Bull. Math. Soc. Sci. Math. Roumanie Tome 60 (108) No. 1, 2017, 3–18
On a coupled system of sequential fractional differential equations with variable coefficients and coupled integral boundary conditions by Bashir Ahmada , Ahmed Alsaedia , Shorog Aljoudia and Sotiris K. Ntouyasb,a Abstract In this paper, we investigate the existence and uniqueness of solutions for a coupled system of Caputo (Liouville-Caputo) type sequential fractional differential equations with variable coefficients supplemented with coupled nonlocal Riemann-Liouville integral boundary conditions. We make use of standard tools of the fixed-point theory to obtain the desired results. Our results are new and give more insight into the study of coupled systems of fractional differential equations with non-constant coefficients. Examples are included for the illustration of main results.
Key Words: Caputo derivative; coupled system; variable coefficients; RiemannLiouville integral boundary conditions; fixed point. 2010 Mathematics Subject Classification: Primary 34A08 Secondary 34A12, 34B15
1
Introduction
Fractional-order derivatives are found to be of great value in modelling many real world phenomena. In contrast to integer-order derivatives, these operators can trace the past history of the processes and materials involved in the phenomena. In particular, fractional Laplacian provides a paradigm of the vast family of nonlocal linear operators, and appears in the formulation of anomalous diffusion process (a diffusion process involving nonlinear relationship to time). Further details on anomalous diffusion phenomena can be found in a recent paper [1] and the text [2]. Due to the widespread applications of fractional calculus in biomedical and chemical processes, control theory, biomathematics, signal and image processing, wave propagation, etc., the mathematical community has shown a great interest in this subject, though its basics were known in the period of Riemann and Liouville. For more examples and explanations, we refer the reader to the texts [3]. The topic of boundary value problems (BVPs) of fractional differential equations supplemented with a variety of boundary conditions has attracted a significant attention in recent years. In particular, the literature on fractional order BVPs involving nonlocal and integral boundary conditions is now much enriched, for instance, see [4]-[9] and the references cited therein. Coupled systems of fractional-order differential equations have also been extensively studied due to their occurrence in several diverse disciplines, for example, nonlocal thermoelasticity [10], synchronization phenomena [11, 12], anomalous diffusion [13], etc. Some recent works on coupled systems of fractional-order differential equations equipped
4
B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
with different kinds of boundary conditions can be found in [14]-[18] and the references cited therein. In a recent work [18], the authors studied the following problem of a coupled system of sequential fractional differential equations and nonlocal uncoupled integral boundary conditions: c q ( D + k c Dq−1 )x(t) = f (t, x(t), y(t)), t ∈ [0, 1], 2 < q ≤ 3, k > 0, c p c p−1 )y(t) = g(t, x(t), y(t)), 2 < p ≤ 3, ( D +k D Rη β−1 x(s)ds, β > 0, 0 < η < ζ < 1, x(0) = 0, x0 (0) = 0, x(ζ) = a 0 (η−s) Γ(β) Rθ γ−1 y(0) = 0, y 0 (0) = 0, y(z) = b 0 (θ−s) Γ(γ) y(s)ds, γ > 0, 0 < θ < z < 1, where c D(.) denotes the Caputo fractional derivatives of order (.), f, g : [0, 1] × R2 → R are given continuous functions and a, b are real constants. To explore further in this direction, we replace the parameter k in the above system by different variable coefficients k1 (t) and k2 (t) in the equations of the system. We also allow the nonlinearities in the system to depend on the unknown functions together with their fractional derivatives. Furthermore, we consider nonlocal coupled integral boundary conditions in contrast to the uncoupled ones in the above system. Precisely, for 2 < p, q ≤ 3, 0 < α, δ < 1, we consider the following coupled system: (c Dq + k1 (t) c Dq−1 )x(t) c
p
c
( D + k2 (t) D
p−1
)y(t)
= =
f (t, x(t), y(t), c Dα y(t)), c
t ∈ [0, 1],
δ
g(t, x(t), D x(t), y(t)),
(1.1)
subject to coupled Riemann-Liouville type integral boundary conditions: x(0) y(0)
= =
0
Z
η
0, x (0) = 0, x(ζ) = a 0, y 0 (0) = 0, y(z) = b
0 Z θ 0
(η − s)β−1 y(s) ds , Γ(β)
β > 0,
(θ − s)γ−1 x(s) ds , Γ(γ)
γ > 0,
(1.2)
where c D(.) denotes the Caputo derivatives of fractional order (.) , f, g : [0, 1] × R3 → R are given continuous functions, k1 (t), k2 (t) are increasing functions with k1 (t), k2 (t) ∈ C([0, 1], R), 0 < η < ζ < 1, 0 < θ < z < 1, and a, b are real constants. In the rest of the paper, we organize the content as follows. We recall some basic facts of fractional calculus and establish an auxiliary lemma in Section 2. The main existence and uniqueness results, relying on contraction mapping principle and Leray-Schauder alternative, are presented in Section 3. Though the methods of proofs are the standard ones, yet their application in the context of problem (1.1)-(1.2) contribute further to the development of the subject of coupled systems of fractional differential equations with Riemann-Liouville integral boundary conditions. We also illustrate the existence and uniqueness result with the aid of an example.
