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.

__TOC__

==Proof==

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

:<math>\mathbf{\nabla}\cdot(\epsilon^2 \mathbf{\nabla}\varphi) = -4 \pi frac{\rho_rho_f}{f\epsilon_0},</math>

Herewhere <math>\varphi</math> is the [[electric potential]] and <math>\mathbf{E}=-\mathbf{\nabla}\varphirho_f</math> is the [[electriccharge fielddensity|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.

Suppose that therewe areclaim to have two solutions <math>\varphi_{1}</math>of andPoisson's <math>\varphi_{2}</math>equation. OneLet canus thencall define <math>\phi=\varphi_{2}-\varphi_{1}</math> which is the difference of thethese two solutions. Given that both <math>\varphi_{1}varphi_1</math> and <math>\varphi_{2}varphi_2</math>. satisfy [[Poisson's Equation]], <math>\phi</math> must satisfyThen

:<math>\mathbf{\nabla}^2 \cdot(\epsilonvarphi_1 = - \mathbffrac{\nablarho_f}{\phi)= 0epsilon_0},</math> and

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

Using the identity

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>\nabla \cdot (\phi \epsilon \, \nabla \phi )=\epsilon \, (\nabla \phi )^2 + \phi \nabla \cdot (\epsilon \, \nabla \phi )</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

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 + \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

Taking the volume integral over all space specified by the boundary conditions gives

:<math>\int_Vnabla \mathbf{\nabla}\cdot (\phivarphi \epsilon, \mathbf{\nabla} \phivarphi ) d^3 \mathbf{r}= \int_V \epsilon, (\mathbf{\nabla} \phivarphi )^2 \, d^3 \mathbf{r}.</math>

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

Applying the [[divergence theorem]], the expression can be rewritten as

:<math>\sum_iint_V \int_mathbf{S_i\nabla} \cdot(\phivarphi \epsilon, \mathbf{\nabla}\phivarphi) \cdot, \mathbfmathrm{dSd}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

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

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

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.

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

This means that the gradient of the solution is unique when

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

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.

The boundary conditions for which the above is true are:

#Second, we consider the case where [[DirichletNeumann boundary condition]]:s are specified as <math>\nabla\varphi = 0</math> ison wellthe definedboundary atof allthe ofregion. If the Neumann boundary surfaces.condition is Assatisfied suchon <math>\varphi_1=\varphi_2S</math> soby atboth thesolutions, boundarythen <math>\phithe =left-hand 0</math>side andof correspondingly({{EquationNote|2}}) theis surface integralzero vanishesagain. Consequently, as before, we find that

# [[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.

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

# Modified [[Neumann boundary condition]] (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.

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]]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

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[[Category:Electrostatics]]

[[Category:Vector calculus]]

[[Category:Uniqueness theorems]]

[[Category:Theorems in calculus]]

[[ar:نظرية الفردية]]