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I came across the following boundary value problem that I can't solve. It's the Laplacian on the upper half of an annulus with radius $1 \leq r \leq 2$ in polar coordinates:

$u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta} =0$

$u(1,\theta)=u(2,\theta)=0, ~~~ 0<\theta<\pi$

$u(r,0)=0, ~~ u(r,\pi)=r, ~~~ 1

The problem mentions that one must show in this case the choice of separation constant $\lambda=-\alpha^2<0$ leads to eigenvalues and eigenfunctions. The problem is that in this case, after separating variables and solving

$ \Theta'' + \lambda \Theta = 0$

$ r^2R'' + rR' - \lambda R =0$

we would get $\Theta(\theta)=c_1 \cosh(\alpha \theta) + c_2 \sinh(\alpha \theta)$ which is not periodic in $\theta$. Usually we always have $\alpha^2>0$ which gives us the periodic sine and cosine solution for $\Theta$. Help would be appreciated!

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    This prescription of boundary values is not continuous in $(1, \pi)$ and $(2,\pi)$. You sure you cited that correctly?2012-07-27
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    @Thomas According to the author of the book, there is no typo. I copied it correctly.2012-07-27
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    @Thomas I see what you mean though, this is a problem.2012-07-27
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    Are you aware that you can use double dollar signs to turn these into displayed equations? That centres them and makes the fractions come out a lot nicer.2012-07-27
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    Also, note that you can't "solve a Laplacian" -- the Laplacian is a differential operator; what you're solving here is the Laplace equation, which results from setting the result of applying the Laplacian to zero.2012-07-27
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    You can take the separation constant as $\alpha^2$ if you like. The only thing is you will fail to get the most simplified form when facing $u(r,0)=0$ and $u(r,\pi)=r$2012-07-30

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