493
ON K-TRANSFORM
C. ASlM
Department of l[athe’matics
The University of Calgary
Calgary, Alberta
Canada
(Received Hatch 12, 1980)
ABSTRACT.
Using a coblnatlon of infinite order linear dlfferentlal operators and
Integral operators, the inversion of K-transform is establlshed.
Inversion
procedures for Laplace transform and Potential transform are derived as special
cases.
KEV
0R$
A PES.
/(-transform, Laplace transform, dlfferentlal operators.
1980 MATHEMATICS SUBJECT CLASSIFICATION COF..S. 0029
i.
INTRODUCTION.
In this paper, we discuss the inversion of the K-transform, in Hilbert space
L2(0,(R)).
The method involves only the real analysls and employs dlfferentlal
operators of infinite order, cf [
43 and [ 6, Chapter vii]
An algorithm for
C. NASlM
494
inverting the bilateral Laplace transform is established as a special case.
Also
some examples are given to illustrate the procedure.
L 2(0,) and
Let y(z)
[5, p.94]
LEMMA i.
0 f(x)xS-I
F(s)
it,
<
t <
then
1
f(x)
+/-m
[1/2+i
t
F(s )x-sds
t 1/2-it
F
f.
is called the Mellin transform of
[5, p.95]
LEMMA 2.
-
If
f
and g
L2(0, )
and have Mellin transforms F and G
respectively, then
0
f
1
g(xt)
Next some operational considerations.
If 8
=-zd
then it is an easy matter to see that
8n[x-8]
snx-s,
pn(8) [z
Pn
+ it,
< t <
Therefore
where
pn(8)
is a polynomial of degree n in 8; consequently
p(e)[x-8]
limpn(e)[x-8]
p(s)x
-8
Also,
n[
x-s
-
495
K-TRANSFORM
in n
e
N
llm
N-,
Z
k=o
(8 in n) k
k’.
[-8
lira
r
(s In n)
k!
x
N-w k=o
-s
s
=nx-s
Wth this understanding, we interpret the operator
-S]
B) [x
lira n
n-,,
e)
a linear dlfferen-
[6, p.234], such that
tial operator of infinite order,
r(1
I
(1
e
[x-s]
i
k--I
I
r (1
s)
a:-S
Similarly
i
r(
S) [
-8
r(
I
-s
(’)
Ss)
Next we define the operator F(
8B) as having the property that
(1.2)
-
This is not a dffferential operator in the above sense, but behaves in the manner
of (1.2) for all
2.
8
c
+ it,
<
t <
, R(8)
THE MAIN RESULT
THEOREM i.
Let the functions
f
and
<
8
(L2(0,).
Define
(xt)K (xt)@(t)dt
f(x)
o
where
II
i
-<%-
> 0 and
K being the usual modified Bessel function of order
C. NASIM
496
If
l
i(R) (mt)SJ (xt)f(t)dt
0
II_<1/2
> 0 then
S
[R()]
PROOF.
(), a.a.
Note chat the integral defining the function
L2(0, (R))
gent, since
and
taKe(t)
L2(0, (R))
(
f
0.
is absolutely conver-
due to the hypotheses.
Now,
R()
0
2
() ()f()d
I(R) (xt)8
J (zt)dt
o
r
0
I
x
0
The change of order of integration is
0
(ut)
o
()
Justified by absolute convergence
(u)
using the asymptotic expansions of the Bessel functions J and K.
The t-integral
can be evaluated, [I(II), p.137] so that (2.1) above gives,
,,,h,.,
Hypergeometric function.
and a
1/2(x-I-t+l-I’-a.+13), b-1/2(u-+l-l-c.), 2F! being the
497
K-TRANSFORM
Let,
()
k ()d
where
k(x)
Now k
since
(L2(0, (R))
2 a+ F (a) F (b)
r(+l)
since 8
+ v
>
-
xa+ 2El (a,b
and
I[
_2
/I
< a
+-
and further
1
(o,(R)),
z
does, hence by Lemma 2, we have
(s)(s)x-sdo
()
1/2-i(R)
where # and K are the Mellin transforms of # and k respectively and [i(I), p.336],
(")
Next,
2a+-I
---V--
[1/2(u+a+l-s) ]r [1/2(v+8+.) ] r [1/2(a-u+l-s) ]
r [1/2(-S+2-) ]
r
we apply the operator
]<(o) [}(z)]
1
Using the results (i.I) and
d
to the function
R(z),
to obtain
#(s)z(s)z-s
z(o)
=
-z
where 8
K(8)
1/2-i-
(s Ix-s]
(.)z(s)
1/2-i(R)
(1.2),
we see that
1
K(8)
[z-8]
1
-8
----V a:
A’8
Thus,
1
K(e) JR(z)]
i
[%+(R)
(z),
as required
1/2-i(R)
z-" d
e(s)K(s) K(s)
1/2_(R)
a.a.
x > 0,
The bringing of the operator
i
K(8)
inside the integral, amounts to
C. NASIM
498
nothing more than differentiating inside the integral sign.
