Energy method for the PDE $\Delta u - u^3 = f$

Question

I wish to prove that the following PDE has at most one solution

\begin{align} \Delta u - u^3 &= f, \quad x \in \Omega \\ u &= \phi, \quad x \in \partial \Omega \end{align}

with $f$ continuous in $\Omega$. I tried to apply the energy method, but it didn't work.

I started by supposing that there are two solutions, $u,v$. Then I set $w = u - v$. Therefore, we have that

$$\Delta w = u^3 - v^3$$

where $w$ equals to zero on the boundary. I tried multiplying by w and integrating over $\Omega$. Then I got

$$\|\nabla w\|^{2}_{L^{2}} = \|u^{2} + uv + v^{2}\| \|w\|^{2}_{L^{2}}$$

Then, I got stuck.

I appreciate any help. Thank you.

Answer 1

You almost have the right argument. You still have invoked Hölder's inequality to early. As you have written, you get $$ 0 = \| \nabla w\|_{L^2}^2 + \int_\Omega(u^2 + u\,v + v^2)\,(u-v)^2\,dx. $$ Now, we have$$u^2+u\,v+v^2 = \frac12 u^2 + \frac12v^2 + \frac12 (u+v)^2 \ge 0.$$ This implies $\nabla w = 0$.