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?
Asked
Active
Viewed 114 times
0
-
It might be necessary to be a bit more specific about what $V$ is (e.g. a closed connected subset of $\mathbb R^n$ with Dirichlet boundary conditions). If $V$ is not connected, for instance we can use a counterexample of two disjoint intervals in $\mathbb R$. – Kajelad Aug 14 '17 at 19:59
-
Thanks. Assume ${\mathbb{R}}^3$ and usual geometry that allows for a unique solution within V given Dirichlet boundary condition on the surface of V. – D_J_S Aug 14 '17 at 20:18
1 Answers
0
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 $V$.
D_J_S
- 151
- 6