Uniqueness Theorem for Boundary-Value Problems

Derivation

Uniqueness Theorem for Boundary-Value Problems

D-044 Home PU-102 Threads fields · symmetry Depends on Scalar Potential, Poisson and Laplace Equations, greens-first-identity

Statement

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.

Why it matters

The theorem converts electrostatics from a search problem into a verification problem. Instead of solving the partial differential equation directly, you may guess any potential that (i) obeys Poisson's equation with the correct sources and (ii) meets the boundary data; uniqueness guarantees the guess is the physical answer. This is exactly the logical licence behind the method of images, conformal mapping, and separation-of-variables expansions.

It also underwrites the physical intuition that a conductor's geometry plus its charge or potential fully determines the exterior field. Two identical laboratory setups must produce identical fields; the theorem is the mathematical statement that nature does not leave the field ambiguous.

Assumptions

The region is bounded by a surface on which data is prescribed.On an unbounded domain the surface integral need not vanish; one must instead impose a fall-off condition (\( \phi \to 0 \) at infinity faster than \( 1/r \)) as a boundary condition at infinity, or uniqueness fails. The source \( \rho \) is identical for the two candidate solutions.If the charge distributions differ, the difference potential no longer satisfies Laplace's equation and the proof collapses at the first step. The potential and its first derivatives are continuous and the fields decay so Green's identity applies.At an idealised surface charge or point charge inside \( V \) the integrand is singular; the domain must be defined to exclude such singularities (or bound them with small surfaces) for the divergence theorem to hold. Either \( \phi \) or \( \partial\phi/\partial n \) — not a mix that leaves the surface term indeterminate — is fixed on every part of \( S \).A Robin (mixed) condition \( \alpha\phi + \beta\,\partial\phi/\partial n \) also gives uniqueness only when \( \alpha/\beta \ge 0 \); the wrong sign lets the surface term be negative and destroys the argument.

Derivation

\[ \nabla^2\phi_1 = -\frac{\rho}{\varepsilon_0}, \qquad \nabla^2\phi_2 = -\frac{\rho}{\varepsilon_0} \]

Suppose two solutions exist; both obey Poisson's equation with the same source. A

\[ W \equiv \phi_1 - \phi_2 \]

Define the difference potential — the object whose vanishing would prove uniqueness. A

\[ \nabla^2 W = \nabla^2\phi_1 - \nabla^2\phi_2 = 0 \]

The Laplacian is linear and the equal sources cancel, so \( W \) obeys Laplace's equation throughout \( V \). A

\[ \int_V \left( \phi\,\nabla^2\psi + \nabla\phi\cdot\nabla\psi \right) dV = \oint_S \phi\,\frac{\partial\psi}{\partial n}\, dA \]

Green's first identity (a prior result), valid for any smooth scalar fields \( \phi, \psi \) on \( V \) with boundary \( S \). B

\[ \int_V \left( W\,\nabla^2 W + \left|\nabla W\right|^2 \right) dV = \oint_S W\,\frac{\partial W}{\partial n}\, dA \]

Set \( \phi = \psi = W \) in the identity — the choice that turns the middle term into a positive-definite gradient magnitude. B

\[ \int_V \left|\nabla W\right|^2 dV = \oint_S W\,\frac{\partial W}{\partial n}\, dA \]

Drop the \( W\,\nabla^2 W \) term because \( \nabla^2 W = 0 \) from Step 3. A

\[ \oint_S W\,\frac{\partial W}{\partial n}\, dA = 0 \]

Boundary data is common to both solutions: Dirichlet gives \( W = 0 \) on \( S \), Neumann gives \( \partial W/\partial n = 0 \) on \( S \). Either way each element of the surface integrand vanishes. B

\[ \int_V \left|\nabla W\right|^2 dV = 0 \quad\Rightarrow\quad \nabla W = \vec{0}\ \text{ in } V \]

The integrand is non-negative and continuous; its integral can vanish only if it is identically zero everywhere in \( V \). C

\[ W = \phi_1 - \phi_2 = \text{const} \]

