Problem: Let $\Omega\subset\mathbb R^2$ be a bounded domain with smooth boundary. Prove that, for all $p>1$ and $1\le q<\infty$, for all $f \in L^p(\Omega)$, there exists a unique $u\in H_0^1(\Omega)$, such that $$\Delta u=\vert u\vert^{q-1}u+f\text{ in }\Omega.$$ $$$$ My intuition of this problem is that this form is not common, and the only similar form I can come up now is the Bôcher's theorem for harmonic functions, in which the harmonic corresponds to the $\Delta u$ here, and regardless $f$ the righthand side is quite similar. But I don't know how to use the $H_0^1(\Omega)$ condition, what property of $H_0^1(\Omega)$ makes this $u$ being unique? And how to use the bounded domain condition? I'm stucking with the two conditions and any hints or solutions are appreciated. Thanks in advance!
Edit 1. Subtract the two equation can cancel $f$, but we also have a $\vert u_1\vert^{q-1}u_1-\vert u_2\vert^{q-1}u_2$ term due to un-linearity, are there any good ways to handle this term?
Suppose that there are two solutions $u_1,u_2$ such that $$ \Delta u_1=|u_1|^{q-1}+f, \Delta u_2=|u_2|^{q-1}+f. $$ Then one has $$ \Delta (u_1-u_2)=|u_1|^{q-1}u_1-|u_2|^{q-1}u_2. $$ Multiplying $u_1-u_2$ and then integrating give $$ \int_{\Omega}|\nabla(u_1-u_2)|^2dx+\int_{\Omega}(|u_1|^{q-1}u_1-|u_2|^{q-1}u_2)(u_1-u_2)dx=0. \tag1$$ Now one has to show the second term above to be nonnegative and hence $u_1=u_2$. This can be done by observing \begin{eqnarray} &&\int_{\Omega}(|u_1|^{q-1}u_1-|u_2|^{q-1}u_2)(u_1-u_2)dx\\ &=&\int_{u_1\ge0,u_2\ge0}(|u_1|^{q-1}u_1-|u_2|^{q-1}u_2)(u_1-u_2)dx\\ &&+\int_{{u_1\ge0,u_2<0}}(|u_1|^{q-1}u_1-|u_2|^{q-1}u_2)(u_1-u_2)dx\\ &&+\int_{{u_1<0,u_2\ge0}}(|u_1|^{q-1}u_1-|u_2|^{q-1}u_2)(u_1-u_2)dx\\ &&+\int_{{u_1<0,u_2<0}}(|u_1|^{q-1}u_1-|u_2|^{q-1}u_2)(u_1-u_2)dx. \end{eqnarray} and it is easy to see each term is nonnegative.