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, thefor equationa haslarge aclass uniqueof [[gradientboundary condition]] ofs, the solutionequation formay ahave largemany classsolutions, but the gradient of [[boundaryevery condition]]ssolution is the same. In the case of [[electrostatics]], this means that ifthere anis a unique [[electric field]] satisfyingderived thefrom boundarya conditions[[Electric ispotential|potential found,function]] thensatisfying itPoisson's isequation under the completeboundary electric fieldconditions.

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==Proof==

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

:<math>\mathbf{\nabla}\cdot(\epsilon\mathbf{\nabla}^2 \varphi) = -4\pifrac{\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(\epsilon\mathbf{\nabla}\varphi)= -4\pi\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 \cdot(\epsilonvarphi_1 = - \mathbffrac{\nablarho_f}{\phi)= 0epsilon_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}^2 \varphi_2 = - \frac{\rho_f}{\epsilon_0}.</math>

Suppose that there are two solutions <math>\varphi_{1}</math> and <math>\varphi_{2}</math>. One can then define <math>\phi=\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>\phi</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(\epsilon \mathbf{\nabla}\phi)= 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 (\phi \epsilonvarphi \, \nabla \phivarphi )=\epsilon \, (\nabla \phivarphi )^2 + \phi \nabla \cdot (\epsilonvarphi \, \nabla^2 \phi )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 (\phivarphi \epsilon, \mathbf{\nabla} \phivarphi )= \epsilon, (\mathbf{\nabla} \phivarphi )^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(\phivarphi \epsilon, \mathbf{\nabla}\phivarphi) d^3\, \mathbfmathrm{rd}V = \int_V \epsilon (\mathbf{\nabla}\phivarphi)^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} (\phivarphi \epsilon, \mathbf{\nabla}\phivarphi) \cdot \mathrm{d}\mathbf{dSS}= \int_V \epsilon (\mathbf{\nabla}\phivarphi)^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} (\phi\epsilon \mathbf{\nabla}\phi) \cdot \mathbf{dS}= \int_V \epsilon (\mathbf{\nabla}\phi)^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>\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0.</math>

Since <math>\epsilon > 0</math> and <math>(\mathbf{\nabla}\phi)^2 \ge 0</math>, then <math>\mathbf{\nabla}\phi</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} (\phi\epsilon \, \mathbf{\nabla}\phi) \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 are:

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>\phi = 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}\phi=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 asis wellspecified at each point of the uniquenessboundary. theoremBoundary holdsconditions ifat theinfinity surfacealso integralhold. vanishes,This whichresults isfrom the casefact (forthat examplethe surface integral in ({{EquationNote|2}}) whenstill vanishes at large distances because the integrand decays faster than the surface area grows.v1-v2=0

==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

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[[Category:Electrostatics]]

[[Category:Vector calculus]]

[[Category:Uniqueness theorems]]

[[Category:Theorems in calculus]]