A field with zero gradient throughout a connected region is constant. A

\[ \text{Dirichlet: } W = 0 \Rightarrow \phi_1 = \phi_2 \qquad \text{Neumann: } \phi_1 = \phi_2 + \text{const} \]

For Dirichlet the constant is fixed to zero by \( W = 0 \) on \( S \); for Neumann the constant is unfixed but \( \vec{E} = -\nabla\phi \) is identical. A

Result

\[ \int_V \left|\nabla W\right|^2 dV = 0 \;\Rightarrow\; \phi_1 \equiv \phi_2 \ (\text{Dirichlet}), \quad \vec{E}_1 \equiv \vec{E}_2 \ (\text{Neumann}) \]

Reading. The energy-like quantity \( \int_V |\nabla W|^2\,dV \) — proportional to the field energy stored in the difference solution — is forced to zero by the shared source and shared boundary data. A non-negative quantity that sums to zero must be zero pointwise, so the two candidate solutions cannot differ in any physically observable way. Guessing a valid solution therefore is solving the problem.

Units check. \( |\nabla W|^2 \) has units \( (\mathrm{V\,m^{-1}})^2 = \mathrm{V^2\,m^{-2}} \); multiplied by \( dV \) (\( \mathrm{m^3} \)) gives \( \mathrm{V^2\,m} \). Multiplying by \( \varepsilon_0/2 \) (\( \mathrm{F\,m^{-1}} = \mathrm{C^2\,J^{-1}\,m^{-1}} \)) converts \( \mathrm{V^2\,m} \) into \( (\mathrm{J^2\,C^{-2}})(\mathrm{C^2\,J^{-1}\,m^{-1}})(\mathrm{m}) = \mathrm{J} \) — a genuine energy, confirming the integral is the field energy of \( W \).

Limiting cases

No sources (\( \rho = 0 \)). The theorem reduces to uniqueness for Laplace's equation — the harmonic function is fixed entirely by its boundary values.
Charge-free grounded cavity. With \( \rho = 0 \) inside and \( \phi = 0 \) on all walls, \( \phi = 0 \) everywhere is a solution, hence the only one; no field can exist in an empty shielded cavity.
Pure Neumann, total flux fixed. The unfixed additive constant is the only residual freedom; grounding one point (setting a reference) removes it and restores full uniqueness.
Single candidate found by images. One valid guess that meets Poisson plus boundary data is promoted to the exact, complete solution.

Breaks when

Unbounded domain without a decay condition. If \( S \) recedes to infinity but no fall-off is imposed, \( \oint_S W(\partial W/\partial n)\,dA \) can stay finite and nonzero — e.g. adding a uniform background field \( W = -E_0 x \) leaves the interior equations satisfied yet changes the field. Uniqueness requires the explicit condition \( \phi \to 0 \) at infinity.
Multiply-connected or disconnected region. On a domain with holes or separate pieces, \( \nabla W = \vec{0} \) permits a different constant on each component; the Neumann solution then carries several independent constants, and topological (circulation) freedom can appear that a single additive constant does not capture.
Mixed / Robin boundary condition of the wrong sign. A condition \( \partial\phi/\partial n = +\kappa\phi \) with \( \kappa > 0 \) makes the surface term \( \kappa\oint_S W^2\,dA \ge 0 \), which cannot be forced to cancel the volume term, and non-trivial solutions (eigenmodes) can exist.
Interior singular sources not excised. If a point or line charge lies in \( V \) and is not surrounded by a bounding surface, Green's identity is invalid there and the "proof" quietly assumes what it should establish.

