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

Tthe 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.

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}}}

\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 (a special case of Poisson's equation which equals $0$ ) because subtracting the two solutions above gives

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

Let us first 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'.

We can now use the vector differential identity

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

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

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

Taking the volume integral over the region gives

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

And after aplying the divergence theorem, the expression above can be rewritten as

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

where $S$ is the boundary surface specified by the boundary conditions.

If the Dirichlet boundary condition is satisified by both solutions (ie, $\varphi =0$ on the boundary) then the left-hand side of $(2)$ is zero thus

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

However, because this is the volume integral of a positive quantity (due to the squared term), we must have

$\nabla \varphi =0$

at all points.

Finally, because the gradient of $\varphi$ is everywhere zero and $\varphi$ is zero on the boundary, $\varphi$ must be zero throughout the whole region. This proves $\varphi _{1}=\varphi _{2}$ and the solutions are identical.

If the Neumann boundary conditions had been specified then the normal component of $\nabla \phi$ on the left-hand side of $(2)$ would be zero on the boundary and we would arrive at the same conclusion. In this case, 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.

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.

Proof

See also

References