Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 595 https://internationalpubls.com Application of Fractional Calculus Operators to Fredholm type Integral Equations Neeti Ghiya 1 , Manjula S 2 1 Department of Mathematics, RV College of Engineering, Bangalore 2 Department of Mathematics, Government PU College, Dakshina kannada 1 neethighiya@rvce.edu.in, 2 manjula.siddesh.banakar@gmail.com Article History: Received: 04-06-2024 Revised: 03-07-2024 Accepted: 30-07-2024 Abstract: The purpose of this article is to study of fractional calculus operators of Riemann- Liouville and Weyl in solving the Fredholm type integral equations. The kernels of these equations are the product of incomplete -functions and generalized polynomials. Further, the Mellin transform technique is also used to find the solution of the Fredholm integral equations. A number of intriguing special cases have also been mentioned. Keywords: Reimann-Liouville fractional integral, Weyl fractional integral, Incomplete -functions, Fredholm type integral equation, Fox’s -function, Mellin transform, General class of polynomials, Generalized polynomials. 1. Introduction: Fractional calculus is one of the branches of applied mathematics that is based on any real or complex order derivatives and integrals. Many advanced applications of fractional calculus are found in various areas of science and engineering. Several authors (refer [1]-[6]) have used fractional derivatives and integrals to solve problems like fluid dynamics, thermonuclear fusion, dynamical systems, control theory, quantum mechanics, image processing and many more. Integral equations usually occur in physics, mechanics and applied mathematics. Applications of Fredholm integral equations include the spectrum concentration problem, linear forward modelling, inverse problems, and signal processing theory. In recent years, several authors have studied integral equations of Fredholm type taking kernel as special functions and polynomials (refer [7]-[13]). 2. Objectives: The following Fredholm type integral equations are to be solved in order to accomplish the goal of this work. 1 1 1 1 2 2,..., , ,..., 1 , 1 10 ( , , ), ( , ),..., ( , ) ,..., ( , ),..., ( , ) k k r r p pM M m n N N k p q q q u U w u U u Uy y y x S w w z v V v Vx x x                                     ( ) ( ) (0 ),x dx g y y   f (1) 1 1 2 2, , , , 1 10 ( , ) ( , , ), ( , ),..., ( , ) ( , ) ( , ),..., ( , ) P P p pU M N m n V P Q p q Q Q q q e E u U w u U u Uy y y x S H T z f F v V v Vx x x                                                mailto:neethighiya@rvce.edu.in mailto:manjula.siddesh.banakar@gmail.com Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 596 https://internationalpubls.com ( ) ( ) (0 ),x dx g y y   f (2) 1 1 1 1 2 2,..., , ,..., 1 , 1 10 ( , , ), ( , ),..., ( , ) ,..., ( , ),..., ( , ) k k r r p pM M m n N N k p q q q u U w u U u Uy y y x S w w z v V v Vx x x                                    ( ) ( ) (0 )x dx g y y   f (3) and 1 1 2 2, , , , 1 10 ( , ) ( , , ), ( , ),..., ( , ) ( , ) ( , ),..., ( , ) P P p pU M N m n V P Q p q Q Q q q e E u U w u U u Uy y y x S H T z f F v V v Vx x x                                               ( ) ( ) (0 ).x dx g y y   f (4) The general class of polynomials is familiarized by Srivastava [14] as:   ' ' ' ' ' ' / ' , 0 ( ) , ! V U U lU l V V l l V S y A y l         (5) where 'V is non-negative integer, 'U Z and the co-efficients ' ' , ( , 0) V l A V l  are constants (real or complex) and are arbitrary in nature. The generalized polynomial is given by Srivastava [15] in the following way:  1 1 ,..., ,..., 1,..., k k M M