Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 17:46, 6 January 2021

Revision as of 17:58, 6 January 2021

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

Further, becauseSince 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 still, 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. Consequently, as before, we find that

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

Again, 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 within <math>V</math>, and <math>\nabla\varphi = 0</math> on the boundary, <math>\varphi</math> must be constant---but not necessarily zero---throughout the whole region. Finally, since <math>\varphi = k</math> throughout the whole region and since <math>\varphi = \varphi_2 - \varphi_1</math> throughout the whole region, therefore <math>\varphi_1 = \varphi_2 - k</math> throughout the whole region. This completes the proof that there is the unique solution up to an additive constant $k$ of Poisson's equation with a Neumann boundary condition.

In the case of the Neumann boundary condition, however, the relationship between the solutions is only constrained to a constant factor <math>k</math>. In other words, <math>\varphi_1 - \varphi_2 = k</math>, because only the normal derivative of <math>\varphi</math> was specified to be zero.