Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers
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12.13 Properties of Hankel Transforms and Applications 491
(a)
˜
f (κ)=H
n
[r
n
H (a − r)] =
a
0
r
n+1
J
n
(κr) dr =
a
n+1
κ
J
n+1
(aκ) .
(b)
˜
f (κ)=H
n
r
n
exp
−ar
2
!
=
∞
0
r
n+1
J
n
(κr)exp
−ar
2
dr =
κ
n
(2a)
n+1
exp
−
κ
2
4a
.
12.13 Properties of Hankel Transforms and
Applications
We state the following properties of the Hankel transforms:
(i) The Hankel transform operator, H
n
is a linear integral operator, that
is,
H
n
{af (r)+bg (r)} = aH
n
{f (r)} + bH
n
{g (r )}
for any constants a and b.
(ii) The Hankel transform satisfies the Parseval relation
∞
0
rf (r) g (r) dr =
∞
0
k
˜
f (k)˜g (k) dk (12.13.1)
where
˜
f (k)and˜g (k) are Hankel transforms of f (r)andg (r) respec-
tively.
To prove (12.13.1), we proceed formally to obtain
∞
0
k
˜
f (k)˜g (k) dk =
∞
0
k
˜
f (k) dk
∞
0
rJ
n
(kr) g (r) dr
=
∞
0
rg (r) dr
∞
0
kJ
n
(kr)
˜
f (k) dk
=
∞
0
rf (r) g (r) dr.
(iii) H
n
{f
′
(r)} =
k
2n
"
(n − 1)
˜
f
n+1
(k) − (n +1)
˜
f
n−1
(k)
#
provided rf (r) vanishes as r → 0andasr →∞.
(iv)
H
n
1
r
d
dr
r
df
dr
−
n
2
r
2
f (r)
= −k
2
˜
f
n
(k) (12.13.2)
provided both
r
df
dr
and rf (r) vanish as r → 0andasr →∞.
492 12 Integral Transform Methods with Applications
We have, by definition,
H
n
1
r
d
dr
r
df
dr
−
n
2
r
2
f (r)
=
∞
0
d
dr
r
df
dr
J
n
(kr) dr −
∞
0
n
2
r
2
rf (r) J
n
(kr) dr
=
r
df
dr
J
n
(kr)
∞
0
−
∞
0
kJ
′
n
(kr) r
df
dr
dr
−
∞
0
n
2
r
2
[rf (r)] J
n
(kr) dr, by partial integration
= −[f (r) krJ
′
n
(kr)]
∞
0
+
∞
0
d
dr
[krJ
′
n
(kr)] f (r) dr
−
∞
0
n
2
r
2
rf (r) J
n
(kr) dr, by partial integration
which is, by the given assumption and Bessel’s differential equation (8.6.1),
= −
∞
0
k
2
−
n
2
r
2
rf (r) J
n
(kr) dr −
∞
0
n
2
r
2
[rf (r)] J
n
(kr) dr
= −k
2
∞
0
rf (r) J
n
(kr) dr
= −k
2
H
n
{f (r)} = −k
2
˜
f
n
(k) .
(v) (Scaling). If H
n
{f (r)} =
˜
f
n
(κ), then
H
n
{f (ar)} =
1
a
2
˜
f
n
κ
a
,a>0. (12.13.3)
Proof. We have, by definition,
H
n
{f (ar)} =
∞
0
rJ
n
(κr) f (ar) dr
=
1
a
2
∞
0
sJ
n
κ
a
s
f (s) ds =
1
a
2
˜
f
n
κ
a
.
These results are used very widely in solving partial differential equa-
tions in the axisymmetric cylindrical configurations. We illustrate this point
by considering the following examples of applications.
Example 12.13.1. Obtain the solution of the free vibration of a l arge circular
membrane governed by the initial-value problem
∂
2
u
∂r
2
+
1
r
∂u
∂r
=
1
c
2
∂
2
u
∂t
2
, 0 <r<∞,t>0, (12.13.4)
u (r, 0) = f (r) ,u
t
(r, 0) = g (r) , 0 ≤ r<∞, (12.13.5)
12.13 Properties of Hankel Transforms and Applications 493
where c
2
=(T/ρ) = constant, T is the tension in the membrane, an d ρ is
the surface density of the membrane.
Application of the Hankel transform of order zero
˜u (k, t)=
∞
0
ru(r, t) J
0
(kr) dr
to the vibration problem gives
d
2
˜u
dt
2
+ k
2
c
2
˜u =0
˜u (k, 0) =
˜
f (k) , ˜u
t
(k, 0) = ˜g (k) .
The general solution of this transformed system is
˜u (k, t)=
˜
f (k) cos (ckt)+
˜g (k)
ck
sin (ckt) .
