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,
p
r = x 2 + y 2 , φ = arctan (y /x), z = z,
with 0 ≤ r ≤ ∞, −π < φ ≤ π, and −∞ ≤ z ≤ ∞.
(1)
(2)
Unit Basis Vectors in Cylindrical Coordinates
Unit Basis vectors
1̂r = 1̂r (φ) = 1̂x cos φ + 1̂y sin φ,
1̂φ = 1̂φ (φ) = −1̂x sin φ + 1̂y cos φ,
(3)
1̂z = 1̂z
where (Orthogonality Relations)
1̂r × 1̂φ = 1̂z ,
1̂φ × 1̂z = 1̂r ,
(4)
1̂z × 1̂r = 1̂φ .
Problem 14. Using the relations in Eq. (3), determine expressions for
the unit vectors 1̂x and 1̂y in terms of the unit vectors 1̂r and 1̂φ .
Vectors in Cylindrical Coordinates
Any vector V may be expressed in cylindrical polar coordinates as
V = 1̂v V = 1̂r Vr + 1̂φ Vφ + 1̂z Vz ,
(5)
where
Vr = 1̂r · V =
1̂x cos φ + 1̂y sin φ · 1̂x Vx + 1̂y Vy + 1̂z Vz
= Vx cos φ + Vy sin φ,
Vφ = 1̂φ · V = −1̂x sin φ + 1̂y cos φ · 1̂x Vx + 1̂y Vy + 1̂z Vz
= −Vx sin φ + Vy cos φ,
Vz = 1̂z · V = 1̂z · 1̂x Vx + 1̂y Vy + 1̂z Vz = Vz ,
(6)
with magnitude
V = |V| =
√
V·V=
q
Vr2 + Vφ2 + Vz2 .
(7)
Position Vector in Cylindrical Coordinates
Position Vector of a point P = P(r , φ, z) is given by
~ = 1̂r (φ)r + 1̂z z,
R = OP
(8)
where
r = Rr = 1̂r · R =
1̂x cos φ + 1̂y sin φ · 1̂x x + 1̂y y + 1̂z z
= x cos φ + y sin φ.
(9)
Although the position vector in cylindrical coordinates does not have
an azimuthal 1̂φ -component, it does indeed depend upon the
azimuthal angle φ through the φ-dependence of the radial unit vector
1̂r = 1̂r (φ).
Cylindrical Coordinates - Differential Elements
Differential elements of length along the 1̂r , 1̂φ , & 1̂z - directions:
d ℓr = dr ,
d ℓφ = rd φ,
d ℓz = dz.
Differential Length, Surface Area, & Volume
The vector differential element of length:
d ~ℓ = 1̂r dr + 1̂φ rd φ + 1̂z dz.
(10)
Fundamental quadratic form or metric form:
d ℓ2 = dr 2 + r 2 d φ2 + dz 2 .
(11)
Differential elements of surface area:
φz-Cylindrical Surface: d sr = 1̂r (d ℓφ d ℓz ) = 1̂r (rd φdz).
rz-Planar Surface:
d sφ = 1̂φ (d ℓr d ℓz ) = 1̂φ (drdz).
r φ-Planar Surface:
d sz = 1̂z (d ℓr d ℓφ ) = 1̂z (rdrd φ).
Differential element of volume:
dV = d ℓr d ℓφ d ℓz = rdrd φdz.
(12)
Distance between Two Points
Coordinates of P1 (x1 , y1 , z1 ) = P1 (r1 , φ1 , z1 ):
x1 = r1 cos φ1 ,
y1 = r1 sin φ1 ,
z1 = z1 .
Coordinates of P2 (x2 , y2 , z2 ) = P2 (r2 , φ2 , z2 ):
x2 = r2 cos φ2 ,
y2 = r2 sin φ2 ,
z2 = z2 .
Distance between P1 & P2 is then given by Pythagorean’s theorem as
1/2
(r2 cos φ2 − r1 cos φ1 )2 + (r2 sin φ2 − r1 sin φ1 )2 + (z2 − z1 )2
1/2
= r12 + r22 − 2r1 r2 cos (φ2 − φ1 ) + (z2 − z1 )2
.
