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International Journal of Trend in Scientific Research and Development (IJTSRD) International Open Access Journal ISSN No: 2456 - 6470 | www.ijtsrd.com | Volume - 2 | Issue – 3

Three-Term Term Linear Fractional Nabla Difference Equation G. Pushpalatha Assistant Professor, Department of Mathematics, Vivekanandha College of Arts and Sciences for Women, Tiruchengode, Namakkal, Tamilnadu, India

R. Ranjitha Research Scholar, Department of Mathematics, Vivekanandha College of Arts and Sciences for Women, Tiruchengode, ode, Namakkal, Tamilnadu, India

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 Three-term linear fractional nabla abla difference equation. A relevant example is proved and justify the proposed notions. Keywords: Fractional difference operator, nabla difference equation, linear fractional, Third Third-term equation.

equations. We shall consider the three term equations, (1) is limited. An equation of the form

iss called a sequential fractional difference equation. In general equation is,

1. Introduction

(4) t  1, 2,3.. ,

In this present paper, we shall use the transform method to obtain solutions of a linear fractional nabla difference equation of the form

Assume

0 x (t )  C1x (t )  C2 x (t )  g (t ),

(1) t  1, 2,3.. ,

If   0 , define the  -term of fractional sum by

(t  (s))1 x(s) (2)  x(t)     sa t

Where  ( s )  s  1 . 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

that

0  1  1  2  2

as

the

only

 x(t )  x(t )  x(t  1) ,

 k x(t )   k 1 x(t ) , k  1, 2, 3.. The raising ising factorial power function is defined below,

th

 a

02 x (t )  C101 x (t )  C2 x(t )  g (t ),

connection between 1 and 2 . The operator of nabla is usually represents the backward difference operator and in this paper (5)

Where 1    2 . The fractional difference operator, 0 is of R-L type and the operator 0 is a RiemannLiouville fractional difference operator, is defined by,

 02  x (t )  C1 0 x(t )  C2 x (t )  g (t ),

(3) t  1, 2,3.. ,

(6)

t 

(t   ) . ( )

Then if 0  m  1    m , define by the RiemannRiemann Liouville fractional difference equation is (7) c x(t )   m c  m x (t ) Where  m denotes the standard m th order nabla (backward) difference.

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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 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 C1 and C2 for convergent of the solutions. 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. For further studying in this previous area, we refer the reader to the article on two-term linear fractional nabla difference equation [7]. 2. Three-term Linear fractional nabla difference equation 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,

Next, consider the term N 2 (x(t )) on (9) and also, we know that the result [7], “If 0    1 ,

N a 1  a f (t )  ( s )  s N a  f (t )  ( s )

(1  s) a 1 f ( a) .” Which implies N 2 (x(t ))  N 2 (x(t ))  x(1)  x(1)  N1 (x(t ))  x (1)  sN1 ( x(t ))  (1  s ) 1 c0  (c1  c0 )

 sN 0 ( x (t )) 

1 (1  s ) c0  c1  c0 (1  s) (1  s )

Where 1    2 .

1 1 s c0  c1  c0  c0 (1  s) (1  s ) (1  s) s (11) N 2 (x(t ))  sN 0 ( x (t ))  c0  c1 (1  s )

Apply the operator N 2 to the equation (8) we get,

Similarly, we consider the last term,

(8)

0 x (t )  C1x (t )  C2 x (t )  0,

x(1)  c1

for

x(0)  c0 ,

t  1, 2,3.. .

 sN 0 ( x (t )) 

N 2 (0 x(t )  C1x(t )  C2 x (t ))  0

(9) N 2 (0 x(t ))  C1 N 2 (x(t ))  C2 N 2 x (t )  0 First, consider a term N 2 (0 x(t )) from (8) and use the result [7] we get, “If 1    2 , N a  2 (a f (t ))( s )  s N a ( f (t ))( s )  s (1  s )

a 1

 1

f (a )  (1  s )  a f (a  1)

a

 s N a ( f (t ))( s )  s (1  s)

a 1

f (a)

(1  s) a ( f ( a  1)  (  1) f ( a)) . ˮ Which implies that, N 2 (0 x(t ))  s N 0 ( x(t ))  s (1  s ) 01 x(0) (1  s ) 0 ( x (1))  (  1) x(0))

1 1 c0  c0 (1  s) (1  s )  N 2 ( x(t ))  c1  (1  s )1 c0  c1  (1  s )1 c0 N 2 ( x(t ))  N 2 ( x (t ))  c1  c1 

