Uniqueness Theorem For Poisson's Equation - Proof

Proof

In Gaussian units, the general expression for Poisson's equation in electrostatics is

Here is the electric potential and is the electric field.

The uniqueness of the gradient of the solution (the uniqueness of the electric field) can be proven for a large class of boundary conditions in the following way.

Suppose that there are two solutions and . One can then define which is the difference of the two solutions. Given that both and satisfy Poisson's Equation, must satisfy

Using the identity

And noticing that the second term is zero one can rewrite this as

Taking the volume integral over all space specified by the boundary conditions gives

Applying the divergence theorem, the expression can be rewritten as

Where are boundary surfaces specified by boundary conditions.

Since and, then must be zero everywhere (and so ) when the surface integral vanishes.

This means that the gradient of the solution is unique when

\sum_i \int_{S_i} (\phi\epsilon \, \mathbf{\nabla}\phi) \cdot \mathbf{dS} = 0

The boundary conditions for which the above is true are:

  1. Dirichlet boundary condition: is well defined at all of the boundary surfaces. As such so at the boundary and correspondingly the surface integral vanishes.
  2. Neumann boundary condition: is well defined at all of the boundary surfaces. As such so at the boundary and correspondingly the surface integral vanishes.
  3. Modified Neumann boundary condition (where boundaries are specified as conductors with known charges): is also well defined by applying locally Gauss's Law. As such, the surface integral also vanishes.
  4. Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann boundary conditions): the uniqueness theorem will still hold.

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