(13)
d =
Gradient Operator in Cylindrical Coordinates
Cylindrical Polar Coordinates
p
r = x 2 + y 2 , φ = arctan (y /x),
z = z,
so that
x
∂φ
1
∂r
= 2
= cos φ,
= − sin φ,
2
1/2
∂x
(x + y )
∂x
r
∂r
y
∂φ
1
= 2
= sin φ,
= cos φ.
2
1/2
∂y
(x + y )
∂y
r
By the Chain Rule
∂f ∂r
∂f ∂φ ∂f ∂z
∂f
sin φ ∂f
∂f
=
+
+
= cos φ
−
,
∂x
∂r ∂x
∂φ ∂x
∂z ∂x
∂r
r ∂φ
∂f ∂r
∂f ∂φ ∂f ∂z
∂f
cos φ ∂f
∂f
=
+
+
= sin φ
+
,
∂y
∂r ∂y
∂φ ∂y
∂z ∂y
∂r
r ∂φ
∂f
∂f ∂r
∂f ∂φ ∂f ∂z
∂f
=
+
+
=
.
∂z
∂r ∂z
∂φ ∂z
∂z ∂z
∂z
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
∂f
∂f
+ 1̂y
+ 1̂z
∇f (r) = 1̂x
∂x
∂y
∂z
∂f
sin φ ∂f
=
1̂r cos φ − 1̂φ sin φ cos φ
−
∂r
r ∂φ
∂f
cos φ ∂f
∂f
+ 1̂z
+
+ 1̂r sin φ + 1̂φ cos φ sin φ
∂r
r ∂φ
∂z
∂f
1 ∂f
∂f
= 1̂r
+ 1̂φ
+ 1̂z .
(14)
∂r
r ∂φ
∂z
The gradient operator in cylindrical polar coordinates is then given by
∇ = 1̂r
1 ∂
∂
∂
+ 1̂φ
+ 1̂z
∂r
r ∂φ
∂z
(15)
Divergence, Curl, & Laplacian Operators in
Cylindrical Polar Coordinates
Vector function of position F(r) = F(r , φ, z) = 1̂r Fr + 1̂φ Fφ + 1̂z Fz
has divergence
∇·F =
1 ∂
1 ∂Fφ ∂Fz
(rFr ) +
+
r ∂r
r ∂φ
∂z
(16)
and curl
∂Fr
∂(rFφ )
∂Fz
1̂z ∂(rFφ ) ∂Fr
1̂r ∂Fz
+ 1̂φ
+
−
−
−
∇×F=
r ∂φ
∂z
∂z
∂r
r
∂r
∂φ
(17)
Scalar function of position f (r) = f (r , φ, z) has Laplacian
1 ∂
∂f
1 ∂2f
∂2f
2
∇ f =
r
+ 2 2+ 2
(18)
r ∂r
∂r
r ∂φ
∂z
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 r0 , as illustrated.
With the z-axis of a cylindrical coordinate system taken along the
cylinder axis, cylindrical symmetry requires that
E(r) = 1̂r 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
ZZZ
1
̺ πr 2 ℓ ; r ≤ r0
3
E · n̂da =
̺d r =
ǫ0
ǫ0 πr02 ℓ ; r ≥ r0
S
V
with
I
S
E · n̂da = E (r )
ZZ
cyl
1̂r · 1̂r d 2 r = 2πr ℓE (r ),
the integration over the two cylinder ends vanishing because
E = 1̂r E (r ) and n̂ are, by construction, orthogonal on those surfaces.