Failure modes

Fixing both \( \phi \) and \( \partial\phi/\partial n \) on the same surface (over-specification). Students impose Cauchy data on a closed surface for an elliptic equation; generically no solution exists, and the theorem does not apply.
Forgetting the Neumann additive constant. Concluding \( \phi_1 = \phi_2 \) exactly for a pure-Neumann problem, then being surprised the numerical potential is offset by a constant.
Applying uniqueness with a placed image charge inside the region of interest. The image must live in the excluded region; putting it in \( V \) changes \( \rho \) and violates the equal-source premise.
Assuming the image construction is unique. Uniqueness is of the field, not of the trick; several image arrangements that reproduce the same boundary data give the same physical field, which is the point, not a contradiction.
Using it on an infinite grounded plane without noting the half-space is the bounded-by-data region. The plane plus the point-at-infinity condition together supply the closed boundary; omitting the infinity condition invites the background-field counterexample above.

Discussion

The heart of the proof is that Laplace's equation admits no interior maxima or minima — a fact the energy integral \( \int_V |\nabla W|^2\,dV \) encodes globally. Because the difference field carries positive-definite energy that is pinned to zero by the boundary and source constraints, there is simply no room for two distinct solutions to coexist. This is the electrostatic face of the maximum principle for elliptic operators, and the same structure recurs in steady heat conduction, incompressible potential flow, and equilibrium diffusion.

Physically, uniqueness expresses determinism: specify the sources and what happens at the edges, and the interior is fixed with no residual gauge or choice (beyond the harmless Neumann constant, which is just the arbitrary zero of potential). The method of images exploits this ruthlessly — a grounded plane is replaced by a mirror charge, a grounded sphere by a scaled interior image — and each replacement is legitimate solely because it reproduces the exact boundary data while leaving the true region's charges untouched.

At the rigorous level the crucial step is Step 8: from \( \int_V |\nabla W|^2\,dV = 0 \) to \( \nabla W = \vec{0} \) pointwise. This requires the integrand to be continuous and non-negative; a merely measurable field could vanish in integral while being nonzero on a set of zero measure. The theorem thus belongs to the classical (smooth) theory. Its Sobolev-space generalisation replaces "pointwise" with "almost everywhere" and demands \( W \in H^1(V) \) with the trace of \( W \) vanishing on \( S \) — the setting in which existence (via the Lax–Milgram theorem or direct minimisation of the Dirichlet energy) and uniqueness are proved together.

Common misconceptions. Uniqueness does not claim the solution is easy to find, nor that the image construction is the only route to it; it claims that whatever route yields a Poisson-satisfying, boundary-matching potential has found the one true field. Nor does it require the boundary to be a conductor — any surface on which \( \phi \) or \( \partial\phi/\partial n \) is known will do.

Worked examples

Point charge above a grounded plane. Charge \( q \) at height \( d \) over an infinite grounded conductor.

Region of interest: the upper half-space, boundary = plane (\( \phi=0 \)) plus infinity (\( \phi\to0 \)) — Dirichlet data. A

\[ \phi(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\left[ \frac{q}{r_+} - \frac{q}{r_-} \right] \]

Guess: real charge \( q \) at \( +d \) and an image \( -q \) at \( -d \) (outside \( V \)). On the plane \( r_+ = r_- \) so \( \phi = 0 \); only \( q \) lies in \( V \), so Poisson holds. Uniqueness \( \Rightarrow \) this is the solution. B

\[ F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2d)^2} = \frac{k\,q^2}{4d^2} \]

Force on \( q \) equals its attraction to the image (the field in \( V \) is genuinely that of the two charges). B

\[ q = 5.0\ \mathrm{nC}, \quad d = 0.10\ \mathrm{m}, \quad k = 8.99\times10^{9}\ \mathrm{N\,m^2\,C^{-2}} \]

Insert numbers. A

\[ F = \frac{(8.99\times10^{9})(5.0\times10^{-9})^2}{4\,(0.10)^2} = \frac{(8.99\times10^{9})(2.5\times10^{-17})}{0.040} \]

Evaluate numerator \( 2.247\times10^{-7} \), divide by \( 0.040 \). A

\[ F \approx 5.6\times10^{-6}\ \mathrm{N} \approx 5.6\ \mu\mathrm{N}, \ \text{directed toward the plane} \]

Reading. The charge is pulled toward the conductor with the force it would feel from a mirror charge; uniqueness is what licenses replacing the induced surface charge by that single image.

