I'm trying to prove that a non-linear Poisson problem has a unique solution. The context is the following:
Let $\Omega \subset \mathbb{R}^n$ be a bounded open subset of class $C^2$. Consider the following non-linear problem: $$\begin{cases} &(\Delta u)(x_0) = f(x_0, u(x_0)), \text{ for every } x_0 \in \Omega \\ &u(x_0) = \varphi(x_0), \text{ for every $x_0 \in \partial \Omega$}\end{cases}$$ where $f(\bullet, u) \in C^{0, \gamma}(\overline{\Omega})$, $f(x, \bullet) \in C^1(\mathbb{R})$ and $f$ is non decreasing in $u$, i.e $\displaystyle{\frac{\partial f}{\partial u}(x) \geq 0}$ for every $x \in \overline{\Omega}$. Prove that this problem has at most one solution in $C^2(\Omega) \cap C(\overline{\Omega})$.
If $u, v$ are both solutions of the problem, ideally, I'd like to use the maximum principle for $w = u -v$. Since the problem is non-linear, the usual maximum principle doesn't apply, so I need to come up with another linear problem that $w$ satisfies on which the maximum principle can be used. I think the non-negativity of the derivative of $f$ will play an essential role, but I haven't been able to come up with anything helpful. Any help will be greatly appreciated. Thanks in advance!
Suppose that $u ,v\in C^2(\Omega) \cap C^0(\overline \Omega)$ are solutions of \begin{align*} \Delta u &= f(x,u(x)), && \text{in } \Omega \\ u &= \varphi, &&\text{on } \partial \Omega. \end{align*} Let $w:=u-v$. On $\partial \Omega$, it is clear that $w=0$. If we write $f=f(x,z)$ then in $\Omega$, \begin{align*} \Delta w &= f(x,u(x)) - f(x,v(x)) \\ &= \int_0^1 \frac{d}{dt} f(x, tu(x)+(1-t) v(x)) \, dt \\ &=\bigg (\int_0^1 \partial_z f(x, tu(x)+(1-t) v(x)) \, dt \bigg )(u(x)-v(x))\\ &=c(x) w(x) \end{align*} where $$ c(x) := \int_0^1 \partial_z f(x, tu(x)+(1-t) v(x)) \, dt. $$ Hence, \begin{align*} -\Delta w+c(x)w(x) &= 0, && \text{in } \Omega \\ w &= 0, &&\text{on } \partial \Omega. \end{align*}Since $\partial_zf \geqslant 0$ by assumption, $c\geqslant 0$, so we can apply the maximum principle to conclude that $w=0$ which implies that $u=v$.