 Open Access
 Total Downloads : 60
 Authors : Anil Tiwari , Dr. Archana Lala , Dr. Chitra Singh
 Paper ID : IJERTV8IS050248
 Volume & Issue : Volume 08, Issue 05 (May 2019)
 Published (First Online): 18052019
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Conversion of Dual Integral Equations into an Integral Equation
Anil Tiwari1
Ph.D. Scholar, Dept. of Mathematics Ravindranath Tagore University Bhopal, India
Dr. Chitra Singp
Department of Mathematics
Dr. Archana Lala2 Department of Mathematics S R Group of Institutions Jhansi, India
Ravindranath Tagore University Bhopal, India
Abstract The purpose of this paper is to reduce a dual
(1,1)
1 (a n 1) , 1
integral equation into an integral equation whose kernel
L(x) H 1, 0 x
r 1
r
involves Generalized Hermite Polynomial by use of mellin transform. We believe that there is some more possible way to reduce such dual integral equations using different transform
and
2,1
(1 n,1)
like those of Henkel, Fourier. For the purpose of illustration we consider a dual integral equation of certain type and by use of fractional operator and mellin transform reduced it to an integral equation.
t (x) h(x), 0 x 1
r x n a1
t(x)
1 (a a )
Keywords Generalized Hermite Polynomial; Mellin Transform; Fractional operators; FoxH function.
r 2 1
1
r r
r ( a2 a1 )1
n a r 1
I. INTRODUCTION
Dual integral equations are often encountered in different
0 (v
x ) v 2
g (v) dv;1 x
branches of mathematical physics and they generally arise while solving a boundary value problem with mixed boundary conditions In the present paper, we attempt to solve the certain dual integral equations by converting

MATHEMATICAL PRELIMINARY
To prove the theorems we shall use Mellin transformer and fractional integral operator.
them into an integral equations. Many attempts have been made in the past to solve such problems. The following integral representation is basic tool for our illustration.
f * (s) M [ f (x); s] f (x)xs 1 dx
0
When s i is a complex variable.
(3.1)
k (x, u) A (u) du (x); 0 x 1
(1.1)
The inverse melling transform f(x) of f*(s) is given by
0 1
M 1 [ f * (s)] f (x) 1
i
f * (s) xs ds
(3.2)
0 k2 (x, u) A(u) du (x); x 1
(1.2)
2 i
i
k1 & k2 are kernels defind over xu plane.
By convolution theorem for mellin transform
r r d
r r d
M k(xy) f ( y) dy; s k* (s) f * (1 s)
n
n
Hr (x, a,

(1)r xaepx Dn [xaepx ] D
dx
0
0
0
a, r, p parameter.
k (xy) f ( y) dy M 1 k* (s) f * (1 s); s
II. THEOREM
1
k* (s) f * (1 s) xs ds
(3.3)
If f is unknown function satisfying the dual integral
2 i L
equation.
r
r
(xy)a1 e( xy ) H r (xy; a ,1)
f ( y) dy h (x); 0 x 1
When L is suitable contour. Fractional integral operator
r x
r x
0 n 1
r r 1
( ; ; r; w(x)) (xr vr ) 1 v w(v) dv
(xy)a2
e( xy ) H r (xy; a ,1)
r
r
(2.1)
f ( y) dy g (x);1 x
( ) 0
(3.4)
0 n 2
(2.2)
r x
1
r r r r 1
When h and g are prescribed function and parameters, then f is giving by
f (x) 1 L(xy) t( y) dy
a1 , a2 and r are
R ; ; r; w(x)
( ) x
(v x ) v w(v) dv
(3.5)
r 0
Where


