Solution of Laplace's equation in an annulus with constant Dirichlet conditions?

Question

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!

Answer 1

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 $C_2$.

Answer 2

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.

Answer 3

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 ${\displaystyle {\ln(r) \over \ln(2)}}$.