# Laplace equation in cylindrical coordinates

## 1 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.

### 1.1 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

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

First expand out the terms

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

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

$$\mathrm{\Phi}(r,\theta ,z)=R(r)P(\theta )Z(z).$$ |

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

$$\frac{PZ}{r}\frac{dR}{dr}+PZ\frac{{d}^{2}R}{d{r}^{2}}+\frac{RZ}{{r}^{2}}\frac{{d}^{2}P}{d{\theta}^{2}}+RP\frac{{d}^{2}Z}{d{z}^{2}}=0.$$ |

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

$$\frac{1}{Rr}\frac{dR}{dr}+\frac{1}{R}\frac{{d}^{2}R}{d{r}^{2}}+\frac{1}{P{r}^{2}}\frac{{d}^{2}P}{d{\theta}^{2}}+\frac{1}{Z}\frac{{d}^{2}Z}{d{z}^{2}}=0.$$ |

“Separating” the z term to the other side gives

$$\frac{1}{Rr}\frac{dR}{dr}+\frac{1}{R}\frac{{d}^{2}R}{d{r}^{2}}+\frac{1}{P{r}^{2}}\frac{{d}^{2}P}{d{\theta}^{2}}=-\frac{1}{Z}\frac{{d}^{2}Z}{d{z}^{2}}.$$ |

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

$$-\frac{1}{Z}\frac{{d}^{2}Z}{d{z}^{2}}=\lambda $$ | (3) |

$$\frac{1}{Rr}\frac{dR}{dr}+\frac{1}{R}\frac{{d}^{2}R}{d{r}^{2}}+\frac{1}{P{r}^{2}}\frac{{d}^{2}P}{d{\theta}^{2}}=\lambda .$$ | (4) |

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

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

Separate the terms

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

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

$$-\frac{1}{P}\frac{{d}^{2}P}{d{\theta}^{2}}=\kappa $$ | (5) |

$$\frac{r}{R}\frac{dR}{dr}+\frac{{r}^{2}}{R}\frac{{d}^{2}R}{d{r}^{2}}-\lambda {r}^{2}=\kappa .$$ | (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.

### 1.2 Axial Solutions ($z$)

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

$$\frac{{d}^{2}Z}{d{z}^{2}}+Z\lambda =0$$ |

is to find the roots of the characteristic polynomial^{}

$$C(r)={r}^{2}+\lambda =0$$ |

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

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

$$\lambda =-{\gamma}^{2}$$ |

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

$$\lambda ={\gamma}^{2},$$ |

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

$${r}_{1}=\gamma $$ |

$${r}_{2}=-\gamma $$ |

$${r}_{3}=0.$$ |

Therefore, the solutions for these roots are

$${h}_{1}(z)={C}_{1}{e}^{\gamma z}$$ |

$${h}_{2}(z)={C}_{2}{e}^{-\gamma z}$$ |

$${h}_{3}(z)={C}_{3}z{e}^{0}={C}_{3}z.$$ |

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

$${Z}_{\gamma}(z)={C}_{1}{e}^{\gamma z}+{C}_{2}{e}^{-\gamma z}+{C}_{3}z+{C}_{4}.$$ | (7) |

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

The roots are

$${r}_{4}=i\gamma $$ |

$${r}_{5}=-i\gamma $$ |

and the corresponding solutions

$${h}_{4}(z)={C}_{5}{e}^{0}\mathrm{cos}(\gamma z)={C}_{5}\mathrm{cos}(\gamma z)$$ |

$${h}_{5}(z)={C}_{6}{e}^{0}\mathrm{sin}(\gamma z)={C}_{6}\mathrm{sin}(\gamma z).$$ |

Combining these into a general solution yields

$${Z}_{\gamma}(z)={C}_{5}\mathrm{cos}(\gamma z)+{C}_{6}\mathrm{sin}(\gamma z)+{C}_{7}.$$ |

### 1.3 Azimuthal Solutions ($\theta $)

Azimuthal solutions for

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

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

$$C(r)={r}^{2}+\kappa =0$$ |

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

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

$${h}_{1}(z)={C}_{1}{e}^{\nu \theta}$$ |

$${h}_{2}(z)={C}_{2}{e}^{-\nu \theta}$$ |

$${h}_{3}(z)={C}_{3}\theta {e}^{0}={C}_{3}\theta .$$ |

Combining these for the general solution,

$${P}_{\nu}(\theta )={C}_{1}{e}^{\nu \theta}+{C}_{2}{e}^{-\nu \theta}+{C}_{3}\theta +{C}_{4}.$$ | (8) |

