Electrostatic Field Problems:
Cylindrical Symmetry
EE 141 Lecture Notes
Topic 6
Professor K. E. Oughstun
School of Engineering
College of Engineering & Mathematical Sciences
University of Vermont
2014

Cylindrical Polar Coordinates
Coordinate Transformations x = r cos φ, y = r sin φ, z = z , r = p x 2 + y 2 , φ = arctan ( y / x ) , z = z ,
(1)
(2) with 0 ≤ r ≤ ∞ , − π < φ ≤ π , and −∞ ≤ z ≤ ∞ .

Unit Basis Vectors in Cylindrical Coordinates
Unit Basis vectors
1
1
1 r
φ z
1 r
( φ ) = ˆ x cos φ + ˆ y sin φ,
1
φ
( φ ) = −
ˆ x sin φ + ˆ y cos φ,
1 z where (Orthogonality Relations)
(3)
1 r
× 1
φ
1 z
,
1
φ
× 1 z
1 r
,
1 z
× 1 r
1
φ
.
(4)
Problem 14. Using the relations in Eq. (3), determine expressions for
1 x
1 y
1 r
1
φ
.

Vectors in Cylindrical Coordinates
Any vector V may be expressed in cylindrical polar coordinates as
V = ˆ v
V = ˆ r
V r
1
φ
V
φ
1 z
V z
, (5) where
V r
1 r
· V = 1 x cos φ + ˆ y sin φ ·
= V x cos φ + V y sin φ,
V
φ
1
φ
· V = −
ˆ x sin φ + ˆ y cos φ ·
1 x
V x
1 x
V x
V z
1 z
· V
= − V x sin φ + V y cos φ,
= ˆ z
· 1 x
V x
1 y
V y
1 z
V z
= V
1 y
V y z
,
1 y
V y
1 z
V z
1 z
V z
(6) with magnitude
V = | V | =
√
V · V = q
V r
2 + V 2
φ
+ V z
2 .
(7)

Position Vector in Cylindrical Coordinates
Position Vector of a point P = P ( r , φ, z ) is given by
R =
~
= ˆ r
( φ ) r + ˆ z z , where r = R r
1 r
· R = 1 x cos φ + ˆ y sin φ ·
= x cos φ + y sin φ.
1 x x + ˆ y y + ˆ z z
(8)
(9)
Although the position vector in cylindrical coordinates does not have
1
φ
-component, it does indeed depend upon the azimuthal angle φ through the φ -dependence of the radial unit vector
1 r
1 r
( φ ).

Cylindrical Coordinates - Differential Elements
Differential elements of length along the ˆ r
1
φ
1 z
- directions: d ℓ r
= dr , d ℓ
φ
= rd φ, d ℓ z
= dz .

Differential Length, Surface Area, & Volume
The vector differential element of length: d ~ℓ = ˆ r dr + ˆ
φ rd φ 1 z dz .
Fundamental quadratic form or metric form: d ℓ 2 = dr 2 + r 2 d φ 2 + dz 2 .
Differential elements of surface area: r
φ z -Cylindrical Surface: d s r
= ˆ r
( d ℓ
φ d ℓ z
) = ˆ r
( rd φ dz ).
rz -Planar Surface:
φ -Planar Surface: d s
φ d s z
Differential element of volume:
1
φ
( d ℓ r
1 z
( d ℓ r d ℓ z d ℓ
φ
1
φ
( drdz ).
1 z
( rdrd φ ).
dV = d ℓ r d ℓ
φ d ℓ z
= rdrd φ dz .
(10)
(11)
(12)

Distance between Two Points
Coordinates of P
1
( x
1
, y
1
, z
1
) = P
1
( r
1
, φ
1
, z
1
): x
1
= r
1 cos φ
1
, y
1
= r
1 sin φ
1
, z
1
= z
1
.
Coordinates of P
2
( x
2
, y
2
, z
2
) = P
2
( r
2
, φ
2
, z
2
): x
2
= r
2 cos φ
2
, y
2
= r
2 sin φ
2
, z
2
= z
2
.
Distance between P
1
& P
2 is then given by Pythagorean’s theorem as d = ( r
2
= r 2
1 cos φ
+ r 2
2
2
− r
1 cos φ
1
) 2 + ( r
2 sin φ
2
− r
1 sin φ
1
) 2
− 2 r
1 r
2 cos ( φ
2
− φ
1
) + ( z
2
− z
1
) 2
1 / 2
.
+ ( z
2
− z
1
) 2
1 / 2
(13)