Units check. \( \mathrm{N\,m^2\,C^{-2}} \times \mathrm{C^2} / \mathrm{m^2} = \mathrm{N} \). \( \checkmark \)

Concentric spherical capacitor. Inner sphere \( r = a \) at \( \phi = V_0 \), outer \( r = b \) grounded, \( \rho = 0 \) between.

Charge-free shell region, Dirichlet data on both surfaces. A

\[ \phi(r) = A + \frac{B}{r} \]

Guess the general spherically-symmetric harmonic function; if it meets both boundary values, uniqueness makes it exact. B

\[ A + \frac{B}{a} = V_0, \quad A + \frac{B}{b} = 0 \;\Rightarrow\; B = \frac{V_0}{\frac{1}{a}-\frac{1}{b}}, \quad A = -\frac{B}{b} \]

Apply the two Dirichlet conditions and solve the linear pair. A

\[ a = 0.020\ \mathrm{m}, \quad b = 0.050\ \mathrm{m}, \quad V_0 = 100\ \mathrm{V} \]

Insert numbers: \( \frac{1}{a}-\frac{1}{b} = 50 - 20 = 30\ \mathrm{m^{-1}} \). A

\[ B = \frac{100}{30} = 3.33\ \mathrm{V\,m}, \quad A = -\frac{3.33}{0.050} = -66.7\ \mathrm{V} \]

Evaluate the constants. A

\[ \phi(0.030) = -66.7 + \frac{3.33}{0.030} = -66.7 + 111.1 = 44.4\ \mathrm{V} \]

Potential at mid-radius \( r = 0.030\ \mathrm{m} \). Field \( E = B/r^2 = 3.33/(0.030)^2 \approx 3.7\times10^{3}\ \mathrm{V\,m^{-1}} \). B

\[ \phi(r) = -66.7\ \mathrm{V} + \frac{3.33\ \mathrm{V\,m}}{r}, \quad \phi(0.030\ \mathrm{m}) \approx 44\ \mathrm{V} \]

Reading. Because the two-constant form already satisfies Laplace and both boundary values, the theorem certifies it is the complete solution — no further terms are hiding.

Units check. \( B/r = (\mathrm{V\,m})/\mathrm{m} = \mathrm{V} \); \( B/r^2 = (\mathrm{V\,m})/\mathrm{m^2} = \mathrm{V\,m^{-1}} \). \( \checkmark \)

