Bessel equation

Bessel equation
Jan Dereziński
Department of Mathematical Methods in Physics
University of Warsaw
Ho˙za 74, 00-682, Warszawa, Poland
March 28, 2013

3
Chapter 1
Modified Bessel equation
1.1 Modified Bessel equation and related equations
The modified Bessel equation has the form
(z 2 ∂ 2 z + z∂z − z 2 − m 2 )v(z) = 0.
One sometimes introduces the d-dimensional modified Bessel equation
Note the special cases
d 1−
z 2
z 2 ∂ 2 z + z∂z − z 2 − (l + d/2 − 1) 2 d −1+
z 2
= (z 2 ∂ 2 z + (d − 1)z∂z + z 2 − l(l + d − 2)).
z −m z 2 ∂ 2 z + z∂z − z 2 − m 2 z m
= z z∂ 2 z + (1 + 2m)∂z − z , (1.1.1)

= z 2
z 1
2 2 2
z ∂z + z∂z − z 2 − m 2 1 −
z 2
∂ 2 z − 1 + 1/4 − m 2 1
z 2
More generally, √ tv(tδ ) = w(t) leads to the Schrödinger equation of the form
∂2 t + (δtδ−1 ) 2 + ( 1
4 − m2δ2 ) 1
t2
+ 1 w(t) = 0. (1.1.2)
Hence, if v 1
ρ+1 is a solution of the modified Bessel equation with the parameter 1
ρ+1 , then
is a solution of
1.2 Integral representations
u(t) = √ tv 1
ρ+1
2 ρ
t1+ 2
ρ + 2
(∂ 2 t + t ρ )u = 0.
Theorem 1.2.1 Bessel–Schläfli type representations Let ]0, 1[∋ τ ↦→ γ(τ) be a contour such that
z
2 (t − t−1
z
) + m exp
2 (t + t−1
) t −m
t=γ(1)
= 0, (1.2.3)
t=γ(0)
Then
γ
.
z
exp
2 (t + t−1
) t −m−1 dt (1.2.4)

Proof. We differentiate the integral with respect to the parameter z:
(z 2 ∂ 2 z + z∂z − z 2 − m 2
z
) exp
γ 2 (t + t−1
) t −m−1 dt
z 2 −1
=
t + t
γ 2
2 z −1
+ t + t
2
− z 2 − m 2
z
exp
2 (t + t−1
) t −m−1 dt
z 2 −1
=
t − t
γ 2
2 z −1
+ t + t
2
− m 2
z
exp
2 (t + t−1
) t −m−1 dt
z
= ∂t
γ 2 (t − t−1
z
) + m exp
2 (t + t−1
) t −m
dt
z
=
2 (t − t−1
z
) + m exp
2 (t + t−1
) t −m
γ(1)
= 0.
γ(0)
✷
Theorem 1.2.2 Poisson type representations Let ]0, 1[∋ τ ↦→ γ(τ) be a contour such that
Then
(1 − t 2 1 m+
)
z m
is a solution of the modified Bessel equation.
γ
2e zt
γ(1)
γ(0)
= 0.
(1 − t 2 1 m−
) 2e zt dt

Proof. We use (1.1.1).
2
z∂z + (1 + 2m)∂z − z
=
γ
= −
1.3 Modified Bessel function
γ
(1 − t 2 1 m−
) 2e zt dt
(1 − t 2 1 m−
) 2(zt 2 + (1 + 2m)t − z)e zt dt
γ
∂t(1 − t 2 1 m−
) 2e zt
dt = 0.
The modified Bessel equation has a regular-singular point at 0 with the indicial equation
Its indices at 0 are equal to ±m.
The modified Bessel function is defined as
λ(λ − 1) + λ − m 2 = 0.
Im(z) =
∞
n=0
z 2n+m
2
n!Γ(m + n + 1) .
It is a solution of the Bessel equation with the parameter ±m. Note that 1
Γ(m+1) = 0 dla m =
−1, −2, . . . For m = −1, −2, . . . the function Im is the unique solution of the Bessel equation satisfying
Im(z) ∼
z
m 1
, z ∼ 0,
2 Γ(m + 1)
which can be treated as a definition of the modified Bessel function. (By f(z) ∼ g(z), z ∼ 0, we
understand that f(z) is analytic around zero and at zero equals 1.

