Positive and negative solutions of a boundary value problem for a fractional -difference equation

Invention Journal of Research Technology in Engineering & Management (IJRTEM) ISSN: 2455-3689 www.ijrtem.com Volume 2 Issue 9 ǁ September 2018 ǁ PP 73-83

Positive and negative solutions of a boundary value problem for a fractional q,  -difference equation Yizhu Wang1 1,2,(

Chengmin Hou1*

Department of Mathematics, Yanbian University, Yanji, 133002, P.R. China)

ABSTRACT : In this work, we study a boundary value problem for a fractional q,  -difference equation. By using the monotone iterative technique and lower-upper solution method, we get the existence of positive or negative solutions under the nonlinear term is local continuity and local monotonicity. The results show that we can construct two iterative sequences for approximating the solutions.

KEYWORDS: fractional q,  -difference equation; positive and negative solutions; lower and upper solutions; iterative method; I.

INTRODUCTION

A quantum calculus substitutes the classical derivative by a difference operator, which allows one to deal with sets of non-differentiable functions. There are many different types of quantum difference operators such as hcalculus, q-calculus, Hahn’s calculus. These operators are also found in many applications of mathematical areas such as orthogonal polynomials, combinatorics and the calculus of variations.

Hahn [1] introduced his difference operator Dq, as follows:

Dq , f (t ) =

f (qt +  ) − f (t ) , t (q − 1) + 

t

 1− q

,

where f is real function, and q  (0,1) and   0 are real fixed numbers. Malinowska and Torres [2,3] introduced the Hahn quamtum variational calculus, while Malinowska and Martins [4] investigated the generalized transversality conditions for the Hahn quantum variational calculus. Recently, Hamza et al. [5,6] studied the theory of linear Hahn difference equations, and studied the existence and uniqueness results of the initial value problems with Hahn difference equations using the method of successive approximations.

Motivated by the aforementioned work, we consider the following nonlinear boundary value problem for a fractional q,  -difference equation:

 a Dq, u (t ) + f (t , u (t )) = 0, t  (a, b),   u (a ) = a Dq , u (a ) = a Dq , u (b) = 0, where q  (0,1), 2    3,

f :[a, b]  [0, +),

(1.1)

a

Dq, is the fractional q,  -derivative of the Riemann -

Liouville type?

II.

BACKGROUND AND DEFINITIONS

To show the main result of this work, we give in the following some basic definitions, lemmas and theorems, which can be found in [7].

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Positive and negative solutions of a boundary value…

Definition 2.1 [7] Let I be a closed interval of

0 , a, b  I .

such that

For f : I →

we define the

q,  -integral of f from a to b by



b

a

b

a

0

0

f (t )d q , t :=  f (t )d q , t −  f (t )d q , t ,

where x



0

with



f (t )d q , t := ( x(1 − q ) −  ) q k f ( xq k + [k ]q , ),

x  I,

k =0

[k ]q , =

 (1 − q k ) 1− q

for k 

0

{0}, provided that the series converges at x = a and

=

x = b. f : I → be continuous at 0 . Define

Lemma 2.2 [7] Assume at

0 .

Futhermore,



b

a

Dq, F ( x) exists for every

and

x

F ( x) :=  f (Then t )d q , tF. is continuous 0

x  IDq, F ( x) = f ( x).Conversely,

Dq , F ( x)d q , t = f (b) − f (a ),

for all a , b  I .

Lemma 2.3 [7] For

 ,

q, ( + 1) = [ ]q q, ( ),

q, (1) = 1.

For any positive integer k,

q, (k + 1) = [k ]!.

Definition 2.4 [7] Let

 0

and

Riemann-Liouville type is given by

( a I q, f )(t ) =

f be a function defined on [a,b]. The Hahn’s fractional integration of ( a I q0, f )(t ) = f (t ) and

t 1 (t − 0  q ( s))(0 −1) f ( s)d q , s,  a  q ( )

  0, t  [a, b].

Definition 2.5 [7] The fractional q,  -derivative of the Riemann -Liouville type is 

 −

( a Dq, f )(t ) = ( Dq, a I q, where

 

f )(t ),

  0,

denotes the smallest integer greater or equal to  .

