Poisson equation on manifolds

Question

Let $(\mathcal{M},g)$ be a compact Riemannian manifold with Levi-Civita connection $\nabla$. It is well-known that the Poisson equation

$$\Delta u=f$$

does have a solution on $C^{\infty}(\mathcal{M})$ for a source $f\in C^{\infty}(\mathcal{M})$ whenever

$$\int_{\mathcal{M}}f\,\mathrm{vol}_{g}=0.$$

Now, if I take $\mathcal{M}$ to be not necessarily compact and a compactly-supported source $f\in C^{\infty}_{c}(\mathcal{M})$ with vanishing integral, i.e. $\int_{\mathcal{M}}f\,\mathrm{vol}_{g}=0$, does there exist a solution to

$$\Delta u=f$$

on $C^{\infty}_{c}(\mathcal{M})$?

($\mathcal{M}$ is assumed to have empty boundary)

If needed, we can assume $(\mathcal{M},g)$ to be geodesically complete.

Answer 1

As alluded to in a comment of @PietroMajer the reason that one can't solve in the space of compactly supported functions is somewhat subtle (and in particular, I think the other answers are not complete).

Basically, for a solution of $\Delta_g f=g$ to exist with both $f,g\in C^\infty_c(M)$ one has to have that $g$ is $L^2$ orthogonal to every harmonic function. There are, in general, a plethora of these when the ambient space is non-compact and these impose serious restrictions on the allowed $g$. In contrast, when the manifold is compact the only harmonic functions are the constant functions which is why the condition of integrating to zero suffices.

Just to give a concrete example in $\mathbb{R}^2$. Let $\phi\geq 0$ be a non-negative compactly supported function with $\phi=1$ on $B_1$. Set $g=(x^2-y^2)\phi$. Now suppose one had a compactly supported solution $f$ to $\Delta f=g$. Integration by parts gives \begin{align*} 0&=\int_{\mathbb{R}^2} f \Delta (x^2-y^2) \\ &=\int_{\mathbb{R}^2} (x^2-y^2) \Delta f \\ &= \int_{\mathbb{R}^2} \phi (x^2-y^2)^2 >0 \end{align*} This is a contradiction so there can be no such solution.

It might also be illustrative to see what happens in $\mathbb{R}$. You can see that to solve $f''=g$ for $f$ and $g$ compactly supported as this is just an integration and the conditions are quite explicit.

Answer 2

In general, it is impossible to find a compactly supported solution $u$ to the equation $\Delta u = f$. For the sake of argument, consider the case $M = \mathbb{R}^n$ and $f$ is supported inside a ball $B$ and suppose such a function $u$ exists. Outside of $B$, the solution $u$ to $\Delta u = f$ is harmonic. If $u$ also has compact support then it follows from unique continuation that $u$ vanishes identically on the complement of $B$. Now on the ball $B$, we see that $u$ is the unique solution to the equation $\Delta u = f$ with $u$ vanishing on $\partial B$. This means our only hope is to take $u$ to be the solution to the problem on the ball $B$ with Dirichlet boundary conditions, extended by zero outside. In general, this won't be smooth however, or even differentiable, because the normal derivative for this solution is not necessarily zero.

Answer 3

Maybe Theorem 4.8 in Thierry Aubin - Some Nonlinear Problems in Riemannian Geometry will help. It said that

let $\bar{W}_n$ be a compact Riemannian manifold with boundary, then there exists a solution $\varphi \in C^{\infty}(\bar{W}_n)$ of $$\Delta \varphi =f$$ here $f$ is smooth and $\varphi$ vanishes on the boundary.