New results for a coupled system of fractional differential equations
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(2) 134. M. Houas and Z. Dahmani. ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩. (1.1). Dα x (t) + f1 t, y (t) , Dδ y (t) = 0, t ∈ J, Dβ y (t) + f2 (t, x (t) , Dσ x (t)) = 0, t ∈ J, = x∗0, y (0)= y ∗0 , x (0) x (0) + x (0) + y (0) + y (0) = 0, x (0) = Jp x η , y (0) = Jq y (ξ) ,. where α, β ∈]3, 4], δ ≤ α−1, σ ≤ β−1, ξ, η ∈]0, 1[, J p, Jq are the Riemann-Liouville fractional integrals, Dα , Dβ , Dδ , Dσ are the Caputo fractional derivatives, J = [0, 1] , x ∗0 , y∗0 are real constants, and f1 and f2 are two functions which will be specified later. The rest of this paper is organized as follows: In section 2, we present some preliminaries and lemmas. Section 3 is devoted to existence of solution of problem (1.1). In section 4, some examples are treated to illustrate our results. 2. Preliminaries The following notations, definitions and preliminary facts will be used throughout this paper. Definition 2.1. The Riemann-Liouville fractional integral operator of order α ≥ 0, for a continuous function f on [0, ∞[ is defined as:. Jα f (t) =. (2.1). 1 Γ (α). t. (t − τ)α−1 f (τ) dτ, α > 0,. 0. J0 f (t) = f (t) ,. (2.2) where Γ (α) :=.
(3) ∞ 0. e−u uα−1 du.. Definition 2.2. The fractional derivative of f ∈ C n ([0, ∞[) in the Caputo’s sense is defined as:. (2.3). Dα f (t) =. 1 Γ (n − α). t. (t − τ)n−α−1 f (n) (τ) dτ, n − 1 < α, n ∈ N ∗ .. 0. For more details about fractional calculus, we refer the reader to [15, 18]. The following lemmas give some properties of Riemann-Liouville fractional integral and Caputo fractional derivative [12, 13]..
(4) 135. Fractional Differential Equations. Lemma 2.1. Let r, s > 0, f ∈ L 1 ([a, b]). Then J r Js f (t) = J r+s f (t), Ds Js f (t) = f (t), t ∈ [a, b] . Lemma 2.2. Let s > r > 0, f ∈ L 1 ([a, b]). Then Dr Js f (t) = J s−r f (t), t ∈ [a, b] . Let us now introduce the spaces X = {x : x ∈ C ([0, 1]) , Dσ x ∈ C ([0, 1])} and Y = {y : y ∈ C ([0, 1]) , Dδ y ∈ C ([0, 1])} endowed respectively with the norms x X = x + Dσ x ; x = sup |x (t)| , Dσ x = sup |Dσ x (t)| t∈J. and. t∈J. y Y = y + Dδ y ; y = sup y (t) , Dδ y = sup Dδ y (t) . t∈J. t∈J. The product . X ) and (Y, . Y ) are two Banach spaces. Obviously, (X, space. . X × Y, x, y X×Y is also a Banach space with norm x, y X×Y = xX + y Y . We give the following lemmas [11]: Lemma 2.3. For α > 0, the general solution of the fractional differential equation D α x (t) = 0 is given by x (t) = c0 + c1 t + c2 t2 + · · · + cn−1 tn−1 ,. (2.4). where ci ∈ R, i = 0, 1, 2, . . . , n − 1, n = [α] + 1. Lemma 2.4. Let α > 0. Then (2.5). Jα Dα x (t) = x (t) + c0 + c1 t + c2 t2 + ... + cn−1 tn−1 ,. for some ci ∈ R, i = 0, 1, 2, ..., n − 1, n = [α] + 1. We need also the following auxiliary result: Lemma 2.5. Let ∈ C ([0, 1]) . The solution of the equation Dα x (t) + (t) = 0, t ∈ J, 3 < α ≤ 4. (2.6) subject to the conditions (2.7). x (0) x (0) + x (0) x (0). is given by:. (2.8). 1 x (t) = − Γ (α) Γ(p+4)t3 + Γ(4)ηp+3 −6Γ p+4 ( ). . t 0.
(5) η 1 Γ(α+p) 0. = x∗0 , = 0, . = Jp x η ,. (t − s)α−1 (s) ds + x∗0 . α+p−1 ηp (s) ds − x∗0 Γ p+1 . η−s ( ).
