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On page 28, line 1 in PDE Evans, 2nd edition, the theorem and proof state

Theorem. Let $g \in C(\partial U), f \in C(U)$. Then there exists at most one solution $u \in C^2(U) \cap C(\bar{U})$ of the boundary-value problem

\begin{cases} -\Delta u = f & \text{ in } U \\ \quad \, \, \, u = g & \text{ on } \partial U \end{cases}

Proof. "If $u$ and $\tilde{u}$ both satisfy the above BVP, apply Theorem 4 (Strong maximum principle, on page 27) to the harmonic functions $w := \pm(u-\tilde{u})$.

What I did was, in applying the Strong maximum principle as it directed me to:

\begin{align} \max_{\bar{U}} w &= \max_{\partial U} w \\ \max_{\bar{U}} \pm(u-\tilde{u}) &= \max_{\partial U} \pm(u-\tilde{u}) \end{align}

But it is at this point that I am stuck. I do not see how the solution $w$ is unique, the only one that exists.

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Note that by applying the maximum principle to $-u$, we obtain the same conclusion with $\max$ replaced with $\min.$ Note also that $w:=u-\tilde{u}$ is harmonic if $u$ and $\tilde{u}$ are by linearity.

Now, we are on a bounded open domain $U$ with smooth boundary and so by the weak maximum principle, any harmonic function $u$ achieves its maximum and minimum on $\partial U$. Since $w\equiv0$ on $\partial U$ (because $u$ and $\tilde{u}$ have the same boundary values by assumption) and $w$ is harmonic, it follows that $\min_{U}w=\max_{U}w=0$ which implies $w\equiv0$ in $U$. Hence $u=\tilde{u}$ in all of $\bar{U}$, which is the uniqueness claim. (Note that the strong maximum principle is not needed here, but you can adapt the argument to use it if you wish.)

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Another way of proving the uniqueness of the solution could be using integration by parts to get:

$$ \int_{\Omega}\nabla w \cdot\nabla w = \int_{\partial \Omega} w \cdot \frac {\partial w} {\partial n} -\int_{\Omega}w \, \Delta w =0 \\\Rightarrow |\nabla w|^2=0 \\\Rightarrow w=Constant \\\Rightarrow u=\tilde{u}+C $$ but because $u|_{\partial \Omega}= \tilde {u}|_{\partial \Omega} = g \,$, $C$ must be $0$

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