In particular (12) N 2 ( x(t ))  N 0 ( x (t ))  c1  (1  s ) 1 c0 Substitute (10), (11) and (12) in (9) we get N 2 (0 x(t ))  C1 N 2 (x(t ))  C2 N 2 x(t )  0 s N 0 ( x (t ))  s (1  s ) 1 c0  (c1  (  1)c0 )

  s C1  sN 0 ( x(t ))  c0  c1  (1  s)   1 C2  N 0 ( x(t ))  c1  (1  s ) c0   0 1 ( s  C1 s  C2 )c0 (1  s) (1  C1  C2 )c1  (1   )c0  0

( s  C1 s  C2 ) N 0 ( x (t )) 

(10) N 2 (0 x(t ))  s N 0 ( x(t ))  s (1  s ) 1 c0  (c1  (  1)c0 )

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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470

( s  C1 s  C2 )c0 (1  s )( s  C1 s  C2 ) (1  C1  C2 )c1  (1   )c0  ( s  C1 s  C2 )

 m  m 1  (1)m  C1mnC2n s((1)m)n (14)  s  C1s  C2 m0 n0  n

(13) N 0 ( x(t )) 

Now take

1  ( s  C1 s  C2 ) 

  t ( 1) m  n  ( 1) Since s  N1    ((  1)m  n   )    (   1) m  n  (   1)   t .  N0   ((  1)m  n   )    ((1 ) m )  n 

1  Cs C  s 1  1  2  s s  

By using the result [7], “ Af (0) AN1 f (t )    AN 0 f (t ) ”. 1 s

1 C   s  1  C1 s1  2  s  

1  ( s  C1 s  C2 ) 

1   C2  s 1  C1 s1  1    s 1  C1 s1    

In

general, n1

 1 1  n  (n1) n   (  1) s C   2 1  s  C1s  C2 n0  1 C1s 

Now, the above equation is re-express N1 as N 0 , and we have,  m 1 m  m  mn n  (  1)    C1 C2 s  C1 s  C2 m 0 n 0 n

   t ( 1) m  n  ( 1)  N1      ((   1) m  n   )    (15)

Note that,

  t ( 1) m  n  ( 1)     ((  1)m  n   ) 

n1

  1  mn  m 1 mn  (  1) C s        1 1 mn  n 1 C1s 

We get    m 1  (1)m   C1mn s(1)mnC2n s (n1)   s  C1s  C2 n0 mn n

 m  (1)  C1mnC2n smmnnn m0 n0 n 

m

m

 m  m  N 0  (1) m C1m  n C2n   m  0 n 0 n

It follows from the result [7] “ N a f (t  1)  (1  s ) 1 N a 1 f (t ) ”. Which implies that, equation (15) we get,  m (1  s) 1 m mn n  m   N (  1) C C2    0 1 s  C1 s  C2 m 0 n 0 n (16)

 (t  1)( 1) m  n  ( 1)      ((  1)m  n   ) 

Moreover, from (14)  m  m 1 s  (1)m   C1mnC2n s(1 )mn .   s s  C1s  C2 m0 n0 n

 m s m  m mn n (1 )mn1  (  1) .   C1 C2 s n s  C1s  C2 m0 n0  

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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 Similarly, we get  m m s  ( 1) m C1m  n C2n   N 0   s  C1 s  C2 m 0 n 0 n

  t ( 1) m  n  (  2)     ((  1) m  n  (  1)) 

m

m

mn 1

n 2

 m  m (t 1)(1)mn(2) (1  C1 )c0 (1)m C1mnC2n   m0 n0  n  (( 1)m  n  ( 1))

and so  m s (1  s ) 1 m mn n  m   N C2   0  ( 1) C1  s  C1 s  C2 m 0 n 0 n (17)

 m (t 1)(1)mn(1) x(t)  C2c0 (1) C C   m0 n0  n  (( 1)m  n ) 

  (t  1)( 1) m  n  (  2)     ((  1) m  n  (  1)) 

 m  m t (1)mn( 1) K (1)m C1mnC2n   m0 n0  n  (( 1)m  n )

(19) Where K  ((1  C1  C2 )c1  (1   )c0 ) . To simplify this representation, note that

We consider an equation (13) we get

N 0 x(t ) 

(18)

( s  C1 s  C2 )c0 (1  s )( s  C1 s  C2 )

(t  1) ( 1) m  n  ( 1) t ( 1) m  n  ( 1)  ((  1)m  n   ) ((  1)m  n   )

