Show that solution has range $[-1,1]$ Let $u$ be a solution of $\Delta u=u^3-u$ on a bounded domain $\Omega$. I want to show that $u\in[-1,1]$ throughout $\Omega$ and find out if $u$ can attain the values $-1,1$.
This is an exercise from Rogers - "An Introduction to PDEs". I assume that the problem can be solved using some form of the maximum principle for elliptic equations as this exercise is from this section.
I wrote the equation in a form for which the maximum principles were stated, that is
$Lu=\Delta u + u =u^3$ 
corresponding to the general form of the equations treated in this chapter
$$Lu=a_{ij}(x)\frac{\partial^2 u}{\partial x_i\partial x_j}+b_i(x)\frac{\partial u}{\partial x_i}+c(x)u$$
where Einstein Summation is used. This means for my equation that $c=1,b=0$ and $a_{ij}=\delta_{ij}$. So we have a semilinear elliptic equation.
I am stuck here, as most of the theorems require that $Lu\le 0$ or $Lu\ge0$ to begin with. And even if a condition like this was satisfied, I am unsure of how to show that the range lies in an interval.
 A: The argument that follows is for classical solutions, i.e. $u\in C^2(\Omega)\cap C(\overline{\Omega})$.
Let us show that $-1\leq u \leq 1 $ in $\Omega$, assuming $u=0$ on $\partial \Omega$. 
We first prove that $u\leq 1$ in $\Omega$. Indeed, if the maximum of $u$ is attained on the boundary, then $u\leq 0$ and we are done, otherwise let $x_0\in \Omega$ be a point of maximum for $u$. Then the Hessian matrix of $u$ at $x_0$, i.e. $D^2(x_0)$, must be non-positive definite. Observe that the Laplacian is the trace of the Hessian matrix, hence we get
$$
0\geq trace(D^2 u(x_0)) = \Delta u(x_0) = u^3(x_0) - u(x_0).
$$
Since $u^3(x_0) \leq u(x_0)$ it follows that $u(x_0) \leq 1$. But $x_0$ was the point of maximum for $u$, hence $u\leq 1$ in $\Omega$.
To see $u\geq -1$ in $\Omega$ proceed similarly. Assuming $u$ attains a minimum in $\Omega$ (as otherwise there is nothing to prove since $u = 0 $ on $\partial \Omega$), we must have $D^2u(x_0) \geq 0$ which translates to $u^3(x_0) - u(x_0) \geq 0$ and hence $u(x_0) \geq -1$. Since $x_0$ was the point of minimum for $u$ it follows that $u\geq -1$ everywhere in $\Omega$.
The last part: observe that without the boundary condition $u=0$ the constant functions $u=1$ or $u=-1$ both satisfy the equation and attain the values $\pm 1$. With the boundary condition $u=0$, there doesn't seem to be an obvious violation of maximum principles if $u$ takes values $\pm 1$. Nevertheless, we have an equation of the form $\Delta u = f(u)$, with $f(x) = x^3 - x$, where $-1\leq x \leq 1$ as we proved already, and $u=0$ on $\partial \Omega$, hence one may apply a maximum principle in narrow domains to obtain an upper bound on $\max|u|$ depending on the size of $\Omega$. For instance, following the book you refer to (check the apriori bounds there)
$$
\max_{\Omega}|u| \leq (e^{d} -1 )\max\limits_{\overline{\Omega}}|f|,
$$
provided $\Omega$ is contained in a strip (region bounded by two parallel planes) of width $d>0$. We have $\max\limits_{\overline{\Omega}}|f| = |f(\pm 1/\sqrt{3})| < 0.385$. This means that for certain "narrow" domains (if the $d$ in the above inequality is small enough) $u$ cannot attain the values of $\pm 1$. Without such assumptions on $\Omega$, if you find something more general, please share it here, it would be interesting to see.
