Laplace equation in cylindrical coordinates

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

∇

2

⁡

Φ

=

1

r

⁢

∂

∂

⁡

r

⁢

(

r

⁢

∂

⁡

Φ

∂

⁡

r

)

+

1

r

2

⁢

∂

2

⁡

Φ

∂

⁡

θ

2

+

∂

2

⁡

Φ

∂

⁡

z

2

=

0

.

(1)

First expand out the terms

∇

2

⁡

Φ

=

1

r

⁢

∂

⁡

Φ

∂

⁡

r

+

∂

2

⁡

Φ

∂

⁡

r

2

+

1

r

2

⁢

∂

2

⁡

Φ

∂

⁡

θ

2

+

∂

2

⁡

Φ

∂

⁡

z

2

=

0

.

(2)

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

Φ

⁢

(

r

,

θ

,

z

)

=

R

⁢

(

r

)

⁢

P

⁢

(

θ

)

⁢

Z

⁢

(

z

)

.

Plug this into (2) and note how we can bring out the

functions

that are not affected by the derivatives

P

⁢

Z

r

⁢

d

⁢

R

d

⁢

r

+

P

⁢

Z

⁢

d

2

⁢

R

d

⁢

r

2

+

R

⁢

Z

r

2

⁢

d

2

⁢

P

d

⁢

θ

2

+

R

⁢

P

⁢

d

2

⁢

Z

d

⁢

z

2

=

0

.

Divide by

R

⁢

(

r

)

⁢

P

⁢

(

θ

)

⁢

Z

⁢

(

z

)

and use short hand notation to get

1

R

⁢

r

⁢

d

⁢

R

d

⁢

r

+

1

R

⁢

d

2

⁢

R

d

⁢

r

2

+

1

P

⁢

r

2

⁢

d

2

⁢

P

d

⁢

θ

2

+

1

Z

⁢

d

2

⁢

Z

d

⁢

z

2

=

0

.

“Separating” the z term to the other side gives

1

R

⁢

r

⁢

d

⁢

R

d

⁢

r

+

1

R

⁢

d

2

⁢

R

d

⁢

r

2

+

1

P

⁢

r

2

⁢

d

2

⁢

P

d

⁢

θ

2

=

-

1

Z

⁢

d

2

⁢

Z

d

⁢

z

2

.

This equation can only be satisfied for all values if both sides are equal to a constant,

λ

, such that

-

1

Z

⁢

d

2

⁢

Z

d

⁢

z

2

=

λ

(3)

1

R

⁢

r

⁢

d

⁢

R

d

⁢

r

+

1

R

⁢

d

2

⁢

R

d

⁢

r

2

+

1

P

⁢

r

2

⁢

d

2

⁢

P

d

⁢

θ

2

=

λ

.

(4)

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

r

2

r

R

⁢

d

⁢

R

d

⁢

r

+

r

2

R

⁢

d

2

⁢

R

d

⁢

r

2

+

1

P

⁢

d

2

⁢

P

d

⁢

θ

2

=

λ

⁢

r

2

.

Separate the terms

r

R

⁢

d

⁢

R

d

⁢

r

+

r

2

R

⁢

d

2

⁢

R

d

⁢

r

2

-

λ

⁢

r

2

=

-

1

P

⁢

d

2

⁢

P

d

⁢

θ

2

.

As before, set both sides to a constant,

κ

-

1

P

⁢

d

2

⁢

P

d

⁢

θ

2

=

κ

(5)

r

R

⁢

d

⁢

R

d

⁢

r

+

r

2

R

⁢

d

2

⁢

R

d

⁢

r

2

-

λ

⁢

r

2

=

κ

.

(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

d

2

⁢

Z

d

⁢

z

2

+

Z

⁢

λ

=

0

is to find the roots of the

characteristic polynomial

C

⁢

(

r

)

=

r

2

+

λ

=

0

r

=

±

-

λ

.