Theorem 2.6 [7] Let

  ( N − 1, N ]. Then for some constants ci  , i = 1, 2,

, N , the following

equality holds:

( a I q, a Dq, f )( x) = f ( x) + c1 ( x − a)(0 −1) + c2 ( x − a)(0 −2) + |Volume 2| Issue 9 |

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Positive and negative solutions of a boundary value…

Lemma 2.7 [8] Assume that X is a Banach space and K is a normal cone in X, T :[u0 , v0 ] → X is a completely

u0  Tu0 ,Tv0  v0 .

continuous increasing operator which satisfies and a maximal fixed point v

*

with

u* = lim T u0 , n

n →

Then T has a minimal fixed point u*

u  u* nv*  v0 . In addition, *0 v = lim T v0 , n →

{T nu0 }n=1 is an increasing sequence, {T n v0 }n=1 is a decreasing sequence.

where

III.

EXISTENCE OF q,  -FRACTIONAL POSITIVE SOLUTIONS FOR PROBLEM

Lemma 3.1

Assume g  C[a, b], then the following boundary value problem:

 a Dq, u (t ) + g (t ) = 0, t  (a, b),   u (a ) = a Dq , u (a ) = a Dq , u (b) = 0, has a unique solution b

u (t ) =  G (t , 0  q ( s ))g ( s )d q , s, a

where

 (t − a )( −1) 0  (b − 0  q ( s ))(0 − 2) − (t − 0  q ( s ))(0 −1) , ( − 2)  (b − a )0 1  G (t , 0  q ( s )) = a  0  q ( s )  t  b,  q ( )  ( −1)  (t − a )0 ( − 2)  (b − a )( − 2) (b − 0  q ( s ))0 , a  t  0 q (s)  b. 0  Proof

In view of Theorem 2.6,

u (t ) = −

t 1 (t − 0  q ( s))(0 −1) g ( s)d q , s + c1 ( x − a)(0 −1) + c2 ( x − a)(0 −2)   q ( ) a

+c3 ( x − a)(0 −3) , Since

u(a) = a Dq , u (a) = 0, we have c2 = c3 = 0. From the boundary condition a Dq , u (b) = 0,

we get

c1 =

1  q ( )(b − a )(0 − 2)



b

a

(b − 0  q ( s ))(0 − 2) g ( s )d q , s.

Hence

(t − a)(0 −1) t 1 ( −1) u (t ) = − (t − 0  q ( s))0 g ( s)d q , s +  q ( ) a (b − a)(0 −2)  q ( ) b

 (b − 0  q ( s))(0 − 2) g ( s)d q , s a

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Positive and negative solutions of a boundary value…

( −1)  t  (t − a ) 1 ( − 2) ( −1) 0 = ( b −  ( s )) − ( t −  ( s ))  g ( s)d q , s  q   q  0 0 0 0  q ( ) a  (b − a)(0 −2)  ( −1)  b  (t − a ) 1 ( − 2) 0 + ( b −  ( s ))  g (s)d q , s  q  (  − 2) 0 0 q ( ) t  (b − a)0 

b

=  G (t , 0  q ( s ))g ( s )d q , s.

□

a

Lemma 3.2 (a) (b)

G(t , 0 q (s))  0, G(t, 0 q (s))  G(b, 0 q (s)), a  t , 0 q (s)  b; G (t , 0  q ( s)) 

Remark 3.3 (1)

The function G(t , 0 q (s)) has the following properties:

(t − a)(0 −1) (b − a)(0 −2)

G (b, 0  q ( s)), a  t ,  q (s)  b.

0

The function G(t , 0 q (s)) has some other properties:

( −1) 1 (t − a)0 1 G (t , 0  q ( s))  (b − 0  q ( s))(0 −2)  (t − a)(0 −1) , ( − 2)  q ( ) (b − a)0  q ( )

a  t , 0 q (s)  b.

