Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 17:25, 6 January 2021

Revision as of 17:29, 6 January 2021

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

HoweverFurther, because this is the volume integral of a positive quantity (due to the squared term), we must have <math>\nabla \varphi = 0</math> at all points.

:<math>\nabla \varphi = 0</math>

at all points.

Finally, because the gradient of <math>\varphi</math> is everywhere zero and <math>\varphi</math> is zero on the boundary, <math>\varphi</math> must be zero throughout the whole region. This proves <math>\varphi_1 = \varphi_2</math> and the solutions are identical.