Consider the wave equation $y_{tt} = c^2 y_{xx}$ with initial conditions $y(x,0) = 0$ and $y_t(x,0)$ on the entire real line.

a) Show that the solution is
\[
y(x, t) = \frac{1}{2c} \int_{x-ct}^{x+ct} v(s) \, \de{s}.
\]

b) If
\[
v(x) = \begin{cases}
cx/a & a < x < 2a \\
-cx/a & -2a < x < -a \\
\text{undefined} & \text{otherwise}
\end{cases}
\]

give a clear sketch of the part (or parts) of the $(x, t)$ plane where $y(x, t)$ can be determined, and obtain an expression for $y(x, t)$ in each such region.

Spoiler

The integral form of the solution follows directly from d'Alembert's solution. Consider an arbitrary point in the upper-half $(x,t)$-plane. The value of the solution at the point $(x_0, t_0)$ is found by first drawing the lines $x \pm ct = x_0 \pm c t_0$ through this point. These two lines will intersect the $x$-axis at two points, $x = x_0 \pm ct_0$. However, in order for the integral to be defined, we need
\[
a \leq x_0 - ct \leq 2a,
\]

or alternatively
\[
-2a \leq x_0 - ct < -a.
\]

This is equivalent to the condition that the point $(x_0, t_0)$ lies underneath the two small wedges illustrated below:

We can verify through direct integration that
\[
y(x,t) = \begin{cases}
-ctx/a & \text{Left wedge} \\
ctx/a & \text{Right wedge}
\end{cases}
\]

2. Axisymmetric solutions of the wave equation

Let $r$ be the distance from the origin in $(x,y,z)$-space, and let $u(r,t)$ be the solution of the spherically symmetric wave equation
\[
u_{tt} = c^2 u_{rr} + \frac{2}{r} u_r, \qquad r > 0,
\]

which satisfies the initial conditions $u(r,0) = 0$ and $u_t(r,0) = e^{-r^2}$ for $r > 0$.

a) Let $w(r,t) = ru(r,t)$. Show that $w$ is a solution of the wave equation
\[
w_{tt} = c^2 w_{rr}, \qquad r > 0
\]

for which $w(r,0) = 0$ and $w_t(r,0) = re^{-r^2}$ for $r > 0$ and $w(0, t) = 0$ for $t \geq 0$.

b) Find $w(r,t)$ and show that
\[
u(r,t) = \frac{\text{sinh}(2crt)}{2cr} e^{-(r^2 + c^2 t^2)}.
\]

c) What is the value of the limit $\lim_{r\to 0} u(r,t)$?

Spoiler

The first part is straightforward substitution. To solve the equation, use d'Alembert's, with
\[
w(r,t) = \frac{1}{2c} \int_{r-ct}^{r+ct} se^{-s^2} \, \de{s} = \frac{e^{-(r-ct)^2} - e^{-(r+ct)^2}}{4c}
= \frac{\text{sinh}(2crt)}{2c} e^{-(r^2 + c^2 t^2)}.
\]

Setting $u = w/r$ then yields the desired result. To obtain the limit as $r \to 0$, we note that that the Taylor series expansion,
\[
\text{sinh}(2crt) = 2crt + \ldots,
\]

Let $T(x,y)$ satisfy Laplace's equation in the square $0 < x < \pi$ and $0 < y < \pi$. Let $T = 1$ on the side $y = \pi$, $0 < x < \pi$, and let $T = 0$ on the other three sides.

a) Show that
\[
T(x, t) = \frac{4}{\pi} \sum_{n=0}^\infty \frac{\text{sinh}[(2n + 1)y]}{\text{sinh}[(2n + 1)\pi]}
\frac{\sin[(2n +1)x]}{(2n+1)}.
\]

Spoiler

This follows easily through separation of variables. The key is that when you separate, $T(x,y) = X(x)Y(y)$ and obtain
\[
\frac{F''}{F} -\frac{G''}{G} = -\lambda^2,
\]

you want the right hand side to be negative. The trick is to note that the condition that $T = 0$ at $y = 0$ and $y = \pi$ hints that $T$ should be a sinusoidal along $y$ (instead of exponentials). The solution then follows in the usual way.

b) By considering three similar problems, show that $T = 1/4$ at the centre of the square.

