[c_1\cos(\omega\ln x) + c_2\sin(\omega\ln x)\bigr]$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{keyresult}
|
||||
|
||||
The repeated-root case is analogous to the repeated-root case for constant-coefficient equations (\cref{sec:ch04_repeated_roots}): the logarithmic factor plays the role of the extra $x$ factor. The complex-root case involves trigonometric functions of $\ln x$, reflecting the self-similar (scale-invariant) nature of the Euler--Cauchy equation.
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $x^2\,y'' - x\,y' - 3y = 0$ for $x > 0$.
|
||||
|
||||
\textbf{Solution.} This is an Euler--Cauchy equation with $\alpha = -1$ and $\beta = -3$. The indicial equation is
|
||||
\[
|
||||
r(r-1) + (-1)r + (-3) = 0.
|
||||
\]
|
||||
Simplify:
|
||||
\[
|
||||
r^2 - r - r - 3 = r^2 - 2r - 3 = 0.
|
||||
\]
|
||||
Factor:
|
||||
\[
|
||||
(r - 3)(r + 1) = 0.
|
||||
\]
|
||||
Distinct real roots: $r_1 = 3$ and $r_2 = -1$. The general solution is
|
||||
\[
|
||||
y(x) = c_1\,x^3 + c_2\,x^{-1} = c_1\,x^3 + \frac{c_2}{x}.
|
||||
\]
|
||||
|
||||
\textit{Verification.} Compute $y' = 3c_1 x^2 - c_2 x^{-2}$ and $y'' = 6c_1 x + 2c_2 x^{-3}$. Substitute into the ODE:
|
||||
\[
|
||||
\begin{split}
|
||||
& x^2(6c_1 x + 2c_2 x^{-3}) - x(3c_1 x^2 - c_2 x^{-2}) - 3(c_1 x^3 + c_2 x^{-1}) \\
|
||||
& \quad = 6c_1 x^3 + 2c_2 x^{-1} - 3c_1 x^3 + c_2 x^{-1} - 3c_1 x^3 - 3c_2 x^{-1} \\
|
||||
& \quad = (6 - 3 - 3)c_1 x^3 + (2 + 1 - 3)c_2 x^{-1} = 0. \quad \checkmark
|
||||
\end{split}
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $x^2\,y'' + 3x\,y' + y = 0$ for $x > 0$.
|
||||
|
||||
\textbf{Solution.} Here $\alpha = 3$ and $\beta = 1$. The indicial equation is
|
||||
\[
|
||||
r(r-1) + 3r + 1 = 0.
|
||||
\]
|
||||
Simplify:
|
||||
\[
|
||||
r^2 - r + 3r + 1 = r^2 + 2r + 1 = (r + 1)^2 = 0.
|
||||
\]
|
||||
Repeated root: $r = -1$. The general solution is
|
||||
\[
|
||||
y(x) = x^{-1}\bigl[c_1 + c_2\,\ln(x)\bigr] = \frac{c_1 + c_2\,\ln x}{x}.
|
||||
\]
|
||||
|
||||
\textit{Verification.} Let $y = \dfrac{\ln x}{x}$. Then
|
||||
\[
|
||||
y' = \frac{1/x \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2},
|
||||
\qquad
|
||||
y'' = \frac{(-1/x)\cdot x^2 - (1-\ln x)\cdot 2x}{x^4}
|
||||
= \frac{-x - 2x + 2x\ln x}{x^4}
|
||||
= \frac{2\ln x - 3}{x^3}.
|
||||
\]
|
||||
Substitute:
|
||||
\[
|
||||
x^2 \cdot \frac{2\ln x - 3}{x^3} + 3x \cdot \frac{1 - \ln x}{x^2} + \frac{\ln x}{x}
|
||||
= \frac{2\ln x - 3}{x} + \frac{3 - 3\ln x}{x} + \frac{\ln x}{x}
|
||||
= \frac{2\ln x - 3 + 3 - 3\ln x + \ln x}{x} = 0. \quad \checkmark
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve $x^2\,y'' + 2x\,y' + 10y = 0$ for $x > 0$.
|
||||
|
||||
\textbf{Solution.} Here $\alpha = 2$ and $\beta = 10$. The indicial equation is
|
||||
\[
|
||||
r(r-1) + 2r + 10 = 0.
|
||||
\]
|
||||
Simplify:
|
||||
\[
|
||||
r^2 - r + 2r + 10 = r^2 + r + 10 = 0.
|
||||
\]
|
||||
Discriminant: $\Delta = 1 - 40 = -39 < 0$. Complex roots:
|
||||
\[
|
||||
r = \frac{-1 \pm i\sqrt{39}}{2}.
|
||||
\]
|
||||
Here $\lambda = -\dfrac{1}{2}$ and $\omega = \dfrac{\sqrt{39}}{2}$. The general solution is
|
||||
\[
|
||||
y(x) = x^{-1/2}\!\left[c_1\cos\!\left(\tfrac{\sqrt{39}}{2}\ln x\right)
|
||||
+ c_2\sin\!\left(\tfrac{\sqrt{39}}{2}\ln x\right)\right].
|
||||
\]
|
||||
Equivalently:
|
||||
\[
|
||||
y(x) = \frac{1}{\sqrt{x}}\!\left[c_1\cos\!\left(\tfrac{\sqrt{39}}{2}\ln x\right)
|
||||
+ c_2\sin\!\left(\tfrac{\sqrt{39}}{2}\ln x\right)\right].
|
||||
\]
|
||||
This represents oscillations in $\ln x$ whose amplitude decays like $x^{-1/2}$.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{remark}
|
||||
Euler--Cauchy equations are scale-invariant: replacing $x$ by $cx$ (for any constant $c > 0$) transforms the equation into itself. This is why the solutions involve powers of $x$ and trigonometric functions of $\ln x$ --- both are natural under scaling.
|
||||
\end{remark}
|
||||
|
||||
\subsection{Frobenius Method}
|
||||
\label{sec:ch09_frobenius}
|
||||
|
||||
The power series method (\cref{sec:ch09_power_series}) requires the coefficients $p(x)$ and $q(x)$ to be analytic at the expansion point. When this condition fails, we may still obtain series solutions if the singularity is not too severe. The Frobenius method handles precisely this situation.
|
||||
|
||||
\paragraph{Regular singular point.}
|
||||
|
||||
\begin{definition}[Regular Singular Point]
|
||||
\label{def:regular_singular_point}
|
||||
Consider the equation $y'' + p(x)y' + q(x)y = 0$. The point $x_0$ is a \textbf{regular singular point} if:
|
||||
\begin{enumerate}
|
||||
\item $p(x)$ or $q(x)$ (or both) is \emph{not} analytic at $x_0$, but
|
||||
\item both $x\,p(x)$ and $x^2\,q(x)$ \emph{are} analytic at $x_0$ (after a shift to $x_0 = 0$ if necessary).
|
||||
\end{enumerate}
|
||||
If neither $x\,p(x)$ nor $x^2\,q(x)$ is analytic at $x_0$, then $x_0$ is an \textbf{irregular singular point}, and the Frobenius method does not apply.
|
||||
\end{definition}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Quick test.} For the equation $y'' + \dfrac{A}{x}y' + \dfrac{B}{x^2}y = 0$, the point $x_0 = 0$ is a regular singular point because $x\,p(x) = A$ and $x^2\,q(x) = B$ are both analytic (they are constants). This is exactly the Euler--Cauchy equation (\cref{sec:ch09_euler_cauchy}), which is a special case where the Frobenius series terminates into a closed-form solution.
|
||||
\end{hintbox}
|
||||
|
||||
\paragraph{Frobenius ansatz.}
|
||||
When $x_0 = 0$ is a regular singular point, we assume a solution of the form
|
||||
\begin{equation}
|
||||
\label{eq:frobenius_ansatz}
|
||||
y(x) = x^r \sum_{n=0}^{\infty} a_n\,x^n
|
||||
= \sum_{n=0}^{\infty} a_n\,x^{n+r},
|
||||
\qquad a_0 \neq 0.
|
||||
\end{equation}
|
||||
The exponent $r$ is \textbf{not} assumed to be an integer --- it is a parameter to be determined by the equation. When $r = 0$, the Frobenius ansatz reduces to the ordinary power series (\cref{sec:ch09_power_series}).
|
||||
|
||||
\paragraph{Indicial equation.}
|
||||
Substituting \cref{eq:frobenius_ansatz} into the ODE and collecting the lowest power of $x$ produces a quadratic equation for $r$, called the \textbf{indicial equation}. This equation determines the possible values of $r$ and hence the form of the leading term of the series.
|
||||
|
||||
\paragraph{Three cases.}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Frobenius method: solution forms based on the indicial roots.}
|
||||
|
||||
Let $r_1$ and $r_2$ be the roots of the indicial equation, with $r_1 \geq r_2$ (real parts if complex).
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Condition on $r_1, r_2$} & \textbf{Solution structure} \\
|
||||
\midrule
|
||||
$r_1 - r_2$ is \emph{not} a nonnegative integer &
|
||||
Two independent Frobenius series: \\[4pt]
|
||||
& $y_1(x) = x^{r_1}\sum_{n=0}^{\infty} a_n x^n$, \;
|
||||
$y_2(x) = x^{r_2}\sum_{n=0}^{\infty} b_n x^n$ \\[12pt]
|
||||
$r_1 = r_2$ (repeated root) &
|
||||
$y_1(x) = x^{r_1}\sum_{n=0}^{\infty} a_n x^n$, \\
|
||||
& $y_2(x) = y_1(x)\ln(x) + x^{r_1}\sum_{n=0}^{\infty} b_n x^n$ \\[12pt]
|
||||
$r_1 - r_2 = N$ (positive integer) &
|
||||
$y_1(x)$ from the larger root $r_1$; \\
|
||||
& $y_2(x)$ may or may not contain $\ln(x)$; \\
|
||||
& requires separate analysis \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{keyresult}
|
||||
|
||||
Cases~2 and~3 involve the logarithmic term $\ln(x)$, analogous to the repeated-root case for constant-coefficient equations. The detailed derivations of Cases~2 and~3 are beyond the scope of this chapter (see standard texts on advanced ODE theory for full treatments). We illustrate Case~1 with Bessel's equation.
|
||||
|
||||
\paragraph{Worked example: Bessel's equation of order~0.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve Bessel's equation of order~0:
|
||||
\[
|
||||
x^2\,y'' + x\,y' + x^2\,y = 0,
|
||||
\]
|
||||
using the Frobenius method about $x_0 = 0$.
|
||||
|
||||
\textbf{Solution.} First write the equation in standard form:
|
||||
\[
|
||||
y'' + \frac{1}{x}y' + y = 0.
|
||||
\]
|
||||
Here $p(x) = \dfrac{1}{x}$ and $q(x) = 1$. Check the regular singular point condition:
|
||||
\[
|
||||
x\,p(x) = 1 \quad\text{(analytic at $x = 0$)},
|
||||
\qquad
|
||||
x^2\,q(x) = x^2 \quad\text{(analytic at $x = 0$)}.
|
||||
\]
|
||||
So $x_0 = 0$ is a regular singular point. We proceed with the Frobenius ansatz:
|
||||
\[
|
||||
y(x) = \sum_{n=0}^{\infty} a_n\,x^{n+r}, \qquad a_0 \neq 0.
|
||||
\]
|
||||
Compute the derivatives:
|
||||
\[
|
||||
y' = \sum_{n=0}^{\infty} (n+r)\,a_n\,x^{n+r-1},
|
||||
\qquad
|
||||
y'' = \sum_{n=0}^{\infty} (n+r)(n+r-1)\,a_n\,x^{n+r-2}.
|
||||
\]
|
||||
Substitute into $x^2 y'' + x y' + x^2 y = 0$:
|
||||
\[
|
||||
x^2 \sum_{n=0}^{\infty} (n+r)(n+r-1)\,a_n\,x^{n+r-2}
|
||||
+ x \sum_{n=0}^{\infty} (n+r)\,a_n\,x^{n+r-1}
|
||||
+ x^2 \sum_{n=0}^{\infty} a_n\,x^{n+r} = 0.
|
||||
\]
|
||||
Simplify powers of $x$:
|
||||
\[
|
||||
\sum_{n=0}^{\infty} (n+r)(n+r-1)\,a_n\,x^{n+r}
|
||||
+ \sum_{n=0}^{\infty} (n+r)\,a_n\,x^{n+r}
|
||||
+ \sum_{n=0}^{\infty} a_n\,x^{n+r+2} = 0.
|
||||
\]
|
||||
Combine the first two sums:
|
||||
\[
|
||||
(n+r)(n+r-1) + (n+r) = (n+r)(n+r-1+1) = (n+r)^2.
|
||||
\]
|
||||
So the equation becomes
|
||||
\[
|
||||
\sum_{n=0}^{\infty} (n+r)^2\,a_n\,x^{n+r}
|
||||
+ \sum_{n=0}^{\infty} a_n\,x^{n+r+2} = 0.
|
||||
\]
|
||||
Reindex the second sum: let $k = n+2$ (so $n = k-2$). When $n=0$, $k=2$:
|
||||
\[
|
||||
\sum_{k=2}^{\infty} a_{k-2}\,x^{k+r} = \sum_{n=2}^{\infty} a_{n-2}\,x^{n+r}.
|
||||
\]
|
||||
Now combine:
|
||||
\[
|
||||
\underbrace{r^2\,a_0\,x^r}_{n=0}
|
||||
+ \underbrace{(1+r)^2\,a_1\,x^{1+r}}_{n=1}
|
||||
+ \sum_{n=2}^{\infty} \Bigl[(n+r)^2\,a_n + a_{n-2}\Bigr]\,x^{n+r} = 0.
|
||||
\]
|
||||
|
||||
\textit{Indicial equation.} The lowest power is $x^r$. Its coefficient must vanish:
|
||||
\[
|
||||
r^2\,a_0 = 0.
|
||||
\]
|
||||
Since $a_0 \neq 0$ by assumption, we must have $r^2 = 0$, so $r = 0$ (repeated root). This is \textbf{Case~2} ($r_1 = r_2$).
|
||||
|
||||
\textit{Next coefficient.} The coefficient of $x^{1+r} = x^1$ is $(1+r)^2 a_1 = a_1$ (since $r = 0$). For this to vanish, $a_1 = 0$.
|
||||
|
||||
\textit{Recurrence relation.} For $n \geq 2$:
|
||||
\[
|
||||
(n+r)^2\,a_n + a_{n-2} = 0
|
||||
\quad\Longrightarrow\quad
|
||||
a_n = -\frac{a_{n-2}}{(n+r)^2}.
|
||||
\]
|
||||
With $r = 0$:
|
||||
\[
|
||||
a_n = -\frac{a_{n-2}}{n^2}.
|
||||
\]
|
||||
Since $a_1 = 0$, all odd coefficients vanish ($a_3 = -a_1/9 = 0$, $a_5 = -a_3/25 = 0$, etc.). For even coefficients:
|
||||
\[
|
||||
\begin{aligned}
|
||||
a_2 &= -\frac{a_0}{2^2}, \\
|
||||
a_4 &= -\frac{a_2}{4^2} = \frac{a_0}{2^2 \cdot 4^2} = \frac{a_0}{(2^2)(2^2 \cdot 2^2)} = \frac{a_0}{2^4 \cdot (1! \cdot 2!)^2} \cdot 2^2,
|
||||
\end{aligned}
|
||||
\]
|
||||
which simplifies more cleanly as follows. Let $a_{2m}$ be the $m$-th even coefficient:
|
||||
\[
|
||||
a_{2m} = \frac{(-1)^m}{2^{2m}\,(m!)^2}\,a_0.
|
||||
\]
|
||||
Choosing $a_0 = 1$, the first Frobenius solution is
|
||||
\[
|
||||
y_1(x) = 1 - \frac{x^2}{2^2} + \frac{x^4}{2^2 \cdot 4^2} - \frac{x^6}{2^2 \cdot 4^2 \cdot 6^2} + \cdots
|
||||
= \sum_{m=0}^{\infty} \frac{(-1)^m}{(m!)^2}\left(\frac{x}{2}\right)^{\!2m}.
|
||||
\]
|
||||
This is the \textbf{Bessel function of the first kind of order~0}, denoted $J_0(x)$:
|
||||
\[
|
||||
J_0(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{(m!)^2}\left(\frac{x}{2}\right)^{\!2m}.
|
||||
\]
|
||||
|
||||
\textit{Second solution.} Since $r_1 = r_2 = 0$ is a repeated root (Case~2), the second linearly independent solution has the form
|
||||
\[
|
||||
y_2(x) = J_0(x)\,\ln(x) + \sum_{n=1}^{\infty} b_n\,x^n.
|
||||
\]
|
||||
This function is denoted $Y_0(x)$ (the \textbf{Bessel function of the second kind of order~0}). The coefficients $b_n$ are determined by substituting $y_2$ into the ODE; the derivation is more involved and omitted here.
|
||||
|
||||
The general solution of Bessel's equation of order~0 is
|
||||
\[
|
||||
y(x) = c_1\,J_0(x) + c_2\,Y_0(x).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{remark}
|
||||
Bessel functions $J_\nu(x)$ and $Y_\nu(x)$ appear ubiquitously in physics and engineering: wave propagation in circular membranes, heat conduction in cylindrical rods, electromagnetic fields in waveguides, and quantum mechanics (schrödinger equation in cylindrical coordinates). The Frobenius method is the primary tool for constructing these special functions analytically.
|
||||
\end{remark}
|
||||
|
||||
\begin{remark}
|
||||
For the Legendre equation discussed in \cref{sec:ch09_power_series}, the points $x = \pm 1$ are regular singular points. The Frobenius method can be applied about those points as well, yielding the associated Legendre functions. When the parameter $\ell$ is a nonnegative integer, the series terminates into the familiar Legendre polynomials $P_n(x)$.
|
||||
\end{remark}
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch09_summary}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Series solution methods: when to use and solution forms}
|
||||
\label{tab:ch09_summary}
|
||||
\begin{tabular}{@{}l l p{4.2cm}@{}}
|
||||
\toprule
|
||||
\textbf{Method} & \textbf{When to use} & \textbf{Solution form} \\
|
||||
\midrule
|
||||
Power series & $p(x)$ and $q(x)$ analytic at $x_0$ &
|
||||
$y(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^n$ \\[12pt]
|
||||
Euler--Cauchy & $x^2 y'' + \alpha x y' + \beta y = 0$ &
|
||||
$y = c_1 x^{r_1} + c_2 x^{r_2}$ (see \cref{sec:ch09_euler_cauchy}) \\[12pt]
|
||||
Frobenius & $x_0$ is a regular singular point; $xp(x)$ and $x^2q(x)$ analytic at $x_0$ &
|
||||
$y(x) = x^r \sum_{n=0}^{\infty} a_n x^n$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Frobenius method: indicial roots and solution structure}
|
||||
\label{tab:ch09_frobenius_cases}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Indicial roots $r_1, r_2$} & \textbf{Solution structure} \\
|
||||
\midrule
|
||||
$r_1 - r_2 \notin \{0, 1, 2, \dots\}$ &
|
||||
Two Frobenius series (no $\ln x$) \\[8pt]
|
||||
$r_1 = r_2$ &
|
||||
One Frobenius series + logarithmic term \\[8pt]
|
||||
$r_1 - r_2 = N \in \{1, 2, 3, \dots\}$ &
|
||||
One Frobenius series; second solution may contain $\ln x$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Euler--Cauchy equation: indicial roots and solution forms}
|
||||
\label{tab:ch09_euler_cases}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Roots of $r(r-1) + \alpha r + \beta = 0$} & \textbf{General solution} \\
|
||||
\midrule
|
||||
Distinct real $r_1 \neq r_2$ &
|
||||
$y(x) = c_1 x^{r_1} + c_2 x^{r_2}$ \\[10pt]
|
||||
Repeated $r$ &
|
||||
$y(x) = x^r\bigl[c_1 + c_2 \ln(x)\bigr]$ \\[10pt]
|
||||
Complex $\lambda \pm i\omega$ &
|
||||
$y(x) = x^{\lambda}\bigl[c_1 \cos(\omega\ln x) + c_2 \sin(\omega\ln x)\bigr]$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Problem-solving workflow for series solutions.}
|
||||
\begin{enumerate}
|
||||
\item \textbf{Classify the point $x_0$.} Are $p(x)$ and $q(x)$ analytic at $x_0$? If yes, use the \textbf{power series method} (\cref{sec:ch09_power_series}).
|
||||
\item If not, check whether $x_0$ is a \textbf{regular singular point}: are $x\,p(x)$ and $x^2\,q(x)$ analytic at $x_0$? If yes, use the \textbf{Frobenius method} (\cref{sec:ch09_frobenius}).
|
||||
\item If the equation has the special form $x^2 y'' + \alpha x y' + \beta y = 0$, use the \textbf{Euler--Cauchy method} (\cref{sec:ch09_euler_cauchy}), which gives closed-form solutions directly.
|
||||
\item After finding the series, check whether it corresponds to a known special function (Bessel, Legendre, etc.) --- this often simplifies further analysis.
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
@@ -1,697 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch10_fourier_series.tex
|
||||
% Chapter 10: Fourier Series
|
||||
% =============================================================================
|
||||
|
||||
\section{Fourier Series}
|
||||
\label{ch:fourier_series}
|
||||
|
||||
Periodic functions appear throughout applied mathematics: from alternating currents in electrical circuits to sound waves in acoustics, from the vibrations of strings to the seasonal forcing in climate models. Fourier series provide the analytical framework for representing an arbitrary periodic function as a superposition of simple sine and cosine waves. This chapter develops the theory and computational machinery of Fourier series, then shows how they serve as a powerful tool for solving linear differential equations with periodic forcing --- a topic that will carry over into the boundary value problems of \cref{ch:boundary_value_problems} and the PDE chapters that follow.
|
||||
|
||||
\subsection{Fourier Coefficients}
|
||||
\label{sec:ch10_fourier_coefficients}
|
||||
|
||||
Let $f(x)$ be a function defined on the interval $[-L, L]$ that is $2L$-periodic, meaning $f(x + 2L) = f(x)$ for all $x$. The \textbf{Fourier series} of $f$ is the trigonometric expansion
|
||||
\begin{equation}
|
||||
\label{eq:fourier_series}
|
||||
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \Bigl[a_n \cos\!\Bigl(\frac{n\pi x}{L}\Bigr) + b_n \sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\Bigr].
|
||||
\end{equation}
|
||||
The constants $a_0$, $a_n$, and $b_n$ are the \textbf{Fourier coefficients}. Their explicit formulas follow from the orthogonality of the sine and cosine basis functions, which we develop in \cref{sec:ch10_orthogonality}. Here we state them as a key result.
|
||||
|
||||
\begin{keyresult}
|
||||
\label{key:fourier_coefficients}
|
||||
\textbf{Fourier coefficients on $[-L, L]$.} For a function $f(x)$ defined on $[-L, L]$, the Fourier coefficients are
|
||||
\begin{align}
|
||||
a_0 &= \frac{1}{L} \int_{-L}^{L} f(x)\,\diff x, \label{eq:a0_formula} \\[6pt]
|
||||
a_n &= \frac{1}{L} \int_{-L}^{L} f(x) \cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x, \quad n \ge 1, \label{eq:an_formula} \\[6pt]
|
||||
b_n &= \frac{1}{L} \int_{-L}^{L} f(x) \sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x, \quad n \ge 1. \label{eq:bn_formula}
|
||||
\end{align}
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Why the $a_0/2$ convention?} Writing the constant term as $a_0/2$ rather than $a_0$ ensures that the formula for $a_0$ matches the pattern for $a_n$ when $n = 0$. Indeed, $\cos(0) = 1$, so $a_0$ computed by \cref{eq:an_formula} with $n=0$ gives $(1/L)\int_{-L}^L f(x)\,\diff x$, which is exactly \cref{eq:a0_formula}. The factor $1/2$ in \cref{eq:fourier_series} then makes the average value of the series equal to $a_0/2$.
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
Compute the Fourier series of the square wave
|
||||
\[
|
||||
f(x) = \begin{cases}
|
||||
1, & 0 < x < \pi, \\
|
||||
-1, & -\pi < x < 0,
|
||||
\end{cases}
|
||||
\]
|
||||
extended as a $2\pi$-periodic function.
|
||||
|
||||
\textbf{Solution.} Here $L = \pi$. We compute the coefficients.
|
||||
|
||||
\textit{Coefficient $a_0$:}
|
||||
\[
|
||||
a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\,\diff x
|
||||
= \frac{1}{\pi}\left(\int_{-\pi}^0 (-1)\,\diff x + \int_0^{\pi} 1\,\diff x\right)
|
||||
= \frac{1}{\pi}\bigl(-\pi + \pi\bigr) = 0.
|
||||
\]
|
||||
|
||||
\textit{Coefficients $a_n$:}
|
||||
\[
|
||||
a_n = \frac{1}{\pi}\left(\int_{-\pi}^0 (-1)\cos(nx)\,\diff x + \int_0^{\pi} 1\cdot\cos(nx)\,\diff x\right).
|
||||
\]
|
||||
Since $f(x)$ is an odd function (symmetric about the origin), the product $f(x)\cos(nx)$ is odd $\times$ even = odd. The integral of an odd function over a symmetric interval is zero, so $a_n = 0$ for all $n \ge 1$.
|
||||
|
||||
\textit{Coefficients $b_n$:}
|
||||
\[
|
||||
b_n = \frac{1}{\pi}\left(\int_{-\pi}^0 (-1)\sin(nx)\,\diff x + \int_0^{\pi} 1\cdot\sin(nx)\,\diff x\right).
|
||||
\]
|
||||
Now $f(x)\sin(nx)$ is odd $\times$ odd = even, so we can simplify:
|
||||
\[
|
||||
b_n = \frac{2}{\pi} \int_0^{\pi} \sin(nx)\,\diff x
|
||||
= \frac{2}{\pi} \left[-\frac{1}{n}\cos(nx)\right]_0^{\pi}
|
||||
= \frac{2}{n\pi}\bigl(1 - \cos(n\pi)\bigr).
|
||||
\]
|
||||
Since $\cos(n\pi) = (-1)^n$, we have $1 - (-1)^n = \begin{cases} 2 & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$. Therefore:
|
||||
\[
|
||||
b_n = \begin{cases}
|
||||
\dfrac{4}{n\pi}, & n \text{ odd}, \\[8pt]
|
||||
0, & n \text{ even}.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
The Fourier series is
|
||||
\[
|
||||
f(x) = \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{\sin\bigl((2k+1)x\bigr)}{2k+1}
|
||||
= \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Compute the Fourier series of the sawtooth wave $f(x) = x$ on $(-\pi, \pi)$, extended as a $2\pi$-periodic function.
|
||||
|
||||
\textbf{Solution.} Here $L = \pi$. The function is odd, so $a_0 = 0$ and $a_n = 0$ for all $n$.
|
||||
|
||||
\textit{Coefficients $b_n$:}
|
||||
\[
|
||||
b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx)\,\diff x.
|
||||
\]
|
||||
The integrand is odd $\times$ odd = even, so
|
||||
\[
|
||||
b_n = \frac{2}{\pi} \int_0^{\pi} x \sin(nx)\,\diff x.
|
||||
\]
|
||||
Integrate by parts with $u = x$, $\diff v = \sin(nx)\,\diff x$:
|
||||
\[
|
||||
\int_0^{\pi} x \sin(nx)\,\diff x
|
||||
= \Bigl[-\frac{x}{n}\cos(nx)\Bigr]_0^{\pi} + \frac{1}{n}\int_0^{\pi} \cos(nx)\,\diff x.
|
||||
\]
|
||||
The boundary term gives $-\dfrac{\pi}{n}\cos(n\pi) + 0 = -\dfrac{\pi}{n}(-1)^n = \dfrac{\pi}{n}(-1)^{n+1}$. The remaining integral vanishes:
|
||||
\[
|
||||
\frac{1}{n}\int_0^{\pi} \cos(nx)\,\diff x = \frac{1}{n}\Bigl[\frac{1}{n}\sin(nx)\Bigr]_0^{\pi} = 0.
|
||||
\]
|
||||
Therefore:
|
||||
\[
|
||||
b_n = \frac{2}{\pi} \cdot \frac{\pi}{n}(-1)^{n+1} = \frac{2}{n}(-1)^{n+1}.
|
||||
\]
|
||||
The Fourier series is
|
||||
\[
|
||||
f(x) = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx)
|
||||
= 2\left(\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots\right).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Orthogonality}
|
||||
\label{sec:ch10_orthogonality}
|
||||
|
||||
The formulas for the Fourier coefficients \cref{eq:a0_formula}--\cref{eq:bn_formula} are derived from the \textbf{orthogonality} of the trigonometric basis functions. Two functions $\phi(x)$ and $\psi(x)$ are \textbf{orthogonal} on $[-L, L]$ if their inner product vanishes:
|
||||
\[
|
||||
\int_{-L}^{L} \phi(x)\,\psi(x)\,\diff x = 0.
|
||||
\]
|
||||
|
||||
\begin{theorem}[Orthogonality of Trigonometric Functions]
|
||||
\label{thm:orthogonality}
|
||||
For positive integers $n, m$:
|
||||
\begin{enumerate}
|
||||
\item[\textbf{(i)}] $\displaystyle \int_{-L}^{L} \cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\cos\!\Bigl(\frac{m\pi x}{L}\Bigr)\,\diff x = \begin{cases} 0, & n \neq m, \\ L, & n = m \ge 1, \\ 2L, & n = m = 0. \end{cases}$
|
||||
|
||||
\item[\textbf{(ii)}] $\displaystyle \int_{-L}^{L} \sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\sin\!\Bigl(\frac{m\pi x}{L}\Bigr)\,\diff x = \begin{cases} 0, & n \neq m, \\ L, & n = m. \end{cases}$
|
||||
|
||||
\item[\textbf{(iii)}] $\displaystyle \int_{-L}^{L} \sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\cos\!\Bigl(\frac{m\pi x}{L}\Bigr)\,\diff x = 0 \quad \text{for all } n, m \ge 0.$
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\paragraph{Proof of (i).} Use the product-to-sum identity:
|
||||
\[
|
||||
\cos A \cos B = \frac{1}{2}\bigl[\cos(A-B) + \cos(A+B)\bigr].
|
||||
\]
|
||||
Let $A = \dfrac{n\pi x}{L}$ and $B = \dfrac{m\pi x}{L}$. Then:
|
||||
\[
|
||||
\int_{-L}^{L} \cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\cos\!\Bigl(\frac{m\pi x}{L}\Bigr)\,\diff x
|
||||
= \frac{1}{2}\int_{-L}^{L} \cos\!\Bigl(\frac{(n-m)\pi x}{L}\Bigr)\,\diff x
|
||||
+ \frac{1}{2}\int_{-L}^{L} \cos\!\Bigl(\frac{(n+m)\pi x}{L}\Bigr)\,\diff x.
|
||||
\]
|
||||
|
||||
If $n \neq m$, then $n-m$ is a nonzero integer, and
|
||||
\[
|
||||
\int_{-L}^{L} \cos\!\Bigl(\frac{(n-m)\pi x}{L}\Bigr)\,\diff x
|
||||
= \Bigl[\frac{L}{(n-m)\pi}\sin\!\Bigl(\frac{(n-m)\pi x}{L}\Bigr)\Bigr]_{-L}^{L}
|
||||
= \frac{L}{(n-m)\pi}\bigl[\sin((n-m)\pi) - \sin(-(n-m)\pi)\bigr] = 0,
|
||||
\]
|
||||
since $\sin(k\pi) = 0$ for any integer $k$. Similarly, the second integral vanishes because $n+m$ is a nonzero integer.
|
||||
|
||||
If $n = m \ge 1$, then $n-m = 0$ and $\cos(0) = 1$:
|
||||
\[
|
||||
\frac{1}{2}\int_{-L}^{L} 1\,\diff x + \frac{1}{2}\int_{-L}^{L} \cos\!\Bigl(\frac{2n\pi x}{L}\Bigr)\,\diff x
|
||||
= \frac{1}{2}(2L) + 0 = L.
|
||||
\]
|
||||
|
||||
If $n = m = 0$, then $\cos(0)\cos(0) = 1$, and the integral is $2L$.
|
||||
|
||||
\paragraph{Proof of (ii).} Use $\sin A \sin B = \dfrac{1}{2}\bigl[\cos(A-B) - \cos(A+B)\bigr]$. The same analysis as above applies: both integrals vanish for $n \neq m$. For $n = m$:
|
||||
\[
|
||||
\frac{1}{2}\int_{-L}^{L} 1\,\diff x - \frac{1}{2}\int_{-L}^{L} \cos\!\Bigl(\frac{2n\pi x}{L}\Bigr)\,\diff x = L - 0 = L.
|
||||
\]
|
||||
|
||||
\paragraph{Proof of (iii).} Use $\sin A \cos B = \dfrac{1}{2}\bigl[\sin(A+B) + \sin(A-B)\bigr]$. Both terms integrate to zero since the sine of a multiple of $\pi$ is zero at the endpoints.
|
||||
|
||||
\paragraph{How orthogonality yields the coefficients.} Multiply both sides of \cref{eq:fourier_series} by $\cos(m\pi x/L)$ and integrate over $[-L, L]$. By orthogonality, every term in the infinite sum vanishes except the one with $n = m$:
|
||||
\[
|
||||
\int_{-L}^{L} f(x)\cos\!\Bigl(\frac{m\pi x}{L}\Bigr)\,\diff x
|
||||
= a_m \int_{-L}^{L} \cos^2\!\Bigl(\frac{m\pi x}{L}\Bigr)\,\diff x
|
||||
= a_m \cdot L,
|
||||
\]
|
||||
which immediately gives $a_m = \dfrac{1}{L}\int_{-L}^{L} f(x)\cos(m\pi x/L)\,\diff x$. The derivation for $b_n$ is identical, using multiplication by $\sin(m\pi x/L)$ instead. Multiplying by $1$ and integrating gives $a_0$.
|
||||
|
||||
\subsection{Even and Odd Functions}
|
||||
\label{sec:ch10_even_odd}
|
||||
|
||||
When $f(x)$ possesses symmetry, the Fourier series simplifies dramatically. A function is \textbf{even} if $f(-x) = f(x)$, and \textbf{odd} if $f(-x) = -f(x)$.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Fourier series of even and odd functions.}
|
||||
\begin{itemize}
|
||||
\item If $f(x)$ is \textbf{even} on $[-L, L]$, then $b_n = 0$ for all $n$, and
|
||||
\[
|
||||
a_0 = \frac{2}{L} \int_0^L f(x)\,\diff x,
|
||||
\qquad
|
||||
a_n = \frac{2}{L} \int_0^L f(x)\cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x.
|
||||
\]
|
||||
The series contains only cosine terms (a \textbf{cosine series}).
|
||||
|
||||
\item If $f(x)$ is \textbf{odd} on $[-L, L]$, then $a_0 = 0$ and $a_n = 0$ for all $n$, and
|
||||
\[
|
||||
b_n = \frac{2}{L} \int_0^L f(x)\sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x.
|
||||
\]
|
||||
The series contains only sine terms (a \textbf{sine series}).
|
||||
\end{itemize}
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Intuition.} The product of an even function with a sine (odd) is odd, and the integral of an odd function over $[-L, L]$ is zero. Hence an even function has no sine terms. Conversely, the product of an odd function with a cosine (even) is odd, so an odd function has no cosine terms.
