Harmonic functions question

Question

Consider a function which is harmonic in V (ie $\nabla^2 \Phi({\bf x}) = 0$ in V). If $\Phi({\bf x})$ is known on an arbitrary closed surface within V is the potential on the surface of V uniquely determined?

Answer 1

I think the proof goes like this. Please check me on this.

First, let's define $V_1$ (with surface $S_1$) to be a subset of $V$ (with surface $S$). Since $\Phi$ is harmonic in $V$ then, by the well known uniqueness theorem for harmonic functions, the solution within $V_1$ is uniquely determined by its value on $S_1$.

Using the following theorem (see Foundations of Potential Theory by Kellog):

$\bf{Theorem}$: If U is harmonic in a domain T, and if U vanishes at all the points of a domain T' in T, then U vanishes at all points of T.

we can show that $\Phi$ is uniquely determined in all $V$ by its value on $S_1$. To do this assume that there are two harmonic functions $\Phi_1$ and $\Phi_2$ which are equal within $V$ (since both have the same boundary condition on $S_1$) and consider the harmonic function $U = \Phi_1 - \Phi_2$. Using the above theorem $U$ is zero at all points in $V$ and therefore $\Phi_1 = \Phi_2$ at all points in $V$. Therefore $\Phi$ is uniquely determined within $V$ by its value on $S_1$.

The final step is achieved by again invoking the usual uniqueness theorem to conclude that there is a uniquely determined value of $\Phi$ on $S$ given $\Phi$ within $V$.