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% =============================================================================
% 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}
\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}
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