---
title: "State and prove uniqueness theorem"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:uniqueness-theorem-state-prove"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/uniqueness-theorem-state-prove
site: BCT Engineering
---

# State and prove uniqueness theorem

Electromagnetism · Chapter 2: Electric field

- Asked in: 2076 Chaitra, 2 marks; 2073 Shrawan, 1+5 marks; 2070 Chaitra, 2 marks; 2069 Chaitra, 2+6 marks (asked 4 times)
- Video: [Uniqueness Theorem: Proof and Derivation Explained](https://www.youtube.com/watch?v=45CAQuxYSX0) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/uniqueness-theorem-state-prove
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

State (define) the uniqueness theorem and prove it (for Laplace's / Poisson's equation).

## Short answer

The solution to Poisson's/Laplace's equation with specified boundary conditions is unique. Proof: assume two solutions V1, V2; their difference Vd satisfies Laplace's equation with zero boundary condition. Using identity ∇·(Vd ∇Vd) = |∇Vd|², integrate over volume, get ∫|∇Vd|² dv=0, so Vd constant, zero on boundary, thus V1=V2.

## Answer

**Statement:** If a solution of Poisson's (or Laplace's) equation satisfies the given boundary conditions, i.e. the potential V is specified on every boundary surface of the region (or the normal derivative, or a mixture), then that solution is **unique**. Two different potential functions cannot satisfy the same equation and the same boundary conditions.

So any solution found by any method (trial, inspection, images, numerical) that satisfies the equation and the boundary conditions is the only correct solution.

#### Proof (by contradiction)

Take a region bounded by a closed surface S (the conductors), containing charge density ρ_v. Suppose there are two different solutions V₁ and V₂ of Poisson's equation with the same boundary values.

1. Both satisfy the equation:
   ```text
   ∇²V₁ = −ρ_v/ε,    ∇²V₂ = −ρ_v/ε
   ```
2. On the boundary S both have the same specified value: V₁ = V₂ on S.
3. Define the difference V_d = V₁ − V₂. Subtracting:
   ```text
   ∇²V_d = ∇²V₁ − ∇²V₂ = 0       (inside the region)
   V_d = 0                        (on S)
   ```
4. Use the vector identity ∇ · (V_d ∇V_d) = V_d ∇²V_d + |∇V_d|². Integrate over the volume and apply the divergence theorem:
   ```text
   ∮_S V_d ∇V_d · dS = ∫_v V_d ∇²V_d dv + ∫_v |∇V_d|² dv
   ```
5. On S, V_d = 0, so the left side is zero. Inside, ∇²V_d = 0, so the first volume term is zero. Hence
   ```text
   ∫_v |∇V_d|² dv = 0
   ```
6. |∇V_d|² is never negative, so its integral is zero only if ∇V_d = 0 everywhere. So V_d is constant in the region. Since V_d = 0 on the boundary, that constant is zero:
   ```text
   V_d = 0   →   V₁ = V₂
   ```

This contradicts the assumption that they were different. Hence the solution is unique. **Proved.**

(If instead the normal derivative ∂V/∂n is specified on S, then ∇V_d · dS = 0 on S, and the same steps give ∇V_d = 0, i.e. the solutions differ at most by a constant, which does not change E.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the uniqueness theorem state?
   A. Many solutions can fit the same boundary conditions
   B. A solution of Laplace's or Poisson's equation satisfying the boundary conditions is the only one
   C. Every potential is zero on the boundary
   D. The potential is always constant
   Answer: B. If V satisfies the equation and the given boundary values, no other different function can, so it is unique. (Key point: Statement)

2. In the proof, what equation and boundary condition does V_d = V1 - V2 satisfy?
   A. Poisson's equation with V_d = V on S
   B. Laplace's equation ∇²V_d = 0 inside, with V_d = 0 on S
   C. ∇²V_d = -ρ_v/ε inside
   D. V_d = V1 + V2 on S
   Answer: B. Subtracting the two Poisson equations cancels the charge term, giving ∇²V_d = 0, and the equal boundary values cancel to give V_d = 0 on S. (Key point: Difference function)

3. How does the proof conclude V_d = 0?
   A. ∫|∇V_d|² dv = 0 forces ∇V_d = 0, so V_d is constant, and it is zero on S
   B. By setting V_d = 1
   C. By differentiating V_d once
   D. By assuming ρ_v = 0
   Answer: A. A non-negative integrand with zero integral must vanish everywhere, so V_d is constant, and zero at the boundary means V1 = V2. (Key point: Conclusion)

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