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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{\nabla}^2 \varphi = -\frac{\rho_f}{\epsilon_0},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is the electric potential and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho_f} is the charge distribution over some region $V$ with boundary surface Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi_2} . Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{\nabla}^2 \varphi_1 = - \frac{\rho_f}{\epsilon_0},} and

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \mathbf {\nabla } ^{2}\varphi _{2}=-{\frac {\rho _{f}}{\epsilon _{0}}}.}

It follows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi=\varphi_2-\varphi_1} is a solution of Laplace's equation, which is a special case of Poisson's equation that equals to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} . Subtracting the two solutions above gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{\nabla}^2 \varphi = \mathbf{\nabla}^2 \varphi_1 - \mathbf{\nabla}^2 \varphi_2 = 0. }

(1)

By applying the vector differential identity we know that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \cdot (\varphi \, \nabla \varphi )= \, (\nabla \varphi )^2 + \varphi \, \nabla^2 \varphi.}

However, from (1) we also know that throughout the region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla^2 \varphi = 0.} Consequently, the second term goes to zero and we find that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \cdot (\varphi \, \nabla \varphi )= \, (\nabla \varphi )^2.}

By taking the volume integral over the region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} , we find that

\int _{V}\mathbf {\nabla } \cdot (\varphi \,\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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{S} (\varphi \, \mathbf{\nabla}\varphi) \cdot \mathrm{d}\mathbf{S}= \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V. }

(2)

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi = 0} on the boundary of the region. If the Dirichlet boundary condition is satisfied on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} by both solutions (i.e., if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi = 0} on the boundary), then the left-hand side of (2) is zero. Consequently, we find that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0.}

Since this is the volume integral of a positive quantity (due to the squared term), we must have $\nabla \varphi =0$ at all points. Further, because the gradient of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is everywhere zero and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is zero on the boundary, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} must be zero throughout the whole region. Finally, since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi = 0} throughout the whole region, and since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi = \varphi_2 - \varphi_1} throughout the whole region, therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\varphi = 0} on the boundary of the region. If the Neumann boundary condition is satisfied on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} by both solutions, then the left-hand side of (2) is zero again. Consequently, as before, we find that

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

As before, since this is the volume integral of a positive quantity, we must have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \varphi = 0} at all points. Further, because the gradient of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is everywhere zero within the volume Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} , and because the gradient of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is everywhere zero on the boundary Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} , therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} must be constant---but not necessarily zero---throughout the whole region. Finally, since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi = k} throughout the whole region, and since $\varphi =\varphi _{2}-\varphi _{1}$ throughout the whole region, therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi_1 = \varphi_2 - k} throughout the whole region. This completes the proof that there is the unique solution up to an additive constant of Poisson's equation with a Neumann boundary condition.

Mixed boundary conditions could be given as long as either the gradient or the potential is specified at each point of the boundary. Boundary conditions at infinity also hold. This results from the fact that the surface integral in (2) still vanishes at large distances because the integrand decays faster than the surface area grows.

References

L.D. Landau, E.M. Lifshitz (1975). The Classical Theory of Fields. Vol. 2 (4th ed.). Butterworth–Heinemann. ISBN 978-0-7506-2768-9.
J. D. Jackson (1998). Classical Electrodynamics (3rd ed.). John Wiley & Sons. ISBN 978-0-471-30932-1.

Uniqueness theorem for Poisson's equation

Contents

Proof

See also

References