Toronto Math Forum

\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin n \theta - \frac{1}{r} \frac{2}{3} \sin n \theta
\end{equation}

Sorry, there aren't supposed to be n's, those are supposed to be just $\sin \theta$.

So the solution should be:
\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin \theta - \frac{1}{r} \frac{2}{3} \sin\theta
\end{equation}

Sorry, I meant n, odd! As in all the odd terms, my bad!

We separate variables. So $u(x,t) = T(t)X(x)$. Our equation is then $\frac{T''}{9T} = \frac{X''}{X} =-  \lambda$.

So we solve the $X$ equation. We have $X'' = - \lambda X$. First consider $\lambda = \omega^2 > 0$. Then the solution is $X(x) = A\cos\omega x + B \sin \omega X$. Using the boundary conditions, we see that $A=0$, and $\omega 2 = n \pi \longrightarrow \omega = \frac{n \pi}{2}$.

Now consider negative eigenvalues. The solution will be $X(x) = A\cosh\omega x + B \sinh \omega X$. The boundary conditions clearly show that there is only a trivial solution, so we have no negative eigenvalues.

Likewise with the zero eigenvalue: the solution will be $X(x) = Ax + B$, and we can see that there is only a trivial solution. So no zero eigenvalues.

So then the eigenvalues are $\frac{n^2 \pi^2}{4}$ with eigenfunctions $X(x) = B \sin ( \frac{n \pi x}{2})$.

We can then solve the $T$ equation.

\begin{equation}
\frac{T''}{9T} = - \frac{n^2 \pi^2}{4} \longrightarrow T'' = - \frac{9 n^2 \pi^2 T}{4}
\end{equation}

So the solution is

\begin{equation}
T(t) = C\cos(3n\pi t/2) + D\ sin(3n\pi t/2)
\end{equation}

So overall the solution is:

\begin{equation}
u(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) \left( A_n\cos(3n\pi t/2) + B_n\sin(3n\pi t/2) \right)
\end{equation}

When we use the initial condition at $u(x,0)$, we see right away that $A_n = 0$. So:

\begin{equation}
u(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) B_n\sin(3n\pi t/2)
\end{equation}

\begin{equation}
u_t(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) B_n \frac{3 n \pi}{2} \cos(3n\pi t/2)
\end{equation}

Then using the other condition:

\begin{equation}
u_t(x,0) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) \frac{3 n \pi}{2} B_n = x(2-x)
\end{equation}

We then solve for the $B_n$'s:

\begin{equation}
B_n = \frac{2}{3 n \pi} \int_0^2 (2x-x^2) \sin( \frac{n \pi x}{2}) dx = -( \frac{2}{3 n \pi} ) \frac{8(\pi \sin(n \pi) + 2\cos(n \pi) - 2)}{\pi^3 n^3} =( \frac{2}{3 n \pi} ) \frac{16 ((-1)^n - 1)}{\pi^3 n^3}
\end{equation}

If n is even we see that this is zero, and if n is odd we see that this is $\frac{-64}{3 \pi^4 n^4}$

So we see that

\begin{equation}
u(x,t) = -\frac{64}{3\pi^4}\sum_{n odd}^{\infty}\frac{1}{n^4}\sin(\frac{n \pi x}{2})\sin(3n\pi t/2)
\end{equation}

$T''=4n^2 T\Rightarrow T=Ce^{2ny}+De^{-2ny}$. However, $y>0$, so for the solution to be bounded we must have $C=0$.
Then the general solution is, after absorbing and redefining some constants:
$$u(x,y)=\sum_{n=1}^{\infty}A_n \sin(2nx)e^{-2ny}$$
$u(x,0)=\cos(x)=\sum_{n=1}^{\infty} A_n\sin(2nx)\Rightarrow A_n=\frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\cos(x)\sin(2nx)dx=\frac{8n}{\pi(4n^2 -1)}$
Therefore the final solution is:
$$u(x,y)=\sum_{n=1}^{\infty}\frac{8n}{\pi(4n^2 -1)}\sin(2nx)e^{-2ny}$$

b) There are only strictly positive eigenvalues, so let $\lambda=\omega^2$. Then $X''=-\omega^2 X\Rightarrow X=A\cos(\omega x)+B\sin(\omega x)$. The boundary conditions imply that $A=0$, and $\omega\frac{\pi}{2}=n\pi\Rightarrow\omega=2n$. The eigenfunctions and eigenvalues are:
$$\lambda_n=-4n^2,~X_n (x)=B_n\sin(2nx)$$

(c) The differential equation that we must solve for $R$ is $r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda$, or $r^2R'' + rR' = \lambda R$, so we assume a solution $r^m$, and then we get:

\begin{equation}
r^2m(m-1)r^{m-2} + rmr^{m-1} - \lambda r^m = 0 \longrightarrow m^2 - m + m - \lambda = 0 \longrightarrow m^2 = n^2
\end{equation}

So we have $m = \pm n$, which means that the solution is $R(r) = r^n + r^{-n}$.

(b) The eigenvalue problem for $P$ (as seen in part a) is:

\begin{equation}
P'' = - \lambda P
\end{equation}

First let $\lambda = \omega^2 > 0$. Then the equation is $P'' = - \omega^2 P$, and the solution is $P(\theta) = A\cos(\omega \theta) + B\sin(\omega \theta)$.

Let $\lambda = - \omega^2 < 0$. Then the equation is $P'' = \omega^2 P$, and the solution is $P(\theta) = C\cosh(\omega \theta) + D\sinh(\omega \theta)$.

Then let $\lambda = 0$. Then the equation is $P'' = 0$, and the solution is $P(\theta) = E \theta + F$.

So the eigenvalues are $n^2$, and the eigenfunctions are $P(\theta) = A\cos(n \theta) + B\sin(n \theta)$.

(a) We solve this question using the two dimensional Laplacian in polar coordinates. So the equation becomes:

\begin{equation}
u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta} = 0
\end{equation}

So the equation is

\begin{equation}
\frac{R''}{R} + \frac{1}{r}\frac{R'}{R} + \frac{1}{r^2}\frac{P''}{P} = 0
\end{equation}

Multiplying through by $r^2$, we see that the equation is:

\begin{equation}
r^2\frac{R''}{R} + r\frac{R'}{R} + \frac{P''}{P} = 0
\end{equation}

The equation is now separable. We have a set of ordinary differential equations:

\begin{gather}
r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda \\
\frac{P''}{P} = - \lambda
\end{gather}

The boundary condition for P is that it is $2\pi$-periodic: $P(\theta) = P(2\pi + \theta)$.