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

The roots are

$${r}_{4}=i\nu $$ |

$${r}_{5}=-i\nu $$ |

and the corresponding solutions

$${h}_{4}(\theta )={C}_{5}{e}^{0}\mathrm{cos}(\nu \theta )={C}_{5}\mathrm{cos}(\nu \theta )$$ |

$${h}_{5}(\theta )={C}_{6}{e}^{0}\mathrm{sin}(\nu \theta )={C}_{6}\mathrm{sin}(\nu \theta ).$$ |

Combining these into a general solution

$${P}_{\nu}(\theta )={C}_{5}\mathrm{cos}(\nu \theta )+{C}_{6}\mathrm{sin}(\nu \theta )+{C}_{7}.$$ |

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

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

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

$$(\kappa ={\nu}^{2})$$ |

but also $\nu $ must be an integer, i.e.

$${P}_{\nu}(\theta )={C}_{5}\mathrm{cos}(\nu \theta )+{C}_{6}\mathrm{sin}(\nu \theta )+{C}_{7}\mathit{\hspace{1em}\hspace{0.5em}\hspace{1em}}\nu =0,1,2,\mathrm{\dots}$$ | (9) |

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

$$P(\theta )={C}_{4}.$$ |

### 1.4 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 $\theta $ and $z$ solutions (excluding non-periodic solutions for $P$)
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: $$ $(\lambda =-{\gamma}^{2})$, $\kappa >0$ $(\kappa ={\nu}^{2})$.

Substitute $\gamma $ and $\nu $ into the radial equation (6) to get

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

Next, use the substitution

$$x=\gamma r$$ |

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

Therefore, the derivatives are

$$dx=\gamma dr$$ |

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

and make a special note that

$$\frac{{d}^{2}}{d{x}^{2}}=\frac{d}{dx}\frac{d}{dx}$$ |

so

$$d{x}^{2}=dx*dx={\gamma}^{2}d{r}^{2}$$ |

$$d{r}^{2}=\frac{d{x}^{2}}{{\gamma}^{2}}.$$ |

Substituting these relationships into (10) gives us

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

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

$$\frac{{d}^{2}R}{d{x}^{2}}+\frac{1}{x}\frac{dR}{dx}+\left(1-\frac{{\nu}^{2}}{{x}^{2}}\right)R=0.$$ | (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 $\nu $ is a positive integer or zero.

$${R}_{\nu}(r)={C}_{1}{J}_{\nu}(\gamma r)+{C}_{2}{Y}_{\nu}(\gamma r)+{C}_{3}$$ | (12) |

Bessel function of the first kind:

$${J}_{\nu}(x)=\sum _{m=0}^{\mathrm{\infty}}\frac{{(-1)}^{m}{(\frac{-1}{2}x)}^{\nu +2m}}{m!(m+\nu )!}.$$ |

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

$${Y}_{\nu}(x)=2{J}_{\nu}(x)\left(\eta +ln\left(\frac{x}{2}\right)\right)-{\left(\frac{x}{2}\right)}^{-\nu}\sum _{m=0}^{\nu -1}\frac{(\nu -m-1)!}{m!}{\left(\frac{x}{2}\right)}^{2m}$$ |

$$-\sum _{m=0}^{\mathrm{\infty}}\frac{{(-1)}^{m}{(\frac{x}{2})}^{\nu +2m}}{m!(\nu +m)!}\left\{\frac{1}{1}+\frac{1}{2}+\mathrm{\dots}+\frac{1}{m}+\frac{1}{1}+\frac{1}{2}+\mathrm{\dots}+\frac{1}{\nu +m}\right\}.$$ |

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

$$\left\{\frac{1}{1}+\frac{1}{2}+\mathrm{\dots}+\frac{1}{\nu}\right\}.$$ |

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: $\lambda >0$ $(\lambda ={\gamma}^{2})$, $\kappa >0$ $(\kappa ={\nu}^{2})$.

