Int. J. Appl. Comput. Math (2018) 4:102 https://doi.org/10.1007/s40819-018-0534-6 ORIGINAL PAPER

Existence Theory to a Coupled System of Higher Order Fractional Hybrid Differential Equations by Topological Degree Theory Mian Bahadur Zada1 · Kamal Shah1

· Rahmat Ali Khan1

© Springer (India) Private Ltd., part of Springer Nature 2018

Abstract In this manuscript, we are concerned with the existence of solutions to a coupled system of higher order fractional hybrid differential equations (FHDEs). By using the concept of topological degree theory, we establish proper conditions under which the considered coupled system of FHDEs has at least one solution. The respective method is very rarely used to handle nonlinear FHDEs. Keywords Coupled systems · Boundary value problems · Fractional differential equations · Hybrid fixed point theorems · Topological degree theory Mathematics Subject Classification 26A33 · 34A08 · 35R11

Introduction Fractional calculus have been found by the mathematicians to be more effective as well as more practical than the corresponding ones of classical calculus in the mathematical modeling of many phenomena. Fractional differential equations are very important and significant part of the mathematics and has various applications into visco-elasticity, electro-analytical chemistry and many physical problems [1–6]. There are several papers which deal with the solvability of nonlinear boundary value problems of fractional differential equations, by using the techniques of nonlinear analysis (fixed-point theorems, Leray–Schauder theory, etc.), see, for instance, [7–9]. The use of coincidence degree theory come very close to examine the existence solutions of fractional differential equations. This approach is new

B

Kamal Shah [email protected] Mian Bahadur Zada [email protected] Rahmat Ali Khan [email protected]

1

Department of Mathematics, University of Malakand, Chakdara Dir(L), Khyber Pakhtunkhwa, Pakistan

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and hardly a couple of results can be found in the literature that deal with nonlinear boundary value problems [10–14]. On the other hand the study of coupled systems of fractional differential equations is important because such systems occur in several problems of applied nature; for instance [15–18]. Bashiri et al. [19] established fixed point theorem and discussed the existence of solution to the following system of fractional hybrid differential equations of order p ∈ (0, 1): ⎧ p α ⎪ ⎨ D [x(t) − f (t, x(t))] = g(t, y(t), I y(t)), a.e. t ∈ (0, T ], T > 0, p D [y(t) − f (t, y(t))] = g(t, x(t), I α x(t)), a.e. t ∈ (0, T ], T > 0, (1) ⎪ ⎩ x(0) = 0, y(0) = 0, where α > 0, and the functions f : (0, T ]×R → R, f (0, 0) = 0 and g : (0, T ]×R×R → R satisfy certain conditions. D p is the Riemann–Liouville fractional derivative. It is important to describe that Caputo fractional derivatives play vital role in applied problems as it provides known physical interpretation for initial and boundary conditions, while the Riemann–Liouville derivatives do not provide physical interpretations in most of the cases for initial and boundary conditions. The existence solutions for fractional differential equations involving the Caputo fractional derivative were studied in [20–22]. For recent advancement in the said derivative and its application, we refer [23–28]. Using topological degree theory, we are concerned with the existence of solutions to threepoint boundary value problem for a coupled system of fractional hybrid differential equations of order p ∈ [n − 1, n) given by ⎧ p D [x(t) − f (t, x(t))] = g(t, y(t), I α y(t)), a.e. t ∈ [0, 1], ⎪ ⎪ ⎪ ⎪ ⎨ D p [y(t) − f (t, y(t))] = g(t, x(t), I α x(t)), a.e. t ∈ [0, 1], (2) ⎪ x(0) = φ1 (x(η1 )) , y(0) = φ1 (y(η1 )) , x(1) = φ2 (x(η2 )) , y(1) = φ2 (y(η2 )) , ⎪ ⎪ ⎪ ⎩ (i) x (0) = y (i) (0) = 0, for i = 1, 2, 3, . . . (n − 2). where α > 0, 0 < η1 < η2 < 1, f : [0, 1] × R → [0, 1], f (0, x(0)) = 0 and g : [0, 1] × R × R → [0, 1], φ1 , φ2 ∈ C([0, 1], [0, 1]) are non-local linear functions such that φ1 (1)  = 1, φ2 (1)  = 1. Also D p is the Caputo fractional derivative of order p.

Preliminaries Let C([0, 1]×R, [0, 1]) denotes the space of all continuous functions f : [0, 1]×R → [0, 1] and let C([0, 1] × R × R, [0, 1]) denote the class of functions g : [0, 1] × R × R → [0, 1] such that • the map t → g(t, x, y) is measurable for each x, y ∈ [0, 1], • the map x → g(t, x, y) is continuous for each x ∈ [0, 1], • the map y → g(t, x, y) is continuous for each y ∈ [0, 1]. Let X be a Banach space and B ⊂ P(X ) denotes the family of all bounded subsets of X . The following is a discussion of some notions from [29] we will need. Definition 2.1 The Kuratowski measure of noncompactness α : B → R+ is defined as α(B) = inf{d > 0, where B ∈ B admits a finite cover by sets of diameter ≤ d}. Definition 2.2 Let F :  → X be continuous bounded map, where  ⊆ X . Then F is

