Show that the solution to a Dirichlet problem for Poisson's equation is unique

Question

Show that the solution to a Dirichlet problem for Poisson's equation is unique

I know that we are supposed to show that the bounding surface is unique but I have no clue how to do it.

Answer 1

Let $U\subset\mathbb{R}^n$ be your domain and $\partial U$ be the boundary. Suppose there are two solutions $u\left(\mathbf{x}\right)$ and $v\left(\mathbf{x}\right)$ on $U$, satisfying boundary condition \begin{equation} Bu=g\left(\mathbf{x}\right),\forall\mathbf{x}\in\partial U \\ Bv=g\left(\mathbf{x}\right),\forall\mathbf{x}\in\partial U \end{equation} for some sufficiently well-behaved function $g\left(\mathbf{x}\right)$.

Then consider the difference $w\left(\mathbf{x}\right) = u\left(\mathbf{x}\right) - v\left(\mathbf{x}\right)$. We see that $w$ solves the completely homogeneous problem, i.e.

\begin{equation} \Delta w=\Delta u-\Delta v=f\left(\mathbf{x}\right)-f\left(\mathbf{x}\right)=0 \\ Bw = Bu - Bv = g\left(\mathbf{x}\right)-g\left(\mathbf{x}\right)=0 \end{equation}

Hence $w$ is harmonic. Every nonconstant harmonic function satisfies a maximum and minimum principle, namely, $w$ achieves its maximum and minimum on its boundary. See page 9 of this.

Applying the max/min principle to $w$, we have, on the one hand,

\begin{equation} w\left(\mathbf{x}\right)\leq\max_{\mathbf{x}\in U}w\left(\mathbf{x}\right)=w\left(\mathbf{x}\right)\mid_{x\in\partial U}=0 \end{equation}

on the other hand,

\begin{equation} 0=w\left(\mathbf{x}\right)\mid_{x\in\partial U}=\min_{\mathbf{x}\in U}w\left(\mathbf{x}\right)\leq w\left(\mathbf{x}\right) \end{equation}

Therefore, $w\left(\mathbf{x}\right)\leq0$ and $w\left(\mathbf{x}\right)\geq0$ at the same time, forcing $w\left(\mathbf{x}\right)=0$. Thus $u\left(\mathbf{x}\right)=v\left(\mathbf{x}\right)$, i.e. the solution to Poisson's equation is unique.