Dirichlet Green's Function for Spherical Surface

As an example of a boundary value problem, suppose that we wish to solve Poisson's equation, subject to Dirichlet boundary conditions, in some domain $V$ that lies between the spherical surfaces $r=a$ and $r=\infty$ , where $r$ is a radial spherical coordinate. Let $S$ and $S'$ denote the former and latter surfaces, respectively. The Green's function for the problem, $G_D({\bf r},{\bf r}')$ , must satisfy

$$\nabla^{\,2}G_D({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$$ (272)

for ${\bf r}$ , ${\bf r}'$ not outside $V$ , and

$$G_D({\bf r},{\bf r}')=0$$ (273)

when ${\bf r}'$ lies on $S$ or on $S'$ . The Green's function also has the symmetry property

$$G_D({\bf r}',{\bf r})=G_D({\bf r},{\bf r}').$$ (274)

Let us try

$$G_D({\bf r},{\bf r}') = -\frac{1}{4\pi\,\vert{\bf r}'-{\bf r}\vert}+ \frac{a}{4\pi\,r'\,\vert a^{\,2}\,{\bf r}'/r'^{\,2}-{\bf r}\vert}.$$ (275)

Note that the above function is symmetric with respect to its arguments, because $r'\,\vert a^{\,2}\,{\bf r}'/r'^{\,2}-{\bf r}\vert= r\,\vert a^{\,2}\,{\bf r}/r^{\,2}-{\bf r}'\vert$ . It follows from Equation (25) that

$$\nabla^{\,2} G_D({\bf r},{\bf r}') = \delta({\bf r}-{\bf r}')- \frac{a}{r'}\,\delta({\bf r}-a^{\,2}\,{\bf r}'/r'^{\,2}).$$ (276)

However, if ${\bf r}$ , ${\bf r}'$ do not lie outside $V$ then the argument of the latter delta function cannot be zero. Hence, for ${\bf r}$ , ${\bf r}'$ not outside $V$ , this function takes the value zero, and the above expression reduces to

$$\nabla^{\,2}G_D({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$$ (277)

as required. Equation (276) can be written

$$G_D({\bf r},{\bf r}') = -\frac{1}{4\pi\,(r^{\,2}-2\,r\,r'\,\cos\g... ...\frac{1}{4\pi\,(r^{\,2}\,r'^{\,2}/a^{\,2}-2\,r\,r'\,\cos\gamma+a^{\,2})^{1/2}},$$ (278)

where $\cos\gamma={\bf r}\cdot{\bf r}'/(r\,r')$ . When written in this form, it becomes clear that $G_D({\bf r},{\bf r}')=0$ when ${\bf r}'$ lies on $S$ (i.e., when $r'=a$ ) or on $S'$ (i.e., when $r'=\infty$ ). We conclude that expression (279) is the unique Green's function for the Dirichlet problem within the domain $V$ .

According to Equation (246), the electrostatic potential within the domain $V$ is written

$$\phi({\bf r})=-\frac{1}{\epsilon_0} \int_V G_D({\bf r},{\bf r}')\... ..._S \phi_S({\bf r}')\,\frac{\partial G_D({\bf r},{\bf r}')}{\partial r'}\, d S'.$$ (279)

Here, $\rho({\bf r})$ is the charge distribution within $V$ (i.e., the region $r>a$ ), $\phi_S({\bf r})$ is the potential on $S$ (i.e., the surface $r=a$ ), and the potential at infinity (i.e., on the surface $S'$ ) is assumed to be zero. Moreover, we have made use of the fact that $\partial/\partial n'=-\partial/\partial r'$ on $S$ , because the unit vector ${\bf n}'$ points radially inward. Finally, it is easily demonstrated that

$$\left.\frac{\partial G_D}{\partial r'}\right\vert _{r'=a}= \frac{a-r^{\,2}/a}{4\pi\,(r^{\,2}-2\,r\,a\,\cos\gamma+a^{\,2})^{3/2}}.$$ (280)

Hence,

$$\phi(r,\theta,\varphi) =\frac{1}{4\pi\,\epsilon_0}\int_a^\infty\int_0^\pi\int_0^{2\pi}\f... ...heta'\,dr'\,d\theta'\,d\varphi'}{(r^{\,2}-2\,r\,r'\,\cos\gamma+r'^{\,2})^{1/2}}$$

$$\phantom{=}-\frac{1}{4\pi\,\epsilon_0}\int_a^\infty\int_0^\pi\int... ...ta'\,d\varphi'}{(r^{\,2}\,r'^{\,2}/a^{\,2}-2\,r\,r'\,\cos\gamma+a^{\,2})^{1/2}}$$

$$\phantom{=}+\frac{1}{4\pi}\int_0^\pi\int_0^{2\pi}\frac{\phi_S(\th... ...\sin\theta'\,d\theta'\,d\varphi'}{(r^{\,2}-2\,r\,a\,\cos\gamma+a^{\,2})^{3/2}},$$ (281)

where $r$ , $\theta$ , $\varphi$ and $r'$ , $\theta'$ , $\varphi'$ are standard spherical coordinates, in terms of which,

$$\cos\gamma = \cos\theta\,\cos\theta'+\sin\theta\,\sin\theta'\,\cos(\varphi-\varphi').$$ (282)

