My teacher made the following to prove that the solution for the Dirichlet problem:
Let $\Omega$ be a bounded open set
Given $f\in C(\partial\Omega)$, find $u\in C^2(\Omega)\cap C(\Omega)$
such that$$\Delta u = 0 \mbox{ in }\Omega\\ u = f\mbox{ in } \partial\Omega$$
is unique.
Suppose that there exists $2$ solutions $u_1,u_2$. Then $\Delta (u_1-u_2) = 0$ in $\Omega$, and $u_1-u_2 = f-f = 0$ in $\partial\Omega\implies u_1=u_2$
Well, it just proves that $u_1=u_2$ in the boundary, not inside. I know that both $u_1$ and $u_2$ have $Delta=0$ inside, but they could be different functions just being equal in the boundary.
Is this proof wrong?
Best Answer
Let $\Omega$ be bounded and let $u \in {C^2({\Omega})} \cup C(\overline{\Omega})$ and $f \in C(\partial \Omega)$ such that \begin{align} (*)\left\{ \begin{array}{ll} \Delta u = 0 & \mbox{in } \Omega \\ \ \ \ u = f & \mbox{on } \partial \Omega \end{array} \right. \end{align} Now suppose that $\widetilde{u}$ is another solution to the Dirichlet problem $(*)$. Define $w := \widetilde{u} - u$ and due to the linearity of the Laplace operator we obtain \begin{align} \left\{ \begin{array}{ll} \Delta w = 0 & \mbox{in } \Omega \\ \ \ \ w = 0 & \mbox{on } \partial \Omega \end{array} \right. \implies w \equiv 0 \implies u \equiv \widetilde{u} \end{align} You can see that this is directly implied from the Strong Maximum Principle. In fact, if for any point $x \in \Omega$ we had $w > 0$ or $w < 0$ then $w$ would attain a maximum/minimun in $\Omega$ and thus from s.m.p. $w$ must be constant. Since the boundary value is $0$, it must be $w \equiv 0$. It is not that $\widetilde{u} = u$ only at the boundary $\partial \Omega$ but that $u$ is identical to $\widetilde{u}$ in $\overline{\Omega}$ because $w$ is identical to zero in $\overline{\Omega}$.
You can see that with the same approach we obtain the uniqueness of the Dirichlet problem for the Poisson equation.