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.

∇^{2}Φ_{1}
= -ρ/ε_{0}, ∇^{2}Φ_{2}
= -ρ/ε_{0}.

Let U = Φ_{1} - Φ_{2},
then ∇^{2}U = 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

Φ_{1}
= constant, Φ_{2} = constant --> U is constant on the surface,

and ∮_{S}** (**∂U/∂n) dS = 0, since

∮_{S}** (**∂Φ_{1}/∂n) dS = ∮_{S}** (**∂Φ_{1}/∂n) dS
= -∮_{S}** E∙n**dS = -Q_{inside}/ε_{0}.

Now apply Green's first identity with Φ = ψ = U.

∫_{V}
(U**∇**^{2}U** **+
**∇**U**∙∇**U)**
**dV = ∮_{S}** **
U**(**∂U/∂n) dS

∇^{2}U
= 0. Therefore

∫_{V}|**∇**U|** ^{2} **dV = ∮

For all boundary conditions the right hand side of this equation is zero.

Therefore ∫

Since |∇U|

Using any technique, if we find one solution Φ(

Green's first identity

∫_{V}
(Φ**∇**^{2}ψ** **+
**∇**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∇^{2}ψ
+ **∇**Φ**∙∇**ψ.**A∙n** = Φ**∇**ψ**∙n** = Φ**(**∂ψ/∂n).