Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 17:02, 6 January 2021

Revision as of 17:06, 6 January 2021

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:<math>\mathbf{\nabla}^2 \varphi_2 = - \frac{\rho_f}{\epsilon_0}</math>.

It follows that <math>\varphi=\varphi_2-\varphi_1</math> is a solution of [[Laplace's equation]], which is (a special case of [[Poisson's equation]] whichthat equals to <math>0</math>). becauseBy subtracting the two solutions above gives

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

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'.