Gradient Operator in Cylindrical Coordinates
Cylindrical Polar Coordinates r = p x 2 + y 2 , φ = arctan ( y / x ) , z = z , so that
∂ r
∂ x
∂ r
∂ y
=
=
( x 2 x
+ y 2 ) 1 / 2 y
( x 2 + y 2 ) 1 / 2
= cos
= sin
φ,
φ,
∂φ
∂ x
∂φ
= −
1 r
∂ y
= sin φ, r
1 cos φ.
By the Chain Rule
∂ f
∂ x
∂
∂ f y
=
=
∂
∂ r f ∂
∂ r x
+
∂ f ∂φ
∂φ ∂ x
+
∂ f ∂ z
∂ z ∂ x
= cos φ
∂ f
∂ r
− sin φ r
∂ f
∂φ
,
∂ f ∂ r
∂ r
∂ f
∂ y
=
∂ z
+
∂
∂ f
∂ r f
∂φ
∂φ
∂ y
∂ r
∂ z
+
+
∂ f
∂
∂
∂φ f z
∂
∂
∂φ
∂ z z y
+
= sin φ
∂ f
∂ r
∂
∂ f z
∂
∂ z z
=
+
∂ f
∂ z
.
cos φ r
∂ f
∂φ
,

Gradient Operator in Cylindrical Coordinates
With the solution to Problem 12, the gradient of a scalar function f ( r ) = f ( x , y , z ) = f ( r , φ, z ) may be expressed in cylindrical polar coordinates as:
∇ f ( r ) = ˆ x
∂ f
∂ x
1 y
∂ f
∂ y
1 z
∂ f
∂ z
= 1 r cos φ − 1
φ sin φ cos φ
∂ f
∂ r
− sin φ r
∂ f
∂φ
+
1 r
∂ f
∂ r
1 r sin φ + ˆ
φ cos φ sin φ
∂ f
∂ r
1
φ
1 r
∂ f
∂φ
1 z
∂ f
∂ z
.
+ cos φ ∂ f r ∂φ
1 z
∂ f
∂ z
(14)
The gradient operator in cylindrical polar coordinates is then given by
∂
∇ = ˆ r
∂ r
1
φ r
1 ∂
∂φ
∂
1 z
∂ z
(15)

Divergence, Curl, & Laplacian Operators in
Cylindrical Polar Coordinates
Vector function of position F ( r ) = F ( r , φ, z ) = ˆ r
F r has divergence
∇ · F = r
1 ∂
∂ r
( rF r
) +
1 ∂ F
φ r ∂φ
+
∂ F z
∂ z and curl
1
φ
F
φ
1 z
F z
(16)
∇ × F =
1 r r
∂ F
∂φ z
−
∂ ( rF
φ
)
∂ z
1
φ
∂
∂
F z r
−
∂ F
∂ r z
+
1 r z
∂ ( rF
φ
)
∂ r
−
∂ F r
∂φ
(17)
Scalar function of position f ( r ) = f ( r , φ, z ) has Laplacian
∇ 2 f =
1 r
∂
∂ r r
∂
∂ f r
+
1 ∂ 2 f r 2 ∂φ 2
∂ 2 f
+
∂ z 2
(18)

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
Consider a uniform, infinitely extended charge density ̺ in a cylinder of radius r
0
, as illustrated.
With the z -axis of a cylindrical coordinate system taken along the cylinder axis, cylindrical symmetry requires that
E ( r ) = ˆ r
E ( r ) .

