# 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

${\displaystyle \mathbf {\nabla } \cdot (\varepsilon \,\mathbf {\nabla } \varphi )=-\rho _{f}}$

Here ${\displaystyle \varphi }$ is the electric potential and ${\displaystyle \mathbf {E} =-\mathbf {\nabla } \varphi }$ 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 ${\displaystyle \varphi _{1}}$ and ${\displaystyle \varphi _{2}}$. One can then define ${\displaystyle \varphi =\varphi _{2}-\varphi _{1}}$ which is the difference of the two solutions. Given that both ${\displaystyle \varphi _{1}}$ and ${\displaystyle \varphi _{2}}$ satisfy Poisson's equation, ${\displaystyle \varphi }$ must satisfy

${\displaystyle \mathbf {\nabla } \cdot (\varepsilon \,\mathbf {\nabla } \varphi )=0}$

Using the identity

${\displaystyle \nabla \cdot (\varphi \varepsilon \,\nabla \varphi )=\varepsilon \,(\nabla \varphi )^{2}+\varphi \,\nabla \cdot (\varepsilon \,\nabla \varphi )}$

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

${\displaystyle \mathbf {\nabla } \cdot (\varphi \varepsilon \,\mathbf {\nabla } \varphi )=\varepsilon (\mathbf {\nabla } \varphi )^{2}}$

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

${\displaystyle \int _{V}\mathbf {\nabla } \cdot (\varphi \varepsilon \,\mathbf {\nabla } \varphi )\,d^{3}\mathbf {r} =\int _{V}\varepsilon (\mathbf {\nabla } \varphi )^{2}\,d^{3}\mathbf {r} }$

Applying the divergence theorem, the expression can be rewritten as

${\displaystyle \sum _{i}\int _{S_{i}}(\varphi \varepsilon \,\mathbf {\nabla } \varphi )\cdot \mathbf {dS} =\int _{V}\varepsilon (\mathbf {\nabla } \varphi )^{2}\,d^{3}\mathbf {r} }$

where ${\displaystyle S_{i}}$ are boundary surfaces specified by boundary conditions.

Since ${\displaystyle \varepsilon >0}$ and ${\displaystyle (\mathbf {\nabla } \varphi )^{2}\geq 0}$, then ${\displaystyle \mathbf {\nabla } \varphi }$ must be zero everywhere (and so ${\displaystyle \mathbf {\nabla } \varphi _{1}=\mathbf {\nabla } \varphi _{2}}$) when the surface integral vanishes.

This means that the gradient of the solution is unique when

${\displaystyle \sum _{i}\int _{S_{i}}(\varphi \varepsilon \,\mathbf {\nabla } \varphi )\cdot \mathbf {dS} =0}$

The boundary conditions for which the above is true include:

1. Dirichlet boundary condition: ${\displaystyle \varphi }$ is well defined at all of the boundary surfaces. As such ${\displaystyle \varphi _{1}=\varphi _{2}}$ so at the boundary ${\displaystyle \varphi =0}$ and correspondingly the surface integral vanishes.
2. Neumann boundary condition: ${\displaystyle \mathbf {\nabla } \varphi }$ is well defined at all of the boundary surfaces. As such ${\displaystyle \mathbf {\nabla } \varphi _{1}=\mathbf {\nabla } \varphi _{2}}$ so at the boundary ${\displaystyle \mathbf {\nabla } \varphi =0}$ 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): ${\displaystyle \mathbf {\nabla } \varphi }$ 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.