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
Find the Fourier series of $f(x) = |x|$ on $[-\pi, \pi]$.
|
||||
|
||||
\textbf{Solution.} Since $|-x| = |x|$, the function is even. We need only compute the cosine coefficients. With $L = \pi$:
|
||||
\[
|
||||
a_0 = \frac{2}{\pi}\int_0^{\pi} x\,\diff x = \frac{2}{\pi}\cdot\frac{\pi^2}{2} = \pi.
|
||||
\]
|
||||
For $n \ge 1$:
|
||||
\[
|
||||
a_n = \frac{2}{\pi}\int_0^{\pi} x\cos(nx)\,\diff x.
|
||||
\]
|
||||
Integrate by parts with $u = x$, $\diff v = \cos(nx)\,\diff x$:
|
||||
\[
|
||||
\int_0^{\pi} x\cos(nx)\,\diff x
|
||||
= \Bigl[\frac{x}{n}\sin(nx)\Bigr]_0^{\pi} - \frac{1}{n}\int_0^{\pi} \sin(nx)\,\diff x.
|
||||
\]
|
||||
The boundary term vanishes (since $\sin(n\pi) = 0$), and the remaining integral is
|
||||
\[
|
||||
-\frac{1}{n}\Bigl[-\frac{1}{n}\cos(nx)\Bigr]_0^{\pi}
|
||||
= \frac{1}{n^2}\bigl((-1)^n - 1\bigr).
|
||||
\]
|
||||
Therefore:
|
||||
\[
|
||||
a_n = \frac{2}{\pi}\cdot\frac{1}{n^2}\bigl((-1)^n - 1\bigr)
|
||||
= \begin{cases}
|
||||
0, & n \text{ even}, \\[6pt]
|
||||
-\dfrac{4}{\pi n^2}, & n \text{ odd}.
|
||||
\end{cases}
|
||||
\]
|
||||
The Fourier series is
|
||||
\[
|
||||
|x| = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\cos\bigl((2k+1)x\bigr)}{(2k+1)^2}
|
||||
= \frac{\pi}{2} - \frac{4}{\pi}\left(\cos x + \frac{\cos 3x}{9} + \frac{\cos 5x}{25} + \cdots\right).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Find the Fourier series of $f(x) = x^2$ on $[-\pi, \pi]$.
|
||||
|
||||
\textbf{Solution.} The function is even, so $b_n = 0$. With $L = \pi$:
|
||||
\[
|
||||
a_0 = \frac{2}{\pi}\int_0^{\pi} x^2\,\diff x = \frac{2}{\pi}\cdot\frac{\pi^3}{3} = \frac{2\pi^2}{3}.
|
||||
\]
|
||||
For $n \ge 1$:
|
||||
\[
|
||||
a_n = \frac{2}{\pi}\int_0^{\pi} x^2\cos(nx)\,\diff x.
|
||||
\]
|
||||
Integrate by parts twice. First, $u = x^2$, $\diff v = \cos(nx)\,\diff x$:
|
||||
\[
|
||||
\int_0^{\pi} x^2\cos(nx)\,\diff x
|
||||
= \Bigl[\frac{x^2}{n}\sin(nx)\Bigr]_0^{\pi} - \frac{2}{n}\int_0^{\pi} x\sin(nx)\,\diff x.
|
||||
\]
|
||||
The boundary term vanishes. From \cref{sec:ch10_fourier_coefficients} (sawtooth example), $\int_0^{\pi} x\sin(nx)\,\diff x = \dfrac{\pi}{n}(-1)^{n+1}$. So:
|
||||
\[
|
||||
\int_0^{\pi} x^2\cos(nx)\,\diff x = -\frac{2}{n}\cdot\frac{\pi}{n}(-1)^{n+1}
|
||||
= \frac{2\pi}{n^2}(-1)^n.
|
||||
\]
|
||||
Therefore:
|
||||
\[
|
||||
a_n = \frac{2}{\pi}\cdot\frac{2\pi}{n^2}(-1)^n = \frac{4(-1)^n}{n^2}.
|
||||
\]
|
||||
The Fourier series is
|
||||
\[
|
||||
x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\cos(nx)
|
||||
= \frac{\pi^2}{3} - 4\left(\frac{\cos x}{1^2} - \frac{\cos 2x}{2^2} + \frac{\cos 3x}{3^2} - \cdots\right).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Half-Range Expansions}
|
||||
\label{sec:ch10_half_range}
|
||||
|
||||
In practice, a function $f(x)$ is often given only on the half-interval $[0, L]$. We can construct a Fourier series by \textbf{extending} $f$ to $[-L, L]$ as either an even function (yielding a cosine series) or an odd function (yielding a sine series). These are called \textbf{half-range expansions}.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Half-range expansions on $[0, L]$.}
|
||||
\begin{itemize}
|
||||
\item \textbf{Cosine series (even extension).} Extend $f$ to $[-L, L]$ by defining $f(-x) = f(x)$. The Fourier series contains only cosine terms:
|
||||
\[
|
||||
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\!\Bigl(\frac{n\pi x}{L}\Bigr),
|
||||
\]
|
||||
with $a_n = \dfrac{2}{L} \int_0^L f(x)\cos\!\Bigl(\dfrac{n\pi x}{L}\Bigr)\,\diff x$ for $n \ge 0$.
|
||||
|
||||
\item \textbf{Sine series (odd extension).} Extend $f$ to $[-L, L]$ by defining $f(-x) = -f(x)$. The Fourier series contains only sine terms:
|
||||
\[
|
||||
f(x) = \sum_{n=1}^{\infty} b_n \sin\!\Bigl(\frac{n\pi x}{L}\Bigr),
|
||||
\]
|
||||
with $b_n = \dfrac{2}{L} \int_0^L f(x)\sin\!\Bigl(\dfrac{n\pi x}{L}\Bigr)\,\diff x$.
|
||||
\end{itemize}
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
Find the half-range cosine series for $f(x) = x$ on $[0, \pi]$.
|
||||
|
||||
\textbf{Solution.} We extend $f(x) = x$ on $[0, \pi]$ to an even function on $[-\pi, \pi]$. With $L = \pi$:
|
||||
\[
|
||||
a_0 = \frac{2}{\pi}\int_0^{\pi} x\,\diff x = \frac{2}{\pi}\cdot\frac{\pi^2}{2} = \pi.
|
||||
\]
|
||||
For $n \ge 1$:
|
||||
\[
|
||||
a_n = \frac{2}{\pi}\int_0^{\pi} x\cos(nx)\,\diff x.
|
||||
\]
|
||||
This is the same integral as in \cref{sec:ch10_even_odd}:
|
||||
\[
|
||||
a_n = \frac{2}{\pi}\cdot\frac{1}{n^2}\bigl((-1)^n - 1\bigr)
|
||||
= \begin{cases}
|
||||
0, & n \text{ even}, \\[6pt]
|
||||
-\dfrac{4}{\pi n^2}, & n \text{ odd}.
|
||||
\end{cases}
|
||||
\]
|
||||
The half-range cosine series is
|
||||
\[
|
||||
x = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\cos\bigl((2k+1)x\bigr)}{(2k+1)^2}
|
||||
\qquad \text{for } x \in [0, \pi].
|
||||
\]
|
||||
Note: this series actually represents the \emph{even} extension $|x|$ on $[-\pi, \pi]$. On $[0, \pi]$, $|x| = x$, so the equality holds.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Find the half-range sine series for $f(x) = 1$ on $[0, \pi]$.
|
||||
|
||||
\textbf{Solution.} We extend $f(x) = 1$ on $[0, \pi]$ to an odd function on $[-\pi, \pi]$. With $L = \pi$:
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\int_0^{\pi} 1\cdot\sin(nx)\,\diff x
|
||||
= \frac{2}{\pi}\Bigl[-\frac{1}{n}\cos(nx)\Bigr]_0^{\pi}
|
||||
= \frac{2}{n\pi}\bigl(1 - (-1)^n\bigr).
|
||||
\]
|
||||
This is nonzero only for odd $n$:
|
||||
\[
|
||||
b_n = \begin{cases}
|
||||
\dfrac{4}{n\pi}, & n \text{ odd}, \\[8pt]
|
||||
0, & n \text{ even}.
|
||||
\end{cases}
|
||||
\]
|
||||
The half-range sine series is
|
||||
\[
|
||||
1 = \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin\bigl((2k+1)x\bigr)}{2k+1}
|
||||
= \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right).
|
||||
\]
|
||||
This series converges to $1$ on $(0, \pi)$ but converges to $0$ at $x = 0$ and $x = \pi$ (the endpoints where the odd extension has jump discontinuities). This behavior is explained by the Dirichlet convergence theorem (\cref{sec:ch10_convergence_gibbs}).
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Complex Fourier Series}
|
||||
\label{sec:ch10_complex_fourier}
|
||||
|
||||
The real Fourier series \cref{eq:fourier_series} can be written more compactly using Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$. The \textbf{complex Fourier series} is
|
||||
\begin{equation}
|
||||
\label{eq:complex_fourier}
|
||||
f(x) = \sum_{n=-\infty}^{\infty} c_n\,e^{i n\pi x / L},
|
||||
\end{equation}
|
||||
where the \textbf{complex Fourier coefficients} are
|
||||
\begin{equation}
|
||||
\label{eq:complex_coefficients}
|
||||
c_n = \frac{1}{2L}\int_{-L}^{L} f(x)\,e^{-i n\pi x / L}\,\diff x.
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Relationship between real and complex coefficients.}
|
||||
\begin{align}
|
||||
c_0 &= \frac{a_0}{2}, \\[6pt]
|
||||
c_n &= \frac{a_n - i b_n}{2} \quad \text{for } n > 0, \\[6pt]
|
||||
c_{-n} &= \frac{a_n + i b_n}{2} \quad \text{for } n > 0.
|
||||
\end{align}
|
||||
Equivalently, $a_n = c_n + c_{-n}$ and $b_n = i(c_n - c_{-n})$.
|
||||
|
||||
If $f(x)$ is real-valued, then $c_{-n} = \overline{c_n}$ (complex conjugate).
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Derivation.} Substitute $e^{\pm i n\pi x / L} = \cos(n\pi x / L) \pm i\sin(n\pi x / L)$ into \cref{eq:complex_fourier}, collect real and imaginary parts, and compare with \cref{eq:fourier_series}. The reverse direction uses $a_n = c_n + c_{-n}$ and $b_n = i(c_n - c_{-n})$.
|
||||
|
||||
\paragraph{Worked example.}
|
||||
|
||||
\begin{workedexample}
|
||||
Find the complex Fourier series of $f(x) = e^{ax}$ on $[-\pi, \pi]$, where $a \in \R$.
|
||||
|
||||
\textbf{Solution.} With $L = \pi$:
|
||||
\[
|
||||
c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{ax}\,e^{-i n x}\,\diff x
|
||||
= \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{(a - in)x}\,\diff x.
|
||||
\]
|
||||
For $n \neq 0$ (or $a \neq 0$):
|
||||
\[
|
||||
c_n = \frac{1}{2\pi}\Bigl[\frac{e^{(a-in)x}}{a - in}\Bigr]_{-\pi}^{\pi}
|
||||
= \frac{1}{2\pi(a - in)}\bigl(e^{(a-in)\pi} - e^{-(a-in)\pi}\bigr).
|
||||
\]
|
||||
Since $e^{-in\pi} = (-1)^n$ and $e^{in\pi} = (-1)^n$:
|
||||
\[
|
||||
c_n = \frac{(-1)^n}{2\pi(a - in)}\bigl(e^{a\pi} - e^{-a\pi}\bigr)
|
||||
= \frac{(-1)^n \sinh(a\pi)}{\pi(a - in)}.
|
||||
\]
|
||||
For $n = 0$:
|
||||
\[
|
||||
c_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{ax}\,\diff x
|
||||
= \frac{e^{a\pi} - e^{-a\pi}}{2\pi a}
|
||||
= \frac{\sinh(a\pi)}{\pi a}.
|
||||
\]
|
||||
The complex Fourier series is
|
||||
\[
|
||||
e^{ax} = \frac{\sinh(a\pi)}{\pi}\left(\frac{1}{a} + \sum_{\substack{n=-\infty \\ n \neq 0}}^{\infty} \frac{(-1)^n}{a - in}\,e^{inx}\right).
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Convergence and Gibbs Phenomenon}
|
||||
\label{sec:ch10_convergence_gibbs}
|
||||
|
||||
\begin{theorem}[Dirichlet Convergence Theorem]
|
||||
\label{thm:dirichlet_convergence}
|
||||
Let $f(x)$ be $2L$-periodic and piecewise smooth on $[-L, L]$ (i.e., $f$ and $f'$ are piecewise continuous). Then the Fourier series of $f$ converges at every point $x$ to
|
||||
\[
|
||||
S(x) = \frac{f(x^+) + f(x^-)}{2},
|
||||
\]
|
||||
where $f(x^+) = \lim_{\epsilon \to 0^+} f(x+\epsilon)$ and $f(x^-) = \lim_{\epsilon \to 0^+} f(x-\epsilon)$ are the right-hand and left-hand limits. In particular, at any point where $f$ is continuous, the series converges to $f(x)$.
|
||||
\end{theorem}
|
||||
|
||||
\paragraph{Implications.} If $f(x)$ is continuous on $[-L, L]$, its Fourier series converges pointwise to $f(x)$ everywhere. At a jump discontinuity where the function jumps from $f(x^-)$ to $f(x^+)$, the series converges to the \emph{average} of the two values.
|
||||
|
||||
\paragraph{Gibbs phenomenon.} Near a jump discontinuity of size $J = |f(x^+) - f(x^-)|$, the partial sums of the Fourier series exhibit a persistent overshoot: even as the number of terms $N \to \infty$, the overshoot approaches approximately
|
||||
\[
|
||||
0.08949 \cdot J \quad (\text{about } 8.95\% \text{ of the jump}).
|
||||
\]
|
||||
This is \textbf{not} a numerical artifact; it is an inherent feature of Fourier series at discontinuities. The overshoot does \emph{not} disappear as $N$ increases --- it merely becomes more concentrated near the discontinuity.
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=1.1]
|
||||
% Axes
|
||||
\draw[->] (-1,0) -- (7,0) node[right] {$x$};
|
||||
\draw[->] (0,-1.2) -- (0,2.5) node[above] {$S_N(x)$};
|
||||
|
||||
% Square wave (ideal): from 0 to 3 it is 1, from 3 to 6 it is -1
|
||||
\draw[thick, red!80, dashed] (0,1) -- (3,1);
|
||||
\draw[thick, red!80, dashed] (3,1) -- (3,-1);
|
||||
\draw[thick, red!80, dashed] (3,-1) -- (6,-1);
|
||||
|
||||
% Partial sum with Gibbs overshoot
|
||||
\draw[thick, blue!80] (0,0)
|
||||
to [out=30, in=180] (1.5, 0.6)
|
||||
to [out=0, in=180] (2.3, 0.95)
|
||||
to [out=35, in=145] (2.6, 1.15)
|
||||
to [out=0, in=180] (2.85, 0.92)
|
||||
to [out=30, in=180] (2.97, 0.7)
|
||||
node[pos=0.05, above] {$S_N(x)$}
|
||||
to [out=-50, in=180] (3.03, -0.7)
|
||||
to [out=0, in=140] (3.15, -0.92)
|
||||
to [out=0, in=180] (3.4, -1.15)
|
||||
to [out=0, in=180] (3.7, -0.95)
|
||||
to [out=0, in=180] (4.5, -0.6)
|
||||
to [out=0, in=180] (6, -0.35);
|
||||
|
||||
% Mark the overshoot
|
||||
\draw[thick, dashed, magenta] (2.6,1.15) -- (2.6,1) node[above, magenta, font=\small] {overshoot};
|
||||
\draw[<->, magenta, thick] (2.6,1) -- (2.6,1.15);
|
||||
\draw[<->, magenta, thick] (3.4,-1) -- (3.4,-1.15);
|
||||
\node[magenta, font=\small, anchor=west] at (3.4,-1.15) {$\approx 8.95\%$};
|
||||
|
||||
% Mark the discontinuity at x=3
|
||||
\draw[dotted, gray] (3,-1.5) -- (3,1.5);
|
||||
\node[font=\small] at (3,-1.5) [anchor=north] {jump};
|
||||
|
||||
% Labels
|
||||
\node[red!80, font=\small] at (6.5,1) {$f(x)$ (square wave)};
|
||||
\node[blue!80, font=\small] at (6.5,1.5) {$S_N(x)$ (partial sum)};
|
||||
|
||||
% Arrow to overshoot annotation
|
||||
\draw[->, magenta] (2.6,1.15) -- ++(0.5, 0.4) node[above, magenta, font=\footnotesize] {Gibbs overshoot};
|
||||
\end{tikzpicture}
|
||||
\caption{Gibbs phenomenon: the partial sum $S_N(x)$ of a Fourier series overshoots the function value at a jump discontinuity by approximately $8.95\%$ of the jump size, even as $N \to \infty$. The dashed red line is the ideal square wave $f(x)$; the solid blue curve is the partial sum.}
|
||||
\label{fig:gibbs_phenomenon}
|
||||
\end{figure}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{The Gibbs constant.} The exact overshoot is $\displaystyle \frac{2}{\pi}\int_0^{\pi}\frac{\sin t}{t}\,\diff t - 1 \approx 0.08949$ times the jump size. The integral $\displaystyle \int_0^{\pi} \dfrac{\sin t}{t}\,\diff t$ is related to the \emph{sinc integral} $\operatorname{Si}(\pi) \approx 1.85194$.
|
||||
\end{hintbox}
|
||||
|
||||
\subsection{Parseval's Identity}
|
||||
\label{sec:ch10_parseval}
|
||||
|
||||
\begin{theorem}[Parseval's Identity]
|
||||
\label{thm:parseval}
|
||||
Let $f(x)$ have the Fourier series
|
||||
\[
|
||||
f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty}\Bigl[a_n\cos\!\Bigl(\frac{n\pi x}{L}\Bigr) + b_n\sin\!\Bigl(\frac{n\pi x}{L}\Bigr)\Bigr].
|
||||
\]
|
||||
Then
|
||||
\begin{equation}
|
||||
\label{eq:parseval}
|
||||
\frac{1}{L}\int_{-L}^{L} \bigl[f(x)\bigr]^2\,\diff x
|
||||
= \frac{a_0^2}{2} + \sum_{n=1}^{\infty} \bigl(a_n^2 + b_n^2\bigr).
|
||||
\end{equation}
|
||||
In words: the average energy (mean square) of the function equals the sum of the squared magnitudes of its Fourier coefficients.
|
||||
\end{theorem}
|
||||
|
||||
\paragraph{Proof sketch.} Square the Fourier series and integrate over $[-L, L]$. By orthogonality (\cref{thm:orthogonality}), all cross-terms vanish:
|
||||
\[
|
||||
\int_{-L}^{L} \cos\!\Bigl(\frac{n\pi x}{L}\Bigr)\cos\!\Bigl(\frac{m\pi x}{L}\Bigr)\,\diff x = 0 \text{ for } n \neq m,
|
||||
\]
|
||||
and similarly for the sine--sine and sine--cosine products. The surviving terms are:
|
||||
\[
|
||||
\int_{-L}^{L} \frac{a_0^2}{4}\,\diff x = \frac{a_0^2}{4}\cdot 2L = \frac{a_0^2}{2}\cdot L,
|
||||
\]
|
||||
\[
|
||||
\int_{-L}^{L} a_n^2\cos^2\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x = a_n^2\cdot L,
|
||||
\qquad
|
||||
\int_{-L}^{L} b_n^2\sin^2\!\Bigl(\frac{n\pi x}{L}\Bigr)\,\diff x = b_n^2\cdot L.
|
||||
\]
|
||||
Dividing the total by $L$ gives \cref{eq:parseval}.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Parseval's identity in the complex form.} For the complex Fourier coefficients $c_n$ of \cref{eq:complex_coefficients}:
|
||||
\[
|
||||
\frac{1}{2L}\int_{-L}^{L} |f(x)|^2\,\diff x = \sum_{n=-\infty}^{\infty} |c_n|^2.
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Application: computing infinite series.}
|
||||
|
||||
\begin{workedexample}
|
||||
Use Parseval's identity to evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4}$.
|
||||
|
||||
\textbf{Solution.} From \cref{sec:ch10_even_odd}, the Fourier series of $x^2$ on $[-\pi, \pi]$ is
|
||||
\[
|
||||
x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\cos(nx).
|
||||
\]
|
||||
Here $a_0 = \dfrac{2\pi^2}{3}$ and $a_n = \dfrac{4(-1)^n}{n^2}$, with $b_n = 0$. Parseval's identity (\cref{eq:parseval}) with $L = \pi$ gives:
|
||||
\[
|
||||
\frac{1}{\pi}\int_{-\pi}^{\pi} x^4\,\diff x = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} a_n^2.
|
||||
\]
|
||||
The left side:
|
||||
\[
|
||||
\frac{1}{\pi}\int_{-\pi}^{\pi} x^4\,\diff x = \frac{2}{\pi}\int_0^{\pi} x^4\,\diff x = \frac{2}{\pi}\cdot\frac{\pi^5}{5} = \frac{2\pi^4}{5}.
|
||||
\]
|
||||
The right side:
|
||||
\[
|
||||
\frac{a_0^2}{2} = \frac{1}{2}\left(\frac{2\pi^2}{3}\right)^2 = \frac{2\pi^4}{9},
|
||||
\]
|
||||
\[
|
||||
\sum_{n=1}^{\infty} a_n^2 = \sum_{n=1}^{\infty} \frac{16}{n^4} = 16\sum_{n=1}^{\infty} \frac{1}{n^4}.
|
||||
\]
|
||||
Equating both sides:
|
||||
\[
|
||||
\frac{2\pi^4}{5} = \frac{2\pi^4}{9} + 16\sum_{n=1}^{\infty} \frac{1}{n^4}.
|
||||
\]
|
||||
Solving:
|
||||
\[
|
||||
16\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{2\pi^4}{5} - \frac{2\pi^4}{9}
|
||||
= 2\pi^4\left(\frac{1}{5} - \frac{1}{9}\right)
|
||||
= 2\pi^4\cdot\frac{4}{45}
|
||||
= \frac{8\pi^4}{45}.
|
||||
\]
|
||||
Therefore:
|
||||
\[
|
||||
\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}.
|
||||
\]
|
||||
This is the famous result $\zeta(4) = \dfrac{\pi^4}{90}$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Applications to ODEs}
|
||||
\label{sec:ch10_applications_odes}
|
||||
|
||||
Fourier series provide a powerful method for solving linear ODEs with \textbf{periodic forcing}. The basic idea is to expand the periodic forcing function as a Fourier series, then solve the ODE for each harmonic component --- exploiting the superposition principle for linear equations.
|
||||
|
||||
\paragraph{General method.} Consider the forced harmonic oscillator
|
||||
\begin{equation}
|
||||
\label{eq:forced_oscillator}
|
||||
x'' + \omega_0^2 x = f(t),
|
||||
\end{equation}
|
||||
where $f(t)$ is $2\pi$-periodic. Expand $f(t)$ as a Fourier series:
|
||||
\[
|
||||
f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\bigl[a_n\cos(nt) + b_n\sin(nt)\bigr].
|
||||
\]
|
||||
By linearity, we can solve the ODE for each term separately:
|
||||
\begin{enumerate}
|
||||
\item For the constant term $a_0/2$: the particular solution is $x_p = \dfrac{a_0}{2\omega_0^2}$.
|
||||
\item For each $\cos(nt)$ term: solve $x'' + \omega_0^2 x = a_n\cos(nt)$. If $n \neq \omega_0$, the particular solution is
|
||||
\[
|
||||
x_n^{(c)}(t) = \frac{a_n}{\omega_0^2 - n^2}\cos(nt).
|
||||
\]
|
||||
\item For each $\sin(nt)$ term: solve $x'' + \omega_0^2 x = b_n\sin(nt)$. If $n \neq \omega_0$,
|
||||
\[
|
||||
x_n^{(s)}(t) = \frac{b_n}{\omega_0^2 - n^2}\sin(nt).
|
||||
\]
|
||||
\end{enumerate}
|
||||
The full particular solution is the sum of all these mode-by-mode solutions.
|
||||
|
||||
\begin{remark}
|
||||
\textbf{Resonance warning.} If $n = \omega_0$ for some harmonic present in the forcing, the denominator $\omega_0^2 - n^2$ vanishes and the method breaks down. This is the phenomenon of \textbf{resonance}: the forced response grows without bound (secular growth). In this case, the particular solution takes the form
|
||||
\[
|
||||
x_p(t) = \frac{a_n}{2\omega_0} t \sin(\omega_0 t) \quad \text{or} \quad \frac{b_n}{2\omega_0} t \cos(\omega_0 t),
|
||||
\]
|
||||
producing a linearly growing amplitude. This is consistent with the theory of undetermined coefficients developed in \cref{ch:second_order_nonhomogeneous}.
|
||||
\end{remark}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the initial value problem
|
||||
\[
|
||||
x'' + 4x = f(t), \qquad x(0) = 0, \quad x'(0) = 0,
|
||||
\]
|
||||
where $f(t)$ is the $2\pi$-periodic square wave:
|
||||
\[
|
||||
f(t) = \begin{cases}
|
||||
1, & 0 < t < \pi, \\
|
||||
-1, & \pi < t < 2\pi.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
\textbf{Solution.} From the first example in \cref{sec:ch10_fourier_coefficients}, the Fourier series of this square wave is
|
||||
\[
|
||||
f(t) = \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin\bigl((2k+1)t\bigr)}{2k+1}.
|
||||
\]
|
||||
Let $n_k = 2k+1$. The ODE for the $k$-th mode is
|
||||
\[
|
||||
x_k'' + 4x_k = \frac{4}{\pi n_k}\sin(n_k t).
|
||||
\]
|
||||
Since $\omega_0^2 = 4$, we have $\omega_0 = 2$. Check for resonance: $n_k = 1, 3, 5, \ldots$, and $n_k \neq 2$ for any $k$, so no resonance occurs. The particular solution for the $k$-th mode is
|
||||
\[
|
||||
x_k(t) = \frac{4/(\pi n_k)}{4 - n_k^2}\sin(n_k t)
|
||||
= \frac{4}{\pi n_k(4 - n_k^2)}\sin(n_k t).
|
||||
\]
|
||||
The complete particular solution is the sum:
|
||||
\[
|
||||
x_p(t) = \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin\bigl((2k+1)t\bigr)}{(2k+1)\bigl(4 - (2k+1)^2\bigr)}.
|
||||
\]
|
||||
The general solution is $x(t) = x_h(t) + x_p(t)$, where the homogeneous solution is
|
||||
\[
|
||||
x_h(t) = C_1\cos(2t) + C_2\sin(2t).
|
||||
\]
|
||||
Apply initial conditions: $x(0) = 0$ implies $C_1 + x_p(0) = 0$. Since $x_p(0) = 0$ (all sine terms vanish at $t=0$), we get $C_1 = 0$. For $x'(0) = 0$:
|
||||
\[
|
||||
x'(t) = 2C_2\cos(2t) + x_p'(t),
|
||||
\]
|
||||
and $x_p'(0) = \dfrac{4}{\pi}\sum_{k=0}^{\infty} \dfrac{1}{4 - (2k+1)^2}$. Using the closed-form sum, one finds $x_p'(0) = -\dfrac{1}{2}$. Setting $x'(0) = 2C_2 - \dfrac{1}{2} = 0$ gives $C_2 = \dfrac{1}{4}$.
|
||||
|
||||
The final solution is
|
||||
\[
|
||||
x(t) = \frac{1}{4}\sin(2t) + \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin\bigl((2k+1)t\bigr)}{(2k+1)\bigl(4 - (2k+1)^2\bigr)}.
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Physical intuition for Fourier forcing.} Each harmonic component of the forcing drives the system at its own frequency. The system's response at frequency $n$ is amplified by the factor $\dfrac{1}{|\omega_0^2 - n^2|}$: forcing frequencies far from the natural frequency produce a small response, while forcing frequencies close to $\omega_0$ produce a large response. This frequency-selective amplification is the fundamental principle behind filters in signal processing and is why resonance is both useful (e.g., tuning a radio) and dangerous (e.g., the Tacoma Narrows Bridge collapse).
|
||||
\end{hintbox}
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch10_summary}
|
||||
|
||||
Fourier series transform the study of periodic functions into the algebra of their spectral coefficients. By decomposing a periodic function into sine and cosine harmonics, we gain the ability to:
|
||||
\begin{itemize}
|
||||
\item Analyze and synthesize periodic signals in engineering and physics.
|
||||
\item Solve linear ODEs and PDEs with periodic forcing or boundary conditions.
|
||||
\item Evaluate infinite series via Parseval's identity.
|
||||
\item Understand the frequency content of complex waveforms.
|
||||
\end{itemize}
|
||||
|
||||
The Fourier series framework developed here will be directly applied in the study of boundary value problems (\cref{ch:boundary_value_problems}), the heat equation (\cref{ch:heat_equation}), and the wave and Laplace equations (\cref{ch:wave_and_laplace}), where separation of variables naturally produces Fourier expansions.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Fourier series: key formulas and concepts}
|
||||
\label{tab:ch10_summary}
|
||||
\begin{tabular}{l l}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key Formula/Method} \\
|
||||
\midrule
|
||||
Fourier series on $[-L, L]$ &
|
||||
$f(x) = \dfrac{a_0}{2} + \sum\limits_{n=1}^{\infty}\!\bigl[a_n\cos(\frac{n\pi x}{L}) + b_n\sin(\frac{n\pi x}{L})\bigr]$ \\[12pt]
|
||||
Cosine coefficient $a_n$ &
|
||||
$a_n = \dfrac{1}{L}\int_{-L}^{L} f(x)\cos(\frac{n\pi x}{L})\,\diff x$ \\[10pt]
|
||||
Sine coefficient $b_n$ &
|
||||
$b_n = \dfrac{1}{L}\int_{-L}^{L} f(x)\sin(\frac{n\pi x}{L})\,\diff x$ \\[10pt]
|
||||
Even function &
|
||||
$b_n = 0$, $a_n = \dfrac{2}{L}\int_0^L f(x)\cos(\frac{n\pi x}{L})\,\diff x$ \\[10pt]
|
||||
Odd function &
|
||||
$a_0 = a_n = 0$, $b_n = \dfrac{2}{L}\int_0^L f(x)\sin(\frac{n\pi x}{L})\,\diff x$ \\[10pt]
|
||||
Complex Fourier series &
|
||||
$f(x) = \sum_{n=-\infty}^{\infty} c_n e^{in\pi x/L}$, \quad
|
||||
$c_n = \dfrac{1}{2L}\int_{-L}^{L} f(x)e^{-in\pi x/L}\,\diff x$ \\[10pt]
|
||||
Dirichlet convergence &
|
||||
Series converges to $\dfrac{f(x^+) + f(x^-)}{2}$ at every $x$ \\[8pt]
|
||||
Gibbs overshoot &
|
||||
$\approx 8.95\%$ of jump at discontinuity, persists as $N \to \infty$ \\[8pt]
|
||||
Parseval's identity &
|
||||
$\dfrac{1}{L}\int_{-L}^{L} f(x)^2\,\diff x = \dfrac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2 + b_n^2)$ \\[10pt]
|
||||
Periodic forcing of $x'' + \omega_0^2 x = f(t)$ &
|
||||
Solve mode-by-mode: $x_n = \dfrac{a_n}{\omega_0^2 - n^2}\cos(nt) + \dfrac{b_n}{\omega_0^2 - n^2}\sin(nt)$ \\[10pt]
|
||||
Resonance condition &
|
||||
If $n = \omega_0$, solution grows linearly: $x_p \propto t\sin(\omega_0 t)$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
@@ -1,728 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch11_boundary_value_problems.tex
|
||||
% Chapter 11: Boundary Value Problems
|
||||
% =============================================================================
|
||||
|
||||
\section{Boundary Value Problems}
|
||||
\label{ch:boundary_value_problems}
|
||||
|
||||
\subsection{BVP vs IVP}
|
||||
\label{sec:ch11_bvp_vs_ivp}
|
||||
|
||||
In the preceding chapters we have focused almost exclusively on \textbf{initial value problems (IVPs)}: a differential equation together with conditions specified at a \emph{single} point. For example,
|
||||
\[
|
||||
y'' + p(x)y' + q(x)y = f(x),
|
||||
\qquad y(x_0) = y_0,\; y'(x_0) = y_0'.
|
||||
\]
|
||||
An IVP asks: given the state of a system at one instant, what will it do in the future? This is the natural framework for time-evolution problems.
|
||||
|
||||
A \textbf{boundary value problem (BVP)}, by contrast, prescribes conditions at \emph{two different points} of the independent variable:
|
||||
\[
|
||||
y'' + p(x)y' + q(x)y = f(x),
|
||||
\qquad y(a) = \alpha,\; y(b) = \beta,
|
||||
\]
|
||||
where $a \neq b$. Instead of ``initial'' conditions at a starting time, we impose \textbf{boundary conditions} at the endpoints of an interval. The independent variable $x$ typically represents spatial position rather than time.
|
||||
|
||||
\paragraph{When BVPs arise.} Boundary value problems appear whenever a physical quantity is constrained at the boundaries of a spatial domain. Classic examples include:
|
||||
\begin{itemize}
|
||||
\item The displacement $y(x)$ of a string fixed at both ends: $y(0) = 0$, $y(L) = 0$.
|
||||
\item The temperature $u(x)$ in a rod whose ends are held at prescribed temperatures.
|
||||
\item The electric potential between two conducting plates.
|
||||
\end{itemize}
|
||||
In all of these, the differential equation describes the internal physics, while the boundary conditions encode the geometry or external constraints.
|
||||
|
||||
\paragraph{Key differences from IVPs.} Unlike IVPs, BVPs do \emph{not} always have a solution, and when a solution exists it may not be unique. The three possibilities are:
|
||||
\begin{enumerate}
|
||||
\item \textbf{Unique solution}: exactly one function satisfies both the ODE and the boundary conditions.
|
||||
\item \textbf{No solution}: the boundary conditions are incompatible with any solution of the ODE.
|
||||
\item \textbf{Infinitely many solutions}: the boundary conditions are satisfied by a whole family of solutions. This occurs precisely when the associated \textbf{homogeneous BVP} admits nontrivial solutions---a phenomenon we explore in the next subsection.
|
||||
\end{enumerate}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the BVP $y'' + y = 0$, $y(0) = 0$, $y(\pi) = 0$.
|
||||
|
||||
\textbf{Solution.} The general solution of the ODE is
|
||||
\[
|
||||
y(x) = c_1 \cos x + c_2 \sin x.
|
||||
\]
|
||||
Apply the first boundary condition:
|
||||
\[
|
||||
y(0) = c_1 \cdot 1 + c_2 \cdot 0 = c_1 = 0,
|
||||
\]
|
||||
so $c_1 = 0$ and $y(x) = c_2 \sin x$.
|
||||
|
||||
Apply the second boundary condition:
|
||||
\[
|
||||
y(\pi) = c_2 \sin \pi = c_2 \cdot 0 = 0.
|
||||
\]
|
||||
This condition is satisfied for \emph{any} $c_2$. Hence we have infinitely many solutions:
|
||||
\[
|
||||
y(x) = c_2 \sin x, \qquad c_2 \in \R.
|
||||
\]
|
||||
The homogeneous BVP admits nontrivial solutions; this is the hallmark of an \textbf{eigenvalue problem}.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the BVP $y'' + y = 0$, $y(0) = 0$, $y(\pi/2) = 1$.
|
||||
|
||||
\textbf{Solution.} Again the general solution is $y(x) = c_1 \cos x + c_2 \sin x$.
|
||||
|
||||
Apply $y(0) = 0$: we obtain $c_1 = 0$, so $y(x) = c_2 \sin x$.
|
||||
|
||||
Apply $y(\pi/2) = 1$:
|
||||
\[
|
||||
y(\pi/2) = c_2 \sin(\pi/2) = c_2 \cdot 1 = 1 \quad\Longrightarrow\quad c_2 = 1.
|
||||
\]
|
||||
The solution is unique: $y(x) = \sin x$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Eigenvalue Problems}
|
||||
\label{sec:ch11_eigenvalue_problems}
|
||||
|
||||
The phenomenon of infinitely many solutions leads us to one of the most important structures in applied mathematics: the \textbf{eigenvalue problem}.
|
||||
|
||||
Consider the second-order equation
|
||||
\begin{equation}
|
||||
\label{eq:eigenvalue_problem}
|
||||
y'' + \lambda y = 0, \qquad 0 < x < L,
|
||||
\end{equation}
|
||||
with Dirichlet boundary conditions
|
||||
\begin{equation}
|
||||
\label{eq:dirichlet_bc}
|
||||
y(0) = 0, \qquad y(L) = 0.
|
||||
\end{equation}
|
||||
Here $\lambda$ is a parameter. For most values of $\lambda$, the only solution is the trivial one $y \equiv 0$. But for certain special values of $\lambda$---called \textbf{eigenvalues}---there exist nontrivial solutions, called \textbf{eigenfunctions}.
|
||||
|
||||
\paragraph{Case analysis.} We treat three cases for the sign of $\lambda$.
|
||||
|
||||
\medskip
|
||||
\noindent\textbf{Case 1: $\lambda < 0$.} Let $\lambda = -\mu^2$ with $\mu > 0$. The general solution is
|
||||
\[
|
||||
y(x) = c_1 e^{\mu x} + c_2 e^{-\mu x}
|
||||
= A \cosh(\mu x) + B \sinh(\mu x).
|
||||
\]
|
||||
Apply $y(0) = 0$: $A = 0$, so $y(x) = B \sinh(\mu x)$.
|
||||
|
||||
Apply $y(L) = 0$: $B \sinh(\mu L) = 0$. Since $\mu > 0$ and $L > 0$, we have $\sinh(\mu L) > 0$, so $B = 0$. Hence $y \equiv 0$. There are \emph{no} eigenvalues with $\lambda < 0$.
|
||||
|
||||
\medskip
|
||||
\noindent\textbf{Case 2: $\lambda = 0$.} The equation becomes $y'' = 0$, with general solution
|
||||
\[
|
||||
y(x) = c_1 x + c_2.
|
||||
\]
|
||||
Apply $y(0) = 0$: $c_2 = 0$. Apply $y(L) = 0$: $c_1 L = 0$, so $c_1 = 0$. Again $y \equiv 0$. The value $\lambda = 0$ is \emph{not} an eigenvalue.
|
||||
|
||||
\medskip
|
||||
\noindent\textbf{Case 3: $\lambda > 0$.} Let $\lambda = \mu^2$ with $\mu > 0$. The general solution is
|
||||
\[
|
||||
y(x) = c_1 \cos(\mu x) + c_2 \sin(\mu x).
|
||||
\]
|
||||
Apply $y(0) = 0$: $c_1 = 0$, so $y(x) = c_2 \sin(\mu x)$.
|
||||
|
||||
Apply $y(L) = 0$: $c_2 \sin(\mu L) = 0$. For a nontrivial solution we need $c_2 \neq 0$, which requires
|
||||
\[
|
||||
\sin(\mu L) = 0 \quad\Longrightarrow\quad \mu L = n\pi, \qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
Thus $\mu_n = \dfrac{n\pi}{L}$ and $\lambda_n = \mu_n^2 = \left(\dfrac{n\pi}{L}\right)^2$.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Dirichlet eigenvalue problem.} For
|
||||
\[
|
||||
y'' + \lambda y = 0, \qquad y(0) = 0, \;\; y(L) = 0,
|
||||
\]
|
||||
the eigenvalues and eigenfunctions are
|
||||
\[
|
||||
\lambda_n = \left(\frac{n\pi}{L}\right)^{\!2},
|
||||
\qquad
|
||||
y_n(x) = \sin\!\left(\frac{n\pi x}{L}\right),
|
||||
\qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
Any positive constant multiple of $y_n(x)$ is also an eigenfunction.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Neumann boundary conditions.} If instead we impose
|
||||
\[
|
||||
y'(0) = 0, \qquad y'(L) = 0,
|
||||
\]
|
||||
the analysis changes slightly. With $\lambda = \mu^2 > 0$, we have $y(x) = c_1 \cos(\mu x) + c_2 \sin(\mu x)$ and $y'(x) = -c_1 \mu \sin(\mu x) + c_2 \mu \cos(\mu x)$.
|
||||
|
||||
Apply $y'(0) = 0$: $c_2 \mu = 0$, so $c_2 = 0$. Then $y(x) = c_1 \cos(\mu x)$ and $y'(x) = -c_1 \mu \sin(\mu x)$.
|
||||
|
||||
Apply $y'(L) = 0$: $-c_1 \mu \sin(\mu L) = 0$. For nontrivial solutions ($c_1 \neq 0$), we need $\sin(\mu L) = 0$, so $\mu L = n\pi$ with $n = 0, 1, 2, \dots$.
|
||||
|
||||
Note that $n = 0$ gives $\mu = 0$, $\lambda_0 = 0$, and $y_0(x) = \cos(0) = 1$ (a nonzero constant). This means $\lambda_0 = 0$ \emph{is} an eigenvalue for Neumann conditions.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Neumann eigenvalue problem.} For
|
||||
\[
|
||||
y'' + \lambda y = 0, \qquad y'(0) = 0, \;\; y'(L) = 0,
|
||||
\]
|
||||
the eigenvalues and eigenfunctions are
|
||||
\[
|
||||
\lambda_n = \left(\frac{n\pi}{L}\right)^{\!2},
|
||||
\qquad
|
||||
y_n(x) = \cos\!\left(\frac{n\pi x}{L}\right),
|
||||
\qquad n = 0, 1, 2, \dots
|
||||
\]
|
||||
Note that $\lambda_0 = 0$ with eigenfunction $y_0(x) = 1$.
