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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Copyright 2001 Computational Physics RuG
Last change: October 09, 2001