### Author Topic: TT2--P4  (Read 1986 times)

#### Victor Ivrii

• Elder Member
• Posts: 2607
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##### TT2--P4
« on: March 23, 2018, 06:12:46 AM »
Consider Laplace equation in the disc with a cut

\tag{1}

with the Dirichlet boundary conditions as $\theta=0$ and $\theta=2\pi$}

u|_{\theta=0}=u|_{\theta=2\pi}=0
\tag{2}

and the Neumann boundary condition as $r=9$

u_r|_{r=9}=1.
\tag{3}

Using separation of variables find solution as a series.

#### Elliot Jarmain

• Jr. Member
• Posts: 6
• Karma: 4
##### Re: TT2--P4
« Reply #1 on: March 23, 2018, 08:43:13 AM »
Making a seperation of variables $u(r, \theta) = R(r)\Theta(\theta)$ yields
\begin{gather*}
R''\Theta + \frac{1}{r}R'\Theta + \frac{1}{r^2}R\Theta'' = 0 \\
r^2\frac{R''}{R} + r\frac{R'}{R} + \frac{\Theta''}{\Theta} = 0 \\
r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda, \quad \frac{\Theta''}{\Theta} = -\lambda
\end{gather*}
Solving the $\Theta$ equation:
\begin{equation*}
\Theta(\theta) =
\begin{cases}
A\theta + B, &\text{if $\lambda = 0$;}\\
Ae^{\sqrt{-\lambda}\theta} + Be^{-\sqrt{-\lambda}\theta} &\text{if $\lambda < 0$;}\\
A \cos{\sqrt{\lambda}\theta} + B \sin{\sqrt{\lambda}\theta} &\text{if $\lambda > 0$.}
\end{cases}
\end{equation*}
Plugging the boundary conditions (2) yields the trivial solution for $\lambda \leq 0$. And for $\lambda > 0$:
\begin{align*}
u|_{\theta=0}=0 &\implies A = 0 \\
u|_{\theta=2\pi}=0 &\implies B \sin{\sqrt{\lambda}2\pi} = 0 \\ &\implies
\sqrt{\lambda} 2\pi = n \pi, \quad n = 1,2,3, \dots
\end{align*}
Therefore $\lambda_n = \frac{n^2}{4}$ and $\Theta_n(\theta) = \sin{\frac{n\theta}{2}}$ for $n = 1,2,3, \dots$
To solve the $R$ equation we first assume that $R$ has the form $R(r) = r^m$ for some $m$. This yields
\begin{equation*}
m(m-1) +m = \lambda \implies m^2 = \lambda
\implies m = \pm \frac{n}{2}
\end{equation*}
So $R(r) = A_n r^{\frac{n}{2}} + B_n r^{-\frac{n}{2}}$ and
\begin{equation*}
u(r, \theta) =
\sum_{n=1}^\infty (A_n r^{\frac{n}{2}} + B_n r^{-\frac{n}{2}}) \sin{\frac{n\theta}{2}}
\end{equation*}
Since $r < 9$, we set $B_n = 0$ for all $n$ to avoid singularities at the origin. The boundary condition (3) yields:
\begin{equation*}
u_r|_{r=9}= \sum_{n=1}^\infty (A_n \left(\frac{n}{2}\right)9^{\frac{n}{2} - 1}) \sin{\frac{n\theta}{2}} =
\sum_{n=1}^\infty \left(\frac{A_n3^{n-2}n}{2}\right) \sin{\frac{n\theta}{2}} = 1
\end{equation*}
So
\begin{equation*}
A_n = \frac{2}{3^{n-2}n\pi}
\int_0^{2\pi}\sin{\frac{n\theta}{2}}
= \frac{2}{3^{n-2}n\pi} \frac{2}{n}
\left. \left(
-\cos{\frac{n\theta}{2}}
\right) \right|_0^{2\pi}
= \frac{4}{3^{n-2}n^2\pi}
(1 - (-1)^n)
\end{equation*}
\begin{equation*}
A_n =
\begin{cases}
\frac{8}{3^{n-2}n^2\pi} &\text{if $n$ odd;}\\
0 &\text{if $n$ even.}
\end{cases}
\end{equation*}
Therefore
\begin{equation*}
u(r, \theta) =
\sum_{n\geq1, \, n \, \text{odd}} \frac{8r^{\frac{n}{2}}}{3^{n-2}n^2\pi}  \sin{\frac{n\theta}{2}}
\end{equation*}
« Last Edit: March 23, 2018, 08:46:14 AM by Elliot Jarmain »