From 9d723da9366fa66edd56038d42752b50c61e0173 Mon Sep 17 00:00:00 2001 From: Worker Agent Date: Thu, 4 Jun 2026 15:34:14 -0500 Subject: [PATCH] ch1: classification, direction fields, existence/uniqueness --- chapters/ch01_introduction.tex | 436 +++++++++++++++++++++++++++++++-- 1 file changed, 415 insertions(+), 21 deletions(-) diff --git a/chapters/ch01_introduction.tex b/chapters/ch01_introduction.tex index cf00bd6..be7cb9e 100644 --- a/chapters/ch01_introduction.tex +++ b/chapters/ch01_introduction.tex @@ -1,34 +1,428 @@ -\section{Introduction} -\label{ch:introduction} - -\subsection{Classification of ODEs} +\section{Classification of Differential Equations} \label{sec:ch01_classification} -% Content goes here +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{Direction Fields} +\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} -% Content goes here +A direction field provides a geometric picture of the solution family of a first-order ODE without solving it analytically. -\subsection{Existence and Uniqueness} +\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} -% Content goes here +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{Summary} +\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 $y(x) = \left(\frac{x + 2}{2}\right)^2 = \frac{(x+2)^2}{4}$. + + This solution is defined and differentiable for all $x \in \R$, so the interval of validity is $(-\infty, \infty)$. 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 $y(0) = 1$ never reaches $y = 0$ for $x > -2$, so uniqueness is preserved along this trajectory. +\end{workedexample} + +\section{Summary} \label{sec:ch01_summary} -% Summary table at end of chapter \begin{table}[htbp] -\centering -\caption{Chapter Summary} -\label{tab:ch01_summary} -\begin{tabular}{l l} -\toprule -\textbf{Concept} & \textbf{Key Formula/Method} \\ -\midrule -TBD & TBD \\ -\bottomrule -\end{tabular} + \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}.