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
Application of Gauss’ law to a coaxial cylindrical surface S of radius r and axial length ℓ yields
I
S
E · ˆ da =
1 ǫ
0
Z Z Z
V
̺ d 3 r =
̺ ǫ
0
π r
π r 2
0
2 ℓ ; r ≤ r
0 ℓ ; r ≥ r
0 with
I
S
E · ˆ da = E ( r )
Z Z cyl
1 r
· 1 r d 2 r = 2 π r ℓ E ( r ) , the integration over the two cylinder ends vanishing because
E = ˆ r
E ( r ) and ˆ are, by construction, orthogonal on those surfaces.
Combination of these results then gives
̺
E ( r ) = ˆ r
2 ǫ
0
( r r r ; r ≤ r
0
2
0 ; r ≥ r
0

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
The absolute potential [Eq. (4.8)] produced by the uniform cylindrical charge distribution of radius r
0 is then given by (with d
~ℓ
= ˆ r dr )
Z
∞
V ( r ) = r
1 r
E ( r ) · 1 r dr =
Z
∞ r
E ( r ) dr .
For r ≥ r
0 one obtains
V ( r ) =
̺ r 2
0
2 ǫ
0
Z
∞ r dr r
=
̺ r 2
0
2 ǫ
0 ln ( ∞ ) − ln ( r ) and for r ≤ r
0 one obtains
V ( r ) =
2
̺ ǫ
0 r 2
0
Z
∞ r
0 dr r
+
Z r r
0 rdr =
̺ r
2 ǫ
2
0
0 ln ( ∞ ) − ln ( r
0
)+
1
2 r 2
1 − r 2
0
The divergent nature of these expressions is simply a manifestation of the nonphysical character of the source charge distribution which extends to ±∞ along the z -axis.

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
Because of the arbitrariness in choosing the reference potential, the constant term ln ( ∞ ) appearing in these expressions may be replaced by some other constant. For convenience, this constant is chosen such that the potential V ( r ) vanishes at r = r
0
(the surface of the cylindrical charge distribution), in which case the potential becomes
V ( r ) =
(
−
̺ r
2
0
2 ǫ
0
̺ r
4 ǫ
2
0
0 ln ( r / r
0
) ; r ≥ r
0
1 − r r
2
2
0
; r ≤ r
0
In the limit as r
0
→ 0 the cylindrical volume charge distribution goes over to a line charge with density (charge per unit length) ̺ ℓ
= π r 2
0
̺ .
The electrostatic field intensity then becomes
E ( r ) = ˆ r
̺ ℓ
2 πǫ
0 r with absolute potential V ( r ) = ( ̺ ℓ
/ 2 πǫ
0
) ln ( ∞ ) − ln ( r ) .

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
ρ r
0
2
/
4 ε
0
ρ r
0
/
2 ε
0
V(r)
E(r) r 0 r
0
2r
0
−ρ r
0
2
/ 4 ε
0

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
This problem can also be solved using Poisson’s & Laplace’s equations as follows:
• For r ≤ r
0
, Poisson’s equation ∇ 2 coordinates becomes
V ( r ) = − ̺/ǫ
0 in cylindrical r
1
∂
∂ r r
∂ V
∂ r
̺
= − ǫ
0
= ⇒
∂
∂ r r
∂ V
∂ r
̺
= − ǫ
0 r which is integrated to yield r
∂
∂
V r
̺
= −
2 ǫ
0 r 2 + A = ⇒
∂
∂
V r
̺
= −
2 ǫ
0 r + which in turn is integrated to yield
̺
V ( r ) = −
4 ǫ
0 r 2 + A ln ( r ) + B
A r where A and B are constants.
Because V ( r ) must remain finite at r = 0, then A must equal zero.

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
Hence
̺
V ( r ) = −
4 ǫ
0 r 2 + B ; r ≤ r
0
.
• For r ≥ r
0
, Laplace’s equation ∇ 2 coordinates becomes
V ( r ) = 0 in cylindrical
∂
∂ r r
∂ V
∂ r
= 0 = ⇒
∂ V
∂ r
=
C r where C is a constant, which in turn is integrated to yield
V ( r ) = C ln ( r ) − ln ( ∞ ) ; r ≥ r
0 with the condition that V ( r ) = 0 at r = ∞ .

