Uniqueness theorem for Poisson's equation: Difference between revisions

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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>. Subtracting 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