Monday Math 127

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, φ1 and φ2. 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, φ1=φ2, 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 φ1=φ2: 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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One Response to “Monday Math 127”

  1. Monday Math 132 « Twisted One 151's Weblog Says:

    […] function for the three-dimensional Laplacian, and the only one which goes to zero as , due to the uniqueness theorem). For n=4, nC_n=2π2, so […]

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