Problems

Prove that for a pure Neumann problem the solution is determined only up to an additive constant, and show explicitly that the electric field is nonetheless unique.
Solution
Let \( \phi_1, \phi_2 \) satisfy the same Poisson equation with the same Neumann data \( \partial\phi/\partial n \) on \( S \). With \( W = \phi_1 - \phi_2 \), Green's first identity with both fields \( = W \) gives \( \int_V |\nabla W|^2\,dV = \oint_S W(\partial W/\partial n)\,dA \). The Neumann condition makes \( \partial W/\partial n = 0 \) on \( S \), so the RHS is zero, forcing \( \nabla W = \vec{0} \) and \( W = c \) (a constant). Hence \( \phi_1 = \phi_2 + c \). Since \( \vec{E} = -\nabla\phi \) and \( \nabla c = \vec{0} \), \( \vec{E}_1 = \vec{E}_2 \) exactly. The constant is the free zero of potential and carries no physics.
For the point charge \( q = 5.0\ \mathrm{nC} \) at height \( d = 0.10\ \mathrm{m} \) above the grounded plane (Worked Example 1), find the induced surface-charge density directly below the charge and the total induced charge.
Solution
The image field gives, on the plane at horizontal distance \( s \) from the foot of the charge, \( \sigma(s) = -\dfrac{qd}{2\pi(s^2+d^2)^{3/2}} \). Directly below (\( s = 0 \)): \( \sigma(0) = -\dfrac{q}{2\pi d^2} = -\dfrac{5.0\times10^{-9}}{2\pi(0.10)^2} = -7.96\times10^{-8}\ \mathrm{C\,m^{-2}} \approx -80\ \mathrm{nC\,m^{-2}} \). Integrating \( \sigma \) over the whole plane, \( Q_{\text{ind}} = \int_0^\infty \sigma(s)\,2\pi s\,ds = -q = -5.0\ \mathrm{nC} \): the total induced charge equals the negative of the real charge, consistent with the image.
A point charge \( q \) sits a distance \( L = 0.30\ \mathrm{m} \) from the centre of a grounded conducting sphere of radius \( R = 0.10\ \mathrm{m} \). Give the image charge magnitude and position, and the potential energy for \( q = 2.0\ \mathrm{nC} \).
Solution
Method of images for a grounded sphere: image \( q' = -qR/L \) located at distance \( b = R^2/L \) from the centre, on the line to \( q \). Numerically \( q' = -(2.0\ \mathrm{nC})(0.10)/(0.30) = -0.667\ \mathrm{nC} \) at \( b = (0.10)^2/0.30 = 0.0333\ \mathrm{m} \). The interaction energy is \( U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\,q'}{L-b} = \dfrac{(8.99\times10^{9})(2.0\times10^{-9})(-0.667\times10^{-9})}{0.30-0.0333} \). Numerator \( = -1.199\times10^{-8} \); divide by \( 0.2667 \Rightarrow U \approx -4.5\times10^{-8}\ \mathrm{J} \). Negative, so the charge is attracted to the grounded sphere. (Note: the interaction energy uses one factor, not the naive self-energy doubling.)
An empty (\( \rho = 0 \)) cavity is enclosed by a conductor held at potential \( V = 12\ \mathrm{V} \). Show that \( \phi = 12\ \mathrm{V} \) everywhere inside and hence \( \vec{E} = \vec{0} \), and state which theorem guarantees this is the only answer.
Solution
Inside, \( \nabla^2\phi = 0 \) (no charge). The constant function \( \phi = 12\ \mathrm{V} \) satisfies Laplace's equation and matches the Dirichlet boundary value \( \phi = 12\ \mathrm{V} \) on the whole wall. By the uniqueness theorem this constant is the only solution, so \( \vec{E} = -\nabla\phi = \vec{0} \) throughout the cavity — the electrostatic shielding of a field-free hollow conductor. Uniqueness (Dirichlet case) is precisely what forbids any other, structured, solution.
Two concentric spheres, \( a = 0.010\ \mathrm{m} \) at \( \phi = 0 \) and \( b = 0.040\ \mathrm{m} \) at \( \phi = 200\ \mathrm{V} \), enclose vacuum. Find \( \phi(r) \), verify uniqueness applies, and evaluate the field at \( r = 0.020\ \mathrm{m} \).
Solution
General harmonic form \( \phi(r) = A + B/r \). Conditions: \( A + B/a = 0 \), \( A + B/b = 200 \). Subtract: \( B(1/b - 1/a) = 200 \), i.e. \( B(25 - 100) = 200 \Rightarrow B = 200/(-75) = -2.667\ \mathrm{V\,m} \); then \( A = -B/a = 2.667/0.010 = 266.7\ \mathrm{V} \). So \( \phi(r) = 266.7 - 2.667/r \) (V). Check: \( \phi(0.010) = 266.7 - 266.7 = 0 \) \( \checkmark \), \( \phi(0.040) = 266.7 - 66.7 = 200 \) \( \checkmark \). It solves Laplace and both Dirichlet values, so uniqueness certifies it. Field \( E_r = -d\phi/dr \); here \( d\phi/dr = +2.667/r^2 \) so \( E_r = -2.667/r^2 \); at \( r = 0.020 \), \( E_r = -2.667/(4.0\times10^{-4}) = -6.67\times10^{3}\ \mathrm{V\,m^{-1}} \) (pointing inward, magnitude \( 6.7\times10^{3}\ \mathrm{V\,m^{-1}} \)).