Combination of these results then gives
(
r ; r ≤ r0
̺
E(r) = 1̂r
r02
2ǫ0
; r ≥ r0
r
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 r0 is then given by (with d ~ℓ = 1̂r dr )
Z ∞
Z ∞
V (r ) =
1̂r E (r ) · 1̂r dr =
E (r )dr .
r
r
For r ≥ r0 one obtains
Z
̺r 2
̺r02 ∞ dr
= 0 ln (∞) − ln (r )
V (r ) =
2ǫ0 r r
2ǫ0
and for r ≤ r0 one obtains
Z ∞
Z r0
1
r2
̺r02
dr
̺
2
r
ln (∞)−ln (r0 )+ 1− 2
+
rdr =
V (r ) =
2ǫ0 0 r0 r
2ǫ0
2
r0
r
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 = r0 (the surface of the
cylindrical charge distribution), in which case the potential becomes
( ̺r 2
ln (r /r
− 2ǫ0
0 ) ; r ≥ r0
0
V (r ) =
2
̺r02
; r ≤ r0
1 − rr 2
4ǫ0
0
In the limit as r0 → 0 the cylindrical volume charge distribution goes
over to a line charge with density (charge per unit length) ̺ℓ = πr02 ̺.
The electrostatic field intensity then becomes
̺ℓ
E(r ) = 1̂r
2πǫ0 r
with absolute potential V (r ) = (̺ℓ /2πǫ0 ) ln (∞) − ln (r ) .
E-Field of a Uniform Infinitely Extended Cylindrical
Charge Distribution (Symmetry about a Line)
2
ρr0 /4ε
0
V(r)
ρr0 /2ε0
E(r)
r
0
r0
2
−ρr0 /4ε0
2r0
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 ≤ r0 , Poisson’s equation ∇2 V (r ) = −̺/ǫ0 in cylindrical
coordinates becomes
∂V
̺
∂
∂V
̺
1 ∂
r
=−
=⇒
r
=− r
r ∂r
∂r
ǫ0
∂r
∂r
ǫ0
which is integrated to yield
̺
∂V
̺
A
∂V
= − r 2 + A =⇒
=− r+
r
∂r
2ǫ0
∂r
2ǫ0
r
which in turn is integrated to yield
̺
V (r ) = − r 2 + A ln (r ) + B
4ǫ0
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 ) = −
̺ 2
r + B;
4ǫ0
r ≤ r0 .
• For r ≥ r0 , Laplace’s equation ∇2 V (r ) = 0 in cylindrical
coordinates becomes
∂V
∂V
C
∂
r
= 0 =⇒
=
∂r
∂r
∂r
r
where C is a constant, which in turn is integrated to yield
V (r ) = C ln (r ) − ln (∞) ; r ≥ r0
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 = r0 requires that
−
C
̺
̺
r0 =
=⇒ C = − r02
2ǫ0
r0
2ǫ0
Continuity of V (r ) at r = r0 requires that
−
̺r 2 ̺ 2
r0 +B = C ln (r0 )−ln (∞) =⇒ B = 0 1−2 ln (r0 )−ln (∞)
4ǫ0
4ǫ0
Hence
V (r ) =
̺r 2
̺
r02 − r 2 − 0 ln (r0 ) − ln (∞) ;
4ǫ0
2ǫ0
V (r ) = −
̺r02
ln (r ) − ln (∞) ;
2ǫ0
r ≥ r0
r ≤ r0
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 ≤ r0
̺r02
∂
̺
̺
2
2
E(r ) = −1̂r
r0 − r −
ln (r0 ) − ln (∞) = 1̂r
r
∂r 4ǫ0
2ǫ0
2ǫ0
and for r ≥ r0
̺r02
̺r 2
∂
−
ln (r ) − ln (∞) = 1̂r 0
E(r ) = −1̂r
∂r
2ǫ0
2ǫ0 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/2πb, 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 = −1̂r b + 1̂z z
in polar cylindrical coordinates, with unit vector
R̂′1 =
−1̂r b + 1̂z z
.
