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
(
|Volume3|Issue5|
)
www.ijrtem.com
(1.1)
| 40 |
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 ), tb − .
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:
|Volume3|Issue5|
www.ijrtem.com
| 41 |
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 tb − ,
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 tb−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 ) =
|Volume3|Issue5|
R \ {..., −2,−1,0}, 0, and (t − a ) : a + → R.
Then
( + 1) (t − a ) + , for t a+ + ; and ( + 1 + )
www.ijrtem.com
| 42 |
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) + c11−−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
|Volume3|Issue5|
www.ijrtem.com
| 43 |
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 − 20 .
Where
(b + − 2 ) −1 − ( ) D = max 1 + s0 ,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|
www.ijrtem.com
| 44 |
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
|Volume3|Issue5|
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 )
www.ijrtem.com
,
for
any
x E
,
| 45 |
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 s0 ,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
( (
(
|Volume3|Issue5|
))
)
www.ijrtem.com
(4.1)
| 46 |
Existence of positive solutions for…
1 , h = t 2 , for t 0.5,2.50.5 , 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 0.5,2.50.5 , 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.
REFERENCES [1]
Ibrahim, RW, Jalab, HA: Existence of a class of fractional difference equations with two-point boundary value problem[J]. Adv. Differ. Equ. 2015: 269.
[2]
Goodrich, CS: Systems of discrete fractional boundary value problems with nonlinear satisfying no growth conditions[J]. J. Differ. Equ. Appl. 2015:21(5), 437-453.
[3]
Xie, Z, Jin, Y, Hou, C: Multiple solutions for a fractional difference boundary value problem variational approach[J].Abstr. Appl. Anal. 2012:2012, 137.
[4]
G.Anastassiou, Nabla discrete fractional calculus and nabla inequalities[J], Math. Comput.Modelling 51(2010)562-571.
[5]
F.M. Atici and P.W. Eloe, Linear systems of fractional nabla difference equations[J], Rocky Mountain Math. 41(2011),pp.353-370
[6]
N.R.O. Bastos, R.A.C. Ferreira, and D.F.M. Torres, Discrete-time fractional variational problems[J], Sigal Process. 91(2011), pp. 513-524.
[7]
C.S. Goodrich, Existence of a positive solution to a system of discrete fractional boundary value problems[J], Appl. Math. Comput. 217(2011), pp. 191-202.
[8]
Malinowska, AB, Torres, DFM: The delta-nabla calculus of variations[J]. Fasc. Math.2009: 44(44), 75-83.
[9]
Liu Huan,Jin,Hou, C: Existence of positive solutions for discrete delta-nabla fractional boundary value problems with p-Laplacian[J]. Springer 2017: 13661-017-0765-1.
[10]
Xie, Z, Hou, C: Properties of right fractional sum and right fractional difference operators and application[J]. Adv. Differ.Equ. 2015: 288 [9].
[11]
Atici, F.M. Eloe, P.W.:Two-point boundary value problems for finite fractional difference equations[J]. Differ. Equ. Appl. 17(4), 445-456(2001).
|Volume3|Issue5|
www.ijrtem.com
| 47 |
Existence of positive solutions for‌ [12]
Han Zhenlai, Lu Hongling, Zhang Chao. Positive solutions for eigenvalue problems of fractional differential equation with generalized p-Laplacian[J]. Applied Mathematics and Computation, 2015, 257(2): 526-536.
[13]
Wang Jinhua, Xiang Hongjun. Upper and lower solutions method for a class of singular fractional boundary value problems with p-Laplacian[J]. Abstract and Applied Analysis, 2010, 2010(1085-3375): 331-336.
[14]
H.G. Schuster (Ed.),Reviews of Nonlinear Dynamics and Complexity, Wiley-VCH,2008.
[15]
Ding, XL, Jiang, YL: Waveform relaxation methods for fractional functional differential equations[J]. Fract. Calc. Appl.Anal. 16(3), 573-594 (2013).
|Volume3|Issue5|
www.ijrtem.com
| 48 |