**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):**18-05-2019 -
**ISSN (Online) :**2278-0181 -
**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; Fox-H 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 x-u 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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No. 5, pp. 10771086

A. Chakrabarti and S. C. Martha, Methods of solution of singular integral equations, Math. Sci. (2012) 29 pages.

N. A. Hoshan, Exact solution of certain dual Integral equations involing heat conduction equation, Far East Journal of Applied Mathematics 35(1)(2009) 81-88.

k* (s) f * (1 s) T*(s)

1

1

(4.9)

S. R. Manam, On the solution of dual integral equations, Appl.

Math. Lett. 20 (2007), no. 4, 391395.

A. Chakrabarti and G. V. Berghe, Approximate solution of singular

s

* r

(s n a1

integral equations, Appl. Math. Lett. 17 (2004), no. 5, 533559.

A. J. Jerry, Introduction to Integral Equations with Applications,

Where k (s)

1 s n

Second Edition, WileyInterscience, New York, 1999

C. Nasim and B. D. Aggarwala, On some dual integral equations,

Indian J. Pure Appl. Math.15 (1984), no. 3, 323340.

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.51-63.

I. N. Sneddon, The elementary solution of dual integral equations,

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

Erdelyi A, Magnus W, Oberhettinger F and Triconi F G, Tables of Integral Transforms (New York: McGraw-Hili) Vol. I, 1954.

(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.