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Next: Exercises Up: Potential Theory Previous: Laplace's Equation in Cylindrical

Poisson's Equation in Cylindrical Coordinates

Let us, finally, consider the solution of Poisson's equation,

$\displaystyle \nabla^{\,2}\phi = -\frac{\rho}{\epsilon_0},$ (442)

in cylindrical coordinates. Suppose that the domain of solution extends over all space, and the potential is subject to the simple boundary condition

$\displaystyle \phi({\bf r}) \rightarrow 0$$\displaystyle \mbox{\hspace{1cm}as $\vert{\bf r}\vert\rightarrow\infty$}$$\displaystyle .$ (443)

In this case, the solution is written (see Section 2.3)

$\displaystyle \phi({\bf r}) = -\int \frac{\rho({\bf r}')}{\epsilon_0}\,G({\bf r},{\bf r}')\,dV',$ (444)

where the integral is over all space, and $ {\bf G}({\bf r},{\bf r}')$ is a symmetric Green's function [i.e., $ {\bf G}({\bf r}',{\bf r})=G({\bf r},{\bf r'})$ --see Equation (143)] that satisfies

$\displaystyle \nabla^{\,2}G({\bf r},{\bf r}') = \delta({\bf r}-{\bf r}'),$ (445)

subject to the constraint [see Equation (143)]

$\displaystyle G({\bf r}, {\bf r}') \rightarrow 0$   $\displaystyle \mbox{\hspace{1cm}as $\vert{\bf r}\vert\rightarrow\infty$}$$\displaystyle .$ (446)

In cylindrical coordinates,

$\displaystyle \delta({\bf r}-{\bf r}') = \frac{1}{r}\,\delta(r-r')\,\delta(\theta-\theta')\,\delta(z-z').$ (447)

This follows because, by definition (see Section 1.5),

$\displaystyle \int_V \delta({\bf r}-{\bf r}')\,dV = \int_V\delta({\bf r}-{\bf r}')\,r\,dr\,d\theta\,dz = 1$ (448)

whenever $ {\bf r}'$ lies within the volume $ V$ . Thus, Equation (446) becomes

$\displaystyle \frac{1}{r}\,\frac{\partial}{\partial r}\left(r\,\frac{\partial G...
...ial z^{\,2}} = \frac{1}{r}\,\delta(r-r')\,\delta(\theta-\theta')\,\delta(z-z').$ (449)

The well-known mathematical identities

$\displaystyle \delta(\theta-\theta')$ $\displaystyle = \frac{1}{2\pi}\sum_{m=-\infty,\infty} {\rm e}^{\,{\rm i}\,m\,(\theta-\theta')},$ (450)
$\displaystyle \delta(z-z')$ $\displaystyle = \frac{1}{2\pi}\int_{-\infty}^\infty {\rm e}^{\,{\rm i}\,k\,(z-z')}\,dk,$ (451)

are conventionally used to invert Fourier series and Fourier transforms, respectively. In the present case, if we write

$\displaystyle G({\bf r},{\bf r}') = \frac{1}{4\pi^2}\sum_{m=-\infty,\infty}\int...
...{\,{\rm i}\,k\,(z-z')}\,{\rm e}^{\,{\rm i}\,m\,(\theta-\theta')}\,g_m(r,r')\,dk$ (452)

then, making use of these identities, Equation (450) becomes

$\displaystyle \frac{1}{r}\,\frac{d}{dr}\!\left(r\,\frac{dg_m}{dr}\right)-\left(k^{\,2} + \frac{m^{\,2}}{r^{\,2}}\right) g_m= \frac{1}{r}\,\delta(r-r').$ (453)

In the general case, when $ r\neq r'$ , the previous equation reduces to the modified Bessel equation,

$\displaystyle \frac{1}{r}\,\frac{d}{dr}\!\left(r\,\frac{dg_m}{dr}\right)-\left(k^{\,2} + \frac{m^{\,2}}{r^{\,2}}\right) g_m= 0.$ (454)

As we saw in Section 3.10, the modified Bessel function $ I_m(k\,r)$ [defined in Equation (435)] is a solution of the modified Bessel equation that is well behaved at $ r=0$ , and badly behaved as $ r\rightarrow \infty$ . On the other hand, the modified Bessel function $ K_m(k\,r)$ , where[*]