(2) According to the property of being non-decreasing of function

(b − 0  q ( s))(0 −2) (b − a)(0 −2)

=

(b − 0 iq+1 ( s))(b − 0  iq+ − 2 (a)) (b − 0 iq+ −1 ( s))(b − 0 iq (a))

on a and non-increasing of (t2 −   q ( s))( −2) on s, We can obtain the following inequalities: 0 0

a  t1  t2  0 q (s)  b, we get

(a) For

G(t2 , 0 q (s)) − G(t1 , 0 q (s)) ( −1) ( −1) 1 (t2 − a)0 1 (t1 − a)0 ( − 2) = (b − 0  q ( s))0 − (b − 0  q ( s))(0 −2) q ( ) (b − a)(0 −2)  q ( ) (b − a)(0 −2)

=  (b) For

( − 2) 1 (b − 0  q ( s))0 (t2 − a)(0 −1) − (t1 − a)(0 −1) ( − 2)  q ( ) (b − a)0

1 [(t2 − a)(0 −1) − (t1 − a)(0 −1) ].  q ( )

a  t1  0 q (s)  t2  b, we get G(t2 , 0 q (s)) − G(t1 , 0 q (s))

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Positive and negative solutions of a boundary value… ( −1)  1  (t2 − a )0 ( − 2) ( −1) = ( b −  ( s )) − ( t −  ( s ))   0 q 0 2 0 q 0  q ( )  (b − a)(0 − 2) 

=

( − 2) 1 (b − 0  q ( s))0 (t2 − a)(0 −1) − (t1 − a)(0 −1)  − (t2 − 0  q ( s))(0 −1) q ( ) (b − a)(0 −2)

−

1 [(t2 − a)(0 −1) − (t1 − a)(0 −1) + (t2 − t1 )(0 −1) ].  q ( )

 (c) For

( −1) 1 (t1 − a)0 (b − 0  q ( s))(0 −2) ( − 2) q ( ) (b − a)0

a  0 q (s)  t1  t2  b, we get

G(t2 , 0 q (s)) − G(t1 , 0 q (s)) ( −1) 1  (t2 − a )0 = (b − 0  q ( s ))(0 − 2) − (t2 − 0  q ( s ))(0 −1)  ( − 2)  q ( )  (b − a)0

−

(t1 − a )(0 −1) ( − 2)

(b − a )0

 (b − 0  q ( s ))(0 − 2) − (t1 − 0  q ( s ))(0 −1)  

( − 2) 1 (b − 0  q ( s))0  ( −1) ( −1)  (t2 − a)0 − (t1 − a)0  q ( ) (b − a)(0 −2)



+ (t1 − 0  q ( s ))(0 −1) − (t2 − 0  q ( s ))(0 −1)  

1 (t2 − a)(0 −1) − (t1 − a)(0 −1) + (t2 − 0  q ( s))(0 −1) − (t1 − 0  q ( s))(0 −1)  .  q ( ) 

(3) G(t , 0 q (s))  0, Let

X  C[a, b],

for the

(t, 0 q (s))  (a, b)  (a, b).

Banach

space

of

all

continuous

functions

on

[a,b],

u = max{ u(t ) : t [a, b]}. In our considerations, we need the standard cone

with

norm

KX

by

K = {u  [a, b] : u (t )  a, a  t  b}. It is clear that the cone K is normal.

Theorem 3.4

Assume that

( F1 ) there exist a real number d  0 and g  L1[a, b], such that (i1 )

f :[a, b]  [a, d ] → [a, +) is continuous, f (t , u )  g (t ) for (t , u )  [a, b]  [a, d ]

and f (t , u )  f (t , v ) for a  t  b, a  u  v  d ;

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Positive and negative solutions of a boundary value…

(i2 ) the following inequality holds: b 1 (b − 0  q ( s))(0 −2)  a  q ( )

 ( s − a)(0 −1)  f  s, d d q , s  d .  (b − a)(0 −2)  

( F2 ) there exist c  (a, d ) such that  ( s − a)(0 −1)  G ( b ,  ( s )) f s , c d s  c.  0 q ( − 2)  a   q , ( b − a )  0   b

Then the problem (1.1) has two positive solutions u , v  D, where *

D = {u  C[a, b] c

In addition, let u0 (t ) = c

(t − a)(0 −1) (b − a)(0 −2)

 u (t )  d

(t − a)(0 −1)

, v0 (t ) = d ( − 2)

