Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 17:29, 6 January 2021

Revision as of 17:35, 6 January 2021

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

Further, 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. 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.

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.

If the [[Neumann boundary condition|Neumann boundary conditions]] had been specified then the normal component of <math>\nabla \phi</math> on the left-hand side of <math>(2)</math> would be zero on the boundary and we would arrive at the same conclusion. In this case, 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.