(b2 + 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 E1 = R̂′1
̺ℓ b (−1̂r b + 1̂z z)
1 ̺ℓ
d φ,
dℓ =
′2
4πǫ0 R1
4πǫ0 (b2 + z 2 )3/2
while from element 2
d E2 = R̂′2
1 ̺ℓ
̺ℓ b (1̂r b + 1̂z z)
d φ.
dℓ =
′2
4πǫ0 R2
4πǫ0 (b2 + z 2 )3/2
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
dφ
̺ℓ bz
.
d E = d E1 + d E2 = 1̂z
2
2πǫ0 (b + z 2 )3/2
The total electrostatic field is then obtained by integrating around
the ring from φ = 0 to φ = π as
Z π
̺ℓ bz
E(0, 0, z) = 1̂z
dφ
2πǫ0 (b2 + z 2 )3/2 0
̺ℓ bz
= 1̂z
2ǫ0 (b2 + z 2 )3/2
Qz
.
= 1̂z
4πǫ0 (b2 + z 2 )3/2
Notice that E(0, 0, 0) = 0 and that E(0, 0, z) → ±1̂z Q/(4πǫ0z 2 ) as
z → ±∞, respectively.
On-Axis Potential of a Circular Charged Annulus
The absolute potential produced by the circular charged annulus is
given by Coulomb’s law as
Z 2π
̺ℓ
1
bd φ
V (0, 0, z) =
4πǫ0 0 (b2 + z 2 )1/2
̺ℓ b
=
.
2ǫ0 (b2 + z 2 )1/2
The electrostatic field along the z-axis is then given by
−1/2
̺ℓ b ∂
b2 + z 2
E(0, 0, z) = −∇V (0, 0, z) = −1̂z
2ǫ0 ∂z
̺ℓ bz
= 1̂z
2ǫ0 (b2 + z 2 )3/2
Qz
,
= 1̂z
4πǫ0 (b2 + z 2 )3/2
in agreement with the previous result.
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 = 2π̺s 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 = 1̂z
z
4πǫ0 (r 2 + z 2 )3/2
2π̺s rdr ,
so that
̺s z
E(0, 0, z) = 1̂z
2ǫ0
Z
̺s
= ±1̂z
2ǫ0
a
0
rdr
(r 2
1−
+ z 2 )3/2
|z|
(a2 + 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 → ±1̂z ̺s /2ǫ0 .
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 + αr 2 ; r ≤ a,
with ̺(r ) = 0 for r > a, where ̺0 and α are constants.
1
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
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).
3
Determine the value of the parameter α for which the
electrostatic field vanishes everywhere in the region outside the
cylinder (r > a). Plot Er (r ) and V (r ) as a function of r for this
value of α.
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, 2π), and z ∈ (−∞, +∞), Laplace’s equation ∇2 φ = 0
assumes the form
1 ∂
∂φ
1 ∂2φ ∂2φ
r
+ 2 2 + 2 = 0.
(19)
r ∂r
∂r
r ∂ϕ
∂z
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 dR
1 d 2Q
1 d 2Z
1 d 2R
+
+
=
−
= −k 2 ,
(20)
R dr 2
rR dr
r 2 Q d ϕ2
Z dz 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 dR
1 d 2Q
r 2 d 2R
2 2
+
+
k
r
=
−
= ν 2,
R dr 2
R dr
Q d ϕ2
(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 2π, 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
ν2
d 2 R 1 dR
2
+
+ k − 2 R = 0.
dr 2
r dr
r
(22)
Under the change of variable x = kr , this equation assumes the
standard mathematical form
1 dR
ν2
d 2R
+
+ 1− 2 R =0
(23)
dx 2
x dx
x
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 ±ν X
2
j=0
x 2j
(−1)j
j!Γ(j + 1 ± ν) 2
(24)
• Bessel functions of the second kind of order ν (Weber, Neumann)
Yν (x) =
Jν (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 Ym (x) = limν→m Yν (x) when ν is
an integer or zero,
• Bessel functions of the third kind (Hankel functions)
Hν(j) = Jν (x) + (−1)j iYν (x)
for j = 1, 2.
(26)
Boundary Value Problems Cylindrical Coordinates
It follows from l’Hôpital’s rule that when ν is an integer or zero,
( )
∂
cos
(νπ)J
(x)
−
J
(x)
ν
−ν
Ym (x) = lim ∂ν
ν→m
π cos (νπ)
∂Jν (x)
1
m ∂J−ν (x)
,
lim
− (−1)
=
π ν→m
∂ν
∂ν
which then leads to the series representations
(
)
j
∞
i
X
2 h x (−1)j x 2j X 1
Y0 (x) =
ln
+ γ J0 (x) −
, (27)
2
π
2
(j!)
