Three Term Linear Fractional Nabla Difference Equation

1. International Research Research and Development (IJTSRD) International Open Access Journal Term Linear Fractional Nabla Difference Equation International Journal of Trend in Scientific Scientific (IJTSRD) International Open Access Journal ISSN No: 2456 ISSN No: 2456 - 6470 | www.ijtsrd.com | Volume 6470 | www.ijtsrd.com | Volume - 2 | Issue – 3 Three-Term Linear Fractional Nabla Difference Equation G. Pushpalatha R. Ranjitha R. Ranjitha Assistant Professor, Department of Mathematics, Vivekanandha College of Arts and Sciences for Women, Tiruchengode, Namakkal, Tamilnadu, India Women, Tiruchengode, Namakkal, Tamilnadu, India Mathematics, Research Scholar, Department of Mathematics, Vivekanandha College of Arts and Sciences for Women, Tiruchengode, Namakkal, Tamilnadu, India ode, Namakkal, Tamilnadu, India Research Scholar, Department of Mathematics, Vivekanandha College of Arts and Sciences for Vivekanandha College of Arts and Sciences for ABSTRACT In this present paper, a study on nabla difference equation and its third order linear fractional difference equation. A new generalized nabla difference equation is investigated from Three fractional nabla difference equation. A relevant example is proved and justify the proposed notions. example is proved and justify the proposed notions. In this present paper, a study on nabla difference equation and its third order linear fractional difference equation. A new generalized nabla difference equation is investigated from Three-term linear abla difference equation. A relevant equations. We shall consider the three term equations, equations. We shall consider the three term equations, (1) is limited. An equation of the form (1) is limited. An equation of the form               2 0 0 (3) t  ( ) x t ( ) x t C C ( ) x t ( ) x t C x t C x t ( ) ( ) ( ), g t ( ), g t 1 1 0 0 2 2 , 1,2,3.. is called a sequential fractional s called a sequential fractional difference equation. Keywords: Fractional difference operator, nabla difference equation, linear fractional, Third equation. Fractional difference operator, nabla difference equation, linear fractional, Third-term In general equation is,               (4) t  ( ) x t ( ) x t C C ( ) x t ( ) x t C x t C x t ( ) ( ) ( ), g t ( ), g t 2 2 1 1 0 0 1 1 0 0 2 2 1.Introduction , 1,2,3.. In this present paper, we shall use the method to obtain solutions of a linear fractional nabla difference equation of the form this present paper, we shall use the transform method to obtain solutions of a linear fractional nabla    1  and          Assume connection between is usually represents the backward difference operator that 0 0 1 1 2 2 as the only 1 1 2 2  2. The operator of nabla 2 is usually represents the backward difference operator            (1) t  and in this paper ( ) x t ( ) x t C C ( ) x t ( ) x t C x t C x t ( ) ( ) ( ), g t ( ), g t 0 0 1 1 2 2 , 1,2,3..     (5) 1) 1) , ( ) x t ( ) ( ) x t ( ) x t ( ( ,    Where 1  is of R-L type and the operator Liouville fractional difference operator, is defined by, Liouville fractional difference operator, is defined by, 2 . The fractional difference operator, The fractional difference operator,     k  k k 1 ( ) x t ( ) ( ) x t ( ) , 1,2,3..  