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
Continuity of ∂ V /∂ r at r = r
0 requires that
̺
−
2 ǫ
0 r
0
=
C r
0
̺
= ⇒ C = −
2 ǫ
0 r 2
0
Continuity of V ( r ) at r = r
0 requires that
̺
−
4 ǫ
0 r 2
0
+ B = C ln ( r
0
) − ln ( ∞ ) = ⇒ B =
̺ r 2
0
4 ǫ
0
1 − 2 ln ( r
0
) − ln ( ∞ )
Hence
V ( r ) =
̺
4 ǫ
0 r 2
0
− r 2
V ( r ) = −
̺ r 2
0
2 ǫ
0
−
̺ r 2
0
2 ǫ
0 ln ( r ) − ln ( r ln (
0
)
∞
−
) ; ln ( ∞ r
)
≥
; r
0 r ≤ r
0

E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
The electrostatic field is then given by the negative gradient of the potential as E = −∇ V , so that for r ≤ r
0
E ( r ) = −
ˆ r
∂
∂ r
̺
4 ǫ
0 r 2
0
− r 2 −
̺ r 2
0
2 ǫ
0 ln ( r
0
) − ln ( ∞ ) = ˆ r
̺
2 ǫ
0 r and for r ≥ r
0
E ( r ) = −
ˆ r
∂
∂ r
−
̺ r 2
0
2 ǫ
0 ln ( r ) − ln ( ∞ ) = ˆ r
̺ r 2
0
2 ǫ
0 r in agreement with the result obtained using Gauss’ law.

On-Axis E-Field of a Circular Charged Annulus
Consider a thin charged ring of radius b with uniform line density
̺ ℓ
= Q / 2 π b , situated in the xy -plane with center at the origin O .
The electrostatic field along the z -axis may then be determined using
Coulomb’s law in the following manner.
Because of the symmetry of any two elements 1 and 2 on opposite sides of the ring, the electrostatic field along the z -axis is directed along the z -axis away from the xy -plane.

On-Axis E-Field of a Circular Charged Annulus
The vector from the charged element 1 to the point P = (0 , 0 , z ) along the z -axis is given by
R
′
1
= −
ˆ r b + ˆ z z in polar cylindrical coordinates, with unit vector
R
′
1
=
−
ˆ r b + ˆ z z
( b 2 + z 2 )
1 / 2
.
The resultant differential field contribution from this differential line charge element 1 is then given by Coulomb’s law as d E
1
R
′
1
1
4 πǫ
0
̺ ℓ
R ′ 2
1 d ℓ =
̺ ℓ b
4 πǫ
0
( −
ˆ r
( b 2 b
+
+ ˆ z 2 ) z z )
3 / 2 d φ, while from element 2 d E
2
R
′
2
1
4 πǫ
0
̺ ℓ
R ′ 2
2 d ℓ =
̺ ℓ b
4 πǫ
0
1
( b 2 r b + ˆ
+ z 2 ) z z )
3 / 2 d φ.

On-Axis E-Field of a Circular Charged Annulus
The total differential field contribution from differential line charge elements on opposite sides of the ring is then given by d E = d E
1
+ d E
2
̺ ℓ bz
1 z
2 πǫ
0
( b 2 d φ
+ z 2 )
3 / 2
.
The total electrostatic field is then obtained by integrating around the ring from φ = 0 to φ = π as
̺
E (0 , 0 , z ) = ˆ z
2 πǫ
0
( b 2 ℓ bz
+ z 2 )
3 / 2
1 z
2 ǫ
0
( b
̺ ℓ bz
2 + z 2 )
3 / 2
1 z
4 πǫ
0
Qz
( b 2 + z 2 )
3 / 2
.
Z
π
0 d φ
Notice that E (0 , 0 , 0) = 0 and that E (0 , 0 , z ) → ± z → ±∞ , respectively.
1 z
Q / (4 πǫ
0 z 2 ) as

