Continuation (extension) of harmonic functions

Question

Suppose $(M,g)$ is a simply connected smooth Riemannian manifold with smooth boundary and suppose that $U \subset \partial M$ is a smooth open connected subset of the boundary. Now my question is the following:

suppose $f,h \in C^{\infty}_c(U)$. Does there exist a harmonic function in $M$ such that it's Dirichlet and Neumann values on $U$ are $f$ and $h$ respectively? More precisely does there exist $u$ such that:

$\triangle_g u=0 $ in $M$.

$u|_{U}=f$.

$ \partial_{\nu} u|_U = h$.

Here $\nu$ represents the normal outward unit vector field on $\partial M$.

Note: the answer might be no in general but maybe there is some relaxation on assumptions that makes this possible?

Answer 1

The answer is no. Let $M$ be a region in the upper half-space $x_1>0$ in $R^n$, (you can take $n=2$) and $\partial M$ contains an open piece $U$ of the plane $x_1=0$. Take $f=0$ in $U$. Then your harmonic function $u$ extends by symmetry, the extended function will be real analytic on $U$, so $h$ must be real-analytic, so you cannot assign arbitrary $C^\infty$ function $h$. And of course you cannot assign any $h\in C^\infty_c$, except zero.

Even if your $f$ and $h$ are real-analytic, the answer is still no. Indeed, if $f$ and $h$ are real analytic, consider them as the data for the Cauchy problem for the Laplace equation. According to Cauchy-Kowalewski theorem, we have uniqueness. But then the analytic continuation of solution to $M$ is not guaranteed. There are no reasonable conditions on $f$ and $h$ which will guarantee such an extension.