Author Topic: TT2--P4N  (Read 901 times)

Victor Ivrii

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TT2--P4N
« on: March 23, 2018, 06:18:18 AM »
Consider Laplace equation in the disc with a cut
\begin{equation}
u_{rr} +\frac{1}{r}u_r +\frac{1}{r^2}u_{\theta\theta}=0 \qquad  r<9,\,0<\theta<2\pi
\label{4-1}\end{equation}
with the Neumann boundary conditions as $\theta=0$ and $\theta=2\pi$}
\begin{equation}
u_\theta|_{\theta=0}=u_\theta|_{\theta=2\pi}=0\label{4-2}
\end{equation}
and the Dirichlet boundary condition as $r=9$
\begin{equation}
u|_{r=9}=\pi-\theta.\label{4-3}
\end{equation}
Using separation of variables find solution as a series.

Jingxuan Zhang

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Re: TT2--P4N
« Reply #1 on: March 23, 2018, 07:57:27 AM »
General solution for \eqref{4-1} is, due to boundary condition and consideration of regularity at zero,
\begin{equation}\label{4-4}
u=\frac{1}{2} A_0+
\sum_n r^n\Bigl( A_n\cos n\theta \Bigr).\end{equation}
Plugging in \eqref{4-3} and using even continuation
\begin{equation}\begin{split}
9^n A_n&=\frac{2}{\pi}\int_0^\pi (\pi-\theta)\cos n\theta\,d\theta
=\frac{2}{\pi}\int_0^\pi\theta'\cos n(\theta'-\pi)\,d\theta'\\
&=\frac{2}{n\pi}\int_0^\pi\sin n(\theta'-\pi)\,d\theta'\\
&=\left\{\begin{split}&0&& \text{ n even},\\ &\frac{4}{\pi n^2}&& \text{ n odd.}\end{split}\right.
\end{split}\label{4-5}\end{equation}
Combining \eqref{4-4}, \eqref{4-5}:
$$u=\frac{4}{\pi}\sum_k \Bigl(\frac{r}{9}\Bigr)^{2k+1} \frac{\cos(2k+1)\theta}{(2k+1)^2}.$$

« Last Edit: March 23, 2018, 02:45:35 PM by Jingxuan Zhang »