2
$\begingroup$

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 $y$.

$\endgroup$
  • $\begingroup$ I believe this is one of those situations that start with the phrase, "Suppose not…" (i.e. that the solution is not unique), and then you go on to show that this leads to a contradiction (e.g. the solutions don't obey the boundary conditions, or they weren't unique in the first place). $\endgroup$ – Sean E. Lake Nov 9 '17 at 22:09
  • 1
    $\begingroup$ Try first showing that if $\nabla^2 V = 0$ and then $V=0$ at the boundary, then $V=0$ everywhere. $\endgroup$ – Gonenc Mogol Nov 9 '17 at 22:11
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.