Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 17:18, 6 January 2021

Revision as of 17:19, 6 January 2021

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Let us first consider the case where [[Dirichlet boundary condition|Dirichlet boundary conditions]] are specified as <math>\varphi = 0</math> on the boundary of the region. These follow because the boundary conditions and the charge distributions are the same for both 'solutions'.

By the application ofapplying the [[Vector calculus identities#Divergence 2|vector differential identity]] we know that

:<math>\nabla \cdot (\varphi \, \nabla \varphi )= \, (\nabla \varphi )^2 + \varphi \, \nabla^2 \varphi.</math>

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By taking the volume integral over the region, we find that

:<math>\int_V \mathbf{\nabla}\cdot(\varphi \, \mathbf{\nabla}\varphi) \, \mathrm{d}V= \int_V \mathbf{(\nabla}\varphi)\cdot( \mathbf{\nabla}\varphi) \, \mathrm{d}V= \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V.</math>

AndBy after aplyingapplying the [[divergence theorem]], we rewrite the expression above can be rewritten as

:<math>\int_{S} (\varphi \, \mathbf{\nabla}\varphi) \cdot \mathrm{d}\mathbf{S}= \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V. \qquad (2)</math>

where <math>S</math> is the boundary surface specified by the boundary conditions.