On-Axis Potential of a Circular Charged Annulus
The absolute potential produced by the circular charged annulus is given by Coulomb’s law as
V (0 , 0 , z ) =
=
1
4 πǫ
0
Z
0
2 π ̺ ℓ
( b 2 + z 2 )
1 / 2 bd φ
2 ǫ
0
̺ ℓ b
( b 2 + z 2 )
1 / 2
.
The electrostatic field along the z -axis is then given by
E (0 , 0 , z ) = −∇ V (0 , 0 , z ) = −
ˆ z
̺ ℓ b
2 ǫ
0
̺ ℓ bz
1 z
2 ǫ
0
( b 2 + z 2 )
3 / 2
∂
∂ z
1 z
4 πǫ
0
( b
Qz
2 + z 2 )
3 / 2
, in agreement with the previous result.
b 2 + z 2
− 1 / 2

On-Axis E-Field of a Circular Charged Disk
The on-axis field due to a uniformly charged disk of radius a and surface charge density ̺ s can then be obtained from the expression for the on-axis field due to a circular charged annulus with the associations
Q → dq = ̺ s d a = 2 π̺ s rdr b → r

On-Axis E-Field of a Circular Charged Disk
The differential contribution to the on-axis field from the annular element of the disk is then given by d E = ˆ z
4 πǫ
0
( r 2 z
+ z 2 )
3 / 2
2 π̺ s rdr , so that
E (0 , 0 , z ) = ˆ z
̺ s z
2 ǫ
0
Z a
0 rdr
( r 2 + z 2 )
3 / 2
= ±
ˆ z
̺ s
2 ǫ
0
1 −
( a 2
| z |
+ z 2 )
1 / 2
!
where the + sign applies when z > 0 and the − sign when z < 0.
In the limit as a → ∞ , the circular disk becomes a plane sheet of charge and the above result becomes E → ± 1 z
̺ s
/ 2 ǫ
0
.

Take Home Exam Problem 1
An infinitely extended cylindrical region of radius a > 0 situated in free space contains a volume charge density given by
̺ ( r ) = ̺
0
1 + α r 2 ; r ≤ a, with ̺ ( r ) = 0 for r > a , where ̺
0
1 and α are constants.
Utilize Gauss’ law together with the inherent symmetry of the problem to derive the resulting electrostatic field vector E ( r ) both inside and outside the cylinder.
2
3
Use both Poisson’s and Laplace’s equations to directly determine the electrostatic potential V ( r ) both inside and outside the cylindrical region. From this potential function, determine the electrostatic field vector E ( r ).
Determine the value of the parameter α for which the electrostatic field vanishes everywhere in the region outside the cylinder ( r > a ). Plot E r value of α .
( r ) and V ( r ) as a function of r for this

Boundary Value Problems Cylindrical Coordinates
In cylindrical polar coordinates ( r , ϕ, z ) defined by the transformation equations x = r cos ϕ , y = r sin ϕ , z = z with r ∈ [0 , ∞ ), ϕ ∈ [0 , 2 π ), and z ∈ ( −∞ , + ∞ ), Laplace’s equation ∇ 2 φ = 0 assumes the form r
1
∂
∂ r r
∂φ
∂ r
+
1 ∂ 2 φ r 2 ∂ϕ 2
∂ 2 φ
+
∂ z 2
= 0 .
(19)
The special case when the potential has either longitudinal invariance alone [ φ = φ ( r , ϕ ) is independent of z ] or both longitudinal and axial invariance [ φ = φ ( r ) is a function of r alone] has been considered in
Topic 5 and is not pursued any further here.

Boundary Value Problems Cylindrical Coordinates
For the general case where φ = φ ( r , ϕ, z ), Laplace’s equation admits a separated solution of the form φ ( r , ϕ, z ) = R ( r ) Q ( ϕ ) Z ( z ), so that
1 d 2 R 1
+ rR dR dr
1
+ r 2 Q d 2 Q d ϕ 2
1
= −
Z d 2 Z dz 2
= − k 2 , (20)
R dr 2 where k 2 is a separation constant. The ode for Z ( z ) has elementary solutions Z ( z ) = e ± kz . The remaining part of Eq. (20) is r
R
2 d 2 R dr 2 r
+
R dR dr
+ k 2 r 2
1
= −
Q d 2 Q d ϕ 2
= ν 2 , (21) where ν 2 is another separation constant. The ode for Q ( ϕ ) has elementary solutions Q ( ϕ ) = e ± i νϕ . In order that the potential be single-valued when the domain of interest covers the full azimuthal range from 0 to 2 π , the separation constant ν must be an integer.
However, unless some boundary condition is imposed in the z -direction, the separation constant k is arbitrary. For the present, it is assumed that k is real and positive.