(b − a)0

b

un +1 =  G (t , 0  q ( s )) f ( s, un ( s ))d q , s, a

n = 0,1, 2,

*

(t − a)(0 −1) (b − a)(0 −2)

(t − a)(0 −1) (b − a)(0 − 2)

, t  [a, b]}.

and construct the following sequences:

b

vn +1 =  G (t , 0  q ( s)) f ( s, vn ( s))d q , s, a

un = u* , lim vn = v* . , one has lim n → n →

Proof From the non-negativeness and continuity of

G and f , we can define an operator T : C[a, b] → C[a, b]

by b

Tu (t ) =  G (t , 0  q ( s)) f ( s, u ( s))d q , s, a

a  t  b.

From Lemma 3.1, we can see that u is the solution of problem (1.1) if and only if u is the fixed point of T. We will show that T has fixed points in the order interval [u0 , v0 ].

We need to show that T :[u0 , v0 ] → C[ a, b] is a completely continuous operator. For u  [u0 , v0 ], we have

ac

(t − a)(0 −1) (b − a)(0 −2)

 u (t )  d

(t − a)(0 −1) (b − a)(0 −2)

only prove T is compact. Let M =



b

a

 d,

a  t  b. Since G(t ,  q (s)) is continuous. So we 0

g ( s )d q , s, then a  M  +. From the hypothesis ( F1 ) − (i2 ) and

lemma 3.2, we get

Tu (t ) = max a t b





b

a

max

G(t , 0  q ( s)) f ( s, u( s))d q , s

a  0  q ( s )b

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b

G(t , 0  q ( s))  f ( s, u ( s))d q , s a

b 1 M (t − a)(0 −1)  g ( s)d q , s = (t − a)(0 −1) . a  q ( )  q ( )

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Positive and negative solutions of a boundary value…

This shows that the set T ([u0 , v0 ]) is uniform bounded in C[ a, b]. After that, for given t1 , t2  [a, b] with

t1  t2 , and u  [u0 , v0 ], we obtain b

Tu1 (t ) − Tu2 (t )   G (t1 , 0  q ( s )) − G (t2 , 0  q ( s )) f ( s, u ( s ))d q , s a



max

a  0  q ( s ) b

=M

G(t1 , 0  q ( s)) − G(t2 , 0  q ( s))

max

a  0 q ( s )b



b

a

g ( s)d q , s

G(t1 , 0 q (s)) − G(t2 , 0  q (s)) .

In view of Remark 3.3(2), one has Tu1 (t ) → Tu2 (t ), as t1 → t2 . So we claim that the set T ([u0 , v0 ]) is equicontinuous in

C[ a, b].By means of the Arzela-Ascoli theorem, T :[u0 , v0 ] → C[a, b] is a completely operator.

By the hypothesis ( F1 ) − (i1 ), T is an increasing operator.

From ( F1 ), ( F2 ) and Lemma 3.2, for any t  [a, b], one can see that b

Tu0 (t ) =  G (t , 0  q ( s )) f ( s, u0 ( s))d q , s a

 ( s − a)(0 −1)  b =  G(t , 0  q ( s)) f  s, c d s ( − 2)  a   q , ( b − a )  0   

b

a

 ( s − a)(0 −1)  G(b, 0  q ( s)) f  s, c  d q , s  (b − a)(0 −2) (b − a)(0 −2)   (t − a)(0 −1)

(t − a)(0 −1)



(b − a)(0 −2)

c = u0 (t )

and b

Tv0 (t ) =  G (t , 0  q ( s )) f ( s, v0 ( s ))d q , s a

 ( s − a)(0 −1)  b =  G(t , 0  q ( s)) f  s, d d s ( − 2)  a   q , ( b − a )  0   



( −1) 1 (t − a)0  q ( ) (b − a)(0 −2)

(t − a)(0 −1) (b − a)(0 −2)

 ( s − a)(0 −1)  ( − 2) ( b −  ( s )) f s , d  a 0 q 0  (b − a)( −2) dq, s 0   b

d = v0 (t ).

