Uniqueness theorem for Poisson's equation: Difference between revisions

Revision as of 17:35, 6 January 2021

Revision as of 17:46, 6 January 2021

Line 23:

We now sequentially consider three distinct boundary conditions: a Dirichlet boundary condition, a Neumann boundary condition, and a mixed boundary condition.

LetFirst, us firstwe consider the case where [[Dirichlet boundary condition|Dirichlet boundary conditions]] are specified as <math>\varphi = 0</math> on the boundary of the region. These follow because the boundary conditions and the charge distributions are the same for both 'solutions'.

By applying the [[Vector calculus identities#Divergence 2|vector differential identity]] we know that

Line 47:

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.

IfSecond, we consider the case where [[Neumann boundary condition|Neumann boundary conditions]] had beenare specified then the normal component ofas <math>\nabla \phi</math>varphi on= the left-hand side of <math>(2)0</math> would be zero on the boundary and we would arrive atof the same conclusionregion. In this case, however,If the relationshipNeumann betweenboundary the solutionscondition is onlysatisfied constrained to a constant factoron <math>kS</math>, inby otherboth solutions (i.e., wordsif <math>\varphi_1 - nabla\varphi_2varphi = k0</math> on the boundary), because onlythen the normalleft-hand derivativeside of <math>\varphi(2)</math> wasis specifiedzero. toConsequently, beas zero.before, we find that

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

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.