EXAMPLE i.
(z)
Let
2
f(x)
0
2
Jn (z).
Then
(xt)Jn (t)Ku (zt)dt
-1/2 -(n+l)
--q-
F[1/2(n++e+l)]F[1/2(n-++l)]
r
2FII1/2(n+++l)’ 1/2(n-u+e+l);
[I(II), p.137].
n(x)
n+l;
And
e
F (c)r (,d)
-V r (.+i) x8
2
0
tS-o-I 2Fl
J (xt)dt
c,d; +1;
where
2a+B-n-I
1/2i-c’
1-d
(S-n+v),
o,
-n,
I
-(S-n-)
[i(II), p.82], where G is Meljer’s G Function.
Now
2-nr [1/2(-S+2-8) ]
i
c(o) R(x)
r
[1/2(++l-e) ] r [1/2(v+s+e) ]r [1/2(-+l-e) ]
(S-n+u),
---2 -n
_r ,,.,10
l,l_. x2
0,
J (x), as predicted.
o,
-n,
-(8-n-v)
I
499
K-TRANSFORM
The operator
(’),
applied on
K(’8)
is evaluated by writing
(’)
in terms of
the complex integral and then the differential operator being applied inside the
integral to give
3.
rl2 ()
0
SPECIAL CASES.
(i)
+ 8
Let
2 "+
r
i
+ v.
B
The inversion of the main theorem
[1/2(.+=+l-O) ]r [1/2(S++e ]
(=),
R(=)
where
(x)
2 ++l
r(u++l)
xB+
I
U
+
o (x+u) u++
giving us an inversion of the generalized Potential transform.
The following special cases provide us with a procedure for inverting Laplace
transform, cf. [3] and [6, p.232].
(ii)
i
B
If
THEOREM 2.
then Theorem i gives
and @ ( L2 (0,).
Let
f(x)
i
and
Define
(t)dt
e
o
the bilateral Laplace transform of @.
2
R(=)
I
cos xt
If
f(t)dt,
0
then
sin
Here
i
8[R(z)]
the operator
(x),
i
K- is
a.a.
x > O.
reduced to the operator sin
pret this operator by considering the product expansion
i
8.
We can inter-
C. NASIM
500
=
Now if S (a)
sin
(ill)
1
H
wa
1
we[za]
ll,
1
Put e
THEOREM 3.
f(z)
f
Let
then, for example
.a
,.S(-c*)m1
v
and 8
-sin
1
s
we m
in Theorem i, then we have
and # { L 2 (0,-).
Define
e-Zt# (t)dt
o
.
the Laplace transform of
w j0
sin
If
mt f(t)dt,
then
cos
1
(z),
we[R(z)]
a.a.
z
Here again the differential operator
> 0.
i
K(’e)
of Theorem i is reduced to
which can be interpreted in a similar way as the operator sln
instance, cos
i
r1
e--
cos
i
wa
i
wS.
me
Hamburger’s formula.
(iv)
EXAMPLE 2.
sin2nm.
Let (m)
slnZnt e-mtdt
f(m)
0
(Hamburger
Then by Theorem 3,
(2n).!
Z(z2+2 z) (zx+4z)... (x2+(2n) z)
s formula).
Now
R(m)
2 (2n)
w
0
sin mt
t t2+2 z )(tz4z) .’: .(tZ+4n z
[2, p.414].
Thus, according to Theorem 3
)dt
cosI wS,
Also, for
501
K-TRANSFORM
e [R(m)]
cos
n-I
(_i)n2_2 n
cos e[e
-0
1
cos,,ee"=
-
1
>.
cosine k=o
+ (-i
k,.
k
k=O
cos
k
m
cos
Zk=0 -.
cos
1
zk
Z
k=0
2k
m
k 2k
(-1) n
(2)’
n%q.
Hence
B[R(=)]
cos
(-l)n2 -2n
2
Z (-i)
k=0
cos[2(k-n)n] + (-i) n
2n0:
sln
[2, p.25], as predicted, by the theorem, verifying the Hamburger formula.
REFERENCES
i.
Erdelyi, A. et al. Tables of integral transforms, Vols. I and II, McGraw-Hill,
New York, 1954.
2.
Gradshteyn, l.S. and Ryzhik, I.M. Table of integral, series and products,
(Translation) Academic Press, New York, 1965.
3.
Haimo, D.T. and Cholewlnski, F.M. Inversion of the reduced Poisson-Hankel
transform, J. d’Analyse Math. 25 (1972).
4.
NASIM, C.
An inversion formula for Hankel transform, Pacific J. of Math. 57
(1)(1975).
5.
Tichmarsh, E.C. Introduction to the theory of Fourier integrals, Second
Edition, Clarendon Press, Oxford, England.
6.
Widder, D.V.
1971.
An introduction to transform theory, Academic Press, New York,