Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random | Template Test |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Talkback

Downloads

Information
Laplace equation in cylindrical coordinates (Definition)

Laplace Equation in Cylindrical Coordinates

Solutions to the Laplace equation in cylindrical coordinates have wide applicability from fluid mechanics to electrostatics. Applying the method of separation of variables to Laplace's partial differential equation and then enumerating the various forms of solutions will lay down a foundation for solving problems in this coordinate system. Finally, the use of Bessel functions in the solution reminds us why they are synonymous with the cylindrical domain.

Separation of Variables

Beginning with the Laplacian in Cylindrical Coordinates, apply the operator to a potential function and set it equal to zero to get the Laplace equation
$\displaystyle \nabla^{2} \Phi = \frac{1}{r} \frac{\partial \Phi}{\partial r}\le... ...{\partial^2\Phi}{\partial \theta^2} + \frac{\partial^2 \Phi}{\partial z^2} = 0.$ (1)

First expand out the terms

$\displaystyle \nabla^{2} \Phi = \frac{1}{r} \frac{\partial \Phi}{\partial r} + ... ...{\partial^2\Phi}{\partial \theta^2} + \frac{\partial^2 \Phi}{\partial z^2} = 0.$ (2)

Then apply the method of separation of variables by assuming the solution is in the form

$\displaystyle \Phi \left ( r,\theta,z \right) = R(r)P(\phi)Z(z).$

Plug this into (2) and note how we can bring out the functions that are not affected by the derivatives

$\displaystyle \frac{P(\phi) Z(z)}{r} \frac{\partial R(r)}{\partial r} + P(\phi)... ...i)}{\partial \theta^2} + R(r) P(\phi) \frac{\partial^2 Z(z)}{\partial z^2} = 0.$

Divide by $R(r) P(\phi) Z(z)$ and use short hand notation to get

$\displaystyle \frac{1}{Rr} \frac{\partial R}{\partial r} + \frac{1}{R} \frac{\p... ...ac{\partial^2P}{\partial \theta^2} + \frac{1}{Z} \frac{\partial^2 Z}{dz^2} = 0.$

“Separating” the z term to the other side gives

$\displaystyle \frac{1}{Rr} \frac{d R}{d r} + \frac{1}{R} \frac{d^2R}{d r^2} + \frac{1}{Pr^2} \frac{d^2P}{d \theta^2} = - \frac{1}{Z} \frac{d^2 Z}{dz^2}.$

This equation can only be satisfied for all values if both sides are equal to a constant, $\lambda$, such that

$\displaystyle -\frac{1}{Z} \frac{d^2 Z}{dz^2} = \lambda$ (3)
$\displaystyle \frac{1}{Rr} \frac{d R}{d r} + \frac{1}{R} \frac{d^2R}{d r^2} + \frac{1}{Pr^2} \frac{d^2P}{d\theta^2} = \lambda.$ (4)

Before we can focus on solutions, we need to further separate (4), so multiply (4) by $r^2$

$\displaystyle \frac{r}{R} \frac{d R}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} + \frac{1}{P} \frac{d^2P}{d\theta^2} = \lambda r^2.$

Separate the terms

$\displaystyle \frac{r}{R} \frac{d R}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} - \lambda r^2 = - \frac{1}{P} \frac{d^2P}{d \theta^2}.$

As before, set both sides to a constant, $\kappa$

$\displaystyle - \frac{1}{P} \frac{d^2P}{d \theta^2}= \kappa$ (5)
$\displaystyle \frac{r}{R} \frac{dR}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} - \lambda r^2 = \kappa.$ (6)

Now there are three differential equations and we know the form of these solutions. The differential equations of (3) and (5) are ordinary differential equations, while (6) is a little more complicated and we must turn to Bessel functions.

