# Three Term Linear Fractional Nabla Difference Equation ```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
R. Ranjitha
Research Scholar, Department of Mathematics,
Vivekanandha College of Arts and Sciences for
Women, Tiruchengode,
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 .
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 ,
“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  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 , “
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 
“ 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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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
and
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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
@ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018
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
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
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