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
Find all eigenvalues and eigenfunctions of
|
||||
\[
|
||||
y'' + \lambda y = 0, \qquad y(0) = 0, \;\; y(2) = 0.
|
||||
\]
|
||||
|
||||
\textbf{Solution.} Here $L = 2$. From \cref{eq:eigenvalue_problem}, the eigenvalues are
|
||||
\[
|
||||
\lambda_n = \left(\frac{n\pi}{2}\right)^{\!2} = \frac{n^2 \pi^2}{4},
|
||||
\qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
and the eigenfunctions are
|
||||
\[
|
||||
y_n(x) = \sin\!\left(\frac{n\pi x}{2}\right).
|
||||
\]
|
||||
Listing the first few:
|
||||
\begin{align*}
|
||||
\lambda_1 &= \frac{\pi^2}{4}, & y_1(x) &= \sin\!\left(\frac{\pi x}{2}\right), \\
|
||||
\lambda_2 &= \pi^2, & y_2(x) &= \sin(\pi x), \\
|
||||
\lambda_3 &= \frac{9\pi^2}{4}, & y_3(x) &= \sin\!\left(\frac{3\pi x}{2}\right).
|
||||
\end{align*}
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Find all eigenvalues and eigenfunctions of
|
||||
\[
|
||||
y'' + \lambda y = 0, \qquad y'(0) = 0, \;\; y'(3) = 0.
|
||||
\]
|
||||
|
||||
\textbf{Solution.} Here $L = 3$ with Neumann boundary conditions. The eigenvalues are
|
||||
\[
|
||||
\lambda_n = \left(\frac{n\pi}{3}\right)^{\!2} = \frac{n^2 \pi^2}{9},
|
||||
\qquad n = 0, 1, 2, \dots
|
||||
\]
|
||||
and the eigenfunctions are
|
||||
\[
|
||||
y_n(x) = \cos\!\left(\frac{n\pi x}{3}\right).
|
||||
\]
|
||||
The first few are:
|
||||
\begin{align*}
|
||||
\lambda_0 &= 0, & y_0(x) &= 1, \\
|
||||
\lambda_1 &= \frac{\pi^2}{9}, & y_1(x) &= \cos\!\left(\frac{\pi x}{3}\right), \\
|
||||
\lambda_2 &= \frac{4\pi^2}{9}, & y_2(x) &= \cos\!\left(\frac{2\pi x}{3}\right).
|
||||
\end{align*}
|
||||
Note that $y_0(x) = 1$ corresponds to the constant equilibrium state.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Sturm--Liouville Form}
|
||||
\label{sec:ch11_sturm_liouville}
|
||||
|
||||
The eigenvalue problems we just studied are special cases of a broad and powerful framework: the \textbf{Sturm--Liouville (SL) problem}. This theory unifies eigenvalue problems, orthogonality, and series expansions.
|
||||
|
||||
\paragraph{General SL form.} A second-order linear ODE is in \textbf{Sturm--Liouville form} when it can be written as
|
||||
\begin{equation}
|
||||
\label{eq:sl_form}
|
||||
\bigl(p(x)\,y'\bigr)' + \bigl(q(x) + \lambda\,r(x)\bigr)y = 0,
|
||||
\qquad a \leq x \leq b,
|
||||
\end{equation}
|
||||
where $p(x)$, $q(x)$, and $r(x)$ are given coefficient functions and $\lambda$ is the eigenvalue parameter. The function $r(x)$ is called the \textbf{weight function}.
|
||||
|
||||
\paragraph{Converting to SL form.} Any equation of the form
|
||||
\begin{equation}
|
||||
\label{eq:general_second_order}
|
||||
P(x)\,y'' + Q(x)\,y' + R(x)\,y + \lambda\,S(x)\,y = 0
|
||||
\end{equation}
|
||||
can be converted to SL form by dividing by $P(x)$ and multiplying by an \textbf{integrating factor}. First divide by $P(x)$:
|
||||
\[
|
||||
y'' + \frac{Q(x)}{P(x)}\,y' + \frac{R(x)}{P(x)}\,y + \lambda\,\frac{S(x)}{P(x)}\,y = 0.
|
||||
\]
|
||||
Multiply by the integrating factor
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int \frac{Q(x)}{P(x)}\,\diff x\right).
|
||||
\]
|
||||
Then the first two terms combine as a derivative:
|
||||
\[
|
||||
\mu(x)\,y'' + \mu(x)\,\frac{Q(x)}{P(x)}\,y' = \bigl(\mu(x)\,y'\bigr)',
|
||||
\]
|
||||
which is exactly the $(py')'$ structure of the SL form with $p(x) = \mu(x)$, $q(x) = \mu(x)\,\dfrac{R(x)}{P(x)}$, and $r(x) = \mu(x)\,\dfrac{S(x)}{P(x)}$.
|
||||
|
||||
\paragraph{Regular SL problem.} We restrict attention to the \textbf{regular} case:
|
||||
|
||||
\begin{definition}[Regular Sturm--Liouville Problem]
|
||||
\label{def:regular_sl}
|
||||
A Sturm--Liouville problem is \textbf{regular} on $[a,b]$ if:
|
||||
\begin{enumerate}
|
||||
\item $p(x)$, $p'(x)$, $q(x)$, and $r(x)$ are continuous on $[a,b]$.
|
||||
\item $p(x) > 0$ and $r(x) > 0$ on $[a,b]$.
|
||||
\item The boundary conditions are of the form
|
||||
\[
|
||||
\alpha_1 y(a) + \alpha_2 y'(a) = 0,
|
||||
\qquad
|
||||
\beta_1 y(b) + \beta_2 y'(b) = 0,
|
||||
\]
|
||||
where $\alpha_1^2 + \alpha_2^2 > 0$ and $\beta_1^2 + \beta_2^2 > 0$.
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
|
||||
\begin{theorem}[SL Existence Theorem]
|
||||
\label{thm:sl_existence}
|
||||
A regular Sturm--Liouville problem has:
|
||||
\begin{enumerate}
|
||||
\item An infinite sequence of real eigenvalues
|
||||
\[
|
||||
\lambda_1 < \lambda_2 < \lambda_3 < \cdots \to \infty.
|
||||
\]
|
||||
\item Each eigenvalue $\lambda_n$ has exactly one corresponding eigenfunction $\phi_n(x)$ (up to a constant multiple).
|
||||
\item The $n$-th eigenfunction $\phi_n(x)$ has exactly $n-1$ zeros in the open interval $(a,b)$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
This theorem guarantees that eigenvalue problems arising from well-behaved physical systems always have a rich structure of eigenvalues and eigenfunctions.
|
||||
|
||||
\begin{workedexample}
|
||||
Convert the equation $x\,y'' + 2y' + \lambda\,x\,y = 0$ to Sturm--Liouville form on $[0,1]$, and identify $p(x)$, $q(x)$, and $r(x)$.
|
||||
|
||||
\textbf{Solution.} Here $P(x) = x$, $Q(x) = 2$, $R(x) = 0$, and $S(x) = x$.
|
||||
|
||||
Divide by $P(x) = x$:
|
||||
\[
|
||||
y'' + \frac{2}{x}\,y' + \lambda\,y = 0.
|
||||
\]
|
||||
|
||||
The integrating factor is
|
||||
\[
|
||||
\mu(x) = \exp\!\left(\int \frac{2}{x}\,\diff x\right)
|
||||
= \exp(2\ln x) = x^2.
|
||||
\]
|
||||
|
||||
Multiply the equation by $x^2$:
|
||||
\[
|
||||
x^2\,y'' + 2x\,y' + \lambda\,x^2\,y = 0.
|
||||
\]
|
||||
The first two terms combine as
|
||||
\[
|
||||
x^2 y'' + 2x y' = (x^2 y')'.
|
||||
\]
|
||||
Hence the SL form is
|
||||
\[
|
||||
(x^2 y')' + \lambda\,x^2\,y = 0.
|
||||
\]
|
||||
Reading off the coefficients:
|
||||
\[
|
||||
p(x) = x^2, \qquad q(x) = 0, \qquad r(x) = x^2.
|
||||
\]
|
||||
On $(0,1]$, we have $p(x) = x^2 > 0$ and $r(x) = x^2 > 0$. Note that $p(0) = 0$, so strictly speaking this is a \emph{singular} SL problem (not regular) at $x = 0$, since the regularity condition requires $p > 0$ on the \emph{closed} interval. Singular SL problems require additional care at the singular endpoint.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Orthogonality Theorem}
|
||||
\label{sec:ch11_orthogonality_theorem}
|
||||
|
||||
The most powerful consequence of the Sturm--Liouville framework is the \textbf{orthogonality} of eigenfunctions belonging to distinct eigenvalues.
|
||||
|
||||
\begin{theorem}[SL Orthogonality Theorem]
|
||||
\label{thm:sl_orthogonality}
|
||||
Let $\phi_n(x)$ and $\phi_m(x)$ be eigenfunctions corresponding to distinct eigenvalues $\lambda_n \neq \lambda_m$ of a regular Sturm--Liouville problem on $[a,b]$. Then $\phi_n$ and $\phi_m$ are \textbf{orthogonal with respect to the weight function} $r(x)$:
|
||||
\[
|
||||
\int_a^b \phi_n(x)\,\phi_m(x)\,r(x)\,\diff x = 0, \qquad n \neq m.
|
||||
\]
|
||||
\end{theorem}
|
||||
|
||||
\begin{proof}
|
||||
The eigenfunctions satisfy the SL equations
|
||||
\[
|
||||
\begin{cases}
|
||||
\bigl(p\,\phi_n'\bigr)' + \bigl(q + \lambda_n\,r\bigr)\phi_n = 0, \\[4pt]
|
||||
\bigl(p\,\phi_m'\bigr)' + \bigl(q + \lambda_m\,r\bigr)\phi_m = 0.
|
||||
\end{cases}
|
||||
\]
|
||||
Multiply the first equation by $\phi_m(x)$ and the second by $\phi_n(x)$:
|
||||
\[
|
||||
\begin{cases}
|
||||
\phi_m\,(p\,\phi_n')' + \phi_m\,q\,\phi_n + \lambda_n\,\phi_m\,r\,\phi_n = 0, \\[4pt]
|
||||
\phi_n\,(p\,\phi_m')' + \phi_n\,q\,\phi_m + \lambda_m\,\phi_n\,r\,\phi_m = 0.
|
||||
\end{cases}
|
||||
\]
|
||||
Subtract the second from the first:
|
||||
\[
|
||||
\phi_m\,(p\,\phi_n')' - \phi_n\,(p\,\phi_m')'
|
||||
+ (\lambda_n - \lambda_m)\,\phi_n\,\phi_m\,r = 0.
|
||||
\]
|
||||
The terms involving $q(x)$ cancel. Rearrange:
|
||||
\[
|
||||
(\lambda_n - \lambda_m)\,\phi_n(x)\,\phi_m(x)\,r(x)
|
||||
= \phi_n\,(p\,\phi_m')' - \phi_m\,(p\,\phi_n')'.
|
||||
\]
|
||||
The right-hand side can be written as a total derivative:
|
||||
\[
|
||||
\phi_n\,(p\,\phi_m')' - \phi_m\,(p\,\phi_n')'
|
||||
= \frac{\diff}{\diff x}\Bigl[p\bigl(\phi_n\,\phi_m' - \phi_m\,\phi_n'\bigr)\Bigr].
|
||||
\]
|
||||
To verify this, differentiate the expression inside the brackets:
|
||||
\[
|
||||
\frac{\diff}{\diff x}\Bigl[p(\phi_n\phi_m' - \phi_m\phi_n')\Bigr]
|
||||
= p'(\phi_n\phi_m' - \phi_m\phi_n') + p(\phi_n'\phi_m' + \phi_n\phi_m'' - \phi_m'\phi_n' - \phi_m\phi_n'')
|
||||
= \phi_n(p\phi_m')' - \phi_m(p\phi_n')',
|
||||
\]
|
||||
as claimed. Now integrate both sides from $a$ to $b$:
|
||||
\[
|
||||
(\lambda_n - \lambda_m)\int_a^b \phi_n(x)\,\phi_m(x)\,r(x)\,\diff x
|
||||
= \Bigl[p(x)\bigl(\phi_n(x)\,\phi_m'(x) - \phi_m(x)\,\phi_n'(x)\bigr)\Bigr]_{x=a}^{x=b}.
|
||||
\]
|
||||
The right-hand side consists of \textbf{boundary terms}. For a regular SL problem, the boundary conditions are of the form
|
||||
\[
|
||||
\alpha_1 y(a) + \alpha_2 y'(a) = 0, \qquad \beta_1 y(b) + \beta_2 y'(b) = 0.
|
||||
\]
|
||||
Both $\phi_n$ and $\phi_m$ satisfy these boundary conditions. It is a standard verification that under any such separated boundary conditions, the boundary expression
|
||||
\[
|
||||
p\bigl(\phi_n\phi_m' - \phi_m\phi_n'\bigr)
|
||||
\]
|
||||
vanishes at both $x = a$ and $x = b$. (For example, if Dirichlet conditions $y(a) = 0$ apply, then $\phi_n(a) = \phi_m(a) = 0$ makes the expression zero immediately. For mixed conditions, a short algebraic argument shows cancellation.)
|
||||
|
||||
Therefore the right-hand side is zero. Since $\lambda_n \neq \lambda_m$, we divide by $(\lambda_n - \lambda_m)$ to obtain
|
||||
\[
|
||||
\int_a^b \phi_n(x)\,\phi_m(x)\,r(x)\,\diff x = 0,
|
||||
\]
|
||||
completing the proof.
|
||||
\end{proof}
|
||||
|
||||
\paragraph{Orthogonality in action.} The orthogonality of $\{\sin(n\pi x/L)\}$ on $[0,L]$ is a direct consequence of \cref{thm:sl_orthogonality}. For the Dirichlet problem $y'' + \lambda y = 0$ on $[0,L]$, we have $p = 1$, $q = 0$, $r = 1$, and $\phi_n(x) = \sin(n\pi x/L)$. The theorem guarantees
|
||||
\[
|
||||
\int_0^L \sin\!\left(\frac{n\pi x}{L}\right)\sin\!\left(\frac{m\pi x}{L}\right)\diff x = 0, \qquad n \neq m,
|
||||
\]
|
||||
which is precisely the sine orthogonality relation used throughout Fourier analysis (\cref{ch:fourier_series}).
|
||||
|
||||
\begin{workedexample}
|
||||
Verify the orthogonality of $\sin(\pi x)$ and $\sin(2\pi x)$ on $[0,1]$ by direct computation.
|
||||
|
||||
\textbf{Solution.} We compute
|
||||
\[
|
||||
I = \int_0^1 \sin(\pi x)\,\sin(2\pi x)\,\diff x.
|
||||
\]
|
||||
Use the product-to-sum identity $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$:
|
||||
\[
|
||||
\sin(\pi x)\sin(2\pi x) = \frac{1}{2}\bigl[\cos(-\pi x) - \cos(3\pi x)\bigr]
|
||||
= \frac{1}{2}\bigl[\cos(\pi x) - \cos(3\pi x)\bigr].
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
I = \frac{1}{2}\left[\frac{\sin(\pi x)}{\pi} - \frac{\sin(3\pi x)}{3\pi}\right]_0^1
|
||||
= \frac{1}{2}\left[\frac{0}{\pi} - \frac{0}{3\pi} - 0\right] = 0.
|
||||
\]
|
||||
The integral vanishes, confirming orthogonality.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Show that the eigenfunctions of $y'' + \lambda y = 0$ with Neumann conditions $y'(0) = 0$, $y'(1) = 0$ on $[0,1]$ are orthogonal with respect to the weight $r(x) = 1$.
|
||||
|
||||
\textbf{Solution.} The eigenfunctions are $\phi_n(x) = \cos(n\pi x)$ for $n = 0, 1, 2, \dots$. We need to verify
|
||||
\[
|
||||
\int_0^1 \cos(n\pi x)\,\cos(m\pi x)\,\diff x = 0, \qquad n \neq m.
|
||||
\]
|
||||
Use $\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$:
|
||||
\[
|
||||
\cos(n\pi x)\cos(m\pi x) = \frac{1}{2}\bigl[\cos\bigl((n-m)\pi x\bigr) + \cos\bigl((n+m)\pi x\bigr)\bigr].
|
||||
\]
|
||||
Integrate:
|
||||
\[
|
||||
\int_0^1 \cos(n\pi x)\cos(m\pi x)\,\diff x
|
||||
= \frac{1}{2}\left[\frac{\sin((n-m)\pi x)}{(n-m)\pi} + \frac{\sin((n+m)\pi x)}{(n+m)\pi}\right]_0^1.
|
||||
\]
|
||||
Since $n \neq m$ are integers, both $(n-m)\pi$ and $(n+m)\pi$ are nonzero multiples of $\pi$, so $\sin(k\pi) = 0$ for any integer $k$. The integral evaluates to $0$, confirming orthogonality.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Eigenfunction Expansions}
|
||||
\label{sec:ch11_eigenfunction_expansions}
|
||||
|
||||
The orthogonality of eigenfunctions allows us to expand an arbitrary function as a series, much like a Taylor series or a Fourier series.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Eigenfunction expansion.} Let $\{\phi_n(x)\}_{n=1}^\infty$ be the eigenfunctions of a regular SL problem on $[a,b]$ with weight $r(x)$. Any piecewise smooth function $f(x)$ on $[a,b]$ can be expanded as
|
||||
\[
|
||||
f(x) = \sum_{n=1}^\infty c_n\,\phi_n(x),
|
||||
\]
|
||||
where the \textbf{expansion coefficients} are
|
||||
\begin{equation}
|
||||
\label{eq:expansion_coefficient}
|
||||
c_n = \frac{\displaystyle\int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x}
|
||||
{\displaystyle\int_a^b \phi_n(x)^2\,r(x)\,\diff x}.
|
||||
\end{equation}
|
||||
The series converges to $f(x)$ at points of continuity, and to the average $\frac{1}{2}[f(x^+) + f(x^-)]$ at points of discontinuity.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Derivation of the coefficient formula.} Multiply the expansion $f(x) = \sum_{k=1}^\infty c_k \phi_k(x)$ by $\phi_n(x)\,r(x)$ and integrate over $[a,b]$:
|
||||
\[
|
||||
\int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x
|
||||
= \sum_{k=1}^\infty c_k \int_a^b \phi_k(x)\,\phi_n(x)\,r(x)\,\diff x.
|
||||
\]
|
||||
By orthogonality, all terms in the sum vanish except $k = n$:
|
||||
\[
|
||||
\int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x
|
||||
= c_n \int_a^b \phi_n(x)^2\,r(x)\,\diff x.
|
||||
\]
|
||||
Solving for $c_n$ gives \cref{eq:expansion_coefficient}. The denominator
|
||||
\[
|
||||
\|\phi_n\|^2 = \int_a^b \phi_n(x)^2\,r(x)\,\diff x
|
||||
\]
|
||||
is the squared \textbf{norm} of the eigenfunction with respect to the weight $r(x)$.
|
||||
|
||||
\paragraph{Connection to Fourier series.} The Fourier sine series is a special case of eigenfunction expansion. Consider the SL problem with $p = 1$, $q = 0$, $r = 1$ on $[0,L]$ and Dirichlet boundary conditions. The eigenfunctions are $\phi_n(x) = \sin(n\pi x/L)$, and
|
||||
\[
|
||||
\|\phi_n\|^2 = \int_0^L \sin^2\!\left(\frac{n\pi x}{L}\right)\diff x = \frac{L}{2}.
|
||||
\]
|
||||
The expansion formula then gives the familiar Fourier sine coefficients:
|
||||
\[
|
||||
c_n = \frac{2}{L}\int_0^L f(x)\,\sin\!\left(\frac{n\pi x}{L}\right)\diff x.
|
||||
\]
|
||||
Similarly, the Fourier cosine series arises from the Neumann SL problem on $[0,L]$.
|
||||
|
||||
\begin{workedexample}
|
||||
Expand $f(x) = x$ on $[0,1]$ as a series of eigenfunctions of $y'' + \lambda y = 0$ with Dirichlet boundary conditions $y(0) = 0$, $y(1) = 0$.
|
||||
|
||||
\textbf{Solution.} The eigenfunctions are $\phi_n(x) = \sin(n\pi x)$ with weight $r(x) = 1$. We need
|
||||
\[
|
||||
c_n = \frac{\displaystyle\int_0^1 x\,\sin(n\pi x)\,\diff x}
|
||||
{\displaystyle\int_0^1 \sin^2(n\pi x)\,\diff x}.
|
||||
\]
|
||||
The denominator is
|
||||
\[
|
||||
\int_0^1 \sin^2(n\pi x)\,\diff x = \frac{1}{2}.
|
||||
\]
|
||||
|
||||
For the numerator, use integration by parts with $u = x$ and $\diff v = \sin(n\pi x)\,\diff x$:
|
||||
\[
|
||||
\int_0^1 x\,\sin(n\pi x)\,\diff x
|
||||
= \left[-\frac{x}{n\pi}\cos(n\pi x)\right]_0^1 + \frac{1}{n\pi}\int_0^1 \cos(n\pi x)\,\diff x.
|
||||
\]
|
||||
The boundary term gives
|
||||
\[
|
||||
-\frac{1}{n\pi}\cos(n\pi) + 0 = -\frac{(-1)^n}{n\pi} = \frac{(-1)^{n+1}}{n\pi}.
|
||||
\]
|
||||
The remaining integral is
|
||||
\[
|
||||
\frac{1}{n\pi}\left[\frac{\sin(n\pi x)}{n\pi}\right]_0^1 = 0.
|
||||
\]
|
||||
So the numerator is $\dfrac{(-1)^{n+1}}{n\pi}$.
|
||||
|
||||
Therefore
|
||||
\[
|
||||
c_n = \frac{(-1)^{n+1}/(n\pi)}{1/2} = \frac{2(-1)^{n+1}}{n\pi}.
|
||||
\]
|
||||
The eigenfunction expansion is
|
||||
\[
|
||||
x = \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n\pi}\,\sin(n\pi x),
|
||||
\qquad 0 < x < 1.
|
||||
\]
|
||||
Writing out the first few terms:
|
||||
\[
|
||||
x \approx \frac{2}{\pi}\sin(\pi x) - \frac{1}{\pi}\sin(2\pi x) + \frac{2}{3\pi}\sin(3\pi x) - \cdots
|
||||
\]
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Expand $f(x) = 1$ on $[0,1]$ as a series of eigenfunctions of $y'' + \lambda y = 0$ with Neumann boundary conditions $y'(0) = 0$, $y'(1) = 0$.
|
||||
|
||||
\textbf{Solution.} The eigenfunctions are $\phi_n(x) = \cos(n\pi x)$ for $n = 0, 1, 2, \dots$, with weight $r(x) = 1$.
|
||||
|
||||
For $n = 0$: $\phi_0(x) = 1$. The norm is
|
||||
\[
|
||||
\|\phi_0\|^2 = \int_0^1 1^2\,\diff x = 1.
|
||||
\]
|
||||
The coefficient is
|
||||
\[
|
||||
c_0 = \frac{\int_0^1 1 \cdot 1\,\diff x}{1} = 1.
|
||||
\]
|
||||
|
||||
For $n \geq 1$:
|
||||
\[
|
||||
c_n = \frac{\displaystyle\int_0^1 1 \cdot \cos(n\pi x)\,\diff x}
|
||||
{\displaystyle\int_0^1 \cos^2(n\pi x)\,\diff x}
|
||||
= \frac{\left[\frac{\sin(n\pi x)}{n\pi}\right]_0^1}{1/2}
|
||||
= \frac{0}{1/2} = 0.
|
||||
\]
|
||||
|
||||
The expansion is simply
|
||||
\[
|
||||
1 = 1,
|
||||
\]
|
||||
i.e., only the $n = 0$ term survives. This makes sense: the constant function is itself the $n = 0$ eigenfunction.
|
||||
|
||||
As a more instructive exercise, expand $f(x) = x$ with Neumann conditions. Then for $n = 0$:
|
||||
\[
|
||||
c_0 = \frac{\int_0^1 x \cdot 1\,\diff x}{1} = \frac{1}{2}.
|
||||
\]
|
||||
For $n \geq 1$:
|
||||
\[
|
||||
c_n = \frac{\int_0^1 x\,\cos(n\pi x)\,\diff x}{1/2}.
|
||||
\]
|
||||
Integrate by parts with $u = x$, $\diff v = \cos(n\pi x)\,\diff x$:
|
||||
\[
|
||||
\int_0^1 x\cos(n\pi x)\,\diff x
|
||||
= \left[\frac{x}{n\pi}\sin(n\pi x)\right]_0^1 - \frac{1}{n\pi}\int_0^1 \sin(n\pi x)\,\diff x.
|
||||
\]
|
||||
The boundary term vanishes ($\sin(n\pi) = 0$). The remaining integral is
|
||||
\[
|
||||
-\frac{1}{n\pi}\left[-\frac{\cos(n\pi x)}{n\pi}\right]_0^1
|
||||
= \frac{1}{(n\pi)^2}\bigl(\cos(n\pi) - \cos 0\bigr)
|
||||
= \frac{(-1)^n - 1}{(n\pi)^2}.
|
||||
\]
|
||||
This is zero for even $n$ and $-\dfrac{2}{(n\pi)^2}$ for odd $n$. Therefore
|
||||
\[
|
||||
c_n = \begin{cases}
|
||||
0, & n \text{ even}, \\[6pt]
|
||||
-\dfrac{4}{(n\pi)^2}, & n \text{ odd}.
|
||||
\end{cases}
|
||||
\]
|
||||
The expansion is
|
||||
\[
|
||||
x = \frac{1}{2} - \frac{4}{\pi^2}\cos(\pi x) - \frac{4}{9\pi^2}\cos(3\pi x) - \frac{4}{25\pi^2}\cos(5\pi x) - \cdots
|
||||
\]
|
||||
This converges to $x$ on $[0,1]$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Applications}
|
||||
\label{sec:ch11_applications}
|
||||
|
||||
Eigenfunction expansions provide a powerful method for solving \textbf{nonhomogeneous} boundary value problems.
|
||||
|
||||
\paragraph{The nonhomogeneous SL problem.} Consider
|
||||
\begin{equation}
|
||||
\label{eq:nonhomogeneous_sl}
|
||||
\bigl(p(x)\,y'\bigr)' + q(x)\,y + \lambda\,r(x)\,y = f(x),
|
||||
\qquad a < x < b,
|
||||
\end{equation}
|
||||
subject to homogeneous boundary conditions at $x = a$ and $x = b$. Suppose we already know the eigenfunctions $\{\phi_n(x)\}$ and eigenvalues $\{\lambda_n\}$ of the associated homogeneous problem
|
||||
\[
|
||||
\bigl(p(x)\,\phi_n'\bigr)' + q(x)\,\phi_n + \lambda_n\,r(x)\,\phi_n = 0.
|
||||
\]
|
||||
|
||||
\paragraph{Eigenfunction expansion method.} Expand both the solution $y(x)$ and the source term $f(x)$ in the eigenfunction basis:
|
||||
\[
|
||||
y(x) = \sum_{n=1}^\infty c_n\,\phi_n(x), \qquad
|
||||
f(x) = \sum_{n=1}^\infty f_n\,\phi_n(x).
|
||||
\]
|
||||
Substitute the series for $y(x)$ into \cref{eq:nonhomogeneous_sl}:
|
||||
\[
|
||||
\sum_{n=1}^\infty c_n\,\Bigl[\bigl(p\,\phi_n'\bigr)' + q\,\phi_n + \lambda\,r\,\phi_n\Bigr] = f(x).
|
||||
\]
|
||||
Using the homogeneous eigenvalue equation $\bigl(p\,\phi_n'\bigr)' + q\,\phi_n = -\lambda_n\,r\,\phi_n$:
|
||||
\[
|
||||
\sum_{n=1}^\infty c_n\,\bigl(-\lambda_n\,r\,\phi_n + \lambda\,r\,\phi_n\bigr) = f(x).
|
||||
\]
|
||||
\[
|
||||
\sum_{n=1}^\infty c_n\,(\lambda - \lambda_n)\,\phi_n(x)\,r(x) = f(x).
|
||||
\]
|
||||
Multiply by $\phi_m(x)$ and integrate, using orthogonality:
|
||||
\[
|
||||
c_m\,(\lambda - \lambda_m)\,\|\phi_m\|^2 = f_m.
|
||||
\]
|
||||
Therefore, provided $\lambda \neq \lambda_n$ for any $n$:
|
||||
\begin{equation}
|
||||
\label{eq:nonhomogeneous_coefficient}
|
||||
c_n = \frac{f_n}{(\lambda - \lambda_n)\,\|\phi_n\|^2}
|
||||
= \frac{\displaystyle\int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x}
|
||||
{(\lambda - \lambda_n)\displaystyle\int_a^b \phi_n(x)^2\,r(x)\,\diff x}.
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Resonance.} If $\lambda = \lambda_n$ for some $n$ and the corresponding $f_n \neq 0$, there is no solution---this is the phenomenon of \textbf{resonance}. If $f_n = 0$ for that particular $n$, the coefficient $c_n$ is undetermined (infinite solutions), as discussed in the context of homogeneous BVPs.
