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What's the solution to Laplace's equation $\nabla^2V=0 $ in the annulus with centre 0, inner radius 1, and outer radius 2, with boundary conditions $V=0$ on the inner boundary and $V=1$ on the outer boundary?

By separating variables, I got the general solution $V(r,\theta)=\Sigma[(C_n r^n + D_n r^{-n})(A_n \cos n\theta + B_n \sin n\theta)]$. The inner boundary condition gives $D_n = -C_n \forall n$. The outer boundary condition gives $\Sigma[C_n(2^n - 2^{-n})(A_n \cos n\theta + B_n \sin n\theta)] = 1.$ But this gives a Fourier series with zero constant term being identically equal to 1. By uniqueness of Fourier series, isn't this a contradiction?

Or is there a better method than using separation of variables, which gives a different general form for the solution?

Many thanks for any help with this!

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Just a guess: have you tried a radially symmetric solution? – Siminore Apr 24 '12 at 9:55
up vote 0 down vote accepted

Look for a solution $V=V(r)=V(|x|)$. Then $\Delta V=0$ can be rewritten as $$V''(r)+\frac{2}{r}V'(r)=0$$ in the interval $(1,2)$. Integrating this equation you find $V(r)=C_1+\frac{C_2}{r}$ for suitable constants $C_1$ and $C_2$. The boundary conditions imply $C_1=2$ and $C_2=-2$. Hence $V(x)=2-\frac{2}{|x|}$ solves your equation in the annulus.

Edit: as pointed out in the comments, this treated the case in dimension $3$. In dimension $2$, radially symmetric harmonic functions have the form $r \mapsto C_1+C_2 \log r$, but the conclusion is similar: just match the boundary conditions and find $C_1$ and $C_2$.

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Your formula for $\Delta V$ applies to ${\mathbb R}^3$. In the plane there is no factor of $2$. – Christian Blatter Apr 24 '12 at 12:53
Sorry, when I read $\nabla^2$ I immediately think of the physical space. – Siminore Apr 24 '12 at 13:01
Your $V(r)$ does not solve the corrected equation. – Christian Blatter Apr 24 '12 at 13:26

I don't know what you did exactly in your computation process. At any rate the "eigenvalue" $n=0$ is a double eigenvalue, and besides the constant solution there is an additional solution, namely $u(r,\theta):=\log r$. This should give you enough manoeuvring space to satisfy the boundary conditions with a rotationally symmetric solution.

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Functions of the form $A + B \ln(r)$ are harmonic, and you can use your boundary conditions to solve for $A$ and $B$. You obtain the solution ${\displaystyle {\ln(r) \over \ln(2)}}$.

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