# Laplace equation in cylindrical coordinates

## 1 Laplace Equation in Cylindrical Coordinates

### 1.1 Separation of Variables

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

First expand out the terms

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

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

 $\Phi\left(r,\theta,z\right)=R(r)P(\theta)Z(z).$
 $\frac{PZ}{r}\frac{dR}{dr}+PZ\frac{d^{2}R}{dr^{2}}+\frac{RZ}{r^{2}}\frac{d^{2}P% }{d\theta^{2}}+RP\frac{d^{2}Z}{dz^{2}}=0.$

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

 $\frac{1}{Rr}\frac{dR}{dr}+\frac{1}{R}\frac{d^{2}R}{dr^{2}}+\frac{1}{Pr^{2}}% \frac{d^{2}P}{d\theta^{2}}+\frac{1}{Z}\frac{d^{2}Z}{dz^{2}}=0.$

“Separating” the z term to the other side gives

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

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

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

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

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

Separate the terms

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

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

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

 $\frac{d^{2}Z}{dz^{2}}+Z\lambda=0$
 $C(r)=r^{2}+\lambda=0$
 $r=\pm\sqrt{-\lambda}.$

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: $\lambda\leq 0$ 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

 $h_{1}(z)=C_{1}e^{\gamma z}$
 $h_{2}(z)=C_{2}e^{-\gamma z}$
 $h_{3}(z)=C_{3}ze^{0}=C_{3}z.$

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

 $h_{4}(z)=C_{5}e^{0}\cos(\gamma z)=C_{5}\cos(\gamma z)$
 $h_{5}(z)=C_{6}e^{0}\sin(\gamma z)=C_{6}\sin(\gamma z).$

Combining these into a general solution yields

 $Z_{\gamma}(z)=C_{5}\cos(\gamma z)+C_{6}\sin(\gamma z)+C_{7}.$

### 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: $\kappa\leq 0$ 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

 $h_{4}(\theta)=C_{5}e^{0}\cos(\nu\theta)=C_{5}\cos(\nu\theta)$
 $h_{5}(\theta)=C_{6}e^{0}\sin(\nu\theta)=C_{6}\sin(\nu\theta).$

Combining these into a general solution

 $P_{\nu}(\theta)=C_{5}\cos(\nu\theta)+C_{6}\sin(\nu\theta)+C_{7}.$

For the first glimpse at simplification, we will note a restriction on $\kappa$ 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.

 $P_{\nu}(\theta)=C_{5}\cos(\nu\theta)+C_{6}\sin(\nu\theta)+C_{7}\,\,\,\,\,\,\,% \,\,\,\,\nu=0,1,2,...$ (9)

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

 $P(\theta)=C_{4}.$

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

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

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

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

so

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

Substituting these relationships into (10) gives us

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

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

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

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

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

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

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

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

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

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

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

 $R_{\nu}(r)=C_{1}I_{\nu}(\gamma r)+C_{2}K_{\nu}(\gamma r)+C_{3}$ (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

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

### 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

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

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

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

 $\Phi\left(r,\theta,z\right)=\sum_{\nu}\sum_{\gamma}\left\{\begin{array}[]{c}e^% {\gamma z}\\ e^{-\gamma z}\end{array}\right.\left\{\begin{array}[]{c}\cos(\nu\theta)\\ \sin(\nu\theta)\end{array}\right.\left\{\begin{array}[]{c}J_{\nu}(\gamma r)\\ Y_{\nu}(\gamma r)\end{array}\right.$ (15)

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

 $\Phi\left(r,\theta,z\right)=\sum_{\nu}\sum_{\gamma}\left\{\begin{array}[]{c}% \cos(\gamma z)\\ \sin(\gamma z)\end{array}\right.\left\{\begin{array}[]{c}\cos(\nu\theta)\\ \sin(\nu\theta)\end{array}\right.\left\{\begin{array}[]{c}I_{\nu}(\gamma r)\\ K_{\nu}(\gamma r)\end{array}\right.$ (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

 $\Phi\left(r,\theta,z\right)=\sum_{\nu}\sum_{\gamma}\left\{\begin{array}[]{c}\,% A_{\nu\gamma}e^{\gamma z}\cos(\nu\theta)J_{\nu}(\gamma r)\\ +B_{\nu\gamma}e^{\gamma z}\cos(\nu\theta)Y_{\nu}(\gamma r)\\ +C_{\nu\gamma}e^{\gamma z}\sin(\nu\theta)J_{\nu}(\gamma r)\\ +D_{\nu\gamma}e^{\gamma z}\sin(\nu\theta)Y_{\nu}(\gamma r)\\ +E_{\nu\gamma}e^{-\gamma z}\cos(\nu\theta)J_{\nu}(\gamma r)\\ +F_{\nu\gamma}e^{-\gamma z}\cos(\nu\theta)Y_{\nu}(\gamma r)\\ +G_{\nu\gamma}e^{-\gamma z}\sin(\nu\theta)J_{\nu}(\gamma r)\\ +H_{\nu\gamma}e^{-\gamma z}\sin(\nu\theta)Y_{\nu}(\gamma r)\\ \par \end{array}\right\}.$ (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 LaplaceEquationInCylindricalCoordinates 2013-03-22 16:24:24 2013-03-22 16:24:24 bloftin (6104) bloftin (6104) 13 bloftin (6104) Derivation msc 35J05 BesselFunction BesselsEquation