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# Boundary Value Problems

Consider a volume bounded by a surface . Suppose that we wish to solve Poisson's equation,

 (238)

throughout , subject to given Dirichlet or Neumann boundary conditions on . The charge density distribution, , is assumed to be known throughout . This type of problem is called a boundary value problem.

Similarly to the approach taken in Section 2.3, we can solve Poisson's equation by means of a Green's function, , that satisfies

 (239)

In fact, it follows from Equation (25) [because , by symmetry] that

 (240)

where

 (241)

throughout . Here, the function is chosen in such a manner as to satisfy the boundary conditions on . Making use of Green's theorem, (220), where , we find that

 (242)

Here, use has been made of Equations (23), (238), and (239). Note, incidentally, that the divergence theorem, combined with Equation (239), yields

 (243)

Consider the Dirichlet problem in which is known on , but is unknown. We can construct an appropriate Green's function for this problem, , where

 (244)

by choosing the function in Equation (240) in such a manner that

 (245)

when lies on . It then follows from Equation (242) that

 (246)

Hence, the potential is specified in terms of integrals over known functions throughout and on .

It is possible to prove that the Dirichlet Green's function is symmetric with respect to its arguments. In other words,

 (247)

Making use of Green's theorem, (220), where and , we find that

However, by definition, , , and when lies on . Hence,

 (248)

which yields

 (249)

It is also possible to demonstrate that the Dirichlet Green's function is unique. Proceeding in the usual fashion, suppose that there are two different functions, and , that both satisfy Equations (244) and (245). It follows that satisfies

 (250)

throughout , subject to the boundary condition

 (251)

when lies on . However, we saw in Section 2.2 that the only solution to this problem is for in or on . Hence, the functions and are identical, and the Dirichlet Green's function is unique. It follows that the potential specified in Equation (246) is also unique.

Consider the Neumann problem in which is known on , but is unknown. In this case, the obvious ansatz, when lies on , is incorrect, because it is inconsistent with the constraint (243). The simplest ansatz that works is a choice of in Equation (240) such that

 (252)

when lies on . Hence, Equation (242) yields

 (253)

where is the average value of the potential on . This average value can be absorbed into the arbitrary constant that can always be added to a scalar potential. Thus, the potential is again specified in terms of integrals over known functions throughout and on .

It is possible to prove that the Neumann Green's function can be chosen in such a manner that it is symmetric with respect to its arguments. In other words,

 (254)

Consider a Neumann Green's function that is not symmetric with respect to its arguments. That is, an asymmetric function which satisfies

 (255)

and

 (256)

when lies on . Consider

 (257)

where is arbitrary. It follows that

 (258)

and

 (259)

when lies on . Hence, is also a valid Neumann Green's function. Making use of Green's theorem, (220), where and , we find that

 (260)

We deduce that is symmetric provided that . We can ensure that this is the case by choosing

 (261)

Thus, given an asymmetric Neumann Green's function, it is always possible to construct a symmetric Green's function that satisfies

 (262) (263)

and

 (264)

when lies on .

We can also show that the symmetric Neumann Green's function is unique. Proceeding in the usual fashion, suppose that there are two different functions, and , that both satisfy Equations (263)--(265). It follows that satisfies

 (265)

throughout , subject to the boundary condition

 (266)

when lies on . Equation (130) can be written

 (267)

Suppose that . It follows that

 (268)

which implies that

 (269)

where is arbitrary. However, also satisfies

 (270)

Hence, , and the Green's function is unique. It follows that the potential specified in Equation (254) is also unique (up to an arbitrary additive constant).

Finally, the fact that the Green's function for Poisson's equation, , is (or can be chosen to be) symmetric implies from Equation (239) that

 (271)

because .

Next: Dirichlet Green's Function for Up: Electrostatic Fields Previous: Green's Theorem
Richard Fitzpatrick 2014-06-27