Uniqueness theorem for Poisson's equation: Difference between revisions

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{{short description|For a large class of boundary conditions, all solutions have the same gradient}}
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==

In [[Gaussian units]], theThe general expression for [[Poisson's equation]] in [[electrostatics]] is

:<math>\mathbf{\nabla}\cdot(\varepsilon\,\mathbf{\nabla}^2 \varphi) = -\frac{\rho_f}{\epsilon_0},</math>

where <math>\varphi</math> is the [[electric potential]] and <math>\rho_f</math> is the [[charge density|charge distribution]] over some region <math>V</math> with boundary surface <math>S</math> .

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 followingas wayfollows.

:<math>\mathbf{\nabla}\cdot(\varepsilon\,\mathbf{\nabla}\varphi)= -\rho_f</math>

Suppose that we claim to have two solutions of Poisson's equation. Let us call these two solutions <math>\varphi_1</math> and <math>\varphi_2</math>. Then

Here <math>\varphi</math> is the [[electric potential]] and <math>\mathbf{E}=-\mathbf{\nabla}\varphi</math> is the [[electric field]].

:<math>\mathbf{\nabla}^2 \varphi_1 = - \frac{\rho_f}{\epsilon_0},</math> and

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.

:<math>\mathbf{\nabla}\cdot(\varepsilon^2 \,varphi_2 = - \mathbffrac{\nablarho_f}{\varphi)= 0epsilon_0}.</math>

Suppose that there are two solutions <math>\varphi_1</math> and <math>\varphi_2</math>. One can then define <math>\varphi=\varphi_2-\varphi_1</math> which is the difference of the two solutions. Given that both <math>\varphi_1</math> and <math>\varphi_2</math> satisfy [[Poisson's equation]], <math>\varphi</math> must satisfy

It follows that <math>\varphi=\varphi_2-\varphi_1</math> is a solution of [[Laplace's equation]], which is a special case of [[Poisson's equation]] that equals to <math>0</math>. Subtracting the two solutions above gives

:<math>\mathbf{\nabla}\cdot(\varepsilon \, \mathbf{\nabla}\varphi)= 0</math>

{{NumBlk||<math display="block">\mathbf{\nabla}^2 \varphi = \mathbf{\nabla}^2 \varphi_2 - \mathbf{\nabla}^2 \varphi_1 = 0. </math>|{{EquationRef|1}}}}

By applying the [[Vector calculus identities#Divergence 2|vector differential identity]] we know that

Using the identity

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

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

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

TakingBy taking the volume integral over allthe spaceregion specified by the<math>V</math>, boundarywe conditionsfind givesthat

:<math>\int_V \mathbf{\nabla}\cdot(\varphi\varepsilon \, \mathbf{\nabla}\varphi) \, d^3 \mathbfmathrm{rd}V = \int_V \varepsilon (\mathbf{\nabla}\varphi)^2 \, d^3 \mathbfmathrm{rd}V.</math>

ApplyingBy applying the [[divergence theorem]], thewe expressionrewrite canthe beexpression rewrittenabove as

:{{NumBlk||<math>\sum_i display="block">\int_{S_iS} (\varphi\varepsilon \, \mathbf{\nabla}\varphi) \cdot \mathrm{d}\mathbf{dSS}= \int_V \varepsilon (\mathbf{\nabla}\varphi)^2 \, d^3 \mathbfmathrm{rd}V. </math>|{{EquationRef|2}}}}

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

:<math>\sum_i \int_{S_i} (\varphi\varepsilon \, \mathbf{\nabla}\varphi) \cdot \mathbf{dS}= \int_V \varepsilon (\mathbf{\nabla}\varphi)^2 \, d^3 \mathbf{r}</math>

First, we consider the case where [[Dirichlet boundary condition]]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

where <math>S_i</math> are boundary surfaces specified by boundary conditions.

:<math>\sum_i \int_{S_i}int_V (\varphi\varepsilon \, \mathbf{\nabla}\varphi)^2 \cdot, \mathbfmathrm{dSd}V = 0 .</math>

Since <math>\varepsilon > 0</math> and <math>(\mathbf{\nabla}\varphi)^2 \ge 0</math>, then <math>\mathbf{\nabla}\varphi</math> must be zero everywhere (and so <math>\mathbf{\nabla}\varphi_1 = \mathbf{\nabla}\varphi_2</math>) when the surface integral vanishes.

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.

This means that the gradient of the solution is unique when

Second, we consider the case where [[Neumann boundary condition]]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>\sum_i \int_{S_i} (\varphi\varepsilon \, \mathbf{\nabla}\varphi) \cdot \mathbf{dS} = 0 </math>

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

The boundary conditions for which the above is true include:

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.

# [[Dirichlet boundary condition]]: <math>\varphi</math> is well defined at all of the boundary surfaces. As such <math>\varphi_1=\varphi_2</math> so at the boundary <math>\varphi = 0</math> and correspondingly the surface integral vanishes.

# [[Neumann boundary condition]]: <math>\mathbf{\nabla}\varphi</math> is well defined at all of the boundary surfaces. As such <math>\mathbf{\nabla}\varphi_1=\mathbf{\nabla}\varphi_2</math> so at the boundary <math>\mathbf{\nabla}\varphi=0</math> and correspondingly the surface integral vanishes.

# Modified [[Neumann boundary condition]] (also called [[Robin boundary condition]] – conditions where boundaries are specified as conductors with known charges): <math>\mathbf{\nabla}\varphi</math> is also well defined by applying locally [[Gauss's Law]]. As such, the surface integral also vanishes.

# Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann boundary conditions): the uniqueness theorem will still hold.

The[[Mixed boundary surfacescondition]]s maycould alsobe includegiven boundariesas atlong infinityas (describing''either'' unboundedthe domains)gradient –''or'' forthe thesepotential theis uniquenessspecified theoremat each holdspoint ifof the surfaceboundary. integralBoundary vanishes,conditions whichat isinfinity also hold. This results from the casefact (forthat examplethe surface integral in ({{EquationNote|2}}) whenstill 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