Axial Solutions ($z$)

Following the guidelines setup in [Etgen] for linear homogeneous differential equations, the first step in solving

$\displaystyle \frac{d^2 Z}{dz^2} + Z \lambda = 0$

is to find the roots of the characteristic polynomial

$\displaystyle C(r) = r^2 + \lambda = 0$

$\displaystyle r = \pm \sqrt{ -\lambda}.$

Although, one can go forward using the square root, here we will introduce another constant, $\gamma$ to imply the following cases. So if we want real roots, then we want to ensure a negative constant

$\displaystyle \lambda = -\gamma^2 $

and if we want complex roots, then we want to ensure a positive constant

$\displaystyle \lambda = \gamma^2 ,$

Case 1: $\lambda \le 0$ and real roots $(\lambda = -\gamma^2)$.

For every real root, there will be an exponential in the general solution. The real roots are

$\displaystyle r_1 = \gamma $

$\displaystyle r_2 = -\gamma $

$\displaystyle r_3 = 0 .$

Therefore, the solutions for these roots are

$\displaystyle h_1(z) = C_1e^{\gamma z} $

$\displaystyle h_2(z) = C_2e^{-\gamma z} $

$\displaystyle h_3(z) = C_3z e^0 = C_3z.$

Combining these using the principle of superposition, gives the general solution,

$\displaystyle Z_{\gamma}(z) = C_1e^{\gamma z} + C_2e^{-\gamma z} + C_3z + C_4.$ (7)

Case 2: $\lambda > 0$ and complex roots $(\lambda = \gamma^2)$.

The roots are

$\displaystyle r_4 = i \gamma $

$\displaystyle r_5 = -i \gamma $

and the corresponding solutions

$\displaystyle h_4(z) = C_5e^0 \cos (\gamma z) = C_5\cos (\gamma z) $

$\displaystyle h_5(z) = C_6e^0 \sin (\gamma z)= C_6\sin (\gamma z) .$

Combining these into a general solution yields

$\displaystyle Z_{\gamma}(z) = C_5\cos (\gamma z) + C_6\sin (\gamma z) + C_7.$

Azimuthal Solutions ($\theta$)

Azimuthal solutions for

$\displaystyle \frac{d^2P}{d \theta^2} + \kappa P = 0$

are in the most general sense obtained similarly to the axial solutions with the characteristic polynomial

$\displaystyle C(r) = r^2 + \kappa = 0$

$\displaystyle r = \pm \sqrt{- \kappa}.$

Using another constant, $\nu$ to ensure positive or negative constants, we get two cases.

Case 1: $\kappa \le 0$ and real roots $(\kappa = -\nu^2)$.

The solutions for these roots are then

$\displaystyle h_1(z) = C_1e^{\nu \theta} $

$\displaystyle h_2(z) = C_2e^{-\nu \theta} $

$\displaystyle h_3(z) = C_3\theta e^0 = C_3\theta.$

Combining these for the general solution,

$\displaystyle P_{\nu}(\theta) = C_1e^{\nu \theta} + C_2e^{-\nu \theta} + C_3\theta + C_4.$ (8)

Case 2: $\kappa > 0$ and complex roots $(\kappa = \nu^2)$.

The roots are

$\displaystyle r_4 = i \nu $

$\displaystyle r_5 = -i \nu $

and the corresponding solutions

$\displaystyle h_4(\theta) = C_5e^0 \cos (\nu \theta) = C_5\cos (\nu \theta) $

$\displaystyle h_5(\theta) = C_6e^0 \sin (\nu \theta)= C_6\sin (\nu \theta) .$

Combining these into a general solution

$\displaystyle P_{\nu}(\theta) = C_5\cos (\nu \theta) + C_6\sin (\nu \theta) + C_7.$

For the first glimpse at simplification, we will note a restriction on that is used when it is required that the solution be periodic to ensure $P$ is single valued

$\displaystyle P(\theta) = P(\theta + 2 n \pi).$

Then we are left with either the periodic solutions that occur with complex roots or the zero case. So not only

$\displaystyle (\kappa = \nu^2)$

but also must be an integer, i.e.