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

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

$${R}_{\nu}(r)={C}_{1}{I}_{\nu}(\gamma r)+{C}_{2}{K}_{\nu}(\gamma r)+{C}_{3}$$ | (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

$${I}_{\nu}(\gamma r)={i}^{-\nu}{J}_{\nu}(i\gamma r)$$ |

$${K}_{\nu}(\gamma r)=\frac{\pi}{2}{i}^{\nu +1}\left({J}_{\nu}(i\gamma r)+i{Y}_{\nu}(i\gamma r)\right).$$ |

### 1.5 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 $(\kappa ={\nu}^{2})$.

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

$$\mathrm{\Phi}(r,\theta ,z)=R(r)P(\theta )Z(z).$$ |

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

Case 1: $$ $(\lambda =-{\gamma}^{2})$, $\kappa >0$ $(\kappa ={\nu}^{2})$.

$$\mathrm{\Phi}(r,\theta ,z)=\sum _{\nu}\sum _{\gamma}\{\begin{array}{c}\hfill {e}^{\gamma z}\hfill \\ \hfill {e}^{-\gamma z}\hfill \end{array}\{\begin{array}{c}\hfill \mathrm{cos}(\nu \theta )\hfill \\ \hfill \mathrm{sin}(\nu \theta )\hfill \end{array}\{\begin{array}{c}\hfill {J}_{\nu}(\gamma r)\hfill \\ \hfill {Y}_{\nu}(\gamma r)\hfill \end{array}$$ | (15) |

Case 2: $\lambda >0$ $(\lambda ={\gamma}^{2})$, $\kappa >0$ $(\kappa ={\nu}^{2})$.

$$\mathrm{\Phi}(r,\theta ,z)=\sum _{\nu}\sum _{\gamma}\{\begin{array}{c}\hfill \mathrm{cos}(\gamma z)\hfill \\ \hfill \mathrm{sin}(\gamma z)\hfill \end{array}\{\begin{array}{c}\hfill \mathrm{cos}(\nu \theta )\hfill \\ \hfill \mathrm{sin}(\nu \theta )\hfill \end{array}\{\begin{array}{c}\hfill {I}_{\nu}(\gamma r)\hfill \\ \hfill {K}_{\nu}(\gamma r)\hfill \end{array}$$ | (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

$$\mathrm{\Phi}(r,\theta ,z)=\sum _{\nu}\sum _{\gamma}\left\{\begin{array}{c}\hfill {A}_{\nu \gamma}{e}^{\gamma z}\mathrm{cos}(\nu \theta ){J}_{\nu}(\gamma r)\hfill \\ \hfill +{B}_{\nu \gamma}{e}^{\gamma z}\mathrm{cos}(\nu \theta ){Y}_{\nu}(\gamma r)\hfill \\ \hfill +{C}_{\nu \gamma}{e}^{\gamma z}\mathrm{sin}(\nu \theta ){J}_{\nu}(\gamma r)\hfill \\ \hfill +{D}_{\nu \gamma}{e}^{\gamma z}\mathrm{sin}(\nu \theta ){Y}_{\nu}(\gamma r)\hfill \\ \hfill +{E}_{\nu \gamma}{e}^{-\gamma z}\mathrm{cos}(\nu \theta ){J}_{\nu}(\gamma r)\hfill \\ \hfill +{F}_{\nu \gamma}{e}^{-\gamma z}\mathrm{cos}(\nu \theta ){Y}_{\nu}(\gamma r)\hfill \\ \hfill +{G}_{\nu \gamma}{e}^{-\gamma z}\mathrm{sin}(\nu \theta ){J}_{\nu}(\gamma r)\hfill \\ \hfill +{H}_{\nu \gamma}{e}^{-\gamma z}\mathrm{sin}(\nu \theta ){Y}_{\nu}(\gamma r)\hfill \end{array}\right\}.$$ | (17) |

## References

- 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.

Title | Laplace equation in cylindrical coordinates |
---|---|

Canonical name | LaplaceEquationInCylindricalCoordinates |

Date of creation | 2013-03-22 16:24:24 |

Last modified on | 2013-03-22 16:24:24 |

Owner | bloftin (6104) |

Last modified by | bloftin (6104) |

Numerical id | 13 |

Author | bloftin (6104) |

Entry type | Derivation |

Classification | msc 35J05 |

Related topic | BesselFunction |

Related topic | BesselsEquation |