laplacian solution

Question

If I have $-\Delta u \ge 0$ and in $\Omega$ and $u=0$ on the boundary. and I have proved that $u=0$ is one of the solutions and I want to prove it's the only solution. based on the information on the question I can say $u \ge 0$ can I use strong maximum principle and say that it is the only solution of $u$ ?

Answer 1

This is not true. Let $\rho(x,y,z)\equiv \rho$ be the mass density over a closed ball $\bar B(0, r)\subset\mathbb R^3$, then the gravitational potential $u$ satisfies the Possion's equation $\nabla u = -4\pi G\rho\le 0$. By Newton's theorem, the gravitational force on the sphere $S(0,r)$ is along the radius, therefore $u$ is a constant $C$ on the shell, then $u-C$ is $0$ on the boundary, but not $0$ inside the ball.

To be more mathematical, $-x^2-y^2-z^2 + 1$ is such a function on the closed ball $\bar B(0, 1)$.

P.S. The question doesn't seem to be related to Laplace Transform.