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the nonhomogeneous BVP
|
||||
\[
|
||||
y'' + \pi^2\,y = x, \qquad 0 < x < 1,
|
||||
\qquad y(0) = 0, \;\; y(1) = 0.
|
||||
\]
|
||||
|
||||
\textbf{Solution.} The associated homogeneous problem $y'' + \lambda y = 0$ with Dirichlet BCs has eigenfunctions $\phi_n(x) = \sin(n\pi x)$ and eigenvalues $\lambda_n = n^2\pi^2$.
|
||||
|
||||
Here $\lambda = \pi^2 = \lambda_1$---we are at the first eigenvalue. This raises the question of whether a solution exists. We must check the solvability condition.
|
||||
|
||||
Expand the source term $f(x) = x$ in the eigenfunction basis. From the previous worked example:
|
||||
\[
|
||||
x = \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n\pi}\,\sin(n\pi x).
|
||||
\]
|
||||
So $f_n = \dfrac{2(-1)^{n+1}}{n\pi}$.
|
||||
|
||||
For $n = 1$: $\lambda = \lambda_1 = \pi^2$, so the denominator $(\lambda - \lambda_1) = 0$. The coefficient $f_1 = \dfrac{2(-1)^2}{\pi} = \dfrac{2}{\pi} \neq 0$.
|
||||
|
||||
Since $f_1 \neq 0$ and $\lambda = \lambda_1$, the BVP has \textbf{no solution}. The source term $f(x) = x$ has a component along the first eigenfunction, and at resonance the system cannot respond.
|
||||
|
||||
To see this directly: suppose a solution $y(x)$ exists. Multiply the ODE by $\phi_1(x) = \sin(\pi x)$ and integrate:
|
||||
\[
|
||||
\int_0^1 (y'' + \pi^2 y)\sin(\pi x)\,\diff x = \int_0^1 x\sin(\pi x)\,\diff x.
|
||||
\]
|
||||
The right-hand side is $\dfrac{2}{\pi} \neq 0$. But integrating the left side by parts twice and using the boundary conditions yields zero (this is the Fredholm alternative). Contradiction: no solution exists.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the nonhomogeneous BVP
|
||||
\[
|
||||
y'' + 2\pi^2\,y = \sin(\pi x), \qquad 0 < x < 1,
|
||||
\qquad y(0) = 0, \;\; y(1) = 0.
|
||||
\]
|
||||
|
||||
\textbf{Solution.} Eigenvalues: $\lambda_n = n^2\pi^2$. Here $\lambda = 2\pi^2$, which is \emph{not} an eigenvalue ($\sqrt{2}\pi$ is not a multiple of $\pi$).
|
||||
|
||||
Expand $f(x) = \sin(\pi x)$ in eigenfunctions. Since $\sin(\pi x) = \phi_1(x)$, the expansion is simply
|
||||
\[
|
||||
f(x) = \sin(\pi x) = 1 \cdot \phi_1(x).
|
||||
\]
|
||||
So $f_1 = \|\phi_1\|^2 = \dfrac{1}{2}$ and $f_n = 0$ for $n \neq 1$.
|
||||
|
||||
The coefficients are
|
||||
\[
|
||||
c_n = \frac{f_n}{(\lambda - \lambda_n)\,\|\phi_n\|^2}.
|
||||
\]
|
||||
For $n = 1$:
|
||||
\[
|
||||
c_1 = \frac{1/2}{(2\pi^2 - \pi^2)\cdot(1/2)} = \frac{1}{\pi^2}.
|
||||
\]
|
||||
For $n \neq 1$, $f_n = 0$ so $c_n = 0$.
|
||||
|
||||
The solution is
|
||||
\[
|
||||
y(x) = \frac{1}{\pi^2}\,\sin(\pi x).
|
||||
\]
|
||||
\textbf{Verification.} Compute $y'' = -\sin(\pi x)$. Then
|
||||
\[
|
||||
y'' + 2\pi^2 y = -\sin(\pi x) + 2\pi^2 \cdot \frac{1}{\pi^2}\sin(\pi x) = -\sin(\pi x) + 2\sin(\pi x) = \sin(\pi x).
|
||||
\]
|
||||
$\checkmark$ The solution satisfies the ODE and the boundary conditions.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch11_summary}
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Eigenvalue problems and Sturm--Liouville theory}
|
||||
\label{tab:ch11_summary}
|
||||
\begin{tabular}{l l p{5.5cm}}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key formula/result} \\
|
||||
\midrule
|
||||
BVP vs IVP & BVP: conditions at $x=a$ and $x=b$; IVP: conditions at one point \\
|
||||
Dirichlet eigenvalues & $\lambda_n = (n\pi/L)^2$, $y_n(x) = \sin(n\pi x/L)$, $n = 1, 2, \dots$ \\
|
||||
Neumann eigenvalues & $\lambda_n = (n\pi/L)^2$, $y_n(x) = \cos(n\pi x/L)$, $n = 0, 1, 2, \dots$ \\
|
||||
SL form & $(py')' + (q + \lambda r)y = 0$ \\
|
||||
Integrating factor & $\mu(x) = \exp\!\left(\int \frac{Q(x)}{P(x)}\,\diff x\right)$ \\
|
||||
SL existence & Infinite real eigenvalues $\lambda_1 < \lambda_2 < \cdots \to \infty$ \\
|
||||
Orthogonality & $\displaystyle\int_a^b \phi_n \phi_m\, r\,\diff x = 0$ for $n \neq m$ \\
|
||||
Expansion coefficients & $\displaystyle c_n = \frac{\int_a^b f\phi_n\,r\,\diff x}{\int_a^b \phi_n^2\,r\,\diff x}$ \\
|
||||
Nonhomogeneous solution & $\displaystyle c_n = \frac{f_n}{(\lambda - \lambda_n)\,\|\phi_n\|^2}$ (if $\lambda \neq \lambda_n$) \\
|
||||
Resonance & If $\lambda = \lambda_n$ and $f_n \neq 0$, no solution exists \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Problem-solving checklist for BVPs.}
|
||||
\begin{enumerate}
|
||||
\item Identify the boundary conditions and classify (Dirichlet, Neumann, mixed).
|
||||
\item Determine the eigenvalues and eigenfunctions of the associated homogeneous problem.
|
||||
\item Verify orthogonality of eigenfunctions with respect to the weight $r(x)$.
|
||||
\item For nonhomogeneous problems, expand $f(x)$ in the eigenfunction basis.
|
||||
\item Check whether $\lambda$ coincides with any eigenvalue (resonance).
|
||||
\item If $\lambda \neq \lambda_n$ for all $n$, compute coefficients using \cref{eq:nonhomogeneous_coefficient}.
|
||||
\item If $\lambda = \lambda_n$, check the solvability condition ($f_n = 0$).
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
@@ -1,841 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch12_heat_equation.tex
|
||||
% Chapter 12: The Heat Equation
|
||||
% =============================================================================
|
||||
|
||||
\section{Heat Equation}
|
||||
\label{ch:heat_equation}
|
||||
|
||||
\subsection{Physical Derivation}
|
||||
\label{sec:ch12_physical_derivation}
|
||||
|
||||
The heat equation is the prototypical parabolic partial differential equation (PDE). It describes how temperature diffuses through a material over time, and it serves as a mathematical model for many other diffusion processes --- from the spread of a pollutant in a river to the flow of electrical charge through a semiconductor.
|
||||
|
||||
\paragraph{Fourier's law of heat conduction.} In 1822, Joseph Fourier established the fundamental law governing heat conduction. Consider a thin rod aligned along the $x$-axis. Let $u(x,t)$ denote the temperature at position $x$ and time $t$. Fourier observed that heat flows from hot regions to cold regions, and that the \textbf{heat flux} $J$ (amount of heat energy flowing per unit area per unit time) is proportional to the temperature gradient:
|
||||
\begin{equation}
|
||||
\label{eq:fouriers_law_1d}
|
||||
J(x,t) = -\kappa\,\pd{u}{x}(x,t).
|
||||
\end{equation}
|
||||
Here $\kappa > 0$ is the \textbf{thermal conductivity} of the material (units: $\mathrm{W/(m{\cdot}K)}$). The minus sign is essential: heat flows in the direction of decreasing temperature, i.e., opposite to the temperature gradient.
|
||||
|
||||
\paragraph{Energy conservation in a rod element.} Now consider a small segment of the rod from $x$ to $x+\Delta x$. The amount of heat energy $E$ contained in this segment is
|
||||
\[
|
||||
E = \rho\,c\,A\,\Delta x \cdot u(x,t),
|
||||
\]
|
||||
where $\rho$ is the mass density of the material, $c$ is the specific heat capacity (energy per unit mass per degree), and $A$ is the cross-sectional area of the rod.
|
||||
|
||||
The rate of change of the energy in this segment must equal the net heat flux into the segment (energy conservation):
|
||||
\[
|
||||
\frac{\diff E}{\diff t} = A\bigl[J(x,t) - J(x+\Delta x, t)\bigr].
|
||||
\]
|
||||
Substituting the expressions:
|
||||
\[
|
||||
\rho\,c\,A\,\Delta x\,\pd{u}{t}(x,t)
|
||||
= A\bigl[J(x,t) - J(x+\Delta x, t)\bigr].
|
||||
\]
|
||||
Divide by $A\,\Delta x$:
|
||||
\[
|
||||
\rho\,c\,\pd{u}{t}(x,t) = -\frac{J(x+\Delta x, t) - J(x,t)}{\Delta x}.
|
||||
\]
|
||||
Taking the limit $\Delta x \to 0$, the right side becomes $-\pd{J}{x}$:
|
||||
\[
|
||||
\rho\,c\,\pd{u}{t} = -\pd{J}{x}.
|
||||
\]
|
||||
|
||||
\paragraph{Derivation of the heat equation.} Substitute Fourier's law \cref{eq:fouriers_law_1d} into the energy conservation equation:
|
||||
\[
|
||||
\rho\,c\,\pd{u}{t} = -\pd{}{x}\bigl(-\kappa\,\pd{u}{x}\bigr)
|
||||
= \kappa\,\frac{\partial^2 u}{\partial x^2}.
|
||||
\]
|
||||
Assuming the material is homogeneous ($\kappa$, $\rho$, and $c$ are constant), we divide by $\rho\,c$ to obtain the \textbf{one-dimensional heat equation}:
|
||||
\begin{equation}
|
||||
\label{eq:heat_equation_1d}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, \qquad 0 < x < L, \;\; t > 0,
|
||||
\end{equation}
|
||||
where
|
||||
\begin{equation}
|
||||
\label{eq:thermal_diffusivity}
|
||||
\alpha = \frac{\kappa}{\rho\,c}
|
||||
\end{equation}
|
||||
is the \textbf{thermal diffusivity} (units: $\mathrm{m^2/s}$).
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Physical interpretation of $\alpha$.} The thermal diffusivity $\alpha$ governs the rate at which temperature disturbances propagate through a material. A large $\alpha$ means heat diffuses quickly (the material is a good conductor relative to its heat capacity). A small $\alpha$ means temperature changes propagate slowly. In the units of the heat equation, the characteristic diffusion time across a distance $L$ is $t_{\mathrm{diff}} \sim L^2/\alpha$.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Higher dimensions.} In three dimensions, Fourier's law becomes $\bm{J} = -\kappa\nabla u$ and energy conservation gives
|
||||
\[
|
||||
\pd{u}{t} = \alpha\,\nabla^2 u = \alpha\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right).
|
||||
\]
|
||||
Throughout this chapter, we focus on the one-dimensional case \cref{eq:heat_equation_1d}, which captures the essential mathematical structure.
|
||||
|
||||
\subsection{Separation of Variables}
|
||||
\label{sec:ch12_separation_of_variables}
|
||||
|
||||
We now solve the heat equation \cref{eq:heat_equation_1d} on a finite rod $[0,L]$ with homogeneous boundary conditions. The primary method is \textbf{separation of variables}.
|
||||
|
||||
\paragraph{The method.} We seek solutions of the form
|
||||
\[
|
||||
u(x,t) = X(x)\,T(t),
|
||||
\]
|
||||
where $X(x)$ depends only on space and $T(t)$ depends only on time. Substitute this ansatz into \cref{eq:heat_equation_1d}:
|
||||
\[
|
||||
X(x)\,T'(t) = \alpha\,X''(x)\,T(t).
|
||||
\]
|
||||
Assuming neither factor vanishes identically, divide by $\alpha\,X(x)\,T(t)$:
|
||||
\begin{equation}
|
||||
\label{eq:separation_step}
|
||||
\frac{T'(t)}{\alpha\,T(t)} = \frac{X''(x)}{X(x)}.
|
||||
\end{equation}
|
||||
The left side depends only on $t$, while the right side depends only on $x$. For this equality to hold for all $x$ and $t$, both sides must equal the same \textbf{separation constant}, which we denote by $-\lambda$.
|
||||
|
||||
\begin{theorem}[Separation of Variables for the Heat Equation]
|
||||
\label{thm:separation_heat}
|
||||
Assuming $u(x,t) = X(x)T(t)$, the heat equation \cref{eq:heat_equation_1d} separates into two ordinary differential equations:
|
||||
\begin{align}
|
||||
T'(t) + \alpha\lambda\,T(t) &= 0, \label{eq:time_ode} \\[6pt]
|
||||
X''(x) + \lambda\,X(x) &= 0, \label{eq:space_ode}
|
||||
\end{align}
|
||||
where $\lambda$ is the separation constant. The choice $\lambda > 0$ is required by the homogeneous boundary conditions and the physical requirement of decay.
|
||||
\end{theorem}
|
||||
|
||||
\paragraph{Justification for the negative sign.} Why do we write the separation constant as $-\lambda$ rather than $+\lambda$? There are three complementary reasons:
|
||||
|
||||
\begin{enumerate}
|
||||
\item \textbf{Physical reasoning:} Temperature disturbances should decay over time, not grow. If we used $+\lambda > 0$, the time equation $T' = \alpha\lambda T$ would yield $T(t) = e^{\alpha\lambda t}$, an exponentially growing solution, which contradicts the second law of thermodynamics.
|
||||
|
||||
\item \textbf{Boundary conditions:} With homogeneous Dirichlet conditions $X(0) = 0$ and $X(L) = 0$, the spatial equation $X'' + \lambda X = 0$ admits nontrivial solutions only for $\lambda > 0$ (as established in the eigenvalue analysis of \cref{ch:boundary_value_problems}). If $\lambda \leq 0$, only the trivial solution $X \equiv 0$ satisfies both boundary conditions.
|
||||
|
||||
\item \textbf{Consistency:} Using $-\lambda$ gives the time equation $T' = -\alpha\lambda T$, yielding $T(t) = e^{-\alpha\lambda t}$, which decays for $\lambda > 0$.
|
||||
\end{enumerate}
|
||||
|
||||
\paragraph{Solving the separated equations.} The time equation \cref{eq:time_ode} is a simple first-order linear ODE:
|
||||
\[
|
||||
T'(t) = -\alpha\lambda\,T(t)
|
||||
\quad\Longrightarrow\quad
|
||||
T(t) = A\,e^{-\alpha\lambda t},
|
||||
\]
|
||||
where $A$ is an arbitrary constant.
|
||||
|
||||
The space equation \cref{eq:space_ode} is exactly the eigenvalue problem studied in \cref{sec:ch11_eigenvalue_problems}. The specific eigenvalues and eigenfunctions depend on the boundary conditions, as we develop in the next subsections.
|
||||
|
||||
\subsection{Dirichlet Boundary Conditions}
|
||||
\label{sec:ch12_dirichlet}
|
||||
|
||||
Consider a rod of length $L$ whose ends are held at zero temperature:
|
||||
\[
|
||||
u(0,t) = 0, \qquad u(L,t) = 0, \qquad t > 0.
|
||||
\]
|
||||
These are \textbf{homogeneous Dirichlet boundary conditions}. Together with an initial temperature distribution
|
||||
\[
|
||||
u(x,0) = f(x), \qquad 0 < x < L,
|
||||
\]
|
||||
we have the initial-boundary value problem:
|
||||
\begin{equation}
|
||||
\label{eq:heat_dirichlet_problem}
|
||||
\begin{cases}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < L, \;\; t > 0, \\[6pt]
|
||||
u(0,t) = 0, \;\; u(L,t) = 0, & t > 0, \\[6pt]
|
||||
u(x,0) = f(x), & 0 < x < L.
|
||||
\end{cases}
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Eigenvalue problem for $X(x)$.} The boundary conditions $u(0,t) = 0$ and $u(L,t) = 0$ imply $X(0) = 0$ and $X(L) = 0$. The spatial ODE is
|
||||
\[
|
||||
X''(x) + \lambda X(x) = 0, \qquad X(0) = 0, \;\; X(L) = 0.
|
||||
\]
|
||||
From \cref{sec:ch11_eigenvalue_problems}, the eigenvalues and eigenfunctions are:
|
||||
\[
|
||||
\lambda_n = \left(\frac{n\pi}{L}\right)^{\!2},
|
||||
\qquad
|
||||
X_n(x) = \sin\!\left(\frac{n\pi x}{L}\right),
|
||||
\qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
|
||||
\paragraph{Product solutions.} For each eigenvalue $\lambda_n$, the corresponding time factor is
|
||||
\[
|
||||
T_n(t) = e^{-\alpha\lambda_n t}
|
||||
= \exp\!\left[-\alpha\left(\frac{n\pi}{L}\right)^{\!2} t\right].
|
||||
\]
|
||||
The product solutions are
|
||||
\[
|
||||
u_n(x,t) = \sin\!\left(\frac{n\pi x}{L}\right)\,
|
||||
\exp\!\left[-\alpha\left(\frac{n\pi}{L}\right)^{\!2} t\right],
|
||||
\qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
Each $u_n(x,t)$ satisfies the PDE and the homogeneous boundary conditions.
|
||||
|
||||
\paragraph{General solution.} By linearity, any linear combination of product solutions is also a solution. We form the \textbf{infinite series} (Fourier sine series in space):
|
||||
\begin{equation}
|
||||
\label{eq:heat_dirichlet_solution}
|
||||
u(x,t) = \sum_{n=1}^{\infty} b_n\,
|
||||
\sin\!\left(\frac{n\pi x}{L}\right)\,
|
||||
\exp\!\left[-\alpha\left(\frac{n\pi}{L}\right)^{\!2} t\right].
|
||||
\end{equation}
|
||||
This series satisfies the PDE and boundary conditions for any choice of coefficients $\{b_n\}$. To determine the coefficients, we apply the initial condition:
|
||||
\[
|
||||
u(x,0) = \sum_{n=1}^{\infty} b_n\,\sin\!\left(\frac{n\pi x}{L}\right)
|
||||
= f(x).
|
||||
\]
|
||||
This is a Fourier sine series for $f(x)$ on $[0,L]$. Using the orthogonality of the sine functions (\cref{sec:ch10_orthogonality}), the coefficients are
|
||||
\[
|
||||
b_n = \frac{2}{L}\int_0^L f(x)\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x.
|
||||
\]
|
||||
|
||||
\begin{keyresult}
|
||||
\label{key:heat_dirichlet}
|
||||
\textbf{Heat equation with homogeneous Dirichlet BCs.} For the problem \cref{eq:heat_dirichlet_problem}, the solution is
|
||||
\[
|
||||
u(x,t) = \sum_{n=1}^{\infty} b_n\,
|
||||
\sin\!\left(\frac{n\pi x}{L}\right)\,
|
||||
e^{-\alpha(n\pi/L)^2 t},
|
||||
\]
|
||||
where
|
||||
\[
|
||||
b_n = \frac{2}{L}\int_0^L f(x)\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x.
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Physical interpretation of the series solution.} Each term in the series corresponds to a \textbf{mode} of the temperature distribution. The $n=1$ mode (the fundamental mode) has the lowest decay rate and dominates the long-time behavior:
|
||||
\[
|
||||
u(x,t) \sim b_1\,\sin\!\left(\frac{\pi x}{L}\right)\,
|
||||
e^{-\alpha(\pi/L)^2 t}
|
||||
\quad\text{as } t \to \infty.
|
||||
\]
|
||||
Higher modes ($n \geq 2$) decay much faster because their decay rates scale as $n^2$. After sufficient time, the temperature profile approaches the shape of the fundamental mode.
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the heat equation on a rod of length $L = \pi$ with zero-temperature ends and initial temperature $f(x) = \sin(2x)$:
|
||||
\[
|
||||
\begin{cases}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
|
||||
u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt]
|
||||
u(x,0) = \sin(2x), & 0 < x < \pi.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
\textbf{Solution.} Here $L = \pi$, so the eigenvalues are $\lambda_n = n^2$ and the eigenfunctions are $X_n(x) = \sin(nx)$. The general solution is
|
||||
\[
|
||||
u(x,t) = \sum_{n=1}^{\infty} b_n\,\sin(nx)\,e^{-\alpha n^2 t}.
|
||||
\]
|
||||
Apply the initial condition:
|
||||
\[
|
||||
u(x,0) = \sum_{n=1}^{\infty} b_n\,\sin(nx) = \sin(2x).
|
||||
\]
|
||||
By orthogonality, $b_n = 0$ for $n \neq 2$ and $b_2 = 1$. Alternatively, compute explicitly:
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\int_0^{\pi} \sin(2x)\,\sin(nx)\,\diff x.
|
||||
\]
|
||||
For $n \neq 2$, the integral vanishes by the orthogonality of sines (\cref{thm:orthogonality}). For $n = 2$:
|
||||
\[
|
||||
b_2 = \frac{2}{\pi}\int_0^{\pi} \sin^2(2x)\,\diff x
|
||||
= \frac{2}{\pi}\cdot\frac{\pi}{2} = 1.
|
||||
\]
|
||||
The solution is
|
||||
\[
|
||||
u(x,t) = e^{-4\alpha t}\,\sin(2x).
|
||||
\]
|
||||
This is a single-mode solution: the temperature profile retains its shape (a half-wave of a sine) and simply decays exponentially in amplitude. The decay rate $4\alpha$ corresponds to the second mode.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the heat equation on a rod of length $L = \pi$ with zero-temperature ends and a triangular initial temperature distribution:
|
||||
\[
|
||||
\begin{cases}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
|
||||
u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt]
|
||||
u(x,0) = x(\pi - x), & 0 < x < \pi.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
\textbf{Solution.} With $L = \pi$, the solution is
|
||||
\[
|
||||
u(x,t) = \sum_{n=1}^{\infty} b_n\,\sin(nx)\,e^{-\alpha n^2 t},
|
||||
\]
|
||||
where
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\int_0^{\pi} x(\pi - x)\,\sin(nx)\,\diff x.
|
||||
\]
|
||||
Expand the integrand:
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\left[\pi\int_0^{\pi} x\sin(nx)\,\diff x
|
||||
- \int_0^{\pi} x^2\sin(nx)\,\diff x\right].
|
||||
\]
|
||||
\textit{First integral.} Integrate by parts with $u = x$, $\diff v = \sin(nx)\,\diff x$:
|
||||
\[
|
||||
\int_0^{\pi} x\sin(nx)\,\diff x
|
||||
= \Bigl[-\frac{x}{n}\cos(nx)\Bigr]_0^{\pi}
|
||||
+ \frac{1}{n}\int_0^{\pi} \cos(nx)\,\diff x
|
||||
= -\frac{\pi}{n}(-1)^n + 0
|
||||
= \frac{\pi}{n}(-1)^{n+1}.
|
||||
\]
|
||||
So the first contribution is
|
||||
\[
|
||||
\pi\cdot\frac{\pi}{n}(-1)^{n+1} = \frac{\pi^2}{n}(-1)^{n+1}.
|
||||
\]
|
||||
|
||||
\textit{Second integral.} Integrate by parts with $u = x^2$, $\diff v = \sin(nx)\,\diff x$:
|
||||
\[
|
||||
\int_0^{\pi} x^2\sin(nx)\,\diff x
|
||||
= \Bigl[-\frac{x^2}{n}\cos(nx)\Bigr]_0^{\pi}
|
||||
+ \frac{2}{n}\int_0^{\pi} x\cos(nx)\,\diff x.
|
||||
\]
|
||||
The boundary term gives $-\dfrac{\pi^2}{n}(-1)^n = \dfrac{\pi^2}{n}(-1)^{n+1}$.
|
||||
For the remaining integral, use integration by parts again with $u = x$, $\diff v = \cos(nx)\,\diff x$:
|
||||
\[
|
||||
\int_0^{\pi} x\cos(nx)\,\diff x
|
||||
= \Bigl[\frac{x}{n}\sin(nx)\Bigr]_0^{\pi}
|
||||
- \frac{1}{n}\int_0^{\pi} \sin(nx)\,\diff x
|
||||
= 0 - \frac{1}{n}\Bigl[-\frac{1}{n}\cos(nx)\Bigr]_0^{\pi}
|
||||
= \frac{1}{n^2}\bigl((-1)^n - 1\bigr).
|
||||
\]
|
||||
So the second integral is
|
||||
\[
|
||||
\int_0^{\pi} x^2\sin(nx)\,\diff x
|
||||
= \frac{\pi^2}{n}(-1)^{n+1} + \frac{2}{n}\cdot\frac{(-1)^n - 1}{n^2}
|
||||
= \frac{\pi^2}{n}(-1)^{n+1} + \frac{2}{n^3}\bigl((-1)^n - 1\bigr).
|
||||
\]
|
||||
|
||||
\textit{Combine.}
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\left[
|
||||
\frac{\pi^2}{n}(-1)^{n+1}
|
||||
- \frac{\pi^2}{n}(-1)^{n+1}
|
||||
- \frac{2}{n^3}\bigl((-1)^n - 1\bigr)
|
||||
\right]
|
||||
= \frac{2}{\pi}\cdot\frac{-2}{n^3}\bigl((-1)^n - 1\bigr)
|
||||
= \frac{4}{\pi n^3}\bigl(1 - (-1)^n\bigr).
|
||||
\]
|
||||
Since $1 - (-1)^n = \begin{cases} 2 & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$:
|
||||
\[
|
||||
b_n = \begin{cases}
|
||||
\dfrac{8}{\pi n^3}, & n \text{ odd}, \\[8pt]
|
||||
0, & n \text{ even}.
|
||||
\end{cases}
|
||||
\]
|
||||
The solution is
|
||||
\[
|
||||
u(x,t) = \frac{8}{\pi}\sum_{k=0}^{\infty}
|
||||
\frac{\sin\bigl((2k+1)x\bigr)}{(2k+1)^3}
|
||||
\exp\!\left[-\alpha(2k+1)^2 t\right].
|
||||
\]
|
||||
Writing out the first few terms:
|
||||
\[
|
||||
u(x,t) = \frac{8}{\pi}\Bigl[
|
||||
\sin(x)\,e^{-\alpha t}
|
||||
+ \frac{\sin(3x)}{27}\,e^{-9\alpha t}
|
||||
+ \frac{\sin(5x)}{125}\,e^{-25\alpha t}
|
||||
+ \cdots\Bigr].
|
||||
\]
|
||||
Notice the rapid decay of higher modes: the $n=3$ term decays $9\times$ faster than the fundamental, and the $n=5$ term decays $25\times$ faster.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
% Axes
|
||||
\draw[->] (0,0) -- (6.5,0) node[right] {$x$};
|
||||
\draw[->] (0,-0.3) -- (0,3.5) node[above] {$u(x,t)$};
|
||||
|
||||
% Rod endpoints
|
||||
\draw[dotted, gray] (0,0) -- (0,3.5);
|
||||
\draw[dotted, gray] (6,0) -- (6,3.5);
|
||||
\node[font=\small, gray] at (0,-0.25) {$0$};
|
||||
\node[font=\small, gray] at (6,-0.25) {$L$};
|
||||
|
||||
% Time t=0: triangular profile
|
||||
\draw[thick, red] (0,0) -- (3,3.2) -- (6,0);
|
||||
\node[font=\small, red, anchor=west] at (4.2,3.3) {$t=0$};
|
||||
|
||||
% Time t=t1: smoothed
|
||||
\draw[thick, orange] (0,0)
|
||||
.. controls (1,2.5) and (2,2.9) .. (3,2.8)
|
||||
.. controls (4,2.7) and (5,1.8) .. (6,0);
|
||||
\node[font=\small, orange, anchor=west] at (4.2,2.9) {$t=t_1$};
|
||||
|
||||
% Time t=t2: further decay
|
||||
\draw[thick, blue!70] (0,0)
|
||||
.. controls (1.5,1.5) and (2.5,1.7) .. (3,1.7)
|
||||
.. controls (3.5,1.6) and (4.5,1.0) .. (6,0);
|
||||
\node[font=\small, blue!70, anchor=west] at (4.2,1.8) {$t=t_2$};
|
||||
|
||||
% Time t=large: nearly zero
|
||||
\draw[thick, gray!60] (0,0)
|
||||
.. controls (1.5,0.5) and (2.5,0.6) .. (3,0.6)
|
||||
.. controls (3.5,0.55) and (4.5,0.3) .. (6,0);
|
||||
\node[font=\small, gray!60, anchor=west] at (4.2,0.7) {$t \to \infty$};
|
||||
|
||||
% Arrow showing time direction
|
||||
\draw[->, >=stealth, thick, teal] (6.3,2.5) -- (6.3,0.5)
|
||||
node[right, font=\footnotesize, teal] {decay};
|
||||
\end{tikzpicture}
|
||||
\caption{Temperature profile evolution for the Dirichlet heat equation. The initial triangular distribution (red) smooths out over time as higher modes decay faster. Eventually all temperature dissipates through the zero-temperature boundaries.}
|
||||
\label{fig:temperature_evolution}
|
||||
\end{figure}
|
||||
|
||||
\subsection{Neumann Boundary Conditions}
|
||||
\label{sec:ch12_neumann}
|
||||
|
||||
Now consider a rod whose ends are \textbf{insulated}, meaning no heat can flow through the endpoints. The boundary conditions are
|
||||
\[
|
||||
\pd{u}{x}(0,t) = 0, \qquad \pd{u}{x}(L,t) = 0, \qquad t > 0.
|
||||
\]
|
||||
These are \textbf{homogeneous Neumann boundary conditions}. Physically, $\pd{u}{x} = 0$ at an endpoint means the temperature gradient vanishes there, so there is no heat flux ($J = -\kappa\,\pd{u}{x} = 0$).
|
||||
|
||||
\paragraph{Eigenvalue analysis.} The spatial ODE with Neumann conditions is
|
||||
\[
|
||||
X''(x) + \lambda X(x) = 0, \qquad X'(0) = 0, \;\; X'(L) = 0.
|
||||
\]
|
||||
As derived in \cref{sec:ch11_eigenvalue_problems}, the eigenvalues and eigenfunctions are:
|
||||
\[
|
||||
\lambda_0 = 0, \;\; X_0(x) = 1,
|
||||
\qquad\text{and}\qquad
|
||||
\lambda_n = \left(\frac{n\pi}{L}\right)^{\!2}, \;\;
|
||||
X_n(x) = \cos\!\left(\frac{n\pi x}{L}\right),
|
||||
\qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
|
||||
\paragraph{The $\lambda = 0$ mode.} The eigenvalue $\lambda_0 = 0$ deserves special attention. The time equation for this mode is
|
||||
\[
|
||||
T_0'(t) + \alpha\cdot 0\cdot T_0(t) = 0
|
||||
\quad\Longrightarrow\quad
|
||||
T_0'(t) = 0
|
||||
\quad\Longrightarrow\quad
|
||||
T_0(t) = \text{constant}.
|
||||
\]
|
||||
This means the $n=0$ mode is a \textbf{steady-state component} that does not decay. Physically, it represents the average temperature of the rod, which is conserved because no heat can escape through the insulated ends.
|
||||
|
||||
\paragraph{General solution.}
|
||||
\begin{equation}
|
||||
\label{eq:heat_neumann_solution}
|
||||
u(x,t) = b_0 + \sum_{n=1}^{\infty} b_n\,
|
||||
\cos\!\left(\frac{n\pi x}{L}\right)\,
|
||||
\exp\!\left[-\alpha\left(\frac{n\pi}{L}\right)^{\!2} t\right].
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Heat equation with homogeneous Neumann BCs.} For the problem
|
||||
\[
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, \qquad
|
||||
u_x(0,t) = 0,\; u_x(L,t) = 0, \qquad
|
||||
u(x,0) = f(x),
|
||||
\]
|
||||
the solution is \cref{eq:heat_neumann_solution} with
|
||||
\[
|
||||
b_0 = \frac{1}{L}\int_0^L f(x)\,\diff x,
|
||||
\qquad
|
||||
b_n = \frac{2}{L}\int_0^L f(x)\,\cos\!\left(\frac{n\pi x}{L}\right)\,\diff x,
|
||||
\quad n \geq 1.
|
||||
\]
|
||||
As $t \to \infty$, the exponentially decaying terms vanish and
|
||||
\[
|
||||
\lim_{t\to\infty} u(x,t) = b_0 = \frac{1}{L}\int_0^L f(x)\,\diff x,
|
||||
\]
|
||||
i.e., the temperature approaches the \textbf{average} of the initial distribution.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Physical interpretation.} With insulated ends, the total heat energy in the rod is conserved. The temperature distribution smooths out as higher-frequency modes decay, but the overall average temperature remains fixed. In the long run, the rod reaches a uniform temperature equal to the initial average.
|
||||
|
||||
\paragraph{Worked example.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the heat equation on a rod of length $L = \pi$ with insulated ends and initial temperature $f(x) = \cos(2x) + 3$:
|
||||
\[
|
||||
\begin{cases}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
|
||||
\pd{u}{x}(0,t) = 0, \;\; \pd{u}{x}(\pi,t) = 0, & t > 0, \\[6pt]
|
||||
u(x,0) = \cos(2x) + 3, & 0 < x < \pi.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
\textbf{Solution.} With $L = \pi$, the solution has the form
|
||||
\[
|
||||
u(x,t) = b_0 + \sum_{n=1}^{\infty} b_n\,\cos(nx)\,e^{-\alpha n^2 t}.
|
||||
\]
|
||||
Compute the coefficients. For $n = 0$:
|
||||
\[
|
||||
b_0 = \frac{1}{\pi}\int_0^{\pi} \bigl(\cos(2x) + 3\bigr)\,\diff x
|
||||
= \frac{1}{\pi}\left[\frac{\sin(2x)}{2} + 3x\right]_0^{\pi}
|
||||
= \frac{1}{\pi}\bigl(0 + 3\pi\bigr) = 3.
|
||||
\]
|
||||
For $n \geq 1$:
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\int_0^{\pi} \bigl(\cos(2x) + 3\bigr)\,\cos(nx)\,\diff x.
|
||||
\]
|
||||
Split the integral:
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\int_0^{\pi} \cos(2x)\cos(nx)\,\diff x
|
||||
+ \frac{2}{\pi}\int_0^{\pi} 3\cos(nx)\,\diff x.
|
||||
\]
|
||||
The second integral vanishes: $\int_0^{\pi} \cos(nx)\,\diff x = 0$ for $n \geq 1$.
|
||||
For the first integral, orthogonality of cosines gives zero for $n \neq 2$ and:
|
||||
\[
|
||||
b_2 = \frac{2}{\pi}\int_0^{\pi} \cos^2(2x)\,\diff x
|
||||
= \frac{2}{\pi}\cdot\frac{\pi}{2} = 1.
|
||||
\]
|
||||
The solution is
|
||||
\[
|
||||
u(x,t) = 3 + e^{-4\alpha t}\cos(2x).
|
||||
\]
|
||||
\textbf{Check.} At $t = 0$: $u(x,0) = 3 + \cos(2x)$. $\checkmark$
|
||||
|
||||
As $t \to \infty$: $u(x,t) \to 3$, which equals $b_0 = \dfrac{1}{\pi}\int_0^{\pi} (\cos(2x) + 3)\,\diff x = 3$. The constant background temperature $3$ is preserved, while the spatially varying $\cos(2x)$ component decays.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Steady-State Solution}
|
||||
\label{sec:ch12_steady_state}
|
||||
|
||||
A \textbf{steady-state solution} is a solution that does not change in time: $\pd{u}{t} = 0$. In the context of the heat equation, this represents the temperature distribution that the system approaches after an infinite amount of time (if such a limit exists).
|
||||
|
||||
\paragraph{Derivation.} Setting $\pd{u}{t} = 0$ in \cref{eq:heat_equation_1d} gives
|
||||
\[
|
||||
0 = \alpha\,\frac{\partial^2 u}{\partial x^2} \quad\Longrightarrow\quad \frac{\partial^2 u}{\partial x^2} = 0.
|
||||
\]
|
||||
Integrating twice:
|
||||
\begin{equation}
|
||||
\label{eq:steady_state_general}
|
||||
u_{\mathrm{ss}}(x) = A\,x + B,
|
||||
\end{equation}
|
||||
where $A$ and $B$ are constants determined by the boundary conditions.
|
||||
|
||||
\paragraph{Dirichlet BCs with nonzero temperatures.} Suppose the ends of the rod are held at fixed (possibly nonzero) temperatures:
|
||||
\[
|
||||
u(0,t) = T_1, \qquad u(L,t) = T_2.
|
||||
\]
|
||||
The steady-state solution satisfies these boundary conditions:
|
||||
\[
|
||||
u_{\mathrm{ss}}(0) = B = T_1,
|
||||
\qquad
|
||||
u_{\mathrm{ss}}(L) = A\,L + T_1 = T_2.
|
||||
\]
|
||||
Solving for $A$:
|
||||
\[
|
||||
A = \frac{T_2 - T_1}{L}.
|
||||
\]
|
||||
Therefore:
|
||||
\begin{equation}
|
||||
\label{eq:steady_state_dirichlet}
|
||||
u_{\mathrm{ss}}(x) = T_1 + \frac{T_2 - T_1}{L}\,x.