On a coupled system of sequential fractional differential equations
2
5
Background material
This section is devoted to some fundamental concepts of fractional calculus [3] and a basic lemma related to the linear variant of the given problem. Definition 1. The fractional integral of order r with the lower limit zero for a function f is defined as Z t f (s) 1 r ds, t > 0, r > 0, I f (t) = Γ(r) 0 (t − s)1−r provided the right hand-side is point-wise defined on [0, ∞), where Γ(·) is the gamma funcR∞ tion, which is defined by Γ(r) = 0 tr−1 e−t dt. Definition 2. The Riemann-Liouville fractional derivative of order r > 0, n − 1 < r < n, n ∈ N , is defined as 1 = Γ(n − r)
r D0+ f (t)
d dt
n Z
t
(t − s)n−r−1 f (s)ds,
0
where the function f (t) has absolutely continuous derivative up to order (n − 1). Definition 3. The Caputo derivative of order r for a function f : [0, ∞) → R can be written as c
Dr f (t) = Dr
f (t) −
! t (k) f (0) , k!
n−1 X k k=0
t > 0,
n − 1 < r < n.
Remark 1. If f (t) ∈ C n [0, ∞), then c
1 D f (t) = Γ(n − r) r
Z 0
t
f (n) (s) ds = I n−r f (n) (t), t > 0, n − 1 < r < n. (t − s)r+1−n
To define the solution for problem (1.1)-(1.2), we consider the following lemma dealing with the linear variant of (1.1)-(1.2). Lemma 1. Let h1 , h2 ∈ C([0, 1], R) and x, y ∈ C 3 ([0, 1], R). Then the integral solution for the linear system of fractional differential equations: (c Dq + k1 (t) c Dq−1 )x(t) = h1 (t), (c Dp + k2 (t) c Dp−1 )y(t) = h2 (t),
(2.1)
6
B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
supplemented with the boundary conditions (1.2) is given by
x(t)
=
!(
t
" Z B2 a
η
(η − s)β−1 × Γ(β) 0 0 # ! Z Z s ζ −(µ2 (s)−µ2 (m)) p−1 −(µ1 (ζ)−µ1 (s)) q−1 e I h1 (s) ds e I h2 (m)dmds − × Z
1 ∆
se
−(µ1 (t)−µ1 (s))
ds
0
0
" Z −B1 b
θ 0
(θ − s) Γ(γ)
s
Z
! −(µ1 (s)−µ1 (m)) q−1
e
I
e−(µ2 (z)−µ2 (s)) I p−1 h2 (s)ds
−
=
t
Z +
0
y(t)
h1 (m)dm ds
0
#)
z
Z
γ−1
e−(µ1 (t)−µ1 (s)) I q−1 h1 (s)ds,
(2.2)
0
!(
t
" Z A1 b
θ
(θ − s)γ−1 × Γ(γ) 0 0 ! # Z z s −(µ1 (s)−µ1 (m)) q−1 −(µ2 (z)−µ2 (s)) p−1 e I h1 (m)dm ds − e I h2 (s)ds Z
1 ∆ Z
se
−(µ2 (t)−µ2 (s))
ds
0
0
" Z −A2 a
η 0
ζ
Z −
β−1
(η − s) Γ(β)
s
Z
! e−(µ2 (s)−µ2 (m)) I p−1 h2 (m)dmds
0
#) −(µ1 (ζ)−µ1 (s)) q−1
e
I
h1 (s) ds
0
Z
t
+
e−(µ2 (t)−µ2 (s)) I p−1 h2 (s)ds,
(2.3)
0
where
Z ∆ = A1 B2 − A2 B1 6= 0,
µi (t) =
t
ki (s)ds, i = 1, 2,
(2.4)
0 ζ
Z
s e−(µ1 (ζ)−µ1 (s)) ds, Z s Z η (η − s)β−1 B1 = −a u e−(µ2 (s)−µ2 (u)) du ds, Γ(β) 0 0 Z s Z θ γ−1 (θ − s) −(µ1 (s)−µ1 (u)) A2 = −b ue du ds, Γ(γ) 0 Z z 0 B2 = s e−(µ2 (z)−µ2 (s)) ds. A1 =
(2.5)
0
0
(2.6) (2.7) (2.8)
On a coupled system of sequential fractional differential equations
7
Proof: As argued in [19], the general solution of the system (2.1) can be written as
t
Z t e−(µ1 (t)−µ1 (s)) ds + b1 s e−(µ1 (t)−µ1 (s)) ds + b2 e−µ1 (t) 0 0 Z s Z t (s − τ )q−2 −(µ1 (t)−µ1 (s)) + e h1 (τ ) dτ ds, Γ(q − 1) 0 0 Z t Z t = c0 e−(µ2 (t)−µ2 (s)) ds + c1 s e−(µ2 (t)−µ2 (s)) ds + c2 e−µ2 (t) 0 0 Z s Z t (s − τ )p−2 + h2 (τ ) dτ ds, e−(µ2 (t)−µ2 (s)) Γ(p − 1) 0 0 Z
x(t)
y(t)
= b0
(2.9)
(2.10)
Rt where µi (t) = 0 ki (s)ds, i = 1, 2 and bj , cj , (j = 0, 1, 2) are unknown arbitrary constants. Using the boundary conditions (1.2) in (2.9) and (2.10), we find that b0 = 0, b2 = 0, c0 = 0, c2 = 0 and
A1 b1 + B1 c1 = J1 , A2 b1 + B2 c1 = J2 ,
(2.11)
where Ai and Bi (i = 1, 2) are respectively given by (2.5)-(2.8), and
η
Z J1
= a 0
Z
J2
= b − 0
e−(µ2 (s)−µ2 (u)) I p−1 h2 (u) du ds
0
e−(µ1 (ζ)−µ1 (s)) I q−1 h1 (s)ds,
0 θ
Z0
s
Z
ζ
− Z
(η − s)β−1 Γ(β)
(θ − s)γ−1 Γ(γ)
Z
(2.12)
s −(µ1 (s)−µ1 (u)) q−1 e I h1 (u) du ds
0
z
e−(µ2 (z)−µ2 (s)) I p−1 h2 (s)ds.
