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

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

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>\varphi=\varphi_2-\varphi_1</math> which is the difference of thethese two solutions. Given that both <math>\varphi_1</math> and <math>\varphi_2</math>. satisfy [[Poisson's equation]], <math>\varphi</math> must satisfyThen

:<math>\mathbf{\nabla}\cdot(\varepsilon^2 \,varphi_1 = - \mathbffrac{\nablarho_f}{\varphi)= 0epsilon_0}</math>

:<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]] (a special case of [[Poisson's equation]] which equals <math>0</math>) because subtracting the two solutions above gives

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

:<math>\mathbf{\nabla}\cdot(^2 \varphi\varepsilon \,= \mathbf{\nabla}\varphi)=^2 \varepsilonvarphi_1 - (\mathbf{\nabla}\varphi)^2 \varphi_2 = 0 \qquad (1)</math>

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

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'.

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

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

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

:<math>\int_Vnabla \mathbf{\nabla}\cdot (\varphi\varepsilon \, \mathbf{\nabla} \varphi )= \, d^3 \mathbf{r}= \int_V \varepsilon (\mathbf{\nabla} \varphi )^2 + \varphi \, d\nabla^32 \mathbf{r}varphi</math>

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

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

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

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

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

Since :<math>\varepsilonint_V > 0</math> and <math>(\mathbf{\nabla}\cdot(\varphi)^2 \ge 0</math>, then <math>\mathbf{\nabla}\varphi</math>) must\, be\mathrm{d}V= zero everywhere\int_V (and so <math>\mathbf{\nabla}\varphi_1varphi)^2 =\, \mathbfmathrm{\nablad}\varphi_2V</math>) when the surface integral vanishes. [The English here is poor. Page in need of improvement for readability.]

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

This means that the gradient of the solution is unique when

:<math>\sum_i \int_{S_iS} (\varphi\varepsilon \, \mathbf{\nabla}\varphi) \cdot \mathrm{d}\mathbf{dSS} = 0\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V \qquad (2)</math>

where <math>S_iS</math> areis the boundary surfacessurface specified by the boundary conditions.

The boundary conditions for which the above is true include:

#If the [[Dirichlet boundary condition]]: is satisified by both solutions (ie, <math>\varphi = 0</math> ison wellthe definedboundary) atthen allthe ofleft-hand theside boundaryof surfaces.<math>(2)</math> Asis suchzero thus<blockquote><math>\varphi_1=int_V (\varphi_2mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0</math></blockquote>However, sobecause atthis is the boundaryvolume integral of a positive quantity (due to the squared term), we must have <blockquote><math>\nabla \varphi = 0</math></blockquote>at and correspondingly the surface integralall vanishespoints.

# [[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.

Finally, 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. This proves <math>\varphi_1 = \varphi_2</math> and the solutions are identical.

The boundary surfaces may also include boundaries at infinity (describing unbounded domains) – for these the uniqueness theorem holds if the surface integral vanishes, which is the case (for example) when at large distances the integrand decays faster than the surface area grows.

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.

[[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.

Boundary conditions at infinity also hold as the surface integral in <math>(2)</math> still vanishes at large distances as the integrand decays faster than the surface area grows.

==See also==