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Linear steady-state profile.} For the heat equation with fixed end temperatures $u(0) = T_1$ and $u(L) = T_2$, the steady-state temperature distribution is a \textbf{linear function}:
|
||||
\[
|
||||
u_{\mathrm{ss}}(x) = T_1 + \frac{T_2 - T_1}{L}\,x.
|
||||
\]
|
||||
This represents a uniform temperature gradient from one end to the other. Heat flows from the hotter end to the colder end at a constant rate.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Physical interpretation.} The linear steady state reflects a balance between the fixed boundary temperatures and the diffusion process. Once the gradient is established, heat flows at a constant rate through the rod (constant flux $J = -\kappa(T_2-T_1)/L$), and the temperature profile no longer changes.
|
||||
|
||||
\paragraph{Relation to Neumann BCs.} For insulated ends ($u_x(0) = 0$, $u_x(L) = 0$), the steady state satisfies $A = 0$, giving $u_{\mathrm{ss}}(x) = B$. The constant $B$ equals the average of the initial temperature (as discussed in \cref{sec:ch12_neumann}).
|
||||
|
||||
\subsection{Nonhomogeneous Boundary Conditions}
|
||||
\label{sec:ch12_nonhomogeneous_bc}
|
||||
|
||||
When the boundary conditions are nonhomogeneous --- for example, one end held at a nonzero constant temperature --- we cannot directly apply the separation of variables method, which requires homogeneous BCs. The standard approach is a \textbf{shifting technique}: we decompose the solution into a steady-state part that satisfies the nonhomogeneous boundary conditions and a transient part that satisfies homogeneous boundary conditions.
|
||||
|
||||
\paragraph{The shifting technique.} Consider the problem
|
||||
\[
|
||||
\begin{cases}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < L, \;\; t > 0, \\[6pt]
|
||||
u(0,t) = T_1, \;\; u(L,t) = T_2, & t > 0, \\[6pt]
|
||||
u(x,0) = f(x), & 0 < x < L.
|
||||
\end{cases}
|
||||
\]
|
||||
We seek a solution of the form
|
||||
\[
|
||||
u(x,t) = v(x,t) + \phi(x),
|
||||
\]
|
||||
where $\phi(x)$ is a time-independent function chosen to satisfy the boundary conditions:
|
||||
\[
|
||||
\phi(0) = T_1, \qquad \phi(L) = T_2.
|
||||
\]
|
||||
The simplest choice is the linear steady-state profile \cref{eq:steady_state_dirichlet}:
|
||||
\[
|
||||
\phi(x) = T_1 + \frac{T_2 - T_1}{L}\,x.
|
||||
\]
|
||||
Substituting $u = v + \phi$ into the heat equation:
|
||||
\[
|
||||
\pd{v}{t} = \alpha\,\frac{\partial^2 v}{\partial x^2} + \alpha\,\phi''(x).
|
||||
\]
|
||||
Since $\phi(x)$ is linear, $\phi''(x) = 0$, so $v$ satisfies the same homogeneous heat equation:
|
||||
\[
|
||||
\pd{v}{t} = \alpha\,\frac{\partial^2 v}{\partial x^2}.
|
||||
\]
|
||||
The boundary conditions for $v$ are homogeneous:
|
||||
\[
|
||||
v(0,t) = u(0,t) - \phi(0) = T_1 - T_1 = 0,
|
||||
\qquad
|
||||
v(L,t) = u(L,t) - \phi(L) = T_2 - T_2 = 0.
|
||||
\]
|
||||
The initial condition for $v$ is
|
||||
\[
|
||||
v(x,0) = u(x,0) - \phi(x) = f(x) - \phi(x).
|
||||
\]
|
||||
|
||||
\paragraph{Complete method.}
|
||||
\begin{enumerate}
|
||||
\item Find $\phi(x)$ satisfying the nonhomogeneous BCs (usually the linear steady state).
|
||||
\item Set $u(x,t) = v(x,t) + \phi(x)$.
|
||||
\item Solve the heat equation for $v(x,t)$ with homogeneous Dirichlet BCs and initial condition $v(x,0) = f(x) - \phi(x)$, using the method of \cref{sec:ch12_dirichlet}.
|
||||
\item Recover $u(x,t) = v(x,t) + \phi(x)$.
|
||||
\end{enumerate}
|
||||
|
||||
\paragraph{Worked example.}
|
||||
|
||||
\begin{workedexample}
|
||||
A metal rod of length $L = \pi$ has one end held at $0^\circ$C and the other at $100^\circ$C. Initially the rod is at a uniform temperature of $50^\circ$C. Find the temperature $u(x,t)$ for $t > 0$.
|
||||
|
||||
\textbf{Solution.} The problem is
|
||||
\[
|
||||
\begin{cases}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
|
||||
u(0,t) = 0, \;\; u(\pi,t) = 100, & t > 0, \\[6pt]
|
||||
u(x,0) = 50, & 0 < x < \pi.
|
||||
\end{cases}
|
||||
\]
|
||||
\textit{Step 1: Find $\phi(x)$.} The steady-state solution satisfying $u(0) = 0$ and $u(\pi) = 100$ is
|
||||
\[
|
||||
\phi(x) = 0 + \frac{100 - 0}{\pi}\,x = \frac{100}{\pi}\,x.
|
||||
\]
|
||||
|
||||
\textit{Step 2: Define $v(x,t) = u(x,t) - \phi(x)$.} Then $v$ satisfies
|
||||
\[
|
||||
\begin{cases}
|
||||
\pd{v}{t} = \alpha\,\frac{\partial^2 v}{\partial x^2}, & 0 < x < \pi, \\[6pt]
|
||||
v(0,t) = 0, \;\; v(\pi,t) = 0, \\[6pt]
|
||||
v(x,0) = 50 - \frac{100}{\pi}\,x.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
\textit{Step 3: Solve for $v(x,t)$.} Using the Dirichlet solution from \cref{sec:ch12_dirichlet}:
|
||||
\[
|
||||
v(x,t) = \sum_{n=1}^{\infty} b_n\,\sin(nx)\,e^{-\alpha n^2 t},
|
||||
\]
|
||||
where
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\int_0^{\pi} \left(50 - \frac{100}{\pi}x\right)\sin(nx)\,\diff x.
|
||||
\]
|
||||
Split the integral:
|
||||
\[
|
||||
b_n = \frac{2}{\pi}\cdot 50\int_0^{\pi}\sin(nx)\,\diff x
|
||||
- \frac{200}{\pi^2}\int_0^{\pi}x\sin(nx)\,\diff x.
|
||||
\]
|
||||
The first integral:
|
||||
\[
|
||||
\int_0^{\pi}\sin(nx)\,\diff x = \Bigl[-\frac{\cos(nx)}{n}\Bigr]_0^{\pi}
|
||||
= \frac{1 - (-1)^n}{n}.
|
||||
\]
|
||||
The second integral (from the triangular example above):
|
||||
\[
|
||||
\int_0^{\pi}x\sin(nx)\,\diff x = \frac{\pi}{n}(-1)^{n+1}.
|
||||
\]
|
||||
Therefore:
|
||||
\[
|
||||
b_n = \frac{100}{\pi}\cdot\frac{1 - (-1)^n}{n}
|
||||
- \frac{200}{\pi^2}\cdot\frac{\pi}{n}(-1)^{n+1}
|
||||
= \frac{100}{\pi n}\Bigl[1 - (-1)^n - 2(-1)^{n+1}\Bigr].
|
||||
\]
|
||||
Simplify the bracket: $1 - (-1)^n + 2(-1)^n = 1 + (-1)^n$. So:
|
||||
\[
|
||||
b_n = \frac{100}{\pi n}\bigl(1 + (-1)^n\bigr).
|
||||
\]
|
||||
This is nonzero only for \textbf{even} $n$: if $n$ is odd, $1 + (-1)^n = 0$; if $n$ is even, $1 + (-1)^n = 2$. Let $n = 2k$:
|
||||
\[
|
||||
b_{2k} = \frac{200}{\pi(2k)} = \frac{100}{\pi k},
|
||||
\qquad k = 1, 2, 3, \dots
|
||||
\]
|
||||
|
||||
\textit{Step 4: Recover $u(x,t)$.}
|
||||
\[
|
||||
u(x,t) = \frac{100}{\pi}\,x
|
||||
+ \frac{100}{\pi}\sum_{k=1}^{\infty}
|
||||
\frac{\sin(2kx)}{k}\,e^{-4\alpha k^2 t}.
|
||||
\]
|
||||
As $t \to \infty$, the transient part vanishes and
|
||||
\[
|
||||
u_{\mathrm{ss}}(x) = \frac{100}{\pi}\,x,
|
||||
\]
|
||||
a linear gradient from $0^\circ$C at $x=0$ to $100^\circ$C at $x=\pi$.
|
||||
|
||||
\textbf{Check at $t=0$:}
|
||||
\[
|
||||
u(x,0) = \frac{100}{\pi}\,x + \frac{100}{\pi}\sum_{k=1}^{\infty}\frac{\sin(2kx)}{k}.
|
||||
\]
|
||||
The series $\sum_{k=1}^{\infty}\dfrac{\sin(2kx)}{k}$ is a known Fourier series that equals $\dfrac{\pi - 2x}{2}$ on $(0, \pi)$. Substituting:
|
||||
\[
|
||||
u(x,0) = \frac{100}{\pi}\,x + \frac{100}{\pi}\cdot\frac{\pi - 2x}{2}
|
||||
= \frac{100x}{\pi} + 50 - \frac{100x}{\pi} = 50.
|
||||
\]
|
||||
$\checkmark$ The initial condition is satisfied.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Source Terms}
|
||||
\label{sec:ch12_source_terms}
|
||||
|
||||
The basic heat equation \cref{eq:heat_equation_1d} assumes no internal heat generation. In many practical situations, there is a \textbf{source term} $Q(x,t)$ representing internal heat production (e.g., electrical heating, chemical reactions, nuclear decay):
|
||||
\begin{equation}
|
||||
\label{eq:heat_with_source}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2} + Q(x,t), \qquad 0 < x < L, \;\; t > 0.
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Eigenfunction expansion method.} We assume the solution can be expanded in the eigenfunctions of the spatial problem:
|
||||
\[
|
||||
u(x,t) = \sum_{n=1}^{\infty} T_n(t)\,\sin\!\left(\frac{n\pi x}{L}\right),
|
||||
\]
|
||||
and similarly expand the source term:
|
||||
\[
|
||||
Q(x,t) = \sum_{n=1}^{\infty} Q_n(t)\,\sin\!\left(\frac{n\pi x}{L}\right),
|
||||
\qquad
|
||||
Q_n(t) = \frac{2}{L}\int_0^L Q(x,t)\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x.
|
||||
\]
|
||||
Substituting into \cref{eq:heat_with_source}:
|
||||
\[
|
||||
\sum_{n=1}^{\infty} T_n'(t)\,\sin\!\left(\frac{n\pi x}{L}\right)
|
||||
= \alpha\sum_{n=1}^{\infty} T_n(t)\left(-\frac{n^2\pi^2}{L^2}\right)\,
|
||||
\sin\!\left(\frac{n\pi x}{L}\right)
|
||||
+ \sum_{n=1}^{\infty} Q_n(t)\,\sin\!\left(\frac{n\pi x}{L}\right).
|
||||
\]
|
||||
By orthogonality, equate coefficients of each eigenfunction:
|
||||
\begin{equation}
|
||||
\label{eq:mode_ode_source}
|
||||
T_n'(t) + \alpha\left(\frac{n\pi}{L}\right)^{\!2} T_n(t) = Q_n(t).
|
||||
\end{equation}
|
||||
This is a first-order linear ODE for each mode $T_n(t)$, solvable by the integrating factor method.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Solution of the modal ODE.} The solution to \cref{eq:mode_ode_source} is
|
||||
\[
|
||||
T_n(t) = e^{-\alpha\lambda_n t}\left[
|
||||
T_n(0) + \int_0^t Q_n(\tau)\,e^{\alpha\lambda_n\tau}\,\diff\tau
|
||||
\right],
|
||||
\]
|
||||
where $\lambda_n = (n\pi/L)^2$ and $T_n(0)$ is determined by the initial condition.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Worked example.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve the heat equation with a spatially uniform source on a rod of length $L = \pi$ with zero-temperature ends and initially zero temperature:
|
||||
\[
|
||||
\begin{cases}
|
||||
\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2} + Q_0, & 0 < x < \pi, \;\; t > 0, \\[6pt]
|
||||
u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt]
|
||||
u(x,0) = 0, & 0 < x < \pi,
|
||||
\end{cases}
|
||||
\]
|
||||
where $Q_0$ is a positive constant representing uniform internal heat generation.
|
||||
|
||||
\textbf{Solution.} With $L = \pi$, the eigenfunctions are $\sin(nx)$ and $\lambda_n = n^2$.
|
||||
|
||||
\textit{Expand the source.} For a constant source $Q(x,t) = Q_0$:
|
||||
\[
|
||||
Q_n(t) = \frac{2}{\pi}\int_0^{\pi} Q_0\,\sin(nx)\,\diff x
|
||||
= \frac{2Q_0}{\pi}\Bigl[-\frac{\cos(nx)}{n}\Bigr]_0^{\pi}
|
||||
= \frac{2Q_0}{n\pi}\bigl(1 - (-1)^n\bigr).
|
||||
\]
|
||||
This is nonzero only for odd $n$:
|
||||
\[
|
||||
Q_n = \begin{cases}
|
||||
\dfrac{4Q_0}{n\pi}, & n \text{ odd}, \\[8pt]
|
||||
0, & n \text{ even}.
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
\textit{Solve the modal ODE.} With zero initial temperature, $T_n(0) = 0$. For odd $n$:
|
||||
\[
|
||||
T_n'(t) + \alpha n^2 T_n(t) = \frac{4Q_0}{n\pi}.
|
||||
\]
|
||||
Using the integrating factor $e^{\alpha n^2 t}$:
|
||||
\[
|
||||
\frac{\diff}{\diff t}\Bigl[e^{\alpha n^2 t} T_n(t)\Bigr]
|
||||
= \frac{4Q_0}{n\pi}\,e^{\alpha n^2 t}.
|
||||
\]
|
||||
Integrate from $0$ to $t$:
|
||||
\[
|
||||
e^{\alpha n^2 t} T_n(t) - T_n(0)
|
||||
= \frac{4Q_0}{n\pi}\int_0^t e^{\alpha n^2\tau}\,\diff\tau
|
||||
= \frac{4Q_0}{n\pi}\cdot\frac{e^{\alpha n^2 t} - 1}{\alpha n^2}.
|
||||
\]
|
||||
Since $T_n(0) = 0$:
|
||||
\[
|
||||
T_n(t) = \frac{4Q_0}{n^3\pi\alpha}\Bigl(1 - e^{-\alpha n^2 t}\Bigr).
|
||||
\]
|
||||
|
||||
\textit{Reconstruct the solution.} Letting $n = 2k+1$ for odd modes:
|
||||
\[
|
||||
u(x,t) = \frac{4Q_0}{\pi\alpha}
|
||||
\sum_{k=0}^{\infty}
|
||||
\frac{1 - e^{-\alpha(2k+1)^2 t}}{(2k+1)^3}\,
|
||||
\sin\bigl((2k+1)x\bigr).
|
||||
\]
|
||||
|
||||
\textit{Long-time behavior.} As $t \to \infty$, the exponential terms vanish:
|
||||
\[
|
||||
u_{\mathrm{ss}}(x) = \frac{4Q_0}{\pi\alpha}
|
||||
\sum_{k=0}^{\infty}
|
||||
\frac{\sin\bigl((2k+1)x\bigr)}{(2k+1)^3}.
|
||||
\]
|
||||
This is the Fourier sine series of a quadratic function. In fact, this series equals $\dfrac{Q_0}{2\alpha}\,x(\pi - x)$, which is the parabolic steady-state profile. This makes physical sense: with uniform heat generation and zero-temperature ends, the temperature reaches a parabolic profile --- hottest in the middle, zero at the ends --- with $\dfrac{\diff^2 u_{\mathrm{ss}}}{\diff x^2} = -\dfrac{Q_0}{\alpha}$, consistent with the steady-state equation $\alpha u'' + Q_0 = 0$.
|
||||
|
||||
The full solution is
|
||||
\[
|
||||
u(x,t) = \frac{Q_0}{2\alpha}\,x(\pi - x)
|
||||
- \frac{4Q_0}{\pi\alpha}\sum_{k=0}^{\infty}
|
||||
\frac{e^{-\alpha(2k+1)^2 t}}{(2k+1)^3}
|
||||
\sin\bigl((2k+1)x\bigr),
|
||||
\]
|
||||
showing the approach to the parabolic steady state from an initially cold rod.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch12_summary}
|
||||
|
||||
The heat equation unifies several concepts developed in \cref{ch:fourier_series} and \cref{ch:boundary_value_problems}: separation of variables, eigenvalue problems, orthogonality, and Fourier series expansion. The choice of boundary conditions determines the eigenfunction basis and the qualitative behavior of the solution.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Heat equation solution methods by boundary condition type}
|
||||
\label{tab:ch12_summary}
|
||||
\begin{tabular}{l l p{4.5cm}}
|
||||
\toprule
|
||||
\textbf{BC type} & \textbf{Eigenfunctions} & \textbf{Key features} \\
|
||||
\midrule
|
||||
Dirichlet: &
|
||||
$\sin(n\pi x/L)$, $n = 1, 2, \dots$ &
|
||||
Temperature forced to zero at both ends; all modes decay; $u \to 0$ as $t \to \infty$ \\[12pt]
|
||||
Neumann: &
|
||||
$1$, $\cos(n\pi x/L)$, $n = 0, 1, 2, \dots$ &
|
||||
Insulated ends; $\lambda_0 = 0$ mode persists; $u \to$ average of $f(x)$ as $t \to \infty$ \\[12pt]
|
||||
Mixed (one Dirichlet, one Neumann): &
|
||||
$\sin\!\bigl((n+\frac{1}{2})\pi x/L\bigr)$, $n = 0, 1, 2, \dots$ &
|
||||
Eigenvalues $\lambda_n = \bigl((n+\frac{1}{2})\pi/L\bigr)^2$; all modes decay \\[12pt]
|
||||
Nonhomogeneous Dirichlet: &
|
||||
Shifting: $u = v + \phi$ &
|
||||
$\phi(x)$ = steady-state profile; $v$ solved with homogeneous BCs \\[12pt]
|
||||
Source term $Q(x,t)$: &
|
||||
Eigenfunction expansion &
|
||||
Mode-by-mode ODEs: $T_n' + \alpha\lambda_n T_n = Q_n(t)$; solved via integrating factor \\[12pt]
|
||||
Steady state: &
|
||||
$u_{\mathrm{ss}}(x) = Ax + B$ &
|
||||
Linear profile for Dirichlet BCs; constant for Neumann BCs \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Problem-solving checklist for the heat equation.}
|
||||
\begin{enumerate}
|
||||
\item Write down the PDE, boundary conditions, and initial condition.
|
||||
\item Classify the boundary conditions (homogeneous/nonhomogeneous, Dirichlet/Neumann).
|
||||
\item If BCs are nonhomogeneous, use the shifting technique to homogenize them.
|
||||
\item Identify the eigenfunctions and eigenvalues for the spatial problem.
|
||||
\item Write the general series solution with undetermined coefficients.
|
||||
\item Compute the coefficients from the initial condition using orthogonality.
|
||||
\item If a source term is present, expand it in eigenfunctions and solve the modal ODEs.
|
||||
\item Interpret the long-time behavior of the solution.
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
@@ -1,989 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch13_wave_and_laplace.tex
|
||||
% Chapter 13: Wave Equation and Laplace's Equation
|
||||
% =============================================================================
|
||||
|
||||
\section{Wave and Laplace Equations}
|
||||
\label{ch:wave_and_laplace}
|
||||
|
||||
This chapter covers two fundamental second-order PDEs: the \textbf{wave equation}, which governs vibrations and wave propagation, and \textbf{Laplace's equation}, which describes steady-state phenomena in electrostatics, fluid flow, and heat conduction. Both arise naturally from the separation of variables method developed in \cref{ch:heat_equation}, but they exhibit qualitatively different behavior --- oscillatory versus decaying.
|
||||
|
||||
\subsection{Wave Equation Derivation}
|
||||
\label{sec:ch13_wave_derivation}
|
||||
|
||||
We derive the one-dimensional wave equation from the physics of a vibrating string.
|
||||
|
||||
\paragraph{Physical setup.} Consider a taut, flexible string stretched along the $x$-axis, fixed at its endpoints. Let $y(x,t)$ denote the transverse displacement of the string at position $x$ and time $t$. We assume:
|
||||
|
||||
\begin{itemize}
|
||||
\item The string is perfectly flexible (no resistance to bending).
|
||||
\item The motion is purely transverse (horizontal displacements are negligible).
|
||||
\item The tension $T$ in the string is uniform and remains constant during vibration.
|
||||
\item The string has a uniform linear mass density $\rho$ (mass per unit length).
|
||||
\item Displacements are small, so the angle $\theta$ between the string and the horizontal satisfies $\tan\theta \approx \sin\theta \approx \theta$.
|
||||
\end{itemize}
|
||||
|
||||
\paragraph{Newton's second law for a string element.} Consider a small segment of the string from $x$ to $x + \Delta x$. The mass of this segment is $m = \rho\,\Delta x$. The forces acting on it are the tensions at the two endpoints, directed tangent to the string. Let $\theta$ and $\theta + \Delta\theta$ denote the angles the string makes with the horizontal at the left and right endpoints of the segment, respectively.
|
||||
|
||||
The net transverse force on the segment is
|
||||
\[
|
||||
F_y = T\sin(\theta + \Delta\theta) - T\sin(\theta).
|
||||
\]
|
||||
Applying Newton's second law in the transverse direction:
|
||||
\[
|
||||
\rho\,\Delta x \cdot \frac{\partial^2 y}{\partial t^2} = T\sin(\theta + \Delta\theta) - T\sin(\theta).
|
||||
\]
|
||||
|
||||
\paragraph{Small-angle approximation.} For small displacements, the slope of the string at any point is the tangent of the angle:
|
||||
\[
|
||||
\tan\theta \approx \sin\theta = \pd{y}{x}(x,t).
|
||||
\]
|
||||
Similarly, at the right endpoint:
|
||||
\[
|
||||
\sin(\theta + \Delta\theta) \approx \pd{y}{x}(x+\Delta x, t).
|
||||
\]
|
||||
Substituting into the force equation:
|
||||
\[
|
||||
\rho\,\Delta x \cdot \frac{\partial^2 y}{\partial t^2}
|
||||
= T\left[\pd{y}{x}(x+\Delta x, t) - \pd{y}{x}(x,t)\right].
|
||||
\]
|
||||
Divide both sides by $\rho\,\Delta x$:
|
||||
\[
|
||||
\frac{\partial^2 y}{\partial t^2}
|
||||
= \frac{T}{\rho}\cdot\frac{\pd{y}{x}(x+\Delta x, t) - \pd{y}{x}(x,t)}{\Delta x}.
|
||||
\]
|
||||
Taking the limit $\Delta x \to 0$, the difference quotient becomes a derivative:
|
||||
\[
|
||||
\frac{\partial^2 y}{\partial t^2} = \frac{T}{\rho}\,\frac{\partial^2 y}{\partial x^2}.
|
||||
\]
|
||||
|
||||
\paragraph{The wave equation.} Define the wave speed
|
||||
\[
|
||||
c = \sqrt{\frac{T}{\rho}},
|
||||
\]
|
||||
which has units of velocity. The one-dimensional wave equation is
|
||||
\begin{equation}
|
||||
\label{eq:wave_equation}
|
||||
\frac{\partial^2 y}{\partial t^2} = c^2\,\frac{\partial^2 y}{\partial x^2}, \qquad -\infty < x < \infty, \;\; t > 0.
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Wave speed.} The speed $c = \sqrt{T/\rho}$ depends only on the physical properties of the string. Increasing the tension $T$ increases the wave speed (tighter strings transmit vibrations faster). Increasing the linear density $\rho$ decreases the wave speed (heavier strings respond more sluggishly).
|
||||
\end{keyresult}
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
% Horizontal axis
|
||||
\draw[->] (-0.5,0) -- (7,0) node[right, font=\small] {$x$};
|
||||
% Equilibrium position (dashed)
|
||||
\draw[dashed, gray] (-0.2,0) -- (6.5,0);
|
||||
% String at time t: a wave profile
|
||||
\draw[thick, blue!70] (0,0)
|
||||
.. controls (0.8,1.2) and (1.5,1.4) .. (2,1.2)
|
||||
.. controls (2.5,0.9) and (3,0) .. (3.5,-1.0)
|
||||
.. controls (4,-1.5) and (4.5,-1.3) .. (5,-0.8)
|
||||
.. controls (5.4,-0.3) .. (6,0);
|
||||
|
||||
% String element segment
|
||||
\draw[thick, red] (2.2,0.6) -- (2.8,0.3);
|
||||
\draw[dotted, red, thin] (2.2,0.6) -- (2.2,0);
|
||||
\draw[dotted, red, thin] (2.8,0.3) -- (2.8,0);
|
||||
\node[font=\footnotesize, red, below] at (2.5,-0.15) {$\Delta x$};
|
||||
|
||||
% Tension vectors at left endpoint of element
|
||||
\draw[-{Stealth[length=4mm]}, thick, red] (2.2,0.6) -- +(0.4,0.5)
|
||||
node[font=\footnotesize, red, right] {$T$};
|
||||
\draw[-{Stealth[length=4mm]}, thick, red] (2.2,0.6) -- +(-0.35,-0.45)
|
||||
node[font=\footnotesize, red, below left] {$T$};
|
||||
|
||||
% Tension vectors at right endpoint of element
|
||||
\draw[-{Stealth[length=4mm]}, thick, red] (2.8,0.3) -- +(0.45,0.35)
|
||||
node[font=\footnotesize, red, above right] {$T$};
|
||||
\draw[-{Stealth[length=4mm]}, thick, red] (2.8,0.3) -- +(-0.4,-0.4)
|
||||
node[font=\footnotesize, red, below left] {$T$};
|
||||
|
||||
% Displacement arrow
|
||||
\draw[->, >=stealth, thick, teal] (1.0,0) -- (1.0,1.3);
|
||||
\node[font=\small, teal, right] at (1.1,0.7) {$y(x,t)$};
|
||||
|
||||
% Angles
|
||||
\draw (2.0,0) arc (0:25:0.5);
|
||||
\node[font=\footnotesize] at (2.5,0.2) {$\theta$};
|
||||
\draw (2.6,0) arc (0:15:0.5);
|
||||
\node[font=\footnotesize] at (3.15,0.15) {$\theta+\Delta\theta$};
|
||||
|
||||
% Label
|
||||
\node[font=\small, anchor=south] at (3,-1.8) {Taut string element};
|
||||
\end{tikzpicture}
|
||||
\caption{Derivation of the wave equation. A small segment of the string of length $\Delta x$ is subject to tension forces at its endpoints. The net transverse force drives the vertical acceleration, leading to the wave equation.}
|
||||
\label{fig:string_derivation}
|
||||
\end{figure}
|
||||
|
||||
\paragraph{Initial conditions.} To obtain a unique solution, the wave equation requires two initial conditions: the initial displacement and the initial velocity:
|
||||
\begin{equation}
|
||||
\label{eq:wave_initial_conditions}
|
||||
y(x,0) = f(x), \qquad \pd{y}{t}(x,0) = g(x),
|
||||
\end{equation}
|
||||
where $f(x)$ and $g(x)$ are given functions. This is natural for a second-order equation in time: we need both the position and velocity at $t=0$.
|
||||
|
||||
\subsection{d'Alembert's Solution}
|
||||
\label{sec:ch13_dalembert}
|
||||
|
||||
For the wave equation on an infinite string ($-\infty < x < \infty$), there is a closed-form solution known as d'Alembert's formula. The key insight is that the wave equation naturally factors into two independent propagation directions.
|
||||
|
||||
\paragraph{Characteristic coordinates.} Define new variables
|
||||
\[
|
||||
\xi = x - ct, \qquad \eta = x + ct.
|
||||
\]
|
||||
These are the \textbf{characteristic coordinates}. The coordinate $\xi$ is constant along lines moving to the right with speed $c$, while $\eta$ is constant along lines moving to the left with speed $c$.
|
||||
|
||||
We express the partial derivatives in terms of $\xi$ and $\eta$. By the chain rule:
|
||||
\begin{align*}
|
||||
\pd{y}{x} &= \pd{y}{\xi}\,\frac{\partial\xi}{\partial x} + \pd{y}{\eta}\,\frac{\partial\eta}{\partial x}
|
||||
= \pd{y}{\xi} + \pd{y}{\eta}, \\[6pt]
|
||||
\pd{y}{t} &= \pd{y}{\xi}\,\frac{\partial\xi}{\partial t} + \pd{y}{\eta}\,\frac{\partial\eta}{\partial t}
|
||||
= -c\,\pd{y}{\xi} + c\,\pd{y}{\eta}.
|
||||
\end{align*}
|
||||
Computing the second derivatives:
|
||||
\begin{align*}
|
||||
\frac{\partial^2 y}{\partial x^2}
|
||||
&= \frac{\partial}{\partial x}\Bigl(\pd{y}{\xi} + \pd{y}{\eta}\Bigr)
|
||||
= \frac{\partial^2 y}{\partial \xi^2} + 2\,\frac{\partial^2 y}{\partial \xi\,\partial \eta}
|
||||
+ \frac{\partial^2 y}{\partial \eta^2}, \\[6pt]
|
||||
\frac{\partial^2 y}{\partial t^2}
|
||||
&= \frac{\partial}{\partial t}\Bigl(-c\,\pd{y}{\xi} + c\,\pd{y}{\eta}\Bigr)
|
||||
= -c\Bigl(-c\,\frac{\partial^2 y}{\partial \xi^2} + c\,\frac{\partial^2 y}{\partial \xi\,\partial \eta}\Bigr)
|
||||
+ c\Bigl(-c\,\frac{\partial^2 y}{\partial \eta\,\partial \xi} + c\,\frac{\partial^2 y}{\partial \eta^2}\Bigr) \\[6pt]
|
||||
&= c^2\,\frac{\partial^2 y}{\partial \xi^2} - 2c^2\,\frac{\partial^2 y}{\partial \xi\,\partial \eta}
|
||||
+ c^2\,\frac{\partial^2 y}{\partial \eta^2}.
|
||||
\end{align*}
|
||||
Substitute into the wave equation $\dfrac{\partial^2 y}{\partial t^2} = c^2\,\dfrac{\partial^2 y}{\partial x^2}$:
|
||||
\[
|
||||
c^2\Bigl(y_{\xi\xi} - 2y_{\xi\eta} + y_{\eta\eta}\Bigr)
|
||||
= c^2\Bigl(y_{\xi\xi} + 2y_{\xi\eta} + y_{\eta\eta}\Bigr).
|
||||
\]
|
||||
Cancel $c^2$ and the matching terms $y_{\xi\xi}$ and $y_{\eta\eta}$:
|
||||
\[
|
||||
-2y_{\xi\eta} = 2y_{\xi\eta}
|
||||
\quad\Longrightarrow\quad
|
||||
4y_{\xi\eta} = 0
|
||||
\quad\Longrightarrow\quad
|
||||
\frac{\partial^2 y}{\partial \xi\,\partial \eta} = 0.
|
||||
\]
|
||||
|
||||
\paragraph{Integration.} The reduced equation $y_{\xi\eta} = 0$ is very simple. Integrate with respect to $\eta$:
|
||||
\[
|
||||
\pd{y}{\xi} = F'(\xi),
|
||||
\]
|
||||
where $F'(\xi)$ is an arbitrary function of $\xi$ alone. Now integrate with respect to $\xi$:
|
||||
\[
|
||||
y(\xi, \eta) = F(\xi) + G(\eta),
|
||||
\]
|
||||
where $G(\eta)$ is another arbitrary function (the ``constant of integration'' when integrating with respect to $\xi$). Converting back to $x$ and $t$:
|
||||
\[
|
||||
y(x,t) = F(x - ct) + G(x + ct).
|
||||
\]
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{General solution of the wave equation.} The general solution of \cref{eq:wave_equation} on an infinite string is
|
||||
\[
|
||||
y(x,t) = F(x - ct) + G(x + ct),
|
||||
\]
|
||||
where $F$ and $G$ are arbitrary twice-differentiable functions. The term $F(x-ct)$ represents a \textbf{right-traveling wave} (shape $F$ propagating to the right at speed $c$), and $G(x+ct)$ represents a \textbf{left-traveling wave} (shape $G$ propagating to the left at speed $c$).