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(1) α-Lipschitz if there exists k ≥ 0 such that α (F(S)) ≤ kα(S), for all bounded S ⊆ ; (2) strict α-contraction if there exists 0 ≤ k < 1 such that α (F(S)) ≤ kα(S), for all bounded S ⊆ ; (3) α-condensing if α (F(S)) < α(S), for all bounded S ⊆  with α(S) > 0. In other words, α (F(S)) ≥ α(S) implies α(S) = 0. Denote the class of all strict α-contractions F :  → X by Cα () and the class of all α-condensing maps F :  → X by Cα (). Then Cα () ⊂ Cα () and every F ∈ Cα () is α-Lipschitz with constant k = 1. Moreover, F :  → X is Lipschitz if there exists k > 0 such that F(x) − F(y) ≤ k|x − y|, for all x, y ∈  and that if k < 1, then F is a strict contraction. Proposition 2.1 If F, G :  → X is Lipschitz with constant k1 and k2 respectively, then F + G is α-Lipschitz with constant k1 + k2 . Proposition 2.2 If F :  → X is Lipschitz with constant k, then F is α-Lipschitz with the same constant k. Proposition 2.3 If F :  → X is compact, then F is α-Lipschitz with constant k = 0. Isaia [30] present the following results using topological degree theory. Theorem 2.1 Let F : X → X be α-condensing and  = {x ∈ X : there exists 0 ≤ λ ≤ 1 such that x = λF x}. If  is a bounded set in X , so there exists r > 0 such that  ⊂ Br (0), then D(I − λF, Br (0), 0) = 1, for all λ ∈ [0, 1]. Consequently, F has at least one fixed point and the set of the fixed points of F lies in Br (0). We need the following definitions which can be found in [31]. Definition 2.3 The Riemann–Liouvillle fractional integral I of order α > 0 of function f ∈ L 1 (R + ) is defined as  t 1 (t − s)α−1 f (s)ds, (3) I α f (t) = (α) 0 provided that the right side is pointwise defined on (0, ∞). Definition 2.4 Let α be a positive real number, such that m −1 ≤ α < m, m ∈ N and f (m) (x) exists a function of class C. Then Caputo fractional derivative of f is defined as  t 1 D α f (t) = (t − s)m−α−1 f (m) (s)ds, (4) (m − α) 0 provided that the right side is pointwise defined on (0, ∞), where m = [α] + 1 and [α] represents the integer part of α. Definition 2.5 The Riemann–Liouvillle fractional integral operator of order α > 0 of continuous function f : R+ → R, f ∈ L 1 (R + ) is defined as  t 1 D α f (t) = (t − s)n−α−1 f (s)ds. (5) (n − α) 0

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Lemma 2.1 The following result holds for fractional differential equations I α [c D α h(t)] = h(t) + c0 + c1 t + c2 t 2 + · · · + cn−1 t n−1 , for arbitrary ci ∈ R, i = 0, 1, 2, . . . , n − 1.

Main Results The space X = C([0, 1], R) of all continuous functions is a Banach space under the topological norm x = sup{x(t) : t ∈ [0, 1]} and the product space X × X is a Banach space under the norms (x, y) = x + y and (x, y) = max{ x , y }. To obtain our existence result, we assume that the mappings f : [0, 1] × R → [0, 1] and g : [0, 1] × R × R → [0, 1] satisfy the following conditions: (C1 ) for every (t, x), (t, y) ∈ [0, 1] × R, there exists λ1 , λ2 ∈ [0, 1) such that | f (t, x) − f (t, y)| ≤ λ1 |x − y| , | f (t, x) − f (t, y)| ≤ λ2 |x − y| ; (C2 ) for every (t, x, y) ∈ R, there exists positive constants l g1 , l g2 and Mg such that |g(t, x, y)| ≤ l g1 x q1 + l g2 y q1 + Mg (C3 ) for every (t, x) ∈ [0, 1] × R, there exists positive constants l f , M f such that | f (t, x(t)) | ≤ l f x q2 + M f ; (C4 ) for every u, v ∈ [0, 1], there exists positive constants lφ1 , lφ2 , Mφ1 and Mφ2 such that |φ1 (u) | ≤ lφ1 u + Mφ1 , |φ2 (v) | ≤ lφ2 v + Mφ2 ; (C5 ) for i = 1, 2 and for every u, v ∈ [0, 1], there exists σ ∈ [0, 1) such that φi (t) ≤ σ t. For brevity, let us set = (1 − φ1 (1))(1 − φ2 (η2n−1 )) + φ1 (η1n−1 )(1 − φ2 (1)).  i  = 0 for i = 1, 2, . . . , (n − 2). Then Lemma 3.1 If f (0, x(0)) = 0 and ∂ f (t,x(t)) t=0 ∂t i integral representation of the system (2) is given by x(t) = f (t, x(t)) 

1 1 − φ1 η2n−1 φ1 ( f (η1 , x(η1 ))) +





+ φ1 η1n−1 φ2 ( f (η2 , x(η2 ))) − φ1 η1n−1 f (1, x(1))  1 + (1 − φ1 (1)) φ2 ( f (η2 , x(η2 )))

