Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the Laplacian $\Delta = d/dx^2 + d/dy^2$ on $\mathbb{R}^2$.

This is true: Let $f$ be a nonnegative function, not identically zero. Then any positive solution of $\Delta u = f u$ is unbounded.

Let $g$ be any function such that $\Delta u = g u$ has a positive solution. Is it the case that any positive solution to $\Delta u = ( f + g ) u$ is unbounded?

share|improve this question

2 Answers 2

The answer to the first question is yes. If $u$ is positive and $f$ non-negative then the RHS is non-negative, thus $u$ is subharmonic. Subharmonic function bounded from above must be constant ("Liouville's theorem" for subharmonic functions). If it is constant then the LHS iz $0$, then RHS is $0$ and this is a contradiction.

share|improve this answer

The second problem can be solved using the following Liouville type theorem, which was first used in works related to De Giorgi conjecture (see for example L. Ambrosio and X. Cabré). Let $u$ be a positive solution of $\Delta u=gu$ and $v$ a bounded solution of $\Delta v=fv+gv$. Then consdier $\psi=v/u$. It satisfies $$div(u^2\nabla \psi)=fu^2\psi.$$ Using $\psi\eta^2$ as a test function with $\eta\in C_0^\infty$, we get $$\int fu^2\psi^2\eta^2+u^2|\nabla\psi|^2\eta^2\leq\int v^2|\nabla\eta|^2.$$ So if $v$ is bounded, you can use a classical technique by choosing a good test function $\eta_R$ with some $\log$ dependence to show the RHS goes to $0$ as the radius $R\to\infty$.

Note that this is something special in dimension $2$. The existence of positive solution $u$ is also related to the stablity of $\Delta-g$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.