4.1: The Uniqueness Theorem

Consider a linear dielectric material where the permittivity may vary with position:

\[\textbf{D} = \varepsilon (r) \textbf{E} = - \varepsilon(r) \nabla V \]

The special case of different constant permittivity media separated by an interface has \(\varepsilon (r)\) as a step function. Using (1) in Gauss's law yields

\[\nabla \cdot [ \varepsilon (r) \nabla V ] = - \rho_{f} \]

which reduces to Poisson's equation in regions where \(\varepsilon (r)\) is a constant. Let us call V_p a solution to (2).

The solution V_L to the homogeneous equation

\[\nabla \cdot [\varepsilon (r) \nabla V] = 0 \]

which reduces to Laplace's equation when \(\varepsilon (r)\) is constant, can be added to V_p and still satisfy (2) because (2) is linear in the potential:

\[\nabla \cdot [\varepsilon (r) \nabla (V_{p} + V_{L})] = \nabla \cdot [ \varepsilon (r) \nabla V_{p}] + \underbrace{\nabla \cdot [ \varepsilon (r) \nabla V_{L}]}_{0} = - \rho_{f} \]

Any linear physical problem must only have one solution yet (3) and thus (2) have many solutions. We need to find what boundary conditions are necessary to uniquely specify this solution. Our method is to consider two different solutions V₁ and V₂ for the same charge distribution

\[\nabla \cdot (\varepsilon \nabla V_{1}) = - \rho_{f}, \: \: \: \: \nabla \cdot (\varepsilon \nabla V_{2}) - \rho_{f} \]

so that we can determine what boundary conditions force these solutions to be identical, V₁ = V₂

The difference of these two solutions V_T = V₁ - V₂ obeys the homogeneous equation

\[\nabla \cdot (\varepsilon \nabla V_{T}) = 0 \]

We examine the vector expansion

\[\nabla \cdot (\varepsilon V_{T} \nabla V_{T}) = V_{T} \underbrace{\nabla \cdot (\varepsilon \nabla V_{T})}_{0} + \varepsilon \nabla V_{T} \cdot \nabla C_{T} = \varepsilon \vert \nabla V_{T} \vert^{2} \]

noting that the first term in the expansion is zero from (6) and that the second term is never negative.

We now integrate (7) over the volume of interest V, which may be of infinite extent and thus include all space

\[\int_{\textrm{V}} \nabla \cdot (\varepsilon V_{T} \nabla V_{t}) d \textrm{V} = \oint_{S}\varepsilon V_{T} \nabla V_{T} \cdot \textbf{dS} = \int_{\textrm{V}} \varepsilon \vert \nabla V_{T} \vert ^{2} d \textrm{V} \]

The volume integral is converted to a surface integral over the surface bounding the region using the divergence theorem. Since the integrand in the last volume integral of (8) is never negative, the integral itself can only be zero if V_T is zero at every point in the volume making the solution unique (\(V_{T} = 0 \Rightarrow V_{1} = V_{2}\)). To force the volume integral to be zero, the surface integral term in (8) must be zero. This requires that on the surface S the two solutions must have the same value (\(V_{1} = V_{2}\)) or their normal derivatives must be equal [\(\nabla V_{1} \cdot \textbf{N} = \nabla V_{2} \cdot \textbf{n}\)]. This last condition is equivalent to requiring that the normal components of the electric fields be equal (\(\textbf{E} = - \nabla V\)).

Thus, a problem is uniquely posed when in addition to giving the charge distribution, the potential or the normal component of the electric field on the bounding surface surrounding the volume is specified. The bounding surface can be taken in sections with some sections having the potential specified and. other sections having the normal field component specified.

If a particular solution satisfies (2) but it does not satisfy the boundary conditions, additional homogeneous solutions where \(\rho_{f} = 0\), must be added so that the boundary conditions are met. No matter how a solution is obtained, even if guessed, if it satisfies (2) and all the boundary conditions, it is the only solution.