 − (1 − φ2 (1)) φ1 ( f (η1 , x(η1 ))) − (1 − φ1 (1)) f (1, x(1)) t n−1

+

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1 ( p)



t 0



(t − s) p−1 g s, y(s), I α y(s) ds

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1 1 − φ1 η2n−1 − (1 − φ2 (1)) t n−1

( p)   η1

(η1 − s) p−1 g s, y(s), I α y(s) ds φ1 0  

1 φ1 η1n−1 + (1 − φ1 (1)) t n−1 +

( p)   η2

(η2 − s) p−1 g s, y(s), I α y(s) ds φ2

+

0

 1 



1 − φ1 η1n−1 + (1 − φ1 (1)) t n−1 (1 − s) p−1 g s, y(s), I α y(s) ds, (6)

( p) 0

and y(t) = f (t, y(t)) 

1 1 − φ1 η2n−1 φ1 ( f (η1 , y(η1 ))) +





n−1 n−1 φ2 ( f (η2 , y(η2 ))) − φ1 η1 f (1, y(1)) + φ 1 η1  1 + (1 − φ1 (1)) φ2 ( f (η2 , y(η2 )))

 − (1 − φ2 (1)) φ1 ( f (η1 , y(η1 ))) − (1 − φ1 (1)) f (1, y(1)) t n−1  t

1 + (t − s) p−1 g s, x(s), I α x(s) ds ( p) 0  

1 1 − φ1 η2n−1 − (1 − φ2 (1)) t n−1 +

( p)  η1 

(η1 − s) p−1 g s, x(s), I α x(s) ds φ1 0  

1 n−1 n−1 + + (1 − φ1 (1)) t φ 1 η1

( p)  η2 

(η2 − s) p−1 g s, x(s), I α x(s) ds φ2 0  

1 − φ1 η1n−1 + (1 − φ1 (1)) t n−1

( p)  1

(1 − s) p−1 g s, x(s), I α x(s) ds.

(7)

0

Proof Applying the operator I p on the first equation of the system (2) and using Lemma (2.1), we obtain x(t) = f (t, x(t)) +

1 ( p)



t



(t − s) p−1 g s, y(s), I α y(s) ds

0

+ C0 + C1 t + C2 t 2 + · · · + Cn−1 t n−1 .

(8)

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Apply the initial conditions x (i) (0) = 0, for i = 1, 2, 3, . . . , (n − 2), we conclude that C1 = C2 = · · · = Cn−2 = 0 and thus Eq. (8) becomes  t

1 (t − s) p−1 g s, y(s), I α y(s) ds + C0 + Cn−1 t n−1 . (9) x(t) = f (t, x(t)) + ( p) 0 Now, to find the values of C0 and Cn+1 . Since x(0) = φ1 (x(η1 )), then we get from equation (9)    η1

1 C0 = φ1 f (η1 , x(η1 )) + (η1 − s) p−1 g s, y(s), I α y(s) ds + C0 + Cn−1 η1n−1 ( p) 0    η1

1 (η1 − s) p−1 g s, y(s), I α y(s) ds = φ1 f (η1 , x(η1 )) + ( p) 0

+ C0 φ1 (1) + Cn−1 φ1 η1n−1 ,

which implies that

(1 − φ1 (1)) C0 − φ1 η1n−1 Cn−1    η1

1 = φ1 f (η1 , x(η1 )) + (η1 − s) p−1 g s, y(s), I α y(s) ds . ( p) 0

(10)

Also, since x(1) = φ2 (x(η2 )), then from equation (9), we have  1

1 (1 − s) p−1 g s, y(s), I α y(s) ds + C0 + Cn−1 f (1, x(1)) + ( p) 0    η2

1 = φ2 f (η2 , x(η2 )) + (η2 − s) p−1 g s, y(s), I α y(s) ds ( p) 0

+ C0 φ2 (1) + Cn−1 φ2 η2n−1 , implies that

(1 − φ2 (1)) C0 + 1 − φ2 η2n−1 Cn−1    η2

1 = φ2 f (η2 , x(η2 )) + (η2 − s) p−1 g s, y(s), I α y(s) ds ( p) 0  1

1 − f (1, x(1)) − (1 − s) p−1 g s, y(s), I α y(s) ds. ( p) 0

(11)

Solving (10) and (11) for C0 , we get  



(1 − φ1 (1)) (1 − φ2 η2n−1 ) + φ1 η1n−1 (1 − φ2 (1)) C0   η1



1 n−1 p−1 α φ1 f (η1 , x(η1 )) + = 1 − φ 2 η2 (η1 − s) g s, y(s), I y(s) ds ( p) 0   

η2

1 + φ1 η1n−1 φ2 f (η2 , x(η2 )) + (η2 − s) p−1 g s, y(s), I α y(s) ds ( p) 0   1



1 − φ1 η1n−1 f (1, x(1)) + (1 − s) p−1 g s, y(s), I α y(s) ds , ( p) 0

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implies that C0 =

   η1



1 1 (η1 − s) p−1 g s, y(s), I α y(s) ds 1 − φ2 η2n−1 φ1 f (η1 , x(η1 )) +

( p) 0   η2



1 n−1 + φ1 η1 (η2 − s) p−1 g s, y(s), I α y(s) ds φ2 f (η2 , x(η2 )) + ( p) 0   1



