📂Partial Differential Equations

Uniqueness of the Solution to the Dirichlet Problem for the Poisson Equation

Theorem¹

Let’s assume that $\Omega \subset \mathbb{R}^n$ is open and bounded. And let $g \in C(\partial \Omega)$, $f \in C(\Omega)$. Then in the Dirichlet problem of the Poisson equation as below, the solution $u \in C^2(\Omega) \cap C(\bar{\Omega})$, if exists, is unique (=at most one exists).

$$ \begin{equation} \left\{ \begin{aligned} -\Delta u &= f && \text{in } \Omega \\ u &= g && \text{on }\partial \Omega \end{aligned} \right. \label{eq1} \end{equation} $$

Proof

Suppose two functions $u,\ \tilde{u} \in C^2(\Omega) \cap C(\bar{\Omega})$ satisfy $\eqref{eq1}$. And define function $w :=u-\tilde{u}$. Then because of $\Omega$, $\Delta w=\Delta u- \Delta \tilde{u}=f-f=0$ implies that $w$ is harmonic in $\Omega$.

Maximum Principle for Harmonic Functions

For a harmonic function $u$, the following holds:

$$ \max \limits_{\bar{\Omega}} u = \max \limits_{\partial \Omega} u \quad \left( \mathrm{or} \quad \min \limits_{\bar{\Omega}} u= \min \limits_{\partial \Omega} u \right) $$

Thus, by the Maximum Principle, the following holds.

$$ \max \limits_{\bar{\Omega}} w = \max \limits_{\partial \Omega} w $$

But given the boundary conditions, $w=g-g=0 \text{ in } \partial \Omega$, hence we get:

$$ \max \limits_{\bar{\Omega}} w = \max \limits_{\partial \Omega} w=0 $$

By the same logic, we also obtain:

$$ \min \limits_{\bar{\Omega}} w = \min \limits_{\partial \Omega} w =0 $$

Therefore, because in $\bar{\Omega}$, $w=u-\tilde{u}=0$,

$$ u=\tilde{u} \text{ in } \bar{\Omega} $$

■