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.

__TOC__

==Proof==

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

:<math>\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 solution can be proven for a large class of boundary conditions as follows.

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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

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

:<math>\mathbf{\nabla}^2 \varphi_2 = - \frac{\rho_f}{\epsilon_0}</math>

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

:<math>\mathbf{\nabla}^2 \varphi = \mathbf{\nabla}^2 \varphi_1 - \mathbf{\nabla}^2 \varphi_2 = 0 \qquad (1)</math>

Let us first consider the case where [[Dirichlet boundary condition|Dirichlet boundary conditions]] are specified as <math>\varphi = 0</math> on the boundary of the region. These follow because the boundary conditions and the charge distributions are the same for both 'solutions'.

We can now use the [[Vector calculus identities#Divergence 2|vector differential identity]]

:<math>\nabla \cdot (\varphi \, mathbf{\nabla}^2 \varphivarphi_2 )= \,- (\nabla frac{\varphi )^2 + \varphi \, \nabla^2 rho_f}{\varphiepsilon_0}.</math>

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]] whichthat equals to <math>0</math>) because. subtractingSubtracting the two solutions above gives

However, from <math>(1)</math> we know that <math>\nabla^2 \varphi = 0</math> throughout the region so the second term goes to zero

:{{NumBlk||<math display="block">\mathbf{\nabla}^2 \varphi = \mathbf{\nabla}^2 \varphi_1varphi_2 - \mathbf{\nabla}^2 \varphi_2varphi_1 = 0. \qquad (1)</math>|{{EquationRef|1}}}}

WeBy can now useapplying the [[Vector calculus identities#Divergence 2|vector differential identity]] we know that

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

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

Taking the volume integral over the region gives

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

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

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

And after aplying the [[divergence theorem]], the expression above can be rewritten as

TakingBy taking the volume integral over the region gives<math>V</math>, we find that

:<math>\int_{S} (\varphi \, \mathbf{\nabla}\varphi) \cdot \mathrm{d}\mathbf{S}= \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V \qquad (2)</math>

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

where <math>S</math> is the boundary surface specified by the boundary conditions.

AndBy after aplyingapplying the [[divergence theorem]], we rewrite the expression above can be rewritten as

If the Dirichlet boundary condition is satisified by both solutions (ie, <math>\varphi = 0</math> on the boundary) then the left-hand side of <math>(2)</math> is zero thus

:{{NumBlk||<math display="block">\int_{S} (\varphi \, \mathbf{\nabla}\varphi) \cdot \mathrm{d}\mathbf{S}= \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V. \qquad (2)</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>\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0</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

However, because this is the volume integral of a positive quantity (due to the squared term), we must have

:<math>\nablaint_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.

at all points.

FinallySecond, becausewe consider the gradientcase ofwhere <math>\varphi</math>[[Neumann isboundary everywherecondition]]s zeroare specified andas <math>\nabla\varphi = 0</math> ison zerothe onboundary of the region. If the Neumann boundary, condition is satisfied on <math>\varphiS</math> mustby beboth zerosolutions, throughoutthen the wholeleft-hand region.side Thisof proves({{EquationNote|2}}) <math>\varphi_1is = \varphi_2</math>zero andagain. theConsequently, solutionsas arebefore, identical.we find that

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

If the [[Neumann boundary condition|Neumann boundary conditions]] had been specified then the normal component of <math>\nabla \phi</math> on the left-hand side of <math>(2)</math> would be zero on the boundary and we would arrive at the same conclusion. In this case, however, the relationship between the solutions is only constrained to a constant factor <math>k</math>, in other words <math>\varphi_1 - \varphi_2 = k</math>, because only the normal derivative of <math>\varphi</math> was specified to be zero.

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]] could be given as long as ''either'' the gradient ''or'' the potential is specified at each point of the proof.

[[Mixed boundary condition]]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. asThis results from the fact that the surface integral in <math>({{EquationNote|2}})</math> still vanishes at large distances asbecause 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