Although, one can go forward using the

square root

, here we will introduce another constant,

γ

to imply
the following cases. So if we want real roots, then we want to ensure a

negative

constant

λ

=

-

γ

2

and if we want

complex roots

, then we want to ensure a positive constant

λ

=

γ

2

,

Case 1

:

λ

≤

0

and real roots

(

λ

=

-

γ

2

)

.

For every real root, there will be an

exponential

in the

general solution

. The real roots are

r

1

=

γ

r

2

=

-

γ

r

3

=

0

.

Therefore, the solutions for these roots are

h

1

⁢

(

z

)

=

C

1

⁢

e

γ

⁢

z

h

2

⁢

(

z

)

=

C

2

⁢

e

-

γ

⁢

z

h

3

⁢

(

z

)

=

C

3

⁢

z

⁢

e

0

=

C

3

⁢

z

.

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

Z

γ

⁢

(

z

)

=

C

1

⁢

e

γ

⁢

z

+

C

2

⁢

e

-

γ

⁢

z

+

C

3

⁢

z

+

C

4

.

(7)

Case 2

:

λ

>

0

and complex roots

(

λ

=

γ

2

)

.

The roots are

r

4

=

i

⁢

γ

r

5

=

-

i

⁢

γ

and the corresponding solutions

h

4

⁢

(

z

)

=

C

5

⁢

e

0

⁢

cos

⁡

(

γ

⁢

z

)

=

C

5

⁢

cos

⁡

(

γ

⁢

z

)

h

5

⁢

(

z

)

=

C

6

⁢

e

0

⁢

sin

⁡

(

γ

⁢

z

)

=

C

6

⁢

sin

⁡

(

γ

⁢

z

)

.

Combining these into a general solution yields

Z

γ

⁢

(

z

)

=

C

5

⁢

cos

⁡

(

γ

⁢

z

)

+

C

6

⁢

sin

⁡

(

γ

⁢

z

)

+

C

7

.

1.3

Azimuthal Solutions (

θ

)

Azimuthal solutions for

d

2

⁢

P

d

⁢

θ

2

+

κ

⁢

P

=

0

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

C

⁢

(

r

)

=

r

2

+

κ

=

0

r

=

±

-

κ

.

Using another constant,

ν

to ensure positive or negative constants, we get two cases.

Case 1

:

κ

≤

0

and real roots

(

κ

=

-

ν

2

)

.

The solutions for these roots are then

h

1

⁢

(

z

)

=

C

1

⁢

e

ν

⁢

θ

h

2

⁢

(

z

)

=

C

2

⁢

e

-

ν

⁢

θ

h

3

⁢

(

z

)

=

C

3

⁢

θ

⁢

e

0

=

C

3

⁢

θ

.

Combining these for the general solution,

P

ν

⁢

(

θ

)

=

C

1

⁢

e

ν

⁢

θ

+

C

2

⁢

e

-

ν

⁢

θ

+

C

3

⁢

θ

+

C

4

.

(8)

Case 2

:

κ

>

0

and complex roots

(

κ

=

ν

2

)

.

The roots are

r

4

=

i

⁢

ν

r

5

=

-

i

⁢

ν

and the corresponding solutions

h

4

⁢

(

θ

)

=

C

5

⁢

e

0

⁢

cos

⁡

(

ν

⁢

θ

)

=

C

5

⁢

cos

⁡

(

ν

⁢

θ

)

h

5

⁢

(

θ

)

=

C

6

⁢

e

0

⁢

sin

⁡

(

ν

⁢

θ

)

=

C

6

⁢

sin

⁡

(

ν

⁢

θ

)

.

Combining these into a general solution

P

ν

⁢

(

θ

)

=

C

5

⁢

cos

⁡

(

ν

⁢

θ

)

+

C

6

⁢

sin

⁡

(

ν

⁢

θ

)

+

C

7

.

For the first glimpse at simplification, we will note a

restriction

on

κ

that is used when it is required that the solution be periodic to ensure

P

is single valued

P

⁢

(

θ

)

=

P

⁢

(

θ

+

2

⁢

n

⁢

π

)

.

Then we are left with either the

periodic solutions

that occur with complex roots or the zero case. So not only

(

κ

=

ν

2

)

but also

ν

must be an

integer

, i.e.