If m ∈ Z, then I−m(z) and Im(z) are linearly independent and span the space of solutions of the
modified Bessel equation.
We have
Im(e ±iπ z) = e ±iπm Im(z).
1.4 Integral representations of modified Bessel function
We start with Bessel-Schläfli-type representations. If Rez > 0, then the basic contour integral representation
of Im is
Im(z) = 1
2πi ]−∞,0 +
z
exp
,−∞[ 2 (t + t−1
) t −m−1 =
dt (1.4.5)
1
z
m
exp s +
2πi 2
z2
s
4s
−m−1 ds. (1.4.6)
]−∞,0 + ,−∞[
To see this note that by (1.4.5), Im is holomorphic around zero. Besides, by the Hankel identity
1 1
= exp (s) s
Γ(m + 1) 2πi
−m−1 ds.
We also have
]−∞,0 + ,−∞[
I−m(z) = 1
2πi [(0−0) +
z
exp
] 2 (t + t−1
) t −m−1 dt. (1.4.7)
Here, the contour starts at 0 from the negative side on the lower sheet, encircling 0 in the positive
direction and ends at 0 from the negative side on the upper sheet,
To see this we make a substitution t = s −1 in (1.4.5) noting that dt = −s −2 ds, and then we change
the orientation of the contour.

Schläfli representation is just a simple consequence of (1.4.6):
2π
Im(z) = 1
2π 0
e z cos φ cos(mφ)dφ − 1
π sin(mπ)
0
e −z cosh β−mβ dβ, Rez > 0.
Next we consider various Poisson-type representations. The first is called the Poisson representation
1
z
m 1
Im(z) = √ (1 − t
πΓ 2 −1
2 1 m−
) 2e zt dt, m > − 1
2 .
m + 1
2
∞
The following Poisson-type representations are due to Hankel:
1
Im(z) =
2πi √ π Γ
1
z
m
− m
2 2 [1,−1− ,1 + 1 m−
(t − 1) 2(t + 1)
]
1
I−m(z) =
2πi √ π Γ
1
z
m 1 m−
− m
(t − 1) 2(t + 1)
2 2
[−∞,1 + ,−∞]
1.5 Relationship to hypergeometric type functions
m− 1
2e zt dt,
m− 1
2e zt dt.
The modified Bessel function and the hypergeometric functions 0F1 and 1F1 are closely related:
m
Im(z) =
1.6 Bessel function for integral parameters
Theorem 1.6.1
=
1
z
0F1 1 + m;
Γ(m + 1) 2
z2
4
1
z
m e
Γ(m + 1) 2
−z
1F1 m + 1
; 2m + 1; 2z .
2

Proof. It is enough to assume that m = 0, 1, . . . .
✷
For m ∈ Z we have
Generating function
Im(z) =
=
=
=
∞
n=0
∞
n=0
∞
n=m
∞
n=0
Im(z) = 1
2πi
= 1
2πi
z 2n+m
2
n!(n + m)!
z 2(n+m)−m
2
(n + m)!(n + m − m)!
z 2n−m
2
n!(n − m)!
(1.6.8)
(1.6.9)
(1.6.10)
z 2n−m
2
n!Γ(n − m + 1) = I−m(z). (1.6.11)
[0 + ]
z
2
z
exp
2 (t + t−1
dt
)
tm+1
exp s + z2
ds
4s sm+1. m
[0 + ]
z
exp
2 (t + t−1
) =
∞
m=−∞
t m Im(z).

Proof.
✷
∞
m=−∞
t m Im(z) =
1.7 MacDonald/Basset function
=
n≥0, n+m≥0
n≥0 n+m≥0
= e z/2t e tz/2 .
t m (z/2) m+2n
n!(n + m)!
(z/2t) n (tz/2) m+n
n!(n + m)!
We define the Macdonald or Basset function by a Bessel-Schláfli-type representation:
Km(z) := 1
∞
exp −
2 0
z
2 (s + s−1
) s −m−1 ds
= 1
z
exp
2
2 (t + t−1
) (−t) −m−1 dt.
Theorem 1.7.1
]−∞,0]
π
K−m(z) = Km(z) =
2 sin πm (I−m(z) − Im(z)), (1.7.12)
Im(z) = 1
Km(e
iπ
−iπ z) − e iπm Km(z) . (1.7.13)
−1