N N kS y y   1 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 1 ( )( ) ,..., ,..., , ;...; , ,..., , ! ! k k k k k k N MN M k MM k k k k NN B N N y y              (6) where 1,..., kN N are non-negative integers, 1( ,..., )kM M Z and the co-efficient  1 1, ;...; ,k kB N N  are constants (real or complex). All are arbitrary in nature. Fox’s H -function is represented in form of series as follows (refer [16]). ' ( , ), , ( , ) 0 1 ( 1) ( ) ! G P P Q Q LGM u UM N G P Q v V G g g L z H z G V          , (7) where 1 1 ' 1 1 ( ) (1 ) ( ) (1 ) ( ) M N j j G j j G j j j G G Q P j j G j j G j M j N v V L u U L L v V L u U L                        (8) and ( ) /G g gL v G V  . Special functions are significant in a wide range of scientific and engineering domains. The familiar H -function has spanned a greater range than the other special functions. However, in the fields of heat conduction and astrophysics, it has been revealed that certain issues cannot be solved using Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 597 https://internationalpubls.com these special functions. During this period, the perception of incomplete Gamma functions existed, and academics researched several articles (refer [17]-[21]) connected to incomplete special functions alongside equivalent higher transcendental special functions. Incomplete Gamma functions ( , )s w and ( , )s w are defined as: 1 0 ( , ) ( ( ) 0; 0) w s ts w t e dt s w      (9) and 1( , ) ( 0; ( ) 0 0)s t w s w t e dt w s when w         (10) respectively, satisfy the following decomposition formula ( , ) ( , ) ( ), ( ( ) 0)s w s w s s     , (11) where ( )s is the familiar Gamma function. Srivastava hosted the incomplete H -functions , , ( )m n p q z and , , ( )m n p q z , are described in Mellin-Barnes type of contour integrals as shown below: 1 1 2,, , , , 1, ( , , ), ( , ) ( ) ( , ) j j pm n m n p q p q j j q u U w u U z z v V           1 1 2 2, , 1 1 ( , , ), ( , ),..., ( , ) ( , ),..., ( , ) p pm n p q q q u U w u U u U z v V v V          (12) '1 ( , ) , 2 s L g s w z ds i   where 1 1 1 2' 1 1 (1 , ) ( ) (1 ) ( , ) (1 ) ( ) m n j j j j j j q p j j j j j m j n u U s w v V s u U s g s w v V s u U s                         (13) and 1 1 2,, , , , 1, ( , , ), ( , ) ( ) ( , ) j j pm n m n p q p q j j q u U w u U z z v V           1 1 2 2, , 1 1 ( , , ), ( , ),..., ( , ) ( , ),..., ( , ) p pm n p q q q u U w u U u U z v V v V          (14) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 598 https://internationalpubls.com '1 ( , ) , 2 s L G s w z ds i   where 1 1 1 2' 1 1 (1 , ) ( ) (1 ) ( , ) . (1 ) ( ) m n j j j j j j q p j j j j j m j n u U s w v V s u U s G s w v V s u U s                         (15) Incomplete H -functions , , ( )m n p q z and , , ( )m n p q z in (12) and (14) exist for all 0w  according to the same contour and set of circumstances as mentioned in [22]. Some special cases of incomplete H -function [23] are as follows: (i) Specializing 0w  in (14), the incomplete H -function , , ( )m n p q z moderates to the Fox’s H - function. 1 1 2 2 1 1, , , , 1 1 1 1 ( , ,0), ( , ),..., ( , ) ( , ),..., ( , ) . ( , ),..., ( , ) ( , ),..., ( , ) p p p pm n m n p q p q q q q q u U u U u U u U u U z H z v V v V v V v V                 (16) The H -function of one variable [24] is defined in the following manner: 1 1, , , , 1 1 ( , ),..., ( , ) [ ] ( , ),..., ( , ) p pm n m n p q p q q q u U u U H z H z v V v V         (17) 1 ( ) , ( 1, \{0}) 2 s L s z ds i z i       , where 1 1 1 1 ( ) (1 ) ( ) , (1 ) ( ) m n j j j j j j q p j j j j j m j n v V s u U s s v V s u U s                       (18) where m lies between [1, ]q and n lies between [0, ]p when ,m q and 0, .n p Here represents positive integers, denote the set of complex numbers and  0 : 0 .  (ii) Substituting m by 1, n by p and changing q to 1q in functions given in (12) and (14), the functions moderates to incomplete Fox-Wright functions (IFWFs) ( ) p q  and ( ) p q  by taking suitable parameters: 1 1 2, 1 1 2,1, ( ) , 1 1, 1, (1 , , ), (1 , ) ( , , ), ( , ) (0,1), (1 , ) ( , ) j j p j j pp p q p q j j q j j q u U w u U u U w u U z z v V v V                    (19) and Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 599 https://internationalpubls.com 1 1 2, 1 1 2,1, ( ) , 1 1, 1, (1 , , ), (1 , ) ( , , ), ( , ) . (0,1), (1 , ) ( , ) j j p j j pp p q p q j j q j j q u U w u U u U w u U z z v V v V                      (20) (iii) Further, specializing 0w  in (20), ( ) p q  (IFWF) moderates to Fox-Wright function (FWF) p q : 1 1 2, 1 1( ) 1, 1 1 ( , ,0), ( , ) ( , ),..., ( , ) . ( , ) ( , ),..., ( , ) j j p p p p q p q j j q p p u U u U u U u U z z v V v V v V                  (21) (iv) Taking 1j jU V  in (19) and (20), the incomplete Fox-Wright functions become incomplete generalized hypergeometric functions (IGHFs) p q and p q [25]: 1 2, 1 2( ) 1, 1 ( ,1, ), ( ,1) ( , ), ,..., ( ,1) ,..., j p p p q p q j q q u w u u w u u z z v v v                  (22) and 1 2, 1 2( ) 1, 1 ( ,1, ), ( ,1) ( , ), ,..., . ( ,1) ,..., j p p p q p q j q q u w u u w u u z z v v v                   (23) (v) Specializing 0w  in (23), p q (IGHF) moderates to the known function generalized hypergeometric (GHF) p qF : 1 2 1 1 1 ( ,0), ,..., ,..., . ,..., ,..., p p p q p q q q u u u u u z F z v v v v                 (24) Let  be a set of all functions f , defined on [0, )R   and satisfies the followings: (i) ( ),C R f (ii) lim[ ( )] 0r y y y  f for all non-negative integers  and r, (iii) ( ) 0(1)y f when 0,y  for correspondence to the space of good functions defined on the whole real line ( , ).  th order Reimann-Liouville fractional integral is given as follows:    0( ) ( )yD y D y  f f 1 0 1 ( ) ( ) , (Re( ) 0 : ) ( ) y y d           f f (25) and thh order Weyl fractional integral is described in the following manner:    ( ) ( )h h yW y D y  f f 11 ( ) ( ) , (Re( ) 0 : ). ( ) h y y d h h           f f (26) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 600 https://internationalpubls.com 3. Main Results: The Fredholm integral equation with kernel, the product of Generalized Polynomials and the incomplete H -functions. Lemma 1: Consider (i)  , , , 0m n p q   non-negative integers thus 1 m q  ; 0 .n p  (ii) ( ) 0, 0 ( 1,..., ). j j v j m V                      (iii) 0, 0w   and .  (iv) 1 arg( ) 2 z   , where 1 1 1 1 0, p qn m j j j j j j j n j m U V U V               Then, 1 1 1 1 2,,..., , ,..., 1 , 1, ( , , ), ( , ) ,..., ( , ) k k r r j j pM M m n N N k p q j j q u U w u Uy y y W x S w w z v Vx x x                                         11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 1 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1 1 2, 1, 1 1, 1 1, 1, 1 ( , , ), 1 , , ( , ) ( , ) , 1 , , ( , ) k i j j p im n p q k j j m i j j m q i u U w r u U y z x v V r v V                                         . (27) Proof: We use the formulation of Weyl fractional integral provided by (26), to prove the lemma 1. Equations (6) and (12) are used to precisely describe the incomplete H -function in form of Mellin- Barnes contour integral and generalized polynomial in series form. In order to get the desired result, change the order of summation and integrals under the aforementioned circumstances, then solve the integral using substitution method, and interpret the resultant contour integral by means of incomplete H -function. Theorem 1: Consider (i)  , , , 0m n p q   non-negative integers thus 1 m q  ; 0 .n p  (ii) 1 0 ( 1,..., ), 0 ( 1,..., ). j j j j u v j n j m U V                                       (iii) 0, 0w   and .  Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 601 https://internationalpubls.com Consequently, the relationship shown below is valid: 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1 1 2, 1, 1 1, 1 1, 1, 1 ( , , ), 1 , , ( , ) ( ) ( , ) , 1 , , ( , ) k i j j p im n p q k j j m i j j m q i u U w r u U y z x dx x v V r v V                                         f 1 1 1 1 2 2,..., , ,..., 1 , 1 10 ( , , ), ( , ),..., ( , ) ,..., ( , ),..., ( , ) k k r r p pM M m n N N k p q q q u U w u U u Uy y y x S w w z v V v Vx x x                                       ( ) ,D x dx  f (28) for f and 0.y  Proof: If  denote the left-hand side of the equation (28), 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                                1 1 2, 1, 1 1, 1 1, 1, 1 ( , , ), 1 , , ( , ) ( ) . ( , ) , 1 , , ( , ) k i j j p im n p q k j j m i j j m q i u U w r u U y z x dx x v V r v V                                         f Utilizing the result of lemma 1 and equation (26), we have 1 1 ,...,1 , ,..., 1 , 0 ( ) ( ) ,..., . ( ) k k r r M M m n N N k p q x x y y y x S w w z d dx                                                            f (29) We get the following equation by altering the order of integration and applying equation (25).  1 1 ,..., , ( ) ,..., 1 , 0 ,..., ( ) .k k r r M M m n N N k p q y y y S w w z D d                                                   f (30) On changing  to x in the equation (30), we attain the required proof of the theorem 1. Lemma 2: Consider (i)  , , , 0m n p q   thus 1 m q  ; 0 .n p  , where  is positive integers. (ii) ( ) 0, 0 ( 1,..., ). j j v j m V                      (iii) 0, 0w   and .  Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 602 https://internationalpubls.com (iv) 1 arg( ) 2 z   , where 1 1 1 1 0. p qn m j j j j j j j n j m U V U V               Then, 1 1 1 1 2,,..., , ,..., 1 , 1, ( , , ), ( , ) ,..., ( , ) k k r r j j pM M m n N N k p q j j q u U w u Uy y y W x S w w z v Vx x x                                        11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 1 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1 1 2, 1, 1 1, 1 1, 1, 1 ( , , ), 1 , , ( , ) . ( , ) , 1 , , ( , ) k i j j p im n p q k j j m i j j m q i u U w r u U y z x v V r v V                                         (31) Proof: Similar lines of the proof of lemma1 and equations (6), (14), (26) are used to reach the result of Lemma 2. Theorem 2: For (i)  , , , 0m n p q   thus 1 m q  ; 0 .n p  (ii) 1 0 ( 1,..., ), 0 ( 1,..., ). j j j j u v j n j m U V                                       (iii) 0, 0w   and .  Consequently, the relationship shown below is valid: 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1 1 2, 1, 1 1, 1 1, 1, 1 ( , , ), 1 , , ( , ) ( ) ( , ) , 1 , , ( , ) k i j j p im n p q k j j m i j j m q i u U w r u U y z x dx x v V r v V                                         f 1 1 1 1 2 2,..., , ,..., 1 , 1 10 ( , , ), ( , ),..., ( , ) ,..., ( , ),..., ( , ) k k r r p pM M m n N N k p q q q u U w u U u Uy y y x S w w z v V v Vx x x                                      ( ) ,D x dx  f (32) when f and 0.y  Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 603 https://internationalpubls.com Proof: Lemma 2 is used to achieve this outcome. The proof resembles the proof of Theorem 1. The Fredholm