Hence, we get Tu0  u0 , Tv0  v0 . We construct the following sequences: b

un +1 =  G (t , 0  q ( s )) f ( s, un ( s ))d q , s, a

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b

vn +1 =  G (t , 0  q ( s)) f ( s, vn ( s))d q , s, a

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Positive and negative solutions of a boundary value…

. From the monotonicity of T, we have un +1  un , vn +1  vn , n = 0,1, 2,

n = 0,1, 2,

2.7, we know that the operator T has two positive solutions u , v  C[a, b] with *

that is,

c

(t − a)(0 −1)

 u* (t )  v* (t )  d

( − 2)

(b − a)0

(t − a)(0 −1) ( − 2)

(b − a)0

a  t  b.

,

*

. By using Lemma

u0  u*  v*  v0 ,

un = u* , lim vn = v* .

In addition, lim n → n →

□

Theorem 3.5 Assume that

( F3 ) there exist a real number d  0 and g  L1[a, b], such that f :[a, b]  [a, d ] →

(i3 )

f (t , u )  g (t ) for (t , u )  [a, b]  [a, d ] and

is continuous,

f (t , u )  f (t , v) for t  [a, b], a  u  v  d ;

(i4 ) the following inequality holds: b 1 (b − 0  q ( s))(0 −2) max{ f  a q ( )

 ( s − a)(0 −1)  s , d , 0}d q , s  ( − 2)    ( b − a )  0  

 ( s − a)(0 −1)  b +  G(b, 0  q ( s)) min{ f  s, d  , 0}d q , s  d . a  (b − a)(0 −2)  

( F4 ) there exist c  [a, d ] such that  ( s − a)(0 −1)  G ( b ,  ( s )) max{ f s , c , 0}d q , s  0 q ( − 2)  a   ( b − a )  0   b

+

b 1 (b − 0  q ( s))(0 −2) min{ f  a  q ( )

 ( s − a)(0 −1)  s , c , 0}d q , s  c.  ( − 2)    ( b − a )  0  

Then the problem (1.1) has two positive solutions u , v  D, where *

D = {u  C[a, b] c

u0 (t ) = c

In addition, let

(t − a)(0 −1) (b − a)(0 −2)

(t − a)(0 −1) (b − a)(0 − 2)

,

*

 u (t )  d

v0 (t ) = d

(t − a)(0 −1) (b − a)(0 −2)

, t  [a, b]}.

(t − a)(0 −1) (b − a)(0 − 2)

b

b

a

a

and construct the following sequences:

un +1 =  G (t , 0  q ( s )) f ( s, un ( s ))d q , s, vn +1 =  G (t , 0  q ( s)) f ( s, vn ( s))d q , s, n = 0,1, 2, Proof

un = u* , lim vn = v* . , one has lim n → n →

Consider the same operator T : C[ a, b] → C[ a, b] as defined in the proof of Theorem 3.4: b

Tu (t ) =  G (t , 0  q ( s)) f ( s, u ( s))d q , s, a

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t  [a, b].

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Positive and negative solutions of a boundary value…

We also show that T has fixed points in the order interval [u0 , v0 ].

Similar to the proof of Theorem 3.4, T : C[ a, b] → C[ a, b] is a completely continuous operator. From the hypothesis ( F3 ) − (i3 ), T is an increasing operator. Further, by using the conditions ( F3 ), ( F4 ), Remark 3.3 and Lemma 3.2, for any t  [a, b], one obtains b

Tu0 (t ) =  G (t , 0  q ( s )) f ( s, u0 ( s))d q , s a

 ( s − a)(0 −1)  =  G(t , 0  q ( s)) max{ f  s, c  , 0}d q , s a  (b − a)(0 −2)   b

 ( s − a)(0 −1)  +  G(t , 0  q ( s)) min{ f  s, c  , 0}d q , s a  (b − a)(0 −2)   b



(t − a )(0 −1)  b  G (b, 0  q ( s )) max{ f (b − a )(0 − 2)  a

 ( s − a )(0 −1)  s, c  (b − a )(0 − 2) 

  , 0}d q , s  

b 1 + (b − 0  q ( s ))(0 − 2) min{ f  a  q ( )

 ( s − a )(0 −1)  s, c  (b − a )(0 − 2) 

