Boundary-Value Problems and Uniqueness
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You should be comfortable with vector calculus identities, specifically the divergence theorem and Green's first identity, and understand the physical meaning of Dirichlet and Neumann boundary conditions.
Why this matters
Uniqueness guarantees that the solution to an electrostatic problem is not just a guess, but the only possible answer.
In solving Poisson's or Laplace's equation, we often employ approximation methods or symmetry arguments to find a candidate potential. Without a rigorous proof of uniqueness, we could not be certain that this candidate is the correct physical solution. The Uniqueness Theorem provides the logical foundation that allows us to discard alternative mathematical solutions that do not satisfy the boundary conditions, ensuring that the electric field configuration in a given region is determined entirely by the charge distribution and the boundary values.
The Intuitive Argument
Imagine two different electric potential fields, \(\phi_1\) and \(\phi_2\), that both satisfy the same Poisson's equation inside a volume and have the same values on the boundary surface. If these two fields were truly different, their difference, let us call it \(W\), would be non-zero somewhere inside. However, because both fields obey the same differential equation with the same source terms, the difference \(W\) must satisfy Laplace's equation (\(\nabla^2 W = 0\)). Furthermore, because the boundary values are identical, \(W\) must be zero on the entire boundary. A function that is zero on the boundary and has no "sources" or "sinks" inside (zero Laplacian) cannot have any peaks or valleys in the interior; it must be flat. Therefore, the difference is zero everywhere, meaning the two original fields are actually identical. This logic holds for Neumann conditions as well, except that the potential is defined only up to an arbitrary constant, while the electric field remains unique.
Mathematical Formulation via Green's Identity
To prove this rigorously, we utilize Green's first identity, which relates the volume integral of the Laplacian to a surface integral. Let \(V\) be a bounded region with boundary surface \(S\). Consider two potentials \(\phi_1\) and \(\phi_2\) satisfying Poisson's equation \(\nabla^2 \phi = -\rho/\varepsilon_0\) in \(V\). Define the difference function \(W = \phi_1 - \phi_2\). Since the charge density \(\rho\) is fixed, \(\nabla^2 W = 0\). We examine the integral of the squared gradient of \(W\):
Using the vector identity \(\nabla \cdot (W \nabla W) = W \nabla^2 W + |\nabla W|^2\) and the divergence theorem, this becomes:
Since \(\nabla^2 W = 0\), the volume integral on the right vanishes. We are left with the surface integral. The sign of this integral depends on the boundary conditions.
Dirichlet Boundary Conditions
For Dirichlet conditions, the potential is prescribed on the boundary. Thus, \(\phi_1 = \phi_2\) on \(S\), which implies \(W = 0\) on \(S\). Consequently, the surface integral \(\int_S W \frac{\partial W}{\partial n} \, dA\) vanishes because the integrand contains the factor \(W\). Therefore:
Since \(|\nabla W|^2\) is non-negative everywhere, the only way for its integral to be zero is if \(\nabla W = 0\) everywhere in \(V\). This implies that \(W\) is a constant throughout the volume. Because \(W = 0\) on the boundary, this constant must be zero. Hence, \(\phi_1 \equiv \phi_2\) everywhere. The solution is unique.
Neumann Boundary Conditions
For Neumann conditions, the normal derivative of the potential is prescribed on the boundary. Thus, \(\frac{\partial \phi_1}{\partial n} = \frac{\partial \phi_2}{\partial n}\) on \(S\), implying \(\frac{\partial W}{\partial n} = 0\) on \(S\). The surface integral again vanishes, leading to \(\int_V |\nabla W|^2 \, dV = 0\). As before, this implies \(\nabla W = 0\) everywhere, so \(W\) is a constant \(C\). Unlike the Dirichlet case, we cannot determine the value of \(C\) from the boundary conditions because the derivative of a constant is zero. Therefore, \(\phi_1 = \phi_2 + C\). The potentials differ by a constant, but the electric field \(\vec{E} = -\nabla \phi\) is identical in both cases. The field is unique, but the potential is unique only up to an additive constant.
The derivation
Uniqueness Theorem for Boundary-Value Problems
Let \( V \) be a bounded region with closed boundary surface \( S \), containing a fixed charge density \( \rho(\vec{r}) \). Any potential \( \phi \) satisfying Poisson's equation \( \nabla^2\phi = -\rho/\varepsilon_0 \) throughout \( V \) and matching prescribed Dirichlet data (\( \phi \) given on \( S \)) is unique. If instead Neumann data (\( \partial\phi/\partial n \) given on \( S \)) is prescribed, the solution is unique up to an additive constant, so the field \( \vec{E} = -\nabla\phi \) is unique.
A common misconception is that the Neumann solution is not unique at all. It is crucial to distinguish between the potential and the field. While the potential is defined up to an arbitrary constant, the physical electric field is uniquely determined. In practical calculations, one often sets the potential to zero at a convenient point to fix this constant, but this choice does not affect the physics. C
Worked example
Problem. Consider a spherical shell of radius \(R = 2.0\ \mathrm{m}\) with a uniform surface charge density \(\sigma = 1.0 \times 10^{-6}\ \mathrm{C/m^2}\). Determine the uniqueness of the potential inside the shell (\(r < R\)) given that the potential is fixed to zero at the surface (\(V(R) = 0\)).
- Identify the region \(V\) as the interior of the sphere, where \(\rho = 0\), so \(\nabla^2 \phi = 0\).
- Note the boundary condition is Dirichlet: \(\phi(R) = 0\).
- By the Uniqueness Theorem for Dirichlet conditions, the solution is unique.
- Since the boundary value is zero and the Laplacian is zero, the only solution is \(\phi(r) = 0\) for all \(r < R\).
Units check. The potential is in volts (\(\mathrm{V}\)). Zero volts is a valid value of a potential; the surface charge only fixes the field outside, not the interior solution once the boundary value is given.
Three quick tests
- If two potentials satisfy the same Poisson equation and the same Neumann boundary conditions, are they identical?
Answer
No, they differ by a constant, but their electric fields are identical. - Why does the integral \(\int_V |\nabla W|^2 dV\) equal zero in the uniqueness proof?
Answer
Because the surface integral vanishes due to boundary conditions and the volume integral of \(W \nabla^2 W\) vanishes because \(\nabla^2 W = 0\). - Can the potential be unique for Neumann conditions?
Answer
No, it is unique only up to an additive constant. The field is unique.
Problem set for PU-102 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.