(2.13)
8
B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
Solving the system (2.11) for b1 and c1 , we get
b1
Z η Z (η − s)β−1 s −(µ2 (s)−µ2 (u)) p−1 1n h = B2 a e I h2 (u) du ds ∆ Γ(β) 0 0 Z ζ i e−(µ1 (ζ)−µ1 (s)) I q−1 h1 (s) ds − 0
h Z θ (θ − s)γ−1 Z s −B1 b e−(µ1 (s)−µ1 (u)) I q−1 h1 (u) du ds Γ(γ) 0 Z z 0 io −(µ2 (z)−µ2 (s)) p−1 e I h2 (s)ds , − 0
c1
=
Z θ Z (θ − s)γ−1 s −(µ1 (s)−µ1 (u)) q−1 1n h A1 b e I h1 (u) du ds ∆ Γ(γ) 0 0 Z z i −(µ2 (z)−µ2 (s)) p−1 − e I h2 (s)ds 0 h Z η (η − s)β−1 Z s −A2 a e−(µ2 (s)−µ2 (u)) I p−1 h2 (u) du ds Γ(β) 0 0 Z ζ io − e−(µ1 (ζ)−µ1 (s)) I q−1 h1 (s) ds ,
(2.14)
0
where ∆ is given by (2.4). Substituting the values of bj , cj (j = 0, 1, 2) in (2.9) and (2.10), we obtain the solution given by (2.2) and (2.3). Note that the converse follows by direct computation. This completes the proof.
3
Main results
n Let X = x : x ∈ C([0, 1], R) and c Dδ x ∈ C([0, 1], R) and Y = y : y ∈ C([0, 1], R) and o c α D y ∈ C([0, 1], R) denote the spaces equipped respectively with the norms kxkX = kxk+ kc Dδ xk = supt∈[0,1] |x(t)|+supt∈[0,1] |c Dδ x(t)| and kykY = kyk+kc Dα yk = supt∈[0,1] |y(t)|+ supt∈[0,1] |c Dα y(t)|. Observe that (X, k.kX ) and (Y, k.kY ) are Banach spaces. In consequence, the product space (X × Y, k.kX×Y ) is a Banach space endowed with the norm k(x, y)kX×Y = kxkX + kykY for (x, y) ∈ X × Y. Using Lemma 1, we introduce an operator H : X × Y → X × Y associated with the problem (1.1)-(1.2) as follows: H(u, v)(t) := (H1 (u, v)(t), H2 (u, v)(t)),
(3.1)
On a coupled system of sequential fractional differential equations
9
where
=
H1 (u, v)(t) ! Z t n h Z η (η − s)β−1 1 −(µ1 (t)−µ1 (s)) se ds B2 a × ∆ Γ(β) 0 0 ! Z ζ Z s i −(µ2 (s)−µ2 (m)) p−1 e−(µ1 (ζ)−µ1 (s)) I q−1 fb(s) ds e I gb(m)dm ds − × 0
0
θ
h Z −B1 b 0 z
Z −
(θ − s)γ−1 Γ(γ)
e
I
e−(µ1 (s)−µ1 (m)) I q−1 fb(m)dm ds
0
−(µ2 (z)−µ2 (s)) p−1
0
=
s
Z
io Z gb(s)ds +
t
e−(µ1 (t)−µ1 (s)) I q−1 fb(s)ds,
(3.2)
0
H2 (u, v)(t) Z n h Z θ (θ − s)γ−1 1 t × s e−(µ2 (t)−µ2 (s)) ds A1 b ∆ 0 Γ(γ) 0 ! Z s Z z i × e−(µ1 (s)−µ1 (m)) I q−1 fb(m)dm ds − e−(µ2 (z)−µ2 (s)) I p−1 gb(s)ds 0
0
η
h Z −A2 a 0 ζ
Z − 0
β−1 Z s
(η − s) Γ(β)
0
e−(µ2 (s)−µ2 (m)) I p−1 gb(m)dm ds
io Z e−(µ1 (ζ)−µ1 (s)) I q−1 fb(s) ds + 0
t
e−(µ2 (t)−µ2 (s)) I p−1 gb(s)ds,
(3.3)
with fb(t) = f (t, u(t), v(t), c Dα v(t)), gb(t) = g(t, u(t), c Dδ u(t), v(t)). Observe that existence of a fixed point of the operator H implies the existence of a solution of the problem (1.1)(1.2). In the forthcoming analysis, we need the following assumptions:
(A1 ) Let f, g : [0, 1] × R3 → R be continuous functions and there exist real constants µj , λj ≥ 0 (j = 1, 2, 3) and µ0 > 0, λ0 > 0 such that |f (t, x1 , x2 , x3 )| ≤ µ0 + µ1 |x1 | + µ2 |x2 | + µ3 |x3 |, |g(t, x1 , x2 , x3 )| ≤ λ0 + λ1 |x1 | + λ2 |x2 | + λ3 |x3 |, ∀xj ∈ R, j = 1, 2, 3.
