Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
Laplace’s equation in Cylindrical Coordinates
1- Circular cylindrical coordinates
The circular cylindrical coordinates ( s , φ , z ) are related to the rectangular Cartesian coordinates
( x , y , z ) by the formulas (see Fig.):
Circular cylindrical coordinates.
x = s cos φ ,
( 0 ≤ s < ∞, 0 ≤ φ ≤ 2π , −∞ < z
y = s sin φ ,
< ∞)
z =z.
The inverse relations are:
y
, z =z.
x
s = x 2 + y 2 , tan φ =
An infinitesimal length d is
d =
An infinitesimal volume element is:
(ds ) + ( sd φ ) + (dz )
2
2
2
d ℑ = s ds d φ dz .
The gradient, divergence, curl and Laplacian become, in cylindrical coordinates are:
Gradient
∇V =
Divergent
∇i v =
∂V
1 ∂V ˆ ∂V
sˆ +
zˆ
φ+
s ∂φ
∂s
∂z
1 ∂
1 ∂
∂
vφ )+
(sv s ) +
(v z )
(
s ∂s
s ∂φ
∂z
Curl
⎛ 1 ∂v z ∂v φ ⎞ ⎛ 1 ∂v s ∂v z ⎞ ˆ 1 ⎡ ∂
∂v s
∇× v = ⎜
−
−
⎟ sˆ + ⎜
⎟ φ + ⎢ ( sv φ ) −
∂z ⎠ ⎝ s ∂z
∂s ⎠
∂φ
s ⎣ ∂s
⎝ s ∂φ
1
⎤
⎥ zˆ
⎦
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
Laplacian
∇ 2V =
1 ∂
s ∂s
2
2
⎛ ∂V ⎞ 1 ∂ V ∂ V
s
+
+
=0
⎜
⎟ 2
2
∂z 2
⎝ ∂s ⎠ s ∂φ
Laplace’s equation in two dimensions (Consult Jackson (page 111) )
Example: Solve Laplace's equation by separation of variables in cylindrical coordinates, assuming
there is no dependence on z (cylindrical symmetry). Make sure that you find all solutions to the
radial equation. Does your result accommodate the case of an infinite line charge?
Answer: For a system with cylindrical symmetry the electrostatic potential does not depend on z.
∂V
This immediately implies that
= 0 . Under this assumption Laplace's equation reads:
∂z
1 ∂
∇V =
s ∂s
2
Consider as a possible solution of V:
2
⎛ ∂V ⎞ 1 ∂ V
=0
⎜s
⎟+ 2
2
⎝ ∂s ⎠ s ∂φ
V (s , φ ) = ℜ(s )Φ (φ )
Substituting this solution into Laplace's equation we obtain
Φ(φ ) ∂
s ∂s
2
⎛ ∂ℜ(s ) ⎞ ℜ(s ) ∂ Φ(φ )
s
+
=0
⎜
⎟
∂s ⎠ s 2
∂φ 2
⎝
Multiplying each term in this equation by s 2 and dividing by ℜ(s )Φ (φ ) we obtain
s ∂ ⎛ ∂ℜ(s ) ⎞
1 ∂ 2Φ(φ )
+
=0
s
⎜
⎟
ℜ(s ) ∂s ⎝
∂s ⎠ Φ(φ ) ∂φ 2
The first term in this equation depends only on s while the second term in this equation depends
only on φ . This equation can therefore be only valid for every s and every φ if each term is equal
to a constant. Thus we require that:
s ∂
ℜ(s ) ∂s
⎛ ∂ℜ(s ) ⎞
⎜s
⎟ = γ ≡ constant
∂s ⎠
⎝
(A)
and
1 ∂ 2 Φ (φ )
= −γ
Φ (φ ) ∂φ 2
(B)
1- consider the case in which γ = − m 2 > 0 . The differential equation for Φ (φ ) can be rewritten as
∂ 2 Φ (φ )
− m 2 Φ (φ ) = 0
2
∂φ
The most general solution of this differential solution is
Φ m (φ ) = C m e m φ + D m e − m φ
However, in cylindrical coordinates the angle φ must be unique, namely, Φ(φ + 2π ) = Φ(φ ) and
therefore the general solution of the equation
d 2Φ
− m 2 Φ = 0 is not satisfied for this solution, and
dφ2
we conclude that γ = m 2 > 0 . The differential equation for Φ (φ ) can be rewritten as
2
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
∂ 2 Φ (φ )
+ m 2 Φ (φ ) = 0
2
∂φ
The most general solution of this differential solution is
Φ m (φ ) = C m cos ( m φ ) + D m sin ( m φ )
The condition that Φ (φ ) = Φ (φ + 2π ) requires that m is an integer. Now consider the radial function
ℜ ( s ) . We will first consider the case in which γ = m 2 > 0 .
