APM346-2015F > Test 2

TT2-P4

(1/1)

Victor Ivrii:
Consider Laplace equation in the half-disk

u_{rr} +\frac{1}{r}u_r +\frac{1}{r^2}u_{\theta\theta}=0 \qquad  r<1,0<\theta<\pi \label{4-1}

with the Dirichlet boundary conditions as $\theta=0$ and $\theta=\pi$

u|_{\theta=0}=u|_{\theta=\pi}=0\label{4-2}

and the Robin boundary condition as $r=1$

(u_r + u)|_{r=1}=1.\label{4-3}

Using separation of variables find solution as a series.

Emily Deibert:
Edit: Note, I accidentally wrote $\phi$ instead of $\theta$ throughout this problem. I do not wish to go back and change them all. Please consider $\phi$ as if I had written $\theta$!

As usual, let's separate variables. We let:

u(x,t) = \Phi(\phi)R(r)

Plugging this into the given equation yields:

\Phi R'' + \frac{1}{r} \Phi R' + \frac{1}{r^2} \Phi'' R = 0 \longrightarrow \left( \frac{r^2R''}{R} + \frac{rR'}{R} \right) + \frac{\Phi''}{\Phi} = 0

Now as usual, we have a term that depends on $R$ and a term that depends on $\Phi$. So we set the $\Phi$ term equal to $\lambda$ and solve the resulting ODE.

\Phi'' - \lambda\Phi = 0

Let's first consider $\lambda = - \omega^2 < 0$. This yields a solution:

\Phi(\phi) = A\cos\omega\phi + B\sin\omega\phi

Using the BCs, we show that:

\Phi(0) = A = 0

\Phi(\pi) = B\sin\omega\pi = 0 \longrightarrow \omega\pi = n\pi \longrightarrow \omega = n

So for negative eigenvalues we conclude that $\Phi_n(\phi) = B_n\sin(n\phi)$ with $\lambda = -n^2$.

Let's check the positive case ($\lambda = \omega^2 > 0$). Wellllll, this will yield:

\Phi(\phi) = C\cosh\omega\phi + D\sinh\omega\phi

Using the boundary conditions here gives us:

\Phi(0) = C = 0

\Phi(\pi) = D\sinh\omega\pi = 0 \longrightarrow \omega\pi = 0 \longrightarrow \omega = 0

But we are checking the case where $\omega$ is a positive (i.e. greater than zero) eigenvalue, so this doesn't make sense. We conclude that there are no positive eigenvalues.

For $\lambda = 0$, the solution will be $\Phi(\phi) = E\phi + F$. The first BC gives $\Phi(0) = F \longrightarrow F = 0$. The second gives $\Phi(\pi) = E\pi = 0 \longrightarrow E = 0$. We conclude that there is no zero eigenvalue.

Now we must consider the $R$ equation. Recall that we have:

\left( \frac{r^2R''}{R} + \frac{rR'}{R} \right) = - \lambda \longrightarrow r^2R'' + rR' + \lambda R = 0

This is a typical Euler equation, so we assume a solution $R = r^m$. Plugging this in gives:

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

The solutions are then:

R(r) = Gr^n + Hr^{-n}

Well where are we? In the half-disk! So we discard the $H$ term, which will blow up at the origin. This leaves us with $R_n(r) = G_nr^n$.

Now let's combine the solutions, as we usually do. We can write our general solution as:

u(r, \phi) = \sum_1^{\infty}a_nr^n\sin(n\phi)

Now we shall make use of this Robin boundary condition. We have that:

u(r, \phi) |_{r=1} = \sum_1^{\infty}a_n\sin(n\phi)

u_r(r, \phi) |_{r=1} = \sum_1^{\infty}a_nn\sin(n\phi)

Altogether, this is:

(u_r + u) |_{r=1} = \sum_1^{\infty} \left(a_nn\sin(n\phi) + a_n\sin(n\phi) \right) = 1 \longrightarrow \sum_1^{\infty} \left(a_n\sin(n\phi)(n+1) \right) = 1

Now we'd better find out what $a_n$ is. So:

a_n = \frac{2}{\pi(n+1)} \int_0^{\pi}1\sin(n\phi)d\phi = \frac{2(1 - \cos(n\pi))}{\pi(n+1)n} = \frac{2(1 - (-1)^n)}{\pi(n+1)n}

So we finally get that:

u(r, \phi) = \sum_1^{\infty} \frac{2(1 - (-1)^n)}{\pi(n+1)n} r^n\sin(n\phi)

If anyone notices any errors, please let me know!

Bruce Wu:
We can go one step further by noting that all $a_n$ for even $n$ are zero. Then $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^n\sin(n\theta)}{n(n+1)}$$

Emily Deibert:

--- Quote from: Fei Fan Wu on November 19, 2015, 12:33:42 AM ---We can go one step further by noting that all $a_n$ for even $n$ are zero. Then $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^n\sin(n\theta)}{n(n+1)}$$

--- End quote ---

Of course! Thank you Fei Fan Wu!