Consider two contours on the boundary of the Riemann surface of the principal branch of t −m−1 ,
each projecting onto ] − ∞, 0], I+ on the upper sheet and I− on the lower sheet, each oriented in the
direction of an increasing t. We have
z
exp
2 (t + t−1
) t −m−1 dt = −e iπm
I−
I+
= −2e iπm Km(z),
z
exp
2 (t + t−1
) t −m−1 dt = −e −iπm
exp
I−
I+
= −2e −iπm Km(z).
z
exp
2 (t + t−1
) (−t) −m−1 dt
z
2 (t + t−1
) (−t) −m−1 dt
Therefore, if I is the contour starting at 0 from the negative side on the lower sheet, encircling 0 in
the positive direction and ending at 0 from the negative side on the upper sheet, we have
−4i sin(πm)Km(z) = −2e iπm Km(z) + 2e −iπm Km(z)
z
= exp
I− 2 (t + t−1
) t −m−1
z
dt − exp
I+ 2 (t + t−1
) t −m−1 dt
z
=
exp
2 (t + t−1
) t −m−1
z
dt − exp
2 (t + t−1
) t −m−1 dt
]−∞,0 + ,−∞[
= 2πi(Im(z) − I−m(z)).
This proves (1.7.12).
Next we write
K(e −iπ π iπm
z) = e I−m(z) − e
2 sin πm
−iπm Im(z)
We subtract from this eiπm times (1.7.12) obtaining
K(e −iπ z) − e iπm π −iπm
K(z) = −e Im(z) + e
2 sin πm
iπm Im(z) = iπIm(z).
I

This proves (1.7.13). ✷
We also have a Poisson-type integral representation:
Km(z) =
Γ(m + 1
2 )(2z)m
√ π
∞
1.8 Asymptotics of the MacDonald/Basset function
Theorem 1.8.1 For | arg z| > π − ɛ,
0
Km(z)
lim
|z|→∞ e−z√ √
π
2z
cos(zt)(t 2 + z 2 1 −m−
) 2dt.
= 1.
Proof. We use the steepest descent method. Set φ(t) := − 1
2 (t + t−1 ). We compute
φ ′ (t) = − 1
2 (1 − t−2 ), φ ′′ (t) = −t −3 .
Hence φ has a critical point at t0 = 1 with φ(t0) = −1 and φ ′′ (t0) = −1. Thus
✷
Km(z) = 1
2
1
2
∞
0 ∞
= 1
2 e−z
t −m−1 exp(zφ(t))dt
exp zφ(t0) + z φ′′ (t0)
(t − t0)
2
2
dt
z
exp (t − 1)2 dt =
2 1
2 e−z
√
2π
√z .
−∞
∞
−∞

Corollary 1.8.2 As x → ∞ we have
Im(x) ∼
1.9 MacDonald/Basset function for integer parameters
Theorem 1.9.1 Set h(n) := n k=1 1
k . Then for m = 0, 1, 2, . . .
Km(z) = (−1) m+1
log z
+ γ Im(z)
2
Proof. Set
Then
Besides,
+ 1
m−1
2
k=0
(−1) k
z
2k−m(m − k − 1)!
+
2 k!
(−1)m
2
φ(z) := d 1
dz Γ(z)
1
√ 2πx e x . (1.8.14)
∞
k=0
= − 1
Γ(z) ∂z log Γ(z).
φ(−n) = (−1) n n!, n = 0, 1, 2, . . . ,
φ(n + 1) =
∂mIm(z) = log
γ − h(n)
, n = 0, 1, 2, . . .
n!
z
Im(z) +
2
∞
k=0
φ(m + k + 1)
k!
h(k) + h(m + k)
k!(m + k)!
z
m+2k .
2
z
2k+m .
2