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Applying the initial conditions.} We now determine $F$ and $G$ from \cref{eq:wave_initial_conditions}:
|
||||
\begin{align}
|
||||
y(x,0) = F(x) + G(x) &= f(x), \label{eq:IC1} \\[6pt]
|
||||
\pd{y}{t}(x,0) = -c\,F'(x) + c\,G'(x) &= g(x). \label{eq:IC2}
|
||||
\end{align}
|
||||
Integrate \cref{eq:IC2} from some fixed point $x_0$ to $x$:
|
||||
\[
|
||||
-c\bigl[F(x) - F(x_0)\bigr] + c\bigl[G(x) - G(x_0)\bigr]
|
||||
= \int_{x_0}^{x} g(s)\,\diff s.
|
||||
\]
|
||||
Divide by $c$:
|
||||
\[
|
||||
-F(x) + G(x) + \bigl[F(x_0) - G(x_0)\bigr] = \frac{1}{c}\int_{x_0}^{x} g(s)\,\diff s.
|
||||
\]
|
||||
Let $C = F(x_0) - G(x_0)$ be a constant. Now we have a system of two equations:
|
||||
\[
|
||||
\begin{cases}
|
||||
F(x) + G(x) = f(x), \\[4pt]
|
||||
-F(x) + G(x) = \dfrac{1}{c}\displaystyle\int_{x_0}^{x} g(s)\,\diff s - C.
|
||||
\end{cases}
|
||||
\]
|
||||
Adding and subtracting:
|
||||
\[
|
||||
G(x) = \frac{1}{2}\,f(x) + \frac{1}{2c}\int_{x_0}^{x} g(s)\,\diff s - \frac{C}{2},
|
||||
\qquad
|
||||
F(x) = \frac{1}{2}\,f(x) - \frac{1}{2c}\int_{x_0}^{x} g(s)\,\diff s + \frac{C}{2}.
|
||||
\]
|
||||
Substitute into $y(x,t) = F(x-ct) + G(x+ct)$:
|
||||
\[
|
||||
y(x,t) = \frac{1}{2}\Bigl[f(x-ct) + f(x+ct)\Bigr]
|
||||
+ \frac{1}{2c}\Biggl[\int_{x_0}^{x+ct} g(s)\,\diff s
|
||||
- \int_{x_0}^{x-ct} g(s)\,\diff s\Biggr].
|
||||
\]
|
||||
The two integrals combine:
|
||||
\[
|
||||
\int_{x_0}^{x+ct} g(s)\,\diff s - \int_{x_0}^{x-ct} g(s)\,\diff s
|
||||
= \int_{x-ct}^{x+ct} g(s)\,\diff s.
|
||||
\]
|
||||
|
||||
\begin{keyresult}
|
||||
\label{key:dalembert}
|
||||
\textbf{d'Alembert's formula.} For the initial value problem
|
||||
\[
|
||||
\frac{\partial^2 y}{\partial t^2} = c^2\,\frac{\partial^2 y}{\partial x^2},
|
||||
\qquad y(x,0) = f(x), \;\; \pd{y}{t}(x,0) = g(x),
|
||||
\]
|
||||
the solution is
|
||||
\begin{equation}
|
||||
\label{eq:dalembert_formula}
|
||||
y(x,t) = \frac{1}{2}\Bigl[f(x-ct) + f(x+ct)\Bigr]
|
||||
+ \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,\diff s.
|
||||
\end{equation}
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Interpretation of the formula.} The first term, $\frac{1}{2}[f(x-ct) + f(x+ct)]$, splits the initial displacement into two copies, each traveling in opposite directions at speed $c$. The second term accounts for the initial velocity: it averages the initial velocity over the interval $[x-ct, x+ct]$ and distributes its effect to point $x$.
|
||||
|
||||
\begin{theorem}[Domain of Dependence]
|
||||
\label{thm:domain_of_dependence}
|
||||
The value of the solution $y(x,t)$ at any point $(x,t)$ depends only on the initial data $f(s)$ and $g(s)$ in the interval
|
||||
\[
|
||||
[x-ct,\; x+ct].
|
||||
\]
|
||||
This interval is called the \textbf{domain of dependence} of the point $(x,t)$. Information travels at the finite speed $c$; disturbances outside this interval have no influence on the solution at $(x,t)$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
% Axes
|
||||
\draw[->] (-0.5,0) -- (7,0) node[right, font=\small] {$x$};
|
||||
\draw[->] (0,-0.5) -- (0,5) node[above, font=\small] {$t$};
|
||||
|
||||
% Characteristic lines from (x*, t*)
|
||||
\coordinate (P) at (3.5,4);
|
||||
\coordinate (L) at (0.5,0);
|
||||
\coordinate (R) at (6.5,0);
|
||||
|
||||
% Characteristic cone (shaded)
|
||||
\fill[red!10] (P) -- (L) -- (R) -- cycle;
|
||||
|
||||
% Characteristic lines
|
||||
\draw[dashed, thick, red] (L) -- (P) node[midway, left, font=\footnotesize, red] {$x-ct$};
|
||||
\draw[dashed, thick, red] (R) -- (P) node[midway, right, font=\footnotesize, red] {$x+ct$};
|
||||
|
||||
% Domain of dependence interval on t=0
|
||||
\draw[thick, red] (L) -- (R);
|
||||
\node[font=\footnotesize, red, below] at (3.5,-0.15) {$[x-ct,\, x+ct]$};
|
||||
|
||||
% Point (x, t)
|
||||
\filldraw[red] (P) circle (2pt);
|
||||
\node[font=\small, red, anchor=south west] at (P) {$\quad(x,t)$};
|
||||
|
||||
% T=0 label
|
||||
\node[font=\small, below left] at (0,0) {$t=0$};
|
||||
|
||||
% Arrows showing wave propagation
|
||||
\draw[->, >=stealth, thick, blue!60] (1.5,0.15) -- (2.5,1.2)
|
||||
node[font=\footnotesize, blue!60, below right] {right wave};
|
||||
\draw[->, >=stealth, thick, blue!60] (5.5,0.15) -- (4.5,1.2)
|
||||
node[font=\footnotesize, blue!60, below left] {left wave};
|
||||
|
||||
% Label
|
||||
\node[font=\small, anchor=south] at (3.5,-1.2) {Characteristic cone and domain of dependence};
|
||||
\end{tikzpicture}
|
||||
\caption{Domain of dependence for d'Alembert's solution. The solution at point $(x,t)$ depends only on initial data in the interval $[x-ct,\, x+ct]$ on the $x$-axis. The dashed red lines are the characteristic lines $x-ct = \text{const}$ and $x+ct = \text{const}$.}
|
||||
\label{fig:characteristic_cone}
|
||||
\end{figure}
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
A string is initially at rest in the shape $f(x) = e^{-x^2}$. Find the solution for $t > 0$.
|
||||
|
||||
\textbf{Solution.} Here $f(x) = e^{-x^2}$ and $g(x) = 0$ (the string is initially at rest). By d'Alembert's formula:
|
||||
\[
|
||||
y(x,t) = \frac{1}{2}\Bigl[e^{-(x-ct)^2} + e^{-(x+ct)^2}\Bigr].
|
||||
\]
|
||||
\textbf{Interpretation.} The initial Gaussian bump splits into two identical bumps of half the original amplitude. One travels to the right at speed $c$, the other to the left. As $t$ increases, the two bumps separate and the amplitude at any fixed point decays to zero (since the bumps move away).
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
A string initially flat ($f(x) = 0$) receives an impulse: the initial velocity is
|
||||
\[
|
||||
g(x) = \begin{cases}
|
||||
v_0, & 0 < x < L, \\
|
||||
0, & \text{otherwise},
|
||||
\end{cases}
|
||||
\]
|
||||
where $v_0$ is a constant. Find $y(x,t)$ using d'Alembert's formula.
|
||||
|
||||
\textbf{Solution.} Since $f(x) = 0$, only the second term survives:
|
||||
\[
|
||||
y(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,\diff s.
|
||||
\]
|
||||
The integral picks up contributions only where $s \in (0,L)$. The intersection of $[x-ct,\, x+ct]$ with $[0,L]$ depends on the position of the interval relative to $[0,L]$. Consider the case where the entire interval $[x-ct,\, x+ct]$ lies within $[0,L]$, i.e., $0 \leq x-ct$ and $x+ct \leq L$. Then:
|
||||
\[
|
||||
y(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} v_0\,\diff s
|
||||
= \frac{v_0}{2c}\cdot 2ct = v_0\,t.
|
||||
\]
|
||||
In this region, the displacement grows linearly with time.
|
||||
|
||||
For the case where $x-ct < 0$ but $x+ct < L$ (left edge of the wave is outside the impulse region):
|
||||
\[
|
||||
y(x,t) = \frac{1}{2c}\int_{0}^{x+ct} v_0\,\diff s
|
||||
= \frac{v_0}{2c}(x+ct).
|
||||
\]
|
||||
|
||||
The full piecewise solution depends on the relative positions of the interval endpoints. The key takeaway: the impulse generates a trapezoidal wave profile that spreads out at speed $c$ in both directions.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Finite String and Standing Waves}
|
||||
\label{sec:ch13_finite_string}
|
||||
|
||||
When the string has finite length $L$ and is fixed at both ends, waves cannot propagate to infinity. Instead, they reflect at the boundaries, and the superposition of right- and left-traveling waves produces \textbf{standing waves} (normal modes).
|
||||
|
||||
\paragraph{Problem statement.} Consider a string of length $L$ fixed at both ends:
|
||||
\begin{equation}
|
||||
\label{eq:finite_string_problem}
|
||||
\begin{cases}
|
||||
\dfrac{\partial^2 y}{\partial t^2} = c^2\,\dfrac{\partial^2 y}{\partial x^2}, & 0 < x < L, \;\; t > 0, \\[10pt]
|
||||
y(0,t) = 0, \;\; y(L,t) = 0, & t > 0, \\[6pt]
|
||||
y(x,0) = f(x), \;\; \pd{y}{t}(x,0) = g(x), & 0 < x < L.
|
||||
\end{cases}
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Separation of variables.} We seek product solutions $y(x,t) = X(x)\,T(t)$. Substituting into the wave equation:
|
||||
\[
|
||||
X(x)\,T''(t) = c^2\,X''(x)\,T(t).
|
||||
\]
|
||||
Divide by $c^2 X(x)\,T(t)$:
|
||||
\[
|
||||
\frac{T''(t)}{c^2\,T(t)} = \frac{X''(x)}{X(x)} = -\lambda.
|
||||
\]
|
||||
This separates into two ODEs:
|
||||
\begin{align}
|
||||
X''(x) + \lambda\,X(x) &= 0, \label{eq:string_space} \\[6pt]
|
||||
T''(t) + c^2\lambda\,T(t) &= 0. \label{eq:string_time}
|
||||
\end{align}
|
||||
|
||||
The boundary conditions $y(0,t) = 0$ and $y(L,t) = 0$ imply $X(0) = 0$ and $X(L) = 0$. The spatial equation \cref{eq:string_space} with these boundary conditions is exactly the Dirichlet eigenvalue problem studied in \cref{sec:ch11_eigenvalue_problems}. The eigenvalues and eigenfunctions are
|
||||
\[
|
||||
\lambda_n = \left(\frac{n\pi}{L}\right)^{\!2},
|
||||
\qquad
|
||||
X_n(x) = \sin\!\left(\frac{n\pi x}{L}\right),
|
||||
\qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
|
||||
For each $\lambda_n$, the temporal equation \cref{eq:string_time} becomes
|
||||
\[
|
||||
T_n''(t) + \omega_n^2\,T_n(t) = 0,
|
||||
\qquad \text{where } \omega_n = c\sqrt{\lambda_n} = \frac{n\pi c}{L}.
|
||||
\]
|
||||
The general solution is
|
||||
\[
|
||||
T_n(t) = A_n\cos(\omega_n t) + B_n\sin(\omega_n t).
|
||||
\]
|
||||
|
||||
\paragraph{General solution.} The product solutions $y_n(x,t) = X_n(x)\,T_n(t)$ are
|
||||
\[
|
||||
y_n(x,t) = \sin\!\left(\frac{n\pi x}{L}\right)\Bigl[A_n\cos\!\left(\frac{n\pi c t}{L}\right)
|
||||
+ B_n\sin\!\left(\frac{n\pi c t}{L}\right)\Bigr].
|
||||
\]
|
||||
By linearity, the general solution is the infinite sum:
|
||||
\begin{equation}
|
||||
\label{eq:finite_string_solution}
|
||||
y(x,t) = \sum_{n=1}^{\infty} \sin\!\left(\frac{n\pi x}{L}\right)
|
||||
\Bigl[A_n\cos\!\left(\frac{n\pi c t}{L}\right)
|
||||
+ B_n\sin\!\left(\frac{n\pi c t}{L}\right)\Bigr].
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Determining the coefficients.} The initial conditions determine $A_n$ and $B_n$. At $t = 0$:
|
||||
\[
|
||||
y(x,0) = \sum_{n=1}^{\infty} A_n\,\sin\!\left(\frac{n\pi x}{L}\right) = f(x).
|
||||
\]
|
||||
This is a Fourier sine series for $f(x)$ on $[0,L]$. By orthogonality:
|
||||
\begin{equation}
|
||||
\label{eq:An_coefficient}
|
||||
A_n = \frac{2}{L}\int_0^L f(x)\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x.
|
||||
\end{equation}
|
||||
|
||||
For the velocity, differentiate \cref{eq:finite_string_solution} with respect to $t$:
|
||||
\[
|
||||
\pd{y}{t}(x,t) = \sum_{n=1}^{\infty} \sin\!\left(\frac{n\pi x}{L}\right)
|
||||
\Bigl[-A_n\frac{n\pi c}{L}\sin\!\left(\frac{n\pi c t}{L}\right)
|
||||
+ B_n\frac{n\pi c}{L}\cos\!\left(\frac{n\pi c t}{L}\right)\Bigr].
|
||||
\]
|
||||
At $t = 0$:
|
||||
\[
|
||||
\pd{y}{t}(x,0) = \sum_{n=1}^{\infty} B_n\,\frac{n\pi c}{L}\,\sin\!\left(\frac{n\pi x}{L}\right) = g(x).
|
||||
\]
|
||||
By orthogonality:
|
||||
\begin{equation}
|
||||
\label{eq:Bn_coefficient}
|
||||
B_n = \frac{2}{cn\pi}\int_0^L g(x)\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x.
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Finite string with fixed ends.} The solution to \cref{eq:finite_string_problem} is \cref{eq:finite_string_solution} with coefficients given by \cref{eq:An_coefficient} and \cref{eq:Bn_coefficient}.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Normal modes and harmonics.} Each term $n$ in the series \cref{eq:finite_string_solution} is a \textbf{normal mode}:
|
||||
\[
|
||||
y_n(x,t) = \sin\!\left(\frac{n\pi x}{L}\right)\Bigl[A_n\cos(\omega_n t) + B_n\sin(\omega_n t)\Bigr].
|
||||
\]
|
||||
Key properties:
|
||||
\begin{itemize}
|
||||
\item The spatial shape $\sin(n\pi x/L)$ is fixed; the amplitude oscillates in time with angular frequency $\omega_n = n\pi c/L$.
|
||||
\item The nodes (points of zero displacement) are at $x = 0, L/n, 2L/n, \dots, L$. The $n$-th mode has exactly $n-1$ interior nodes.
|
||||
\item The \textbf{fundamental frequency} (first harmonic) is $\omega_1 = \pi c/L$. The $n$-th harmonic has frequency $\omega_n = n\,\omega_1$. This integer relationship is responsible for the musical harmonics we hear from strings.
|
||||
\end{itemize}
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=0.6]
|
||||
% Mode 1: Fundamental
|
||||
\begin{scope}[shift={(0,0)}]
|
||||
\draw[thick, blue!70] (0,0) .. controls (0.5,3) and (1.5,3) .. (2,0);
|
||||
\draw[dashed, gray] (0,0) -- (2,0);
|
||||
\node[font=\small] at (1,-0.5) {\textbf{Mode 1:} $\sin(\pi x/L)$};
|
||||
\node[font=\footnotesize, gray] at (1,-1.0) {1 antinode};
|
||||
\end{scope}
|
||||
|
||||
% Mode 2: Second harmonic
|
||||
\begin{scope}[shift={(0,-2)}]
|
||||
\draw[thick, blue!70] (0,0) .. controls (0.25,2) and (0.75,2) .. (1,0)
|
||||
.. controls (1.25,-2) and (1.75,-2) .. (2,0);
|
||||
\draw[dashed, gray] (0,0) -- (2,0);
|
||||
\node[font=\small] at (1,-0.5) {\textbf{Mode 2:} $\sin(2\pi x/L)$};
|
||||
\node[font=\footnotesize, gray] at (1,-1.0) {2 antinodes, 1 interior node};
|
||||
\end{scope}
|
||||
|
||||
% Mode 3: Third harmonic
|
||||
\begin{scope}[shift={(0,-4)}]
|
||||
\draw[thick, blue!70] (0,0)
|
||||
.. controls (0.17,1.8) and (0.5,1.8) .. (0.67,0)
|
||||
.. controls (0.83,-1.8) and (1.17,-1.8) .. (1.33,0)
|
||||
.. controls (1.5,1.8) and (1.83,1.8) .. (2,0);
|
||||
\draw[dashed, gray] (0,0) -- (2,0);
|
||||
\node[font=\small] at (1,-0.5) {\textbf{Mode 3:} $\sin(3\pi x/L)$};
|
||||
\node[font=\footnotesize, gray] at (1,-1.0) {3 antinodes, 2 interior nodes};
|
||||
\end{scope}
|
||||
|
||||
% Mode 4: Fourth harmonic
|
||||
\begin{scope}[shift={(0,-6)}]
|
||||
\draw[thick, blue!70] (0,0)
|
||||
.. controls (0.125,1.5) and (0.375,1.5) .. (0.5,0)
|
||||
.. controls (0.625,-1.5) and (0.875,-1.5) .. (1,0)
|
||||
.. controls (1.125,1.5) and (1.375,1.5) .. (1.5,0)
|
||||
.. controls (1.625,-1.5) and (1.875,-1.5) .. (2,0);
|
||||
\draw[dashed, gray] (0,0) -- (2,0);
|
||||
\node[font=\small] at (1,-0.5) {\textbf{Mode 4:} $\sin(4\pi x/L)$};
|
||||
\node[font=\footnotesize, gray] at (1,-1.0) {4 antinodes, 3 interior nodes};
|
||||
\end{scope}
|
||||
|
||||
% Axis labels
|
||||
\node[font=\small, rotate=90, anchor=south] at (-0.5,-3) {$y$};
|
||||
\node[font=\small, anchor=north] at (2.5,-0.15) {$x$};
|
||||
\end{tikzpicture}
|
||||
\caption{Normal mode shapes for a string of length $L$ fixed at both ends. The $n$-th mode has the shape $\sin(n\pi x/L)$ with $n$ antinodes and $n-1$ interior nodes. Higher modes oscillate at higher frequencies $\omega_n = n\omega_1$.}
|
||||
\label{fig:normal_modes}
|
||||
\end{figure}
|
||||
|
||||
\paragraph{Worked examples.}
|
||||
|
||||
\begin{workedexample}
|
||||
A guitar string of length $L$ is plucked into a triangular shape:
|
||||
\[
|
||||
f(x) = \begin{cases}
|
||||
\dfrac{2h}{L}\,x, & 0 \leq x \leq \dfrac{L}{2}, \\[10pt]
|
||||
\dfrac{2h}{L}\,(L-x), & \dfrac{L}{2} \leq x \leq L,
|
||||
\end{cases}
|
||||
\]
|
||||
and released from rest ($g(x) = 0$). Find the solution $y(x,t)$.
|
||||
|
||||
\textbf{Solution.} Since $g(x) = 0$, all $B_n = 0$. The solution reduces to
|
||||
\[
|
||||
y(x,t) = \sum_{n=1}^{\infty} A_n\,\sin\!\left(\frac{n\pi x}{L}\right)\cos\!\left(\frac{n\pi c t}{L}\right),
|
||||
\]
|
||||
with
|
||||
\[
|
||||
A_n = \frac{2}{L}\int_0^L f(x)\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x.
|
||||
\]
|
||||
Split the integral at $x = L/2$:
|
||||
\[
|
||||
A_n = \frac{2}{L}\left[\frac{2h}{L}\int_0^{L/2} x\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x
|
||||
+ \frac{2h}{L}\int_{L/2}^{L} (L-x)\,\sin\!\left(\frac{n\pi x}{L}\right)\,\diff x\right].
|
||||
\]
|
||||
Let $k = n\pi/L$ for brevity. Compute the first integral by parts ($u = x$, $\diff v = \sin(kx)\,\diff x$):
|
||||
\[
|
||||
\int_0^{L/2} x\sin(kx)\,\diff x
|
||||
= \Bigl[-\frac{x}{k}\cos(kx)\Bigr]_0^{L/2} + \frac{1}{k}\int_0^{L/2}\cos(kx)\,\diff x
|
||||
= -\frac{L}{2k}\cos\!\left(\frac{n\pi}{2}\right) + \frac{1}{k^2}\sin\!\left(\frac{n\pi}{2}\right).
|
||||
\]
|
||||
For the second integral, substitute $u = L-x$ ($\diff u = -\diff x$):
|
||||
\[
|
||||
\int_{L/2}^{L}(L-x)\sin(kx)\,\diff x
|
||||
= \int_0^{L/2} u\sin(k(L-u))\,\diff u.
|
||||
\]
|
||||
Since $\sin(k(L-u)) = \sin(n\pi - ku) = \sin(n\pi)\cos(ku) - \cos(n\pi)\sin(ku) = (-1)^{n+1}\sin(ku)$, the second integral is $(-1)^{n+1}$ times the first. For a symmetric triangular pluck, the result simplifies to:
|
||||
\[
|
||||
A_n = \frac{8h}{n^2\pi^2}\sin\!\left(\frac{n\pi}{2}\right).
|
||||
\]
|
||||
This is nonzero only for \textbf{odd} $n$ (since $\sin(n\pi/2) = 0$ for even $n$). For $n$ odd, $\sin(n\pi/2) = (-1)^{(n-1)/2}$. Therefore:
|
||||
\[
|
||||
A_n = \begin{cases}
|
||||
\dfrac{8h}{n^2\pi^2}\,(-1)^{(n-1)/2}, & n \text{ odd}, \\[10pt]
|
||||
0, & n \text{ even}.
|
||||
\end{cases}
|
||||
\]
|
||||
The final solution is
|
||||
\[
|
||||
y(x,t) = \frac{8h}{\pi^2}\sum_{k=0}^{\infty}
|
||||
\frac{(-1)^k}{(2k+1)^2}\,
|
||||
\sin\!\left(\frac{(2k+1)\pi x}{L}\right)
|
||||
\cos\!\left(\frac{(2k+1)\pi c t}{L}\right).
|
||||
\]
|
||||
Notice that only odd harmonics are present. This is a general property: symmetric initial conditions excite only odd modes. The amplitude of the $n$-th mode decays as $1/n^2$, so the fundamental dominates strongly.
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
A string of length $L = \pi$ is fixed at both ends. At $t = 0$ it is at its equilibrium position but given an initial velocity $g(x) = \sin(3x)$. Find $y(x,t)$.
|
||||
|
||||
\textbf{Solution.} Here $f(x) = 0$ and $g(x) = \sin(3x)$, so $A_n = 0$ for all $n$. We only need $B_n$:
|
||||
\[
|
||||
B_n = \frac{2}{cn\pi}\int_0^{\pi} \sin(3x)\,\sin(nx)\,\diff x.
|
||||
\]
|
||||
By the orthogonality of sines, this integral is zero unless $n = 3$. For $n = 3$:
|
||||
\[
|
||||
B_3 = \frac{2}{3c\pi}\int_0^{\pi} \sin^2(3x)\,\diff x
|
||||
= \frac{2}{3c\pi}\cdot\frac{\pi}{2} = \frac{1}{3c}.
|
||||
\]
|
||||
The solution is a single mode:
|
||||
\[
|
||||
y(x,t) = \frac{1}{3c}\,\sin(3x)\,\sin(3ct).
|
||||
\]
|
||||
\textbf{Interpretation.} The initial velocity profile already has the shape of the third normal mode, so only the third mode is excited. The string oscillates at the third harmonic frequency $\omega_3 = 3c$ with amplitude $1/(3c)$. The two nodes at $x = 0$ and $x = \pi$ are fixed, and there is one additional node at $x = \pi/3$ and $x = 2\pi/3$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Two-Dimensional Wave Equation}
|
||||
\label{sec:ch13_2d_wave}
|
||||
|
||||
The wave equation generalizes to two spatial dimensions to describe the transverse vibration of a membrane (e.g., a drumhead). Let $u(x,y,t)$ be the displacement of the membrane at position $(x,y)$ and time $t$.
|
||||
|
||||
\begin{equation}
|
||||
\label{eq:wave_2d}
|
||||
\frac{\partial^2 u}{\partial t^2} = c^2\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right)
|
||||
= c^2\,\nabla^2 u,
|
||||
\end{equation}
|
||||
where $\nabla^2$ is the two-dimensional Laplacian.
|
||||
|
||||
\paragraph{Separation on a rectangular membrane.} Consider a rectangular membrane $0 < x < a$, $0 < y < b$ with fixed edges:
|
||||
\[
|
||||
u(0,y,t) = u(a,y,t) = u(x,0,t) = u(x,b,t) = 0.
|
||||
\]
|
||||
We seek product solutions $u(x,y,t) = X(x)\,Y(y)\,T(t)$. Substituting into \cref{eq:wave_2d}:
|
||||
\[
|
||||
X\,Y\,T'' = c^2\,(X''\,Y\,T + X\,Y''\,T).
|
||||
\]
|
||||
Divide by $c^2 X Y T$:
|
||||
\[
|
||||
\frac{T''}{c^2\,T} = \frac{X''}{X} + \frac{Y''}{Y}.
|
||||
\]
|
||||
Since the left side depends only on $t$ and the right side depends only on $x$ and $y$, both must equal a separation constant, which we take as $-\lambda$:
|
||||
\[
|
||||
T'' + c^2\lambda\,T = 0,
|
||||
\qquad
|
||||
\frac{X''}{X} + \frac{Y''}{Y} = -\lambda.
|
||||
\]
|
||||
The spatial equation further separates:
|
||||
\[
|
||||
\frac{X''}{X} = -\mu, \qquad \frac{Y''}{Y} = -\nu, \qquad \mu + \nu = \lambda.
|
||||
\]
|
||||
With fixed edges, we have:
|
||||
\begin{align*}
|
||||
X'' + \mu X &= 0, \;\; X(0) = X(a) = 0
|
||||
\quad\Longrightarrow\quad
|
||||
\mu_m = \left(\frac{m\pi}{a}\right)^{\!2}, \;\; X_m(x) = \sin\!\left(\frac{m\pi x}{a}\right), \\[8pt]
|
||||
Y'' + \nu Y &= 0, \;\; Y(0) = Y(b) = 0
|
||||
\quad\Longrightarrow\quad
|
||||
\nu_n = \left(\frac{n\pi}{b}\right)^{\!2}, \;\; Y_n(y) = \sin\!\left(\frac{n\pi y}{b}\right),
|
||||
\end{align*}
|
||||
for $m, n = 1, 2, 3, \dots$.
|
||||
|
||||
The eigenvalues are
|
||||
\[
|
||||
\lambda_{mn} = \mu_m + \nu_n = \pi^2\left(\frac{m^2}{a^2} + \frac{n^2}{b^2}\right),
|
||||
\]
|
||||
and the angular frequencies are
|
||||
\[
|
||||
\omega_{mn} = c\sqrt{\lambda_{mn}} = c\pi\sqrt{\frac{m^2}{a^2} + \frac{n^2}{b^2}}.
|
||||
\]
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Rectangular membrane.} The general solution is
|
||||
\[
|
||||
u(x,y,t) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
|
||||
\sin\!\left(\frac{m\pi x}{a}\right)\sin\!\left(\frac{n\pi y}{b}\right)
|
||||
\Bigl[A_{mn}\cos(\omega_{mn}t) + B_{mn}\sin(\omega_{mn}t)\Bigr],
|
||||
\]
|
||||
where $\omega_{mn} = c\pi\sqrt{m^2/a^2 + n^2/b^2}$ and the coefficients are determined by the initial displacement and velocity.
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
A rectangular membrane of size $a \times b$ is fixed on all edges. The initial displacement is $u(x,y,0) = \sin(\pi x/a)\sin(\pi y/b)$ and the initial velocity is $u_t(x,y,0) = 0$. Find the solution $u(x,y,t)$.
|
||||
|
||||
\textbf{Solution.} The general solution for a rectangular membrane with fixed edges is
|
||||
\[
|
||||
u(x,y,t) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
|
||||
\sin\!\left(\frac{m\pi x}{a}\right)\sin\!\left(\frac{n\pi y}{b}\right)
|
||||
\Bigl[A_{mn}\cos(\omega_{mn}t) + B_{mn}\sin(\omega_{mn}t)\Bigr],
|
||||
\]
|
||||
where $\omega_{mn} = c\pi\sqrt{m^2/a^2 + n^2/b^2}$.
|
||||
|
||||
The initial velocity condition gives
|
||||
\[
|
||||
u_t(x,y,0) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
|
||||
\omega_{mn}\,B_{mn}\,
|
||||
\sin\!\left(\frac{m\pi x}{a}\right)\sin\!\left(\frac{n\pi y}{b}\right) = 0,
|
||||
\]
|
||||
so by the orthogonality of sine products, $B_{mn} = 0$ for all $m, n$.
|
||||
|
||||
The initial displacement condition gives
|
||||
\[
|
||||
u(x,y,0) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
|
||||
A_{mn}\,\sin\!\left(\frac{m\pi x}{a}\right)\sin\!\left(\frac{n\pi y}{b}\right)
|
||||
= \sin\!\left(\frac{\pi x}{a}\right)\sin\!\left(\frac{\pi y}{b}\right).
|
||||
\]
|
||||
Again by orthogonality, $A_{11} = 1$ and $A_{mn} = 0$ for all $(m,n) \neq (1,1)$.
|
||||
|
||||
The solution is a single mode:
|
||||
\[
|
||||
u(x,y,t) = \sin\!\left(\frac{\pi x}{a}\right)\sin\!\left(\frac{\pi y}{b}\right)
|
||||
\cos\!\left(c\pi\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}\,t\right).
|
||||
\]
|
||||
|
||||
\textbf{Interpretation.} The initial displacement already has the shape of the fundamental mode $(m,n) = (1,1)$, so only the fundamental mode is excited. The membrane oscillates at its lowest eigenfrequency $\omega_{11} = c\pi\sqrt{1/a^2 + 1/b^2}$ with unit amplitude. If the initial displacement had contained higher modes (e.g., a sum of several products of sines), each mode would oscillate independently at its own frequency.
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Discussion.} Unlike the one-dimensional string, the frequencies of a rectangular membrane are \emph{not} integer multiples of a fundamental. The ratio $\omega_{mn}/\omega_{11} = \sqrt{m^2 a^2 + n^2 b^2}\,/\,\sqrt{a^2 + b^2}$ is generally irrational. This is why a drum produces a sound with no clear fundamental pitch --- a \textbf{non-harmonic} spectrum.
|
||||
|
||||
\subsection{Laplace's Equation in Rectangles}
|
||||
\label{sec:ch13_laplace_rectangles}
|
||||
|
||||
We now turn to \textbf{Laplace's equation}, the prototypical elliptic PDE:
|
||||
\begin{equation}
|
||||
\label{eq:laplace}
|
||||
\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.
|
||||
\end{equation}
|
||||
Solutions of Laplace's equation are called \textbf{harmonic functions}. They describe steady-state temperature distributions (no time dependence in the heat equation), electrostatic potentials (no charges), and incompressible, irrotational fluid flow.
|
||||
|
||||
\paragraph{Rectangular domain.} Consider Laplace's equation on a rectangle $0 < x < a$, $0 < y < b$. We need four boundary conditions, one on each edge. A common and pedagogically useful case has three edges held at zero and one edge given by a prescribed function:
|
||||
\begin{equation}
|
||||
\label{eq:laplace_rect_problem}
|
||||
\begin{cases}
|
||||
\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0, & 0 < x < a, \;\; 0 < y < b, \\[10pt]
|
||||
u(0,y) = 0, \;\; u(a,y) = 0, & 0 < y < b, \\[6pt]
|
||||
u(x,0) = 0, & 0 < x < a, \\[6pt]
|
||||
u(x,b) = f(x), & 0 < x < a.
|
||||
\end{cases}
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Separation of variables.} Seek $u(x,y) = X(x)\,Y(y)$. Substituting into Laplace's equation:
|
||||
\[
|
||||
X''\,Y + X\,Y'' = 0
|
||||
\quad\Longrightarrow\quad
|
||||
\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda.
|
||||
\]
|
||||
The spatial ODEs are:
|
||||
\begin{align}
|
||||
X'' + \lambda X &= 0, \qquad X(0) = 0, \;\; X(a) = 0, \label{eq:laplace_X} \\[6pt]
|
||||
Y'' - \lambda Y &= 0, \qquad Y(0) = 0. \label{eq:laplace_Y}
|
||||
\end{align}
|
||||
|
||||
Equation \cref{eq:laplace_X} is the familiar Dirichlet eigenvalue problem on $[0,a]$. The eigenvalues and eigenfunctions are
|
||||
\[
|
||||
\lambda_n = \left(\frac{n\pi}{a}\right)^{\!2},
|
||||
\qquad
|
||||
X_n(x) = \sin\!\left(\frac{n\pi x}{a}\right),
|
||||
\qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
|
||||
For each $\lambda_n$, equation \cref{eq:laplace_Y} becomes
|
||||
\[
|
||||
Y_n'' - \left(\frac{n\pi}{a}\right)^{\!2} Y_n = 0.
|
||||
\]
|
||||
The general solution is
|
||||
\[
|
||||
Y_n(y) = A_n\cosh\!\left(\frac{n\pi y}{a}\right) + B_n\sinh\!\left(\frac{n\pi y}{a}\right).
|
||||
\]
|
||||
The boundary condition $Y(0) = 0$ forces $A_n = 0$, so
|
||||
\[
|
||||
Y_n(y) = B_n\sinh\!\left(\frac{n\pi y}{a}\right).
|
||||
\]
|
||||
|
||||
\paragraph{General solution.} The product solutions are $u_n(x,y) = \sin(n\pi x/a)\,\sinh(n\pi y/a)$. By linearity:
|
||||
\begin{equation}
|
||||
\label{eq:laplace_rect_solution}
|
||||
u(x,y) = \sum_{n=1}^{\infty} A_n\,\sin\!\left(\frac{n\pi x}{a}\right)\sinh\!\left(\frac{n\pi y}{a}\right).
|
||||
\end{equation}
|
||||
Apply the remaining boundary condition $u(x,b) = f(x)$:
|
||||
\[
|
||||
f(x) = u(x,b) = \sum_{n=1}^{\infty} A_n\sinh\!\left(\frac{n\pi b}{a}\right)\sin\!\left(\frac{n\pi x}{a}\right).
|
||||
\]
|
||||
This is a Fourier sine series for $f(x)$ on $[0,a]$. Define $C_n = A_n\sinh(n\pi b/a)$:
|
||||
\[
|
||||
C_n = \frac{2}{a}\int_0^a f(x)\,\sin\!\left(\frac{n\pi x}{a}\right)\,\diff x.
|
||||
\]
|
||||
Therefore:
|
||||
\begin{equation}
|
||||
\label{eq:laplace_rect_coefficients}
|
||||
A_n = \frac{2}{a\,\sinh(n\pi b/a)}\int_0^a f(x)\,\sin\!\left(\frac{n\pi x}{a}\right)\,\diff x.
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Laplace's equation on a rectangle.} For the problem \cref{eq:laplace_rect_problem}, the solution is \cref{eq:laplace_rect_solution} with coefficients given by \cref{eq:laplace_rect_coefficients}.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Worked example.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve Laplace's equation on the square $0 < x < \pi$, $0 < y < \pi$, with
|
||||
\[
|
||||
u(0,y) = u(\pi,y) = u(x,0) = 0, \qquad u(x,\pi) = \sin(2x).
|
||||
\]
|
||||
|
||||
\textbf{Solution.} Here $a = \pi$, $b = \pi$, and $f(x) = \sin(2x)$. The solution is
|
||||
\[
|
||||
u(x,y) = \sum_{n=1}^{\infty} A_n\,\sin(nx)\sinh(ny).
|
||||
\]
|
||||
The coefficients are
|
||||
\[
|
||||
A_n = \frac{2}{\pi\,\sinh(n\pi)}\int_0^{\pi} \sin(2x)\,\sin(nx)\,\diff x.
|
||||
\]
|
||||
By orthogonality of sines, the integral vanishes for $n \neq 2$. For $n = 2$:
|
||||
\[
|
||||
\int_0^{\pi} \sin^2(2x)\,\diff x = \frac{\pi}{2}.
|
||||
\]
|
||||
So $A_2 = \dfrac{2}{\pi\sinh(2\pi)}\cdot\dfrac{\pi}{2} = \dfrac{1}{\sinh(2\pi)}$, and $A_n = 0$ for $n \neq 2$. The solution is
|
||||
\[
|
||||
u(x,y) = \frac{\sinh(2y)}{\sinh(2\pi)}\,\sin(2x).
|
||||
\]
|
||||
\textbf{Check.} At $y = \pi$: $u(x,\pi) = \dfrac{\sinh(2\pi)}{\sinh(2\pi)}\sin(2x) = \sin(2x)$. $\checkmark$\\
|
||||
At $y = 0$: $u(x,0) = \dfrac{\sinh(0)}{\sinh(2\pi)}\sin(2x) = 0$. $\checkmark$\\
|
||||
At $x = 0$ and $x = \pi$: $\sin(2x) = 0$, so $u = 0$. $\checkmark$
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Laplace's Equation in Polar Coordinates}
|
||||
\label{sec:ch13_laplace_polar}
|
||||
|
||||
When the domain has circular geometry (e.g., a disk), polar coordinates are the natural choice. The two-dimensional Laplacian in polar coordinates $(r, \theta)$ is
|
||||
\begin{equation}
|
||||
\label{eq:laplace_polar}
|
||||
\nabla^2 u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\,\frac{\partial u}{\partial r}
|
||||
+ \frac{1}{r^2}\,\frac{\partial^2 u}{\partial \theta^2} = 0.
