A (nontrivial) nonlinear Poisson equation

Question

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.

Answer 1

I don't have a definitive answer, but this makes it a little more tractable. The differential operator is uniquely invertible and has a Green's function $G(x,y)$ that is positive. This allows us to reformulate our equation as $$f(x) = \int_U G(x,y)e^{-f(y)}dy = \Phi(f).$$ If you can show that the integral operator on the RHS is a contraction, then the solution exists and is unique by the fixed point theorem.

Let us use the $\infty$-norm for this analysis. Then we have $$\|\Phi(f)-\Phi(g)\| \leq \int_U\|G(\cdot,y)\|\|e^{-f(y)}-e^{-g(y)}\|dy.$$ Now notice that any solution of this integral equation must be positive, so the exponentials are both bounded above by 1. Now we have $$\|\Phi(f)-\Phi(g)\| \leq \int_U\|G(\cdot,y)\|dy.$$

If this value can be bounded above by 1, then there is a unique solution. However, this depends (only) on the domain. This value can also be bounded (in a different norm, so you must be more careful), by the reciprocal of the smallest eigenvalue of the Laplace operator on this domain, which (I believe) is generally inversely proportionally to the volume of the region $U$. Therefore, we expect this problem to have a unique solution on small domains rather than large domains.

Answer 2

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.