The inverse Hankel transformation gives
u (r, t)=
∞
0
k
˜
f (k) cos (ckt) J
0
(kr) dk
+
1
c
∞
0
˜g (k)sin(ckt) J
0
(kr) dr. (12.13.6)
This is the desired solution.
In particular, we consider the following initial conditions
u (r, 0) = f (r)=
A
1+
r
2
a
2
1
2
,u
t
(r, 0) = g (r)=0
so that ˜g (k)=0and
˜
f (k)=Aa
∞
0
rJ
0
(kr) dr
(a
2
+ r
2
)
1
2
=
Aa
k
e
−ak
by means of Example 12.12.1(a).
Thus, solution (12.13.6) becomes
u (r, t)=Aa
∞
0
e
−ak
J
0
(kr) cos (ckt) dk
= Aa Re
∞
0
e
−k(a+ict)
J
0
(kr) dk
= Aa Re
(
r
2
+(a + ict)
2
)
−
1
2
. (12.13.7)
494 12 Integral Transform Methods with Applications
Example 12.13.2. Obtain the steady-state solution of the axisymmetric
acoustic radiation problem governed by the wave equation in cylindrical
polar coordinates (r, θ, z):
c
2
∇
2
u = u
tt
, 0 <r<∞,z>0,t>0 (12.13.8)
u
z
= f (r, t)onz =0, (12.13.9)
where f (r, t) is a given function and c is a constant. We also assume that
the solution is bounded and behaves as outgoing spherical waves. This is
referred to as the Sommerfeld radiation condition.
We seek a solution of the acoustic radiation potential u = e
iω t
φ (r, z)so
that φ satisfies the Helmholtz equation
φ
rr
+
1
r
φ
r
+ φ
zz
+
ω
2
c
2
φ =0, 0 <r<∞,z>0 (12.13.10)
with the boundary condition representing the normal velocity prescrib ed
on the z = 0 plane
φ
z
= f (r)onz =0, (12.13.11)
where f (r) is a known function of r.
We solve the problem by means of the zero-order Hankel transformation
˜
φ (k, z )=
∞
0
rJ
0
(kr) φ (r, z) dr
so that the given differenti al system becomes
˜
φ
zz
= κ
2
˜
φ, z > 0,
˜
φ
z
=
˜
f (k)onz =0
where κ =
k
2
−
ω
2
/c
2
!
1
2
.
The solution of this system is
˜
φ (k, z )=−κ
−1
˜
f (k) e
−κz
, (12.13.12)
where κ is real and positive for k>ω/c, and purely imaginary for k<ω/c.
The inverse transformation yields the solution
φ (r, z)=−
∞
0
κ
−1
˜
f (k) kJ
0
(kr) e
−κz
dk. (12.13.13)
Since the exact evaluation of this integral is difficult, we choose a simple
form of f (r)as
f (r)=AH(a − r) , (12.13.14)
12.14 Mellin Transforms and their Operational Properties 495
where A is a constant and H (x) is the Heaviside unit step function so that
˜
f (k)=
a
0
kJ
0
(ak) dk =
a
k
J
1
(ak) .
Then the solution for this special case is given by
φ (r, z)=−Aa
∞
0
κ
−1
J
1
(ak) J
0
(kr) e
−κz
dk. (12.13.15)
For an asymptotic evaluation of this integral, we express it in terms of the
spherical polar coordinates (R, θ, φ), (x = R sin θ cos φ, y = R sin θ sin φ,
z = R cos θ), combined with the asymptotic result
J
0
(kr) ∼
2
πkr
1
2
cos
kr −
π
4
as r →∞
so that the acoustic potential u = e
iω t
φ is
u ∼−
Aa
√
2 e
iω t
√
πR sin θ
∞
0
J
1
(ka)cos
kR sin θ −
π
4
e
−kz
dk,
where z = R cos θ.
This integral can be evaluated asymptotically for R →∞by using the
stationary phase approximation formula (12.7.8) to obtain
u ∼−
Aac
ωR sin θ
J
1
(k
1
a) e
i(ωt−ωR/c)
, (12.13.16)
where k
1
= ω/csin θ is the stationary point. This solution represents the
outgoing spherical waves with constant velocity c and decaying amplitude
as R →∞.
12.14 Mellin Transforms and their Operational
Properties
If f (t) is not necessarily zero for t<0, it is possible to define the two-sided
(or bilateral) Laplace transform
f (p)=
∞
−∞
e
−pt
f (t) dt. (12.14.1)
Then replacing f (x)withe
−cx
f (x) in Fourier integral formula (6.13.9),
we obtain
e
−cx
f (x)=
1
2π
∞
−∞
e
−ikx
dk
∞
−∞
f (t) e
−t(c−ik)
dt,
496 12 Integral Transform Methods with Applications
or
f (x)=
1
2π
∞
−∞
e
x(c−ik)
dk
∞
−∞
f (t) e
−t(c−ik)
dt.