Hence for m = 0, 1, 2, . . .
∂nIn(z) =
n=m
∂nIn(z) =
n=−m
The last sum can be written as
✷
log z
+ γ Im(z) −
2
∞
k=0
k=0
h(m + k)
z
m+2k , (1.9.15)
(m + k)!k! 2
log z
m−1
+ γ I−m(z) + (−1)
m−k−1(m − k − 1)!
z
2k−m 2 k! 2
−
We use the De L’Hopital rule:
∞
k=m
Km(z) = π
2
(1.9.16)
h(−m + k)
z
−m+2k . (1.9.17)
(−m + k)!k! 2
−
∞
k=m
= (−1)m
2
h(k)
z
m+2k .
k!(k + m)! 2
d
dm (I−m − Im(z))
d
dm
sin πm
− d
dn In(z)
−
n=−m d
dn In(z)
n=m

Corollary 1.9.2 As x → 0, we have
1
x
m Im(x) ∼
, m = −1, −2, . . . ; (1.9.18)
Γ(m + 1) 2
⎧
Re
⎪⎨
Km(x) ∼
⎪⎩
Γ(m)
2 m
x if Rem = 0, m = 0;
− ln
x
2 − γ if m = 0,
Γ(m)
2 m
(1.9.19)
2 x if Rem > 0;
m if Rem < 0.
1.10 Recurrence relations
Theorem 1.10.1
Proof.
✷
2∂zIm(z) =
0 = 2
z
∂t exp
γ
= −2m
z
exp
+z
γ
γ
exp
Γ(−m)
2
x
2
2∂zIm(z) = Im−1(z) + Im+1(z),
2mIm(z) = zIm−1(z) − zIm+1(z).
γ
z
exp
2 (t + t−1
) (t −m + t −m−2 )dt.
t −m dt
2 (t + t−1 )
2 (t + t−1
) t −m−1 dt
z
2 (t + t−1
) t −m
dt − z
γ
z
exp
2 (t + t−1
) t −m−2 dt.

Corollary 1.10.2
Hence
1
z ∂z (z m Im(z)) = z m−1 Im−1(z), or
1
z ∂z
−m
z Im(z) = z −m−1 Im+1(z), or
Analoguous identities hold for Km(z).
1.11 Half-integral parameters
1
z ∂z
n z m Im(z) = z m−n Im−n(z),
∂z + m
Im(z) = Im−1(z),
z
1
z ∂z
n z −m Im(z) = z −m−n Im+n(z).
I 1(z)
=
2
I 1 − (z) =
2
K 1(z) = K 1(z) =
∂z − m
Im(z) = Im+1(z).
z
z
1
2 1
2 Γ(1 + 1
2 )
sinh z
z =
1
2 2
sinh z
πz
z
1 − 2 1
2 Γ(1 − 1
1
2 2
cosh z = cosh z,
2 ) πz
π
1
2
e −z .

One way to derive these identities is
1.12 Wronskian
1 2n+
2 2n!Γ(1/2 + n + 1) =
1 2n−
2 2n!Γ(−1/2 + n + 1) =
π
2 2n n!2 n+1 (1/2)n+1 =
π
2 2n n!2 n (1/2)n =
The Wronskian of two solutions of the modified Bessel equation satisfies
∂z + 1
W (z) = 0.
z
Hence W (z) is proportional to 1
z . Using
I±m(z) ∼
1
Γ(±m + 1)
we can compute the Wronskian of Im and I−m:
W (Im, I−m) = −
z
±m
, I
2
′ ±m(z) ∼
π
(2n + 1)!,
2
π
2 (2n)!.
1
Γ(±m)
2 sin πm
, W (Im, Km) =
πz
1
z .
z
±m−1
,
2

19
Chapter 2
Standard Bessel equation
2.1 Bessel equation
Replacing z with ±iz in the modified Bessel equation leads to the standard Bessel equation:
2.2 Integral representations
(z 2 ∂ 2 z + z∂z + z 2 − m 2 )v(z) = 0.
Theorem 2.2.1 Bessel–Schläfli representations Let γ be a contour satisfying
z
2 (t + t−1
z
) + m exp
2 (t − t−1
1
)
tm
γ(1)
= 0, (2.2.1)
Then
is a solution of the Bessel equation.
C
γ
z
exp
2 (t − t−1
dt
)
tm+1 γ(0)
(2.2.2)

