I have been thinking a lot lately about Poisson's equation:
$$ \nabla^2 \phi = f $$
where $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ is given, and we solve for $\phi: \mathbb{R}^3 \rightarrow \mathbb{R}$. Specifically, I've been thinking about when this equation has solutions. I care about weak solutions to this equation. This means that $f \in L^1_{loc}(\mathbb{R}^3)$ and we solve for $\phi \in L^1_{loc}(\mathbb{R}^3)$ such that for any test function $\psi \in C^\infty_c(\mathbb{R}^3)$, we have
$$ \int_{\mathbb{R}^3} f(x) \psi(x) \mathrm{d}x = - \int_{\mathbb{R}^3} \phi(x) \nabla^2 \psi(x)\mathrm{d}x $$
My question is whether there is any known condition on $f$ that determines whether Poisson's equation can be solved?
For example, if $f$ decays quickly enough, then we can solve for $\phi$ using Green's functions. Define
$$ G(x) = \frac{1}{4\pi |x|} $$
Then, $\phi = G * f$ solves Poisson's equation, where $*$ is the convolution. Since $G \in L^1_{loc}(\mathbb{R}^3)$, the convolution $G * f$ exists if $f$ has compact support, in which case $G * f$ does solve Poisson's equation. But if $f$ does not have compact support, then this approach may not work. For example, if $f(x) = \frac{1}{1 + |x|^2}$, then the convolution $f * G$ is infinite.
I haven't been able to prove that any function does not satisfy Poisson's equation.
Thank you!
This is Theorem 2.2 on page 76 of "Introduction To Partial Differential Equations" by Folland:
Suppose that $f \in L^1(\mathbb{R}^n)$ and that either $n > 2$, or $n = 2$ and $\int |f(y)| \log(|y|) \,dy < \infty$. Then $f * G$ is well defined as a locally integrable function, and $\Delta(f * G) = f$.