(6) 136. M. Houas and Z. Dahmani. Proof. For ci ∈ R, i = 0, 1, 2, 3, and by lemma 2.3 and Lemma 2.4, the general solution of (1.1) is given by. (2.9). 1 x (t) = − Γ (α). t. (t − s)α−1 (s) ds − c0 − c1 t − c2 t2 − c3 t3 .. 0. Thanks to Lemma 2.1, we get (2.10). Jp x (t) =. 1. Γ p+α. t. (t − s)p+α−1 (s) ds +. 0. x ∗ tp Γ (4) tp+3 0 − . . Γ p+1 Γ p+4. Using the conditions (2.7), we get c1 = c2 = 0 and c0 = −x∗0 .. For c3 , we have (2.11) c3 =. . η ηp Γ p+4. p+α−1 1 ∗. (s). x − η − s ds . 0 Γ (4) ηp+3 − 6Γ p + 4 Γ p+1 Γ p+α 0. Substituting the values of c0 and c3 in (2.9), we get (2.8). 3.. Main Results. Let us introduce the quantities:. Γ p + 4 ηp+α 1 N1 = +. . , Γ (α + 1) Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1 N2 =. 1 Γ(α−σ+1). N3 = (3.1) N4 =. +. 1 Γ(β+1). 1 Γ(β−δ+1). |. Γ(p+4)ηp+α. (p+4)|Γ(p+α+1)Γ(4−σ). Γ(4)ηp+3 −6Γ. +. +. |. |. Γ(q+4)ξq+β. (q+4)|Γ(q+β+1). Γ(4)ξq+3 −6Γ. ,. Γ(q+4)ξq+β. (q+4)|Γ(q+β+1)Γ(4−δ). Γ(4)ξq+3 −6Γ. ,. ,. M1 = x∗0 +. |x∗0 |ηp Γ(p+4) |x∗0 |ηp Γ(p+4) , + |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+1) |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+1)Γ(4−σ). M2 = y∗0 +. | y∗0 |ξq Γ(q+4) | y∗0 |ξq Γ(q+4) . + |Γ(4)ξq+3 −6Γ(q+4)|Γ(q+1) |Γ(4)ξq+3 −6Γ(q+4)|Γ(q+1)Γ(4−δ). We list also the following hypotheses: (H1) : The functions f1 , f2 : [0, 1] × R2 → R are continuous..
(7) 137. Fractional Differential Equations. (H2) : There exist non-negative. . functions ai , bi ∈ C ([0, 1]) , i = 1, 2, continuous such that for all t ∈ [0, 1] and all x1 , y1 , x2 , y2 ∈ R2 , we have . f1 t, x1 , y1 − f1 t, x2 , y2 ≤ a1 (t) |x1 − x2 | + b1 (t) y1 − y2 , (3.2) . f2 t, x1 , y1 − f2 t, x2 , y2 ≤ a2 (t) |x1 − x2 | + b2 (t) y1 − y2 , with ω1 = sup a1 (t) , ω2 = sup b1 (t) , 1 = sup a2 (t) , 2 = sup b2 (t) . t∈J. t∈J. t∈J. t∈J. (H3) : There exist non-negative continuous functions l1 and l2 , such that . . f1 t, x, y ≤ l1 (t) , f2 t, x, y ≤ l2 (t) for each t ∈ J and all x, y ∈ R, with θ1 = sup l1 (t) , θ2 = sup l2 (t) . t∈J. t∈J. Our first result is based on Banach contraction principle. We have: Theorem 3.1. Suppose that η p+3 (H2) holds. If (3.3). 6Γ(p+4) q+3 Γ(4) , ξ. . 6Γ(q+4) Γ(4). and assume that the hypothesis. (N1 + N2 ) (ω1 + ω2 ) + (N3 + N4 ) ( 1 + 2 ) < 1,. then the boundary value problem (1.1) has a unique solution on J. Proof. Consider the operator φ : X × Y → X × Y defined by: (3.4) where. and. . φ x, y (t) := φ1 y (t) , φ2 x (t) , t 1 (t − s)α−1 f1 s, y (s) , Dδ y (s) ds + x∗0 φ1 y (t) = − Γ (α)
(8) η 0 p. p+α−1 Γ(p+4)t3 1 δ (s) ∗ η (s) f s, y , D y ds − x η − s 0 Γ(p+1) , Γ(4)ηp+3 −6Γ(p+4) Γ(p+α) 0 t 1 (t − s)β−1 (s, x (s) , Dσ x (s)) ds + y∗0 φ2 x (t) = − . Γ β 0
(9) η. q+β−1 Γ(q+4)t3 1 δ ∗ ξq (s) (s) η − s + Γ(4)ξq+3 −6Γ q+4 Γ q+β f s, y , D y ds − y 0 Γ(q+1) . ( ) ( ) 0.