Theorem 2.2.2 Poisson type representations Let γ be a contour satisfying
(1 − t 2 1 m+
) 2e izt
= 0.
Then
is a solution of the Bessel equation.
2.3 Bessel function
The Bessel function is defined as
z m
m iπ
Jm(z) = e 2 Im(−iz)
m −iπ
= e 2 Im(iz)
=
∞
n=0
= 1
iπ
γ
γ(1)
γ(0)
(1 − t 2 1 m−
) 2e izt dt
2n+m
(−1) n z
2
n!Γ(m + n + 1)
−iπ
e m
m
2 iπ
K(−iz) − e 2 K(iz)
Jm(e ±iπ z) = e ±imπ Jm(z).
Theorem 2.3.1 If Rez > 0, then
Jm(z) = 1
2πi ]−∞,0 +
z
exp
,−∞[ 2 (t − t−1
dt
)
tm+1 = 1 z m
exp s − z2 ds
m+1 .

1
Jm(z) = √
πΓ m + 1
z
m 1
(1 − t
2 −1
2
2 1 m−
) 2e izt dt, m > − 1
2 ,
1
Jm(z) =
2πi √ π Γ
1
z
m
− m
2 2 [1,−1− ,1 + 1
1
m−
(t − 1) 2(t
m−
+ 1) 2,
]
J−m(z) = e −iπm 1
2πi √ π Γ
1
z
m 1 m−
− m
(t − 1) 2(t + 1)
2 2
[i∞,−1 + ,1 + ,i∞]
2.4 Relationship to hypergeometric type functions
Jm(z) =
2.5 Bessel function for integral parameters
Let m ∈ Z.
Theorem 2.5.1
=
1
z
m
0F1 1 + m; −
Γ(m + 1) 2
z2
4
1
z
m e
Γ(m + 1) 2
−iz
1F1 m + 1
; 2m + 1; 2iz .
2
Jm(z) = (−1) m J−m(z), m ∈ Z.
m− 1
2,

Theorem 2.5.2
2.6 Hankel functions
Jm(z) = 1
2πi
= 1
=
z
2πi 2
1
π
π
0
[0 + ]
z
exp
2 (t − t−1
dt
)
tm+1
exp s − z2
ds
4s sm+1 m
[0 + ]
cos(z sin φ − mφ)dφ.
There are two Hankel functions. Both are analytic continuations of the MacDonald function – one to
the lower and the other to the upper part of the complex plane:
Note the identities
H ± m(z) = ±2 m
e∓iπ 2 Km(∓iz),
iπ
Km(z) = ± iπ
2 e±imπH ± iπ ±
(e 2 z).
H ± −m(z) = e ±mπi H ± m(z),
Jm(z) = 1
2 (H+ m(z) + H− m(z)) ,
mπi + e H m(z) + e−mπiH − m(z) ,
J−m(z) = 1
2
H ± (z) = ± ie∓mπi Jm(z)−iJ−m(z) ,

Theorem 2.6.1 The following asymptotic formulas are true for −π + δ < arg z < 2π − δ, δ > 0:
lim
z→∞
lim
z→∞
2
πz
2
πz
H + m(z)
1
2 eize− imπ iπ
2 − 4
H − m(z)
1
imπ iπ
2 e−ize 2 + 4
2.7 Integral representations of Hankel functions
For Rez > 0,
H (1)
m (z) = H + m(z) = − 1
πi
H (2)
m (z) = H − m(z) = 1
πi
]−∞,(0+1·0) − [
]−∞,(0+1·0) + [
= 1,
= 1.
z
exp
2 (t − t−1
dt
)
tm+1,
z
exp
2 (t − t−1
dt
)
tm+1. By ] − ∞, (0 + 1 · 0) − [ we understand the contour starting at −∞, encyrcling 0 clockwise and reaching
zero from the positive direction. Similarly, by ] − ∞, (0 + 1 · 0) + [ we understand the contour starting
at −∞, encyrcling 0 counterclockwise and reaching zero from the positive direction.
Note that
z
lim
t→0+1·0 2 (t + t−1 ) + m
z
exp
2 (t − t−1 )
1
= 0,
tm where by t → 0 + 1 · 0 we denote the convergence to zero through positive values of t (sometimes
denoted by t → 0 + ). Hence the contours ] − ∞, (0 + 1 · 0) + [ i ] − ∞, (0 + 1 · 0) − [ satisfy (2.2.1).
If 0 < arg z < π, then a good contour in the representation of H − m is [i∞, 0]. If −π < arg < 0,
then for H + m one can use [−i∞, 0].

