
Proof of the uniqueness theoremSuppose there were two functions V_{1}(x,y,z) and V_{2}(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: ; ; V_{1}(S) = V_{2}(S) Define: V_{3}(x,y,z) = V_{2}(x,y,z)  V_{1}(x,y,z) Then:
because of the linearity of Laplace's equation. Since V_{1}(S) = V_{2}(S) is valid on the boundary: V_{3}(S) = 0 Since the maxima and minima of a harmonic function lie on the boundary, V_{3}(x,y,z) = 0 everywhere inside V. This means that V_{2}(x,y,z) = V_{1}(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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