Solution of Dual Integral Equations Involving Generalized Function
Solving Dual Integral Equations through Mellin Transform and Polynomial Function Kernel
by Anil Kumar Tiwari*, Dr. Archana Lala, Dr. Chitra Singh,
- Published in Journal of Advances and Scholarly Researches in Allied Education, E-ISSN: 2230-7540
Volume 16, Issue No. 9, Jun 2019, Pages 455 - 457 (3)
Published by: Ignited Minds Journals
ABSTRACT
This article's main goal is to solve a dual integral equation by lowering it to an integral equation through the use of mellin transform whose kernel includes generalized polynomial function. We assume that there are definitely many ways to reduce these dual integral formulas by using various transformations such as Fourier, Henkel, etc. For the reason of illustration we pick a dual integral equation of particular type and reduced it, by use of fractional operators and Mellin transform, to an integral equation.
KEYWORD
dual integral equations, generalized function, mellin transform, kernel, integral equation, polynomial function, Fourier transform, Henkel transform, fractional operators
1. INTRODUCTION
In several fields of mathematical physics, dual integral equations are often found and usually occur when solving a boundary value problem with mixed boundary conditions. The present paper attempts to solve such dual integral equations that involve generalized polynomial function as a kernel by reducing them into integral equations. Many attempts have been made in the past to solve such problems. The following integral equations are basic tool for our illustration. k1 & k2 are kernels defind over xuplane.
2. THEOREM
If f is unknown function satisfying the dual integral equation. When h and g are prescribed function and 12 , aaandr are parameters, then f is giving by Where and
3. MATHEMATICAL PRELIMINARY
To prove the theorems we shall use Mellin transformer and fractional integral operator. When siis a complex variable. Fractional integral operator
4. SOLUTION
Now taking Then from Erdeeyi [11] We get Hence by use of convolution theorem of Mellin transform,(2.1) & (2.2) can be written as Now operating a (4.2) by the operator (3.5) we get Now putting and simplifying we get In equation (4.4), we put 1(1)kna and ),(112aar so that (4.4) Changes to Now we write and Now from (4.2), (4.5), (4.6) we get Again using convolution theorem of Mellin transform, (4.1) & (4.6) becomes When 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 – Where and T*(s) is the mellin transform of t(x).
By use of definition of H – function, we get the inverse transform ()Lxat *()Lsas Taking inverse mellin transform of (4.10) Hence using (4.11) we get When t(y) is given by (4.6). Hence proved the theorem.
REFERENCE
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Corresponding Author Anil Kumar Tiwari*
Ph.D. Scholar, Department of Mathematics, RNT University, Bhopal
