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

The general expression for Poisson's equation in electrostatics is

$\mathbf {\nabla } ^{2}\varphi =-{\frac {\rho _{f}}{\epsilon _{0}}},$

where $\varphi$ is the electric potential and $\rho _{f}$ is the charge distribution over some region $V$ with boundary surface $S$ .

The uniqueness of the solution can be proven for a large class of boundary conditions as follows.

Suppose that we claim to have two solutions of Poisson's equation. Let us call these two solutions $\varphi _{1}$ and $\varphi _{2}$. Then

$\mathbf {\nabla } ^{2}\varphi _{1}=-{\frac {\rho _{f}}{\epsilon _{0}}}$, and

$\mathbf {\nabla } ^{2}\varphi _{2}=-{\frac {\rho _{f}}{\epsilon _{0}}}$.

It follows that $\varphi =\varphi _{2}-\varphi _{1}$ is a solution of Laplace's equation, which is a special case of Poisson's equation that equals to $0$. By subtracting the two solutions above gives

$\mathbf {\nabla } ^{2}\varphi =\mathbf {\nabla } ^{2}\varphi _{1}-\mathbf {\nabla } ^{2}\varphi _{2}=0.\qquad (1)$

We now sequentially consider three distinct boundary conditions: a Dirichlet boundary condition, a Neumann boundary condition, and a mixed boundary condition.

First, we consider the case where Dirichlet boundary conditions are specified as $\varphi =0$ on the boundary of the region. These follow because the boundary conditions and the charge distributions are the same for both 'solutions'.

By applying the vector differential identity we know that

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

However, from $(1)$ we also know that throughout the region $\nabla ^{2}\varphi =0.$ Consequently, the second term goes to zero and we find that

$\nabla \cdot (\varphi \,\nabla \varphi )=\,(\nabla \varphi )^{2}.$

By taking the volume integral over the region $V$, we find that

$\int _{V}\mathbf {\nabla } \cdot (\varphi \,\mathbf {\nabla } \varphi )\,\mathrm {d} V=\int _{V}\mathbf {(\nabla } \varphi )\cdot (\mathbf {\nabla } \varphi )\,\mathrm {d} V=\int _{V}(\mathbf {\nabla } \varphi )^{2}\,\mathrm {d} V.$

By applying the divergence theorem, we rewrite the expression above as

$\int _{S}(\varphi \,\mathbf {\nabla } \varphi )\cdot \mathrm {d} \mathbf {S} =\int _{V}(\mathbf {\nabla } \varphi )^{2}\,\mathrm {d} V.\qquad (2)$

If the Dirichlet boundary condition is satisfied on $S$ by both solutions (i.e., if $\varphi =0$ on the boundary), then the left-hand side of $(2)$ is zero. Consequently, we find that

$\int _{V}(\mathbf {\nabla } \varphi )^{2}\,\mathrm {d} V=0.$

Further, because this is the volume integral of a positive quantity (due to the squared term), we must have $\nabla \varphi =0$ at all points. Further still, because the gradient of $\varphi$ is everywhere zero and $\varphi$ is zero on the boundary, $\varphi$ must be zero throughout the whole region. Finally, since $\varphi =0$ throughout the whole region and since $\varphi =\varphi _{2}-\varphi _{1}$ throughout the whole region, therefore $\varphi _{1}=\varphi _{2}$ throughout the whole region. This completes the proof that there is the unique solution of Poisson's equation with a Dirichlet boundary condition.

Second, we consider the case where Neumann boundary conditions are specified as $\nabla \varphi =0$ on the boundary of the region. If the Neumann boundary condition is satisfied on $S$ by both solutions (i.e., if $\nabla \varphi =0$ on the boundary), then the left-hand side of $(2)$ is zero. Consequently, as before, we find that

$\int _{V}(\mathbf {\nabla } \varphi )^{2}\,\mathrm {d} V=0.$

In the case of the Neumann boundary condition, however, the relationship between the solutions is only constrained to a constant factor $k$. In other words, $\varphi _{1}-\varphi _{2}=k$, because only the normal derivative of $\varphi$ was specified to be zero.

Mixed boundary conditions could be given as long as either the gradient or the potential is specified at each point of the proof.

Boundary conditions at infinity also hold as the surface integral in $(2)$ still vanishes at large distances as the integrand decays faster than the surface area grows.

See also

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

References

• L.D. Landau, E.M. Lifshitz (1975). The Classical Theory of Fields. Vol. Vol. 2 (4th ed.). Butterworth–Heinemann. ISBN 978-0-7506-2768-9. {{cite book}}: |volume= has extra text (help)
• J. D. Jackson (1998). Classical Electrodynamics (3rd ed.). John Wiley & Sons. ISBN 978-0-471-30932-1.