Uniqueness theorem for Poisson's equation: Difference between revisions

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Since 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. Further, 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. Finally, since <math>\varphi = 0</math> throughout the whole region, and since <math>\varphi = \varphi_2 - \varphi_1</math> throughout the whole region, therefore <math>\varphi_1 = \varphi_2</math> throughout the whole region. This completes the proof that there is the unique solution of Poisson's equation with a Dirichlet boundary condition.

Second, we consider the case where [[Neumann boundary condition|Neumann boundary conditions]] are specified as <math>\nabla\varphi = 0</math> on the boundary of the region. If the Neumann boundary condition is satisfied on <math>S</math> by both solutions (i.e., if <math>\nabla\varphi = 0</math> on the boundary), then the left-hand side of <math>(2)</math> is zero again. Consequently, as before, we find that

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