(A2 ) There exist positive constants l, l1 such that |f (t, u1 , u2 , u3 )−f (t, v1 , v2 , v3 )| ≤ l(|u1 − v1 | + |u2 − v2 | + |u3 − v3 |), |g(t, u1 , u2 , u3 ) − g(t, v1 , v2 , v3 )| ≤ l1 (|u1 − v1 | + |u2 − v2 | + |u3 − v3 |), ∀t ∈ [0, 1], uj , vj ∈ R.
10
B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
For computational convenience, we set the following notations: ρi
=
sup |ki (t)|, i = 1, 2,
(3.4)
t∈[0,1]
M1
=
M2
=
c1 M
=
c2 M
=
N1
=
N2
=
c1 N
=
c2 N
=
σ1
=
σ2
σ3
1 θq+γ q |B2 |ζ + |b||B1 | + 2|∆| , 2|∆|Γ(q + 1) Γ(γ + 1) η p+β 1 |a||B2 | + |B1 |z p , 2|∆|Γ(p + 1) Γ(β + 1) 1 ρ1 |B2 |ζ q |b||B1 |θq+γ ρ1 + q 1+ + + , |∆| 2 Γ(q + 1) Γ(γ + 1)Γ(q + 1) Γ(q + 1) ρ1 η p+β zp 1 1+ |a||B2 | + |B1 | , |∆| 2 Γ(β + 1)Γ(p + 1) Γ(p + 1) 1 θq+γ + |A2 |ζ q , |b||A1 | 2|∆|Γ(q + 1) Γ(γ + 1) η p+β 1 |a||A2 | + |A1 |z p + 2|∆| , 2|∆|Γ(p + 1) Γ(β + 1) 1 ρ2 θq+γ ζq 1+ + |b||A1 | |A2 | , |∆| 2 Γ(q + 1) Γ(γ + 1)Γ(q + 1) 1 ρ2 |a||A2 |η p+β |A1 |z p ρ2 + p 1+ + , + |∆| 2 Γ(β + 1)Γ(p + 1) Γ(p + 1) Γ(p + 1) ! c1 c1 M N µ0 M1 + + N1 + Γ(2 − δ) Γ(2 − α) ! c2 c2 M N +λ0 M2 + + N2 + , Γ(2 − δ) Γ(2 − α)
c1 c1 M N + Γ(2 − δ) Γ(2 − α) c2 c2 M N +max{λ1 , λ2 } M2 + N2 + + , Γ(2 − δ) Γ(2 − α) c2 c2 M N = λ3 M2 + N2 + + Γ(2 − δ) Γ(2 − α) c2 c1 M N + +max{µ1 , µ2 } M1 + N1 + . Γ(2 − δ) Γ(2 − α)
(3.5) (3.6) (3.7) (3.8) (3.9) (3.10) (3.11) (3.12)
(3.13)
= µ1 M1 + N1 +
(3.14)
(3.15)
Now we present our first result which deals with the existence of solution of the problem at hand and is based on Leray-Schauder alternative [20]. Lemma 2. (Leray-Schauder alternative) Let G : E → E be a completely continuous operator. Let ε(G) = {x ∈ E : x = λGx for some 0 < λ < 1}. Then either ε(G) is unbounded or G has at least one fixed point.
