Let $\alpha >0$, and let $\Omega\subset \mathbb{R}^N$ be and open domain. I want to prove that the following problem has at most one solution.
$$\Delta u=F \quad \text{in } \Omega$$ $$u=f \quad \text{in } C_1\subset\partial\Omega$$ $$\frac{\partial u}{\partial n}+\alpha u=g \quad \text{in } C_2=\partial\Omega\backslash C_1$$ where $n$ is the normal verctor. So usually here I would take $u_1$, $u_2$ two solutions to the problem, and then $w=u_1-u_2$, and I'd try to prove that $w\equiv0$. The problem that $w$ satisfies is
$$\Delta w=0 \quad \text{in } \Omega$$ $$w=0 \quad \text{in } C_1\subset\partial\Omega$$ $$\frac{\partial w}{\partial n}+\alpha w=0 \quad \text{in } C_2=\partial\Omega\backslash C_1$$ The maximum principle tells me that the $\max$ and $\min$ of $w$ are reached in $\partial \Omega$. If both of them are in $C_1$, then $w\equiv 0$. But if $\max$ is in $C_1$ and $\min$ is in $C_2$, I don't know what to do. Can someone help me?
You are almost done.
Suppose that the maximum occurs on $C_1$ and the minimum occurs on $C_2$. Then at the minimum point, we have $\omega \leq 0$. Actually if it were $\omega = 0$, it would mean that $\omega = 0$ everywhere which is what we want so let's just assume $\omega < 0$.
At the minimum point then, we have $\frac{\partial \omega}{\partial n} = - \alpha \omega > 0$. The derivative of $\omega$ along the normal vector is positive but this means that we are not at a minimum. We get a contradiction so our initial assumption is false.