Uniqueness theorem for Poisson's equation: Difference between revisions

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

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

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

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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.

Suppose that there are two solutions <math>\varphi_{1}varphi_1</math> and <math>\varphi_{2}varphi_2</math>. One can then define <math>\phivarphi=\varphi_{2}varphi_2-\varphi_{1}varphi_1</math> which is the difference of the two solutions. Given that both <math>\varphi_{1}varphi_1</math> and <math>\varphi_{2}varphi_2</math> satisfy [[Poisson's Equationequation]], <math>\phivarphi</math> must satisfy

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

Using the identity

:<math>\nabla \cdot (\phivarphi \epsilonvarepsilon \, \nabla \phivarphi )=\epsilonvarepsilon \, (\nabla \phivarphi )^2 + \phivarphi \, \nabla \cdot (\epsilonvarepsilon \, \nabla \phivarphi )</math>

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

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

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

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

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

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

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

Since <math>\epsilonvarepsilon > 0</math> and <math>(\mathbf{\nabla}\phivarphi)^2 \ge 0</math>, then <math>\mathbf{\nabla}\phivarphi</math> must be zero everywhere (and so <math>\mathbf{\nabla}\varphi_{1}varphi_1 = \mathbf{\nabla}\varphi_{2}varphi_2</math>) when the surface integral vanishes.

This means that the gradient of the solution is unique when

:<math>\sum_i \int_{S_i} (\phivarphi\epsilonvarepsilon \, \mathbf{\nabla}\phivarphi) \cdot \mathbf{dS} = 0 </math>

0</math>

The boundary conditions for which the above is true include:

# [[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>\phivarphi = 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}\phivarphi=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.