# Dirichlet-type condition on Riemannian manifold

Let $$M$$ be a Riemannian manifold and $$S \subset M$$ a compact submanifold of strictly lower dimension. Does every smooth function on $$S$$ extend to a harmonic function on a neighborhood of $$S$$?

This is not possible in general, for example because of constraints related to the analyticity of the extension.

For a concrete example, let $$M = \mathbf{R}^2$$, and $$S \subset \mathbf{R}^2$$ be a smooth, simple closed curve in the plane so that in the unit disc $$$$S \cap D_1 = \{ x_2 = 0 \} \cap D_1 = (-1,1) \times \{ 0 \}.$$$$

On $$S$$ define a function $$u_0$$ so that on the portion lying inside $$D_1$$, $$$$u_0: x_1 \in D_1 \cap S \mapsto \mathrm{e}^{-1/x_1^2}.$$$$ This is not analytic, whereas a harmonic extension to a neighbourhood of $$S$$, say $$u$$ would be.

The existence of such a harmonic extension $$u$$ is therefore absurd. In conclusion: no matter how small $$\delta > 0$$ is chosen, there is no harmonic function $$u: D_\delta \to \mathbf{R}$$ extending $$u_0$$, let alone a function defined on a neighbourhood of $$S$$.

• (+1) Thanks for your answer; that's very interesting. In the examples that I had in mind, the submanifold $S$ was actually real-analytic (in fact holomorphic). Do you think there can still be an issue? Oct 28 at 16:15
• I think the issue would basically be the same: you can't extend non-analytic 'initial data'. That being said---and perhaps you're already aware of this---if the given function $u: S \to \mathbf{R}$ is analytic, then some form of Cauchy-Kovalevskaya should give you the desired harmonic extension. Oct 28 at 16:21
• That's interesting. In my setup, I can actually assume that $u : S \to \mathbf{R}$ is analytic. So you say the answer to my question is 'yes' if $S$ and $u$ are analytic? Oct 28 at 16:25
• The answer should indeed be 'yes', at least if $S$ has codimension one. The result should follow from Cauchy-Kovalevskaya theory. Unfortunately I don't know the best reference, but I think Evans covers some of this. Whether there are some issues if $S$ is not embedded or $\dim S \leq \dim M - 2$ I do not know. Oct 28 at 16:40
• Thanks so much! All you said is very helpful. Oct 28 at 17:13