# Existence of the Sobolev-solution of the equation $- \Delta u + {u^3} = 1$

I have the following problem from my Calculus of Variation class.

Problem. Let $\Omega$ be an open, bounded subset of $\mathbb{R}^3$. Prove or disprove that there exists $u \in W_0^{1,2}\left( \Omega \right)$ such that $- \Delta u + {u^3} = 1$ in $\Omega$.

I started with variational formulation of this equation $$\int_\Omega {\nabla u \cdot \nabla v} + \int_\Omega {{u^3}v} = \int_\Omega v \mbox{ for all }v \in W_0^{1,2}\left( \Omega \right)$$ but the form on the left-hand side is not linear with respect to $u$. So I can't apply Stampacchia or Lax-Milgram theorem.

Let us consider the operator $A u=-\Delta$ and $B=u^3$ over $V=H_0^1(\Omega)$. If we treat them as operators from $V$ into $V'$, then it is well-known that $A$ is a maximal monotone. If we prove that $B$ is demi-continuous and monotone, then $A+B$ is also maximal monotone.
Since $V=H^1_0(\Omega)$ is embedded into $L^6(\Omega)$, then $Bu\in L^2(\Omega)$. The Dominate convergence theorem can imply that
$$\langle B(u_n-u),w\rangle=\int_\Omega (u^3_n-u^3)(x) w(x)dx\to 0$$ if $u_n\to u$ in $V$. This implies the demi-continuity. On the other hand, $\langle (A+B)u,u\rangle\geq \int |\nabla u|^2 dx$. Hence, we have $$\frac{\langle (A+B)u,u\rangle}{\|u\|_V}\to \infty$$ as $\|u\|_V\to \infty$. Therefore, the operator $A+B$ is maximal monotone and coercive. Then Minty's Theorem yields that this it is surjective from $V$ onto $V'$. Then the equation admits a unique solution in $V$.