(10) 138. M. Houas and Z. Dahmani. We shall prove that φ is contraction mapping :. Let x, y , x1 , y1 ∈ X × Y. Then, for each t ∈ J, we have: φ1 y (t) − φ1 y1 (t)
(11) t 1 Γ(α) 0. ≤. (t − s)α−1 f s, y (s) , Dδ y (s) − f s, y1 (s) , Dδ y1 (s) ds +. ×.
(12) η 0. Γ(p+4)t3. |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α). . p+α−1 δ δ η−s f s, y (s) , D y (s) − f s, y1 (s) , D y1 (s) ds.. Thanks to (H2) , we obtain φ1 y (t) − φ1 y1 (t) ≤ (3.5). ω1 y − y1 + ω2 Dδ y − Dδ y1 Γ (α + 1) . p+α ω1 y − y1 + ω2 Dδ y − Dδ y1 Γ p+4 η . +. . Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1. Consequently, φ1 y (t) − φ1 y1 (t) (3.6) ≤. . (ω1 +ω2 ) Γ(α+1). +. Γ(p+4)ηp+α (ω1 +ω2 ). |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α+1). y − y1 + Dδ y − Dδ y1 ,. which implies that . . φ1 y − φ1 y1 ≤ N1 (ω1 + ω2 ) y − y1 + Dδ y − Dδ y1 ,. (3.7) and. Dσ φ1 y (t) − Dσ φ1 y1 (t) ≤.
(13) t 1 Γ(α−σ) 0. (t − s)α−σ−1 f s, y (s) , Dδ y (s) − f s, y1 (s) , Dδ y1 (s) ds +. ×.
(14) η 0. Γ(p+4)t3−σ |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α)Γ(4−σ). . p+α−1 δ δ η−s f s, y (s) , D y (s) − f s, y1 (s) , D y1 (s) ds..
(15) 139. Fractional Differential Equations. By (H2), we have (ω1 + ω2 ) y − y1 + Dδ y − Dδ y1 Dσ φ1 y (t) − Dσ φ1 y1 (t) ≤ Γ (α − σ + 1) +. Γ(p+4)ηp+α (ω1 +ω2 )( y−y1 +Dδ y−Dδ y1 ). |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α+1)Γ(4−σ). .. Hence, Dσ φ1 y (t) − Dσ φ1 y1 (t) ≤. (3.8). . (ω1 +ω2 ) Γ(α−σ+1). +. |. Γ(p+4)ηp+α (ω1 +ω2 ). . (p+4)|Γ(p+α+1)Γ(4−σ) × y − y1 + Dδ y − Dδ y1 . Γ(4)ηp+3 −6Γ. Therefore, (3.9). Dσ φ1 y (t) − Dσ φ1 y1 (t) ≤ N2 (ω1 + ω2 ) y − y1 + Dδ y − Dδ y1 .. And consequently,. (3.10). . . Dσ φ1 y − Dσ φ1 y1 ≤ N2 (ω1 + ω2 ) y − y1 + Dδ y − Dδ y1 .. By (3.7) and (3.10) , we can write . . φ1 y − φ1 y1 X (3.11) ≤ (N1 + N2 ) (ω1 + ω2 ) y − y1 + Dδ y − Dδ y1 . With the same arguments as before, we have. φ2 (x) − φ2 (x1 ) Y (3.12) ≤ (N3 + N4 ) ( 1 + 2 ) (x − x1 + Dσ x − Dσ x1 ) . And by (3.11) and (3.12) , we obtain . . (N1 + N2 ) (ω1 + ω2 ) x − x1 , y − y1 (3.13) φ x, y − φ x1 , y1 X×Y ≤ . X×Y + (N3 + N4 ) ( 1 + 2 ) Tanks to (3.3) , we conclude that φ is contraction. As a consequence of Banach fixed point theorem, we deduce that φ has a fixed point which is a solution of the problem (1.1) ..
