Internat. J. Math. & Math. Sci.
Vol. 9 No. 2 (1986) 293-300
293
ON DUAL INTEGRAL EQUATIONS WITH
HANKEL KERNEL AND AN ARBITRARY WEIGHT FUNCTION
C. NASIM
Department of Mathematics and Statistics
The University of Calgary
Calgary, Alberta
Canada
T2N 1N4
(Received July 3, 1985)
In this paper we deal with dual integral equations with an
ABSTRACT.
functlon
and
Hankel
of
procedure, which depends on exploiting the properties
readily
reduces
function.
Most
the
Mellin
type,
Fredholm
weight
transforms,
and
This then can be inverted by
the dual equations to a single equation.
the Hankel inversion to give us an equation of
arbitrary
We propose an operational
kernels of distinct and general order.
involving
unknown
the
of the known results are then derived as special cases of our general
result.
KEY WORDS AND PHRASES.
transforms,
the
Dual integral equation, Bessel functions of first
kind,
Mellin
Parseval theorem, Banach space L(k-i,k+i(R)), Hankel transforms, Hankel
inversion, Fredholm equation.
AMS CLASSIFICATION CODE.
I.
45FI0, 45E]0
INTRODUCTION.
We shall consider dual integral equations of the type
22 t-2 J
(xt)[l+w(t)]@(t)dt
f(x)
O
<
x
(1 i)
1
o
22 t-25
J (xt)@(t)dt
g(x)
x
I.
o
Such equations arise in the discussion of mixed boundary value problems.
O, then (I.I) become dual equation of Titchmarsh
type,
[I].
Different
If w(t)
methods
for
solving dual integral equations have been proposed by various authors, notably, Tranter,
[2],
[3],
[4]
and Cook [5].
More recently, Erdelyi and
Sneddon gaw a solution using fractional integral operators [6]. Basically, all these
methods, make use of some form of integral operator to reduce the system (l.l) to a
single equation, which is then solved using standard techniques.
In this paper we use a different approach. We develop an operational procedure,
Lebedev
and
Uflyand
Noble
which consists in exploiting the properties of the Mellin transforms.
the
By this technique
dual equations are readily reduced to a single integral equation, which in turn can
be solved using the usual Hankel inversion.
A somewhat similar method has been used for
solving ordinary dual integral equations by Williams [7], Tanno, [8] and Nasim and
Sneddan [9].
294
2.
C. NASIM
KNOWN RESULTS.
Lemma
First, we wrlte down some detinltlons and results which are used later.
e L(O,) and F(s) e L(k-i, k+i), then
[i p.94]. Let
xk-lf(x)
f(x)xS-ldx,
F(s)
k + it,
s
<
r
o
and
1
f(x)
F(s)x-Sds.
lim
1 r-
"k-it
F is then the Mellin transform of f written as
F(s)
l[f(x);s]
and f is the inverse Mellin transform of F, written as
-1
f(x).
l [F(s);x]
Lemma 2 (The Parseval theorem) 1 p. 60].
L(O,) and F(s)
L(k-i,k+i),
If x -kg (x)
then
F(s)G(I-s)x-Sds,
f(xt)g(t)dt
i(R)
o
where ,F and G are the Mellin transforms of f nd g respectively.
3.
EOUATIONS.
Now, we write the system (I.I) as
DUAI, INTEGRAL
o
x-2tXf(x),
hl(Xt){l+w(t)}(t)dt
f h2(xt)(t)dt
x
-2g (x),
<
0
x
<
(3.1a)
1
> I
x
(3.1b)
o
h
where
and h
2
22ax-ZaJ
l(x)
(x)
22x-2J
h2(x)
and
(x).
Then the Mellin transfor of h
are respecatively,
H (s)
2
s-1
F(1/2s+
1
2a-v
1
< Re(s) <
2a
r(l-2s +u+a)
and
1
2- < Re(s) < 2,
r(1- 1 s + i
+)
2
k+ir, [10]. The left hand sides of the equations
(3.1b) represent functions for all values of x and we shall denote these by
both belonging to L(k-i(R),k+i(R)), s
(3.1a)
fl(x)
the
and
and
gl(X),
Fl(S)
having ellin transform
Now we set
Fl(S)
and
functions
w(x)
and
Gl(S)
@(x).
and
Gl(S),
respectively.
both e L(k-i,k+i(R)) and put
Then,
to
due
lemma
appropriate
2,
conditions
on
the equations (3.1) give,
respectively
Hl(S){#(l-s
+
F(l-s)}
Fl(S)
(3.2)
H2(s)#(l-s
G
l(s>
DUAL INTEGRAL EQUATIONS WITH HANKEL KERNEL
-,
v
k + iv,
a.e. on s
where,
Next, consider the functions
F(I
,
s +
1
+a)
u
1 s +
+ )
r(
s
a)
r(1/2
s
r(l
H2*(s)
(s).