$\displaystyle P_{\nu}(\theta) = C_5 \cos (\nu \theta) + C_6 \sin (\nu \theta) + C_7 \,\,\,\,\,\,\,\,\,\,\, \nu = 0, 1, 2, ...$ (9)

Note, that $\nu = 0$, is still a solution, but to be periodic we can only have a constant

$\displaystyle P(\theta) = C_4.$

Radial Solutions ($r$)

The radial solutions are the more difficult ones to understand for this problem and are solved using a power series. The two types of solutions generated based on the choices of constants from the and solutions (excluding non-periodic solutions for ) leads to the Bessel functions and the modified Bessel functions. The first step for both these cases is to transform (6) into the Bessel differential equation.

Case 1: $\lambda < 0$ , .

Substitute and into the radial equation (6) to get

$\displaystyle \frac{r}{R} \frac{dR}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} + \gamma^2 r^2 - \nu^2 = 0.$ (10)

Next, use the substitution

$\displaystyle x = \gamma r$

$\displaystyle r = \frac{x}{\gamma}.$

Therefore, the derivatives are

$\displaystyle dx = \gamma dr$

$\displaystyle dr = \frac{dx}{\gamma}$

and make a special note that

$\displaystyle \frac{d^2}{dx^2} = \frac{d}{dx} \frac{d}{dx}$

so

$\displaystyle dx^2 = dx*dx = \gamma^2 dr^2$

$\displaystyle dr^2 = \frac{dx^2}{\gamma^2}.$

Substituting these relationships into (10) gives us

$\displaystyle \frac{x \gamma}{\gamma R} \frac{dR}{dx} + \frac{x^2 \gamma^2}{\gamma^2 R} \frac{d^2R}{dx^2} + x^2 - \nu^2 = 0.$

Finally, multiply by $R/x^2$ to get the Bessel differential equation

$\displaystyle \frac{d^2R}{dx^2} + \frac{1}{x} \frac{dR}{dx} + \left(1 - \frac{\nu^2}{x^2} \right )R = 0.$ (11)

Delving into all the nuances of solving Bessel's differential equation is beyond the scope of this article, however, the curious are directed to Watson's in depth treatise [Watson]. Here, we will just present the results as we did for the previous differential equations. The general solution is a linear combination of the Bessel function of the first kind $J_{\nu}(r)$ and the Bessel function of the second kind $Y_{\nu}(r)$. Remebering that is a positive integer or zero.

$\displaystyle R_{\nu}(r) = C_1 J_{\nu}(\gamma r) + C_2 Y_{\nu}(\gamma r) + C_3$ (12)

Bessel function of the first kind:

$\displaystyle J_{\nu}(x) = \sum_{m=0}^{\infty} \frac{ (-1)^m ( \frac{-1}{2}x)^{\nu + 2m} }{ m! (m + \nu)! }.$

Bessel function of the second kind (using Hankel's formula):

$\displaystyle Y_{\nu}(x) = 2J_{\nu}(x) \left (\eta + ln \left( \frac{x}{2}\righ... ...} \sum_{m=0}^{\nu-1} \frac{(\nu - m - 1)!}{m!} \left ( \frac{x}{2} \right)^{2m}$

$\displaystyle \,\,\,\,\, -\sum_{m=0}^{\infty} \frac{ (-1)^m (\frac{x}{2})^{\nu ... .... + \frac{1}{m}+ \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{\nu + m} \right \}.$

For the unfortunate person who has to evaluate this function, note that when $m = 0$, the singularity is taken care of by replacing the series in brackets by

$\displaystyle \left \{ \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{\nu} \right \}.$

Some solace can be found since most physical problems need to be analytic at $x = 0$ and therefore $Y_{\nu}(x)$ breaks down at $ln(0)$. This leads to the choice of constant $C_2$ to be zero.

Case 2: , .