$\displaystyle K_m(p) = \int_0^\infty {\rm e}^{-\,p\,\cosh t}\,\cosh(m\,t)\,dt,$ (455)

is a solution that is badly behaved at $ r=0$ , and well behaved as $ r\rightarrow \infty$ . In fact,

$\displaystyle K_m(p)$ $\displaystyle \rightarrow \infty$   $\displaystyle \mbox{\hspace{2.4cm}as $p\rightarrow 0$}$$\displaystyle ,$ (456)
$\displaystyle K_m(p)$ $\displaystyle \rightarrow \sqrt{\frac{\pi}{2\,p}}\,{\rm e}^{-p}$$\displaystyle \mbox{\hspace{1cm}as $p\rightarrow \infty$}$$\displaystyle .$ (457)

We are searching for a solution of Equation (454) that is well behaved at $ r=0$ (because there is no reason for the potential to be infinite at $ r=0$ ) and goes to zero as $ r\rightarrow \infty$ , in accordance with the constraint (447). It follows that

$\displaystyle g_m(r,r')=\left\{\begin{array}{ccc}\alpha(r')\,I_m(k\,r)&\mbox{\hspace{1cm}}& r<r'\\ [0.5ex] \beta(r')\,K_m(k\,r)&&r>r'\end{array}\right..$ (458)

However, given that $ G({\bf r},{\bf r'})$ is a symmetric function, we expect $ g_m(r,r')$ to also be symmetric: that is, $ g_m(r',r)=g_m(r,r')$ . Consequently,

$\displaystyle g_m(r,r') = A\,I_m(k\,r_<)\,K_m(k\,r_>),$ (459)

where $ r_<$ is the lesser of $ r$ and $ r'$ , and $ r_>$ the greater. Integration of Equation (454) across $ r=r'$ yields

$\displaystyle \left[\frac{d g_m}{dr}\right]_{r=r'_-}^{r=r'_+} = \frac{1}{r'},$ (460)

which implies that

$\displaystyle A\,k\left[K_m'(k\,r')\,I_m(k\,r')-K_m(k\,r')\,I_m'(k\,r')\right] = \frac{1}{r'},$ (461)

where $ \phantom{K}'$ denotes differentiation with respect to argument. However, the modified Bessel functions $ I_m(p)$ and $ K_m(p)$ satisfy the well-known mathematical identity[*]

$\displaystyle K_m(p)\,I_m'(p)-K_m'(p)\,I_m(p) = \frac{1}{p}.$ (462)

Hence, we deduce that $ A=-1$ . Thus, our general Green's function becomes

$\displaystyle G({\bf r},{\bf r}') =-\frac{1}{4\pi^2}\sum_{m=-\infty,\infty}\int...
...-z')}\,{\rm e}^{\,{\rm i}\,m\,(\theta-\theta')} \,I_m(k\,r_<)\,K_m(k\,r_>)\,dk.$ (463)

The previous expression for the Green's function, in combination with Equation (445), leads to the following expressions for the general solution to Poisson's equation in cylindrical geometry, subject to the boundary condition (444):

$\displaystyle \phi(r,\theta,z)$ $\displaystyle =\sum_{m=-\infty,\infty}\phi_m(r,z)\,{\rm e}^{\,{\rm i}\,m\,\theta},$ (464)
$\displaystyle \phi_m(r,z)$ $\displaystyle = \int_{-\infty}^\infty {\mit\Phi}_m(r,k)\,{\rm e}^{\,{\rm i}\,k\,z}\,dk,$ (465)
$\displaystyle {\mit\Phi}_m(r,k)$ $\displaystyle = K_m(k\,r)\int_0^r R_m(r',k)\,I_m(k\,r')\,r'\,dr'+I_m(k\,r)\int_r^\infty R_m(r',k)\,K_m(k\,r')\,r'\,dr',$ (466)
$\displaystyle R_m(r',k)$ $\displaystyle = \frac{1}{2\pi}\int_{-\infty}^\infty \rho_m(r',z')\,{\rm e}^{-{\rm i}\,k\,z'}\,dz',$ (467)
$\displaystyle \rho_m(r',z')$ $\displaystyle = \frac{1}{2\pi}\oint \frac{\rho(r',\theta',z')}{\epsilon_0}\,{\rm e}^{-{\rm i}\,m\,\theta'}d\theta'.$ (468)

