Author Topic: TT2--P4  (Read 1807 times)

Victor Ivrii

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TT2--P4
« on: March 23, 2018, 06:12:46 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
\tag{1}
\end{equation}
with the Dirichlet boundary conditions as $\theta=0$ and $\theta=2\pi$}
\begin{equation}
u|_{\theta=0}=u|_{\theta=2\pi}=0
\tag{2}
\end{equation}
and the Neumann boundary condition as $r=9$
\begin{equation}
u_r|_{r=9}=1.
\tag{3}
\end{equation}
Using separation of variables find solution as a series.

Elliot Jarmain

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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 »