I am trying to solve the following nonlinear PDE problem.
Let $D$ be a bounded simply-connected domain on $\mathbb R^2$ and $u(x,y)$ be a classical solution of the nonliner elliptic problem:
$$\left\lbrace\begin{aligned} \Delta u &= u^3 - u & \text{ in } D \\ u &= 0 & \text{ on } \partial D \end{aligned}\right.$$
Show that $-1\le u(x,y) \le 1$ in $D$. Can $u(x,y) = \pm 1$ be achieved in $D$?
Since the equation is nonlinear, we cannot apply maximum principle directly. The only idea coming to my mind is to incapsulate points in $D$ where $u>1$ into closed curve, and to try to justify max principle for subharmonic functions within the curve-bounded portion of the domain. However I could not complete solution following this approach. Any help is appreciated.
You can use similar techniques to those used to prove the maximum principle. For example, suppose that $u > 1$ at any point in $D$. Then $\max_{\overline D} u > 1$. This maximum must be achieved in the interior of $D$ since $u=0$ on the boundary. At the point $x_0$ where the maximum is achieved, we have $$\Delta u(x_0) \le 0.$$ On the other hand, since $u(x_0) > 1$, we have $u(x_0)^3 - u(x_0) > 0$. Thus we have $$\Delta u(x_0)\le 0 < u(x_0)^3 - u(x_0);$$ a contradiction. Hence $u \le 1$ in $D$.
Likewise, if $u < -1$ at some point in $D$, then $\text{min}_{\overline D} u < -1$. At the point $x_1$ where the minimum is achived (which again, is in the interior), we have $$\Delta u(x_1) \ge 0 > u(x_1)^3 - u(x_1)$$ which is a contradiction. Hence $u \ge -1$. This shows that $-1 \le u \le 1$ in $D$.