On a coupled system of sequential fractional differential equations
11
Theorem 1. Assume that (A1 ) holds and that max{σ2 , σ3 } < 1, where σ2 and σ3 are given by (3.14) and (3.15) respectively. Then the boundary value problem (1.1)-(1.2) has at least one solution [0, 1]. Proof: In the first step, we show that the operator H : X × Y → X × Y is completely continuous. By continuity of the functions f and g, it follows that the operators H1 and H2 are continuous. In consequence, the operator H is continuous. Next we show that the operator H is uniformly bounded. For that, let Ω ⊂ X×Y be a bounded set. Then there exist positive constants L1 and L2 such that |f (t, u(t), v(t), c Dα v(t))| ≤ L1 , |g(t, u(t), c Dδ u(t), v(t))| ≤ L2 , ∀(u, v) ∈ Ω. Then, for any (u, v) ∈ Ω, we have θq+γ L1 q |B2 |ζ + |b||B1 | + 2|∆| |H1 (u, v)(t)| ≤ 2|∆|Γ(q + 1) Γ(γ + 1) L2 η p+β p + + |B1 |z |a||B2 | 2|∆|Γ(p + 1) Γ(β + 1) ≤ L1 M1 + L2 M2 , (3.16) which, on taking the norm for t ∈ [0, 1], yields kH1 (u, v)k ≤ L1 M1 + L2 M2 , where M1 , M2 are respectively given by (3.5) and (3.6). Using (3.2) together with (3.9) and (3.10), we obtain ρ1 |b||B1 |θq+γ L1 1 1+ |B2 |ζ q + |H10 (u, v)(t)| ≤ + ρ1 + q Γ(q + 1) |∆| 2 Γ(γ + 1) p+β L2 ρ1 |a||B2 |η + |B1 |z p + 1+ |∆|Γ(p + 1) 2 Γ(β + 1) c1 + L2 M c2 . ≤ L1 M By definition of Caputo fractional derivative with 0 < δ < 1, we get Z t (t − s)−δ 0 |H1 (u, v)(s)|ds |c Dδ H1 (u, v)(t)| ≤ 0 Γ(1 − δ) Z t (t − s)−δ c1 + L2 M c2 ds ≤ L1 M 0 Γ(1 − δ) 1 c1 + L2 M c2 . ≤ L1 M Γ(2 − δ) Hence kH1 (u, v)kX
= kH1 (u, v)k + kc Dδ H1 (u, v)k 1 c1 + L2 M c2 . L1 M ≤ L1 M1 + L2 M2 + Γ(2 − δ)
(3.17)
Similarly, we can find that kH2 (u, v)kY
= ≤
kH2 (u, v)k + kc Dα H2 (u, v)k 1 c1 + L2 N c2 , L1 N1 + L2 N2 + L1 N Γ(2 − α)
(3.18)
12
B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
c1 , N c2 are respectively given by (3.9), (3.10), (3.11) and (3.12). From the where N1 , N2 , N inequalities (3.17) and (3.18), we deduce that H1 and H2 are uniformly bounded and hence the operator H is uniformly bounded. Next, we show that H is equicontinuous. Let t1 , t2 ∈ [0, 1] with t1 < t2 . Then we have
≤
|H1 (u, v)(t2 ) − H1 (u, v)(t1 )| Z t1 Z t2 1 s e−(µ1 (t2 )−µ1 (s)) ds × s e−µ1 (s) e−µ1 (t2 ) − eµ1 (t1 ) ds + |∆| 0 t1 Z s Z ζ n h q−2 (s − τ ) −(µ1 (ζ)−µ1 (s)) e × L1 |B2 | dτ ds Γ(q − 1) 0 0 Z m Z s i Z θ (m − τ )q−2 (θ − s)γ−1 e−(µ1 (s)−µ1 (m)) +|B1 ||b| dτ dm ds Γ(γ) Γ(q − 1) 0 0 0 Z s Z z h p−2 (s − τ ) dτ ds +L2 |B1 | e−(µ2 (z)−µ2 (s)) Γ(p − 1) 0 0 Z s io Z m Z η (η − s)β−1 (m − τ )p−2 dτ dm ds e−(µ2 (s)−µ2 (m)) +|B2 ||a| Γ(β) Γ(p − 1) 0 0 0 Z s h Z t2 q−2 (s − τ ) +L1 dτ ds e−(µ1 (t2 )−µ1 (s)) Γ(q − 1) t1 0 i Z Z t1 s (s − τ )q−2 µ1 (s) −µ1 (t2 ) −µ1 (t1 ) + dτ ds . e e −e Γ(q − 1) 0 0
Evidently, |H1 (u, v)(t2 ) − H1 (u, v)(t1 )| → 0 independent of (u, v) as t2 → t1 . Also |c Dδ H1 (u, v)(t2 ) − c Dδ H1 (u, v)(t1 )| Z t1 |(t1 − s)δ − (t2 − s)δ | 0 1 |H1 (u, v)(s)|ds ≤ Γ(1 − δ) 0 (t1 − s)δ (t2 − s)δ Z t2 1 + |(t2 − s)−δ ||H10 (u, v)(s)|ds Γ(1 − δ) t1 Z t2 c1 + L2 M c2 Z t1 L1 M |(t1 − s)δ − (t2 − s)δ | −δ ≤ ds + |(t2 − s) |ds → 0, Γ(1 − δ) (t1 − s)δ (t2 − s)δ 0 t1 independent of (u, v) as t2 → t1 . In a similar manner, one can obtain that |H2 (u, v)(t2 ) − H2 (u, v)(t1 )| → 0, |c Dα H2 (u, v)(t2 ) − c Dα H2 (u, v)(t1 )| → 0, independent of (u, v) as t2 → t1 . Thus the operator H is equicontinuous in view of equicontinuity of H1 and H2 . Therefore, by Arzel´a-Ascoli’s theorem, we deduce that the operator H is completely continuous. Finally, it will be established that the set ε(H) = {(u, v) ∈ X×Y : (u, v) = λH(u, v) ; 0 ≤ λ ≤ 1} is bounded. Let (u, v) ∈ ε(H). Then (u, v) = λH(u, v). For any t ∈ [0, 1], we have u(t) = λH1 (u, v)(t), v(t) = λH2 (u, v)(t). Using (A1 ) in (3.2), we can find that kuk
≤
(µ0 + µ1 kukX + max{µ2 , µ3 }kvkY )M1 +(λ0 + max{λ1 , λ2 }kukX + λ3 kvkY )M2 .
