Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 00:17, 7 January 2021

Revision as of 00:22, 7 January 2021

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:<math>\mathbf{\nabla}^2 \varphi = \mathbf{\nabla}^2 \varphi_1 - \mathbf{\nabla}^2 \varphi_2 = 0. \qquad (1)</math>

We now sequentially consider three distinct boundary conditions: a Dirichlet boundary condition, a Neumann boundary condition, and a mixed boundary condition.

First, we 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 applying the [[Vector calculus identities#Divergence 2|vector differential identity]] we know that

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

We now sequentially consider three distinct boundary conditions: a Dirichlet boundary condition, a Neumann boundary condition, and a mixed boundary condition.

If the Dirichlet boundary condition is satisfied on <math>S</math> by both solutions (i.e., if <math>\varphi = 0</math> on the boundary), then the left-hand side of <math>(2)</math> is zero. Consequently, we find that

First, we 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 becauseIf the Dirichlet boundary conditionscondition andis thesatisfied chargeon distributions<math>S</math> areby both solutions (i.e., if <math>\varphi = 0</math> on the sameboundary), forthen boththe 'solutions'left-hand side of <math>(2)</math> is zero. Consequently, we find that

:<math>\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0.</math>