0is a Riemann-  0 0 The raising factorial power function is defined ising factorial power function is defined below,  -term offractional   , define the th If fractional sum by 0    (  t ) t  (6) . ( )     ( )) s    1 t ( t      ( ) x t ( ) x s (2)      Then if 0 Liouville fractional difference equation is Liouville fractional difference equation is m 1 m , define by the Riemann , define by the Riemann- a s a     . Where ( ) s s 1       m m (7) ( ) x t ( ) ( ) x t ( ) c c The aim for this paper is to develop and preserve the theory of linear fractional nabla difference equations as a corresponds of the theory of linear difference as a corresponds of the theory of linear difference The aim for this paper is to develop and preserve the theory of linear fractional nabla difference equations  denotes the standard denotes the standard m th Where (backward) difference. m order nabla @ IJTSRD | Available Online @ www.ijtsrd.com @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Apr 2018 Page: 594

2. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 ( )) x t  In section 2, we shall use the transform method to (1) and we find out the solutions. And the same time we shall expressed as a sufficient condition as a function of 1 C and 2 C for convergent of the solutions. Next, consider the term know that the result [7], N 2( on (9) and also, we        ,      “If 0 1 N ( ) ( ) f t s s N ( ) ( ) f t s a 1 a a In section 3, we apply the algorithm in the case of a solution is 2t and verified independently that the series represents the known function.    a 1 (1 s ) ( ) f a .” Which implies For further studying in this previous area, we refer the reader to the article on two-term linear fractional nabla difference equation [7].      N ( ( )) x t N ( ( )) x t x (1) x (1) 2 2    N 1( ( )) x t x (1) 2.Three-term Linear fractional nabla difference equation       1 sN x t ( ( )) (1 s ) c ( c c ) 1 0 1 0   1  (1 (1 s s ) ) In this section, we describe an algorithm to form a solution of an initial value problem for a three-term linear fractional nabla difference equation of the form,     sN x t ( ( )) c c c 0 0 1 0 (1 1 (1 ) s  s c c s  s ) 1  s          c     sN x t ( ( )) c c c c (8) (1) x x (0) c , ( ) x t for C ( ) x t t  C x t 1,2,3.. ( ) 0, 0 0 1 0 0 0 0 1 2 (1 s ) (1 s )  . 1     (11) N ( ( )) x t sN x t ( ( )) 2 0 0 1 (1 )    Where 1 2 . Similarly, we consider the last term, Apply the operator ( ( ) N x t  N to the equation (8) we get, ( ) ( )) 0 x t C x t  2     C 2 0 1 2 1  1       N x t ( ( )) N x t ( ( )) c c c c       (9) First, consider a term the result [7] we get, N ( ( )) x t C N ( N ( )) x t (  C N x t ( )) x t from (8) and use ( ) 0 2 2 1 1 0 0 (1 s )  (1 c s ) 2 0 1 2 2 2     In particular (12) 2 ( ( )) N x t      1 1 N ( ( )) x t c (1 s ) c c (1 s ) 2 0 2 1 0 1 0    “If 1 2 ,      1 N ( ( )) x t c (1 s ) c 0 1 0      N 2( ( ))( ) f t s s N ( ( ))( ) f t s a a a Substitute (10), (11) and (12) in (9) we get 2 0 1 2 ( ( )) ( N x t C N    1 0 ( ( )) (1 ) s N x t s s   s C sN x t       2 0 1 ( ( )) (1 C N x t c      c  ( )) x t c  C N x t       0 c  ( ) 1) ) c 0          a 1 a 1 s (1 s ) ( ) f a (1 s ) f a ( 1) 2 ( 2  a   ( 0 1 0      a 1 s N ( ( ))( ) f t s s (1 s ) ( ) f a  a  ( ( )) c c 1 0 0 1 (1 s )        a (1 s ) ( ( f a 1) ( 1) ( )) f a . ˮ  1 s ) 0 Which implies that, ( ( )) N x t    0 1     s N ( ( )) x t s (1 s ) x (0) 1  2 0 0      ( s C s C c ) ( s C s C N ) ( ( )) x t 1 2 0 1 2 0 (1 s )      0 (1 s ) ( (1)) x ( 1) (0)) x        (1 C C c ) (1 ) c 0 1 2 1 0     (10) N ( ( )) x t s N ( ( )) x t 2 0 0        1 s (1 s ) c ( c ( 1) ) c 0 1 0 @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 595

3. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470       (  (1 C s C s  C c C s )  mC n    m 1   1 2 0 C N x t ( ( )) (13) (14) m n  ) ) m n       ( 1)  m n ((1 C s 0  (1   s s )(  ) 1 2    s Cs C 1 2   m 0 n 0 1 2    (1 C C c  ) ) c            m n     1 2 C s 1 0 ( 1) ( 1) t       ((1 ) ) m n Since  s N ( s ) 1       1 2 (( 1) m n ) 1 1           m n     ( 1) ( 1) t  Now take  N .          C s s C s ( s C s C ) 0       (( 1) m n )    1  2 s 1 1 2  By using the result [7], “ 1  Af (0) s        C s    ”. AN f t ( ) AN f t ( )      1 2 s 1 C s 1 0 1 1  Now, the above equation is re-express we have, N as N , and 1 0 1 ( s C s  1 In            C ) C          m n  m 1 2 1       1 2 s 1 C s 1 m n   ( 1)  m n C C   1    1 s 1 C s 1 2    s C s C 1       m 0 n 0 1 2 general,               m n     ( 1) ( 1) t N 1       (( 1) m n ) 1 s Cs C    n 1        1      ( 1)  n ( n 1) n s C       m n  m  2    m n    ( 1)  1 m n (15) N C C 1 Cs  n 0 0 1 2 1 2 1   m 0 n 0 Note that,           m n     ( 1) ( 1) t           (( 1) m n )  n 1          m n  1   m n   m n     ( 1)  1 Cs It follows from the result [7] “ ( 1) a N f t   Which implies that, equation (15) we get, 1 (1 ) s N s C s C    1   1 1 Cs   1 ”. (1 s ) N f t ( ) m n  1  a 1 We get         m n  m 1 s Cs C    m n   ( 1)  m n C C 0 1 2      mC n       m 0 n 0 1 2 m n    m n      ( 1)m  (1 ) n ( n 1) s C s 1 2           m n    m n     n 0  ( 1) ( 1) 1 2 ( t 1)  (16)      (( 1) m n ) m m mC n     Moreover, from (14)       m n    n n         ( 1)  n m m C s n 1 2       m n        m 1 s s  m n    m n     ( 1)  m n (1 ) m 0 n 0 C C s . 1 2    s Cs C    m 0 n 0 1 s 2 m n  m m n    m n      ( 1)  m n (1 ) 1 C C s . 1 2    s Cs C   m 0 n 0 1 2 @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 596

4. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 Similarly, we get   m n            m n  ( 1) ( 1) m ( t 1)         m n  m n  ( 1)   m n m s ( ) x t C c C C  m n   ( 1)  m n C C N 2 0 1 2  m n    (( 1) ) 1 2 0    s C s C   m 0 n 0   m 0 n 0 1 2               m n     ( 1) ( 2) t   m n             m n ( 1) ( 2) m ( t 1) 1)        (( 1) m n ( 1)) m n  ( 1)  m n (1 C c ) C C 1 0 1 2  m n   (   (( 1))   m 0 n 0         m n     m n  ( 1) ( 1) and so (1 s s C s  m t  m n   ( 1)  m n K C C         m n  1 1 2 m   m n    s )  (( 1) )  m n   ( 1)  m n   N C C m 0 n 0 0 1 2  C   m 0 n 0 1 2 (19)           m n       ( 1) ( 2) ( t 1) 1)      ) ) c  Where K ((1 C C c ) (1 . (17) 1 2 1 0       (( m n ( 1)) To simplify this representation, note that We consider an equation (13) we get ( 1) ((    ( ((    (  s C s  C c C s )    m n       m n      ( 1) ( 1) ( 1) ( 1) t t  1 2 0 C N x t ( )  0           (1 s s )( ) 1)   m 1) 1) n ) (( 1) m n ) 1 2   m n     ( 1) ( 2) t        (1 C C c  ) (1 C ) c  1 2 C s 1 0     m n 1)  ( s ) 1 2 Since ( ((   C c    m n                ( 1) ( 1)  t 1)  (    n m  t 1 ( 1) (( ( n 1) m  n ( 1)) ) 2 0  N x t ( ) (18)  0   (1 s s )( C s C )      1))    1)      m ( ( n ( 1) (( ) 1)  ( t  1) m n 1 2      t m    t s (1 C c  )  1 C s 0 t 1) )    (1 s s )( C ) 1 2             ( ( t ( 1) (( 1)  m  n m ( 1))               (1 C C c  ) (1 C ) c (( 1) m n ( 1))  1 2 C s 1 0  t 1) n )  ( s ) 1 2 Thus, (19) can be expressed as Substitute (15), (16) and (17) directly into (18) we