integral equations with kernel, the product of the general polynomial, one variable H -function and the incomplete H -functions: Lemma 3: Consider (i)  , , , 0m n p q   thus 0 ,0 ; 0( 1,..., ),jm q n p U j p      0( 1,..., ); ( 1,..., )j jV j q u j p   and ( 1,..., )jv j q are complex parameters. (ii) ( ) ( ); 0; j G j v K L V                        where ( 1,..., ); 0j m   (iii) 1 arg( ) , 2 z   where 1 1 1 1 0, p qn m j j j j j j j j j j j n j m u U v V u U v V               Then 1 1 2,, , , , 1, ( , ) ( , , ), ( , ) ( , ) ( , ) P P j j pU M N m n V P Q p q Q Q j j q e E u U w u Uy y y W x S H T z f F v Vx x x                                                     '[ / ] , 0 0 1 ( ) ( 1) ( ) ! ! G G K LGV U M LKUK G V K K G g g V L y x A T K G V x                          1 1 2,, 1 1, 1 1, 1, ( , , ), 1 , , ( , ) . ( , ) , 1 , , ( , ) G j j pm n p q j j m G j j m q u U w K L u Uy z v V K L v Vx                              (33) Proof: We utilize the Weyl fractional integral given by (26) to prove the lemma 3. Express the general polynomial, the Fox’s H -function in series and the incomplete H -function in contour integral form with the help of equations (5), (7) and (12). To achieve the desired outcome, swap the order of summation and integrals under the aforementioned requirements, then evaluate the integral using substitution method, and resultant contour type integral is represented as incomplete H - function. Theorem 3: With the usual states (i), (ii) and (iii) mentioned in lemma 3, the following relation holds: '[ / ] , 0 0 10 ( ) ( 1) ( ) ! ! G G K LGV U M LKUK G V K K G g g V L y x A T K G V x                      Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 604 https://internationalpubls.com     1 1 2,, 1 1, 1 1, 1, ( , , ), 1 , , ( , ) ( ) ( , ) , 1 , , ( , ) G j j pm n p q j j m G j j m q u U w K L u Uy z x dx v V K L v Vx                              f 1 1 2 2, , , , 1 10 ( , ) ( , , ), ( , ),..., ( , ) ( , ) ( , ),..., ( , ) P P p pU M N m n V P Q p q Q Q q q e E u U w u U u Uy y y x S H T z f F v V v Vx x x                                                  ( ) ,D x dx  f (34) Provided that f and 0.y  Proof: Lemma 3 and equation (26) along with following the steps outlined in the demonstration of theorem 1 are used to establish theorem 3. Lemma 4: Let (i)  , , , 0m n p q   thus 0 ,0 ; 0( 1,..., ),jm q n p U j p      0( 1,..., ); ( 1,..., )j jV j q u j p   and ( 1,..., )jv j q are complex parameters. (ii) ( ) ( ); 0; j G j v K L V                        where ( 1,..., ); 0.j m   (iii) 1 arg( ) , 2 z   where 1 1 1 1 0, p qn m j j j j j j j j j j j n j m u U v V u U v V               Then 1 1 2,, , , , 1, ( , ) ( , , ), ( , ) ( , ) ( , ) P P j j pU M N m n V P Q p q Q Q j j q e E u U w u Uy y y W x S H T z f F v Vx x x                                                    '[ / ] , 0 0 1 ( ) ( 1) ( ) ! ! G G K LGV U M LKUK G V K K G g g V L y x A T K G V x                          1 1 2,, 1 1, 1 1, 1, ( , , ), 1 , , ( , ) . ( , ) , 1 , , ( , ) G j j pm n p q j j m G j j m q u U w K L u Uy z v V K L v Vx                              (35) Proof: The proof is comparable to lemma 3's proof. Equations (5), (7) and (14) can be used to reach this conclusion. Theorem 4: With the usual conditions (i), (ii) and (iii) of lemma 4, the following relation holds: Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 605 https://internationalpubls.com '[ / ] , 0 0 10 ( ) ( 1) ( ) ! ! G G K LGV U M LKUK G V K K G g g V L y x A T K G V x                          1 1 2,, 1 1, 1 1, 1, ( , , ), 1 , , ( , ) ( ) ( , ) , 1 , , ( , ) G j j pm n p q j j m G j j m q u U w K L u