I nternat. J. Math. & Math. Sci.
Vol. 5 No. 3 (1982) 485-496
485
INVERSION OF THE POISSON-HANKEL TRANSFORM
C. NASIM
Department of Mthematics
The University of Calgary
Calgary, Alberta
T2N IN4 CANADA
(Received July 15, 1980 and in revised form November 3, 1981)
ABSTRACT.
The Poisson-Hankel transform is defined as an integral transform of the
initial temperature function, with the kernel as the source solution of the general-
ized heat equation.
In this paper a technique involving integral and differential
operators has been used to effect the inversion of the Poisson-Hankel transform.
KEY WORDS AND PHRASES. Heat equation, Poisson-Hankel transform, Gamma functn,
Bess functns, Whittaker funtn.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
i.
44A15.
INTRODUCT ION.
The Poisson-Hankel transform has as its kernel the fundamental solution of the
generalized heat equation.
A special case of the Poisson-Hankel transform, called
the reduced Poisson-Hankel transform has been studied in [i], where a differential
operator of Laguerre-Polya class
[2] has been used
to effect its inversion.
A
general theory of these type of operators has been developed by Widder [2], but
can not be applied to the more general Poisson-Hankel transform.
Our object in
this paper is to establish a procedure for the inversion of this transform in its
general form.
Our technique consists of applying an integral operator and a diff-
erential operator on the transform successively to retrieve the unknown function,
cf [3].
The differential operator is of the Laguerre-Polya class.
We shall also deduce the inversions of the Weierstrass Hankel transform and
the reduced Poisson-Hankel transform as special cases of our inversion algorithm.
486
C. NASIM
In the end we give an example to illustrate the result of the main theorem.
2.
DEFINITIONS AND PRELIMINARIES.
The generalized heat equation is
32u +
3x
u
2
8
u(x,t)
-"
x 8x
2
>
O.
A C 2 solution of (2.1) is called a generalized temperature.
(2.1)
The fundamental solut-
ion of (2.1) is the function
(2t)’-I/2e-x2/4t
G(x;t)
We define the function associated with G(x;t) by
-3/2 F(
2
G(x,y;t)
I
t-i
+
x
(xy) i/2- e
2
+ y2
4t
I-I’2/
zu
I (z) being the Bessel function of imaginary argument and order u.
I,
0,(22)
The function
G(x,y;t) is the source solution of the generalized heat equation (2.1).
G(x,0;t)
9 >
Note that
G(x;t).
The Poisson-Hankel transform is defined by
U(x,t)
G(x,y;t)(y)d(y),
0 < t <
(2.3)
0
where
21 / 2-u
d (,y)
r(
y
+})
2
dy
The convergent Poisson-Hankel transform defines a generalized temperature U(x,t)
with initial temperature
(x).
U(x,0+)
Next, some operational considerations.
From the Euler product of the gamma function
r(z)
3
2
1
lim
z(z + i)
n
(n
z
1)
n
(z + n- i)
one can easily show that
n
1
F(- 8z)
lira
n
lim
8
(n- i)!
n+o (n
8
n
n
k=l
n
i)
-+Sz
n
-e+Sz
Pn (z)
INVERSION OF THE POISSON-ANKEL TRANSFORM
pn(Z)
Now we define the operator
being a polynomial in z of order n.
1
where O =-x
d
x
ential operator.
-
n
--I/2
n
lira
+-
r(v
487
(n
Except for the factor n
i)!2
n
n
8/2
8/2
Pn (8)
1
r(
is the Euler differ-
I
+- 1
To obtain the intended interpretation of the operator n
0/2
we
write
n
8/2
e
N
8in n/2
Z
k=O
llm
qN(8)
N
qN’
a polynomial in @ of degree N.
in._____n_n
lira
k
2
5 0k
To see the effect of n
8/2 on a
function
where a is a constant, first note that
X
and hence
Pn(8)[x]
a polynomial of degree n.
8/2
Pn(-)x
where
Pn
is
Now,
[x=]
llm
N/oo
llm
N-+o
n
qN(e)[x]
qN(-)x
-/2 x
With this understanding, one can readily see that
r(
Thus,
r(
1
i
+7- 2
1
1
+- 2
a
x
l
1
x
a
(2.4)
r(++)
will be called a linear differential operator of infinite
order and the effect of this operator on a function x
constant factor.