An appropriate choice of a contour yields the representations
H − m(z) = −
H + m(z) =
H − m(z) = −
H + m(z) =
imπ
ie− 2
π
imπ
2ie 2
π
−imπ
2ie 2
π
imπ
ie 2
π
∞
−∞ eiz cosh t−mtdt, 0 < arg z < π,
∞
0 eiz cosh t cosh(mt − imπ)dt − i π
0 e−iz cos t
cos mtdt , 0 < arg z < π,
∞
0 e−iz cosh t cosh(mt + imπ)dt + i π
0 eiz cos t
cos mtdt , −π < arg z < 0,
∞
−∞ e−iz cosh t−mtdt, −π < arg z < 0
Poisson type integral representations:
H − m(z) = Γ(1 2 − m)
πi √
z
m
π 2
H + m(z) = Γ(1 2 − m)
πi √
z
m
π 2
]i∞,1 + ,i∞[
]i∞,−1 − ,i∞[
e izt 1
1
m−
(t − 1) 2(t
m−
+ 1) 2dt,
e izt 1
1
m−
(t − 1) 2(t
m−
+ 1) 2dt.
Poisson type representations valid for m ≥ −1 2 .
H − 2
m(z) = −√
1 πΓ(m + 2 )
z
m
e
2 ]1,i∞[
izt (1 − t 2 1 m−
) 2dt
m π
2
2 − 4
= zmei(z−π
π )
Γ(m + 1
2 )
∞
e
0
−zs
1 m−
s 2 1 + is
1 m− 2
dt,
2
H + 2
m(z) = √ 1 πΓ(m + 2 )
z
m
e
2 ]−1,i∞[
izt (1 − t 2 1 m−
) 2dt
m π
2
2 − 4
= zme−i(z−π
π )
Γ(m + 1 ∞
e
)
−zs
1 m−
s 2 1 − is
1 m− 2
ds.
2
0

2.8 Neumann function
Neumann function is defined as
We then have
Ym(z) = 2
π
Ym(z) = 1
(1)
2i H m (z) − H (2)
m (z)
= cos πmJm(z)−J−m(z)
sin πm .
H (1)
m (z) = Jm(z) + iYm(z), H (2)
m (z) = Jm(z) − iYm(z).
Theorem 2.8.1 For m ∈ Z we have
Jm(z)
where h(k) := k
j=1
1
k .
− 1
m−1
π
k=0
2.9 Recurrence relations
log( z
2 ) + γ
(m−k−1)!
k! ( z
2 )2k−m − 1
π
Theorem 2.9.1 We have the identities
∞
k=0
2∂zJm(z) = Jm−1(z) − Jm+1(z),
2mJm(z) = zJm−1(z) + zJm+1(z).
(−1) k
k!(m+k)! (z
2 )m+2k h(k) + h(m + k) ,
Sometimes, more convenient are the following forms of the recurrence relations:

Corollary 2.9.2
1
z ∂z (z m Jm(z)) = z m−1
Jm−1(z), czyli ∂z + m
Jm(z) = Jm−1(z),
z
− 1
z ∂z
−m
z Jm(z) = z −m−1
Jm+1(z), czyli −∂z + m
Jm(z) = Jm+1(z).
z
Besides, n 1
− 1
z ∂z
Analogous identities hold for H ± m(z), and Ym(z).
2.10 Half-integral parameters
z ∂z z m Jm(z) = z m−n Jm−n(z),
n z −m Jm(z) = z −m−n Jm+n(z).
2
J 1(z)
= =
2
J 1 − (z) =
2
H ± 1 (z) =
2
H ±
− 1
2
(z) =
πz
1
2
sin z,
1
2 2
cos z,
πz
1
2 2
π ±i(z−
e 2
πz
) ,
1
2 2
e
πz
±iz

