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Next: Poisson's Equation in Cylindrical Up: Potential Theory Previous: Newmann Problem in Spherical


Laplace's Equation in Cylindrical Coordinates

Suppose that we wish to solve Laplace's equation,

$\displaystyle \nabla^{\,2}\phi = 0,$ (392)

within a cylindrical volume of radius $ a$ and height $ L$ . Let us adopt the standard cylindrical coordinates, $ r$ , $ \theta$ , $ z$ . Suppose that the curved portion of the bounding surface corresponds to $ r=a$ , while the two flat portions correspond to $ z=0$ and $ z=L$ , respectively. Suppose, finally, that the boundary conditions that are imposed at the bounding surface are

$\displaystyle \phi(r,\theta,0)$ $\displaystyle =0,$ (393)
$\displaystyle \phi(a,\theta,z)$ $\displaystyle =0,$ (394)
$\displaystyle \phi(r,\theta,L)$ $\displaystyle = {\mit\Phi}(r,\theta),$ (395)

where $ {\mit\Phi}(r,\theta)$ is a given function. In other words, the potential is zero on the curved and bottom surfaces of the cylinder, and specified on the top surface.

In cylindrical coordinates, Laplace's equation is written

$\displaystyle \frac{1}{r}\,\frac{\partial}{\partial r}\left(r\,\frac{\partial\p...
...{\,2}\phi}{\partial\theta^{\,2}}+\frac{\partial^{\,2}\phi}{\partial z^{\,2}}=0.$ (396)

Let us try a separable solution of the form

$\displaystyle \phi(r,\theta,z) = R(r)\,Q(\theta)\,Z(z).$ (397)

Proceeding in the usual manner, we obtain

$\displaystyle \frac{d^{\,2} Z}{dz^{\,2}} - k^{\,2} \,Z = 0,$ (398)
$\displaystyle \frac{d^{\,2} Q}{d\theta^{\,2}}+m^{\,2}\,Q=0,$ (399)
$\displaystyle \frac{d^{\,2} R}{dr^{\,2}} + \frac{1}{r}\,\frac{dR}{dr}+\left(k^{\,2}-\frac{m^{\,2}}{r^{\,2}}\right) R = 0.$ (400)

Note that we have selected exponential, rather than oscillating, solutions in the $ z$ -direction [by writing $ -k^{\,2}\,Z$ , instead of $ +k^{\,2}\,Z$ , in Equation (399)]. As will become clear, this implies that the radial solutions oscillate, which is the appropriate choice for the particular set of boundary conditions under consideration. The solution to Equation (399), subject to the constraint that $ Z(0) = 0$ [which follows from the first boundary condition, (394)] is

$\displaystyle Z(z)= \sinh(k\,z).$ (401)

The most general solution to Equation (400) is

$\displaystyle Q(\theta) = \sum_{m=0,\infty}\left[A_m\,\cos (m\,\theta) + B_m\,\sin(m\,\theta)\right].$ (402)

Note that, to ensure that the potential is single-valued in $ \theta$ , the constant $ m$ is constrained to be an integer. Finally, if we write $ p=k\,r$ then Equation (401) becomes

$\displaystyle \frac{d^{\,2} R}{dp^{\,2}} + \frac{1}{p}\,\frac{dR}{dp}+\left(1-\frac{m^{\,2}}{p^{\,2}}\right) R = 0.$ (403)

This equation is known as Bessel's equation. The standard solution of this equation that is well behaved at $ r=0$ is[*]

$\displaystyle J_m(p) = \frac{1}{\pi}\int_0^\pi \cos(p\,\sin\theta-m\,\theta)\,d\theta.$ (404)

This solution, which is known as a Bessel function, has the properties that

$\displaystyle J_m(p)$ $\displaystyle \rightarrow \frac{1}{m!}\left(\frac{p}{2}\right)^m$   $\displaystyle \mbox{\hspace{4.2cm}as $p\rightarrow 0$}$$\displaystyle ,$ (405)
$\displaystyle J_m(p)$ $\displaystyle \rightarrow \left(\frac{2}{\pi\,p}\right)^{1/2}\,\cos\left(p - m\,\frac{\pi}{2}-\frac{\pi}{4}\right)$$\displaystyle \mbox{\hspace{1cm}as $p\rightarrow \infty$}$$\displaystyle .$ (406)

In other words, at small arguments the function has a power-law behavior, whereas at large arguments it takes the form of an oscillation of slowly decaying amplitude. It follows that

$\displaystyle R(r) = J_m(k\,r).$ (407)

Let $ j_{mn}$ denote the $ n$ th zero of the Bessel function $ J_m(p)$ . In other words, $ j_{mn}$ is the $ n$ th root (in order, as $ p$ increases from zero) of $ J_m(p)=0$ . The values of the $ j_{mn}$ can be looked up in standard reference books.[*] (For example, $ j_{01}=2.405$ and $ j_{02}=5.520$ .) We can satisfy our second boundary condition, (395), by making the choice $ k=k_{mn}$ , where