  , 0}d q , s    



(t − a)(0 −1) (b − a)(0 −2)

c = u0 (t )

and b

Tv0 (t ) =  G (t , 0  q ( s )) f ( s, v0 ( s ))d q , s a

 ( s − a)(0 −1)  b =  G(t , 0  q ( s)) max{ f  s, d , 0}d q , s ( − 2)  a   ( b − a )  0    ( s − a)(0 −1)  +  G(t , 0  q ( s)) min{ f  s, d  , 0}d q , s a  (b − a)(0 −2)   (t − a)(0 −1)  d = v0 (t ). (b − a)(0 −2) b

Hence, we get Tu0  u0 , Tv0  v0 . We construct the following sequences: b

un +1 =  G (t , 0  q ( s )) f ( s, un ( s ))d q , s, a

n = 0,1, 2,

b

vn +1 =  G (t , 0  q ( s)) f ( s, vn ( s))d q , s, a

. According to the monotonicity of T, we get

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un +1  un , vn +1  vn ,

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n = 0,1, 2,

.

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Positive and negative solutions of a boundary value…

By using Lemma 2.7, we know that the operator T has two positive solutions u , v  C[a, b] with *

u0  u*  v*  v0 ,

that is, a  c

(t − a)(0 −1)

 u* (t )  v* (t )  d

(b − a)(0 −2)

(t − a)(0 −1) (b − a)(0 −2)

*

 d,

a  t  b.

* * In addition, lim un = u , lim vn = v . n →

n →

By using the same proof as Theorem 3.5, we can easily obtain the following conclusion.

Theorem 3.6 Assume that

( F5 ) there exist a real number c  0 and g  L1[a, b], such that f :[a, b]  [c, a] →

(i5 )

f (t , u )  g (t )

is continuous,

for

(t , u )  [a, b]  [c, a]

and

f (t , u )  f (t , v) for t  [a, b], c  u  v  a.

In addition, there exist d  (c, a ) such that

( F3 ) − (i4 )and ( F4 ) in Theorem 3.5 are also satisfied. Then

the problem (1.1) has two negative solutions u , v  D, where *

D = {u  C[a, b] c

Let

u0 (t ) = c

(t − a)(0 −1) (b − a)(0 − 2)

,

(t − a)(0 −1) (b − a)(0 −2) v0 (t ) = d

*

 u (t )  d

(t − a)(0 −1) (b − a)(0 − 2)

(b − a)(0 −2)

, t  [a, b]}

and we construct the following sequences: b

vn +1 =  G (t , 0  q ( s)) f ( s, vn ( s))d q , s,

b

un +1 =  G (t , 0  q ( s )) f ( s, un ( s ))d q , s,

a

a

n = 0,1, 2,

(t − a)(0 −1)

un = u* , lim vn = v* . , we can obtain lim n → n →

REFERENCES [1]

Hahn,W: Über Orthogonalpolynome, die q-Differenzenlgleichungen genügen. Math. Nachr. 2, 434(1949)

[2]

Malinowska, AB, Torres, DFM: The Hahn quantum variational calculus. J. Optim. Theory Appl. 147, 419-442(2010)

[3]

Malinowska, AB, Torres, DFM: Quantum Variational Calculus. Springer Berlin (2014)

[4]

Malinowska, AB, Martins, N: Generalized transversality conditions for the Hahn quantum variational calculus. Optimization 62(3),323-344(2013)

[5]

Hamza, AE, Ahmed, SM: Theory of linear Hahn difference equations. J. Adv. Math. 4(2), 441-461(2013)

[6]

Hamza, AE, Ahmed, SM: Existence and uniqueness of solutions of Hahn difference equations. Adv.

[7]

Differ. Equ. 2013, 316(2013) Yizhu Wang, Yiding Liu, Chengmin Hou: New concepts of fractional Hahn’s q,  -derivative of Riemann-Liouville type and Caputo type and applications. Adv. Differ. Equ. 292(2018)

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Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones. Academic Press, New York(1988)

Yizhu Wang , Positive and negative solutions of a boundary value problem for a fractional difference equation. Invention Journal of Research Technology in Engineering & Management (IJRTEM),2(9), 73-82. Retrieved September 20, 2018, from www.ijrtem.com. |Volume 2| Issue 9 |

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