1
This operator is of
Laguerre-P61ya
is to reproduce it with a
class and further properties
of the operator of this class are well known, cf [5].
Next we give two applications of this operator for future reference.
First,
C. NASIM
488
2 2
e
-a x
(_2 x2) k
i
r( +
k=0
x2k].
k!
k=0
(_2)k
E
k!
k=0
r( +
i/2-
(x)
Also, for v
2k
J9-1/2
1
+ k)
(2ox)
(2.5)
i >
0,
+
2
(i/
T )-0)+1/2)
r( +
E
k=0
I
)
I
k
r(
(_1)
k
k!
I
k)
++
1
++
F(v
+-I
P(v
r(v
k
r(v
k! F(v
k=O
-
i
i
+-f-
i
+E + k) i) k
1
+
x
r(v
2
k
(-7)
i) k
k=O
(-i)
k)
i
[x
2k
2k
i
++
k)
2
1
r(v+
(2.6)
We shall now consider some properties of the function G(s,y;t), s
o
+
iT,
defined as
29-3/2
G(s,y:t)
Y(
i
+
t-i
-(
e
based on the equation (2.2) above, where
LEMMA 2.1.
(2.7), and
A
and
If
B9
G(s,y;t)
2
+
y2
4t
(sy)l/2-v iv_i/2
sY2t
(2.7)
> 0, t > 0, y > 0.
G(s,y,t) is the function defined in the equation
are some constants, then
Y
and
s
(02 + T2)
e
4t
(2.8)
489
INVERSION OF THE POISSON-HANKEL TRANSFORM
(ii)
2
s
s
G(s,y;t)
Y
2
2
+
1
G+I
G(s y;t)]
(s,y;t)
(o- y)
t-3/2 lyl-V (02 + 2) -V/2
<
2
2
4t
By using the asymptotic expansion of the Bessel function
PROOF.
z
I(z)
and definition (2.7)
conclusion (i) follows immediately.
(2z)
Conclusion (ii) follows by direct differentiation and making use of conclusion
(i).
As direct consequences of the above lemma, we have that
(x- y)
<- Au t -I/2 (xy) -v
G(x,y;t)
(i)
G(s,y;t) is a continuous function of the variables s and y.
(ii)
LEMMA 2.2.
(2.10)
e
’o
Let
y
e
-oy
l(y)
dy < oo, for positive a and
> O.
0
F
U(x,t)
Then
G(x,y;t)(y)d(y)
0
exists for 0
x <
and can be analytically extended into the complex plane so
that U(s,t) is analytic for o
PROOF.
Re(s)
>_
0.
Using the estimate (2.10) and the value
2
d] (y)
r(
i/2-)
y
1
2)
dy
+y)
we have
IU(x,t)
IG(x,y;t)qb(Y)
<
di(Y)
0
(x- y)
<
A(x,t)
y
4t
e
0
Since
(x
y)
2
1
"
y
2
x
2
0 -< y
co,
(2.11)
C. NASIM
490
(x- y)
2
x
4t
therefore,
<
2
y
8t
2
e4t
2
U(x,t)
and
B (x, t)
<
y
(y)[
e
dy <
0
1
due to the hypothesis with a
t > 0.
Hence, the integral defining the
Now we consider
function U(x,t) exists and is, in fact, absolutely convergent.
U(s,t)
I G(s,y;t)
(y)d(y),
o + iT.
s
0
Now using the estimate (2.8) of G(s,y:t), we have
IU(s,t)l flG(s,Y;
_
s
t) @(Y)d(Y)
0
(o- y)
(o,’r,t)
y
2
,(y) dy
e
0
2
A(O,I,t)
<
y
[)(y) ]dy
8t
e
<
0
Hence, the function U(s,t) exists
using the inequality (2.11) and the hypothesis.
Now to prove that U(s,t) is
and is defined by an absolutely convergent integral.
analytic in the half-plane o > 0, we need to show that
-s G(s,y;t)
(y) d(y)
0
converges uniformly in the region o
0.