#(s), and *[w(x)(x);s)
*[(x);s]
HI(s)
295
,
Re(s)
<
2+2a+u
e(s)
>
2= u.
s +
2
u
)
Note that
H I(s)H*l(s)
-=
Then
2s-if(
H2(s)H2*(s)
r(1
1
-2s
a)
1
+ 2, +)
K(s), say.
10 p. 326]
T
where
(2x)-TJA(x),
1/2
*-l[K(s);x]
k(x)
1
u + a + / and A
U
We now write equations (3.2) as
Multiply the above
Hl(S)#(l-s) Fl(S) Hl(S)%*(l--s
H2(s)O(l-s) Gl(S).
(s)
equations by the function Hi(s) and
the
respectively and using
definition of K(s), we obtain the following pair of functional equations,
H(s){Fl(S) Hl(S)(l-s)}
H*(S)Gl(S
K(s)(1-s)
2
(3.3a)
K(s)#(l-s)
k + iv,
a.e. on s
v
-.
First we consider the equation (3.3a), whence, for x
1
2,
(3 3b)
k+i- K(s)#(1-s)x-Sds
O,
*
-[1 Ik-i" Hl(S){Fl(S)
Hl(S),(1-s)}x-Sds
k-i"
Here K(s) e L(k-i-,k+i’) if 2a-
>
k+i"
1
1
--u+a+ and if
k
we let
x-k/(x)
(3.4)
e L(O,-), then by
applying Lelama 2, the left-hand side of (3.4) gives
f
(3.5)
k(xt)/(t)dt.
o
The right-hand side of (3.4) is
1
4i
x
I
1-2a-u d
d--
1/2 s + 1/2 u +
a)..
1
1
F(I
k+i
{FI (s)-Hl (s)?(l-s)
fk+i(R) {Fl(s) -Ht(s)i*(1-s)} r(=
k-i(R)
r(1
s +
u)
,
1
i
2 s +
x-Sds
x
u+2a
Sds.
+ )
(3.6)
C. NASIM
296
We define,
l-l[Fl(S)-Hl(S)(1-s);x l-l[Fl(S);x l-l[Hl(S)F(l_s);x
a(x),
f(x)
I
1
x
whet e
I
Here
tl :s) e L(k-i=,k+i(R))
if
k+i(R)
HI
k--i(R)
l-s x
s
-s
ds
(3.7)
hl(xu)w(u)(u)du
o
2a-u
2a
k
and
if we let
x-kw(x)
/(x) e L(O,(R)), the
r(a
result
from
follows
then
2
lemma
above.
The
function
1
r(l
L(k-i(R),ki(R))
k
if
and
2a+
1/2-u;
a-D
and if we assume that x
-kfl
both both belong to L(o,(R)), then by applying Lema 2 on the s-integral,
1
(x) and x
the
(x)
expression
(3.6) yields
1. *.x.
x 1-2a-[x 2a+u: [fl(t)-Q(t)]nlt)dt
0
1,
x
o
where,
O, [I0:p.350].
and T--2al
r(r-
za
x
On simplifying the last expression, (3.6) then becomes
1-2a-u
+ 1)
d
u--+2a-2D
] x
m
l(x),
x
I [fl(t)-(t)] t2a+u-1 x2 t2 $-2adt}
o
<
0
x
< I.
(3.8)
Hence combining (3.5) and (3.8), the equation (3.4), finally becomes
I
The
above
analysis
condition that a--/]
is
k(xt)(t)dt
ml(X),
o
justified
t-.
we
if
x
0
consider
< I,
(3.9)
the strip 2a--u
The formula is actually valid for a-/
k
<
<
2a, with the
I, and it can
be extended to the full range by analytic continuation.
