Uniqueness theorem for Poisson's equation

The uniqueness theorem for Poisson's equation states that the equation has a unique gradient of the solution for a large class of boundary conditions. In the case of electrostatics, this means that if an electric field satisfying the boundary conditions is found, then it is the complete electric field.

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Proof

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

\mathbf{\nabla}\cdot(\epsilon \mathbf{\nabla}\varphi)= -4 \pi \rho_{f}

Here \varphi is the electric potential and \mathbf{E}=-\mathbf{\nabla}\varphi is the electric field.

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 \varphi_{1} and \varphi_{2}. One can then define \phi=\varphi_{2}-\varphi_{1} which is the difference of the two solutions. Given that both \varphi_{1} and \varphi_{2} satisfy Poisson's Equation, \phi must satisfy

\mathbf{\nabla}\cdot(\epsilon \mathbf{\nabla}\phi)= 0

Using the identity

\nabla \cdot (\phi \epsilon \, \nabla \phi )=\epsilon \, (\nabla \phi )^2 %2B \phi \nabla \cdot (\epsilon \, \nabla \phi )

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

\mathbf{\nabla}\cdot(\phi\epsilon \mathbf{\nabla}\phi)= \epsilon (\mathbf{\nabla}\phi)^2

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

\int_V \mathbf{\nabla}\cdot(\phi\epsilon \mathbf{\nabla}\phi) d^3 \mathbf{r}= \int_V \epsilon (\mathbf{\nabla}\phi)^2 \, d^3 \mathbf{r}

Applying the divergence theorem, the expression can be rewritten as

\sum_i \int_{S_i} (\phi\epsilon \mathbf{\nabla}\phi) \cdot \mathbf{dS}= \int_V \epsilon (\mathbf{\nabla}\phi)^2 \, d^3 \mathbf{r}

Where S_i are boundary surfaces specified by boundary conditions.

Since \epsilon > 0 and (\mathbf{\nabla}\phi)^2 \ge 0, then \mathbf{\nabla}\phi must be zero everywhere (and so \mathbf{\nabla}\varphi_{1} = \mathbf{\nabla}\varphi_{2}) when the surface integral vanishes.

This means that the gradient of the solution is unique when

\sum_i \int_{S_i} (\phi\epsilon \, \mathbf{\nabla}\phi) \cdot \mathbf{dS} = 0

The boundary conditions for which the above is true are:

  1. Dirichlet boundary condition: \varphi is well defined at all of the boundary surfaces. As such \varphi_1=\varphi_2 so at the boundary \phi = 0 and correspondingly the surface integral vanishes.
  2. Neumann boundary condition: \mathbf{\nabla}\varphi is well defined at all of the boundary surfaces. As such \mathbf{\nabla}\varphi_1=\mathbf{\nabla}\varphi_2 so at the boundary \mathbf{\nabla}\phi=0 and correspondingly the surface integral vanishes.
  3. Modified Neumann boundary condition (where boundaries are specified as conductors with known charges): \mathbf{\nabla}\varphi is also well defined by applying locally Gauss's Law. As such, the surface integral also vanishes.
  4. Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann boundary conditions): the uniqueness theorem will still hold.

See also

References