2.11 Integral of Sonine and Schafheitlin
([1] Exercise 4.14, p. 236):
=
=
∞
Jm(xξ)Jk(yξ)dξ
0 ξλ ykΓ
1+m+k−λ
2
2λx1+k−λΓ(k + 1)Γ
1 − m + k − λ
F
, 1+m−k−λ
2
2
1 + m + k − λ
; k + 1;
2
y2
x2
, y < x;
xmΓ
1+m+k−λ
2
2λy1+m−λΓ(m + 1)Γ
1 + m − k − λ
F
, 1−m+k−λ
2
2
1 + m + k − λ
; m + 1;
2
x2
y2
, x < y.
2.12 Wronskians of solutions of the Bessel equation
The Wronskian of two solutions of the Bessel equation satisfies
Hence W (z) is proportional to 1
z . Using
J±m(z) ∼
∂z + 1
W (z) = 0.
z
we can compute the Wronskian of Jm(z) and J−m(z):
W (Jm, J−m) = −
1
Γ(±m+1) (z
2 )±m , J ′ ±m(z) ∼ 1
Γ(±m) (z
2 )±m−1 ,
2 sin πm
, W (H
πz
− m, H + m) = − 4i
πz , W (Jm, Ym) = 2
πz .

29
Chapter 3
Separation of variables in the Laplacian
and the Bessel equation
3.1 Euclidean Lie algebra in dimension two
In R2 we introduce the polar coordinates
x = r cos φ = r
2 (w + w−1 ),
y = r sin φ = r
2i (w − w−1 ),
where w = eiφ . The polar coordinates can be interpreted as a unitary map
or
U1 : L 2 (R 2 ) → L 2 ([0, ∞[, r 1/2 ) ⊗ L 2 ([−π, π]),
U1f(r, φ) = f(r cos φ, r sin φ).
U2 : L 2 (R 2 ) → L 2 ([0, ∞[, r 1/2 ) ⊗ L 2 ([−1, 1], (1 − w 2 ) 1/2 ),
U2f(r, w) = f(rw, r(1 − w 2 ) 1/2 ).

Note that
We will use
Note the relations
The equation
∂x = cos φ∂r − r sin φ∂φ,
∂y = cos φ∂r + r cos φ∂φ.
L := x∂y − y∂x = ∂φ = iw∂w,
A + := ∂x + i∂y = w(∂r + ir −1 ∂φ) = w(∂r − r −1 w∂w),
A − := ∂x − i∂y = w −1 (∂r − ir −1 ∂φ) = w −1 (∂r + r −1 w∂w),
∆ := ∂ 2 x + ∂ 2 y = ∂ 2 r + r −1 ∂r + r −2 ∂ 2 φ = ∂2 r + r −1 ∂r − r −2 (w∂w) 2 .
[A + , A − ] = 0,
[A ± , L] = ±A ± ,
∆ = A + A − = A − A + ,
[∆, L] = [∆, A ± ] = 0.
−∆ = −1,
−∆ = = 1
restricted to the subspace L = m coincides respectively with the modified/unmodified Bessel equation.
3.2 Helmholtz equation in dimension 2
The Helmholtz equation

has a solution in the form of a plane wave propagating at the angle ψ, in the Catesian coordinates
equal
and in polar coordinates equal
i(x cos ψ+y sin ψ)
fψ(x, y) := e
fψ(r, φ) = e ir cos(φ−ψ) .
L commutes with ∆, therefore we can look for a solution of
(∆ + 1)f = 0, Lf = imf.
It is solved by a circular wave equal in the polar coordinates
fm(r, φ) = Jm(r)e imφ .
Theorem 3.2.1 A plane wave can be decomposed into circular waves:
∞
fψ(r, φ) =
m=−∞
A circular wave can bedecomposed into plane waves:
i m fm(r, φ) = 1
2π
i m fm(r, φ)e −imψ . (3.2.1)
2π
Proof. To obtain (3.2.1) we take the formula for the generating function
∞
t m Jm(r)
π i(φ−ψ+ and insert t = e 2 ) .
e r
2 (t−t−1 ) =
0
m=−∞
fψ(r, φ)e imψ dψ. (3.2.2)