$\displaystyle k_{mn} = \frac{j_{mn}}{a}.$ (408)

Thus, our separable solution becomes

$\displaystyle \phi(r,\theta,z) = \sum_{m=0,\infty}\sum_{n=1,\infty} \sinh(j_{mn...
...\,J_m(j_{mn}\,r/a)\left[A_{mn}\,\cos(m\,\theta)+B_{mn}\,\sin(m\,\theta)\right].$ (409)

It is convenient to express the specified function $ {\mit\Phi}(r,\theta)$ in the form of a Fourier series: that is,

$\displaystyle {\mit\Phi}(r,\theta) =\sum_{m=0,\infty}\left[C_m(r)\,\cos(m\,\theta)+S_m(r)\,\sin(m\,\theta)\right].$ (410)

Our final boundary condition, (396), then yields

$\displaystyle C_m(r)$ $\displaystyle = \sum_{n=1,\infty}A_{mn}\,\sinh(j_{mn}\,L/a)\,J_m(j_{mn}\,r/a),$ (411)
$\displaystyle S_m(r)$ $\displaystyle = \sum_{n=1,\infty}B_{mn}\,\sinh(j_{mn}\,L/a)\,J_m(j_{mn}\,r/a).$ (412)

It remains to invert the previous two expressions to obtain the coefficients $ A_{mn}$ and $ B_{mn}$ . In fact, it is possible to demonstrate that if

$\displaystyle f(p) = \sum_{n=1,\infty} a_{mn}\,J_m(j_{mn}\,p)$ (413)

then

$\displaystyle a_{mn} = \frac{2}{J_{m+1}^{\,2}(j_{mn})}\int_0^1 p\,f(p)\,J_m(j_{mn}\,p)\,dp.$ (414)

Hence,

$\displaystyle A_{mn}$ $\displaystyle = \frac{2}{a^2\,J_{m+1}^{\,2}(j_{mn})\,\sinh(j_{mn}\,L/a)}\,\int_0^a r\,C_m(r)\,J_m(j_{mn}\,r/a)\,dr,$ (415)
$\displaystyle B_{mn}$ $\displaystyle = \frac{2}{a^2\,J_{m+1}^{\,2}(j_{mn})\,\sinh(j_{mn}\,L/a)}\,\int_0^a r\,S_m(r)\,J_m(j_{mn}\,r/a)\,dr,$ (416)

and our solution is fully determined.

Consider the limit that $ a\rightarrow\infty$ . In this case, according to Equation (409), the allowed values of $ k$ become more and more closely spaced. Consequently, the sum over discrete $ k$ -values in (410) morphs into an integral over a continuous range of $ k$ -values. For instance, suppose that we wish to solve Laplace's equation in the region $ z\geq 0$ , subject to the boundary condition that $ \phi\rightarrow 0$ as $ z\rightarrow \infty$ and $ r\rightarrow \infty$ , with $ \phi(r,\theta,0)={\mit\Phi}(r,\theta)$ , where $ {\mit\Phi}(r,\theta)$ is specified. In this case, we would choose $ Z(z)={\rm e}^{-k\,z}$ in order to satisfy the boundary condition at large $ z$ . The choice $ R(r)=J_m(k\,r)$ ensures that the potential is well behaved at $ r=0$ , and automatically satisfies the boundary condition at large $ r$ . Hence, our general solution becomes

$\displaystyle \phi(r,\theta,z) = \sum_{m=0,\infty} \int_0^\infty {\rm e}^{-k\,z}\,J_m(k\,r)\,\left[A_m(k)\,\cos (m\,\theta)+B_m(k)\,\sin(m\,\theta)\right]\,dk.$ (417)

If we write

$\displaystyle {\mit\Phi}(r,\theta) = \sum_{m=0,\infty}\left[C_m(r)\,\cos(m\,\theta)+ S_m(r)\,\sin(m\,\theta)\right]$ (418)

then the final boundary condition implies that

$\displaystyle C_m(r)$ $\displaystyle = \int_0^\infty J_m(k\,r)\,A_m(k)\,dk,$ (419)
$\displaystyle S_m(r)$ $\displaystyle = \int_0^\infty J_m(k\,r)\,B_m(k)\,dk.$ (420)

We can invert the previous two expressions by means of the identity

$\displaystyle \int_0^\infty r\,J_m(k\,r)\,J_m(k'\,r)\,dr = \frac{1}{k}\,\delta(k-k').$ (421)

Hence, we obtain

$\displaystyle A_m(k)$ $\displaystyle = \int_0^\infty k\,r\,J_m(k\,r)\,C_m(r)\,dr,$ (422)
$\displaystyle B_m(k)$ $\displaystyle = \int_0^\infty k\,r\,J_m(k\,r)\,S_m(r)\,dr,$ (423)

and our solution is fully defined.