1 − φ1 η1n−1 f (1, x(1)) + (1 − s) p−1 g s, y(s), I α y(s) ds . ( p) 0

Similarly, solving (10) and (11) for Cn−1 , we get Cn−1 =

    η2

1 1 (η2 − s) p−1 g s, y(s), I α y(s) ds (1 − φ1 (1)) φ2 f (η2 , x(η2 )) +

( p) 0    η1

1 − (1 − φ2 (1)) φ1 f (η1 , x(η1 )) + (η1 − s) p−1 g s, y(s), I α y(s) ds ( p) 0    1

1 (1 − s) p−1 g s, y(s), I α y(s) ds . − (1 − φ1 (1)) f (1, x(1)) + ( p) 0

Substituting the values of C0 and C1 in Eq. (9), we can write

 t

1 x(t) = f (t, x(t)) + (t − s) p−1 g s, y(s), I α y(s) ds ( p) 0    η1



1 1 1 − φ2 η2n−1 φ1 f (η1 , x(η1 )) + + (η1 − s) p−1 g s, y(s), I α y(s) ds

( p) 0   η2



1 + φ1 η1n−1 φ2 f (η2 , x(η2 )) + (η2 − s) p−1 g s, y(s), I α y(s) ds ( p) 0   1



1 − φ1 η1n−1 (1 − s) p−1 g s, y(s), I α y(s) ds f (1, x(1)) + ( p) 0     η2

1 1 + (η2 − s) p−1 g s, y(s), I α y(s) ds (1 − φ1 (1)) φ2 f (η2 , x(η2 )) +

( p) 0    η1

1 − (1 − φ2 (1)) φ1 f (η1 , x(η1 )) + (η1 − s) p−1 g s, y(s), I α y(s) ds ( p) 0    1

1 (1 − s) p−1 g s, y(s), I α y(s) ds t n−1 . − (1 − φ1 (1)) f (1, x(1)) + ( p) 0

Finally, we have x(t) = f (t, x(t)) 

1 1 − φ2 η2n−1 φ1 ( f (η1 , x(η1 ))) +





n−1 n−1 φ2 ( f (η2 , x(η2 ))) − φ1 η1 f (1, x(1)) + φ 1 η1  1 + (1 − φ1 (1)) φ2 ( f (η2 , x(η2 )))

 − (1 − φ2 (1)) φ1 ( f (η1 , x(η1 ))) − (1 − φ1 (1)) f (1, x(1)) t n−1  t

1 + (t − s) p−1 g s, y(s), I α y(s) ds ( p) 0  

1 + 1 − φ2 η2n−1 − (1 − φ2 (1)) t n−1

( p)

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φ1 (η1 − s) g s, y(s), I y(s) ds 0  

1 + φ1 η1n−1 + (1 − φ1 (1)) t n−1

( p)   η2

φ2 (η2 − s) p−1 g s, y(s), I α y(s) ds η1

p−1



α

0

  1



1 − φ1 η1n−1 + (1 − φ1 (1)) t n−1 (1 − s) p−1 g s, y(s), I α y(s) ds.

( p) 0 Similarly, following the same steps in the above process with the second equation of the system (2), we obtain Eq. (7).  Define operators F, G, T : X × X → X × X by F (x, y) (t) = (F1 x(t), F2 y(t)) , G (x, y) (t) = (G 1 x(t), G 2 y(t)) , T (x, y) (t) = F (x, y) (t) + G (x, y) (t), where F1 , F2 , G 1 , G 2 , : X → X are F1 x(t) = f (t, x(t)) 

1 1 − φ2 η2n−1 φ1 ( f (η1 , x(η1 ))) +





+ φ1 η1n−1 φ2 ( f (η2 , x(η2 ))) − φ1 η1n−1 f (1, x(1))  1 + (1 − φ1 (1)) φ2 ( f (η2 , x(η2 )))

 − (1 − φ2 (1)) φ1 ( f (η1 , x(η1 ))) − (1 − φ1 (1)) f (1, x(1)) t n−1 ,

F2 y(t) = f (t, y(t)) 

1 + 1 − φ2 η2n−1 φ1 ( f (η1 , y(η1 )))





+ φ1 η1n−1 φ2 ( f (η2 , y(η2 ))) − φ1 η1n−1 f (1, y(1))  1 + (1 − φ1 (1)) φ2 ( f (η2 , y(η2 )))

 − (1 − φ2 (1)) φ1 ( f (η1 , y(η1 ))) − (1 − φ1 (1)) f (1, y(1)) t n−1 ,

G 1 x(t) =

123

 t

1 (t − s) p−1 g s, x(s), I α x(s) ds ( p) 0  

1 + 1 − φ2 η2n−1 − (1 − φ2 (1)) t n−1

( p)  η  1

φ1 (η1 − s) p−1 g s, x(s), I α x(s) ds 0  

1 φ1 η1n−1 + (1 − φ1 (1)) t n−1 +

( p)  η  2

(η2 − s) p−1 g s, x(s), I α x(s) ds φ2 0  

1 φ1 η1n−1 + (1 − φ1 (1)) t n−1 −

( p)

Int. J. Appl. Comput. Math (2018) 4:102  1 0

G 2 y(t) =

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(1 − s) p−1 g s, x(s), I α x(s) ds,

 t

1 (t − s) p−1 g s, y(s), I α y(s) ds ( p) 0  

1 + 1 − φ2 η2n−1 − (1 − φ2 (1)) t n−1

( p)  η  1

(η1 − s) p−1 g s, y(s), I α y(s) ds φ1 0  

1 φ1 η1n−1 + (1 − φ1 (1)) t n−1 +

( p)  η  2

(η2 − s) p−1 g s, y(s), I α y(s) ds φ2 0

  1



1 φ1 η1n−1 + (1 − φ1 (1)) t n−1 − (1 − s) p−1 g s, y(s), I α y(s) ds.

