Existence of positive solutions for fractional differential involving the discrete delta−nabla fract

Invention Journal of Research Technology in Engineering & Management (IJRTEM) ISSN: 2455-3689 www.Ijrtem. com Volume 3 Issue 5 ǁ July –August 2019 ǁ PP 40-48

Existence of positive solutions for

fractional differential

involving the discrete delta− nabla fractional boundary value problem with non-homogeneous boundary conditions 1,

Dong Qiang, 2,Chengmin Hou （Yanbian University, Jilin Yanji 133002）

ABSTRACT: In this paper, we consider a nonlinear fractional boundary value problem with non-homogeneous boundary conditions:

( (

))

 b    p  − 2 x(t ) + h(t +  −  − 1) f (x(t +  −  − 1)) = 0, t  T ,  1−   x( − 2) = 0, x(b +  − 2) =  −1 x(t ) t = +  − 2 ,    x(t ) = 0. b  p  −2

 (

Where b  Z

)

+





, T =  − 1, b +  − 2 -1 , 0    1    2, 2   +   3, we invert the problem and

construct and analyse the corresponding Green’s function. We obtain sufficient conditions for the existence of positive solutions for p − Laplacian fractional differential involving the discrete

delta− nabla fractional

boundary value problem with non-homogeneous boundary conditions.

KEYWORDS: boundary value problem; p − Laplacian operator; discrete delta− nabla fractional; non-homogeneous boundary conditions; positive solutions

I.

INTRODUCTION

The study of boundary value problems in the setting of discrete fractional calculus has received a great attention in the last decade[1-3]. In particular, several recent papers by the theory’s powerful and versatile applications to almost all areas of science, engineering and technology, which teem with discrete phenomena, see[4-7].

Therefore, scientific advancements in the area of difference equations are naturally motivated and are of significant interest. However, to the best of our knowledge, there are very few papers on the discrete

delta− nablafractional boundary value problems. Inspired by the above literature, we study existence of solutions for p − Laplacian fractional discrete involving the discrete

delta− nabla fractional boundary

value problem with non-homogeneous boundary conditions:

( (

))

 b    p  − 2 x(t ) + h(t +  −  − 1) f (x(t +  −  − 1)) = 0, t  T ,  1−   x( − 2) = 0, x(b +  − 2) =  −1 x(t ) t = +  −2 ,    x(t ) = 0. b  p  −2

 (

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(1.1)

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Existence of positive solutions for…

Where b  Z

+

, T =  − 1, b +  − 2 −1 , 0    1    2, 2   +   3,  −2 , b 

are left and right fractional difference operators, respectively.  p (s ) =

II.

s

p −2

s,

.

PRELIMINARIES

We first wish to collect some basic lemmas that will be important to us in what follows. We begin with some basic properties regarding the discrete fractional difference. These results will play a decisive role in our proofs later in this paper.

For any real number  , let   

We define t =

=  ,  + 1,  + 2,...,   = ...,  − 2,  −1,  .

(t + 1) ,for any t and  for which the right-hand side is defined. We also appeal to the (t + 1 −  )

t +1 −  is a pole of the gamma function and t + 1 is not a pole, then t  = 0 .

common convention that, if

Definition 2.1[8] Let f : N a → R and

−a f (t ) =

t −

 (t − s − 1) ( ) 1

 −1

s =a

 − f (t ) =

b

 (s − t − 1) ( )  1

 −1

s =t +

Lemma 2.1[8] Let f :  a → R and 

for the left fractional difference a f

left fractional sum of f is given by

f ( s ) , t   a + .

  0 be given. The th

Definition 2.2[9] Let f :b N → R and

b

  0 be given. The th

right fractional sum of f is given by

f (s ), tb − .

  0 be given with N −1    N.

The following two definitions

:  a+ N − → R

are equivalent:

a f (t ) = N −a( N − ) f (t ),  1 t + − −1 ( t − s − 1) f (s ), N − 1    N ,    a f (t ) =  (−  ) s =a N   = N.   f (t ), 

Lemma 2.2[9]Let f :b  → R and

  0 be given with N −1    N.

