I'm trying to solve the following problem:
Discuss the uniqueness of the following problems using energy methods:
\begin{cases} \Delta u -u{\displaystyle \int_{\Omega}}u^2(y)dy=f \quad \mbox{in } \Omega\\ u=\varphi \quad \quad \quad \quad \quad \quad \quad \ \ \mbox{on } \partial\Omega \end{cases}
with $\Omega \subset {\rm I\!R}^n$ bounded, $f$ and $\varphi$ continuous and $u\in C^2(\Omega)\cap C^1(\overline{\Omega})$.
I started considering $w=u-v$. I have $$\Delta w=\Delta u - \Delta v=u||u||^2_{L^2}+f-v||v||^2_{L^2}-f$$ and then, multiplying by $w$ and integrating by parts, I get $$||\nabla w||^2+{\displaystyle \int_{\Omega}}(u-v)(u||u||^2_{L^2}-v||v||^2_{L^2})=0.$$ And then I get stuck. Do you have some suggestion? Thanks!
Close to solution, you are. FOIL inside of the integral, you must. Then you get, $$||\nabla w||_{L^2}^2+\int_{\Omega}(u-v)\left(u||u||_{L^2}^2-v||v||_{L^2}^2\right) = ||\nabla w||_{L^2}^2+\frac{1}{2}\left(||u||_{L^2}^2-||v||_{L^2}^2\right)^2+\frac{1}{2}||w||_{L^2}^2\left(||u||_{L^2}^2+||v||_{L^2}^2\right)=0$$ after applying the fact that $$||w||_{L^2}^2 = ||u||_{L^2}^2+||v||_{L^2}^2 - 2\int_{\Omega}uv$$ and doing some algebra. So now we have $$||\nabla w||_{L^2}^2+\frac{1}{2}\left(||u||_{L^2}^2-||v||_{L^2}^2\right)^2=-\frac{1}{2}||w||_{L^2}^2\left(||u||_{L^2}^2+||v||_{L^2}^2\right)$$ Notice the left hand side is positive while the right hand side is negative. The only way this equation can be satisfied is if both sides are equal to zero, and that can only happen when $u$ and $v$ are equivalent. Thus uniqueness, we have.