How do I show that the Laplace equation has a unique solution under the Dirichlet closed-surface boundary condition?

Question

In other words, when the the potential is specified at a finite boundary, how can I show the solution to $\nabla ^{2} V = 0$ exists and is unique? It is fine to show it for two dimensional Cartesian coordinates $x$ and $y$.

Answer 1

The existence problem is very difficult and it strongly depends on the regularity you require on your solutions. Classical ($C^2$) solutions are guaranteed to exist under suitable hypotheses on the domain: it must be an open non-empty set $\Omega$ whose closure $\overline{\Omega}$ is compact and the boundary $\partial \Omega$ is sufficiently regular as a surface. The method of Green functions permits to exhibit a solution.

Instead, uniqueness is relatively easier. It is based on a well-known theorem called maximum principle for harmonic functions. I henceforth denote by $\Delta$ the Laplacian operator sometime indicated by $\nabla^2$.

THEOREM (weak maximum principle for harmonic functions)
Let $\Omega \subset \mathbb R^n$ be an open non-empty set whose closure $\overline{\Omega}$ is compact (i.e. closed and bounded).

If $f : \overline{\Omega} \to \mathbb R$ satisfies

(a) $f$ is continuous on $\overline{\Omega}$,

(b) $f \in C^2(\Omega)$,

(c) $f$ is harmonic in $\Omega$, that is $\Delta f =0$ in $\Omega$,

then $$\max_{\overline{\Omega}} |f| = \max_{\partial\Omega} |f|$$ and furthermore $$\max_{\overline{\Omega}} f = \max_{\partial\Omega} f\:, \quad \min_{\overline{\Omega}} f = \min_{\partial\Omega} f\:.$$

The said theorem has the following consequence regarding the uniqueness issue in OP's question.

COROLLARY. Let $\Omega \subset \mathbb R^n$ be an open non-empty set whose closure $\overline{\Omega}$ is compact (i.e. closed and bounded).

There exist at most one continuous function $g : \Omega \to \mathbb R$ such that it is $C^2(\Omega)$ and satisfies Poisson equation in $\Omega$: $$\Delta g(x) = \rho(x)\:, \quad x \in \Omega\tag{1}$$ for a given continuous function $\rho : \Omega \to \mathbb R$ and Dirichlet boundary conditions $$g(x) = \psi(x)\:, x \in \partial \Omega\tag{2}$$ for a given continuous function $\psi : \partial \Omega \to \mathbb R$.

Proof. Suppose that both $g, g' : \overline{\Omega} \to \mathbb R$ are continuous, $C^2(\Omega)$ and satisfy (1) and (2) for the same $\rho$ and $\psi$. As a consequence, $f:= g-g'$ satisfies the hypotheses of THEOREM. Hence, for every $x\in \overline{\Omega}$, $$0 \leq |g(x)-g'(x)| \leq \max_{\overline{\Omega}}|g-g'|= \max_{\partial \Omega}|g-g'| = \max_{\partial \Omega}|\psi-\psi| = \max_{\partial \Omega} 0 =0\:.$$ Thus $g(x)= g'(x)$ for every $x \in \overline{\Omega}$.

QED

Answer 2

As the comments said, the solution in proving uniqueness lies in presuming two solutions to the Laplace equation $\phi_1$ and $\phi_2$ satisfying the same Dirichlet boundary conditions. Then, we prove that $\phi = \phi_1 - \phi_1$ is zero everywhere in the volume bounded by the boundary, which implies that $\phi_1 = \phi_2$. Note that by definition $\phi$ is zero on the boundary.

To that effect, we use the identity $$\nabla\cdotp (\psi \nabla\phi) = \psi \nabla^2\phi +\nabla\psi \cdotp\nabla\phi $$ with $\psi = \phi$ to obtain $$ \int_V \phi \nabla^2\phi dV = \int_V \nabla\cdotp (\phi \nabla\phi) dV- \int_V \nabla\phi \cdotp\nabla\phi dV$$
The left hand term is zero, because by definition $\nabla^2\phi=0$ over the whole volume. We are then left with $$ \int_V \nabla\cdotp (\phi \nabla\phi) dV= \int_V \nabla\phi \cdotp\nabla\phi dV=\int_V (\nabla\phi)^2 dV$$
We can use the divergence theorem to obtain $$ \int_V \nabla\cdotp (\phi \nabla\phi) dV=\oint_C (\phi \nabla\phi)\cdotp dA$$ where C is the boundary of V. Since $\phi$ is zero on the boundary by definition, we have $$ \int_V (\nabla\phi)^2 dV= 0 $$
Since the integrand is never negative, this implies that $\nabla\phi = 0$ everywhere in V, which implies that $\phi$ is a constant everywhere in V. In turn, since $\phi$ is zero on the boundary of V, the constant is zero and $\phi$ must be zero everywhere, which implies that $\phi_1=\phi_2$, and therefore a unique solution.

Of course, the solution must satisfy all the required "good behavior" conditions that allow us to use the vector calculus identity I used and the divergence theorem. If I remember correctly, similar techniques can be used to prove the uniqueness of solutions to the diffusion equation and some other partial derivative equations.