Uniqueness theorem for Poisson's equation

Summary

The uniqueness theorem for Poisson's equation states that, for a large class of boundary conditions, the equation may have many solutions, but the gradient of every solution is the same. In the case of electrostatics, this means that there is a unique electric field derived from a potential function satisfying Poisson's equation under the boundary conditions.

Proof

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The general expression for Poisson's equation in electrostatics is

 

where   is the electric potential and   is the charge distribution over some region   with boundary surface   .

The uniqueness of the solution can be proven for a large class of boundary conditions as follows.

Suppose that we claim to have two solutions of Poisson's equation. Let us call these two solutions   and  . Then

  and
 

It follows that   is a solution of Laplace's equation, which is a special case of Poisson's equation that equals to  . Subtracting the two solutions above gives

  (1)

By applying the vector differential identity we know that

 

However, from (1) we also know that throughout the region   Consequently, the second term goes to zero and we find that

 

By taking the volume integral over the region  , we find that

 

By applying the divergence theorem, we rewrite the expression above as

  (2)

We now sequentially consider three distinct boundary conditions: a Dirichlet boundary condition, a Neumann boundary condition, and a mixed boundary condition.

First, we consider the case where Dirichlet boundary conditions are specified as   on the boundary of the region. If the Dirichlet boundary condition is satisfied on   by both solutions (i.e., if   on the boundary), then the left-hand side of (2) is zero. Consequently, we find that

 

Since this is the volume integral of a positive quantity (due to the squared term), we must have   at all points. Further, because the gradient of   is everywhere zero and   is zero on the boundary,   must be zero throughout the whole region. Finally, since   throughout the whole region, and since   throughout the whole region, therefore   throughout the whole region. This completes the proof that there is the unique solution of Poisson's equation with a Dirichlet boundary condition.

Second, we consider the case where Neumann boundary conditions are specified as   on the boundary of the region. If the Neumann boundary condition is satisfied on   by both solutions, then the left-hand side of (2) is zero again. Consequently, as before, we find that

 

As before, since this is the volume integral of a positive quantity, we must have   at all points. Further, because the gradient of   is everywhere zero within the volume  , and because the gradient of   is everywhere zero on the boundary  , therefore   must be constant---but not necessarily zero---throughout the whole region. Finally, since   throughout the whole region, and since   throughout the whole region, therefore   throughout the whole region. This completes the proof that there is the unique solution up to an additive constant of Poisson's equation with a Neumann boundary condition.

Mixed boundary conditions could be given as long as either the gradient or the potential is specified at each point of the boundary. Boundary conditions at infinity also hold. This results from the fact that the surface integral in (2) still vanishes at large distances because the integrand decays faster than the surface area grows.

See also

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References

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  • L.D. Landau, E.M. Lifshitz (1975). The Classical Theory of Fields. Vol. 2 (4th ed.). Butterworth–Heinemann. ISBN 978-0-7506-2768-9.
  • J. D. Jackson (1998). Classical Electrodynamics (3rd ed.). John Wiley & Sons. ISBN 978-0-471-30932-1.