2- Consider the following solution for ℜ ( s ) :
ℜ ( s ) = As k ,
A = constant
Substituting this solution into equation (A) we obtain
s ∂ ⎛ ∂
k ⎞
2
2
⎜ s ( As ) ⎟ = k = m
k
As ∂s ⎝ ∂s
⎠
Therefore, the constant k can take on the following two values:
k + = m,
k − = −m
The most general solution for ℜ ( s ) under the assumption that m 2 > 0 is therefore
Bm
sm
2
Now consider the solutions for ℜ ( s ) when m = 0 . In this case we require that
ℜ (s ) = Am s m +
∂ ⎛ ∂ℜ ( s ) ⎞
⎜s
⎟=0
∂s ⎝
∂s ⎠
This equation can be rewritten as
⇒
s
∂ℜ ( s )
= a0 = constant
∂s
∂ℜ ( s ) a0
=
s
∂s
If a0 = 0 then the solution of this differential equation is
ℜ ( s ) = b0 = constant
If a0 ≠ 0 then the solution of this differential equation is
ℜ ( s ) = a0 ln(s ) + b 0
Combining the solutions obtained for m 2 = 0 with the solutions obtained for m 2 > 0 we conclude
that the most general solution for ℜ ( s ) is given by
∞
B ⎤
⎡
ℜ ( s ) = a0 ln(s ) + b0 + ∑ ⎢ A m s m + mm ⎥
s ⎦
m =1 ⎣
Therefore, the most general solution of Laplace's equation for a system with cylindrical symmetry is
∞
⎡⎛
B ⎞
⎤
V ( s , φ ) = a0 ln(s ) + b0 + ∑ ⎢⎜ A m s m + mm ⎟ (C m cos ( m φ ) + D m sin ( m φ ) ) ⎥
s ⎠
m =1 ⎣⎝
⎦
3
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
Laplace’s equation in three dimensions
Laplace's equation in cylindrical coordinates takes the form:
∇ 2V =
Consider as a possible solution of V:
1 ∂
s ∂s
2
2
⎛ ∂V ⎞ 1 ∂ V ∂ V
s
+
+
=0
⎜
⎟ 2
2
∂z 2
⎝ ∂s ⎠ s ∂φ
(1)
V (s , φ , z ) = ℜ(s )Φ(φ ) Z (z )
(2)
Substituting (2) into (1) we obtain
1 ∂ ⎛ ∂ℜ(s ) ⎞ 1 ∂ 2 Φ(φ ) ∂ 2 Z (z )
⎜s
⎟
2
∇ 2V
∂s ⎠ s 2 ∂φ 2
s ∂s ⎝
=
+
+ ∂z
=0
ℜ(s )
Φ(φ )
V
Z (z )
(3)
Taking ( ∂ 2 Z / ∂z 2 ) / Z to the right-hand side of the equation we have an expression independent of
z on the left, from which we conclude that either expression (on the right or on the left) must be a
constant. Explicitly putting in the sign (which must still be determined from the boundary
conditions) of the separation constant, we have:
1- Azimuthal direction:
1 ∂ 2 Φ (φ )
= −γ .
(3a)
Φ (φ ) ∂φ 2
First consider the case in which γ = − m 2 > 0 . The differential equation for Φ (φ ) can be rewritten
as
∂ 2 Φ (φ )
− m 2 Φ (φ ) = 0
2
∂φ
The most general solution of this differential solution is
Φ m (φ ) = C m e m φ + D m e − m φ .
(3b)
However, in cylindrical coordinates the angle φ must be unique, namely, Φ(φ + 2π ) = Φ(φ ) and
d 2Φ
therefore the general solution of the equation
− m 2 Φ = 0 is not satisfied for this solution, and
dφ2
we conclude that γ = m 2 > 0 . The differential equation for Φ(φ ) can be rewritten as
The most general solution of (4) is
∂ 2 Φ (φ )
+ m 2 Φ (φ ) = 0
2
∂φ
(4)
Φ (φ ) = A sin ( m φ ) + B cos ( m φ )
(5)
The condition that Φ(φ ) = Φ(φ + 2π ) requires that m is an integer.
