Existence and uniqueness of $-\Delta u + u^3 =f$ I want to show the existence and uniqueness of the equation of the form
$-\Delta u+u^3=f\in L^2(\Omega),\quad u=0~\mbox{on}~ \partial\Omega,~$
$\Omega\subset\mathbb{R}^3$-bounded
I know one proof using Minty-Browder theorem and by direct method of calculus of variations, I'm looking for other proofs. Any hints?
What about $\mathbb{R}^2$.
I would be really thankful
 A: Only for $n=3$. Let the norm of $H_0^1(\Omega)$ and $L^p(\Omega)$ be $\|\cdot\|_2$ and $|\cdot|_p$. Note $H_0^1(\Omega)\hookrightarrow L^6(\Omega)$ is continuous. First let us get a priori estimates. Multiplying both sides of the equation by $u$ gives
$$ \|u\|_2^2+|u|_4^4=\int fudx\le \frac1{2}|f|^2_2+\frac1{2}|u|_2^2\le\frac{1}{4\epsilon}|f|^2_2+\frac\epsilon{2}\sqrt{|\Omega|}|u|_4. \tag{1}$$
From this, we have $\|u\|_2\le C,|u|_4\le C, |u|_6\le C$ for some $C$ depending on $\Omega, f$. Let $M=\{u\in H^1_0(\Omega): \|u\|_2\le C\}$. 
Now we use the Schauder fixed point theorem to prove the existence of a solution. Note that the problem
$$ \left\{\begin{array}{ll}-\Delta u=h,&x\in\Omega\\
u=0,&x\in\partial\Omega\end{array}\right.
$$
has a unique solution for $h\in L^2(\Omega)$. Define $u=Kh$ and then $K: L^2(\Omega)\to L^2(\Omega)$ is compact. The equation can be written as
$$ \left\{\begin{array}{ll}-\Delta u=-u^3+f,&x\in\Omega\\
u=0,&x\in\partial\Omega\end{array}\right.
$$
which can be written as $u=K(-u^3+f)$. Let $Tu=K(-u^3+f)$. Now we first show $T: M\to H^1_0(\Omega)$ is compact. In fact, let $\{u_n\}\subset M$. Then there is a subsequence, still denoted by $\{u_n\}$ such that
$$ u_n\to u \text{ weakly in }H^1_0(\Omega)\text{ and strongly in } L^4(\Omega). $$
Let $u_n^*=Tu_n$, namely $u_n^*\in H^1_0(\Omega)$ satisfies
$$ -\Delta u_n^*+u_n^3=f. \tag{2}$$
Similarly we have $\|u_n^*\|_2\le C$ which implies that there is $u^*\in H_0^1(\Omega)$ such that
$$ u_n^*\to u^* \text{ weakly in }H^1_0(\Omega)\text{ and strongly in } L^q(\Omega). $$
Using $u_n^*-u$ as a test function in (2), we have
$$ \int_{\Omega}|\nabla(u_n^*-u^*)|^2dx+\int_{\Omega}\nabla u^*\nabla(u_n^*-u^*)dx+\int_{\Omega}u_n^3(u_n^*-u^*)dx=\int_{\Omega}f(u_n^*-u^*)dx. $$
Letting $n\to\infty$, we have $u_n^*\to u^*$ in $H^1_0(\Omega)$. So $T$ is compact. It is easy to show that $T$ is continuous. Next we show that the set
$$ \{x\in H_0^1(\Omega): x=\lambda Tx \text{ for }\lambda\in[0,1]\} $$
is bounded. It is equivalent to showing the solution of 
$$ -\Delta u=\lambda(-u^3+f) $$
is bounded in $H_0^1(\Omega)$. Using the same argument in (1), it is easy to see $\|u\|_2\le C$. Thus by the Schauder fixed point theorem, $T$ has a fixed point.
