Boundary Value Problems

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

$$\nabla^{\,2}\phi = -\frac{\rho}{\epsilon_0},$$ (238)

throughout $V$ , subject to given Dirichlet or Neumann boundary conditions on $S$ . The charge density distribution, $\rho({\bf r})$ , is assumed to be known throughout $V$ . 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, $G({\bf r},{\bf r}')$ , that satisfies

$$\nabla'^{\,2}G({\bf r},{\bf r}') = \delta({\bf r}-{\bf r}').$$ (239)

In fact, it follows from Equation (25) [because $\nabla^{\,2}(\vert{\bf r}-{\bf r}'\vert^{-1})=\nabla'^{\,2}(\vert{\bf r}-{\bf r}'\vert^{-1})$ , by symmetry] that

$$G({\bf r},{\bf r}') = -\frac{1}{4\pi\,\vert{\bf r}-{\bf r}'\vert} + F({\bf r},{\bf r}'),$$ (240)

where

$$\nabla'^{\,2} F({\bf r}, {\bf r}')=0$$ (241)

throughout $V$ . Here, the function $F({\bf r},{\bf r}')$ is chosen in such a manner as to satisfy the boundary conditions on $S$ . Making use of Green's theorem, (220), where $\psi({\bf r}')=G({\bf r},{\bf r}')$ , we find that

$$\phi({\bf r})=-\frac{1}{\epsilon_0} \int_V G({\bf r},{\bf r}')\,\... ...'}-\phi({\bf r}')\,\frac{\partial G({\bf r},{\bf r}')}{\partial n'}\right]d S'.$$ (242)

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

$$\int_S \frac{\partial G({\bf r},{\bf r}')}{\partial n'}\,dS'=1.$$ (243)

Consider the Dirichlet problem in which $\phi({\bf r})$ is known on $S$ , but $\partial\phi({\bf r})/\partial n$ is unknown. We can construct an appropriate Green's function for this problem, $G_D({\bf r},{\bf r}')$ , where

$$\nabla'^{\,2} G_D({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$$ (244)

by choosing the function $F({\bf r},{\bf r}')$ in Equation (240) in such a manner that

$$G_D({\bf r},{\bf r}')=0$$ (245)

when ${\bf r}'$ lies on $S$ . It then follows from Equation (242) that

$$\phi({\bf r})=-\frac{1}{\epsilon_0} \int_V G_D({\bf r},{\bf r}')\... ...nt_S \phi({\bf r}')\,\frac{\partial G_D({\bf r},{\bf r}')}{\partial n'}\, d S'.$$ (246)

Hence, the potential $\phi({\bf r})$ is specified in terms of integrals over known functions throughout $V$ and on $S$ .

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

$$G_D({\bf r},{\bf r}')= G_D({\bf r}',{\bf r}).$$ (247)

Making use of Green's theorem, (220), where $\psi({\bf r}')=G_D({\bf r},{\bf r}')$ and $\phi({\bf r}')=G_D({\bf r}'',{\bf r}')$ , we find that

$$\begin{multline} \int_V\left[G_D({\bf r},{\bf r}')\,\nabla'^{\,2}G_D({\bf r}'',{... ...')\,\frac{\partial G_D({\bf r},{\bf r}')}{\partial n'}\right]dS'. \end{multline}$$

However, by definition, $\nabla'^{\,2}G_D({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}')$ , $\nabla'^{\,2}G_D({\bf r}'',{\bf r}')=\delta({\bf r}''-{\bf r}')$ , and $G_D({\bf r},{\bf r}')=G_D({\bf r}'',{\bf r}')=0$ when ${\bf r}'$ lies on $S$ . Hence,

$$\int_V\left[G_D({\bf r},{\bf r}')\,\delta({\bf r}''-{\bf r}')-G_D({\bf r}'',{\bf r}')\,\delta({\bf r}-{\bf r}')\right]dV'=0,$$ (248)

which yields

$$G_D({\bf r},{\bf r}'')=G_D({\bf r}'',{\bf r}).$$ (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, $G_1({\bf r},{\bf r}')$ and $G_2({\bf r},{\bf r}')$ , that both satisfy Equations (244) and (245). It follows that $G_3({\bf r},{\bf r}') =G_1({\bf r},{\bf r}')-G_2({\bf r},{\bf r}')$ satisfies

$$\nabla'^{\,2} G_3({\bf r},{\bf r}')=0$$ (250)

throughout $V$ , subject to the boundary condition

$$G_3({\bf r},{\bf r}')= 0$$ (251)

when ${\bf r}'$ lies on $S$ . However, we saw in Section 2.2 that the only solution to this problem is $G_3({\bf r},{\bf r}')=0$ for ${\bf r}'$ in $V$ or on $S$ . Hence, the functions $G_1({\bf r},{\bf r}')$ and $G_2({\bf r},{\bf r}')$ 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 $\partial\phi({\bf r})/\partial n$ is known on $S$ , but $\phi({\bf r})$ is unknown. In this case, the obvious ansatz, $\partial G_N({\bf r},{\bf r}')/\partial n'=0$ when ${\bf r}'$ lies on $S$ , is incorrect, because it is inconsistent with the constraint (243). The simplest ansatz that works is a choice of $F({\bf r},{\bf r}')$ in Equation (240) such that

$$\frac{\partial G_N({\bf r},{\bf r'})}{\partial n'} = 1\left/\int_S dS'\right.$$ (252)

when ${\bf r}'$ lies on $S$ . Hence, Equation (242) yields

$$\phi({\bf r}) = \langle\phi\rangle_S -\frac{1}{\epsilon_0} \int_V... ... -\int_S G_N({\bf r},{\bf r}')\,\frac{\partial\phi({\bf r}')}{\partial n'}d S',$$ (253)

where $\langle\phi\rangle_S = \int_S \phi({\bf r}')\,dS'\left/\int_S dS'\right.$ is the average value of the potential on $S$ . 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 $V$ and on $S$ .