Making a change of variable p = c − ik and using definition (12.14. 1),
we obtain the formal inverse transform after replacing x by t as
f (t)=
1
2πi
c+i∞
c−i∞
e
pt
f (p) dp, c > 0. (12.14.2)
If we put e
−t
= x into (12.14.1) with f (−log x)=g (x)and
f (p) ≡ G (p),
then (12.14.1)–(12.14.2) become
G (p)=M{g (x)} =
∞
0
x
p−1
g (x) dx, (12.14.3)
g (x)=M
−1
{G (p)} =
1
2πi
c+i∞
c−i∞
x
−p
G (p) dp. (12.14.4)
The function G (p) is called the Mellin transform of g (x) defined by
(12.14.3). The inverse Mellin transformation is given by (12.14.4).
We state the following operational properties of the Mellin transforms:
(i) Both M and M
−1
are linear integral operators,
(ii) M[f (ax)] = a
−p
F (p),
(iii) M[x
a
f (x)] = F (p + a),
(iv) M[f
′
(x)] = −(p − 1) F (p − 1), provided that
f (x) x
p−1
!
∞
0
=0,
M[f
′′
(x)]=(p − 1) (p − 2) F (p − 2),
··· ··· ··· ··· ··· ··· ,
M
f
(n)
(x)
!
=
(−1)
n
Γ (p)
Γ (p−n)
F (p − n),
provided lim
x→0
x
p−r−1
f
(r)
(x)=0,r =0,1,2,...,(n − 1),
(v) M{xf
′
(x)} = −pM{f (x)} = −pF (p), provided that [x
p
f (x)]
∞
0
=0,
M
&
x
2
f
′′
(x)
'
=(−1)
2
p + p
2
F (p),
··· ··· ··· ··· ··· ··· ,
M
&
x
n
f
(n)
(x)
'
=(−1)
n
Γ (p+n)
Γ (p)
F (p).
(vi) M
(
x
d
dx
n
f (x)
)
=(−1)
n
p
n
F (p), n =1,2,....
12.14 Mellin Transforms and their Operational Properties 497
(vii) Convolution Property
M
∞
0
f (xξ) g (ξ) dξ
= F (p) G (1 − p),
M
∞
0
f
x
ξ
g (ξ)
dξ
ξ
= F (p) G (p).
(viii) If F (p)=M(f (x)) and G (p)=M(g (x)), then, the following con-
volution result holds:
M[f (x) g (x)] =
1
2πi
c+i∞
c−i∞
F (s) G (p − s) ds.
In particular, when p = 1, we obtain the Parseval formula
∞
0
f (x) g (x) dx =
1
2πi
c+i∞
c−i∞
F (s) G (1 − s) ds.
The reader is referred to Debnath (1995) for other properties of the
Mellin transform.
Example 12.14.1. Show that the Mellin transform of (1 + x)
−1
is π cosec πp,
0 < Re p<1.
We consider the standard definite integral
1
0
(1 − t)
m−1
t
p−1
dt =
Γ (m) Γ (p)
Γ (m + p)
, Re p>0, Re m>0,
and then change the variable t =
x
1+x
to obtain
∞
0
x
p−1
dx
(1 + x)
m+p
=
Γ (m) Γ (p)
Γ (m + p)
.
If we replace m + p by α,thisgives
M
"
(1 + x)
−α
#
=
Γ (p) Γ (α − p)
Γ (α)
.
Setting α = 1 and using the result
Γ (p) Γ (1 − p)=π cosec πp, 0 < Re p<1,
we obtain
M
"
(1 + x)
−1
#
= π cosec πp, 0 < Re p<1.
498 12 Integral Transform Methods with Applications
Example 12.14.2. Obtain the solution of the boundary-value problem
x
2
u
xx
+ xu
x
+ u
yy
=0, 0 ≤ x<∞, 0 <y<1,
u (x , 0) = 0, and u (x, 1) =
⎧
⎨
⎩
A, 0 ≤ x ≤ 1
0,x>1
,
where A is constant.
We apply the Mellin transform
U (p, y)=
∞
0
x
p−1
u (x , y) dx
to reduce the system to the form
U
yy
+ p
2
U =0, 0 <y<1,
U (p, 0) = 0, and U (p, 1) = A
1
0
x
p−1
dx =
A
p
.
The solution of this differential system is
U (p, y)=
A
p
sin (py)
sin p
, 0 < Re p<1.