( p) 0

Then the integral equations in Lemma 3.1 can be written as an operator equation (x, y) = T (x, y) = F(x, y) + G(x, y), and fixed points of the operator equation are solutions of the integral equations in Lemma 3.1. Theorem 3.1 The operator F : X × X → X × X is Lipschitz with constant λ. Consequently F is α−Lipschitz with the same constant λ and satisfies the following growth condition F(x1 , x2 ) ≤ C F (x1 , x2 ) q2 + M F , Proof To show that the operator F is Lipschitz with constant λ. Let x 1 , x2 ∈ X , then we have |F1 x1 (t) − F1 x2 (t)|   =  f (t, x1 (t)) − f (t, x2 (t)) 

1 + 1 − φ2 η2n−1 φ1 ( f (η1 , x1 (η1 )))





+ φ1 η1n−1 φ2 ( f (η2 , x1 (η2 ))) − φ1 η1n−1 f (1, x1 (1)) +

 1 (1 − φ1 (1)) φ2 ( f (η2 , x1 (η2 )))

 − (1 − φ2 (1)) φ1 ( f (η1 , x1 (η1 ))) − (1 − φ1 (1)) f (1, x1 (1)) t n−1  





1 1 − φ2 η2n−1 φ1 ( f (η1 , x2 (η1 ))) + φ1 η1n−1 φ2 ( f (η2 , x2 (η2 ))) − φ1 η1n−1 f (1, x2 (1))

 1 − (1 − φ1 (1)) φ2 ( f (η2 , x2 (η2 ))) − (1 − φ2 (1)) φ1 ( f (η1 , x2 (η1 )))

   − (1 − φ1 (1)) f (1, x2 (1)) t n−1  +



   1 − φ ηn−1  2   2   |φ1 ( f (η1 , x1 (η1 ))) − φ1 ( f (η1 , x2 (η1 )))| ≤ | f (t, x1 (t)) − f (t, x2 (t))| +  

 

   φ ηn−1   1 1   |φ2 ( f (η2 , x1 (η2 ))) − φ2 ( f (η2 , x2 (η2 )))| +  

 

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+ + + +

Int. J. Appl. Comput. Math (2018) 4:102

   φ ηn−1   1 1    | f (1, x1 (1)) − f (1, x2 (1))|  

     (1 − φ1 (1))    |φ2 ( f (η2 , x1 (η2 ))) − φ2 ( f (η2 , x2 (η2 )))|  

   1 − φ2 (1)    |φ1 ( f (η1 , x1 (η1 ))) − φ1 ( f (η1 , x2 (η1 )))|  

   1 − φ1 (1)   | f (1, x1 (1)) − f (1, x1 (1))| .   

Using condition (C5 ), we get |F1 x1 (t) − F1 x2 (t)| ≤ | f (t, x1 (t)) − f (t, x2 (t))|

   1 − φ ηn−1  2   2  |σ1 ( f (η1 , x1 (η1 )) − f (η1 , x2 (η1 )))| +  

 

   φ ηn−1   1 1   |σ2 ( f (η2 , x1 (η2 )) − f (η2 , x2 (η2 )))| +  

 

   φ ηn−1   1 1   | f (1, x1 (1)) − f (1, x2 (1))| +  

     (1 − φ1 (1))   |σ2 ( f (η2 , x1 (η2 )) − f (η2 , x2 (η2 )))| +  

   1 − φ2 (1)   |σ1 ( f (η1 , x1 (η1 )) − f (η1 , x2 (η1 )))| +  

   1 − φ1 (1)   | f (1, x1 (1)) − f (1, x1 (1))| +  

 2 | f (η1 , x1 (η1 )) − f (η1 , x2 (η1 ))| ≤ | f (t, x1 (t)) − f (t, x2 (t))| + | |

 + | f (η2 , x1 (η2 )) − f (η2 , x2 (η2 ))| + | f (1, x1 (1)) − f (1, x2 (1))| .

Using condition (C1 ), we can write

  2 F1 x1 (t) − F1 x2 (t) ≤ λ1 x1 − x2 + λ1 x1 − x2 + λ1 x1 − x2 + λ1 x1 − x2 | |     | | + 3 (| | + 3)λ1 x1 − x2 ≤ x1 − x2 ≤ | | | | ≤ λ x1 − x2 ,

where λ =

| |+3 | | .