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Separation of variables.} Assume $u(r,\theta) = R(r)\,\Theta(\theta)$. Substitute into \cref{eq:laplace_polar}:
|
||||
\[
|
||||
R''\Theta + \frac{1}{r}\,R'\Theta + \frac{1}{r^2}\,R\,\Theta'' = 0.
|
||||
\]
|
||||
Divide by $R\Theta/r^2$:
|
||||
\[
|
||||
\frac{r^2 R'' + r R'}{R} = -\frac{\Theta''}{\Theta} = \lambda.
|
||||
\]
|
||||
This gives two ODEs:
|
||||
\begin{align}
|
||||
\Theta'' + \lambda\,\Theta &= 0, \label{eq:angular} \\[6pt]
|
||||
r^2 R'' + r R' - \lambda R &= 0. \label{eq:radial}
|
||||
\end{align}
|
||||
|
||||
\paragraph{Angular equation.} The solution $\Theta(\theta)$ must be \textbf{periodic} with period $2\pi$ (since $\theta$ and $\theta + 2\pi$ represent the same physical point). This periodicity condition restricts $\lambda$ to be a non-negative integer square.
|
||||
|
||||
\textbf{Case 1: $\lambda = n^2$ with $n = 1, 2, 3, \dots$.} The general solution is
|
||||
\[
|
||||
\Theta_n(\theta) = A_n\cos(n\theta) + B_n\sin(n\theta).
|
||||
\]
|
||||
Periodicity is automatically satisfied since $\cos(n(\theta+2\pi)) = \cos(n\theta)$ and similarly for sine.
|
||||
|
||||
\textbf{Case 2: $\lambda = 0$ ($n = 0$).} The angular equation becomes $\Theta'' = 0$, with solution
|
||||
\[
|
||||
\Theta_0(\theta) = C_0 + D_0\,\theta.
|
||||
\]
|
||||
Periodicity $\Theta_0(\theta + 2\pi) = \Theta_0(\theta)$ requires $D_0 = 0$, so $\Theta_0(\theta) = C_0$ (a constant).
|
||||
|
||||
\paragraph{Radial equation.} Equation \cref{eq:radial} is an \textbf{Euler--Cauchy equation} (also called an equidimensional equation). The substitution $R(r) = r^k$ yields:
|
||||
\[
|
||||
r^2\,k(k-1)r^{k-2} + r\,k\,r^{k-1} - \lambda\,r^k = 0
|
||||
\quad\Longrightarrow\quad
|
||||
k^2 - \lambda = 0
|
||||
\quad\Longrightarrow\quad
|
||||
k = \pm\sqrt{\lambda}.
|
||||
\]
|
||||
|
||||
\textbf{Case 1: $\lambda = n^2$ with $n \geq 1$.} The roots are $k = \pm n$, so
|
||||
\[
|
||||
R_n(r) = C_n\,r^n + D_n\,r^{-n}.
|
||||
\]
|
||||
If the domain includes the origin ($r = 0$), we must require that the solution remains \textbf{bounded} there. Since $r^{-n} \to \infty$ as $r \to 0$, we set $D_n = 0$. Thus
|
||||
\[
|
||||
R_n(r) = C_n\,r^n, \qquad n = 1, 2, 3, \dots
|
||||
\]
|
||||
|
||||
\textbf{Case 2: $\lambda = 0$ ($n = 0$).} The radial equation becomes
|
||||
\[
|
||||
r^2 R'' + r R' = 0.
|
||||
\]
|
||||
Let $S = R'$, then $r^2 S' + r S = 0$, which is separable:
|
||||
\[
|
||||
\frac{S'}{S} = -\frac{1}{r}
|
||||
\quad\Longrightarrow\quad
|
||||
\ln|S| = -\ln r + \text{const}
|
||||
\quad\Longrightarrow\quad
|
||||
S = \frac{K}{r}.
|
||||
\]
|
||||
Integrating once more:
|
||||
\[
|
||||
R_0(r) = C_0 + D_0\ln r.
|
||||
\]
|
||||
Boundedness at $r = 0$ requires $D_0 = 0$, so $R_0(r) = C_0$ (a constant).
|
||||
|
||||
\paragraph{General bounded solution on a disk.} Combining the angular and radial parts:
|
||||
\begin{equation}
|
||||
\label{eq:laplace_disk_solution}
|
||||
u(r,\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty} r^n\Bigl[A_n\cos(n\theta) + B_n\sin(n\theta)\Bigr].
|
||||
\end{equation}
|
||||
The factor $A_0/2$ (rather than $A_0$) is a notational convention that makes the $n=0$ coefficient formula consistent with the Fourier series formula.
|
||||
|
||||
\paragraph{Dirichlet problem on a disk.} Suppose the boundary condition on a disk of radius $a$ is
|
||||
\[
|
||||
u(a,\theta) = f(\theta), \qquad 0 \leq \theta < 2\pi,
|
||||
\]
|
||||
where $f(\theta)$ is a given $2\pi$-periodic function. Then:
|
||||
\[
|
||||
f(\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty} a^n\Bigl[A_n\cos(n\theta) + B_n\sin(n\theta)\Bigr].
|
||||
\]
|
||||
This is precisely the Fourier series of $f(\theta)$ on $[0, 2\pi]$. The coefficients are:
|
||||
\begin{align}
|
||||
A_0 &= \frac{1}{\pi}\int_0^{2\pi} f(\theta)\,\diff\theta, \label{eq:A0_disk} \\[6pt]
|
||||
A_n &= \frac{1}{\pi a^n}\int_0^{2\pi} f(\theta)\cos(n\theta)\,\diff\theta, \qquad n \geq 1, \label{eq:An_disk} \\[6pt]
|
||||
B_n &= \frac{1}{\pi a^n}\int_0^{2\pi} f(\theta)\sin(n\theta)\,\diff\theta, \qquad n \geq 1. \label{eq:Bn_disk}
|
||||
\end{align}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Dirichlet problem on a disk.} For Laplace's equation $\nabla^2 u = 0$ on the disk $r < a$ with boundary condition $u(a,\theta) = f(\theta)$, the bounded solution is
|
||||
\[
|
||||
u(r,\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^{\!n}
|
||||
\Bigl[a^n A_n\cos(n\theta) + a^n B_n\sin(n\theta)\Bigr],
|
||||
\]
|
||||
where $a^n A_n$ and $a^n B_n$ are the standard Fourier coefficients of $f(\theta)$ on $[0, 2\pi]$. Equivalently,
|
||||
\[
|
||||
u(r,\theta) = \frac{1}{2\pi}\int_0^{2\pi} f(\phi)\,\diff\phi
|
||||
+ \frac{1}{\pi}\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^{\!n}
|
||||
\int_0^{2\pi} f(\phi)\cos\bigl(n(\phi-\theta)\bigr)\,\diff\phi.
|
||||
\]
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Poisson kernel.} The solution can be written as a single integral using the \textbf{Poisson kernel}:
|
||||
\begin{equation}
|
||||
\label{eq:poisson_kernel}
|
||||
u(r,\theta) = \frac{1}{2\pi}\int_0^{2\pi} P(r,\theta - \phi)\,f(\phi)\,\diff\phi,
|
||||
\end{equation}
|
||||
where the Poisson kernel is
|
||||
\begin{equation}
|
||||
\label{eq:poisson_kernel_formula}
|
||||
P(r,\psi) = \frac{a^2 - r^2}{a^2 - 2ar\cos\psi + r^2}.
|
||||
\end{equation}
|
||||
The Poisson kernel has several important properties: $P(a,\psi) = 0$ for $\psi \neq 0$ (the boundary data is sharply localized), and $P(0,\psi) = 1$ (the value at the center is the average of the boundary data).
|
||||
|
||||
\paragraph{Worked example.}
|
||||
|
||||
\begin{workedexample}
|
||||
Solve Laplace's equation on the disk $r < 2$ with boundary condition
|
||||
\[
|
||||
u(2,\theta) = 3 + 5\cos(2\theta) - 4\sin(3\theta).
|
||||
\]
|
||||
|
||||
\textbf{Solution.} Here $a = 2$ and $f(\theta) = 3 + 5\cos(2\theta) - 4\sin(3\theta)$. This is already in the form of a Fourier series, so we can read off the coefficients directly.
|
||||
|
||||
Compare with the general solution on the boundary:
|
||||
\[
|
||||
f(\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty} 2^n\Bigl[A_n\cos(n\theta) + B_n\sin(n\theta)\Bigr].
|
||||
\]
|
||||
|
||||
From the constant term: $A_0/2 = 3$, so $A_0 = 6$.
|
||||
|
||||
From the $\cos(2\theta)$ term: $2^2 A_2 = 5$, so $A_2 = 5/4$.
|
||||
|
||||
From the $\sin(3\theta)$ term: $2^3 B_3 = -4$, so $B_3 = -4/8 = -1/2$.
|
||||
|
||||
All other coefficients vanish. The solution is
|
||||
\[
|
||||
u(r,\theta) = 3 + \frac{5}{4}\,r^2\cos(2\theta) - \frac{1}{2}\,r^3\sin(3\theta).
|
||||
\]
|
||||
|
||||
\textbf{Check.} At $r = 2$:
|
||||
\[
|
||||
u(2,\theta) = 3 + \frac{5}{4}\cdot 4\,\cos(2\theta) - \frac{1}{2}\cdot 8\,\sin(3\theta)
|
||||
= 3 + 5\cos(2\theta) - 4\sin(3\theta).
|
||||
\]
|
||||
$\checkmark$ The boundary condition is satisfied.
|
||||
|
||||
\textbf{Physical interpretation.} The solution is a superposition of a constant (average value $3$) and two harmonic modes. The $r^2$ and $r^3$ factors cause higher modes to be suppressed near the center --- the temperature (or potential) is smoothest at the origin and picks up spatial variation as you move outward toward the boundary.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch13_summary}
|
||||
|
||||
This chapter has covered two fundamental second-order PDEs: the wave equation (hyperbolic) and Laplace's equation (elliptic). Together with the heat equation (parabolic) from \cref{ch:heat_equation}, these form the three classical types of PDEs.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Chapter summary: wave equation and Laplace's equation}
|
||||
\label{tab:ch13_summary}
|
||||
\begin{tabular}{l l p{5.5cm}}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key formula/method} \\
|
||||
\midrule
|
||||
Wave equation (1D) &
|
||||
$\displaystyle \frac{\partial^2 y}{\partial t^2} = c^2\,\frac{\partial^2 y}{\partial x^2}$, \quad $c = \sqrt{T/\rho}$ \\[14pt]
|
||||
d'Alembert's formula &
|
||||
$\displaystyle y(x,t) = \tfrac{1}{2}[f(x-ct)+f(x+ct)] + \tfrac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds$ \\[14pt]
|
||||
Domain of dependence &
|
||||
Solution at $(x,t)$ depends only on data in $[x-ct,\, x+ct]$ \\[14pt]
|
||||
Finite string (fixed ends) &
|
||||
$\displaystyle y(x,t) = \sum_{n=1}^{\infty}\sin\!\left(\frac{n\pi x}{L}\right)\!\bigl[A_n\cos(\omega_n t)+B_n\sin(\omega_n t)\bigr]$ \\[14pt]
|
||||
Normal mode frequencies &
|
||||
$\omega_n = \dfrac{n\pi c}{L} = n\,\omega_1$ (harmonic series) \\[14pt]
|
||||
2D wave equation &
|
||||
$\displaystyle u_{tt} = c^2(u_{xx}+u_{yy})$; frequencies $\omega_{mn} = c\pi\sqrt{m^2/a^2+n^2/b^2}$ \\[14pt]
|
||||
Laplace's equation &
|
||||
$\nabla^2 u = 0$; solutions are harmonic functions \\[14pt]
|
||||
Laplace on rectangle &
|
||||
$\displaystyle u(x,y) = \sum_{n=1}^{\infty} A_n\sin\!\left(\frac{n\pi x}{a}\right)\sinh\!\left(\frac{n\pi y}{a}\right)$ \\[14pt]
|
||||
Laplace in polar &
|
||||
$\displaystyle u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0$; Euler--Cauchy radial equation \\[14pt]
|
||||
Dirichlet on disk &
|
||||
$\displaystyle u(r,\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty} r^n[A_n\cos(n\theta)+B_n\sin(n\theta)]$ \\[14pt]
|
||||
Poisson kernel &
|
||||
$\displaystyle P(r,\psi) = \dfrac{a^2-r^2}{a^2-2ar\cos\psi+r^2}$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Problem-solving checklist.}
|
||||
\begin{enumerate}
|
||||
\item \textbf{Wave equation, infinite domain:} Use d'Alembert's formula directly.
|
||||
\item \textbf{Wave equation, finite domain:} Use separation of variables to get the standing wave series. Compute Fourier sine coefficients from initial data.
|
||||
\item \textbf{Wave equation, 2D rectangle:} Double separation; the eigenfrequencies $\omega_{mn}$ are generally non-harmonic.
|
||||
\item \textbf{Laplace's equation, rectangle:} Separate in the direction with homogeneous BCs on both ends (gives trig functions); the other direction gives sinh/cosh. Match the nonhomogeneous boundary with a Fourier series.
|
||||
\item \textbf{Laplace's equation, disk:} Use polar coordinates. Angular equation gives integer $n$ from periodicity. Radial equation is Euler--Cauchy; enforce boundedness at $r=0$. Match boundary data with a Fourier series.
|
||||
\item \textbf{Poisson's equation ($\nabla^2 u = f$):} Use eigenfunction expansion (as with the heat equation with a source), expanding both $u$ and $f$ in the appropriate eigenfunction basis.
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
@@ -1,899 +0,0 @@
|
||||
% =============================================================================
|
||||
% ch14_nonlinear_systems.tex
|
||||
% Chapter 14: Nonlinear Systems
|
||||
% =============================================================================
|
||||
|
||||
\section{Nonlinear Systems}
|
||||
\label{ch:nonlinear_systems}
|
||||
|
||||
Linear systems admit a powerful theory built on superposition and eigenvalue analysis
|
||||
(\cref{ch:systems}). When the equations become nonlinear, the principle of superposition
|
||||
breaks down: the sum of two solutions is generally \emph{not} a solution. No universal
|
||||
closed-form solution formula exists for nonlinear systems. Instead, we rely on
|
||||
\textbf{qualitative methods} --- studying the structure of solutions without finding
|
||||
explicit formulas.
|
||||
|
||||
\subsection{Nonlinear Autonomous Systems}
|
||||
\label{sec:ch14_autonomous}
|
||||
|
||||
A \textbf{nonlinear autonomous system} in two variables takes the form
|
||||
\begin{equation}
|
||||
\label{eq:nonlinear_autonomous}
|
||||
\frac{\mathrm{d}x}{\mathrm{d}t} = f(x,y), \qquad
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = g(x,y),
|
||||
\end{equation}
|
||||
where $f$ and $g$ are nonlinear functions. The system is \emph{autonomous} because
|
||||
$f$ and $g$ do not depend explicitly on $t$.
|
||||
|
||||
\paragraph{Why analytical methods fail.}
|
||||
For linear systems $\mathbf{x}' = A\mathbf{x}$, we can construct the general solution
|
||||
from eigenvalues and eigenvectors. For nonlinear systems:
|
||||
\begin{itemize}
|
||||
\item \textbf{No superposition:} If $\mathbf{x}_1(t)$ and $\mathbf{x}_2(t)$ are solutions,
|
||||
$c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t)$ is \emph{not} generally a solution.
|
||||
\item \textbf{No general formula:} There is no algorithm that produces a closed-form
|
||||
solution for an arbitrary nonlinear system.
|
||||
\item \textbf{Rich behavior:} Nonlinear systems exhibit phenomena absent in linear
|
||||
systems, including multiple equilibria, limit cycles, bifurcations, and chaos.
|
||||
\end{itemize}
|
||||
|
||||
The strategy is to understand the system's \textbf{phase portrait} --- a qualitative
|
||||
map of all possible solution trajectories in the $(x,y)$-plane --- by combining
|
||||
local linear analysis near equilibria with global theorems.
|
||||
|
||||
\paragraph{Equilibrium points.}
|
||||
As in \cref{sec:ch03_autonomous} and \cref{sec:ch08_phase_plane}, equilibrium
|
||||
(singular) points are constant solutions where the system comes to rest.
|
||||
|
||||
\begin{definition}[Equilibrium point]
|
||||
An \textbf{equilibrium point} (or \textbf{fixed point}, or \textbf{critical point})
|
||||
of \cref{eq:nonlinear_autonomous} is a point $(x^*,y^*)$ satisfying
|
||||
\[
|
||||
f(x^*,y^*) = 0 \quad\text{and}\quad g(x^*,y^*) = 0.
|
||||
\]
|
||||
At an equilibrium, both derivatives vanish and the solution is stationary:
|
||||
$x(t) \equiv x^*$, $y(t) \equiv y^*$.
|
||||
\end{definition}
|
||||
|
||||
Finding equilibria is an algebraic problem: solve the system of two nonlinear
|
||||
equations $f = 0$, $g = 0$. Unlike the one-dimensional case (\cref{ch:qualitative}),
|
||||
there may be zero, one, or many equilibrium points.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Find all equilibrium points of the system}
|
||||
\[
|
||||
\frac{\mathrm{d}x}{\mathrm{d}t} = x - x^2 - xy, \qquad
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = y - y^2 - xy.
|
||||
\]
|
||||
|
||||
\textbf{Solution.} Set both equations to zero simultaneously:
|
||||
\begin{align}
|
||||
x(1 - x - y) &= 0, \label{eq:ex_eq1} \\
|
||||
y(1 - x - y) &= 0. \label{eq:ex_eq2}
|
||||
\end{align}
|
||||
From \cref{eq:ex_eq1}, either $x = 0$ or $1 - x - y = 0$.
|
||||
|
||||
\textit{Case 1:} $x = 0$. Substituting into \cref{eq:ex_eq2}: $y(1 - 0 - y) = 0$,
|
||||
so $y = 0$ or $y = 1$. This gives equilibria $(0,0)$ and $(0,1)$.
|
||||
|
||||
\textit{Case 2:} $1 - x - y = 0$, i.e.\ $y = 1 - x$. Substituting into
|
||||
\cref{eq:ex_eq2}: $y(0) = 0$, which is satisfied for any $y$. But we must
|
||||
also have $x = 1 - y$ from the same relation. Setting $x = 0$ gives $y = 1$
|
||||
(already found), and setting $y = 0$ gives $x = 1$, yielding the equilibrium $(1,0)$.
|
||||
|
||||
\textit{Verification:} Check $(1,0)$: $f(1,0) = 1 - 1 - 0 = 0$, $g(1,0) = 0 - 0 - 0 = 0$. $\checkmark$
|
||||
|
||||
The system has three equilibrium points: $(0,0)$, $(0,1)$, and $(1,0)$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Jacobian Linearization}
|
||||
\label{sec:ch14_jacobian}
|
||||
|
||||
The central technique for analyzing nonlinear systems near an equilibrium is
|
||||
\textbf{linearization}. Just as in \cref{sec:ch03_stability}, we approximate the
|
||||
nonlinear functions by their first-order Taylor expansions near the equilibrium.
|
||||
This reduces the nonlinear system to a linear one, whose behavior we understand
|
||||
completely from \cref{ch:systems}.
|
||||
|
||||
\paragraph{Derivation via Taylor expansion.}
|
||||
Let $(x^*,y^*)$ be an equilibrium. Introduce deviation variables
|
||||
$u = x - x^*$ and $v = y - y^*$. Expanding $f$ and $g$ in a
|
||||
two-variable Taylor series about $(x^*,y^*)$:
|
||||
\begin{align*}
|
||||
f(x,y) &= f(x^*,y^*) + \pd{f}{x}\Big|_{(x^*,y^*)} (x-x^*)
|
||||
+ \pd{f}{y}\Big|_{(x^*,y^*)} (y-y^*) + O\!\bigl(|u|^2 + |v|^2\bigr), \\
|
||||
g(x,y) &= g(x^*,y^*) + \pd{g}{x}\Big|_{(x^*,y^*)} (x-x^*)
|
||||
+ \pd{g}{y}\Big|_{(x^*,y^*)} (y-y^*) + O\!\bigl(|u|^2 + |v|^2\bigr).
|
||||
\end{align*}
|
||||
Since $(x^*,y^*)$ is an equilibrium, $f(x^*,y^*) = g(x^*,y^*) = 0$.
|
||||
Discarding the higher-order terms gives the \textbf{linearized system}
|
||||
\begin{equation}
|
||||
\label{eq:linearized_system}
|
||||
\begin{pmatrix} u' \\ v' \end{pmatrix}
|
||||
=
|
||||
\underbrace{
|
||||
\begin{pmatrix}
|
||||
\pd{f}{x}\big|_{(x^*,y^*)} & \pd{f}{y}\big|_{(x^*,y^*)} \\[6pt]
|
||||
\pd{g}{x}\big|_{(x^*,y^*)} & \pd{g}{y}\big|_{(x^*,y^*)}
|
||||
\end{pmatrix}
|
||||
}_{\displaystyle = \; J}
|
||||
\begin{pmatrix} u \\ v \end{pmatrix}.
|
||||
\end{equation}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Jacobian matrix and linearization.}
|
||||
The \textbf{Jacobian matrix} of the system $x' = f(x,y)$, $y' = g(x,y)$ is
|
||||
\[
|
||||
J(x,y) =
|
||||
\begin{pmatrix}
|
||||
\pd{f}{x} & \pd{f}{y} \\[6pt]
|
||||
\pd{g}{x} & \pd{g}{y}
|
||||
\end{pmatrix}.
|
||||
\]
|
||||
Evaluated at an equilibrium $(x^*,y^*)$, the linearized system is
|
||||
$\begin{pmatrix} u' \\ v' \end{pmatrix} = J(x^*,y^*) \begin{pmatrix} u \\ v \end{pmatrix}$,
|
||||
where $u = x-x^*$, $v = y-y^*$. Classify $(x^*,y^*)$ by the eigenvalues
|
||||
of $J(x^*,y^*)$ using the phase plane classification from \cref{ch:systems}.
|
||||
\end{keyresult}
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Eigenvalue classification at an equilibrium of a nonlinear system.}
|
||||
|
||||
Let $\lambda_1, \lambda_2$ be the eigenvalues of $J(x^*,y^*)$.
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{l l p{5cm}}
|
||||
\toprule
|
||||
\textbf{Eigenvalues} & \textbf{Type} & \textbf{Stability} \\
|
||||
\midrule
|
||||
Real, both $> 0$ & Unstable node (source) & Unstable \\[4pt]
|
||||
Real, both $< 0$ & Stable node (sink) & Asymptotically stable \\[4pt]
|
||||
Real, opposite signs & Saddle point & Unstable \\[4pt]
|
||||
Complex $\alpha \pm i\beta$, $\alpha > 0$ & Unstable spiral & Unstable \\[4pt]
|
||||
Complex $\alpha \pm i\beta$, $\alpha < 0$ & Stable spiral & Asymptotically stable \\[4pt]
|
||||
Purely imaginary $\pm i\beta$ & Center (inconclusive) & \textit{Linearization inconclusive} \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
The center case is the only one where the linearization does \emph{not}
|
||||
determine the true nonlinear behavior. See \cref{sec:ch14_hartman_grobman} for details.
|
||||
\end{keyresult}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Linearize and classify the equilibria of}
|
||||
\[
|
||||
\frac{\mathrm{d}x}{\mathrm{d}t} = y, \qquad
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = x - x^2.
|
||||
\]
|
||||
\textit{(This is the undamped nonlinear pendulum in phase-plane form.)}
|
||||
|
||||
\textbf{Step 1: Find equilibria.}
|
||||
Set $y = 0$ and $x - x^2 = 0 \implies x(1-x) = 0$.
|
||||
\[
|
||||
\text{Equilibria: } (0,0) \text{ and } (1,0).
|
||||
\]
|
||||
|
||||
\textbf{Step 2: Jacobian.}
|
||||
With $f(x,y) = y$ and $g(x,y) = x - x^2$:
|
||||
\[
|
||||
\pd{f}{x} = 0, \quad \pd{f}{y} = 1, \quad
|
||||
\pd{g}{x} = 1 - 2x, \quad \pd{g}{y} = 0.
|
||||
\]
|
||||
\[
|
||||
J(x,y) =
|
||||
\begin{pmatrix}
|
||||
0 & 1 \\
|
||||
1 - 2x & 0
|
||||
\end{pmatrix}.
|
||||
\]
|
||||
|
||||
\textbf{Step 3: Classify $(0,0)$.}
|
||||
\[
|
||||
J(0,0) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
|
||||
\]
|
||||
Characteristic equation: $\lambda^2 - 1 = 0 \implies \lambda = \pm 1$.
|
||||
Opposite signs $\implies$ \textbf{saddle point} (unstable).
|
||||
|
||||
\textbf{Step 4: Classify $(1,0)$.}
|
||||
\[
|
||||
J(1,0) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
|
||||
\]
|
||||
Characteristic equation: $\lambda^2 + 1 = 0 \implies \lambda = \pm i$.
|
||||
Purely imaginary eigenvalues $\implies$ \textbf{center by linearization}.
|
||||
The linearization predicts closed orbits, but the nonlinear behavior
|
||||
requires further analysis (confirmed below to be a true center).
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Linearize and classify the equilibria of the competing species model}
|
||||
\[
|
||||
\frac{\mathrm{d}x}{\mathrm{d}t} = 2x - x^2 - xy, \qquad
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = y - y^2 - xy.
|
||||
\]
|
||||
|
||||
\textbf{Step 1: Find equilibria.}
|
||||
\begin{align}
|
||||
x(2 - x - y) &= 0, \label{eq:comp1} \\
|
||||
y(1 - x - y) &= 0. \label{eq:comp2}
|
||||
\end{align}
|
||||
From \cref{eq:comp1}, either $x = 0$ or $x + y = 2$.
|
||||
|
||||
\textit{Case 1:} $x = 0$. From \cref{eq:comp2}: $y(1-y) = 0$, giving $y = 0$ or $y = 1$.
|
||||
Equilibria: $(0,0)$ and $(0,1)$.
|
||||
|
||||
\textit{Case 2:} $x + y = 2$, so $y = 2 - x$. From \cref{eq:comp2}: $y(1 - x - y) = y(1 - 2) = -y = 0$,
|
||||
so $y = 0$, which gives $x = 2$. Equilibrium: $(2,0)$.
|
||||
|
||||
The three equilibria are $(0,0)$, $(0,1)$, and $(2,0)$.
|
||||
|
||||
\textbf{Step 2: Jacobian.}
|
||||
With $f(x,y) = 2x - x^2 - xy$ and $g(x,y) = y - y^2 - xy$:
|
||||
\[
|
||||
J(x,y) =
|
||||
\begin{pmatrix}
|
||||
2 - 2x - y & -x \\[4pt]
|
||||
-y & 1 - 2y - x
|
||||
\end{pmatrix}.
|
||||
\]
|
||||
|
||||
\textbf{Step 3: Classify $(0,0)$.}
|
||||
\[
|
||||
J(0,0) = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix},
|
||||
\quad \lambda_1 = 2,\; \lambda_2 = 1.
|
||||
\]
|
||||
Both positive $\implies$ \textbf{unstable node (source)}. Both species die out is unstable:
|
||||
a small introduction of either population grows.
|
||||
|
||||
\textbf{Step 4: Classify $(0,1)$.}
|
||||
\[
|
||||
J(0,1) = \begin{pmatrix} 2-1 & 0 \\ -1 & 1-2 \end{pmatrix}
|
||||
= \begin{pmatrix} 1 & 0 \\ -1 & -1 \end{pmatrix}.
|
||||
\]
|
||||
Since $J$ is lower triangular, eigenvalues are the diagonal entries:
|
||||
$\lambda_1 = 1$, $\lambda_2 = -1$.
|
||||
Opposite signs $\implies$ \textbf{saddle point} (unstable).
|
||||
Species~$y$ alone at carrying capacity is unstable to invasion by species~$x$.
|
||||
|
||||
\textbf{Step 5: Classify $(2,0)$.}
|
||||
\[
|
||||
J(2,0) = \begin{pmatrix} 2-4 & -2 \\ 0 & 1-2 \end{pmatrix}
|
||||
= \begin{pmatrix} -2 & -2 \\ 0 & -1 \end{pmatrix}.
|
||||
\]
|
||||
Upper triangular: $\lambda_1 = -2$, $\lambda_2 = -1$.
|
||||
Both negative $\implies$ \textbf{stable node (sink)}.
|
||||
Species~$x$ alone at carrying capacity is stable; species~$y$ cannot invade.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Trace-Determinant Classification}
|
||||
\label{sec:ch14_trace_determinant}
|
||||
|
||||
Just as for linear systems (\cref{sec:ch08_trace_det}), the trace and determinant
|
||||
of the Jacobian at an equilibrium provide a quick classification without computing
|
||||
eigenvalues explicitly.
|
||||
|
||||
For $J = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ at an equilibrium:
|
||||
\[
|
||||
\tau = \tr(J) = a + d, \qquad
|
||||
\Delta = \det(J) = ad - bc.
|
||||
\]
|
||||
The eigenvalues satisfy $\lambda^2 - \tau\lambda + \Delta = 0$ with
|
||||
discriminant $D = \tau^2 - 4\Delta$.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Trace-determinant classification for nonlinear equilibria.}
|
||||
Evaluate $\tau = \tr(J)$ and $\Delta = \det(J)$ at the equilibrium point.
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{l l l}
|
||||
\toprule
|
||||
\textbf{Region} & \textbf{Conditions} & \textbf{Classification} \\
|
||||
\midrule
|
||||
$\Delta < 0$ & Below $\tau$-axis & Saddle (unstable) \\[4pt]
|
||||
$\Delta > 0$, $\tau > 0$, $D > 0$ & Right, above parabola & Unstable node \\[4pt]
|
||||
$\Delta > 0$, $\tau < 0$, $D > 0$ & Left, above parabola & Stable node \\[4pt]
|
||||
$\Delta > 0$, $\tau > 0$, $D < 0$ & Right, below parabola & Unstable spiral \\[4pt]
|
||||
$\Delta > 0$, $\tau < 0$, $D < 0$ & Left, below parabola & Stable spiral \\[4pt]
|
||||
$\Delta > 0$, $\tau = 0$ & On positive $\Delta$-axis & Center (inconclusive) \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{The trace-determinant plane.}
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.85]
|
||||
% Axes
|
||||
\draw[->, thick] (-5,0) -- (5,0) node[right] {$\tau$ (trace)};
|
||||
\draw[->, thick] (0,-1) -- (0,5) node[above] {$\Delta$ (det)};
|
||||
|
||||
% Parabola \Delta = \tau^2 / 4
|
||||
\draw[thick, dashed, red!80] plot[domain=-4.5:4.5, samples=100, smooth, variable=\t]
|
||||
({\t}, {\t*\t/4});
|
||||
\node[red!80, font=\footnotesize, anchor=west] at (3.8, 4.3) {$\Delta = \tau^2/4$};
|
||||
|
||||
% Fill saddle region (below \tau-axis)
|
||||
\fill[pattern=north west lines, pattern color=red!25] (-5,0) rectangle (5,-0.45);
|
||||
|
||||
% Stability boundary (\tau = 0 line)
|
||||
\draw[dashed, thick, blue!60] (0,0) -- (0,5);
|
||||
|
||||
% Labels for regions
|
||||
\node[font=\footnotesize, align=center] at (0,-0.25) {\textbf{Saddle}\\$(\Delta < 0)$};
|
||||
|
||||
\node[font=\footnotesize, align=center] at (-3,3.5) {\textbf{Stable node}\\$(\tau < 0,\; D > 0)$};
|
||||
\node[font=\footnotesize, align=center] at (3,3.5) {\textbf{Unstable node}\\$(\tau > 0,\; D > 0)$};
|
||||
|
||||
\node[font=\footnotesize, align=center] at (-2,1.1) {\textbf{Stable spiral}\\$(\tau < 0,\; D < 0)$};
|
||||
\node[font=\footnotesize, align=center] at (2,1.1) {\textbf{Unstable spiral}\\$(\tau > 0,\; D < 0)$};
|
||||
|
||||
\node[font=\footnotesize, align=center] at (0,4.5) {\textbf{Center}\\$(\tau = 0,\; \Delta > 0)$};
|
||||
|
||||
% Origin
|
||||
\filldraw[black] (0,0) circle (1.5pt);
|
||||
\node[font=\scriptsize, below left] at (0,0) {$(0,0)$};
|
||||
|
||||
% Parabola tip label
|
||||
\node[font=\scriptsize, red!80, align=center] at (0,0.15) {parabola};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
The parabola $\Delta = \tau^2/4$ separates real from complex eigenvalues.
|
||||
The $\Delta$-axis ($\tau = 0$) separates stable from unstable systems.
|
||||
The $\tau$-axis ($\Delta = 0$) separates saddles from nodes/spirals/centers.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Classify the equilibria of the competing species model from
|
||||
the previous section using the trace-determinant plane.}
|
||||
|
||||
Recall the Jacobian:
|
||||
\[
|
||||
J(x,y) =
|
||||
\begin{pmatrix}
|
||||
2 - 2x - y & -x \\[4pt]
|
||||
-y & 1 - 2y - x
|
||||
\end{pmatrix}.
|
||||
\]
|
||||
|
||||
\textbf{At $(0,0)$:}
|
||||
\[
|
||||
\tau = 2 + 1 = 3 > 0, \qquad \Delta = (2)(1) - (0) = 2 > 0.
|
||||
\]
|
||||
Discriminant: $D = 9 - 8 = 1 > 0$.
|
||||
Region: $\tau > 0$, $\Delta > 0$, $D > 0$ $\implies$ \textbf{unstable node}. Consistent with $\lambda = 2, 1$.
|
||||
|
||||
\textbf{At $(0,1)$:}
|
||||
\[
|
||||
\tau = 1 + (-1) = 0, \qquad \Delta = (1)(-1) - (0) = -1 < 0.
|
||||
\]
|
||||
Region: $\Delta < 0$ $\implies$ \textbf{saddle point}. Consistent with $\lambda = \pm 1$.
|
||||
|
||||
\textbf{At $(2,0)$:}
|
||||
\[
|
||||
\tau = (-2) + (-1) = -3 < 0, \qquad \Delta = (-2)(-1) - (0) = 2 > 0.
|
||||
\]
|
||||
Discriminant: $D = 9 - 8 = 1 > 0$.
|
||||
Region: $\tau < 0$, $\Delta > 0$, $D > 0$ $\implies$ \textbf{stable node}. Consistent with $\lambda = -2, -1$.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Hartman--Grobman Theorem}
|
||||
\label{sec:ch14_hartman_grobman}
|
||||
|
||||
The linearization tells us about the behavior of the nonlinear system in a
|
||||
\emph{small neighborhood} of the equilibrium. The Hartman--Grobman theorem
|
||||
provides the rigorous justification for this connection.
|
||||
|
||||
\begin{theorem}[Hartman--Grobman]
|
||||
\label{thm:hartman_grobman}
|
||||
Let $(x^*,y^*)$ be an equilibrium point of the autonomous system
|
||||
$x' = f(x,y)$, $y' = g(x,y)$, where $f$ and $g$ are $C^1$ (continuously differentiable).
|
||||
Let $J$ be the Jacobian matrix evaluated at $(x^*,y^*)$.
|
||||
|
||||
If $J$ has \textbf{no eigenvalues with zero real part} (i.e., the equilibrium is
|
||||
\textbf{hyperbolic}), then there exists a neighborhood of $(x^*,y^*)$ in which
|
||||
the nonlinear system is \textbf{topologically conjugate} to its linearization
|
||||
$u' = a u + b v$, $v' = c u + d v$.
|
||||
|
||||
In particular, the local phase portrait of the nonlinear system near $(x^*,y^*)$
|
||||
has the same topological structure as that of the linearized system.
|
||||
\end{theorem}
|
||||
|
||||
\paragraph{What topological conjugacy means.}
|
||||
Two systems are topologically conjugate if there exists a continuous, invertible
|
||||
change of coordinates (a homeomorphism) that maps the trajectories of one system
|
||||
onto the trajectories of the other, preserving the direction of time. Practically,
|
||||
this means:
|
||||
\begin{itemize}
|
||||
\item A stable node of the linearization corresponds to a stable node of the nonlinear system.
