A (nontrivial) nonlinear Poisson equation Let $U \subset \mathbb R^3$ be bounded, with smooth boundary. Does the equation
$$ -\Delta f = e^{-f} \text{ in } U, \quad f = 0 \text{ on } \partial U$$
has a solution? Is there a name for this equation? Any idea or hint or reference is highly appreciated.

I want to provide an addendum to whpowell96's excellent answer. We know that the Green's function of the Laplacian operator in $\mathbb{R}^3$ is
$$G(x)=-\frac{1}{4\pi}\frac{1}{|x|}$$
which means our solution is
$$f(x)=(G\ast e^{-f})(x) = \int_U G(x-y)e^{f(y)}\,dy = \int_U -\frac{1}{4\pi}\frac{e^{f(y)}}{|x-y|}\,dy.$$
Call the integral operator above $\Phi(f)$, so we're looking for the solution to $\Phi(f)=f$. To use the contraction principle, we need $\Phi$ to be a function on a closed subset of some complete normed vector space. Well, the space of continuous functions on $U$, denoted $C^0(\overline{U})$, is a complete normed vector space under the supremum norm,
$$\|u(x)\| = \sup_{x\in U}|u(x)|$$
so we might as well assume $f$ is continuous. To show that $\Phi$ is a contraction, we need to show that
$$\|\Phi(u)-\Phi(v)\| < \|u-v\|$$
for any $u,v \in C^0(\overline{U})$. Provided all this is true, there must be a unique solution. The theorem that shows this also gives a method for approximating the solution. Simply pick any function $u\in C^0(\overline{U})$ and it turns out that
$$\lim_{n\rightarrow \infty} \Phi^n(u) = f$$
meaning that if you start with any function in that space, the repeated action of the integral operator on it will converge to the solution.
EDIT: I realized I did not account for the boundary conditions. This only modifies the Green's function, but the principle is the same.