To obtain (3.2.2) we take
π −i(ψ− substitute t = e 2 ) and multiply by ieimφ . ✷
3.3 Graf addition formula
Jm(r) = 1
2πi [0] +
e ir(t−t−1 ) −m−1
t dt,
The following formula can be interpreted as follows: a circular wave in one polar coordinate can be
decomposed into circular waves in another polar coordinate system:
Theorem 3.3.1 Assume that R, r, ρ and Φ, φ, ψ are related as
Then
R =
(reiφ + ρeiψ )(re−iφ + ρe−iψ ), e iΦ
=
Jm(R)e imΦ =
∞
n=−∞
Jm−n(r)e i(m−n)φ Jn(ρ)e inψ .
reiφ + ρeiψ re−iφ .
+ ρe−iψ If m ∈ Z, then there are no restrictions on the parameters in the formula. If m is nonintegral and all
variables real, then one has to assume that ρ < r (or, equivalently, |Φ − φ| < π).
We can then replace
(i)
2

Proof. We put ˜ ψ = ψ − φ, ˜ Φ = Φ − φ. Then the problem is reduced to the case φ = 0.
∞ n=−∞
= 1 ∞ 2πi n=−∞
= 1
2πi
= 1
2πi
inψ
Jm−n(r)Jn(ρ)e
γ exp r
2 (t − t−1 ) t−m−1Jn(ρ)(teiψ ) n
γ exp
r
2 (t − t−1 ) + ρ
iψ iψ −1
2 te − (te )
t−m−1dt
γ exp R
2 (s − s−1 ) s−m−1dseimΦ = eimΦJm(R), where in the last step we used r + ρe iψ = Re iΦ and turned the contour. ✷
Substituting
we obtain
and the addition formula can be rewritten as
Jm( x 2 + y 2 )
x+iy
√ x 2 +y 2
m
x1 = r cos φ, y = r sin φ,
x2 = ρ cos ψ, y2 = ρ sin ψ,
x = R cos Φ, y = R sin Φ,
(x1, y1) + (x2, y2) = (x + y)
=
n∈Z Jm−n( x 2 1 + y2 1 )
Define an operator on L 2 (Z) given by the matrix
Um,n(x, y) := Jm−n( x 2 + y 2 )
x1+iy1
√ x 2 1 +y 2 1
x+iy
√ x 2 +y 2
Then U(x, y) −1 = U(−x, −y), U(x, y) is a unitary matrix, that is
Un,m(x, y) = Um,n(−x, −y),
m−n Jn( x2 2 + y2 2 )
x2+iy2 √
x2 2 +y2 n .
2
m−n

and R2 ∋ (x, y) ↦→ U(x, y) is a group representation, that means
∞
Uk,n(x2 + x1, y2 + y1) =
m=−∞
Uk,m(x2, y2)Um,n(x1, y1).
Extend this representation to the affine orthogonal group on R 2 . This group can be parametrized
by R 2 × S 1 . The action is defined as
(x2, y2, θ2)(x1, y1, θ1) = (x2 cos θ1 + y2 sin θ1 + x1, −x2 sin θ1 + y2 sin θ1 + y1, θ2 + θ1).
The representation is defined as follows:
R 2 × S 1 ∋ (x, y, θ) ↦→ U(x, y, θ),
Um,n,θ(x, y) := e imθ Jm−n( x 2 + y 2 )
x+iy
√ x 2 +y 2
m−n
.
Then U(x, y, θ) is a unitary matrix and it is a representation, that means
(x2, y2, θ2)(x1, y1, θ1) =
∞
Uk,m(x2, y2, θ2)Um,n(x1, y1, θ1).
3.4 ...
Identity
Uk,n
(2m + κ) (1 − w2 )
∂w + (2m + κ)w∂r =
r
m=−∞
m + κ
mw 2
+ ∂r + (1 − w )∂w
r
m
2
+ − ∂r − (m + κ)w + (1 − w )∂w .
r

35
Bibliography
[1] Andrews, G.E., Askey, R., Roy, R.: Special functions, Cambridge University Press, 1999
[2] Everitt, W.N. and Kalf, H.: The Bessel differential equation and the Hankel transform, to appear
in J. Comput. Appl. Math.
[3] Watson, G.N. A treatise on the theory of Bessel functions, Cambridge University Press, 2nd ed.
1948
[4] Titchmarsh, E.C.: Eigenfunction expansions I, Oxford University Press, 2nd ed. 1962