|
||||
\item A saddle of the linearization corresponds to a saddle of the nonlinear system.
|
||||
\item The qualitative flow pattern (arrows, attraction, repulsion) is preserved.
|
||||
\end{itemize}
|
||||
|
||||
The actual shapes of trajectories may differ --- a nonlinear trajectory near
|
||||
a stable spiral may not be a perfect logarithmic spiral --- but the
|
||||
essential features (spiraling inward, counterclockwise direction, etc.) are identical.
|
||||
|
||||
\paragraph{Limitations: non-hyperbolic equilibria.}
|
||||
The theorem \textbf{does not apply} when the Jacobian has eigenvalues with zero real part:
|
||||
\begin{itemize}
|
||||
\item \textbf{Centers} ($\lambda = \pm i\beta$): The nonlinear system near a center
|
||||
can be a center, a stable spiral, or an unstable spiral. The linearization
|
||||
alone cannot distinguish between these cases. Higher-order terms in the
|
||||
Taylor expansion must be examined.
|
||||
\item \textbf{Non-hyperbolic points} with real zero eigenvalues ($\lambda = 0$):
|
||||
The dynamics can be extremely sensitive to nonlinear terms.
|
||||
Examples include semi-stable equilibria and bifurcation points.
|
||||
\end{itemize}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Hartman--Grobman applies: stable spiral.}
|
||||
|
||||
Consider $x' = -x + y + x(x^2+y^2)$, $y' = -x - y + y(x^2+y^2)$.
|
||||
|
||||
\textit{Equilibrium:} $(0,0)$ (the only one easily found).
|
||||
|
||||
\textit{Jacobian at $(0,0)$:}
|
||||
\[
|
||||
J(0,0) = \begin{pmatrix} -1 & 1 \\ -1 & -1 \end{pmatrix},
|
||||
\quad \tau = -2,\; \Delta = 2,\; D = 4-8 = -4 < 0.
|
||||
\]
|
||||
Eigenvalues: $\lambda = -1 \pm i$. Both have nonzero real part
|
||||
($\Re(\lambda) = -1 \neq 0$).
|
||||
|
||||
\textbf{Conclusion:} The equilibrium is hyperbolic. By the Hartman--Grobman
|
||||
theorem, the nonlinear system near $(0,0)$ is a \textbf{stable spiral},
|
||||
topologically equivalent to its linearization.
|
||||
|
||||
(In fact, the nonlinear terms $x(x^2+y^2)$ and $y(x^2+y^2)$ cause
|
||||
outward spiraling for large radii, creating a stable limit cycle ---
|
||||
but near the origin, the local portrait is exactly a stable spiral.)
|
||||
\end{workedexample}
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Hartman--Grobman does not apply: center vs.\ spiral.}
|
||||
|
||||
Consider the two systems:
|
||||
\begin{align}
|
||||
\text{(A)} \quad x' &= y, \quad y' = -x, \\
|
||||
\text{(B)} \quad x' &= y + x(x^2+y^2), \quad y' = -x + y(x^2+y^2).
|
||||
\end{align}
|
||||
|
||||
\textit{Jacobian at $(0,0)$ for both systems:}
|
||||
\[
|
||||
J(0,0) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},
|
||||
\quad \lambda = \pm i.
|
||||
\]
|
||||
Purely imaginary eigenvalues $\implies$ the equilibrium is \textbf{not hyperbolic}.
|
||||
Hartman--Grobman \textbf{does not apply}.
|
||||
|
||||
\textit{System (A):} The linear system. Trajectories are circles $x^2 + y^2 = C$.
|
||||
This is a true \textbf{center}.
|
||||
|
||||
\textit{System (B):} Convert to polar coordinates ($x = r\cos\theta$, $y = r\sin\theta$):
|
||||
\[
|
||||
r' = r^3, \quad \theta' = -1.
|
||||
\]
|
||||
Since $r' = r^3 > 0$ for $r > 0$, the radius strictly increases.
|
||||
The origin is actually an \textbf{unstable spiral}, \emph{not} a center,
|
||||
despite the linearization suggesting a center.
|
||||
|
||||
\textbf{Lesson:} When $\lambda = \pm i\beta$, the nonlinear terms determine
|
||||
the true behavior. Always use additional tools (Lyapunov functions,
|
||||
polar coordinates, first integrals) to resolve non-hyperbolic equilibria.
|
||||
\end{workedexample}
|
||||
|
||||
\subsection{Limit Cycles}
|
||||
\label{sec:ch14_limit_cycles}
|
||||
|
||||
\begin{definition}[Limit cycle]
|
||||
A \textbf{limit cycle} is a \textbf{closed, isolated periodic orbit}
|
||||
of an autonomous system in the phase plane.
|
||||
\end{definition}
|
||||
|
||||
The key word is \emph{isolated}: a limit cycle is a single closed trajectory
|
||||
that is not part of a continuous family of closed orbits. Nearby trajectories
|
||||
either spiral toward the limit cycle (stable limit cycle) or spiral away from
|
||||
it (unstable limit cycle).
|
||||
|
||||
This contrasts with the center in linear systems, where every trajectory near
|
||||
the equilibrium is a closed orbit forming a continuous family.
|
||||
|
||||
\begin{theorem}[Poincar\'{e}--Bendixson]
|
||||
\label{thm:poincare_bendixson}
|
||||
Let $R$ be a closed, bounded (compact) region in the plane containing no
|
||||
equilibrium points, and let a trajectory enter $R$ and remain there for
|
||||
all future time $t > 0$. Then the trajectory either:
|
||||
\begin{enumerate}
|
||||
\item approaches a periodic orbit (a closed trajectory) as $t \to \infty$, or
|
||||
\item is itself a periodic orbit.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
|
||||
\paragraph{Implications.}
|
||||
The Poincar\'{e}--Bendixson theorem is a powerful tool for proving the
|
||||
existence of limit cycles in two-dimensional systems:
|
||||
\begin{itemize}
|
||||
\item If you can construct a trapping region $R$ that contains no equilibria,
|
||||
any trajectory entering $R$ must approach a periodic orbit.
|
||||
\item In the plane, the only possible long-term behaviors of bounded
|
||||
trajectories are: equilibrium points, periodic orbits (limit cycles),
|
||||
or trajectories that approach limit cycles.
|
||||
\item \textbf{No chaos in 2D autonomous systems:} Strange attractors and
|
||||
chaotic behavior require at least three dimensions.
|
||||
\end{itemize}
|
||||
|
||||
\paragraph{Physical examples.}
|
||||
Limit cycles model sustained oscillations in physical systems:
|
||||
\begin{itemize}
|
||||
\item \textbf{Electrical circuits:} The van der Pol oscillator models self-sustained
|
||||
oscillations in vacuum tube circuits.
|
||||
\item \textbf{Chemical reactions:} The Belousov--Zhabotinsky reaction exhibits
|
||||
periodic color changes due to a chemical limit cycle.
|
||||
\item \textbf{Biological rhythms:} Heartbeat, neural firing, and circadian
|
||||
rhythms can be modeled as limit cycles.
|
||||
\item \textbf{Mechanical oscillators:} A clock pendulum with a periodic
|
||||
driving force (the escapement mechanism) exhibits a limit cycle.
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Lotka--Volterra Predator--Prey Model}
|
||||
\label{sec:ch14_lotka_volterra}
|
||||
|
||||
The Lotka--Volterra model is the classic example of a nonlinear system with
|
||||
closed orbits, first introduced in the 1920s to model predator--prey
|
||||
population dynamics.
|
||||
|
||||
\paragraph{Biological motivation and model derivation.}
|
||||
Consider two interacting populations:
|
||||
\begin{itemize}
|
||||
\item $x(t)$: number of \textbf{prey} (e.g., rabbits).
|
||||
\item $y(t)$: number of \textbf{predators} (e.g., foxes).
|
||||
\end{itemize}
|
||||
|
||||
We make four biological assumptions:
|
||||
\begin{enumerate}
|
||||
\item In the absence of predators, prey grow exponentially at rate $\alpha$:
|
||||
$x' = \alpha x$.
|
||||
\item Predators eat prey at a rate proportional to encounters $\alpha xy$.
|
||||
Combined with assumption~1: $x' = \alpha x - \beta xy$.
|
||||
\item In the absence of prey, predators die off at rate $\gamma$:
|
||||
$y' = -\gamma y$.
|
||||
\item Predators reproduce at a rate proportional to food consumption $\beta xy$.
|
||||
Combined with assumption~3: $y' = \delta xy - \gamma y$.
|
||||
\end{enumerate}
|
||||
|
||||
This yields the \textbf{Lotka--Volterra predator--prey equations}:
|
||||
\begin{equation}
|
||||
\label{eq:lotka_volterra}
|
||||
\boxed{
|
||||
\frac{\mathrm{d}x}{\mathrm{d}t} = \alpha x - \beta xy, \qquad
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = \delta xy - \gamma y,
|
||||
}
|
||||
\end{equation}
|
||||
where $\alpha, \beta, \gamma, \delta > 0$ are parameters.
|
||||
|
||||
\begin{definition}[Lotka--Volterra model]
|
||||
The Lotka--Volterra predator--prey system is the autonomous nonlinear
|
||||
system \cref{eq:lotka_volterra} with $\alpha, \beta, \gamma, \delta > 0$.
|
||||
It has two equilibrium points:
|
||||
\[
|
||||
(0,0) \text{ (extinction)}, \qquad
|
||||
\left(\frac{\gamma}{\delta},\, \frac{\alpha}{\beta}\right) \text{ (coexistence)}.
|
||||
\]
|
||||
\end{definition}
|
||||
|
||||
\paragraph{Equilibrium analysis.}
|
||||
Setting the right-hand sides to zero:
|
||||
\begin{align}
|
||||
x(\alpha - \beta y) &= 0, \label{eq:lv_eq1} \\
|
||||
y(\delta x - \gamma) &= 0. \label{eq:lv_eq2}
|
||||
\end{align}
|
||||
From \cref{eq:lv_eq1}, $x = 0$ or $y = \alpha/\beta$.
|
||||
|
||||
\textit{Case 1:} $x = 0$. From \cref{eq:lv_eq2}, $y(\,-\gamma) = 0 \implies y = 0$.
|
||||
Equilibrium: $(0,0)$.
|
||||
|
||||
\textit{Case 2:} $y = \alpha/\beta$. From \cref{eq:lv_eq2},
|
||||
$\delta x - \gamma = 0 \implies x = \gamma/\delta$.
|
||||
Equilibrium: $(\gamma/\delta, \alpha/\beta)$.
|
||||
|
||||
\paragraph{Jacobian and classification.}
|
||||
The Jacobian of the system is
|
||||
\[
|
||||
J(x,y) =
|
||||
\begin{pmatrix}
|
||||
\alpha - \beta y & -\beta x \\[6pt]
|
||||
\delta y & \delta x - \gamma
|
||||
\end{pmatrix}.
|
||||
\]
|
||||
|
||||
\textbf{At $(0,0)$:}
|
||||
\[
|
||||
J(0,0) = \begin{pmatrix} \alpha & 0 \\ 0 & -\gamma \end{pmatrix},
|
||||
\quad \lambda_1 = \alpha > 0,\; \lambda_2 = -\gamma < 0.
|
||||
\]
|
||||
Opposite signs $\implies$ \textbf{saddle point} (unstable).
|
||||
Both populations at zero is unstable: a small number of prey grows,
|
||||
triggering predator response.
|
||||
|
||||
\textbf{At $(\gamma/\delta, \alpha/\beta)$:}
|
||||
\[
|
||||
J\!\left(\frac{\gamma}{\delta}, \frac{\alpha}{\beta}\right)
|
||||
=
|
||||
\begin{pmatrix}
|
||||
0 & -\beta\gamma/\delta \\[6pt]
|
||||
\delta\alpha/\beta & 0
|
||||
\end{pmatrix}
|
||||
=
|
||||
\begin{pmatrix}
|
||||
0 & -\dfrac{\beta\gamma}{\delta} \\[8pt]
|
||||
\dfrac{\delta\alpha}{\beta} & 0
|
||||
\end{pmatrix}.
|
||||
\]
|
||||
Trace: $\tau = 0$. Determinant: $\Delta = (0) - (-\beta\gamma/\delta)(\delta\alpha/\beta) = \alpha\gamma > 0$.
|
||||
Characteristic equation: $\lambda^2 + \alpha\gamma = 0$, so
|
||||
\[
|
||||
\lambda = \pm i\sqrt{\alpha\gamma}.
|
||||
\]
|
||||
Purely imaginary eigenvalues $\implies$ \textbf{center} by linearization.
|
||||
Hartman--Grobman does not apply. We must analyze the nonlinear system.
|
||||
|
||||
\paragraph{First integral (conservation law).}
|
||||
The Lotka--Volterra system admits an exact first integral, proving that
|
||||
the orbits are indeed closed. Divide the two equations:
|
||||
\[
|
||||
\frac{\mathrm{d}y}{\mathrm{d}x}
|
||||
= \frac{\delta xy - \gamma y}{\alpha x - \beta xy}
|
||||
= \frac{y(\delta x - \gamma)}{x(\alpha - \beta y)}.
|
||||
\]
|
||||
Separate variables:
|
||||
\[
|
||||
\frac{\alpha - \beta y}{y}\,\mathrm{d}y
|
||||
= \frac{\delta x - \gamma}{x}\,\mathrm{d}x.
|
||||
\]
|
||||
Rewrite and integrate:
|
||||
\[
|
||||
\left(\frac{\alpha}{y} - \beta\right)\mathrm{d}y
|
||||
= \left(\delta - \frac{\gamma}{x}\right)\mathrm{d}x.
|
||||
\]
|
||||
\[
|
||||
\alpha \ln y - \beta y = \delta x - \gamma \ln x + C.
|
||||
\]
|
||||
Rearranging gives the \textbf{first integral}:
|
||||
\begin{equation}
|
||||
\label{eq:lv_first_integral}
|
||||
V(x,y) = \delta x - \gamma \ln x + \beta y - \alpha \ln y = \text{constant}.
|
||||
\end{equation}
|
||||
Each value of $C$ defines a closed orbit in the phase plane.
|
||||
The function $V(x,y)$ has a strict global minimum at the coexistence equilibrium
|
||||
$(\gamma/\delta, \alpha/\beta)$, and level curves $V(x,y) = C$ for
|
||||
$C > V_{\min}$ are closed curves surrounding this point.
|
||||
|
||||
\begin{keyresult}
|
||||
\textbf{Lotka--Volterra closed orbits.}
|
||||
The coexistence equilibrium $(\gamma/\delta, \alpha/\beta)$ is a true
|
||||
center for the nonlinear system. Every solution starting in the
|
||||
first quadrant $(x > 0, y > 0)$ traces a closed orbit around
|
||||
the coexistence point, determined by the initial conditions through
|
||||
the first integral \cref{eq:lv_first_integral}.
|
||||
\end{keyresult}
|
||||
|
||||
\paragraph{Nullclines.}
|
||||
The \textbf{nullclines} are curves in the phase plane where one of the
|
||||
derivatives vanishes:
|
||||
\begin{itemize}
|
||||
\item \textbf{$x$-nullcline} ($x' = 0$): $x = 0$ or $y = \alpha/\beta$.
|
||||
On $x = 0$, the prey population is zero.
|
||||
On $y = \alpha/\beta$ (horizontal line), the prey population
|
||||
is momentarily stationary.
|
||||
\item \textbf{$y$-nullcline} ($y' = 0$): $y = 0$ or $x = \gamma/\delta$.
|
||||
On $y = 0$, the predator population is zero.
|
||||
On $x = \gamma/\delta$ (vertical line), the predator population
|
||||
is momentarily stationary.
|
||||
\end{itemize}
|
||||
The two nontrivial nullclines intersect at the coexistence equilibrium
|
||||
$(\gamma/\delta, \alpha/\beta)$. The nullclines divide the first quadrant
|
||||
into four regions, each with a characteristic direction of motion:
|
||||
\begin{itemize}
|
||||
\item Region I ($x > \gamma/\delta, y < \alpha/\beta$): $x' > 0$, $y' > 0$
|
||||
$\implies$ motion up and right.
|
||||
\item Region II ($x < \gamma/\delta, y < \alpha/\beta$): $x' > 0$, $y' < 0$
|
||||
$\implies$ motion down and right.
|
||||
\item Region III ($x < \gamma/\delta, y > \alpha/\beta$): $x' < 0$, $y' < 0$
|
||||
$\implies$ motion down and left.
|
||||
\item Region IV ($x > \gamma/\delta, y > \alpha/\beta$): $x' < 0$, $y' > 0$
|
||||
$\implies$ motion up and left.
|
||||
\end{itemize}
|
||||
The resulting flow is counterclockwise around the equilibrium.
|
||||
|
||||
\begin{workedexample}
|
||||
\textbf{Full analysis of the Lotka--Volterra system with parameters}
|
||||
$\alpha = 1.5$, $\beta = 1$, $\delta = 3$, $\gamma = 1$.
|
||||
|
||||
\textbf{System:}
|
||||
\[
|
||||
\frac{\mathrm{d}x}{\mathrm{d}t} = 1.5\,x - x y, \qquad
|
||||
\frac{\mathrm{d}y}{\mathrm{d}t} = 3\,x y - y.
|
||||
\]
|
||||
|
||||
\textbf{Equilibria:}
|
||||
\[
|
||||
(0,0) \text{ and } \left(\frac{\gamma}{\delta}, \frac{\alpha}{\beta}\right)
|
||||
= \left(\frac{1}{3}, \frac{1.5}{1}\right) = \left(\frac{1}{3}, 1.5\right).
|
||||
\]
|
||||
|
||||
\textbf{Jacobian:}
|
||||
\[
|
||||
J(x,y) =
|
||||
\begin{pmatrix}
|
||||
1.5 - y & -x \\[6pt]
|
||||
3y & 3x - 1
|
||||
\end{pmatrix}.
|
||||
\]
|
||||
|
||||
\textbf{At $(0,0)$:}
|
||||
\[
|
||||
J(0,0) = \begin{pmatrix} 1.5 & 0 \\ 0 & -1 \end{pmatrix},
|
||||
\quad \lambda_1 = 1.5 > 0,\; \lambda_2 = -1 < 0.
|
||||
\]
|
||||
\textbf{Saddle point} (unstable).
|
||||
|
||||
\textbf{At $(1/3, 1.5)$:}
|
||||
\[
|
||||
J\!\left(\tfrac{1}{3}, 1.5\right)
|
||||
= \begin{pmatrix} 0 & -1/3 \\ 4.5 & 0 \end{pmatrix}.
|
||||
\]
|
||||
\[
|
||||
\tau = 0, \quad \Delta = (0) - (-1/3)(4.5) = 1.5.
|
||||
\]
|
||||
\[
|
||||
\lambda = \pm i\sqrt{1.5} \approx \pm 1.225\,i.
|
||||
\]
|
||||
Purely imaginary eigenvalues. The coexistence point is a \textbf{center}
|
||||
(confirmed by the first integral, not just linearization).
|
||||
|
||||
\textbf{Nullclines:}
|
||||
\begin{itemize}
|
||||
\item $x$-nullcline: $x = 0$ or $y = 1.5$ (horizontal line).
|
||||
\item $y$-nullcline: $y = 0$ or $x = 1/3$ (vertical line).
|
||||
\end{itemize}
|
||||
|
||||
\textbf{First integral:}
|
||||
\[
|
||||
V(x,y) = 3x - \ln x + y - 1.5\,\ln y = \text{constant}.
|
||||
\]
|
||||
|
||||
\textbf{Period of oscillation:} Near the equilibrium, the period is
|
||||
approximately $T \approx 2\pi/\sqrt{\alpha\gamma} = 2\pi/\sqrt{1.5}
|
||||
\approx 5.13$ time units.
|
||||
\end{workedexample}
|
||||
|
||||
\paragraph{Phase portrait.}
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
% Axes
|
||||
\draw[->, thick] (-0.3,0) -- (5,0) node[right] {$x$ (prey)};
|
||||
\draw[->, thick] (0,-0.3) -- (0,5) node[above] {$y$ (predator)};
|
||||
|
||||
% Nullclines
|
||||
\draw[dashed, thick, blue!60] (0.333,0) -- (0.333,4.5) node[above, font=\scriptsize, blue!60] {$x=1/3$};
|
||||
\draw[dashed, thick, red!60] (0,1.5) -- (4.8,1.5) node[right, font=\scriptsize, red!60] {$y=1.5$};
|
||||
|
||||
% Closed orbits (ellipses centered at (1/3, 1.5))
|
||||
% Inner orbit
|
||||
\draw[thick, purple!80] (0.333,1.5) ellipse (0.6 and 0.6);
|
||||
% Middle orbit
|
||||
\draw[thick, purple!80] (0.333,1.5) ellipse (1.2 and 1.2);
|
||||
% Outer orbit
|
||||
\draw[thick, purple!80] (0.333,1.5) ellipse (1.9 and 1.8);
|
||||
|
||||
% Arrows on orbits (counterclockwise)
|
||||
% Inner orbit arrows
|
||||
\draw[thick, purple!80, ->] (0.933,1.5) -- (0.983,1.68);
|
||||
\draw[thick, purple!80, ->] (0.333,2.1) -- (0.163,2.08);
|
||||
\draw[thick, purple!80, ->] (-0.267,1.5) -- (-0.267,1.28);
|
||||
\draw[thick, purple!80, ->] (0.333,0.9) -- (0.503,0.92);
|
||||
|
||||
% Middle orbit arrows
|
||||
\draw[thick, purple!80, ->] (1.533,1.5) -- (1.613,1.75);
|
||||
\draw[thick, purple!80, ->] (0.333,2.7) -- (0.093,2.68);
|
||||
\draw[thick, purple!80, ->] (-0.867,1.5) -- (-0.867,1.18);
|
||||
\draw[thick, purple!80, ->] (0.333,0.3) -- (0.573,0.32);
|
||||
|
||||
% Equilibrium points
|
||||
\filldraw[black] (0,0) circle (2.5pt) node[below left=1pt] {$(0,0)$};
|
||||
\filldraw[black] (0.333,1.5) circle (2.5pt) node[below right=1pt] {$(\gamma/\delta,\,\alpha/\beta)$};
|
||||
|
||||
% Region labels
|
||||
\node[font=\scriptsize, align=center, blue!60] at (2.5,2.5) {Region I\\($x' > 0,\; y' > 0$)};
|
||||
\node[font=\scriptsize, align=center, blue!60] at (2.5,0.5) {Region II\\($x' > 0,\; y' < 0$)};
|
||||
\node[font=\scriptsize, align=center, blue!60] at (-0.7,2.5) {Region III\\($x' < 0,\; y' < 0$)};
|
||||
\node[font=\scriptsize, align=center, blue!60] at (-0.7,0.5) {Region IV\\($x' < 0,\; y' > 0$)};
|
||||
|
||||
% Legend
|
||||
\node[font=\footnotesize, align=left, anchor=west] at (1.2,4.2) {
|
||||
\textcolor{blue!60}{\rule{6pt}{1pt}} $x$-nullcline \\
|
||||
\textcolor{red!60}{\rule{6pt}{1pt}} $y$-nullcline \\
|
||||
\textcolor{purple!80}{\rule{6pt}{1pt}} Closed orbits
|
||||
};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\paragraph{Interpretation of the oscillations.}
|
||||
The closed orbits correspond to periodic predator--prey cycles:
|
||||
\begin{enumerate}
|
||||
\item Prey population grows (few predators) $\to$ predators have abundant food.
|
||||
\item Predator population grows $\to$ predation pressure increases.
|
||||
\item Prey population declines $\to$ predators face food shortage.
|
||||
\item Predator population declines $\to$ prey can recover.
|
||||
\item Cycle repeats.
|
||||
\end{enumerate}
|
||||
The amplitude of the oscillation is determined by the initial conditions:
|
||||
the further the initial point from the equilibrium, the larger the orbit.
|
||||
|
||||
\paragraph{Limitations of the model.}
|
||||
The Lotka--Volterra model has several simplifications:
|
||||
\begin{itemize}
|
||||
\item Unlimited prey growth in the absence of predators (no carrying capacity).
|
||||
\item Linear functional response (predation rate proportional to $xy$).
|
||||
\item No time delays in predator reproduction.
|
||||
\item Homogeneous mixing (well-mixed populations).
|
||||
\end{itemize}
|
||||
Realistic extensions include logistic prey growth, Holling-type functional
|
||||
responses, and more complex age-structured models. These modifications can
|
||||
create stable limit cycles (rather than neutral centers) and richer dynamics.
|
||||
|
||||
\subsection{Summary}
|
||||
\label{sec:ch14_summary}
|
||||
|
||||
Nonlinear systems cannot be solved by superposition, but their qualitative
|
||||
behavior near equilibria is accessible through linearization. The Jacobian
|
||||
matrix, combined with the trace-determinant classification, provides a
|
||||
systematic framework for understanding local dynamics. The Hartman--Grobman
|
||||
theorem guarantees that this linear picture is topologically correct for
|
||||
hyperbolic equilibria. Global phenomena like limit cycles require additional
|
||||
tools: the Poincar\'{e}--Bendixson theorem, first integrals, and Lyapunov
|
||||
functions.
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\caption{Chapter 14 Summary: Nonlinear Systems}
|
||||
\label{tab:ch14_summary}
|
||||
\begin{tabular}{l p{8.5cm}}
|
||||
\toprule
|
||||
\textbf{Concept} & \textbf{Key Formula/Method} \\
|
||||
\midrule
|
||||
Nonlinear autonomous system & $\displaystyle x' = f(x,y), \;\; y' = g(x,y)$, no superposition \\
|
||||
Equilibrium point & Solve $f(x^*,y^*) = 0$, $g(x^*,y^*) = 0$ \\
|
||||
Jacobian matrix & $J = \begin{pmatrix} \pd{f}{x} & \pd{f}{y} \\[4pt] \pd{g}{x} & \pd{g}{y} \end{pmatrix}$ \\
|
||||
Linearized system & $\begin{pmatrix} u' \\ v' \end{pmatrix} = J(x^*,y^*) \begin{pmatrix} u \\ v \end{pmatrix}$, $u=x-x^*$, $v=y-y^*$ \\
|
||||
Trace-determinant & $\tau = \tr(J)$, $\Delta = \det(J)$, $D = \tau^2 - 4\Delta$ \\
|
||||
Hartman--Grobman theorem & Hyperbolic equilibria ($\Re(\lambda) \neq 0$): nonlinear $\cong$ linearization \\
|
||||
Non-hyperbolic case & $\Re(\lambda) = 0$: linearization inconclusive; use first integrals, Lyapunov functions \\
|
||||
Limit cycle & Closed, isolated periodic orbit \\
|
||||
Poincar\'{e}--Bendixson theorem & Bounded trajectory in 2D with no equilibria $\implies$ periodic orbit \\
|
||||
Lotka--Volterra model & $x' = \alpha x - \beta xy$, $y' = \delta xy - \gamma y$ \\
|
||||
LV equilibria & $(0,0)$ saddle; $(\gamma/\delta, \alpha/\beta)$ center (closed orbits) \\
|
||||
LV first integral & $V(x,y) = \delta x - \gamma\ln x + \beta y - \alpha\ln y = C$ \\
|
||||
LV nullclines & $x$-nullcline: $y = \alpha/\beta$; $y$-nullcline: $x = \gamma/\delta$ \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
\begin{hintbox}
|
||||
\textbf{Problem-solving workflow for nonlinear autonomous systems.}
|
||||
\begin{enumerate}
|
||||
\item Find all equilibrium points by solving $f(x,y) = 0$, $g(x,y) = 0$.
|
||||
\item Compute the Jacobian matrix $J(x,y)$ and evaluate it at each equilibrium.
|
||||
\item Classify each equilibrium using eigenvalues or the trace-determinant plane.
|
||||
\item Apply the Hartman--Grobman theorem: if the equilibrium is hyperbolic,
|
||||
the local phase portrait matches the linear classification.
|
||||
\item For non-hyperbolic equilibria ($\Re(\lambda) = 0$), use additional tools:
|
||||
first integrals, Lyapunov functions, or polar coordinates.
|
||||
\item Draw nullclines to understand the flow direction in each region.
|
||||
\item Look for limit cycles using the Poincar\'{e}--Bendixson theorem
|
||||
(requires constructing a trapping region).
|
||||
\item Sketch the full phase portrait combining local and global information.
|
||||
\end{enumerate}
|
||||
\end{hintbox}
|
||||
@@ -1,40 +0,0 @@
|
||||
\input{preamble.tex}
|
||||
|
||||
\title{\textbf{Differential Equations Handbook}\\[0.5em]
|
||||
\normalsize A Complete Reference for ODE and Introductory PDE}
|
||||
\author{}
|
||||
\date{}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
\thispagestyle{empty}
|
||||
|
||||
\tableofcontents
|
||||
\newpage
|
||||
|
||||
% ========== CHAPTERS ==========
|
||||
\input{chapters/ch01_introduction}
|
||||
\input{chapters/ch02_first_order}
|
||||
\input{chapters/ch03_qualitative}
|
||||
\input{chapters/ch04_second_order_homogeneous}
|
||||
\input{chapters/ch05_second_order_nonhomogeneous}
|
||||
\input{chapters/ch06_mechanical_applications}
|
||||
\input{chapters/ch07_laplace_transforms}
|
||||
\input{chapters/ch08_systems}
|
||||
\input{chapters/ch09_series_solutions}
|
||||
\input{chapters/ch10_fourier_series}
|
||||
\input{chapters/ch11_boundary_value_problems}
|
||||
\input{chapters/ch12_heat_equation}
|
||||
\input{chapters/ch13_wave_and_laplace}
|
||||
\input{chapters/ch14_nonlinear_systems}
|
||||
|
||||
% ========== APPENDICES ==========
|
||||
\input{appendices/appA_solution_summary}
|
||||
\input{appendices/appB_transform_tables}
|
||||
\input{appendices/appC_integral_tables}
|
||||
\input{appendices/appD_notation}
|
||||
|
||||
\printnomenclature
|
||||
|
||||
\end{document}
|
||||
-189
@@ -1,189 +0,0 @@
|
||||
% =============================================================================
|
||||
% preamble.tex — Differential Equations Handbook
|
||||
% =============================================================================
|
||||
% Complete preamble: packages, theorem styles, tcolorbox environments,
|
||||
% custom macros, and configuration for a 14-chapter handbook.
|
||||
% =============================================================================
|
||||
|
||||
\documentclass[10pt,a4paper]{article}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Geometry
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage[margin=0.75in]{geometry}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Math packages
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage{mathtools}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{bm}
|
||||
\usepackage{empheq}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Colors and boxes
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage[svgnames]{xcolor}
|
||||
\usepackage[most]{tcolorbox}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Nomenclature (notation glossary)
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage{nomencl}
|
||||
\makenomenclature
|
||||
\setlength{\nomlabelwidth}{2.5cm}
|
||||
\setlength{\nomitemsep}{2pt}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Tables
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage{booktabs}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Graphics
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage{graphicx}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% TikZ
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{calc}
|
||||
\usetikzlibrary{arrows.meta}
|
||||
\usetikzlibrary{patterns}
|
||||
\usetikzlibrary{decorations.markings}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Hyperref (loaded before cleveref)
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage[colorlinks=true,
|
||||
linkcolor=MidnightBlue,
|
||||
urlcolor=Navy,
|
||||
citecolor=RoyalBlue,
|
||||
breaklinks=true]{hyperref}
|
||||
|
||||
% ---------------------------------------------------------------------------
|
||||
% Cross-referencing (must be after hyperref)
|
||||
% ---------------------------------------------------------------------------
|
||||
\usepackage[capitalize]{cleveref}
|
||||
|
||||
% =============================================================================
|
||||
% Theorem styles (amsthm)
|
||||
% =============================================================================
|
||||
% Number theorems, definitions, and remarks per chapter (section).
|
||||
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem{proposition}[theorem]{Proposition}
|
||||
\newtheorem{corollary}[theorem]{Corollary}
|
||||
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}[theorem]{Definition}
|
||||
|
||||
\theoremstyle{remark}
|
||||
\newtheorem{remark}[theorem]{Remark}
|
||||
|
||||
% =============================================================================
|
||||
% tcolorbox environments
|
||||
% =============================================================================
|
||||
|
||||
% Worked Example — blue frame
|
||||
\newtcolorbox{workedexample}{
|
||||
breakable,
|
||||
colframe=SteelBlue,
|
||||
colback=LightSteelBlue!15,
|
||||
coltitle=white,
|
||||
fonttitle=\bfseries,
|
||||
title=Worked Example,
|
||||
boxrule=0.6pt,
|
||||
arc=2pt,
|
||||
left=6pt, right=6pt, top=4pt, bottom=4pt
|
||||
}
|
||||
|
||||
% Exercise — gray frame
|
||||
\newtcolorbox{exercise}{
|
||||
breakable,
|
||||
colframe=Gray,
|
||||
colback=LightGray!10,
|
||||
coltitle=white,
|
||||
fonttitle=\bfseries,
|
||||
title=Exercise,
|
||||
boxrule=0.5pt,
|
||||
arc=2pt,
|
||||
left=6pt, right=6pt, top=4pt, bottom=4pt
|
||||
}
|
||||
|
||||
% Key Result — gold/yellow frame
|
||||
\newtcolorbox{keyresult}{
|
||||
breakable,
|
||||
colframe=Goldenrod,
|
||||
colback=LemonChiffon!25,
|
||||
coltitle=black,
|
||||
fonttitle=\bfseries,
|
||||
title=Key Result,
|
||||
boxrule=0.8pt,
|
||||
arc=2pt,
|
||||
left=6pt, right=6pt, top=4pt, bottom=4pt
|
||||
}
|
||||
|
||||
% Hint — green frame
|
||||
\newtcolorbox{hintbox}{
|
||||
breakable,
|
||||
colframe=ForestGreen,
|
||||
colback=Honeydew!40,
|
||||
coltitle=white,
|
||||
fonttitle=\bfseries,
|
||||
title=Hint,
|
||||
boxrule=0.6pt,
|
||||
arc=2pt,
|
||||
left=6pt, right=6pt, top=4pt, bottom=4pt
|
||||
}
|
||||
|
||||
% =============================================================================
|
||||
% Custom macros
|
||||
% =============================================================================
|
||||
|
||||
% Differentials
|
||||
\newcommand{\diff}{\mathrm{d}}
|
||||
\newcommand{\dd}{\mathrm{d}}
|
||||
|
||||
% Partial derivative shorthand
|
||||
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
|
||||
|
||||
% Number sets
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
|
||||
% Operators
|
||||
% Note: \Re and \Im are already provided by amsmath; no need to redefine.
|
||||
\DeclareMathOperator{\tr}{tr}
|
||||
\DeclareMathOperator{\detop}{det}
|
||||
\DeclareMathOperator{\spanop}{span}
|
||||
\DeclareMathOperator{\kerop}{ker}
|
||||
\DeclareMathOperator{\rankop}{rank}
|
||||
|
||||
% =============================================================================
|
||||
% Additional setup
|
||||
% =============================================================================
|
||||
|
||||
% Table of contents depth
|
||||
\setcounter{tocdepth}{2}
|
||||
|
||||
% Cleveref naming
|
||||
\crefname{theorem}{Theorem}{Theorems}
|
||||
\crefname{lemma}{Lemma}{Lemmas}
|
||||
\crefname{proposition}{Proposition}{Propositions}
|
||||
\crefname{corollary}{Corollary}{Corollaries}
|
||||
\crefname{definition}{Definition}{Definitions}
|
||||
\crefname{remark}{Remark}{Remarks}
|
||||
\crefname{equation}{equation}{equations}
|
||||
\crefname{section}{section}{sections}
|
||||
\crefname{figure}{figure}{figures}
|
||||
\crefname{table}{table}{tables}
|
||||
|
||||
% =============================================================================
|
||||
% End of preamble.tex
|
||||
% =============================================================================
|
||||
Reference in New Issue
Block a user