Similarly     | | + 3 (| | + 3)λ2 F1 y1 (t) − F1 y2 (t) ≤ y1 − y2 ≤ y1 − y2 ≤ λ y1 − y2 . | | | | Now, F(x1 , y1 ) − F(x2 , y2 ) = max { F1 x1 (t) − F1 x2 (t) , F2 y1 (t) − F2 y2 (t) } ≤ max {λ x1 − x2 , λ y1 − y2 } = λ max { x1 − x2 , y1 − y2 } = λ (x1 , y1 ) − (x2 , y2 ) .

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By Proposition 2.2, F is α-Lipschitz with constant λ. Moreover, for growth condition, we have |F1 x1 (t)| ≤ | f (t, x1 (t))| 



1  n−1 n−1 1 − φ f η φ η φ2 ( f (η2 , x1 (η2 ))) + , x (η ))) + φ ( (η 2 1 1 1 1 1 2 1 | |  

 − φ1 η1n−1 ( f (1, x1 (1)))   1  + (1 − φ1 (1)) φ2 ( f (η2 , x1 (η2 ))) − (1 − φ2 (1)) φ1 ( f (η1 , x1 (η1 ))) | |    − (1 − φ1 (1)) ( f (1, x1 (1))) |t n−1 |

 ⎞ ⎛   1 − φ ηn−1   2   1 − φ2 (1)   2 + ⎠ |φ1 ( f (η1 , x1 (η1 )))| ≤ | f (t, x1 (t))| + ⎝   



 

 ⎛ ⎞  φ ηn−1      (1 − φ1 (1))   1 1 + ⎠ |φ2 ( f (η2 , x1 (η2 )))| + ⎝   



 

 ⎞ ⎛   φ ηn−1     1 − φ1 (1)   1 1 + ⎠ | f (1, x1 (1))| + ⎝   



 

 ⎞ ⎛  1 − φ ηn−1    2    1 − φ2 (1)  2 + ⎠ |σ1 f (η1 , x1 (η1 ))| ≤ | f (t, x1 (t))| + ⎝   



 

 ⎞ ⎛   φ ηn−1     (1 − φ1 (1))   1 1 + ⎠ |σ2 f (η2 , x1 (η2 ))| + ⎝   



 

 ⎞ ⎛   φ ηn−1     1 − φ1 (1)   1 1 + ⎠ | f (1, x1 (1))| + ⎝   



    2 | f (η1 , x1 (η1 ))| + | f (η2 , x1 (η2 ))| + | f (1, x1 (1))| , ≤ | f (t, x1 (t))| + | | with the help of condition (C3 ) it follows that   2 q2 q2 q2 q2 |F1 x1 (t)| ≤ l f x1 + M f + l f x1 + M f + l f x1 + M f + l f x1 + M f | | M f (| | + 6) l f (| | + 6) x1 q2 + ≤ | | | |

≤ C F x1 q2 + M F , where C F =

l f (| |+6) | |

and M F =

M f (| |+6) . | |

Similarly

|F2 x2 (t)| ≤ C F x2 q2 + M F .

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Hence it follows that F(x1 , x2 ) = (F1 (x1 ), F2 (x2 )) = F1 (x1 ) + F2 (x2 ) ≤ C F (x1 , x2 ) q2 + M F ,

where M F = 2M F .



Theorem 3.2 The operator G : X × X → X × X is continuous and satisfies the following growth condition under the assumption (C5 ): G(x(t), y(t)) ≤ C g (x, y) q1 + Mg .

(12)

Proof Consider a bounded subset of X × X as: B = {(x, y) ∈ X × X : (x, y) ≤ L} ⊆ X × X . Let {(xn , yn )} be a sequence in B such that (xn , yn ) → (x, y) ∈ X × X as n → ∞. Then we show that G is continuous. For this, consider |G 1 (xn ) − G 1 (x)| ≤

1 ( p)



t





 (t − s) p−1 g s, xn (s), I α xn (s) ds − g s, x(s), I α x(s)  ds

0

 

1   +  1 − φ2 η2n−1 − (1 − φ2 (1)) t n−1  | | ( p)   η  1 

p−1 α  φ1 (η − s) g s, x (s), I x (s) ds 1 n n  0  η1  

(η1 − s) p−1 g s, x(s), I α x(s) ds  − φ1 0

 

1   + φ1 η1n−1 + (1 − φ1 (1)) t n−1  | | ( p)   η2  

p−1 α  φ2 (η − s) g s, x (s), I x (s) ds 2 n n  0  η2  

− φ2 (η2 − s) p−1 g s, x(s), I α x(s) ds  0

 

1   + φ1 η1n−1 + (1 − φ1 (1)) t n−1  | | ( p)  1 



 (1 − s) p−1 g s, xn (s), I α xn (s) ds − g s, x(s), I α x(s)  ds 0







 (t − s) p−1 g s, xn (s), I α xn (s) ds − g s, x(s), I α x(s)  ds 0   η1 

2 (η1 − s) p−1 g s, xn (s), I α xn (s) ds + σ1  | | ( p) 0   η1

 p−1 − (η1 − s) g s, x(s), I α x(s) ds  0   η2 

2 + (η2 − s) p−1 g s, xn (s), I α xn (s) ds σ2  | | ( p) 0   η2

 − (η2 − s) p−1 g s, x(s), I α x(s) ds  0  1 



 2 (1 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds, + | | ( p) 0

1 ≤ ( p)