Spoiler

Let $T^{(i)}(x,y)$ be the solution for the four cases where
\[
T^{(1)}(x, \pi) = 1, \quad
T^{(2)}(\pi, y) = 1, \quad
T^{(3)}(0, y) = 1, \quad
T^{(4)}(x, 0) = 1.
\]

and $T^{(i)} = 0$ on all the other boundaries. All of the solutions are obtained by symmetry and coordinate shifts:
\begin{align*}
T^{(1)}(x, y) &= \frac{4}{\pi} \sum_{n=0}^\infty \frac{\text{sinh}[(2n + 1)y]}{\text{sinh}[(2n + 1)\pi]}
\frac{\sin[(2n +1)x]}{(2n+1)} \\
T^{(2)}(x, y) &= \frac{4}{\pi} \sum_{n=0}^\infty \frac{\text{sinh}[(2n + 1)x]}{\text{sinh}[(2n + 1)\pi]}
\frac{\sin[(2n +1)y]}{(2n+1)} \\
T^{(3)}(x, y) &= \frac{4}{\pi} \sum_{n=0}^\infty \frac{\text{sinh}[(2n + 1)(1-y)]}{\text{sinh}[(2n + 1)\pi]}
\frac{\sin[(2n +1)x]}{(2n+1)} \\
T^{(2)}(x, y) &= \frac{4}{\pi} \sum_{n=0}^\infty \frac{\text{sinh}[(2n + 1)(1-x)]}{\text{sinh}[(2n + 1)\pi]}
\frac{\sin[(2n +1)y]}{(2n+1)}.
\end{align*}

Note that by symmetry, $T^{i}(1/2, 1/2)$ is the same for all the solutions. However, the linearity of the boundary conditions and PDE implies that we can add all four to obtain the solution to the case where $T = 1$ on all boundaries. We can verify that this solution must in fact be $T \equiv 1$ (it is the steady-state solution). Thus
\[
4T^{(i)}(1/2,1/2) = 1,
\]

leading to the result.

4. Axisymmetric Laplace's equation

a) (i) If $u(x, y)$ is a twice-differentiable function of the Cartesian coordinates $x$ and $y$, and if $v(r, \theta) = u(r\cos\theta, r\sin\theta)$, where $r$ and $\theta$ are plane polar coordinates, show
that
\[
u_{xx} + u_{yy} = v_{rr} + \frac{1}{r} v_r + \frac{1}{r^2} v_{\theta\theta}
\]

Using these values gives us the result immediately.

(ii) Hence show that the functions
\[
1 \qquad \log r \qquad r^n\cos(n\theta) \qquad r^{-n}\cos(n\theta),
\]

where $n$ is a positive integer, are all solutions of Laplace's equation.

Spoiler

You can either check the solutions directly (by plugging them into Laplace's equation), or you can derive them through separation of variables, with $v(r,theta) = R(r)T(\theta)$ in the usual fashion.

b) A conductor occupies the region $r \geq a$, i.e. exterior to the disk whose centre is the origin and whose radius is $a$. If the steady state temperature, $T(r, \theta)$, satisfies the boundary condition $T(a, \theta) = \cos^2 \theta$ for $0 \leq \theta \leq 2\pi$ and $T$ remains bounded as $r \to \infty$, show that
\[
T(r,\theta) = \frac{1}{2} + \frac{a^2}{2r^2} \cos(2\theta).
\]

Spoiler

In order to satisfy the boundary conditions, consider a Fourier series made up of the four stated forms:
\[
T(r,\theta) = \frac{a_0}{2} + \sum_{n=1}^\infty r^{-n} \left[a_n \cos(n\theta) + b_n\sin(n\theta)\right]
\]

where note we have gotten rid of the mode with $R(r) \sim r^{n}$ because the solutions must be bounded at infinity. Applying the boundary condition then gives the result.