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,

$$G_N({\bf r},{\bf r}')= G_N({\bf r}',{\bf r}).$$ (254)

Consider a Neumann Green's function that is not symmetric with respect to its arguments. That is, an asymmetric function $G_N({\bf r},{\bf r}')$ which satisfies

$$\nabla'^{\,2}G_N({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}')$$ (255)

and

$$\frac{\partial G_N({\bf r},{\bf r}')}{\partial n'} = 1\left/\int_S dS'\right.$$ (256)

when ${\bf r}'$ lies on $S$ . Consider

$$\widetilde{G}_N({\bf r},{\bf r}')= G_N({\bf r},{\bf r}')+F({\bf r}),$$ (257)

where $F({\bf r})$ is arbitrary. It follows that

$$\nabla'^{\,2}\widetilde{G}_N({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}')$$ (258)

and

$$\frac{\partial \widetilde{G}_N({\bf r},{\bf r}')}{\partial n'} = 1\left/\int_S dS'\right.$$ (259)

when ${\bf r}'$ lies on $S$ . Hence, $\widetilde{G}({\bf r},{\bf r}')$ is also a valid Neumann Green's function. Making use of Green's theorem, (220), where $\psi({\bf r}')=\widetilde{G}_N({\bf r},{\bf r}')$ and $\phi({\bf r}')=\widetilde{G}_N({\bf r}'',{\bf r}')$ , we find that

$$\widetilde{G}_N({\bf r},{\bf r}'')-\widetilde{G}_N({\bf r}'',{\bf... ...{\bf r}')-\widetilde{G}_N({\bf r}'',{\bf r}')\right]dS'\left/\int_S dS'\right..$$ (260)

We deduce that $\widetilde{G}_N({\bf r},{\bf r}')$ is symmetric provided that $\int_S \widetilde{G}_N({\bf r},{\bf r}')\,dS'\left/\int_S dS'\right.=0$ . We can ensure that this is the case by choosing

$$F({\bf r}) = -\left.\int_S G_N({\bf r},{\bf r}')\,dS'\right/\int_S dS'.$$ (261)

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

$$\nabla'^{\,2}G_N({\bf r},{\bf r}') =\delta({\bf r}-{\bf r}'),$$ (262)

$$\int_S G_N({\bf r},{\bf r}')\,dS'\left/\int_S dS'\right. =0,$$ (263)

and

$$\frac{\partial G_N({\bf r},{\bf r}')}{\partial n'} = 1\left/\int_S dS'\right.$$ (264)

when ${\bf r}'$ lies on $S$ .

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, $G_1({\bf r},{\bf r}')$ and $G_2({\bf r},{\bf r}')$ , that both satisfy Equations (263)--(265). It follows that $G_3({\bf r},{\bf r}') =G_1({\bf r},{\bf r}')-G_2({\bf r},{\bf r}')$ satisfies

$$\nabla'^{\,2} G_3({\bf r},{\bf r}')=0$$ (265)

throughout $V$ , subject to the boundary condition

$$\frac{\partial G_3({\bf r},{\bf r}')}{\partial n'}=0$$ (266)

when ${\bf r}'$ lies on $S$ . Equation (130) can be written

$$\int_S \phi({\bf r}')\,\frac{\partial\phi}{\partial n'}\,dS' = \int_V\left[\phi({\bf r}')\,\nabla'^{\,2}\phi+\nabla'\phi\cdot\nabla'\phi\right]dV'.$$ (267)

Suppose that $\phi({\bf r}')=G_3({\bf r},{\bf r}')$ . It follows that

$$\int_V\vert\nabla' G_3({\bf r},{\bf r}')\vert^{\,2}\,dV'=0,$$ (268)

which implies that

$$G_3({\bf r},{\bf r}')=F({\bf r}),$$ (269)

where $F({\bf r})$ is arbitrary. However, $G_3({\bf r},{\bf r}')$ also satisfies

$$\int_S G_3({\bf r},{\bf r}')\,dS'\left/\int_S dS' \right.= F({\bf r}) = 0.$$ (270)

Hence, $F({\bf r})=G_3({\bf r},{\bf r}')=0$ , 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, $G({\bf r},{\bf r}')$ , is (or can be chosen to be) symmetric implies from Equation (239) that

$$\nabla'^{\,2}G_D({\bf r},{\bf r}')=\nabla^{\,2}G_D({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$$ (271)

because $\delta({\bf r}'-{\bf r})=\delta({\bf r}-{\bf r}')$ .