As a specific example, suppose that $S$ represents the surface of a grounded spherical conductor. It follows that $\phi_S=0$ . Suppose that there is a single charge, $q$ , in the domain $V$ , located on the $z$ -axis at $r=b>a$ , $\theta=0$ , and $\varphi=0$ . It follows that

$$\rho({\bf r}')\,dV' \rightarrow q\,\delta(r'-b)\,\delta(\theta')\,\delta(\varphi')\,dr'\,d\theta'\,d\varphi'.$$ (283)

Thus, Equations (282) and (283) yield

$$\phi(r,\theta) = \frac{q}{4\pi\,\epsilon_0}\left[\frac{1}{(r^{\,2... ...- \frac{1}{(r^{\,2}\,b^{\,2}/a^{\,2}-2\,r\,b\,\cos\theta+a^{\,2})^{1/2}}\right]$$ (284)

in the region $r>a$ . Of course, $\phi=0$ in the region $r<a$ . It follows from Equations (154) and (203) that there is a charge sheet on the surface of the conductor (because the normal electric field is non-zero just above the surface, but zero just below) whose density is given by

$$\sigma(\theta)= -\epsilon_0\left.\frac{\partial\phi}{\partial r}\... ...\pi\,a}\,\frac{(b^{\,2}-a^{\,2})}{(a^{\,2}-2\,a\,b\,\cos\theta+b^{\,2})^{3/2}}.$$ (285)

Thus, the net charge induced on the surface of the conductor is

$$q' = \int\sigma\,dS= 2\pi\,a^{\,2}\int_0^\pi\sigma(\theta)\,d\theta = - q\,\frac{a}{b}.$$ (286)

In Equation (285), the first term inside the square brackets clearly represents the electric potential generated by the charge $q$ , which suggests that

$$\phi'(r,\theta) = -\frac{q}{4\pi\,\epsilon_0} \frac{1}{(r^{\,2}\,b^{\,2}/a^{\,2}-2\,r\,b\,\cos\theta+a^{\,2})^{1/2}}$$ (287)

is the potential generated by the charges induced on the surface of the conductor. Hence, the attractive force acting on the charge $q$ due to the induced charges is

$${\bf F}' = -\left.q\,\nabla\phi'\right\vert _{r=b,\theta=0} =- \frac{q^2}{4\pi\,\epsilon_0}\,\frac{b\,a}{(b^{\,2}-a^{\,2})^2}\,{\bf e}_z.$$ (288)

Of course, an equal and opposite force acts on the conductor.

As a final example, suppose that there are no charges in the domain $V$ , but that $\phi_S$ is prescribed on the surface $S$ . It follows from Equation (282) that the electric potential on the $z$ -axis (i.e., $r=z$ , $\theta=0$ , and $\varphi=0$ ) is

$$\phi(z) = \frac{1}{4\pi}\int_0^\pi\int_0^{2\pi}\frac{\phi_S(\thet... ...\sin\theta'\,d\theta'\,d\varphi'}{(z^{\,2}-2\,z\,a\,\cos\theta'+a^{\,2})^{3/2}}$$ (289)

for $\vert z\vert\geq a$ . Let $S$ correspond to the surface of two conducting hemispheres (separated by a thin insulator). Suppose that the upper (i.e., $0\leq \theta<\pi/2$ ) hemisphere is held at the potential $+V$ , whereas the lower (i.e., $\pi/2<\theta\leq \pi$ ) hemisphere is held at the potential $-V$ . It follows that

$$\phi(z) = \frac{V}{2}\,(z^{\,2}-a^{\,2})\,a\left[\int_0^1 \frac{d... ...2})^{3/2}}-\int_{-1}^0\frac{d\mu}{(z^{\,2}-2\,z\,a\,\mu+a^{\,2})^{3/2}}\right],$$ (290)

where $\mu=\cos\theta$ , which yields

$$\phi(z) = {\rm sgn}(z)\,V\left[1-\frac{z^{\,2}-a^{\,2}}{\vert z\vert\,(z^{\,2}+a^{\,2})^{1/2}}\right]$$ (291)

for $\vert z\vert\geq a$ .