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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.

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

$\nabla \cdot (\phi \epsilon \,\nabla \phi )=\epsilon \,(\nabla \phi )^{2}+\phi \nabla \cdot (\epsilon \,\nabla \phi )$

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 $(\mathbf {\nabla } \phi )^{2}\geq 0$, 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 (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.

See also

• Poisson's equation
• Gauss's law
• Coulomb's law
• Method of images
• Green's function
• Uniqueness theorem
• Spherical harmonics

References

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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534