123

t

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102

implies that |G 1 (xn ) − G 1 (x)| ≤







 (t − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds 0  η1 



 2 + (η1 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds | | ( p) 0  η2 



 2 + (η2 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds | | ( p) 0  1 



 2 + (1 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds. | | ( p) 0 1 ( p)

t

Now, from the continuity of g, it follows that



g s, xn (s), I α xn (s) ds → g s, x(s), I α x(s) as n → ∞. In view of (C2 ), we obtain the following relations   



 (t − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ≤ (t − s) p−1 (l g1 + l g2 )L + M ,   



 (η1 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ≤ (η1 − s) p−1 (l g1 + l g2 )L + M ,   



 (η2 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ≤ (η2 − s) p−1 (l g1 + l g2 )L + M ,   



 (1 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ≤ (1 − s) p−1 (l g1 + l g2 )L + M , which implies that each term on the left is integrable. By Lebesgue Dominated convergent theorem, we obtain  t 



 (t − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds → 0, as n → ∞,  0 η1 



 (η1 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds → 0, as n → ∞,  0 η2 



 (η2 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds → 0, as n → ∞, 

0 1





 (1 − s) p−1 g s, xn (s), I α xn (s) − g s, x(s), I α x(s)  ds → 0, as n → ∞.

0

Thus G 1 (xn ) − G 1 (x) → 0, as n → ∞,

(13)

and following the same steps as above one can easily obtain that G 2 (yn ) − G 2 (y) → 0, as n → ∞.

(14)

From (13) and (14), we get G(xn , yn ) − G(x, y) → 0, as n → ∞.

(15)

This shows that G is continuous. Now, for growth condition on G, we have  

 1  t p−1 α |G 1 x(t)| ≤ (t − s) g s, x(s), I x(s) ds  ( p)  0  

  1  1 − φ2 ηn−1 − (1 − φ2 (1)) t n−1  + 2   | |( p)

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   φ1 

 

(η1 − s) g s, x(s), I x(s) ds  0  

  1 φ1 ηn−1 + (1 − φ1 (1)) t n−1  + 1   | |( p)   η  2  

p−1 α φ2  (η − s) g s, x(s), I x(s) ds 2   0  

  1 φ1 ηn−1 + (1 − φ1 (1)) t n−1  + 1  | |( p)   1  

  (1 − s) p−1 g s, x(s), I α x(s) ds    0

1 ≤ ( p)



t

η1

p−1



α



 (t − s) p−1 g s, x(s), I α x(s)  ds

0

 1 

 2 + (1 − s) p−1 g s, x(s), I α x(s)  ds | |( p) 0   η  1  

2 p−1 α φ1  (η − s) g s, x(s), I x(s) ds + 1   | |( p) 0   η  2  

2 φ2 + (η2 − s) p−1 g s, x(s), I α x(s) ds  , | |( p)  0

using conditions (C2 ) and (C5 ), we get

1 tp 1 l g x q1 + l g2 I α x q1 + Mg × ( p) p

2 1 1 l g x q1 + l g2 I α x q1 + Mg + × | |( p) p   η  1

 2σ p−1 α   (η − s) g s, x(s), I x(s) ds + 1  | |( p)  0   η 2 

 2σ p−1 α  + (η2 − s) g s, x(s), I x(s) ds  | |( p)  0

1 tp 1 ≤ × l g x q1 + l g2 I α x q1 + Mg ( p) p

2 1 1 l g x q1 + l g2 I α x q1 + Mg + × | |( p) p

2 η1 p 1 l g x q1 + l g2 I α x q1 + Mg + × | |( p) p

2 η2 p 1 l g x q1 + l g2 I α x q1 + Mg + × | |( p) p   2 2 2 1 + + + ≤ ( p + 1) | |( p + 1) | |( p + 1) | |( p + 1)

l g1 x q1 + l g2 I α x q1 + Mg ,

|G 1 x(t)| ≤

that is

 |G 1 x(t)| ≤

123



| | + 6 l g1 x q1 + l g2 I α x q1 + Mg . | |( p + 1)

(16)

Int. J. Appl. Comput. Math (2018) 4:102

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| | + 6 |G 2 y(t)| ≤ l g1 y q1 + l g2 I α y q1 + Mg . | |( p + 1)

(17)

Now, using (16) and (17), we have G(x(t), y(t)) = (G 1 x(t), G 2 y(t)) ≤ C g (x, y) q1 + Mg , where C g =

| |+6 | |( p+1) .

Which is the required growth condition.



