I am trying to prove the following:
Let $u \in C^2(\Omega) \cap C^0(\bar{\Omega})$ be a solution of $\Delta u=u^3-u$ in a bounded domain $\Omega$ with $u=0$ on $\delta \Omega$. Choose a suitable test function to prove that $u(x) \in [-1,1]$ for all $x \in \Omega$.
I am really lost how to choose a test function that will help me show my claim. Any help would be appreciated!
We set $v(x) = \max(u(x)-1,0)$. Then, $v \in H_0^1(\Omega)$. Moreover, we have $\nabla v(x) = \chi_{\{x:u(x)>1\}}(x) \, \nabla u(x)$. Thus, \begin{align*} u(x) \, v(x) &= (v(x) + 1) \, v(x) \\ u(x)^3 \, v(x) &= (v(x) + 1)^3 \, v(x) \\ \nabla u(x) \cdot \nabla v(x) &= \|\nabla v(x)\|^2 \end{align*} Now, we multiply our pde pointwise with $v$ and integrate. This gives $$0 = \int_\Omega (-\Delta u + u^3 - u ) \, v \, \mathrm{d}x.$$ Since $v$ has zero boundary values, we can use integration by parts and obtain $$0 = \int_\Omega \nabla u \cdot \nabla v + (u^3 - u ) \, v \, \mathrm{d}x.$$ Now, we apply the above relations for the products between $u$ and $v$ and get $$0 = \int_\Omega \|\nabla v\|^2 + ((v+1)^3 - (v+1) ) \, v \, \mathrm{d}x.$$ Some simple calculations lead to $$0 = \int_\Omega \|\nabla v\|^2 + v^4 + 3 \, v^3 + 2 \, v^2 \, \mathrm{d}x.$$ Since $v \ge 0$ we obtain $$0 = \int_\Omega \|\nabla v\|^2 + v^4 + 3 \, v^3 + 2 \, v^2 \, \mathrm{d}x \ge \int_\Omega \|\nabla v\|^2 + v^2 \, \mathrm{d}x.$$ Hence $v \equiv 0$, which yields $u \le 1$.
I would guess, $u \ge -1$ can be shown analogously.