Using the previous method of substitution, we just get the change of sign

$\displaystyle \frac{d^2R}{dx^2} + \frac{1}{x} \frac{dR}{dx} - \left(1 + \frac{\nu^2}{x^2} \right)R = 0.$ (13)

This leads to the modified Bessel functions as a solution, which are also known as the pure imaginary Bessel functions. The general solution is denoted

$\displaystyle R_{\nu}(r) = C_1 I_{\nu}(\gamma r) + C_2 K_{\nu}(\gamma r) + C_3$ (14)

where $I_{\nu}$ is the modified Bessel function of the first kind and $K_{\nu}$ is the modified Bessel function of the second kind

$\displaystyle I_{\nu}(\gamma r) = i^{-\nu} J_{\nu}(i \gamma r)$

$\displaystyle K_{\nu}(\gamma r) = \frac{\pi}{2} i^{\nu+1} \left ( J_{\nu}(i \gamma r) + iY_{\nu}(i \gamma r) \right ) .$

Combined Solution

Keeping track of all the different cases and choosing the right terms for boundary conditions is a daunting task when one attempts to solve Laplace's equation. The short hand notation used in [Kusse] and [Arfken] will be presented here to help organize the choices as a reference. It is important to remember that these solutions are only for the single valued azimuth cases .

Once the separate solutions are obtained, the rest is simple since our solution is separable

$\displaystyle \Phi \left ( r,\theta,z \right) = R(r)P(\theta)Z(z).$

so we just combine the individual solutions to get the general solutions to the Laplace equation in cylindrical coordinates.

Case 1: , .

$\displaystyle \Phi \left ( r,\theta,z \right) = \sum_{\nu} \sum_{\gamma} \left ... ...{ \begin{array}{c} J_{\nu}(\gamma r) \ Y_{\nu}(\gamma r) \end{array} \right .$ (15)

Case 2: , .

$\displaystyle \Phi \left ( r,\theta,z \right) = \sum_{\nu} \sum_{\gamma} \left ... ...{ \begin{array}{c} I_{\nu}(\gamma r) \ K_{\nu}(\gamma r) \end{array} \right .$ (16)

Interpreting the short hand notation is as simple as expanding terms and not forgetting the linear solutions, i.e. $(\gamma = 0)$ . As an example, case 1, expanded out while ignoring the linear terms would give

$\displaystyle \Phi \left ( r,\theta,z \right) = \sum_{\nu} \sum_{\gamma} \left ... ...-\gamma z} \sin (\nu \theta) Y_{\nu}(\gamma r) \ \par \end{array} \right \} .$ (17)

Bibliography

1
Arfken, George, Weber, Hans, Mathematical Physics. Academic Press, San Diego, 2001.
2
Etgen, G., Calculus. John Wiley & Sons, New York, 1999.
3
Guterman, M., Nitecki, Z., Differential Equations, 3rd Edition. Saunders College Publishing, Fort Worth, 1992.
4
Jackson, J.D., Classical Electrodynamics, 2nd Edition. John Wiley & Sons, New York, 1975.
5
Kusse, Bruce, Westwig, Erik, Mathematical Physics. John Wiley & Sons, New York, 1998.
6
Lebedev, N., Special Functions & Their Applications. Dover Publications, New York, 1995.
7
Watson, G.N., A Treatise on the Theory of Bessel Functions. Cambridge University Press, New York, 1995.



"Laplace equation in cylindrical coordinates" is owned by bloftin.

View style:


Cross-references: boundary, formula, radial equation, types, power, square, ordinary differential equations, differential equations, function, operator, Laplacian in Cylindrical Coordinates, domain, Bessel functions, system, partial differential equation, separation of variables, mechanics

This is version 5 of Laplace equation in cylindrical coordinates, born on 2006-11-14, modified 2008-02-17.
Object id is 234, canonical name is LaplaceEquationInCylindricalCoordinates.
Accessed 6190 times total.

Classification:
Physics Classification: 02.30.Jr (Partial differential equations)
 41.20.Cv (Electrostatics; Poisson and Laplace equations, boundary-value)
 02.30.Gp (Special functions)
 02.30.Em (Potential theory)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "