Given the non-linear Poisson equation
$\Delta u=f(u(x,y)), \text{in } \Omega$
$u=0, \text{on } \partial \Omega$
in a bounded region $\Omega$,
show that if f is Lipschitz-continuous with a sufficiently small Lipschitz constant, then the problem only has one solution.
Well, Lipschitz continuous means that $|f(u_1(x,y))-f(u_2(x,y))| \leq L|u_1(x,y)-u_2(x,y)|$ for a constant $L>0$. How could I show that $u_1=u_2$? Do I need to use the maximum principle somehow?
Yes, this is an application of the maximum principle, but is not immediately obvious. Suppose that $u,v\in C^2(\Omega) \cap C(\overline \Omega )$ both solve the PDE and let $ w:= u -v$. If we define $c:\Omega \to \mathbb R$ by $$ \tag{$\ast$}c(x) = \begin{cases} -\frac{f(u(x))-f(v(x))}{u(x)-v(x)}, &\text{if } u(x) \neq v(x) \\ 0, &\text{if } u(x) = v(x) . \end{cases} $$ then, $$ \Delta w(x) = f(u(x))-f(v(x))=-c(x)w(x) \qquad \text{for all } x\in \Omega.$$ Thus, $$\begin{cases}\Delta w +cw=0 &\text{in }\Omega \\ w=0, &\text{on } \partial \Omega. \end{cases} $$ Of course the standard maximum principle still doesn't tell us anything about $w$ since it requires $c\leqslant 0$ (which we cannot guarantee). Fortunately for us, there more general maximum principles which relax the assumption that $c\leqslant 0$, but instead (usually) impose either that $\| c\|_{L^\infty (\Omega)}$ is small in some sense or $\Omega$ is small in some sense.
In this case, we can't say anything about the 'size' of $\Omega$, but, since $f$ is Lipschitz continuous with constant $L>0$, it follows immediately from ($\ast$) that $$\| c\|_{L^\infty(\Omega)} \leqslant L $$ and we are allowed to take $L$ as small as we like. By Corollary 3.8 in Gilbarg and Trudinger, for example, if $L$ is sufficiently small then we must have that $w=0$ in $\Omega$ which completes the proof.