Now simplify the expression (3.8), by making use of the definition of the functions
fl(t) and (t) from (3.7), we get
ml(x)
rT:
r(r-2a+l)
x
2-a+l I ) l-2a-d[ xU--.+
x
xU-+2a-2/ tu-l(x2-t2)’-2af(t)dt
I
2a- 2
o
Ix
t
u 1
(x2_t2)T_2adt
-12,
On changing the order of integration in
12,
u2a(u)w(u)du
o
I(R)u_2aJu
o
o
t
o
(ut)/ (u)e (u)
du]
say.
we can write the double integral as,
u-
(x2_t2)Y--2aju(ut)dt
DUAL INTEGRAL EQUATIONS WITH HANKEL KERNEL
xZ(b,+’l"-2a):uU-2a/(u)(u)1F2(u.,+1
,8(’r-2:+I,)
u+l
r(w+l )2
10" p. 327]. Then,
297
22
---)du
u+T- 2a+I;
o
1-2a-u
D(T-2a+l,u)
x
]
u+l
r(’r-2a+l)r(u.+ 1)2
d[x2U oUU-2a/(u)w(u)
12
22
which on differenting inside the integral sign and simplifying gives,
12 22a--Ux
Hence,
ml(x)
r(’r- 2a + I) x
22a-Yx-Y
-
T u-T/(u)w(u)JA
(ux)du.
o
x u--+2<x-2
u-Y#(u)(u)JA(ux)du"
-.
o
tu-l(x2-t2)T-2af(t)dt
O
(3.10)
Next we consider the equation (3.3b), whence
K(s)(l-s)x--Sds
--ico
As be fore
H (s)G
K(s)#(1-s)x-Sds
k--ira
2x-u
where
1
k
1
vI
]u
+ a +
(s)x-Sds
(3.11)
--i(R)
k(xt)@(t)dt,
(3.12)
o
And
k+i(R)
I
2.--
Ik_i(R) (S)Gl(S)x-Sds
x
]
r(s +
Ik_i(R) r(s + 1/2.
k+i"
2-
F( s +
-i(R)
x2_U+2a
a
:)
G (s) x-Sds
1
)
I)
G (s)x
1
-Sds]
(3.13)
r( s*
Here
r(21-s
1)
2u-a-
L (k-i,k+i(R)) if k
e
)
1
2+2a-u and
1
+
Further if
.-)
x -kg l(X)
e L(O,#), then by La 2, the above expression gives
x
x
2-u+
gl(t)
h
2
where
1
(s+-a
-1[ (21_s+._)
T--2/"’
O
---I <
_.e.
2
-1)
];x h(x)_
,r(-:2+l)
x
p-2a-2
(l-x
2
0
x
,x> I,
0
-/, [10, p.349].
Hence on simplifying, we have from (3.13),
.k + i(R)
Jk--i(R)
..,2
(s)
l(s)x-sds
x
v-- 2a- 1
F(T-2D+I)
m2(x),
x
I.
(3.14)
298
C. NASIM
From the results (3.12) and (3.14), the equation (3.11), then gives
k(xt)(t)dt
m2(x)
o
-
(3.15)
l,
x
The above analysis is justified if we consider the strip
1 1
2a-v+2
k
2 and
The formula is actually valid for
< a-,
--i
by analytic continuation.
--.
and it ca extended to the
range
full
Now combining the results (3.9) nd (3.15), we have,
;
mr(x),
m2(x),
m(x)
where
k(xt)(t)dt
m(x),
0
x
<
(R),
o
0
x
ml(X)
x
and
m2(x)
defined by (3.10) and
(3.14) respectively,
and
I=-1
or,
2
2,
+ 1.
-rJA(xt)0(t)
(2t)
o
m(),
which by the usual Hankel inversion gives,
<x)
2
7‘
x
rl
t T+l
o
t
T
JA(xt)ml(t)dt
o
Now substituting the values of
(x)
27‘
F(T-2a+i)
r
(-27‘2/+
x
T+l
ml(t)
_2-2u+A
.[o :
xT‘+]
I)
and
22T-2axT+ 1
m2(t)
JA(xt)d
JA(xt)2(t)dt
/uU-1 (t2-u2) 7‘-2af(
u)du
o
ul--(u2-t2)7‘-2/g(u) du
t
7"-1
12 22T-2ax
t7‘+1
above, and simplifying, we obtain,
(xt)d
12
xT*I
t u-+2a-2/
;1 t J
;1 Jh(xt)dt f
o
i1
27‘
+
JA(xt)m(t)dt
o
u-T(ulw(U)JA(Ut)du
13,
say
j’ u-TO (u)w(u) I
t
du
o
o
JA (xt)JA(ut)dt
(3.16)
0
where L(u,x)
uJA+I(U)JA(x)
x
U --X
JA+I(X)JA(u),
A
1/2M + u
and
I--1
1
<2.-+
+
, "M
7‘
1.
This is the integral equation of the Fredholm type and
can be written as
/(x)
A(x) +
E(,u)(u)du.
o
I
+ a
,
DUAL INTEGRAL EQUATIONS WITH HANKEL KERNEL
299
Thus the solution of this single integral equation gives us the
value
of
the
unknown
function O(x), which is the solution of the system (3.1), as well.