2
n
n=1
j=1
Boundary Value Problems Cylindrical Coordinates
i
h x + γ Jm (x)
πYm (x) = 2 ln
2
m
x −m m−1
X (m − j − 1)! x 2j x m 1 X
1
−
−
2
j!
2
2
m! n=1 n
j=0
"
#
j
m+j
∞
x m X
X
(−1)j x 2j X 2
1
−
,
+
2
j!(m
+
j)!
2
n
n
n=1
j=1
n=j+1
(28)
where m > 0 is an integer, and where
1
1 1
γ ≡ lim 1 + + + · · · + − ln (n)
n→∞
2 3
n
is Euler’s constant.
(29)
Boundary Value Problems Cylindrical Coordinates
Because of the symmetry relations J−m (x) = (−1)m Jm (x) and
Y−m (x) = (−1)m Ym (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 Jm (x) and Ym (x) of
the first and second kind of integer order m follows from their
respective limiting behaviors as x → 0. For m = 0, J0 (x) → 1 while
Y0 (x) → −∞ as ln(x/2). When m > 0, Jm (x) → 0 as x m while
Ym (x) → −∞ as −x −m . Therefore, the general solution of Bessel’s
equation (23) when ν = m is an integer has the form
R(x) = AJm (x) + BYm (x),
(30)
where A and B are arbitrary constants. It is then seen that the
(1)
(2)
Hankel functions Hν (x) and Hν (x) defined in Eq. (26) form a
linearly independent set of solutions to Bessel’s equation for all ν.
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 x1 dR
which has a
dx
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
r
2
π π
,
(31)
cos x − ν −
Jν (x) ∼
πx
2
4
r
2
π π
,
(32)
Yν (x) ∼
sin x − ν −
πx
2
4
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 which
leads to the equation d 2 Z /dz 2 + k 2 Z = 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 2π. The radial part of the equation then
becomes [cf. Eq. (22)]
ν2
d 2 R 1 dR
2
(33)
+
− k + 2 R = 0.
dr 2
r dr
r
The particular solutions of this equation are seen to be the Bessel
functions Jν (ikr ) and Yν (ikr ) of an imaginary argument. Under the
change of variable x = kr this equation assumes the standard form
d 2r
1 dR
ν2
(34)
+
− 1+ 2 R =0
dx 2 x dx
x
Boundary Value Problems Cylindrical Coordinates
Real-valued solutions of this equation are the modified Bessel
functions defined as
Iν (x) ≡ e
−i νπ/2
Jν (ix) =
∞
x ν X
2
j=0
x 2j
1
j!Γ(j + ν + 1) 2
(35)
where I−n (x) = In (x) with n an integer, and
Kν (x) ≡
where K−ν (x) = Kν (x).
π I−ν (x) − Iν (x)
2
sin (νπ)
(36)
Boundary Value Problems Cylindrical Coordinates
The limiting forms of the modified Bessel functions as x → 0 are
1
( x2 )ν , K0 (x) → −I0 (x) ln ( x2 ), and
given by Iν (x) → Γ(ν+1)
Kν (x) → 12 Γ(ν)( x2 )ν for ν > 0.
In addition, their large argument asymptotic approximations are given
by
1
ex ,
Iν (x) ∼ √
2πx
r
π −x
Kν (x) ∼
e ,
2x
(37)
(38)
as x → ∞. It is then seen that the modified Bessel functions Iν (r )
are appropriate for boundary value problem solutions that remain
bounded at r = 0, while the modied Bessel functions Kν (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
φνn (r , ϕ, z) = Rνn (r )Qν (ϕ)Zνn (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
XX
Aνn φνn (r , ϕ, z),
(40)
φ(r , ϕ, z) =
ν
n
where the coefficients Aνn are uniquely determined by the boundary
conditions.