The inverse Mellin transform gives
u (x , y)=
A
2πi
c+i∞
c−i∞
x
−p
p
sin (py)
sin p
dp,
where U (p, y) is analytic in a vertical strip 0 < Re p<πand hence,
0 <c<π. The integrand has simple poles at p = rπ, r =1,2,3,... which
lie inside a semi-circular contour in the right half-plane. Application of the
theory of residues gives the solution for x>1
u (x , y)=
A
π
∞
r=1
(−1)
r
x
−rπ
r
sin (rπy) .
Example 12.14.3. Find the Mellin transform of the Weyl fractional integral
ω (x, α)=W
α
[f (ξ)] =
1
Γ (α)
∞
x
f (ξ)(ξ − x)
α−1
dξ.
We rewrite the Weyl integral by setting
k (x)=x
α
f (x) ,g(x)=
1
Γ (α)
(1 − x)
α−1
H (1 − x) ,
so that
12.15 Finite Fourier Transforms and Applications 499
ω (x, α)=
∞
0
k (ξ) g
x
ξ
dξ
ξ
.
The Mellin transform of this result is obtained by the convolution property
(vii):
Ω (p, a)=K (p) G (p) ,
where K (p)=M[k (x)] = M[x
α
f (x)] = F (p + α)and
G (p)=
1
Γ (α)
1
0
(1 − x)
α−1
x
p−1
dx =
Γ (p)
Γ (p + α)
.
Thus,
Ω (p, a)=M[W
α
f (ξ)] =
Γ (p)
Γ (p + α)
F (p + α) . (12.14.5)
If α is complex with a positive real part such that n −1 < Re α<nwhere
n is a positive integer, the fractional derivative of order α of a function
f (x) is defined by the formula
D
α
∞
f (x)=
d
n
dx
n
W
n−α
f (x)=
1
Γ (n − α)
d
n
dx
n
∞
x
f (ξ)(ξ − x)
n−α−1
dξ.
(12.14.6)
The Mellin transform of this fractional derivative can be given by using
operational property (iv) and (12.14.5)
M[D
α
∞
f (x)] =
(−1)
n
Γ (p)
Γ (p − n)
Ω (p − n, n − α) ,
=
(−1)
n
Γ (p)
Γ (p − a)
F (p − α) .
This is an obvious generalization of the third result listed under (iv).
12.15 Finite Fourier Transforms and Applications
The finite Fourier transforms are often used in determining solutions of
nonhomogeneous problems. These finite transforms, namely the sine and
cosine transforms, follow immediately from the theory of Fourier series.
Let f (x) be a piecewise continuous function in a finite interval, say,
(0,π). This interval is introduced for convenience, and the change of interval
can be made without difficulty.
The finite Fourier sine transform denoted by F
s
(n) of the function f (x)
may be defined by
500 12 Integral Transform Methods with Applications
F
s
(n)=F
s
[f (x)] =
2
π
π
0
f (x)sinnx dx, n =1, 2, 3,..., (12.15.1)
and the inverse of the transform follows at once from the Fourier sine series;
that is
f (x)=
∞
n=1
F
s
(n)sinnx. (12.15.2)
The finite Fourier cosine transform F
c
(n)off (x) may be defined by
F
c
(n)=F
c
[f (x)] =
2
π
π
0
f (x)cosnx dx, n =0, 1, 2,.... (12.15.3)
The inverse of the Fourier cosine transform is given by
f (x)=
F
c
(0)
2
+
∞
n=1
F
c
(n)cosnx. (12.15.4)
Theorem 12.15.1. Let f
′
(x) be continuous and f
′′
(x) be piecewise con-
tinuous in [0,π].IfF
s
(n) is the finite Fourier sine transform of f (x), then
F
s
[f
′′
(x)] =
2n
π
[f (0) − (−1)
n
f (π)] − n
2
F
s
(n) . (12.15.5)
Proof.Bydefinition
F
s
[f
′′
(x)] =
2
π
π
0
f
′′
(x)sinnx dx
=
2
π
[f
′
(x)sinnx]
π
0
−
2n
π
π
0
f
′
(x)cosnx dx
= −
2n
π
[f (x)cosnx]
π
0
−
2n
2
π
π
0
f (x)sinnx dx
= −
2n
π
[f (π)(−1)
n
− f (0)] − n
2
F
s
(n) .
The transforms of higher-order derivatives can be derived in a similar man-
ner.
Theorem 12.15.2. Let f
′
(x) be continuous and f
′′
(x) be piecewise con-
tinuous in [0,π].IfF
c
(n) is the finite Fourier cosine transform of f (x),
then
F
c
[f
′′
(x)] =
2
π
[(−1)
n
f
′
(π) − f
′
(0)] − n
2
F
c
(n) . (12.15.6)
The proof is left to th e reader.