4.
SPECIAL CASES.
In particular if
u, then the solution of the system
2
2cx
t-2cx J
(xt)[l,(t)](t)dt
f(x)
<
0
I
x
(4.1)
o
2 2 t -2 J (xt)(t)dt
g(x)
x
o
is the solution of the equation,
/(x)
t
r(-=+l)
o
2 a+fl
a+fl+l
F(a--/ I) x
-u-+Ju_a+(xt)d t 2a--2
f
u-a--// (u)w (u) L (u’ x)
O
L(u,x)
[a-/[
and
If
2
U -X
Ju_++l(U)Ju_a+(x)
u
)-af(u)du
(u2-t2 g(
u-+$
-22flxa+/ 1
where
u
o
x
(4.2)
2’
Ju_a++l(X) Ju_+(u),
1, derived as a special case from (3.16).
a--
1, then the differentiation under the integral sign in the
second term of (4.2) can be carried out, and we have
I
(x)
(4.3)
13,
I
we
consider
0
12
where
2a++l
I2
The special case
discussion
of
xa++l[-i tu-++Ijv-a+$tx)dtftul-U( u2-t2) a-- Ig(u)du"
F(a-)
O,
v
contact
certain
1
O, a
of
is
interest,
problems in elasiticity.
since
it
arises
in
the
The dual equations (4.1) now
become
t -I J (xt)[1-(t)](t)dt
f
Jo(xt)(t)dt
o
where the unknown function
f(x)
x
0
g(x),
x
> I,
satisfies the Fredholm equation, which can be derived
from
(4.3) to give,
/(x)
W
cos(xt)d
x
o
cos(xt)dt
x
du
o
)t 2 -u2
du +
t
42_t2
K(x,u)/(u)(u)du
o
with
K(x,u)
On
the
other
hand,
if
x[sin(x+u)
-[.
we
+
consider-I
sin(x--u)]
x
u
< a- < O,
[6; 4.6.28].
then the differentiation under the
integral sign in the first term of (4.2) can carried out, and then we have
,(x)
11
12 I3,
(4.4)
300
C. NASIM
where
2a++Ir
++I
y(_) x
1
f
1
o
tl-u+-Ju_a+(xt)dt
O,
One can, now deduce the special case when u
t
uU+l (t 2 u2)--lf (u)du.
o
-.
1
0 and
easily.
In this case
the solution of the dual equations
o
t J (xt)[l+(t)]#(t)dt
o
o
J (xt)/(t)dt
o
f(x)
x
0
g(x)
1
x
is the solution of the equation, from (4.4),
@(x)
sin(xt)dt
o
du
o
ug(u)
sin(xt)d
Jt2_u2
du
K(x ,u)(u)du,
/
Ju2_t 2
1
o
with
K(x,u)
ACKNOWLEDGEMENT:
1
;t[sin(u4x)
sin(u-X)]u
x
[6; 4,6,40].
This research is partially supprted by a grant from
Natural
Sciences
and Engineering Research Council of Canada.
REFERENCES
I.
E.C. Titchmarsh, Introduction to the Theory of Fourier Integrals, Second Edition,
Clarendar Press, Oxford, 1948.
2.
C.J. Tranter, g further note on dual integral equations and an application to the
diffraction of electromagnetic waves, Quart. J. Mech. and Appl. Math, 7_ (1954),
318.
3.
N.N. Lebedev, Ya. C. Uflyand, Appl. Math. Mech. 22, 442 (English translation).
4.
B. Noble, The solution of Bessel function dual integral equations
multiplying-factor method, Proc. Cambridge Phil Soc. 59 (1963a) 351--362.
5.
J.C. Cooke, A solution of Tranter’s dual integral equation problem, Quart. J. Mech.
Appl. Math., _9 (1956), 103.
6.
I.N. Sneddon, Mixed boundarvablems in potential theory, John Wiley & Sons,
Inc., New York, 1966.
7.
W.E.
8.
Y. Tanno, On dual integral equations as convolution
_(., 2_9_0, (1968), 554-566.
9.
C.
by
a
Williams, The solution of certain dual integral equations, Proc. Edinburgh
Math. Soc. (2), 12 (1961), 213--216.
transforms,
Tohoku Math.
J.
Nasim and I.N. Sneddon, A general procedure for deriving solutions of dual
integral equations, J_._Engg. Sc. Vol. 12(3), (1978), I15-128.
I0. A. Erdelyi et al., Tables of integral transforms, vol. I,
1958.
MO3raw-Hill,
New
York,