The following two definitions for

the right fractional difference

a f :  a+ N − → R are equivalent:

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Existence of positive solutions for…

 f (t ) = (− 1) bN  −( N − ) f (t ), N

b

b  1 (s − t − 1)− −1 f (s ), N − 1    N ,   b  f (t ) =  (−  ) s =t −  (− 1)N  N f (t ),  = N. 



Let f :  a → R be given and suppose k   0 and

Lemma 2.3[9] Then for

 0.

t   a+ M − + ,  j f (a ) (t − a ) −k + j . j = 0  ( − k + j + 1)

k −1

−a k f (t ) = ka− f (t ) −  Moreover, if

  0 with M − 1    M , then for t   a + ,  ja− M +  f (a + M −  ) (t − a − M +  ) − M + j . ( − M + j + 1) j =0

M −1

−a+ M −  a f (t ) = a− f (t ) − 

Lemma 2.4[9] Let f :b  → R be given, and suppose k   0 and

 0.

Then for tb − ,

b



−

 f (t )= b  k

b

Moreover, if

k −

 j f (b ) (b − t ) −k + j . f (t ) −  j = 0  ( − k + j + 1) k −1

b

  0 with M − 1    M , then for tb−M +  − , M −1

−   − f (t ) −  b b−M +   b  f (t )= b  j =0

 j − M +  f (b − M +  ) (b − M +  − t ) − M + j . ( − M + j + 1)

j

In the following paragraphs, we define

 y(t ) = 0 for

j i.

t =i

Lemma 2.5[10] Let f be a real-value function defined on  a and let







 ,  0 ,Then



−a+  −a f (t ) = −a( + ) f (t ) = −a+ −a f (t ) .

Lemma 2.6[10]Let a  R,

−a+  (t − a ) = 

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   R \ {..., −2,−1,0},   0, and (t − a ) :  a +  → R.

Then

( + 1) (t − a ) + , for t  a+ + ; and ( + 1 +  )

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Existence of positive solutions for…

( + 1) (t − a ) − , for t   a+ + N − . ( + 1 −  )

a +  (t − a ) = 

Theorem2.1 Let h :

0, b

0

→ R be given, the fractional value boundary problem:

   −2 y(t ) + h(t +  − 1) = 0,  1−    y ( − 2) = 0, y (b +  − 2) =  −1 y (t ) t = +  −2 ,





(2.1)

has a unique solution

y (t ) =

1

b−2

 G (t , s )h(s +  − 1)

( ) s =0

where

 (t ) −1 (b +  − s − 3) −2  −1 + (t − s − 1) ,   −1  (b +  − 2) − ( ) G(t , s ) =   −1  −2  (t ) (b +  − s − 3) ,  −1   (b +  − 2) − ( )

0  s  t −  + 1  b − 1, 0  t −  + 1  s  b − 1.

Proof by Lemma 2.3, we have

y (t ) =

− 1 t − (t − s − 1) −1h(s +  − 1) + c1 (t ) −1 + c2 (t ) −2 ,  ( ) s =0

form y ( − 2 ) = 0,

y (b +  − 2 ) =





y (b +  − 2 ) = 1−−1 y (t ) t = +  − 2 , we have c2 = 0 , we also have

− 1 b−2 (b +  − s − 3) −1h(s +  − 1) + c1 (b +  − 2) −1 ,  ( ) s =0

and

1−−1 y (t ) = − 1−−1−0 h(t +  − 1) + c11−−1 (t )

 −1

( )(t )  h(t +  − 1) + c1 ( +  − 1)

− −  −1 0

= −

 +  −2

t − ( +  )  1 (t − s − 1) +  −1h(s +  − 1) + c1 ( ) (t ) +  −2 = −   ( +  − 1)  ( +  ) s =0 

