Author Topic: FE Problem 4  (Read 844 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
FE Problem 4
« on: April 14, 2015, 07:43:41 PM »
Consider the Laplace equation in the third of the disk
\begin{align}
&u_{xx} +  u_{yy} =0\qquad &&\text{in   } r = \sqrt{x^2+y^2} < a, \; x> |y|/\sqrt{3}
\label{P3-1}
\end{align}
with the boundary conditions
\begin{align}
& u =y\qquad &&\text{for  }  r=a,\ x>0,\  y>0, \label{P3-2}\\
& u=0 \qquad &&\text{for   }  \ y>0, \; x= y/\sqrt{3}\label{P3-3}\\
& u=0 \qquad &&\text{for  }  y<0,\; x=-y/\sqrt{3}. \label{P3-4}
\end{align}

a  Look for solutions $u$ in the form of  $u(r,\theta)= R(r) P(\theta)$  (in polar coordinates) and derive a set of  ordinary differential equations for $R$ and $P$. Write the correct  boundary conditions for $P$.

b Solve the eigenvalue problem for $P$ and find all eigenvalues.

c  Solve the differential equation  for $R$.

d  Find the solution $u$ of (\ref{P3-1})--(\ref{P3-4}).  Write the answer in terms of  Fourier series.
« Last Edit: April 14, 2015, 08:23:22 PM by Victor Ivrii »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: FE Problem 4
« Reply #1 on: April 17, 2015, 08:19:32 AM »
 Domain is a sector $-\frac{\pi}{3}<\theta < \frac{\pi}{3}$:


Then
 \begin{equation*}
 P''+\lambda P =0,\qquad P(-\frac{\pi}{3})=P(\frac{\pi}{3})=0.
 \end{equation*}
 Since our problem is symmetric with respect to $y=0$ and $u(a,\theta)=a\sin (\theta)$ is an odd function we can replace this problem by
 \begin{equation*}
 P''+\lambda P =0,\qquad P(0)=P(\frac{\pi}{3})=0
 \end{equation*}
and   $\lambda_n= 9 n^2$, $P_n = \sin  (3n\theta )$ with $n=1,2,\ldots$

Then $r^2R_n'' + rR_n'+9n^2 R_n=0$ and $R_n=A_nr^{3n}+Br_n^{-3n}$ and we need to take $B_n=0$. 
 
 
 So,
 \begin{equation*}
 u(r,\theta)=\sum_{n=1}^\infty A_n r^{3n } \sin (3 n \theta ).
 \end{equation*}
 We need to satisfy
  \begin{equation*}
 u(a,\theta)=\sum_{n=1}^\infty A_n a^{3n} \sin (3 n \theta )= a\sin (\theta)
 \end{equation*}
 and therefore
 \begin{multline*}
A_n a^{3n} =\frac{6a}{\pi }\int_{0}^{\frac{\pi}{3}} \sin (3n \theta )\sin (\theta)\,d\theta=
\frac{3a}{\pi }\int_{0}^{\frac{\pi}{3}} \Bigl[ \cos ((3n-1)\theta)-\cos ((3n+1)\theta) \Bigr]\,d\theta=\\[3pt]
\frac{3a}{\pi } \Bigl[ \frac{\sin ((3n-1)\theta)}{3n-1}-\frac{\sin ((3n+1)\theta)}{3n+1} \Bigr]_{\theta=0}^{\theta=\pi/3}=
\frac{3a}{\pi } \Bigl[ \frac{\sin (\pi n- \frac{\pi}{3})}{3n-1}-\frac{\sin (\pi n +\frac{\pi}{3}}{3n+1} \Bigr]=\\[3pt]
\frac{3a}{2\pi } (-1)^{n+1}\Bigl[ \frac{1}{3n-1}+\frac{1}{3n+1} \Bigr]= (-1)^{n+1}\frac{9n a}{(9n^2-1)\pi }
\end{multline*}
 and
\begin{equation*}
 u(r,\theta)=\sum_{n=1}^\infty (-1)^{n+1}\frac{9n }{(9n^2-1)\pi } a^{1-3n}  r^{3n } \sin (3 n \theta ).
 \end{equation*}