PDE in S.-T. Yau College Student Mathematics Contests 2019 This problem is from S.-T. Yau College Student Mathematics Contests 2019.
I don't have enough pre-knowledge, but I want to learn.
Let $\Omega \subset \mathbb{R}^2$ be a bounded domain with smooth
boundary. Prove that, for all $p>1$ and $1\leq q<\infty$, for all
$f\in L^p(\Omega)$, there exists a unique $u\in H^1_0(\Omega)$, such
that
$$\Delta u = |u|^{q-1} u+f  \  \text{in} \ \Omega.$$
 A: *

*For the existence, note that the corresponding functional:
\begin{align}
E(u)&=\int \frac{1}{2}|\nabla u|s^2 + \frac{1}{q+1}|u|^{q+1}+fu\\
&\geq \int\frac{1}{2} |\nabla u|^2 +\frac{1}{q+1}|u|^{q+1}-\epsilon \|u\|_{L^{p'}} -C(\epsilon)\|f\|_{L^p}.
\end{align}
The Sobolev embedding yields that $E$ is bounded from below, then the direct method in variational methods (see, for instance, Variational Methods-Michael Struwe, Theorem 1.2) yields that the problem admits a solution.

*For the uniqueness, if $u_1, u_2$ are two solutions, then we have
$$\Delta u_{1}-\Delta u_{2}=|u_{1}|^{p-1}u_{1}-|u_{2}|^{p-1}u_{2},$$
multiplying it by $u_{1}-u_{2}$ and intergral by parts, we obtain
$$\int |\nabla (u_{1}-u_{2})|^2+ (|u_{1}|^{p-1}u_{1}-|u_{2}|^{p-1}u_{2})(u_{1}-u_{2})=0.$$
If $u_{1}>u_{2}$, $(|u_{1}|^{p-1}u_{1}-|u_{2}|^{p-1}u_{2})(u_{1}-u_{2})>0,$
otherwise, $u_{1}\leq u_{2}$, we still have  $$(|u_{1}|^{p-1}u_{1}-|u_{2}|^{p-1}u_{2})(u_{1}-u_{2})\geq 0,$$
which means that $u_{1}=u_{2}.$

*I must point out that, the mountain pass theorem is used to deal with the problem
$$-\Delta u= |u|^{q-1}u+f,$$
which is totally different, since it admits infinitely many solutions by papers of Rabinowitz. And the problem is similar to Variational Methods-Michael Struwe, Theorem 1.3.