=



1−   −1

t − ( +  −1) −1 (t − s − 1) +  −2 h(s +  − 1) + c1 ( ) (t ) +  −2  ( +  − 1) s =0 ( +  − 1)



y (t ) t = +  − 2 = c1( )

we obtain

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Existence of positive solutions for…

c1 =

b−2

)  (b +  − s − 3)

1  −1 ( ) (b +  − 2) − ( )

(

 −2

h(s +  − 1)

s =0

Then

y (t ) =

t −

(t − s − 1) h(s +  − 1)  ( ) 1

−1

s =0

 −1 b−2 ( t) (b +  − s − 3) −2 h(s +  − 1). +   −1 ( )((b +  − 2) − ( )) s =0

We denote s = s +  −  − 1 ,   =  +  −  − 1 we assert that the boundary problem (1.1) has a unique solution

x(t ) =

 G(t , s )  ( ) 1

b−2 s =0

=

q b +  −1



 −  h(s) f ( x(s))

 1 b +  −1  ( ) ( − s − 1) −1h( ) f (x( )).

G t , s    q ( ) s =0   ( )  = s +   1

b−2

Theorem2.2 We suppose that 0   −   1 . The function

(i)

0  G(t , s ) 

min

(ii)

G(t , s ) have the following properties:

D(b +  ) G(s +  − 1, s ), (t , s )   − 1, b +  − 2 −1  0, b − 2 0 ; (s +  − 1) −1  −1

t −1,b + − 2  −1

(  − 1) −1 G(t , s )  G(s +  − 1, s )  0, (s +  − 1) −1

for

s  0, b − 20 .

Where

 (b +  − 2 ) −1 − ( ) D = max 1 +  s0 ,b − 2 0 (b +  − s − 3) −2   =

(b +  − 2) −1 . ( ) III.

MAIN RESULTS

Next we shall devote our attention to finding the existence and uniqueness of solutions for fractional boundary value problem (1.1).

Lemma 3.1[9] (1) If 1  p  2, uv  0 and

u , v  m  0 , then

 p (v ) −  p (u )  ( p − 1)m p−2 v − u .

(2) If p  2 and

u , v  M , then

 p (v ) −  p (u )  ( p − 1)M p−2 v − u . |Volume3|Issue5|

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Existence of positive solutions for…

Define Banach spaces operator

E = x :  −1, b +  − 2 −1 → R. In order to get main results, we introduce a

A : E → E by

 1 b +  −1  ( − s − 1) −1h( ) f (x( )).

Ax(t ) = G (t , s )q    ( ) s =0  ( )  = s +   1

b−2

We shall appeal to the contraction mapping theorem to get a unique solution of boundary value problem (1.1) when p  2 .

Theorem3.1 Suppose that

f (x ) is Lipschitz in x , that is , there exists constant L  0 such that

f (x1 ) − f (x2 )  L x1 − x2 whenever x1 , x2  R, t   −1, b +  − 2 −1 , and there exists a function

A(t ) such that

M2 =

M=

f (x )  A(t ) , for any x E , let M 1 =

b +  −1

( − s − 1) h( )L , if  ( )   1

−1

b +  −1

( − s − 1) h( )A( ) and  ( )   1

−1

=s+

p  2 and

=s+

(q − 1) b−2 D(b +  ) −1 G(s +  − 1, s )M (q−2 )M  1 , then the boundary value problem (1.1) has a  1 2 ( ) s =0 (s +  − 1) −1

unique solution.

Proof By Lemma3.1 for any

x1 , x2  E , we can get that

Ax1 (t ) − Ax2 (t )  (q − 1)

 for

max

t −1,b + − 2  −1

b−2

G (t , s )M (  ( ) 1

q−2 )

1

s =0

M 2 x1 − x2

(q − 1) b−2 D(b +  ) −1 G(s +  − 1, s )M (q−2 )M  1 2 ( ) s =0 (s +  − 1) −1

x1 − x2

t   −1, b +  − 2 −1 , we conclude that Ax1 (t ) − Ax2 (t )  M x1 − x2 ,

(3.1)

whence by M  1 , we find that (1.1) has a unique solution, this completes the proof.

