Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 17:58, 6 January 2021

Revision as of 18:01, 6 January 2021

Line 33:

:<math>\nabla \cdot (\varphi \, \nabla \varphi )= \, (\nabla \varphi )^2.</math>

By taking the volume integral over the region $<math>V$</math>, we find that

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

Line 51:

:<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 $<math>k$</math> 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.

[[Mixed boundary condition|Mixed boundary conditions]] could be given as long as ''either'' the gradient ''or'' the potential is specified at each point of the proof.