(1  C1  C2 )c1  (1   )c0  ( s  C1 s  C2 )

C2 c0 N 0 x(t )   (1  s )( s  C1 s  C2 )

(t  1) ( 1) m  n  (  2) ((  1) m  n    1)

Since (t  1) ( 1) m n  ( 1) (t  1  (  1)m  n  (  1))  ((  1)m  n   ) (t  1) ((  1)m  n   )

t

s (1  C1 )c0  (1  s )( s  C1 s  C2 ) 

(1  C1  C2 )c1  (1   )c0 ( s  C1 s  C2 )

Substitute (15), (16) and (17) directly into (18) we get  m  m (t 1)(1)mn(1) N0 x(t)  C2c0 N0 (1)m C1mnC2n   m0 n0  n  (( 1)m  n )

(t  (  1)m  n  (  1)) (t  1)((  1)m  n   )

(t  (  1)m  n  (  1)) (t  1)((  1)m  n   ) Thus, (19) can be expressed as  ((  1)m  n  (  1))

 m  m (t 1)(1)mn(2) x(t)  K1 (1)m C1mnC2n   m0 n0  n  (( 1)m  n  ( 1))

 m  m t(1)mn(1) K2 (1)mC1mnC2n   m0 n0  n  (( 1)m n )

(20) (1)mn(2)

 m  m (t 1) Where (1 C1 )c0 N0 (1)m C1mnC2n   m0 n0  n  (( 1)m  n  ( 1)) (21) K1  c0 (1  C1  C2 ) , K 2  K  c0 C2  (1  C1  C2 )c1  (1    C2 )c0 Note that (1)mn(1)  m

 m t KN0 (1)m C1mnC2n   m0 n0  n  (( 1)m  n )

(t 1)(1)mn( 2) m mn n  m (  1) C C (22)  1 2   m0 n0  n  (( 1)m  n  ( 1)) 

m

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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470

 m t (1)mn(1) (1) C C  n  (( 1)m  n ) m0 n0   

m

m

mn 1

n 2

(23) are two linear independent solutions of (8). We give the details to obtain conditions for absolute convergence in (23) for fixed t . First note that for each t  1 ,

t

( 1) m  n  ( 1)

is ((  1)m  n   ) an increasing function in n . Now consider the ratio t ( 1) m  n 1 ( 1) t ( 1) m  n ( 1) ((  1)m  n  1   )  ((  1)m  n   ) Then (t  (  1)m  n  1  (  1)) (t )((  1) m  n  1   ) (t )((  1) m  n   ) (t  (  1)m  n  (  1)) t  (  1) m  n  (  1) 1 (  1) m  n   and the inequality is strict if t  1 . 

Thus, compare (23) to

1 , 10   2, C1 

C2 

1 , 5

c0  1,

c1  2,

To obtain

( s  (1 10) s  (1 5))1 (1  s )( s 2  ( 1 10) s  ( 1 5)) (1  (1 10)  (1 5))2  (1  2)1  ( s 2  ( 1 10) s  ( 1 5)) N 0 x (t ) 

1 1 2 2 s 2   1 10 5 10 5 N 0 x(t )   1 1 1 1    (1  s )  s 2  s    s 2  s   10 5  10 5  9 1 4 s 10 5 10 N 0 x(t )   1  2 1 1  2 1 (1  s)  s  s    s  s   10 5 10 5    s

1 1 7  From (21), K1  1 1      10 5  10 1 1 1   K 2  1    2  1  2  1 5  10 5   1  . 5

6 7    2  5  10 

and

t (m(1)) m mn n  m  C1 C2  n  m0 (m  ) n0   

mn

n

7  m 1  1  x(t )      10 m0 n0  5   10 

(t)(m(1)) m  C1  C2 m0 (m ) 

n

and apply the ratio test. Thus, each of (22) and (23) are absolutely convergent if C1  C2  1 . Description of known functions: 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 x(t )  2t , thus we obtain a series representation of 2t as a linear combination of the forms (22) and (23). Consider the initial value problem (24) 10 2 x(t )  x(t )  2 x (t )  0 , x (0)  1 , x (1)  2 .