Theorem3.2 Suppose that

f (x1 ) − f (x2 )  L x1 − x2 non-negative

function

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f (x ) is Lipschitz in x , that is , there exists constant L  0 such that whenever

B(t )

x1 , x2  R, satisfying

t   −1, b +  − 2 −1 , and there exists a f (x )  B(t )

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,

for

any

x E

,

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Existence of positive solutions for…

let

K1 =

b +  −1

( − s − 1) h( )B( )  ( )   1

−1

and

=s+

M2 =

b +  −1

( − s − 1) h( )L ,  ( )   1

−1

if 1  p  2

=s+

and  −1 ( q − 1) b−2 D(b +  ) K= G(s +  − 1, s )K1(q −2 )M 2  1 , then the boundary value problem (1.1) has a   −1 ( ) s =0 (s +  − 1)

unique solution.

Proof Define Banach spaces P =

Tx (s ) =

x : 0, b − 2

0

b +  −1

→ R, we introduce a operator T : P → P by

( − s − 1) h( ) f (x( )).  ( )   1

−1

=s+

By Lemma3.1 for all s0 ,b − 2 0

b +  −1

( − s − 1) h( )Ty ( )  ( )   1

Ty = max 

x  P , we can get that −1

=s+

b +  −1

( − s − 1) h( )B( ).  ( )   1

−1

=s+

What’s more, for any

x1 , x2  E , we have

Ax1 (t ) − Ax2 (t )  (q − 1)

t −1,b + − 2  −1

b−2

G (t , s )K (  ( ) 1

max

s =0

q −2 )

1

M 2 x1 − x2

 −1 ( q − 1) b−2 D(b +  )  G(s +  − 1, s )K1(q −2 )M 2   −1 ( ) s =0 (s +  − 1)

for

x1 − x2

t   −1, b +  − 2 −1 , we conclude that Ax1 (t ) − Ax2 (t )  K x1 − x2 ,

(3.2)

hence (1.1) has a unique solution, this completes the proof.

IV.

EXAMPLES

In this section, we will present some examples to illustrate main results.

Example4.1 Consider boundary value problem of discrete fractional equation

 0.8 1  2 2 1.5  3  3  −0.5 x(t ) + (t − 0.3) (t + 0.7 ) + 800 = 0,    0.2 ( ) ( ) ( ) x − 0 .

5 = 0 , x 2 .

5 =  x t ,  0.5 t = 0.3   1.5 x(t ) = 0, 3  3 −0.5 

( (

 (

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Existence of positive solutions for…

1 , h = t 2 , for t  0.5,2.50.5 , x  R , is Lipschitz with Lipschitz 800 1 1 2 constants L = . When p = 3 , for this choice of L and A(t ) = (t + 1) + , inequality (3.1) is 100 500 satisfied with M  0.038  1 . Therefore, we deduce from Theorem3.1 that problem (4.1) has a unique f ( x ) = (t + 1) + 2

where

solution.

Example4.2 Consider boundary value problem of discrete fractional equation

 0.8   1 .5  3    3  −0.5 x(t )  + (t − 0.3)(t + 0.7 ) = 0,  2    0 .2  x(− 0.5) = 0, x(2.5) =  0.5 x(t ) t =0.3 ,     3 1−.50.5 x(t )  = 0,   2 3

(

)





(4.2)

f (x ) = t + 1, h = t , for t  0.5,2.50.5 , x  R , is Lipschitz with Lipschitz constants L =

1 .

700

(

where

When p =

)

3 , for this choice of L and A(t ) = t , inequality (3.2) is satisfied with M  0.642  1 .

2

Therefore, we deduce from Theorem3.2 that problem (4.2) has a unique solution.

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