Suppose that we wish to solve Poisson's equation within a finite cylindrical volume, $ V$ , bounded by the surfaces $ z=0$ , $ z=L$ , and $ r=a$ . Let the boundary conditions imposed at the surface be

$\displaystyle \phi(r,\theta,0)$ $\displaystyle =0,$ (469)
$\displaystyle \phi(r,\theta,L)$ $\displaystyle =0,$ (470)
$\displaystyle \phi(a,\theta,z)$ $\displaystyle = {\mit\Phi}(\theta,z),$ (471)

where $ {\mit\Phi}(r,\theta)$ is a specified function. According to Section 2.10, the solution to this Dirichlet problem is written

$\displaystyle \phi({\bf r}) = - \int_VG({\bf r},{\bf r}')\,\frac{\rho({\bf r}')...
... + \int_S\phi({\bf r}')\,\frac{\partial G({\bf r},{\bf r}')}{\partial n'}\,dS',$ (472)

where $ S$ represents the bounding surface. Here, the Green's function is the symmetric solution to

$\displaystyle \nabla^{\,2} G({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}')$ (473)

that satisfies

$\displaystyle G({\bf r},{\bf r}')=0$ (474)

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

As before, in cylindrical coordinates, Equation (474) is written

$\displaystyle \frac{1}{r}\,\frac{\partial}{\partial r}\left(r\,\frac{\partial G...
...ial z^{\,2}} = \frac{1}{r}\,\delta(r-r')\,\delta(\theta-\theta')\,\delta(z-z').$ (475)

If we search for a separable solution of the form $ R(r)\,Q(\theta)\,Z(z)$ then it is clear that

$\displaystyle Z(z) = \sum_{n=1,\infty} Z_n\,\sin(k_n\,z),$ (476)


$\displaystyle k_n = n\,\frac{\pi}{L},$ (477)

is the appropriate expression for $ Z(z)$ that satisfies the constraint $ Z=0$ when $ z=0$ and $ z=L$ . The Fourier series (477) can be inverted in the usual fashion to give

$\displaystyle Z_n = \frac{2}{L}\int_0^L Z(z)\,\sin(k_n\,z)\,dz,$ (478)

which implies that

$\displaystyle \delta(z-z') = \frac{2}{L}\sum_{n=1,\infty} \sin(k_n\,z)\,\sin(k_n\,z').$ (479)

Thus, searching for a Green's function of the form

$\displaystyle G({\bf r}, {\bf r}') = \frac{1}{L\,\pi}\sum_{m=-\infty,\infty}\su...
..._n\,z)\,\sin(k_n\,z')\,{\rm e}^{\,{\rm i}\,m\,(\theta-\theta')}\,g_{m,n}(r,r'),$ (480)

Equation (476) reduces to

$\displaystyle \frac{1}{r}\,\frac{d}{dr}\!\left(r\,\frac{d g_{m,n}}{dr}\right) -...
...t(k_n^{\,2}+\frac{m^{\,2}}{r^{\,2}}\right) g_{m,n} = \frac{1}{r}\,\delta(r-r').$ (481)

Of course, $ g_{m,n}(r,r')$ must be well behaved at $ r=0$ . Moreover, the constraint $ G({\bf r}, {\bf r}')=0$ when $ r=a$ implies that $ g_{m,n}(a,r')=0$ . Hence,

$\displaystyle g_{m,n}(r,r')=\left\{\begin{array}{ccc}\alpha(r')\,I_m(k_n\,r)&\m...
...m(k_n\,r)\,K_m(k_n\,a)-I_m(k_n\,a)\,K_m(k_n\,r)\right]&&r>r'\end{array}\right..$ (482)

Now, the Green's function must be continuous when $ r=r'$ (otherwise, it would not be a symmetric function of $ {\bf r}$ and $ {\bf r}'$ ): that is,

$\displaystyle g_{m,n}(r=r'_{+},r')= g_{m,n}(r=r_{-}',r').$ (483)