Uy z x dx v V K L v Vx                              f 1 1 2 2, , , , 1 10 ( , ) ( , , ), ( , ),..., ( , ) ( , ) ( , ),..., ( , ) P P p pU M N m n V P Q p q Q Q q q e E u U w u U u Uy y y x S H T z f F v V v Vx x x                                                 ( ) ,D x dx  f (36) provided that f and 0.y  Proof: This result can be obtained using lemma 4 and the proof resembles from the proof given for the theorem 3. Mellin transform method for the outcome of Fredholm integral equation which contains the product of general polynomial, one variable H -function in series and incomplete H -function as kernel: Theorem 5: When , , { ( )}g D y f f exists 1 0, 0, arg( ) , 0 2 y z     ( is same as mentioned in lemma 3), ( ) ( ) 0,    then the following integral equation , , , , 0 ( , ) ( ) ( ) ( , ) P PU M N m n V P Q p q Q Q e Ey y y g y x S H T z x dx f Fx x x                                                 f (37) has solution given by '[ / ] 1 , 0 0 1 ( ) ( 1) ( ) ( ) lim 2 ! ! G c i GV U M Ls KUK G V K K G g gc i V L y y y A T i K G V                       f 1 ' ( ) , Gs K L Gs K L z g s ds                          (38) provided that 1 1 max min . j jG j n j j u vs K L U V                                           Proof: Changing f by { }D  f in (37), we have '[ / ] , 0 0 10 ( ) ( 1) ( ) ( ) ! ! G G K LGV U M LKUK G V K K G g g V L y g y x A T K G V x                           1 1 2,, 1 1, 1 1, 1, ( , , ), 1 , , ( , ) { ( )} . ( , ) , 1 , , ( , ) G j j pm n p q j j m G j j m q u U w K L u Uy z D x dx v V K L v Vx                                 f Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 606 https://internationalpubls.com Now, multiply both sides the above equation by 1sy  then integrate it w.r.t. y from 0 to  , we reach 1 0 ( ) ( )ss y g y dy      '[ / ] , 0 0 10 ( ) ( 1) ( ) 1 ( ) ! ! G G K LGV U M LKUK G V K K G g g V L x D x A T K G V x                          f     1 1 2,1 , 1 1, 1 1, 1,0 ( , , ), 1 , , ( , ) ( , ) , 1 , , ( , ) G G j j ps K L m n p q j j m G j j m q u U w K L u Uy y z dy dx v V K L v Vx                                          . Utilizing the result of Mellin transform of incomplete H -function, the above result moderates to '[ / ] , 0 0 1 ( ) ( 1) ( )1 ( ) ! ! G G s K LGV U M LKUK G V K K G g g V L s A T z K G V                       ' 0 ( ) ( ) . ( ) sGs K L s g x D x dx s                      f Inverting the above equation by applying theorem of Mellin inversion, we have '[ / ] , 0 0 1 ( ) ( 1) ( ) { ( )} lim 2 ! ! G c i GV U M LKUK G V K K G g gc i V L D x A T i K G V                       f 1 ' 1( ) ( ) . ( ) Gs K L sGs K L s z g x s ds s                               Operating both sides of the above equation by ,D  which gives '[ / ] , 0 0 1 ( ) ( 1) ( ) ( ) lim 2 ! ! G c i GV U M LKUK G V K K G g gc i V L x D A T i K G V                          f 1 ' 1( ) ( ) , ( ) Gs K L sGs K L s z g x s ds s                                this finally yields '[ / ] 1 , 0 0 1 ( ) ( 1) ( ) ( ) lim 2 ! ! G c i GV U M Ls KUK G V K K G g gc i V L y y y A T i K G V                       f 1 ' ( ) , Gs K L Gs K L z g s ds                          which is required solution. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 607 https://internationalpubls.com Theorem 6: If , , { ( )}g D y f f exists 1 0, 0, arg( ) , 0 2 y z     ( is same as mentioned in lemma 3), ( ) ( ) 0,    then the following integral equation '[ / ] , 0 0 10 ( ) ( 1) ( ) ( ) ! ! G G K LGV U M LKUK G V K K G g g V L y g y x A T K G V x                            1 1 2,, 1 1, 1 1, 1, ( , , ), 1 , , ( , ) ( ) ( , ) , 1 , , ( , ) G j j pm n p q j j m G j j m q u U w K L u Uy z D x dx v V K L v Vx                                 f (39) has a solution given by '[ / ] 1 , 0 0 1 ( ) ( 1) ( ) ( ) lim 2 ! ! c i GV U M s KUK G V K K G g gc i V L y y y A i K G V                       f 1 ' ( ) , G G s K L L Gs K L T z G s ds                          (40) provided that 1 1 max min . j jG j n j j u vs K L U V                                           Proof: Result of this theorem is obtained on the similar lines of proof presented for theorem. 