Computational Physics

Proof of the uniqueness theorem

Suppose there were two functions V₁(x,y,z) and V₂(x,y,z), each of them satisfying the Laplace equation and having the same values on the boundary S but being different inside V.  

We have: ; ; V₁(S) = V₂(S)

Define: V₃(x,y,z) = V₂(x,y,z) - V₁(x,y,z)

Then: because of the linearity of Laplace's equation.
It then follows:                   
Thus, V₃ also satisfies the Laplace equation and is thus a harmonic function itself.

Since V₁(S) = V₂(S) is valid on the boundary: V₃(S) = 0

Since the maxima and minima of a harmonic function lie on the boundary, V₃(x,y,z) = 0 everywhere inside V.  

This means that V₂(x,y,z) = V₁(x,y,z) everywhere in V.  

In other words: For given boundary conditions the solution of the Laplace equation is unique.

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Last change: October 09, 2001