SOLUTION
Now taking
r
r
kk ((xx)) xxaaii ex Hr (x;a ; 1), i 1, 2
1
(s)
1 (s n a
2
2
r x s
i n i
Then from Erdeeyi [10] We get
2r i L
s n
s 1 (s n a )
r i
1 ( s)

r
f * (1 s) ds
k * (s)
, i 1, 2
(4.1)
i r s n
. 1 1 s
r r
Hence by use at (3.3),(2.1) & (2.2) can be written as
1
rx (vr xr ) 1 v r r 1 g(v)dv
1
s (s n a x
r i
r i
f * (1 s) x s ds
2r i L
s n
In equation (4.4), we put n a1 and
h(x); 0 x 1
(4.2)
1 (a
r 2

a1 ), so that (4.4) Changes to
2
2
(s) 1 (s n a
r
1
1
f * (1 s) x s ds
(s)
r (s n a2
2r i L
s n
1
x s
g(x); 1 x (4.3)
Now operating a (4.2) by the operator (3.5) we get
2r i L
s n
1 (s n a )
1 r 1
s r (s n a2 )
a a 1 1
f * (1 s) ds
rx . 1
f * (1 s) x s ds
2 1
(n a ) s
2r i
r
L
1
(s n)
r r 1
r r r 1 r
r xn a1
(v

xr )
v g(v)dv
x
rx
(vr
xr ) 1
v r r 1 g(v)dv
1 (a a )
2 1
2 1
r
x
1
2 1 na r 1
2 1 na r 1
(a a )1
r (vr xr ) r v g(v)dv 1 x
Now putting
vr x and simplifying we get
t
2
x
1
1
1
1
(s) r (s n a1
1
(s)
(s n a
r 2
r 2
x s
.
2r i L
s n
xs
f * (1 s)ds
2r i L
1
s n
s 1
r xn a1
1 r *
(1 t) .t dt f
(1 s)ds
1
0
rx
(vr xr ) 1 v r r 1 g(v)dv
(4.4)
r (a2 a1 )
1
x r
r (a2 a1 )1 na r 1
x (v x ) v
r 2
g(v)dv 1 x
(4.5)
2
2
(s) 1 (s n a
r s
Now we write
1
. x
t (x) h(x), 0 x 1
2r i L
s n
r xna1
and t(x)
1 ( S )
1 (a a )
r 2 1
r f * (1 s) ds
1
1 1
r r
r (a2 a1 )1
na r 1
r r
r r
s
x
(v x ) v 2
g (v) dv;1 x
(4.6)
rx
Now from (4.2), (4.5), (4.6) we get
x
(vr xr ) 1 v r r 1 g(v)dv
(s)
1
1 (s n a
1
1
r
xs
f * (1 s)ds t(x)
2r i L
s n
Again using (3.3), (4.1) & (4.6) becomes
k (xy) f ( y) dy t (x); 0 x
(4.7)
(4.8)
REFERENCE

S. Ahdiaghdam, K. Ivaz and S. Shahmorad, Approximate solution
0 1
r
r
1 x 1
1 x 1
When k (x) xa1 ex Hr (x; a ;1)
Thus pair at dual integral equation (1.1) &(1.2) we have been reduced to single integral equation (4.8). Hence by mellin transform(4.8) can be written as
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k* (s) f * (1 s) T*(s)
1
1
(4.9)

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s
* r
(s n a1
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A. J. Jerry, Introduction to Integral Equations with Applications,
Where k (s)
1 s n
Second Edition, WileyInterscience, New York, 1999

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and T*(s) is the mellin transform of t(x). Now replacing s by (1s) in (4.9)
F* (s) L* (s) T* (1 s)
L* (s) 1
k* (1 s)
1 s n
(4.10)

Gould H Wand Hopper A T, Operational formulas connected with two generalizations of Hermite polynomials, Duke Math. J. Vol 29, 1962, PP.5163.

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Proc. Glasgow Math. Assoc. 4 (1960) 108110.

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(1 s) 1 (1 s n a )
r 1
By use of definition of H function, we get the inverse transform L(x) at L*(S) as
(1,1) 1 (a n 1) , 1
L(x) H 1,0 x
r 1 r
(4.11)
2,1
(1 n,1)
Taking inverse mellin transform of (4.10)
f (x) 0 L(xy) t( y) dy
Hence using (4.11) we get
1
1
f (x) 1
H 1,0 xy
(1,1) r (a1 n 1) , r t ( y)dy
r 0
2,1
(1 n,1)
When t (y) is given by (4.6). Hence proved the theorem.