2

Consider the Laplace equation over a bounded domain $\Omega \subset \mathbb R^n$ with smooth boundary:

$-\Delta u = 0$.

This equation has a solution space of infinite dimension if $n > 1$. Now let $\Gamma_0$ and $\Gamma_n$ be subsets of the boundary, say, smooth submanifolds of $\partial\Omega$. We consider the homogeneous boundary conditions

$u = 0$ along $\Gamma_{0}$ (vanishing function values)

$\partial_n u = 0$ along $\Gamma_{n}$ (vanishing normal derivative)

The Laplace equation with these two boundary conditions is well-understood if

  • $\Gamma_0$ and $\Gamma_n$ are disjoint and essentially cover the whole boundary of $\partial\Omega$. Then the solution space has dimension one or zero.

But what if

  • $\Gamma_0$ and $\Gamma_n$ are disjoint but do not cover the boundary? Then there is a free boundary part where no boundary conditions are imposed.

  • $\Gamma_0$ and $\Gamma_n$ cover the boundary but are not disjoint? Then the boundary values seem to be overdetermined.

  • $\Gamma_0$ and $\Gamma_n$ have no restrictions?

I am particularly interested in which of these cases still allows for an infinite-dimensional space of solutions.

shuhalo
  • 7,485

1 Answers1

0
  • $\Gamma_0$ and $\Gamma_n$ are disjoint but do not cover the boundary

Then the set of solutions is infinite, because one can prescribe some values on the free boundary part, reducing this to a standard mixed boundary vvalue problem. There is an infinite-dimensional space of possible boundary values.

  • $\Gamma_0$ and $\Gamma_n$ are not disjoint

By this I understand that their intersection contains a relatively open subset of the boundary. Then the problem is overdetermined, regardless of whether $\Gamma_0$ and $\Gamma_n$ cover the entire boundary. No matter how small an open subset of $\partial \Omega$ is, prescribing both the values and normal derivative on that set allows for at most one harmonic function: see A PDE exercise from Gilbarg Trudinger. problem 2.2. Hence, the boundary data would have to be very special to be consistent.

(Looks like all cases are covered by the above)

  • Thank you. Yes, my implicit assumption was that their intersection could be a relatively open subset, nothing too pathological. – shuhalo Jul 24 '17 at 19:51