Then, apply (13) with

1  m  1    5 m0 n0  5 

mn

 m  t  .    n   m  n  2 mn

(25) The unique solution of (24) is 2t and the series given in (25) absolutely convergent for all t  0,1, 2,.. Write 7 1 (25) as x (t )  C (t )  D (t ) . 10 5 

n

1  1  Where C (t )       m  0 n  0  5   10  m

mn

 m   t  1    n    m  n  1 mn

n mn t   1   1  m D(t )         To m  0 n  0  5   10   n    m  n  2 prove x (t  1)  2 x (t ) , or 

t  2, 3,.. ,

mn

1    10 

 m  t 1    n    m  n 1

m

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mn

Page: 598


International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 7 1 1 7  C (t  1)  D(t  1)  2  C (t )  D(t )  10 5 5  10 

mn

n

mn

n

 1  1       m0 n0  5   10  m

 m  t 1    n   m  n  2 mn1

 m   t  1  m  n  1    n    t  1   m  n  2

m

  t  1   m  n  2 

n

1    m0 n0  5  m

mn

1    10 

 m   t  m  n  1  t  n   t  1   m  n  2

mn

n

 m 1  1        m  0 n  0  5   10    t  m  n  1

m   (m  n  1) n

  t  1   m  n  2  D (t  1)  D (t )  C (t ) .

We have yet to obtain a direct approach to take C (t  1) . We begin by showing directly that D (t ) satisfies 10 2 x(t )  x(t )  2 x (t )  0, t  2, 3,. Apply the power rule and 

mn

n

 1  1  D(t)      m0 n0  5   10  m

 m

n

 m t mn1    n  (m  n  2)

mn

 1  1  D(t)      m0 n0  5  10 

 m t mn1    n  (m  n)

m1n

 m tmn  n   (m n 1)

1   10 

t mn 1    ( m  n  1)  5  n

m 1

t 2 m 1 )  (2m  2)

mn

 m t mn  n   (m  n 1) n mn mn  m1  1  1   m  t       m0 n0  5  10   n 1 (m  n 1) 

1 m  1     m0 10 n0  5 

1    10 

mn

n

 1   1  m          t  m  n  1 m  0 n  0  5   10  n   t  m  n  1 

mn

1    10 

 m 1 t mn    n  (m  n 1) n mn  1 m  1   1   m  m    (          m  0 10 n  0  5   10    n   n  1   1  D(t)    m0 n0  5 

Now consider D (t  1) ,

1  1  D(t 1)      m0 n0  5   10 

n

 m1

2

D (t  1)  C (t )  D (t ) and 7 1 6  C (t  1)   C (t )  D (t )  . 10 5 5 

m

n

1  D(t)    m1 n0  5  m

Thus,

It is sufficient to show that

2

n

mn

1  1  1  1  D (t )      5 m0 n0  5   10  10 m



 2 D (t ) 

 m t mn1    n  (m  n  2)

1 1 D (t )  D (t ) 10 5

A similar calculation shows that C (t ) satisfies 10 2 x(t )  x(t )  2 x (t )  0, t  2, 3,. We close by arguing that 7 1 6  C (t  1)   C (t )  D (t )  . 10 5 5  Simplify 10 2 C (t  1)  C (t  1)  2C (t  1)  0 10(C (t  1)  2C (t )  C (t  1))  (C (t  1)  C (t ))  2C (t  1)  0

7 19 C (t  1)  C (t )  C (t  1) 10 10 6  7   C (t )   C (t )  C (t  1)  5  10 

To obtain

Now it is sufficient to show that  7  1  C (t )  C (t  1)   D (t ) .  10  5

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Page: 599


International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 Employ C (t )  D (t  1)  D (t ) and C (t  1)  D (t )  D (t  1) to obtain  7  7  C (t )  C (t  1)   ( D (t  1)  10  10  D (t ))  ( D (t )  D (t  1))

7 17 ( D (t  1)  D(t )  D (t  1) 10 10

19  7  2   ( D (t  1)  D (t )  D (t  1)   D (t ) 10  10  10  7  1  C (t )  C (t  1)   D (t ) .  10  5

References 1. Agarwal, R.P., Difference Equations and Inequalities, Marcel Dekker, New York,1992. 2. Jagan Mohan Jonnalagadda, Asymptotic Behaviour of Linear Fractional Nabla Difference Equations, Inter. Jour. of Differ. Equ., (2017) 255265. 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 Jonnalagadda, Analysis of 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.

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Page: 600

Three-Term Linear Fractional Nabla Difference Equation  

In this present paper, a study on nabla difference equation and its third order linear fractional difference equation. A new generalized nab...

Three-Term Linear Fractional Nabla Difference Equation  

In this present paper, a study on nabla difference equation and its third order linear fractional difference equation. A new generalized nab...

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