4. Special cases: [a] Considering (19) and (20), we get the following results from theorem 1 and 2. 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1 1 2, 1( )1 1 1 1, 1 1 ( , , ), , , ( , )( ) ( ) ( ) , , ( , ) kk i j j pi ii p qk k i i j j q i i u U w r u Ur y z x dx x r r v V                                               f 1 1 1 1 2 2,..., ( ) ,..., 1 1 10 ( , , ), ( , ),..., ( , ) ,..., ( , ),..., ( , ) k k r r p pM M N N k p q q q u U w u U u Uy y y x S w w z v V v Vx x x                                       ( )D x dx  f (41) and 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 608 https://internationalpubls.com 1 1 2, 1( )1 1 1 1, 1 1 ( , , ), , , ( , )( ) ( ) ( ) , , ( , ) kk i j j pi ii p qk k i i j j q i i u U w r u Ur y z x dx x r r v V                                              f 1 1 1 1 2 2,..., ( ) ,..., 1 1 10 ( , , ), ( , ),..., ( , ) ,..., ( , ),..., ( , ) k k r r p pM M N N k p q q q u U w u U u Uy y y x S w w z v V v Vx x x                                        ( ) .D x dx  f (42) [b] Setting 0w  in (42) and utilizing the result of (21), we attain the required result. 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1, 11 1 1 1, 1 1 , , ( , )( ) ( ) ( ) , , ( , ) kk i j j pi ii p qk k i i j j q i i r u Ur y z x dx x r r v V                                              f 1 1 1 1,..., ,..., 1 1 10 ( , ),..., ( , ) ,..., ( , ),..., ( , ) k k r r p pM M N N k p q q q u U u Uy y y x S w w z v V v Vx x x                                       ( ) .D x dx  f (43) [c] Put 1j jU V  in (41) and (42), we accomplish the following result from (22) and (23). 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1 2 11 1 1 1 1 1 ( ,1, ), , , ( ,1),..., ( ,1)( ) ( ) ( ) , , ( ,1),..., ( ,1) kk i pi ii p qk k i i q i i u w r u ur y z x dx x r r v v                                              f  1 1 1 2,..., ,..., 1 10 ( , ), ,..., ; ,..., ( ) ,..., ; k k r r pM M N N k p q q u w u uy y y x S w w z D x dx v vx x x                                        f (44) and 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 609 https://internationalpubls.com 1 2 11 1 1 1 1 1 ( ,1, ), , , ( ,1),..., ( ,1)( ) ( ) ( ) , , ( ,1),..., ( ,1) kk i pi ii p qk k i i q i i u w r u ur y z x dx x r r v v                                              f  1 1 1 2,..., ,..., 1 10 ( , ), ,..., ; ,..., ( ) . ,..., ; k k r r pM M N N k p q q u w u uy y y x S w w z D x dx v vx x x                                        f (45) [d] Setting 0w  in (45) and utilizing the result of (24), we obtain the following result. 11 1 1 1 1 1 [ / ][ / ] 1 1 1 1 0 0 10 ( )( ) , , ,..., [ , ;...; , ] ,..., ! ! kk k k k k k r rN MN M k MM k k k k NN y y x B N N w w x x                              1 11 1 1 1 1 1 , , ( ,1),..., ( ,1)( ) ( ) ( ) , , ( ,1),..., ( ,1) kk i pi ii p qk k i i q i i r u ur y z x dx x r r v v                                              f  1 1 1,..., ,..., 1 10 ,..., ; ,..., ( ) . ,..., ; k k r r pM M N N k p q q u uy y y x S w w F z D x dx v vx x x                                        f (46) Similarly, we can find special cases for theorem 3 and 4. 5. 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