P

ν

⁢

(

θ

)

=

C

5

⁢

cos

⁡

(

ν

⁢

θ

)

+

C

6

⁢

sin

⁡

(

ν

⁢

θ

)

+

C

7

ν

=

0

,

1

,

2

,

…

(9)

Note, that

ν

=

0

, is still a solution, but to be periodic we can only have a constant

P

⁢

(

θ

)

=

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

θ

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

:

λ

<

0

(

λ

=

-

γ

2

)

,

κ

>

0

(

κ

=

ν

2

)

.

Substitute

γ

and

ν

into the radial equation (6) to get

r

R

⁢

d

⁢

R

d

⁢

r

+

r

2

R

⁢

d

2

⁢

R

d

⁢

r

2

+

γ

2

⁢

r

2

-

ν

2

=

0

.

(10)

Next, use the substitution

x

=

γ

⁢

r

r

=

x

γ

.

Therefore, the derivatives are

d

⁢

x

=

γ

⁢

d

⁢

r

d

⁢

r

=

d

⁢

x

γ

and make a special note that

d

2

d

⁢

x

2

=

d

d

⁢

x

⁢

d

d

⁢

x

so

d

⁢

x

2

=

d

⁢

x

*

d

⁢

x

=

γ

2

⁢

d

⁢

r

2

d

⁢

r

2

=

d

⁢

x

2

γ

2

.

Substituting these relationships into (10) gives us

x

⁢

γ

γ

⁢

R

⁢

d

⁢

R

d

⁢

x

+

x

2

⁢

γ

2

γ

2

⁢

R

⁢

d

2

⁢

R

d

⁢

x

2

+

x

2

-

ν

2

=

0

.

Finally, multiply by

R

/

x

2

to get the

Bessel differential equation

d

2

⁢

R

d

⁢

x

2

+

1

x

⁢

d

⁢

R

d

⁢

x

+

(

1

-

ν

2

x

2

)

⁢

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

ν

⁢

(

r

)

and the Bessel function of the

second kind

Y

ν

⁢

(

r

)

. Remebering that

ν

is a

positive integer

or zero.

R

ν

⁢

(

r

)

=

C

1

⁢

J

ν

⁢

(

γ

⁢

r

)

+

C

2

⁢

Y

ν

⁢

(

γ

⁢

r

)

+

C

3

(12)

Bessel function of the first kind:

J

ν

⁢

(

x

)

=

∑

m

=

0

∞

(

-

1

)

m

⁢

(

-

1

2

⁢

x

)

ν

+

2

⁢

m

m

!

⁢

(

m

+

ν

)

!

.

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

Y

ν

⁢

(

x

)

=

2

⁢

J

ν

⁢

(

x

)

⁢

(

η

+

l

⁢

n

⁢

(

x

2

)

)

-

(

x

2

)

-

ν

⁢

∑

m

=

0

ν

-

1

(

ν

-

m

-

1

)

!

m

!

⁢

(

x

2

)

2

⁢

m

-

∑

m

=

0

∞

(

-

1

)

m

⁢

(

x

2

)

ν

+

2

⁢

m

m

!

⁢

(

ν

+

m

)

!

⁢

{

1

1

+

1

2

+

…

+

1

m

+

1

1

+

1

2

+

…

+

1

ν

+

m

}

.

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

{

1

1

+

1

2

+

…

+

1

ν

}

.

Some solace can be found since most physical problems need to be analytic at

x

=

0

and therefore

Y

ν

⁢

(

x

)

breaks down at

l

⁢

n

⁢

(

0

)

.
This leads to the choice of constant

C

2

to be zero.

Case 2

:

λ

>

0

(

λ

=

γ

2

)

,

κ

>

0

(

κ

=

ν

2

)

.

