Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 12:34, 28 January 2022

Latest revision as of 09:52, 1 April 2025

(One intermediate revision by one other user not shown)

Line 18:

:<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]] that equals to <math>0</math>. By subtractingSubtracting the two solutions above gives

{{NumBlk||<math display="block">\mathbf{\nabla}^2 \varphi = \mathbf{\nabla}^2 \varphi_1varphi_2 - \mathbf{\nabla}^2 \varphi_2varphi_1 = 0. </math>|{{EquationRef|1}}}}

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