Next, consider the second part, i.e.:
∂ 2 Z (z )
∂z 2 = λ 2 ⇒ Z (z ) = A sinh ( λ z ) + B cosh ( λ z )
Z (z )
Or, alternatively,
4
(4)
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
∂ 2 Z (z )
∂z 2 = −λ 2
Z (z )
22-Oct-13
⇒ Z (z ) = A sin ( λ z ) + B cos ( λ z )
For the choice (4), Eq (3) reduces to:
∂ 2 Φ (φ )
∂ ⎛ ∂ℜ(s ) ⎞
s
⎜s
⎟
∂φ 2
∂s ⎝
∂s ⎠
+ λ 2s 2 = −
= m2
ℜ(s )
Φ (φ )
Finally, consider the radial function ℜ ( s ) , in the form:
∂ ⎛ ∂ℜ(s ) ⎞
2 2
2
⎜s
⎟ + ( λ s − m ) ℜ(s ) = 0
∂s ⎝
∂s ⎠
is the Bessel’s equation, having solutions
ℜ(s ) = EJ m (λs ) + FN m (λs )
where J m and N m are Bessel and Neumann functions of order m. Had we picked the negative
separation constant as in equation (7), we would have obtained for ℜ(s ) the modified Bessel
equation:
s
s
∂
∂s
⎛ ∂ℜ(s ) ⎞
2 2
2
⎜s
⎟ + ( −λ s − m ) ℜ(s ) = 0
∂s ⎠
⎝
(5)
(6)
(7)
(8)
(9)
having as solutions the modified Bessel functions I m (λs ) and K m (λ s ) .
ℜ(s ) = EI m (λs ) + FK m (λs )
(11)
Note: K m (λs ) and N m (λs ) diverge at r = 0 and are therefore excluded from problems where the
region of interest includes r = 0 , while J m (λs ) and I m (λs ) diverges as r → ∞ and will therefore
be excluded from any exterior solution.
The complete solution is then of the form:
⎧ J (λ s ) ⎫ ⎧⎪ sin ( m φ ) ⎫⎪ ⎧⎪ sinh ( λ z ) ⎫⎪
V (s , φ , z ) = ∑ ⎨ m
⎬⋅⎨
⎬⋅⎨
⎬
λ , m ⎩ N m (λ s ) ⎭ ⎪
⎩cos ( m φ ) ⎪⎭ ⎪⎩cosh ( λ z ) ⎪⎭
(12)
⎧ I m (λs ) ⎫ ⎧⎪ sin ( m φ ) ⎫⎪ ⎧⎪ sin ( λ z ) ⎫⎪
+ ∑⎨
⎬⋅⎨
⎬⋅⎨
⎬
⎪cos ( m φ ) ⎭⎪ ⎩⎪cos ( λ z ) ⎭⎪
λ , m ⎩ K m (λ s ) ⎭ ⎩
where the braces
{}
stand for the arbitrary linear combination of the two terms within.
5
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
Example (6): A cylinder of radius “a” and height L, is placed
parallel to the z axis. Its basis at z = 0 is grounded, and so is
its face at s = a . The basis at z = L is held at a given
potential V o (s , ϕ ) (a given function). Find the potential
everywhere within the cylinder.
Solution: Let us first consider the general solutions of the three
equations, with the boundary conditions:
(a) B. Cs.
V (s , φ , 0) = 0
(1a )
V (s , φ , L ) =V 0
(1b )
V (a , φ , z ) = 0
V (0, φ , z ) = finite
(2)
(3)
V =V o (s , ϕ )
V =0
i- The angle φ must be unique, namely, Φ(φ + 2π ) = Φ(φ ) and therefore the general solution of the
2
equation d Φ2 + m 2 Φ = 0 will be:
dφ
Φ (φ ) = A m cos ( m φ ) + B m sin ( m φ )
with m an integer.
