Uniqueness theorem for Poisson's equation: Difference between revisions

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The '''uniqueness theorem''' for [[Poisson's equation]] states that, for a large class of [[boundary condition]]s, 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 [[Electric potential|potential function]] satisfying Poisson's equation under the boundary conditions.

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==Proof==

The general expression for [[Poisson's equation]] in [[electrostatics]] is

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:<math>\nabla \cdot (\varphi \, \nabla \varphi )= \, (\nabla \varphi )^2 + \varphi \, \nabla^2 \varphi.</math>

However, from ({{EquationNote|1}}) we also know that throughout the region <math>\nabla^2 \varphi = 0.</math> Consequently, the second term goes to zero and we find that

:<math>\nabla \cdot (\varphi \, \nabla \varphi )= \, (\nabla \varphi )^2.</math>

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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 condition|Dirichlet boundary conditions]]s are specified as <math>\varphi = 0</math> on the boundary of the region. If the Dirichlet boundary condition is satisfied on <math>S</math> by both solutions (i.e., if <math>\varphi = 0</math> on the boundary), then the left-hand side of ({{EquationNote|2}}) is zero. Consequently, we find that

:<math>\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0.</math>

Since this is the volume integral of a positive quantity (due to the squared term), we must have <math>\nabla \varphi = 0</math> at all points. Further, because the gradient of <math>\varphi</math> is everywhere zero and <math>\varphi</math> is zero on the boundary, <math>\varphi</math> must be zero throughout the whole region. Finally, since <math>\varphi = 0</math> throughout the whole region, and since <math>\varphi = \varphi_2 - \varphi_1</math> throughout the whole region, therefore <math>\varphi_1 = \varphi_2</math> 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 condition|Neumann boundary conditions]]s are specified as <math>\nabla\varphi = 0</math> on the boundary of the region. If the Neumann boundary condition is satisfied on <math>S</math> by both solutions, then the left-hand side of ({{EquationNote|2}}) is zero again. Consequently, as before, we find that

:<math>\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0.</math>

As before, since this is the volume integral of a positive quantity, we must have <math>\nabla \varphi = 0</math> at all points. Further, because the gradient of <math>\varphi</math> is everywhere zero within the volume <math>V</math>, and because the gradient of <math>\varphi</math> is everywhere zero on the boundary <math>S</math>, therefore <math>\varphi</math> must be constant---but not necessarily zero---throughout the whole region. Finally, since <math>\varphi = k</math> throughout the whole region, and since <math>\varphi = \varphi_2 - \varphi_1</math> throughout the whole region, therefore <math>\varphi_1 = \varphi_2 - k</math> 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 condition|Mixed boundary conditions]]s 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 ({{EquationNote|2}}) still vanishes at large distances because the integrand decays faster than the surface area grows.

==See also==

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|year=1975

|title=The Classical Theory of Fields

|edition=4th |volume=Vol. 2

|publisher=[[Butterworth–Heinemann]]

|isbn=978-0-7506-2768-9