This implies that

$\displaystyle \alpha(r')\,I_m(k_n\,r') = \beta(r')\left[I_m(k_n\,r')\,K_m(k_n\,a)-I_m(k_n\,a)\,K_m(k_n\,r')\right].$ (484)

Integration of Equation (476) across $ r=r'$ again gives (461), which leads to

$\displaystyle \beta(r') = \frac{I_m(k_n\,r')}{I_m(k_n\,a)},$ (485)

where use has been made of Equations (463) and (485). It follows that

$\displaystyle g_{m,n}(r,r') = -\left[I_m(k_n\,a)\,K_m(k_n\,r_>)- I_m(k_n\,r_>)\,K_m(k_n\,a)\right]\,\frac{I_m(k_n\,r_<)}{I_m(k_n\,a)}.$ (486)

Our general expression for the Dirichlet Green's function becomes

$\displaystyle G({\bf r},{\bf r}')$ $\displaystyle = -\frac{1}{L\,\pi}\sum_{m=-\infty,\infty}\sum_{n=1,\infty}\sin(k_n\,z)\,\sin(k_n\,z')\,{\rm e}^{\,{\rm i}\,m\,(\theta-\theta')}$    
  $\displaystyle \phantom{==}\left[I_m(k_n\,a)\,K_m(k_n\,r_>)- I_m(k_n\,r_>)\,K_m(k_n\,a)\right]\,\frac{I_m(k_n\,r_<)}{I_m(k_n\,a)}.$ (487)

It is easily demonstrated that

$\displaystyle \left.r'\,\frac{\partial G({\bf r},{\bf r}')}{\partial r'}\right\...
...z')\,{\rm e}^{\,{\rm i}\,m\,(\theta-\theta')}\,\frac{I_m(k_n\,r)}{I_m(k_n\,a)}.$ (488)

Hence, making use of Equation (473), in combination with the previous two expressions, our general solution to the problem under discussion is specified by the following set of equations:

$\displaystyle \phi(r,\theta,z)$ $\displaystyle =\sum_{m=-\infty,\infty}\phi_m(r,z)\,{\rm e}^{\,{\rm i}\,m\,\theta},$ (489)
$\displaystyle \phi_m(r,z)$ $\displaystyle = \sum_{n=0,\infty}\phi_{m,n}(r)\,\sin(k_n\,z),$ (490)
$\displaystyle \phi_{m,n}(r)$ $\displaystyle = \left[K_m(k_n\,r)-\frac{I_m(k_n\,r)\,K_m(k_n\,a)}{I_m(k_n\,a)}\right]\int_0^r R_{m,n}(r')\,I_m(k_n\,r')\,r'\,dr'$    
  $\displaystyle \phantom{=}+I_m(k_n\,r)\int_r^\infty R_{m,n}(r')\,\left[K_m(k_n\,r')-\frac{I_m(k_n\,r')\,K_m(k_n\,a)}{I_m(k_n\,a)}\right]r'\,dr'$    
  $\displaystyle \phantom{=}+ \frac{I_m(k_n\,r)}{I_m(k_n\,a)}\,{\mit\Phi}_{m,n},$ (491)
$\displaystyle R_{m,n}(r')$ $\displaystyle = \frac{2}{L}\int_0^L \rho_m(r',z')\,\sin(k_n\,z')\,dz',$ (492)
$\displaystyle \rho_m(r',z')$ $\displaystyle = \frac{1}{2\pi}\oint \frac{\rho(r',\theta',z')}{\epsilon_0}\,{\rm e}^{-{\rm i}\,m\,\theta'}d\theta',$ (493)
$\displaystyle {\mit\Phi}_{m,n}$ $\displaystyle = \frac{2}{L}\int_0^L {\mit\Phi}_m(z')\,\sin(k_n\,z')\,dz',$ (494)
$\displaystyle {\mit\Phi}_m(z')$ $\displaystyle = \frac{1}{2\pi}\oint {\mit\Phi}(\theta',z')\,{\rm e}^{-{\rm i}\,m\,\theta'}d\theta'.$ (495)

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Next: Exercises Up: Potential Theory Previous: Laplace's Equation in Cylindrical
Richard Fitzpatrick 2014-06-27