ii- In our case, Z must vanish at z = 0 , but not at z = L , which means we have the equation
d 2Z
− k 2 Z = 0 and the Z function is of the form:
dz 2
Z (z ) = C sinh ( kz ) + D cosh ( kz ) ,
iii- Due to the above items i and ii, ℜ must be the solution of the equation
d 2ℜ 1 d ℜ ⎛ 2 m 2 ⎞
+
+ ⎜ k − 2 ⎟ ℜ = 0 and taken to be of the form:
ds 2 s ds ⎝
s ⎠
ℜ(s ) = EJ m (k m s ) + FN m (k m s )
6
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
(b) The general solution of the Laplace’s equation for the problem in cylindrical coordinates
consists of a sum (superposition) of terms of the form:
V ( s , φ , z ) = ℜ(s )Φ (φ ) Z (z )
∞
= ∑ [ EJ m ( k m s ) + FN m ( k m s ) ] ⎡⎣ A m sin ( m φ ) + B m cos ( m φ ) ⎤⎦ ⎡⎣C sinh ( kz ) + D cosh ( kz ) ⎤⎦
m =0
B.C. 1a and 1b implies D = 0
B.C. 3 implies F = 0
III-
x mn
,
n = 1, 2,3,
a
x mn is the nth root of J m (k mn a ) . Remember that, Bessel function has an infinite number of roots,
and therefore κ a takes an infinite number of discrete values, all of them are roots of the mth Bessel
function. Namely, k mn a is the nth root of the mth Bessel function.
B.C. 2 implies J m (k mn a ) = 0 ⇒ k mn =
III-
It follows that the general solution of our problem is
∞
∞
V (s , φ , z ) = ∑ ∑ J m ( k mn s ) sinh(k mn z ) {A mn sin ( m φ ) + B mn cos ( m φ )}
n =1 m = 0
We now impose the boundary condition at z = L:
∞
∞
V (s , φ ) = ∑ ∑ J m ( k mn s ) sinh(k mn L ) {A mn sin ( m φ ) + B mn cos ( m φ )}
n =1 m = 0
This is a Fourier series in φ and a Fourier-Bessel series in s.
We now use this property with the boundary condition at z = L to determine all the coefficients in
terms of the given function V o (s , ϕ ) .
First, we use (Fourier trick) the delta functions of the trigonometric functions in the form:
if m ≠ n
⎧0
a
⎪
nπ y
mπ y
a
⎪a
∫0 cos( a ) cos( a )dy = 2 δ mn = ⎨ 2 if m = n ≠ 0
⎪
⎪⎩ a if m = n = 0
if m ≠ n
⎧0
nπ y
mπ y
a
⎪
∫0 sin( a )sin( a )dy = 2 δ mn = ⎨ a if m = n ≠ 0
⎪⎩ 2
a
a
∫ sin(
0
nπ y
mπ y
) cos(
)dy = 0
a
a
⎧ 0
nπ y
⎪
∫0 sin( a )dy = ⎨ 2a
⎪⎩ n π
a
to obtain
7
if n is even
if n is odd
Prof. Dr. I. Nasser
Cylinder_coordinates
∑A
mn
Phys 571, T-131
sinh(k mn L )J m (k mn s ) =
m
∑B
mn
1
π
sinh(k mn L )J m (k mn s ) =
m
22-Oct-13
2π
∫ V (s ,ϕ ) sin ( mϕ ) d ϕ
0
1
2π
∫ V (s ,ϕ ) cos ( mϕ ) d ϕ
π
0
Secondly, we use the orthonormal property of the Bessel function, which can be written in the form
a
∫ Jν (x ν n '
0
s
s
a2 2
) J ν ( x ν n ) sds =
J ν + 1 ( x ν n )δ nn ' .
2
a
a
Then
A mn
2
= 2 2
a π J m +1 (x mn ) sinh(k mn L )
2π
a
0
0
2π
a
∫ d ϕ ∫ sJ
m
(k mn s )V (s , φ )sin ( m φ ) ds ,
2
d ϕ ∫ sJ m (k mn s )V (s , φ ) cos ( m φ ) ds
2
2
a π J m +1 (x mn ) sinh(k mn L ) ∫0
0
which completes the solution.
B mn =
For the special but important case of azimuthal symmetry, for which V is independent of φ , i.e. m = 0 , we
obtain:
A mn = 0
B mn =
4δ m ,0
a
a 2J 12 (x 0 n ) sinh(k 0 n L ) ∫0
sJ 0 (k 0 n s )V (s )ds
The reason we obtained discrete values for k was the demand that φ vanish at s = a . If we let a → ∞ , then
k will be a continuous variable, and instead of a sum over k , we will obtain an integral. This is completely
analogous to the transition from a Fourier series to a Fourier transform, but we will not pursue it
further.