By making use of the estimate (2.9), we obtain
y)
(0
G(s,y:t)(y)d(y)
A(O,I,t)
0
4t
e
y
2
0
(o- y)
+
By (O,,t)
y
0
4t
491
INVERSION OF THE POISSON-HANKEL TRANSFOKM
Now due to the hypothesis and using the inequality (2.11), both the integrals on
the right hand side above, converge for all s and for t > 0, giving us the desired
result and hence the lemma.
As corollaries of Lemma 2.2, we have
o2 + T 2
IU(s,t)
where s
<
Av(t)
(0
2
T2)
-v/2
+
(2.12)
o + iT and t > 0; and
O(e
U(ix,t)
x2
4t
),
x
0(i),
3.
4t
e
(2.13)
0
x
THE INVERSION.
We give below a lemma which is a direct consequence of a general result,
[2; Theorem 2.1].
LEMMA 3.1.
2 /
0
f(x)
If
t)
i
t
--x
29+i
e
-t
2/x2
dt, x > 0,
then
r(
PROOF.
1
+-
If(x)]
1
0 < x <
(x),
> 0,
.
We write the above integral as
[ (t)
f(x)
1
x
k(
dt,
0
where
k(x)
2e
-I/x2 x-(2 + i).
Now the Mellin transform of k(x) is k*(s)
is of
Laguerre-P61ya class.
Thus
1
k*(O) f(x)
or
i%e Main Theorem:
#(x)
i
i
1
i
F(v + A-- -s),
o < 2v + i and
f(x)
1
#(x),
rC+y-y)
Let
e
-ay
(y) dy <
,
> O,
0
and
U(x,t)
f
0
be the Polsson-Hankel transform. If
G(x,y:t) q(y) dl(y)
> 0
x > 0
i
k*(s)
492
C. NASIM
2 2
r( +
R(x;t)
J
)
16t
e
2
G(v;t) U(iv,t) d(v)
(3.1)
0
then
i
r(
R(x;t)
+
e
2
-x / 4t
x > 0, t >
(x),
O,
where the functions G(v;t), G(x,y;t) and d(v) are defined above.
PROOF.
d(v),
From the result (2.8) and the definitions of the functions G(v:t) and
it is clear that the integral defining R(x;t) exists.
/
U(iv,t)
Also note that
G(iv,y;t) (y) d(y)
0
exists due to Lemma 2.2.
Then we can write
2 2
R(x,t)
r(u +
)
e
16t
2
I
G(v;t) d(v)
0
G(iv,y;t) (y) d(y)
0
2 2
F( +
)
(y) d(y)
0
e
16t
2
G(v;t) G(iv,y;t) d(v),
(3.2)
0
the change of order of integration can be justified by absolute convergence; we
need only to observe that
2 2
v x
le
16t2
IG(iv
G(v;t) d(v)
0
,y;t) (y) d(y)
0
2 2
<Kt
-( + i)
e
0
16t
v
v
y
dv
/4t
(Y) dY
<
,
0
by hypothesis.
From the definitions of the functions G(v;t), G(iv,y;t) and d(v), the v-integral in (3.2) can be written as
493
INVERSION OF THE POISSON-HANKEL TRANSFORM
2 2
(2t)
(u + 3/2)
e-y
/4t
y
i/2-u
2
16t
e
v
+I/2
J-i/2
(2t)
dv
0
2+i/2 e -y2/4t-y2/x2
and we then obtain,
-(2 + i)
2
2
e-y /4t-y2/ x y2X)
-(2x) + l) j.
R(x,t)
x
2x
[2, p.29).
(y) dy
0
e-y
2 i
0
2
/4t
(y) e
2 2
-y / x i
y
+
2
y
i
x
dy
Now the Lemma (3.1) is applicable and hence
r(
1
1
R(x,t)
1
e
-x
2
/ 4t
(3.3)
(x)
+-)
establishing the inversion of the Poisson-Hankel transform.
It is to be noted that the transforming function R(x,t) defined by (3.1) is
This can be recognized if we
in fact the modified Laplace transform of U(ix,t).
simplify and write
(v+i/2)
(4t)
R(x,t)
I
py(x,t)
e
P
-i/2
U(2p
1/2
,t) dp
0
x
where y(x,t)
2
16t
2
+
4
Also note that the above inverion algorithm is
t > 0
valid for the entire function
having a series expansion.
The condition on
simply restricts its growth.
Next we shall discuss some special cases.
