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Let $u$ be a solution of $Delta u=u^3-u$ on a bounded domain $Omega$. I want to show that $uin[-1,1]$ 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
$$Lu=a_{ij}(x)frac{partial^2 u}{partial x_ipartial x_j}+b_i(x)frac{partial u}{partial x_i}+c(x)u$$

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 $Lule 0$ or $Luge0$ 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.

The argument that follows is for classical solutions, i.e. $uin C^2(Omega)cap C(overline{Omega})$.

Let us show that $-1leq u leq 1 $ in $Omega$, assuming $u=0$ on $partial Omega$.

We first prove that $uleq 1$ in $Omega$. Indeed, if the maximum of $u$ is attained on the boundary, then $uleq 0$ and we are done, otherwise let $x_0in Omega$ 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
$$
0geq trace(D^2 u(x_0)) = Delta u(x_0) = u^3(x_0) - u(x_0).
$$
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 $uleq 1$ in $Omega$.

To see $ugeq -1$ 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 $ugeq -1$ 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 $-1leq x leq 1$ 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)
$$
max_{Omega}|u| leq (e^{d} -1 )maxlimits_{overline{Omega}}|f|,
$$
provided $Omega$ is contained in a strip (region bounded by two parallel planes) of width $d>0$. We have $maxlimits_{overline{Omega}}|f| = |f(pm 1/sqrt{3})| < 0.385$. 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.