Uniqueness Theorem

There is a uniqueness theorem for Laplace's equation such that if a solution is found, by whatever means, it is the solution. The proof follows a proof by contradiction.

Suppose that, in a given finite volume $V$ bounded by the closed surface $S$, we have

$$\nabla^{2}F = 0, \hbox{ in the volume }V,$$

and that

$$F \hbox{ is given on the surface }S.$$

Now we assume that there are two different functions $F_{1}$ and $F_{2}$ satisfying these conditions. Then,

$$F^{*} = F_{1} - F_{2},$$

satisfies

$$\nabla^{2}F^{*} = 0, \hbox{ in }V,$$

and, since both $F_{1}$ and $F_{2}$ both satisfy the same boundary conditions,

$$F^{*} = 0, \hbox{ on }S.$$

Now we calculate the integral of $\vert\nabla F^{*}\vert^{2}$ over the volume, $V$, of the plasma. This integral must be positive and we will show that it must, in fact, be zero. Hence,

$$\int_{V}\vert\nabla F^{*}\vert^{2}dV = \int_{V}\nabla F^{*}\cdot \nabla F^{*}dV.$$

We now use a vector identity to rewrite the right hand side in a form suitable for the divergence theorem. Thus, the right hand side becomes

$$\int_{V} \nabla \cdot\left (F^{*}\nabla F^{*}\right ) - F^{*}\nabla^{2}F^{*} dV.$$

The second term is automatically zero since $\nabla^{2}F^{*} = 0$. Using the divergence theorem we obtain

$$\int_{V} \nabla \cdot\left (F^{*}\nabla F^{*}\right ) dV = ... ...a F^{*}dS = \int_{S}F^{*}{\partial F^{*}\over \partial n}dS,$$

where $\hat{{\bf n}}$ is a unit vector normal to the surface, $S$. This integral is equal to zero since $F^{*} = 0$ on the surface $S$. Finally, since the origininal integrand is positive, the ony way that the integral can be zero is if $F^{*}$ is a constant. However, the condition on the surface $S$ tells us that the constant must in fact be zero. Hence,

$$F* = 0, \qquad \Rightarrow \qquad F_{1} = F_{2}.$$

Hence, there is a unique solution to Laplace's equation.