From 0ffdeee003ce03d69ba9a865f5b2410cf4f03328 Mon Sep 17 00:00:00 2001 From: Worker Agent Date: Thu, 4 Jun 2026 17:56:41 -0500 Subject: [PATCH] ch12: heat equation, separation, multiple BCs, source terms --- chapters/ch12_heat_equation.tex | 814 +++++++++++++++++++++++++++++++- 1 file changed, 803 insertions(+), 11 deletions(-) diff --git a/chapters/ch12_heat_equation.tex b/chapters/ch12_heat_equation.tex index 5e94c42..7794d09 100644 --- a/chapters/ch12_heat_equation.tex +++ b/chapters/ch12_heat_equation.tex @@ -1,49 +1,841 @@ +% ============================================================================= +% ch12_heat_equation.tex +% Chapter 12: The Heat Equation +% ============================================================================= + \section{Heat Equation} \label{ch:heat_equation} \subsection{Physical Derivation} \label{sec:ch12_physical_derivation} -% Content goes here +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: +\[ + \rho\,c\,\pd{u}{t} = -\pd{}{x}\bigl(-\kappa\,\pd{u}{x}\bigr) + = \kappa\,\pd[2]{u}{x}. +\] +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\,\pd[2]{u}{x}, \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 +\[ + \pd{u}{t} = \alpha\,\nabla^2 u = \alpha\left(\pd[2]{u}{x} + \pd[2]{u}{y} + \pd[2]{u}{z}\right). +\] +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} -% Content goes here +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} -% Content goes here +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\,\pd[2]{u}{x}, & 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 $f(x) = \sin(2x)$: + \[ + \begin{cases} + \pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt] + u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt] + u(x,0) = \sin(2x), & 0 < x < \pi. + \end{cases} + \] + + \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: + \[ + \begin{cases} + \pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt] + u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt] + u(x,0) = x(\pi - x), & 0 < x < \pi. + \end{cases} + \] + + \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} -% Content goes here +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 + \[ + \pd{u}{t} = \alpha\,\pd[2]{u}{x}, \qquad + u_x(0,t) = 0,\; u_x(L,t) = 0, \qquad + u(x,0) = f(x), + \] + 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 $f(x) = \cos(2x) + 3$: + \[ + \begin{cases} + \pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt] + \pd{u}{x}(0,t) = 0, \;\; \pd{u}{x}(\pi,t) = 0, & t > 0, \\[6pt] + u(x,0) = \cos(2x) + 3, & 0 < x < \pi. + \end{cases} + \] + + \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} -% Content goes here +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 +\[ + 0 = \alpha\,\pd[2]{u}{x} \quad\Longrightarrow\quad \pd[2]{u}{x} = 0. +\] +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 +\[ + \begin{cases} + \pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < L, \;\; t > 0, \\[6pt] + u(0,t) = T_1, \;\; u(L,t) = T_2, & t > 0, \\[6pt] + u(x,0) = f(x), & 0 < x < L. + \end{cases} +\] +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: +\[ + \pd{v}{t} = \alpha\,\pd[2]{v}{x} + \alpha\,\phi''(x). +\] +Since $\phi(x)$ is linear, $\phi''(x) = 0$, so $v$ satisfies the same homogeneous heat equation: +\[ + \pd{v}{t} = \alpha\,\pd[2]{v}{x}. +\] +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 + \[ + \begin{cases} + \pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt] + u(0,t) = 0, \;\; u(\pi,t) = 100, & t > 0, \\[6pt] + u(x,0) = 50, & 0 < x < \pi. + \end{cases} + \] + \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 + \[ + \begin{cases} + \pd{v}{t} = \alpha\,\pd[2]{v}{x}, & 0 < x < \pi, \\[6pt] + v(0,t) = 0, \;\; v(\pi,t) = 0, \\[6pt] + v(x,0) = 50 - \frac{100}{\pi}\,x. + \end{cases} + \] + + \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} -% Content goes here +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\,\pd[2]{u}{x} + 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: + \[ + \begin{cases} + \pd{u}{t} = \alpha\,\pd[2]{u}{x} + Q_0, & 0 < x < \pi, \;\; t > 0, \\[6pt] + u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt] + u(x,0) = 0, & 0 < x < \pi, + \end{cases} + \] + 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} -% Summary table at end of chapter +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{Chapter Summary} +\caption{Heat equation solution methods by boundary condition type} \label{tab:ch12_summary} -\begin{tabular}{l l} +\begin{tabular}{l l p{4.5cm}} \toprule -\textbf{Concept} & \textbf{Key Formula/Method} \\ +\textbf{BC type} & \textbf{Eigenfunctions} & \textbf{Key features} \\ \midrule -TBD & TBD \\ +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}