Boundary Value Problems Cylindrical Coordinates
The remaining radial part of Eq. (21) may be written as d 2 R dr 2
+ r
1 dR dr
+ k 2 −
ν 2 r 2
R = 0 .
Under the change of variable x = kr , this equation assumes the standard mathematical form
(22) d 2 R dx 2
1 dR
+ x dx
ν 2
+ 1 − x 2
R = 0 (23) which is Bessel’s equation (first defined by Daniel Bernoulli).

Friedrich Wilhelm Bessel (1784–1846)

Boundary Value Problems Cylindrical Coordinates
• Bessel functions of the first kind of order ± ν with series representation (obtained by the method of Frobenius)
J
± ν
( x ) = x
2
± ν
∞
X j =0
( − 1) j j !Γ( j + 1 ± ν ) x
2
2 j
(24)
• Bessel functions of the second kind of order ν (Weber, Neumann)
Y
ν
( x ) =
J
ν
( x ) cos ( νπ ) − J
− ν
( x ) sin ( νπ )
(25) where the right-hand side of this equation becomes indeterminate and is replaced by its limiting value an integer or zero,
Y m
( x ) = lim
ν → m
Y
ν
( x ) when ν is
• Bessel functions of the third kind (Hankel functions)
H ( j )
ν
= J
ν
( x ) + ( − 1) j iY
ν
( x ) (26) for j = 1 , 2.

Boundary Value Problems Cylindrical Coordinates
It follows from l’Hˆopital’s rule that when ν is an integer or zero,
Y m
( x ) = lim
ν → m
( ∂
∂ν
=
1
π lim
ν → m cos ( νπ ) J
ν
( x ) − J
− ν
( x )
π cos ( νπ )
)
∂ J
ν
( x )
∂ν
− ( − 1) m
∂ J
− ν
( x )
∂ν
, which then leads to the series representations
Y
0
( x ) =
2
π
( h ln x
2
+ γ i
J
0
( x ) −
∞
X j =1
( − 1) j
( j !) 2 x
2
2 j j
X
1
)
, (27) n n =1

Boundary Value Problems Cylindrical Coordinates
π Y m
( x ) = 2 h ln
−
− x
2 x
2
+ γ i
J m
( x ) x
2
− m m − 1
X
( m − j − 1)!
j !
j =0 m
∞
X
( − 1) j j !( m + j )!
j =1 x
2 x
2
2 j
− x
2 m 1 m !
m
X
1 n =1 n
2 j
" j
X
2 n =1 n
+ m + j
X n = j +1
1
#
, n
(28) where m > 0 is an integer, and where
γ ≡ lim n →∞
1 +
1
2
+
1
3
+ · · · +
1 n
− ln ( n ) is Euler’s constant .
(29)

Boundary Value Problems Cylindrical Coordinates
Because of the symmetry relations J
− m
Y
− m
( x ) = ( − 1) m Y m
( x ) = ( − 1) m J m
( x ) and
( x ) when m is an integer, Eqs. (24) and (28) can then be used for all positive and negative integer indices.
The linear independence of the Bessel functions J m
( x ) and Y m
( x ) of the first and second kind of integer order m follows from their respective limiting behaviors as x → 0. For m = 0, J
0
( x ) → 1 while
Y
0 m while
Y m
( x ) → −∞ as ln( x / 2). When m > 0, J m
( x ) → 0 as x
( x ) → −∞ as − x − m . Therefore, the general solution of Bessel’s equation (23) when ν = m is an integer has the form
R ( x ) = AJ m
( x ) + BY m
( x ) , where A and B are arbitrary constants. It is then seen that the
Hankel functions H
(1)
ν
( x ) and H
(2)
ν
( x ) defined in Eq. (26) form a linearly independent set of solutions to Bessel’s equation for all ν .
(30)