On a coupled system of sequential fractional differential equations
13
Similarly, we can obtain ku0 k
c1 ≤ (µ0 + µ1 kukX + max{µ2 , µ3 }kvkY )M c2 , +(λ0 + max{λ1 , λ2 }kukX + λ3 kvkY )M
which leads to kc Dδ uk
≤
n 1 c1 (µ0 + µ1 kukX + max{µ2 , µ3 }kvkY )M Γ(2 − δ) o c2 . +(λ0 + max{λ1 , λ2 }kukX + λ3 kvkY )M
Thus we have kukX
= kuk + kc Dδ uk ≤ (µ0 + µ1 kukX + max{µ2 , µ3 }kvkY )M1 +(λ0 + max{λ1 , λ2 }kukX + λ3 kvkY )M2 n 1 c1 + (µ0 + µ1 kukX + max{µ2 , µ3 }kvkY )M Γ(2 − δ) o c2 . +(λ0 + max{λ1 , λ2 }kukX + λ3 kvkY )M
(3.19)
Likewise, we can have kvkY ≤ (µ0 + µ1 kukX + max{µ2 , µ3 }kvkY )N1 +(λ0 + max{λ1 , λ2 }kukX + λ3 kvkY )N2 1 c1 {(µ0 + µ1 kukX + max{µ2 , µ3 }kvkY )N + Γ(2 − α) c2 }. +(λ0 + max{λ1 , λ2 }kukX + λ3 kvkY )N
(3.20)
From (3.19) and (3.20) together with the notations (3.13)-(3.15), we find that k(u, v)kX×Y = kukX + kvkY ≤ σ1 + max{σ2 , σ3 }k(u, v)kX×Y ,
(3.21)
σ1 . This shows that ε(H) is bounded. Thus, the 1 − max{σ2 , σ3 } conclusion of Lemma 2 applies and the operator H has at least one fixed point. Consequently, the problem (1.1) and (1.2) has at least one solution on [0, 1]. This completes the proof. which yields k(u, v)kX×Y ≤
Example 1. Consider the following coupled system of fractional differential equations c
D4/3 (D + t)x(t) t c 6/5 D D+ y(t) 4
= =
1 7 1 c 1/3 sin x(t) + |y(t)| + | D y(t)|, t ∈ [0, 1] 11 30 120 1 1 4|y(t)| |x(t)| + |c D1/3 x(t)| + , (3.22) t2 + 16 21 61 + |y(t)|
14
B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
supplemented with nonlocal coupled integral boundary conditions: Z 1/2 x(0) = 0, x0 (0) = 0, x(3/4) = 2 y(s) ds, 0 0
Z
1/5
y(0) = 0, y (0) = 0, y(1/4) = (−1/3)
x(s) ds.
