Maximum Principle for a Poisson Equation?

Question

Let $u$ be a $C^2$-solution of $\Delta u=u^3-u$ on a bounded domain $\Omega$ with $u=0$ on $\partial\Omega$. How can one show that $-1\leq u\leq 1$ for every $x\in\Omega$? Also is it possible for $u$ to achieve values $\pm1$? I am think of using maximum principle but have no clue. Could someone help me with this? Thank you in advance.

Answer 1

Partial answer: If $u(x_0)=\max u\,.$ Notice that $x_0$ is necessarily in the interior of the domain unless the function is $\le 0\,,$ in which case of course $u(x)\le 1\,.$ Otherwise there are points where $u$ is positive. Then $0\ge\Delta u(x_0)=u(x_0)^3-u(x_0),$ hence $u(x_0)\le1\,.$ An analogous argument shows that $\min u\ge -1\,.$

Answer 2

I am assuming that you are doing this out of Renardy and Rogers Introduction to Partial Differential Equations, Second Edition, Problem 4.1. It looks like you already got the first part, so I will show how the next part is done.

I show that $u$ cannot achieve the values $\pm 1$. Note that if I can show that $u$ cannot achieve the value $1$, then $-u$ (also being a solution to the PDE) cannot achieve the value $1$ and thus $u$ cannot achieve the value $-1$. Let $w=u-1$. Then $-2 \leq w \leq 0$ on $\Omega$. Now then $$ \Delta w = \Delta u = u^3 - u = (w+1)^3 -(w+1) = w^3 + 3w^2 + 2w $$ Thus $\Delta w - 2w = w^2(w+3)$. Now since $-2 \leq w \leq 0$ the previous equation gives that $\Delta w - 2w \geq 0$ on $\Omega$. Since $c(\textbf{x}) = -2 \leq 0$, the Strong Maximum Principle (Theorem 4.10, p.105) applies and we get that $w$ cannot achieve a nonnegative maximum in $\Omega$. Thus $w<0$ on $\Omega$, which then implies that $u <1$ on $\Omega$.