Boundary Value Problems Cylindrical Coordinates
Notice that the major difference between Bessel’s differential equation (23) and the simple harmonic equation for the sinusoidal functions cos x and sin x is due to the term 1 x dR dx which has a significant impact on the behavior of the solution as x → 0.
However, Bessel’s equation becomes similar to the simple harmonic equation as x → ∞ . In particular, the Bessel functions of the first kind of order ν possess the asymptotic approximations
J
ν
( x ) ∼
Y
ν
( x ) ∼ r
2
π x r
2
π x cos x − ν
π sin x − ν
π
2
2
−
−
π
4
π
4
,
, (31)
(32) as x → ∞ . The transition from the small x behavior given by the first few terms of Eqs. (24) and (28) to the large x asymptotic behavior given in Eqs. (31) and (32), respectively, occurs in the region x ∼ ν .

Bessel Functions of the First & Second Kind of
Integer Order m

Boundary Value Problems Cylindrical Coordinates
The alternate choice of separation constant in Eq. (20) is + k 2 leads to the equation d 2 Z / dz 2 + k 2 Z which
= 0 with elementary solutions
Z ( z ) = e ± ikz . The resultant separated ode for Q ( ϕ ) remains unaltered with this change so that the elementary solutions
Q ( ϕ ) = e ± i νφ still apply with ν an integer when ϕ covers the full azimuthal domain from 0 to 2 π . The radial part of the equation then becomes [cf. Eq. (22)] d 2 R dr 2
+ r
1 dR dr
− k 2 +
ν 2 r 2
R = 0 .
(33)
The particular solutions of this equation are seen to be the Bessel functions J
ν
( ikr ) and Y
ν
( ikr ) of an imaginary argument. Under the change of variable x = kr this equation assumes the standard form d 2 r dx 2
1 dR
+ x dx
− 1 +
ν 2 x 2
R = 0 (34)

Boundary Value Problems Cylindrical Coordinates
Real-valued solutions of this equation are the modified Bessel functions defined as
I
ν
( x ) ≡ e − i νπ/ 2 J
ν
( ix ) = x
2
ν
∞
X
1 j !Γ( j + ν + 1) j =0 x
2
2 j where I
− n
( x ) = I n
( x ) with n an integer, and
K
ν
( x ) ≡
π
2
I
− ν
( x ) − I
ν
( x ) sin ( νπ ) where K
− ν
( x ) = K
ν
( x ).
(35)
(36)

Boundary Value Problems Cylindrical Coordinates
The limiting forms of the modified Bessel functions as x → 0 are given by I
ν
( x ) → 1 ( x , K
0
), and
K
ν
Γ( ν )( 2
Γ( ν +1)
) ν
2
) ν for ν > 0.
( x ) → − I
0
( x ) ln ( x
2
( x ) → 1
2 x
In addition, their large argument asymptotic approximations are given by
I
ν
( x ) ∼
K
ν
( x ) ∼
√
1
2 π x e x , r π
2 x e − x ,
(37)
(38) as x → ∞ . It is then seen that the modified Bessel functions I
ν
( r ) are appropriate for boundary value problem solutions that remain bounded at r = 0, while the modied Bessel functions K
ν
( r ) are appropriate for solutions that remain bounded as r → ∞ .

Boundary Value Problems Cylindrical Coordinates
The elementary solutions of Laplace’s equation in cylindrical coordinates are then given in separated form as
φ
ν n
( r , ϕ, z ) = R
ν n
( r ) Q
ν
( ϕ ) Z
ν n
( z ) , (39) where the indices appearing in each of the separated solutions are related to the separation constants introduced in obtaining this elementary solution. Both the solution domain and the imposed boundary conditions then determine the appropriate from of each separated solution as well as the allowed values of the indices ν and n . The full solution of Laplace’s equation in the specified solution domain that satisfies all of the boundary conditions is then obtained through an appropriate superposition of these selected elementary solutions as
φ ( r , ϕ, z ) =
X X
A
ν n
φ
ν n
( r , ϕ, z ) , (40) where the coefficients conditions.
A
ν n
ν n are uniquely determined by the boundary