The solution of $\Delta u=u^3$ with zero boundary values is identically zero

Question

My question:

My attempt:

I tried to use the Representation using Green's formula: Since $u=0$ on the boundary and $f(x)=x^3$, then the formula becomes:

$$u(x)=\int_\Omega y^3G(x,y)dy \quad (x\in\Omega)$$

Then I got stuck, how to proceed to get $u=0$?

Can anyone give me some clues or answers?

Answer 1

Green's idenitity says that $$\int_{\Omega}u\Delta u+\int_\Omega |\nabla u|^2=\int_{\partial\Omega} u(\nabla u\cdot n).$$ Since $u=0$ on $\partial\Omega$ by assumption, the right hand side is zero. On the other hand, $\Delta u=u^3$ in $\Omega$ by assumption, the left hand side is equal to $$\int_{\Omega}u^4+\int_\Omega |\nabla u|^2.$$ Combining all these, we have $$\int_{\Omega}u^4+\int_\Omega |\nabla u|^2=0$$ which implies that $u\equiv 0$.