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

d

2

⁢

R

d

⁢

x

2

+

1

x

⁢

d

⁢

R

d

⁢

x

-

(

1

+

ν

2

x

2

)

⁢

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

ν

⁢

(

r

)

=

C

1

⁢

I

ν

⁢

(

γ

⁢

r

)

+

C

2

⁢

K

ν

⁢

(

γ

⁢

r

)

+

C

3

(14)

where

I

ν

is the modified Bessel function of the first kind and

K

ν

is the modified Bessel function of the
second kind

I

ν

⁢

(

γ

⁢

r

)

=

i

-

ν

⁢

J

ν

⁢

(

i

⁢

γ

⁢

r

)

K

ν

⁢

(

γ

⁢

r

)

=

π

2

⁢

i

ν

+

1

⁢

(

J

ν

⁢

(

i

⁢

γ

⁢

r

)

+

i

⁢

Y

ν

⁢

(

i

⁢

γ

⁢

r

)

)

.

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

(

κ

=

ν

2

)

.

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

separable

Φ

⁢

(

r

,

θ

,

z

)

=

R

⁢

(

r

)

⁢

P

⁢

(

θ

)

⁢

Z

⁢

(

z

)

.

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

Case 1

:

λ

<

0

(

λ

=

-

γ

2

)

,

κ

>

0

(

κ

=

ν

2

)

.

Φ

⁢

(

r

,

θ

,

z

)

=

∑

ν

∑

γ

{

e

γ

⁢

z

e

-

γ

⁢

z

⁢

{

cos

⁡

(

ν

⁢

θ

)

sin

⁡

(

ν

⁢

θ

)

⁢

{

J

ν

⁢

(

γ

⁢

r

)

Y

ν

⁢

(

γ

⁢

r

)

(15)

Case 2

:

λ

>

0

(

λ

=

γ

2

)

,

κ

>

0

(

κ

=

ν

2

)

.

Φ

⁢

(

r

,

θ

,

z

)

=

∑

ν

∑

γ

{

cos

⁡

(

γ

⁢

z

)

sin

⁡

(

γ

⁢

z

)

⁢

{

cos

⁡

(

ν

⁢

θ

)

sin

⁡

(

ν

⁢

θ

)

⁢

{

I

ν

⁢

(

γ

⁢

r

)

K

ν

⁢

(

γ

⁢

r

)

(16)

Interpreting the short hand notation is as simple as expanding terms and not forgetting the linear solutions, i.e.

(

γ

=

0

)

.
As an example, case 1,

expanded

out while ignoring the linear terms would give

Φ

⁢

(

r

,

θ

,

z

)

=

∑

ν

∑

γ

{

A

ν

⁢

γ

⁢

e

γ

⁢

z

⁢

cos

⁡

(

ν

⁢

θ

)

⁢

J

ν

⁢

(

γ

⁢

r

)

+

B

ν

⁢

γ

⁢

e

γ

⁢

z

⁢

cos

⁡

(

ν

⁢

θ

)

⁢

Y

ν

⁢

(

γ

⁢

r

)

+

C

ν

⁢

γ

⁢

e

γ

⁢

z

⁢

sin

⁡

(

ν

⁢

θ

)

⁢

J

ν

⁢

(

γ

⁢

r

)

+

D

ν

⁢

γ

⁢

e

γ

⁢

z

⁢

sin

⁡

(

ν

⁢

θ

)

⁢

Y

ν

⁢

(

γ

⁢

r

)

+

E

ν

⁢

γ

⁢

e

-

γ

⁢

z

⁢

cos

⁡

(

ν

⁢

θ

)

⁢

J

ν

⁢

(

γ

⁢

r

)

+

F

ν

⁢

γ

⁢

e

-

γ

⁢

z

⁢

cos

⁡

(

ν

⁢

θ

)

⁢

Y

ν

⁢

(

γ

⁢

r

)

+

G

ν

⁢

γ

⁢

e

-

γ

⁢

z

⁢

sin

⁡

(

ν

⁢

θ

)

⁢

J

ν

⁢

(

γ

⁢

r

)

+

H

ν

⁢

γ

⁢

e

-

γ

⁢

z

⁢

sin

⁡

(

ν

⁢

θ

)

⁢

Y

ν

⁢

(

γ

⁢

r

)

}

.

(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

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