8
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
Example: A hollow right circular cylinder of radius “a” has its
axis coincident with the z axis and its ends at z = 0 and z = L.
The potential on the end faces is zero, while the potential on
the cylindrical surface is given as a constant V 0 .Using the
appropriate separation of variables in polar coordinates;
(a) Write down the boundary condition (conditions).
(b) Use the physical principal to write down the general
solution.
(c) Use the boundary conditions in (a) to simplify the general
solution in the separate coordinates. Write your reasons for
dropping any term or terms.
(d) Find a series solution for the potential anywhere inside the
cylinder.
Solution:
(a) B. Cs.
V (s , φ , 0) = 0
(1a )
V (s , φ , L ) = 0
(1b )
V (a , φ , z ) =V 0
(2)
V (0, φ , z ) = finite
(3)
i- The angle φ must be unique, namely, Φ(φ + 2π ) = Φ (φ ) and therefore the general solution of
d 2Φ
+ ν 2 Φ = 0 will be:
the equation
2
dφ
Φ(φ ) = A cos (νφ ) + B sin (νφ )
with ν an integer.
ii- In our case, Z must vanish at z = 0 and z = L , which means we have the equation
d 2Z
+ k 2 Z = 0 and the Z function is of the form:
dz 2
Z (z ) = C sin ( kz ) + D cos ( kz ) ,
iii- Due to the above items i and ii, ℜ must be the solution of the equation
d 2ℜ 1 d ℜ ⎛ 2 m 2 ⎞
+
+ ⎜ k − 2 ⎟ ℜ = 0 and taken to be of the form:
ds 2 s ds ⎝
s ⎠
ℜ(s ) = EI n (k n s ) + FK n (k n s )
(b) The general solution of the Laplace’s equation for the problem in cylindrical coordinates
consists of a sum (superposition) of terms of the form:
V ( s , φ , z ) = ℜ(s )Φ (φ ) Z (z )
∞
= ∑ [ Eν I ν (k s ) + Fnν K ν (k s )] ⎡⎣ Aν sin ( m φ ) + Bν cos ( m φ ) ⎤⎦ ⎡⎣Cν sin ( k s ) + Dν cos ( k s ) ⎤⎦
ν =0
For Z-direction:
B1- Boundary condition (1a) implies D = 0 .
B2- Boundary condition (1b) implies
9
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
Z (L ) = C sin ( kL ) = 0
⇒ kL = nπ
⇒ kn =
.
nπ
,
L
n = 1, 2,3,
n = 0 gives trivial solution.
For Z-direction:
Since we're looking for the potential inside the cylinder and there is no charge at the origin, the
solution must be finite (B. C. 3) as s → 0 , which requires F = 0. Then the potential expansion
becomes
∞
∞
V ( s , φ , z ) = ∑∑ ⎡⎣ A nν sin (νφ ) + B nν cos (νφ ) ⎤⎦ sin ( k n z ) I ν (k n s )
n =1 ν = 0
(4)
at s = a
∞
∞
V ( a , φ , z ) =V 0 = ∑∑ ⎡⎣ A nν sin (νφ ) + B nν cos (νφ ) ⎤⎦ sin ( k n z ) I ν (k n a )
n =1 ν = 0
(5)
Multiplying (5) by sin ( k n ' z ) and integrate, we find:
∞
L
∫V
0
0
∞
L
sin ( k n ' z )dz = ∑∑ I ν (k n a ) ⎡⎣ A nν sin (νφ ) + B nν cos (νφ ) ⎤⎦ ∫ sin ( k n z ) sin ( k n ' z ) dz
n =1 ν = 0
−V 0
0
cos( k n ' z )
L
δ nn '
2
kn'
So:
⎡
⎤
LV 0 ⎢
L ∞
