Consider Poisson’s equation in three dimensions:

, where is some function. Let us investigate the solution on a volume *V*.

Suppose their are two solutions, *φ*_{1} and *φ*_{2}. We can then define . Hence,

.

Now, recall that Green’s first identity states that for scalar fields *f* and *g*,

,

where *S* is the surface bounding the volume *V*, with outward normal . Letting *f*=*g*=*ψ*, we have

,

where is the normal derivative of the function, and we have used the fact that .

Now, suppose we have the Dirichlet boundary condition, so that the value of *φ* is specified on our boundary *S*, Thus, *φ*_{1}=*φ*_{2}, and *ψ*=0, on this surface; so then on the surface *S*, and so

,

but the norm square is positive definite, so the left hand integral is zero if and only if on all of *V*, which requires that *ψ* be constant on *V*; and since *ψ* is zero on the boundary of *V*, we have *ψ*=0 for all points in *V*, and so *φ*_{1}=*φ*_{2}: the solution to Poisson’s equation with the Dirichlet boundary condition is unique.

Suppose instead that we have the Neumann boundary condition, where is specified on all of the boundary *S*. Then

on *S*.

Thus on the surface *S*, and, as in the previous case, , and so *ψ* is constant on *V*, and so our solutions are unique up to addition by a constant.

In fact, there are other boundary equations that lead to similar results, where is unique (and *φ* is thus either unique, or unique up to a constant); including in an infinite domain, with an appropriate boundary at infinity (see more here).

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Tags: Dirichlet Problem, Math, Monday Math, Neumann Boundary Condition, Poisson's Equation, Uniqueness

This entry was posted on July 19, 2010 at 12:04 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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