Suppose that we wish to solve Laplace's equation in a cylindrical volume of radius $ a$ and height $ L$ , subject to the boundary conditions

$\displaystyle \phi(r,\theta,0)$ $\displaystyle =0,$ (424)
$\displaystyle \phi(r,\theta,L)$ $\displaystyle =0,$ (425)
$\displaystyle \phi(a,\theta,z)$ $\displaystyle = {\mit\Phi}(\theta,z),$ (426)

where $ {\mit\Phi}(\theta,z)$ is specified. In other words, the potential is zero on the two flat portions of the bounding surface, and given on the curved portion. We can again look for a separable solution of the form (398). Proceeding in the usual manner, we obtain

$\displaystyle \frac{d^{\,2} Z}{dz^{\,2}} + k^{\,2}\, Z = 0,$ (427)
$\displaystyle \frac{d^{\,2} Q}{d\theta^{\,2}}+m^{\,2}\,Q=0,$ (428)
$\displaystyle \frac{d^{\,2} R}{dr^{\,2}} + \frac{1}{r}\,\frac{dR}{dr}-\left(k^{\,2}+\frac{m^{\,2}}{r^{\,2}}\right) R = 0.$ (429)

Note that we have selected oscillating, rather than exponential solutions in the $ z$ -direction [by writing $ +k^{\,2}\,Z$ , instead of $ -k^{\,2}\,Z$ , in Equation (428)]. This is the appropriate choice for the particular set of boundary conditions under consideration. The solution to Equation (428), subject to the constraints that $ Z(0)=Z(L)=0$ [which follow from the boundary conditions (425) and (426)] is

$\displaystyle Z(k) = \sin(k_n\,z),$ (430)

where

$\displaystyle k_n = n\,\frac{\pi}{L}.$ (431)

Here, $ n$ is a positive integer. The single-valued solution to Equation (429) is again

$\displaystyle Q(\theta) = \sum_{m=0,\infty}\left[A_m\,\cos (m\,\theta) + B_m\,\sin(m\,\theta)\right].$ (432)

Finally, writing $ p=k_n\,r$ , Equation (430) takes the form

$\displaystyle \frac{d^{\,2} R}{dp^{\,2}} + \frac{1}{p}\,\frac{dR}{dp}-\left(1+\frac{m^{\,2}}{p^{\,2}}\right) R = 0.$ (433)

This equation is known as the modified Bessel equation. The standard solution of this equation that is well behaved at $ r=0$ is[*]

$\displaystyle I_m(p) = \frac{1}{\pi}\int_0^\pi {\rm e}^{\,p\,\cos\theta}\,\cos(m\,\theta)\,d\theta.$ (434)

This solution, which is known as a modified Bessel function, has the properties that

$\displaystyle I_m(p)$ $\displaystyle \rightarrow \frac{1}{m!}\left(\frac{p}{2}\right)^m$   $\displaystyle \mbox{\hspace{1.7cm}as $p\rightarrow 0$}$$\displaystyle ,$ (435)
$\displaystyle I_m(p)$ $\displaystyle \rightarrow \frac{{\rm e}^{\,p}}{\sqrt{2\pi\,p}}$$\displaystyle \mbox{\hspace{2cm}as $p\rightarrow \infty$}$$\displaystyle .$ (436)

In other words, at small arguments the function has a power-law behavior, whereas at large arguments it grows exponentially. It follows that

$\displaystyle R(r) = I_m(k_n\,r).$ (437)

Thus, our separable solution becomes

$\displaystyle \phi(r,\theta,z) =\sum_{m=0,\infty} \sum_{n=1,\infty}\sin(k_n\,z)\,I_m(k_n\,z)\left[A_{mn}\,\cos(m\,\theta)+B_{mn}\,\sin(m\,\theta)\right].$ (438)

If we express the function $ {\mit\Phi}(\theta,z)$ as a Fourier series in $ \theta$ and $ z$ , so that

$\displaystyle {\mit\Phi}(\theta,z) = \sum_{m=0,\infty}\sum_{n=1,\infty}\sin(k_n\,z)\left[C_{mn}\,\cos(m\,\theta)+S_{mn}\,\sin(m\,\theta)\right],$ (439)

then the boundary condition (427) yields

$\displaystyle A_{mn}$ $\displaystyle = \frac{C_{mn}}{I_m(k_n\,a)},$ (440)
$\displaystyle B_{mn}$ $\displaystyle = \frac{S_{mn}}{I_m(k_n\,a)}.$ (441)

Hence, our solution is fully specified.


next up previous
Next: Poisson's Equation in Cylindrical Up: Potential Theory Previous: Newmann Problem in Spherical
Richard Fitzpatrick 2014-06-27