(3.23)
0
Here, k1 (t) = t, k2 (t) = t/4, ρ1 = 1 , ρ2 = 1/4, q = 7/3 , a = 2, b = −1/3 , p = 11/5 , ζ = 3/4, η = 1/2, β = 1, z = 1/4, θ = 1/5, γ = 1, α = 1/3, δ = 1/2, 1 7 1 c 1/3 f (t, x(t), y(t), c Dα y(t)) = 11 sin x(t)+ 30 |y(t)|+ 120 | D y(t)| and g(t, x(t), c Dδ x(t), y(t)) = 4|y(t)| 1 1 c 1/3 x(t)| + 61+|y(t)| . Clearly, the functions f and g satisfy the condition t2 +16 |x(t)| + 21 | D 1 7 1 1 1 4 (A1 ) with µ0 = 0, µ1 = 11 , µ2 = 30 , µ3 = 120 , λ0 = 0, λ1 = 16 , λ2 = 21 , λ3 = 61 . Using the given data, we find that A1 ' 0.24516, A2 ' 0.000442, |B1 | ' 0.041279, B2 ' c1 ' 2.32382, M c2 ' 0.388205 , 0.031128, ∆ ' 0.00765, M1 ' 0.734612, M2 ' 0.129402, M c c N1 ' 0.0142834, N2 ' 0.727439, N1 ' 0.0321376, N2 ' 1.71925, σ2 ' 0.509645, σ3 ' 0.495052. With max{σ2 , σ3 } < 1, all the conditions of Theorem 1 are satisfied. Therefore, the problem (3.22)-(3.23) has a solution on [0, 1]. Notice that x = 0, y = 0 is also a solution of the problem (3.22)-(3.23). In the following result, we establish the uniqueness of solutions for the problem (1.1) and (1.2) by means of Banach’s contraction mapping principle. In the sequel, we use the notations: ¯ = r1 M1 + r2 M2 , Λ1 = lM c1 + l1 M c2 , M ¯ 1 = r1 M c1 + r2 M c2 , Λ = lM1 + l1 M2 , M 0 0 ¯ c c ¯ c c Λ = lN1 + l1 N2 , N = r1 N1 + r2 N2 , Λ1 = lN1 + l1 N2 , N1 = r1 N1 + r2 N2 , r1 = supt∈[0,1] |f (t, 0, 0, 0)| < ∞, r2 = supt∈[0,1] |g(t, 0, 0, 0)| < ∞. (3.24) Theorem 2. Assume that (H2 ) holds. Further, we suppose that Λ+
Λ01 Λ1 + Λ0 + < 1, Γ(2 − δ) Γ(2 − α)
(3.25)
where Λ, Λ1 , Λ0 and Λ01 are given by (3.24). Then the problem (1.1)-(1.2) has a unique solution on [0, 1]. !" ¯1 ¯1 M N Λ1 ¯ + ¯ + Proof: Let us define r ≥ M +N 1− Λ+ + Λ0 + Γ(2 − δ) Γ(2 − α) Γ(2 − δ) Λ01 i−1 ¯, M ¯1 , N ¯ and N¯1 are given by (3.24). Then we , where Λ, Λ1 , Λ0 , Λ01 and M Γ(2 − α) show that HBr ⊂ Br , where Br = {(u, v) ∈ X × Y : k(u, v)kX×Y ≤ r} . For (u, v) ∈ Br , we have |f (t, u(t), v(t), c Dα v(t))|
≤
|f (t, u(t), v(t), c Dα v(t)) − f (t, 0, 0, 0)| + |f (t, 0, 0, 0)|
≤
l[|u(t)| + |v(t)| + |c Dα v(t)|] + r1
≤
l[kukX + kvkY ] + r1 6 lk(u, v)kX×Y + r1 ≤ lr + r1 ,
On a coupled system of sequential fractional differential equations
15
where r1 is defined by (3.24). Similarly, we have |g(t, u(t), c Dδ u(t), v(t))| ≤ l1 r + r2 , where r2 is defined by (3.24). Then |H1 (u, v)(t)| 6 (lr + r1 )M1 + (l1 r + r2 )M2 = (lM1 + l1 M2 )r + r1 M1 + r2 M2 ¯, ≤ Λr + M and c1 + (l1 r + r2 )M c2 = (lM c1 + l1 M c2 )r + r1 M c1 + r2 M c2 |H10 (u, v)(t)| 6 (lr + r1 )M ¯ ≤ Λ 1 r + M1 , which implies that c
Z
δ
| D H1 (u, v)(t)| 6 0
t
1 (t − s)−δ ¯ ). |H10 (u, v)(s)| ds 6 (Λ1 r + M Γ(1 − δ) Γ(2 − δ)
Therefore, kH1 (u, v)kX
= kH1 (u, v)k + kc Dδ H1 (u, v)k ¯1 M Λ1 ¯ ≤ Λ+ r+ M + . Γ(2 − δ) Γ(2 − δ)
(3.26)
¯ , |H0 (u, v)(t)| ≤ Λ0 r + N¯1 and In similar manner, we obtain |H2 (u, v)(t)| ≤ Λ0 r + N 2 1 Z t (t − s)−α 1 ¯ ). |c Dα H2 (u, v)(t)| ≤ |H20 (u, v)(s)| ds ≤ (Λ01 r + N Γ(1 − α) Γ(2 − α) 0 In consequence, we get kH2 (u, v)kY
= kH2 (u, v)k + kc Dα H2 (u, v)k Λ01 N¯1 ¯+ ≤ Λ0 + r+ N . Γ(2 − α) Γ(2 − α)
(3.27)
Thus, it follows from (3.26) and (3.27) that kH(u, v)kX×Y = kH1 (u, v)kX +kH2 (u, v)kY 6 r, which implies HBr ⊂ Br . Now we prove that the operator H is a contraction. For ui , vi ∈ Br ; i = 1, 2 and for each t ∈ [0, 1], by virtue of the condition (H2 ), we obtain h i |H1 (u1 , v1 )(t) − H1 (u2 , v2 )(t)| ≤ Λ ku1 − u2 kX + kv1 − v2 kY , |H10 (u1 , v1 )(t) − H10 (u2 , v2 )(t)| ≤ Λ1 [ku1 − u2 kX + kv1 − v2 kY ] , which implies that |c Dδ H1 (u1 , v1 )(t) − c Dδ H1 (u2 , v2 )(t)| Z t (t − s)−δ 0 |H1 (u1 , v1 )(s) − H10 (u2 , v2 )(s)|ds ≤ 0 Γ(1 − δ) 1 ≤ Λ1 [ku1 − u2 kX + kv1 − v2 kY ]. Γ(2 − δ)
