Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 18:10, 6 January 2021

Revision as of 18:14, 6 January 2021

Line 53:

As before, since this is the volume integral of a positive quantity, we must have <math>\nabla \varphi = 0</math> at all points. Further, because the gradient of <math>\varphi</math> is everywhere zero within the volume <math>V</math>, and because the gradient of <math>\varphi</math> is everywhere zero on the boundary <math>S</math>, therefore <math>\varphi</math> must be constant---but not necessarily zero---throughout the whole region. Finally, since <math>\varphi = k</math> throughout the whole region, and since <math>\varphi = \varphi_2 - \varphi_1</math> throughout the whole region, therefore <math>\varphi_1 = \varphi_2 - k</math> throughout the whole region. This completes the proof that there is the unique solution up to an additive constant <math>k</math> of Poisson's equation with a Neumann boundary condition.

[[Mixed boundary condition|Mixed boundary conditions]] could be given as long as ''either'' the gradient ''or'' the potential is specified at each point of the proofboundary. Boundary conditions at infinity also hold. This results from the fact asthat the surface integral in <math>(2)</math> still vanishes at large distances asbecause the integrand decays faster than the surface area grows.

==See also==