Theorem 3.3 The operator G : X × X → X × X is compact. Consequently, G is λ−Lipschitz with zero constant. Proof Consider a bounded set  ⊂ B ⊆ X × X and a sequence {(xn , yn )} in . Then with the help of growth condition (12) we can writ G(xn (t), yn (t)) ≤ C g L q1 + Mg . which shows that G() is uniformly bounded. Now, for equi-continuity take 0 ≤ τ ≤ t ≤ 1, then  t

1 G 1 xn (t) − G 1 xn (τ ) = (t − s) p−1 g s, xn (s), I α xn (s) ds ( p) 0  

1 + 1 − φ2 η2n−1 − (1 − φ2 (1)) t n−1

( p)   η1

p−1 α φ1 (η1 − s) g s, xn (s), I xn (s) ds 0  

1 φ1 η1n−1 + (1 − φ1 (1)) t n−1 +

( p)   η2

φ2 (η2 − s) p−1 g s, xn (s), I α xn (s) ds 0  

1 n−1 n−1 − + (1 − φ1 (1)) t φ 1 η1

( p)  1

(1 − s) p−1 g s, xn (s), I α xn (s) ds 0  τ

1 − (τ − s) p−1 g s, xn (s), I α xn (s) ds ( p) 0  

1 n−1 n−1 1 − φ 2 η2 − (1 − φ2 (1)) τ −

( p)  η1 

φ1 (η1 − s) p−1 g s, xn (s), I α xn (s) ds 0  

1 − φ1 η1n−1 + (1 − φ1 (1)) τ n−1

( p)  η2 

p−1 α φ2 (η2 − s) g s, xn (s), I xn (s) ds 0  

1 φ1 η1n−1 + (1 − φ1 (1)) τ n−1 +

( p)

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1



(1 − s) p−1 g s, xn (s), I α xn (s) ds.

0

Which implies that

    G 1 xn (t) − G 1 xn (τ )    t

1  ≤ (t − s) p−1 g s, xn (s), I α xn (s) ds ( p)  0   τ

 (τ − s) p−1 g s, xn (s), I α xn (s) ds . − 0

From condition (C5 ) it follows that     G 1 xn (t) − G 1 xn (τ )   

1  t ≤ (t − s) p−1 g s, xn (s), I α xn (s) ds ( p)  0   t

 − (τ − s) p−1 g s, xn (s), I α xn (s) ds  0  

 1  τ p−1 α  + (τ − s) g s, x (s), I x (s) ds n n  ( p)  t  t

   1 ≤ (t − s) p−1 − (τ − s) p−1 g s, xn (s), I α xn (s)  ds ( p) 0  τ 

 1 (τ − s) p−1 g s, xn (s), I α xn (s)  ds + ( p) t  t   |g (s, xn (s), I α xn (s))| ≤ (t − s) p−1 − (τ − s) p−1 ds ( p) 0   τ (τ − s) p−1 ds . + t

Using condition (C2 ), we get G 1 xn (t) − G 1 xn (τ ) ≤

l g1 xn q1 + l g2 I α xn q1 + Mg ( p + 1)

  t p − τ p + (τ − t) p + (τ − t) p . (18)

Similarly G 2 yn (t) − G 2 yn (τ ) ≤

l g1 yn q1 + l g2 I α yn q1 + Mg ( p + 1)

  t p − τ p + (τ − t) p + (τ − t) p .

Taking limit as t → τ in equations (18) and (19), we get G 1 xn (t) − G 1 xn (τ ) → 0, and G 2 yn (t) − G 2 yn (τ ) → 0, which implies that G(xn , yn )(t) − G(xn , yn )(τ ) → 0.

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That is G(x, y) is equi-continuous. By Arzelá–Ascoli theorem G(x, y) is compact and hence by Proposition (2.3) G is λ−Lipschitz with zero constant.  Theorem 3.4 Assume that (C1 )–(C5 ) hold, then the coupled system (2) of fractional hybrid differential equations of higher order has at least one solution (x, y) ∈ X × X and the set of solutions is bounded in X × X . Proof By Theorem 3.1 F is Lipschitz with constant λ and by Theorem 3.3 G is Lipschitz with constant 0. Consequently, T is Lipschitz with constant λ. Now consider the following set S = {(x, y) ∈ X × X : ∃ κ ∈ [0, 1]  (x, y) = κ T (x, y)}. To show that S is bounded in X × X . For this, let (x, y) ∈ S, then using the growth conditions of Theorems 3.1 and 3.2, we have (x, y) = κ T (x, y) = κ (F (x, y) + G (x, y)) , which implies that (x, y) ≤ κ ( F (x, y) + G (x, y) )   ≤ κ C F (x, y) q2 + M F + C g (x, y) q1 + Mg , where q1 , q2 ∈ [0, 1). Thus S is bounded in X × X . Hence T has at least one fixed point, that is the coupled system (2) of fractional hybrid differential equations of higher order has at least one solution (x, y) ∈ X × X and the set of solutions is bounded in X × X . 

Conclusion and Discussion Due to the importance applications of hybrid differential equations which are increasingly used in modeling of various dynamical process, it is necessary that whether the mathematical model exists or not? This fundamental questions led us to the existence theory of the considered nonlinear boundary value problems of fractional differential equations . Here we have taken a random system of fractional hybrid differential equations for which, we have successfully developed qualitative theory for existence of solutions to a coupled system of fractional hybrid differential equations. Thanks to the coincidence degree theory of topological type we have been obtained the appropriate conditions for existence of at least one solutions to the proposed system. In future the established theory can be extended to more general system of nonlinear fractional hybrid differential equations. Acknowledgements We are really thankful to the reviewers for their useful suggestions which improved this paper very well.

Compliance with ethical standards Competing interest It is declared that no competing interest exists regarding this manuscript. Authors contribution All authors have equal contribution in this manuscript. Funding Information There is no funding source to support this manuscript financially.

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