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Solve the following boundary value problems for the Laplace equation on the semi-annular domain:

$ 1 < x^2 + y^2 < 2, y > 0 $

$ u(x, y) = 0, x^2 + y^2 = 1, u(x, y) = 1, x^2 + y^2 = 2, u(x, 0) = 0 $

Effort:

We move to polar coordinate and need to solve $u(r,\theta)$ such that:

$$ \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}=0 $$

with

$$ u(1,\theta) = 0, 0 < \theta < \pi, u(\sqrt{2},\theta) = 1, 0 < \theta < \pi,u(r,0) =0,u(r,\pi) =0, 0 \le r \le 2 $$

The general solution as we learned in class is :

$$ u(r, \theta) = \frac{a_0}{2}+ \sum_{n=1}^{\infty} \left( a_n r^n \cos(n\theta) + b_n r^n \sin(n\theta) \right) $$

The only boundary condition that yields non trivial solution is $u(\sqrt{2},\theta) = 1, 0 < \theta < \pi $ and we get

$$ u(\sqrt{2}, \theta) = \frac{a_0}{2}+ \sum_{n=1}^{\infty} \left( a_n 2^{\frac{n}{2}} \cos(n\theta) + b_n 2^{\frac{n}{2}} \sin(n\theta) \right) =1 $$

But now I get a solution of $ u(r,\theta) =1$ which I don't think is correct.

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1 Answer 1

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The solution form you have written is for a disk not an annulus or half-annulus Separation of variables can be used in your half-annulus too, and the form you will get is $$A + B\ln r + \sum_{n=1}^{\infty} (A_n r^n \cos n\theta + B_n r^n \sin n\theta ) $$ $$+ \sum_{n=1}^{\infty} (C_n r^{-n}\cos n\theta + D_n r^{-n} \sin n\theta )$$ You then have to use the boundary values to find the $A_n, B_n, C_n,$ and $D_n$.

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  • $\begingroup$ $ \ln(r) $ and $ r^{-n} $ are not possible functions because of the singularity at $r=0$ $\endgroup$
    – Tomer
    Commented May 5 at 21:31
  • $\begingroup$ The origin isn't part of the semi-annulus so you can have a singularity there. $\endgroup$
    – Zarrax
    Commented May 5 at 23:51
  • $\begingroup$ OK I see - $r^{-n}$ should be included but I don't think that $\lambda=0$ is a valid eigenvalue for those initial conditions(no linearity with $\theta$ is possible) so I think $A=B=0$ $\endgroup$
    – Tomer
    Commented May 6 at 0:22
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    $\begingroup$ Yes, $A$ and $B$ are zero. You can also show the $A_n$ and $C_n$ are all zero by looking at the solution on the real axis where the function are zero. So what you need to do is find the $B_n$ and $D_n$. $\endgroup$
    – Zarrax
    Commented May 6 at 14:57
  • $\begingroup$ Should we assume $2\pi$ periodicity with $\theta$ on this semi-annular domain ? $\endgroup$
    – Tomer
    Commented May 10 at 21:56

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