−
cos ( n ' π ) − 1⎥ = ∑ I ν (k n 'a ) ⎡⎣ A n 'ν sin (νφ ) + B n 'ν cos (νφ ) ⎤⎦
⎥ 2 ν =0
n 'π ⎢
⎣⎢ ( −1)n '
⎦⎥
Relabeling n = n ' , then for n odd,
2LV 0 L ∞
= ∑ I ν (k n a ) ⎡⎣ A nν sin (νφ ) + B nν cos (νφ ) ⎤⎦
nπ
2 ν =0
Equation (7) implies ν = 0 (No terms contain sin (νφ ) or cos (νφ ) in the left hand side), then
(6)
(7)
A nν = 0 ,
Bn0 =
4V 0
1
nπ I 0 (k n a )
And finally,
V (s ,φ , z ) =
4V 0
π
∞
∑
n =1,3,5,
10
1 sin ( k n z ) I 0 (k n s )
n
I 0 (k n a )
(8)
Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
Appendix (Repeated)
Helmholtz’s equation in cylindrical coordinates
In this appendix we will disuses the general solution of Helmholtz’s equation ∇ 2ψ + K 2ψ = 0 , in
cylindrical coordinates. Starting with the
1 ⎡ ∂ ⎛ ∂ψ ⎞ ∂ ⎛ 1 ∂ψ ⎞ ∂ ⎛ ∂ψ ⎞ ⎤
2
(1)
⎢ ⎜s
⎟+
⎜
⎟+
⎜s
⎟ ⎥ + k sψ = 0
s ⎣ ∂s ⎝ ∂s ⎠ ∂φ ⎝ s ∂φ ⎠ ∂z ⎝ ∂z ⎠ ⎦
Expand equation (1) and multiply by s, we have:
∂ 2ψ ∂ψ 1 ∂ 2ψ
∂ 2ψ
s 2 +
+
+
s
+ K 2s 2ψ = 0
(2)
2
2
∂s
∂s s ∂φ
∂z
Using the method of separation of variables,
ψ = ℜ(s )Φ(φ )Z (z )
(3)
eq. (2) will be reduced to, after dividing by s ℜΦZ ,
1 ∂ 2ℜ 1 ∂ℜ
1 ∂ 2Φ
1 ∂2Z
2
sK
+
+
+
=
−
Z ∂z 2
ℜ ∂s 2 s ℜ ∂s s 2Φ ∂φ 2
Since the left hand side of (4) is a function of s and φ and the right hand side is a function of
Z only, we may write
1 ∂2Z
2
−
=λ
2
Z ∂z
with solution
Z (z ) = A cos ( λ z ) + B sin ( λ z )
(4)
(6)
(6)
where λ is a constant and
s 2 ∂ 2ℜ s ∂ℜ 2
1 ∂ 2Φ
2
2
⎡
⎤
+
+ s ⎣K − λ ⎦ = −
ℜ ∂s 2 ℜ ∂s
Φ ∂φ 2
On separating the variables in (7), we obtain
1 ∂ 2Φ
−
= m2
2
Φ ∂φ
with solution
Φ (φ ) = C cos ( m φ ) + D sin ( m φ )
(7)
(8)
(9)
At this stage, λ and m are unknown constants. However, we will be interested in problems in
which the dependence on the angle is uniquely defined, Φ (φ + 2π ) = Φ (φ ) and therefore m = n ,
where n is an integer. Clearly we know the general solution for the differential equations of Z ( z )
and Φ (φ ) . What about the function ℜ ? The third equation then reads
{
}
∂ 2ℜ
∂ℜ
2
+s
+ s 2 ⎣⎡ K 2 − λ ⎦⎤ − m 2 ℜ = 0
2
∂s
∂s
2
For K 2 − λ = α 2 , where α 2 = constant , equation (10) becomes
s2
11
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Prof. Dr. I. Nasser
Cylinder_coordinates
Phys 571, T-131
22-Oct-13
∂ 2ℜ
∂ℜ
+s
+ s 2α 2 − m 2 P = 0
2
∂s
∂s
We now make the following change of the dimensionless variable ξ :
ξ =α s
So that
d
d dξ
d
=
=α
,
ds d ξ ds
dξ
{
s2
}
2
d2
2 d
=
α
ds 2
dξ2
By use of this change of variable, eq. (10) reduces to:
∂ 2ℜ 1 ∂ℜ ⎧ m 2 ⎫
+
+ ⎨1 −
⎬ℜ = 0
∂ξ 2 ξ ∂ξ ⎩ ξ 2 ⎭
Which is the Bessel’s equation, and its solutions are called Bessel (1784-1846) functions (or
cylindrical functions)
12
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