16
B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
From the above inequalities, we get kH1 (u1 , v1 ) − H1 (u2 , v2 )kX = kH1 (u1 , v1 ) − H1 (u2 , v2 )k + kc Dδ H1 (u1 , v1 ) − c Dδ H1 (u2 , v2 )k Λ1 ≤ Λ+ [ku1 − u2 kX + kv1 − v2 kY ] . Γ(2 − δ)
(3.28)
Similarly, we can find that kH2 (u1 , v1 ) − H2 (u2 , v2 )kY ≤ Λ0 +
Λ01 [ku1 − u2 kX + kv1 − v2 kY ] . Γ(2 − α)
(3.29)
Consequently, it follows from (3.28) and (3.29) that kH(u1 , v1 ) − H(u2 , v2 )kX×Y Λ01 Λ1 0 + [ku1 − u2 kX + kv1 − v2 kY ] . ≤ Λ+Λ + Γ(2 − δ) Γ(2 − α) By the assumption (3.25), it follows that the operator H is a contraction. Hence, by Banach’s fixed point theorem, the operator H has a unique fixed point, which corresponds to a unique solution of problem (1.1)-(1.2). This completes the proof. Example 2. Consider the coupled system of fractional differential equations: c
D5/2 (D + 2t)x(t) t c 9/4 y(t) D D+ 2
= f (t, x(t), y(t), c D1/2 y(t)), = g(t, x(t), c D1/3 x(t), y(t)),
(3.30)
equipped with nonlocal coupled integral boundary conditions: Z 1/4 0 x(0) = 0, x (0) = 0, x(3/4) = y(s) ds, y(0) = 0, y 0 (0) = 0, y(1/2) =
Z
0 1/3
x(s) ds.
(3.31)
0
Here, k1 (t) = 2t, k2 (t) = t/2, ρ1 = 2 , ρ2 = 1/2, q = 5/2 , a = b = 1 , p = 9/4 , ζ = c α 3/4, η= 1/4, β = 1, z = 1/2, θ = 1/3, γ = 1, α = 1/2, δ = 1/3, f (t, x(t), y(t), D y(t)) = |y(t)|+|c D 1/2 y(t)| 1 −1 −t c δ (|x(t)|) + e−t +|y(t)|+|c D1/2 y(t)| + e and g(t, x(t), D x(t), 32+t2 tan 1 y(t)) = 20π+t2 sin(x(t)) + sin(c D1/3 x(t)) + |y(t)| + log(t + 1). Clearly l = 1/32, and l1 = 1/(20π). Using the given data, we find that A1 ' 0.214619, A2 ' 0.0059724, B1 ' c1 ' 0.002592, B2 ' 0.120816, ∆ ' 0.0259138, M1 ' 0.642917, M2 ' 0.0142273, M c 2.72212, M2 ' 0.0569091, Λ ' 0.0203176, Λ1 ' 0.085972, N1 ' 0.0435357, N2 ' 0.734259 , c1 ' 0.108839, N c2 ' 1.93371, Λ0 ' 0.0341772 and [Λ + Λ1 /Γ(3/2) + Λ0 + Λ0 ' 0.0130466, N 1 0 Λ1 /Γ(7/4)] ' 0.168233 < 1. Thus all the conditions of Theorem 2 are satisfied. In consequence, by the conclusion of Theorem 2, there exists a unique solution for the problem (3.30)-(3.31) on [0, 1].
On a coupled system of sequential fractional differential equations
17
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B. Ahmad, A. Alsaedi, S. Aljoudi and S. K. Ntouyas
[16] B. Ahmad, S.K. Ntouyas, A fully Hadamard type integral boundary value problem of a coupled system of fractional differential equations, Fract. Calc. Appl. Anal. 17 (2014), no. 2, 348-360. [17] J. Henderson, R. Luca, A. Tudorache, On a system of fractional differential equations with coupled integral boundary conditions, Fract. Calc. Appl. Anal. 18 (2015), 361-386. [18] B. Ahmad, S.K. Ntouyas, Existence results for a coupled system of Caputo type sequential fractional differential equations with nonlocal integral boundary conditions, Appl. Math. Comput. 266 (2015), 615-622. [19] B. Ahmad, J.J. Nieto, Boundary value problems for a class of sequential integrodifferential equations of fractional order, J. Funct. Spaces Appl. 2013, Art. ID 149659, 8 pp. [20] A. Granas, J. Dugundji, Fixed Point Theory, Springer-Verlag, New York, 2003. Received: 5.08.2016 Revised: 18.11.2016 Accepted: 1.12.2016 a
Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group Department of Mathematics Faculty of Science, King Abdulaziz University P.O. Box 80203, Jeddah 21589, Saudi Arabia E-mail: bashirahmad−
[email protected] (B. Ahmad)
[email protected] (A. Alsaedi)
[email protected] (S. Aljoudi) b,a
Department of Mathematics University of Ioannina 451 10 Ioannina, Greece E-mail:
[email protected] 