get       m n           m n     m n ( 1) ( 2)  m ( t 1) 1)  m n   ( 1)  m n ( ) x t K C C 1 1 2  m n   (   (( 1))           m n     m 0 n 0  m n ( 1) ( 1) m ( t 1)  m n  m n N x t ( ) C c N ( 1) C C     ( 1)   m n    ( 1)   m t  0 2 0 0 1 2  m n    m n   ( 1)  (( 1) ) m n K C C   m 0 n 0 2 1 2   m n    (( 1) )   m 0 n 0 (20) Where (21) K K   that   m n             m n ( 1) ( 2)  m ( t 1) 1)  m n    ( 1)  m n (1 C c N ) C C 1 0 0 1 2  m n   (   (( 1))  (1   K c (1 C C  C ) , (1   m 0 n 0 1  0 1 2         c C C c ) C c ) Note 2 0 2 1 2 1 2 0   m n           m n ( 1) ( 1) m t m n  ( 1)  m n KN C C         m n     0 1 2   m n   m n     ( 1) ( 2) m ( t 1) 1) (( 1) )  m n  ( 1)    m n m 0 n 0 C C (22) 1 2  m n   (   (( 1))   m and 0 n 0 @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 597

5. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 Then, apply (13) with         m n     m n  ( 1) ( 1) m t  m n  ( 1)  m n C C 1 2   m n    (( 1) )   1, 1, 5   m (23) are two linear independent solutions of (8). We give the details to obtain conditions for absolute convergence in (23) for fixed t. 0 n 0 c  c    1, 2, C C 0 1 1 2 10   2, To obtain     ( 110) s s ( 1 5))2 ( 110)   1 10      9 10     ( s ( 1 5))1 ( 1 5)) s   (1 2)1 ( 1 5))   1 2 5 1 10 5   1 5 1 1 10 5  1 1 1 1 10 5          s  N x t ( )   m n     ( 1) ( 1) t 0  ( 110)   2 (1 )(   t  , 1 First note that for each is       (( 1) m n )     (1 ( 110) ( s  an increasing function in n. Now consider the ratio ( 1) 1 ( 1) m n t m n         2 s          m n     ( 1) ( 1) t 2 2 5      s s 1      (( 1) 1 ) (( 1) m n ) 10   N x t ( ) 0        1 1 1 5    2 2 (1 s ) s s s s Then 10 4  s            (  t ( 1) m  n 1 ( n      1)) )   10 1 10 7 10   N x t ( ) ( ) (( t    m  1)  m ( ) (( (  1 0            1 5     2 2 (1 s ) s s s s    t 1) m n ) 1))      ( t 1) n (     From (21), K 1 ( 1) ( 1) m    and the inequality is strict if        t m n (   1)             1 1 5 1 5 7 6 5   1       K 1 2 1 2 1 2 n 2 10 10 t  . 1 1 5  . Thus, compare (23) to and           m n  ( m ( 1)) m t   m n  n C C 1 2    (     m n  m ) m n      n       t m n 1   m n           m 7 10 1 5 1    m 0 n 0  ( ) x t   10 1      ( m ( 1)) ( ) t     m 0 n 0 m   C C   t m n   1 2 (    m n  m ) m n     n       m n           m 1 5 1 5 1   m 0  .    10 2 and apply the ratio test. Thus, each of (22) and (23) are absolutely convergent if   m 0 n 0   . C C 1 1 2 (25) Description of known functions: The unique solution of (24) is 2t and the series given in (25) absolutely convergent for all 7 1 ( ) ( ) 10 5 t  Write 0,1,2,.. We represent this method with an initial value problem for a classical second order finite difference equation. The unique solution of the initial value problem is ( ) 2t x t  , thus we obtain a series representation of 2t as a linear combination of the forms (22) and (23).   (25) as x t C t D t ( ) .    n m n  m n   n       t m 1  m n   t m              m 1 5 1   Where C t ( )    10 1   m 0 n 0 m n  m n  Consider the initial value problem n       m n           2 ( )    m 1 5  1   D t ( ) To      10 n 2  x  .   t  2   1) (24) (0) x 10 1 ( ) x t  ( ) x t 2 ( ) x t 0 , , 2,3,.. m 0 n 0  prove ( , or x t x t  , (1) 2 @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 598

6. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 m n           7 1 5 7 1 5 n       m n   m n  1          m 1 5 1 t       C t ( 1) D t ( 1) 2 C t ( ) D t ( )   2 D t ( ) 10 10  m n  10 ( )   m 1 n 0 Thus, It is sufficient to show that   n m 1 n  m n           m 1                           m 1 1 5 1 t    2 Dt ( )    and 1 ( ) 5  D t 7 10 ( 1) C t ( )   D t ( )  m n   n 10 ( 1)      m 0 n 0   6 5   m n  C t ( 1) C t ( ) D t n             m n m     . m 1 1 5 1     ( n 1 10 10   m 0 n 0 D t  Now consider , ( 1)  m 1 m n   2 m 1       1 5    t 1 5 t )       ( m n 1) (2       m n  m m n 2)   m n   1 m n      n       t 1   m n           m 1 5 1  n m n       m n         m   1 10 1 10 t D t ( 1)       10 m n 2  m n     m 0 n 0 ( 1)   m 0 n 0 1 5 m n      1 5 m n      m n     n    m n             m 1 1 5 1 10 t  t  2  m n             n       t 1 m n m n   1 m n         m 1    m n   1 ( 1)    1    10 m 0 n 0 0 0 m n     n m n         m n  1          m 1 5 1 5 1 10 t  1  m n  n       m n              m 1 D t ( )         m n   t m n 1 ( 2) 10   m 0 n 0 10   0 0     t m n  1  1 1 5     2 D t ( ) D t ( ) D t ( )    1    10 t m n 2 1 5 m n      1 5 m n          1 t m    ( D t A similar calculation shows that ( ) 10 ( ) ( ) x t x t    C t satisfies t    m n     t m n        n     1 mt n          m 1   2 2 ( ) x t 0, 2,3,.     1  m n   10 t 2 0 0 We close by arguing that 7 ( 1) 10       6 5 1 5 m n  n       m n                 m 1 . C t C t ( ) D t ( )     ( m n 1) 10 n   1)  0 0    Simplify 2 10 ( C t   t m 1       1) C t ( 1) 2 ( C t 1) 0 n 2 D t  ( ) C t ( ) .     10( ( C t 1) 2 ( ) C t C t ( 1)) We have yet to obtain a direct approach to take ( 1) C t  . We begin by showing directly that satisfies 10 ( ) ( ) 2 ( ) x t x t x t    Apply the power rule and 1 1 ( ) 5 10 m n       1 1 ( ) 5 10 m n       ( (       C t 1) C t ( )) 2 ( C t 1) 0 D t ( ) To obtain 7 19 10     C t ( 1) C t ( ) C t ( 1)  t  2 0, 2,3,. 10       6 5 7     C t ( ) C t ( ) C t ( 1) m n    n       m n   m n  1       10 m t   Dt  m n   ( 2) 0 0 Now it is sufficient to show that 7 ( ) ( 1) 10       1 5    C t C t D t ( ) . m n    n       m n  m n        m t    Dt  m n   ( 1) 0 0 @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 599

7. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470    Employ ( ) ( 1) C t  7 ( ) 10  and References C t D t  D t D t  ( 1)  D t to obtain ( ) ( ) (    1) 7 10 1.Agarwal, Inequalities, Marcel Dekker, New York,1992. 2.Jagan Mohan Jonnalagadda, Behaviour of Linear Fractional Nabla Difference Equations, Inter. Jour. of Differ. Equ., (2017) 255- 265. R.P., Difference Equations and       C t C t ( 1) ( ( D t 1) Asymptotic     D t ( )) ( ( ) D t D t ( 1)) 7 17 10 3.Jagan Mohan, J., Variation of Parameters for Nabla Fractional Difference Equations, Novi. Sad J. Math., (2014) 149-159. 4.Jagan Mohan Jonnalagadda., Numerical Solutions of Linear Fractional Nabla Difference Equations, Math. CA, (2016). 5.Jagan Mohan, J., Deekshitulu, G.V.S.R., Solutions of Nabla Fractional Difference Equations Using N-Transforms, Commun. Math. Stat. (2014) 1-16. 6.Jagan Mohan Nonlinear Fractional Nabla Difference Equations, Inter. Jour. of Analy. and App., (2015) 79-95. 7.Paul Eloe, Zi Ouyang., Multi-term Linear Fractional Nabla Difference Equations with Constant Coefficients, Inter. Jour. of Differ. Eqn, (2015) 91-106.      ( ( D t 1) D t ( ) D t ( 1) 10 7 ( ( 1) 10  7 ( ) ( 1) 10       19 10 2       D t D t ( ) D t ( 1) D t ( ) 10      1 5    C t C t D t ( ) . Jonnalagadda, Analysis of @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 600