State and prove uniqueness theorem

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:

   ∇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:

   ∇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:

   ∮_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

   ∫_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:

   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.)