When solving electrostatic problems, we often rely on the uniqueness theorem. 

Assume you have a volume V, bounded by a collection of surfaces.
Assume you know the charge density ρ(r) everywhere inside the volume V, and on the surfaces of the volume you either know
(a)  the potential Φ (Dirichlet boundary conditions),
(b)  E∙n , i.e the normal derivatives of the potential (Neuman boundary conditions),
(c) the surface is a conductor which carries a total charge Q,
then the solution Φ(r) inside V is unique except for possibly an arbitrary constant C fixing Φ(r) at a point.

Proof:

Assume that two solutions for Φ exist that satisfy the same boundary conditions.
∇²Φ₁ = -ρ/ε₀,  ∇²Φ₂ = -ρ/ε₀.
Let U = Φ₁ - Φ₂, then ∇²U = 0 inside V.

For Dirichlet boundary conditions U = 0 on the surfaces.
For Dirichlet boundary conditions ∂U/∂n  = 0 on the surfaces.
For a conductor with total charge Q
Φ₁ = constant, Φ₂ = constant --> U is constant on the surface,
and ∮_S (∂U/∂n) dS = 0, since
∮_S (∂Φ₁/∂n) dS = ∮_S (∂Φ₁/∂n) dS = -∮_S E∙ndS = -Q_inside/ε₀.

Now apply Green's first identity with Φ = ψ = U.
∫_V (U∇²U + ∇U∙∇U) dV = ∮_S U(∂U/∂n) dS
∇²U = 0.  Therefore 
∫_V|∇U|² dV = ∮_S U(∂U/∂n) dS.
For all boundary conditions the right hand side of this equation is zero.
Therefore ∫_V|∇U|² dV = 0.
Since |∇U|² is a non-negative everywhere, this implies ∇U = 0 everywhere inside V and U is constant.  Φ₁ and Φ₂ differ at most by a constant. 
Using any technique, if we find one solution Φ(r) inside V satisfying the boundary conditions (fixed at one point), we have found the only solution.

Green's first identity
∫_V (Φ∇²ψ + ∇Φ∙∇ψ) dV = ∮_S Φ(∂ψ/∂n) dS
is derived starting with Green's theorem.
∫_V∇∙A dV = ∮_S A∙n dS .
Set A = Φ∇ψ, with Φ, ψ arbitrary scalar fields.
∇∙A = ∇∙(Φ∇ψ) =  Φ∇²ψ  +  ∇Φ∙∇ψ.
A∙n = Φ∇ψ∙n = Φ(∂ψ/∂n).