Let
lira
U(x,t)
f(x).
Then the
t+l
Poisson-Hankel transform (2.3) becomes the Weierstrass-Hankel transform and is
given by
I G(x,y;l) #(y) d(y)
f(x)
0
Now write R(x,l)
R(x), so from (3.1)
2 2
R(x)
F(X) +
)
G(v;l) f(iv) d(v).
e
0
which on simplifying gives
R(x)
1
4v+i/2
o
J
0
v
2
e
2
2 2
v x
v
16
4
f (iv) dv.
C. NASIM
494
According to the inversion algorithm (3.3), we have
e
-x2/4
i
(x)
r(
R(x)
1
+ _)
2 2
i
v
16
4
e
4+I/2
+
r(u
2
v x
f(iv) dv
2 2
v
4+i/2
2
-v
e
2/4
f (iv) dv
r(
0
formlly.
+
Now using the result (2.5), we obtain,
e
2
-x /4
I
(x)
4
]o
e
_2 /4
+i/2
v
x
i/2-
f(iv) dv.
J9-1/2
0
Thus,
I
(x)
C(ix,v;1) f(iv) dp(v),
0
giving the inversion of the Weierstrass-Hankel transform and agreeing with the
inversion given in [4].
Now if we write G(0,y;t)
G(y;t) and U(0,t)
f(t), then the Polsson-Hankel
transform given by (2.3) becomes
f(t)
f
G(y;t) (y) du(y),
0
We can write it, using the
and is called the reduced Poisson-Hankel transform.
definitions of G and dp and making a suitable change of variable, as
f(t2/4)
2
r(u +
i
)
I
e-y2/t 2 l_y
)t -(29+i)
(y)dy.
0
Hence by Lemma (3.1), we have
r(
or,
r( +
2
i
i
+- }el
i
i
f()
x
i
)
2
U(0 ’T
1
r( +
i
1
i
r( +
(x)
i
)
U(x,0),
INVERSION OF THE POISSON-HANKEL TRANSFORM
495
This establishes the inversion of the reduced Polsson-Hankel transform as in
Ill.
Next we shall illustrate the inversion procedure for the Polsson-Hankel transform by an example.
Let
(x)
+ e >
2
x
i
>
-i,
The function satisfies the condition of the main theorem, and
G(ix,y;t)y
U(ix,t)
e
d(y)
0
(2t)-i e x2/4t
x
F e-y2/4t
i/2-
y
e+v+I/2
xy
2t
J9-i/2
dy
0
r( +
(4t)
i
I/2 (++i/2) -i/2- e
x
i)
x2/8tMl
/ 2 (++i/2), i/2 (-i/2)
(x2/4t)
[5, p. 185], M being the Whittaker function.
Now
R(x,t)
F(v +
r(v
)
-v2x2 / 16t2
G(v;t) U(iv:t) d(v)
0
2
I
+2- +)
i
r( + )
x
2
2
(a--i/2)
(4t) i12
+!)
8t
e
0
9-I/2MI /2(e+v+I/2), i/2(9-i/2) (v2/4t)dv
v
r(
1
+
1
e
+)
x
F( +
1
(
x
2
+7
-(+e/2 + 1/2)
[5, p. 215].
Hence,
r(
+-
[R(x,t)]
1
+)
2
r(
+
2
e
e
-x / 4t
-x2/4t
x
by (2.3)
(x),
-(+e/2 + 1/2)
C. NASI
496
according to the main theorem, whence, as predicted,
<x
x
REFERENCES
i.
HAIMO, D.T. and CHOLEWINSKI, F.M.
Transform, J.
2.
WIDDER, D.V.
d’Analvse
Inversion of the reduced Poisson-Hankel
Math. 25 (1972), 323-343.
Inversion of a Convolution Transform by use of Series, J.
d’Analyse Math. 21 (1968), 293-312.
3.
NASIM, C.
4.
CHOLEWINSKI, F.M: and HAIMO, D.T. The Weierstrass Hankel Convolution Transform, J. d’Analyse Math. 17 (1966), 1-58.
5.
ERDELYI, A.
An Inversion Formula for Hankel Transform, Pacific J. of Math.,
Vol. 57, No. i, 1975.
et al.
Tables of Integral Transform, Vol. 2, New York, 1954.