# Uniqueness theorem for Poisson's equation

The uniqueness theorem for Poisson's equation states that, for a large class of boundary conditions, 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 potential function satisfying Poisson's equation under the boundary conditions.

## Proof

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

$\mathbf {\nabla } \cdot (\varepsilon \,\mathbf {\nabla } \varphi )=-\rho _{f}$ 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 $\varphi =\varphi _{2}-\varphi _{1}$ which is the difference of the two solutions. Given that both $\varphi _{1}$ and $\varphi _{2}$ satisfy Poisson's equation, $\varphi$ must satisfy

$\mathbf {\nabla } \cdot (\varepsilon \,\mathbf {\nabla } \varphi )=0$ Using the identity

$\nabla \cdot (\varphi \varepsilon \,\nabla \varphi )=\varepsilon \,(\nabla \varphi )^{2}+\varphi \,\nabla \cdot (\varepsilon \,\nabla \varphi )$ And noticing that the second term is zero, one can rewrite this as

$\mathbf {\nabla } \cdot (\varphi \varepsilon \,\mathbf {\nabla } \varphi )=\varepsilon (\mathbf {\nabla } \varphi )^{2}$ Taking the volume integral over all space specified by the boundary conditions gives

$\int _{V}\mathbf {\nabla } \cdot (\varphi \varepsilon \,\mathbf {\nabla } \varphi )\,d^{3}\mathbf {r} =\int _{V}\varepsilon (\mathbf {\nabla } \varphi )^{2}\,d^{3}\mathbf {r}$ Applying the divergence theorem, the expression can be rewritten as

$\sum _{i}\int _{S_{i}}(\varphi \varepsilon \,\mathbf {\nabla } \varphi )\cdot \mathbf {dS} =\int _{V}\varepsilon (\mathbf {\nabla } \varphi )^{2}\,d^{3}\mathbf {r}$ where $S_{i}$ are boundary surfaces specified by boundary conditions.

Since $\varepsilon >0$ and $(\mathbf {\nabla } \varphi )^{2}\geq 0$ , then $\mathbf {\nabla } \varphi$ 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

$\sum _{i}\int _{S_{i}}(\varphi \varepsilon \,\mathbf {\nabla } \varphi )\cdot \mathbf {dS} =0$ The boundary conditions for which the above is true include:

1. Dirichlet boundary condition: $\varphi$ is well defined at all of the boundary surfaces. As such $\varphi _{1}=\varphi _{2}$ so at the boundary $\varphi =0$ 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 $\mathbf {\nabla } \varphi =0$ and correspondingly the surface integral vanishes.
3. Modified Neumann boundary condition (also called Robin boundary condition – conditions 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.

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.