Laplace equation in an annulus

Question

Suppose we fix $0 < R_{0} < R$ and $f: \mathbb{R}^{3} \to \mathbb{R}$ is the solution of the Laplace equation $-\Delta f = 0$, in the annulus $|x| \in (R_{0},R]$. Suppose $f$ is radial. Is it true that $f$ is necessarily of the form $f(r) = A+br^{-1}$, with $r = |x|$? In other words, all smooth radial solutions of the Laplace equation inside this annulus are of this form?

Answer 1

Yes.

The assumption that your solution is radial implies the boundary values on the inner ring are constant, likewise on the outer ring. So boundary values are determined up to two constants, and can be matched by appropriately setting the two constants $A$ and $b$ in $f(r) = A + br^{-1}$.

By the uniqueness of the Dirichlet problem, this is therefore the only solution.

Evans also proves this fact directly (and more generally in any dimension) in §2.2.1 of Partial Differential Equations. Adapted here:

Let $u:\mathbb{R}^n\to \mathbb{R}$ be a radial solution to $-\Delta u = 0$. That is, there exists a well-defined $v:\mathbb{R}\to\mathbb{R}$ where $u(x) = v(|x|)$. Letting $r = |x|$ we compute

$$\frac{\partial r}{\partial x_i} = \frac{1}{2}\left(x_1^2 + \cdots + x_n^2\right)^{-1/2}2x_i = \frac{x_i}{r} \qquad (x\not = 0)$$

Substituting this into the chain rule we get

$$u_{x_i} = v'(r)\frac{x_i}{r}.$$ Differentiating and substituting again gives

$$u_{x_ix_i} = v''(r)\frac{x_i^2}{r^2} + v'(r)\left(\frac{1}{r}- \frac{x_i^2}{r^3}\right).$$

Summing over $i$, the Laplace operator becomes $$\Delta u = v''(r) + \frac{n-1}{r} v'(r).$$ Applying the fact that $\Delta u = 0$ reduces this to the ODE

$$v'' + \frac{n-1}{r}v' = 0.$$

Whenever $v' \not = 0$ we can rewrite this as

$$\log (|v'|)' = \frac{v''}{v'} = \frac{1-n}{r}$$

which after integrating and simplifying is $$v'(r) = ar^{1-n},\qquad a\in \mathbb{R}.$$

Integrating once again then gives the general radial solution for $r>0$:

$$v(r) = \begin{cases}b\log r + A & \text{if $n=2$}\\ br^{2-n} + A & \text{if $n\geq 3$} \end{cases}$$