(16) 140. M. Houas and Z. Dahmani. The second main result is the following theorem: 6Γ(p+4) 6Γ(q+4) Theorem 3.2. Suppose that η p+3 Γ(4) , ξq+3 Γ(4) and assume that the hypotheses (H1)and (H3) are satisfied. Then, the coupled system (1.1) has at least a solution on J.. Proof. We shall use Scheafer’s fixed point theorem to prove that φ has at least a fixed point on X × Y. It is to note that φ is continuous on X × Y in view of the continuity of f1 and f2 (hypothesis (H1)). (1∗ :) : We shall prove that φ maps bounded sets into bounded sets in X × Y : . . Taking ρ > 0, and x, y ∈ Bρ := { x, y ∈ X × Y; x, y X×Y ≤ ρ}, then for each t ∈ J, we have: φ1 y (t) ≤ +. |.
(17) t 1 Γ(α) 0. (t − s)α−1 f s, y (s) , Dδ y (s) ds + x∗0
(18) η. Γ(p+4)t3. (p+4)|Γ(p+α). Γ(4)ηp+3 −6Γ. +. |. 0. . p+α−1 δ η−s f s, y (s) , D y (s) ds. | |. x∗0 ηp Γ p+3 Γ(4)η −6Γ. Tanks to (H3), we can write φ1 y (t) ≤. sup l1 (t) Γ p + 4 ηp+α. sup l1 (t) t∈J. t∈J. Γ (α + 1). + . Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1. + x∗0 + ≤ sup l1 (t) t∈J. . (3.14). +. Γ(p+4)ηp+α. . |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α+1) | |. x∗0 ηp Γ p+4 p+3 Γ(4)η −6Γ p+4. |. (. (. ) . )|Γ(p+1). . x∗0 ηp Γ p + 4 ∗ φ1 y (t) ≤ θ1 N1 + x0 + . . . Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1. Hence, we have (3.15). |x∗0 |ηp Γ(p+4). |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+1). 1 Γ(α+1). + x∗0 + Therefore,. (p+4)t3 . (p+4)|Γ(p+1). . x∗0 ηp Γ p + 4 ∗ φ1 y ≤ θ1 N1 + x0 + . . . Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1.
(19) 141. Fractional Differential Equations. On the other hand, Dσ φ1 y (t) ≤ +. 1 Γ (α − σ). t 0. Γ(p+4)t3−σ. (t − s)α−σ−1 f s, y (s) , Dδ y (s) ds + x∗0 . |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α)Γ(4−σ) +.
(20) η 0. | |. . p+α−1 δ ds (s) (s) η−s y f s, y , D . x∗0 ηp Γ p+4 t3−σ p+3 Γ(4)η −6Γ p+4 Γ p+1. |. (. (. ) )| (. )Γ(4−σ). .. By (H3), we have, ⎡ ⎢ Dσ φ1 y (t) ≤ sup l1 (t)⎢⎢⎢⎣ t∈J. ⎤. ⎥⎥ Γ p + 4 ηp+α 1 ⎥⎥ +. . Γ (α − σ + 1) Γ (4) ηp+3 − 6Γ p + 4 Γ p + α Γ (4 − σ) ⎦ +. |x∗0 |ηp Γ(p+4) . |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+1)Γ(4−σ). Consequently we obtain, . x∗0 ηp Γ p + 4 σ D φ1 y (t) ≤ θ1 N2 + .. . Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1 Γ (4 − σ). (3.16) Therefore, (3.17). . x∗0 ηp Γ p + 4. σ D φ1 y ≤ θ1 N2 + .. . Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1 Γ (4 − σ). Combining (3.15) and (3.17), yields φ1 y X (3.18). ≤ θ1 (N1 + N2 ) + x∗0 +. |x∗0 |ηp Γ(p+4). |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+1). +. |x∗0 |ηp Γ(p+4). |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+1)Γ(4−σ). .. Similarly, it can be shown that, . φ2 (x) Y (3.19). ≤ θ2 (N3 + N4 ) + y∗0 +. | y∗0 |ξq Γ(q+4) | y∗0 |ξq Γ(q+4) + . |Γ(4)ξq+3 −6Γ(q+4)|Γ(q+1) |Γ(4)ξq+3 −6Γ(q+4)|Γ(q+1)Γ(4−δ). It follows from (3.18) and (3.19) that φ x, y ≤ θ1 (N1 + N2 ) + θ2 (N3 + N4 ) + M1 + M2 . (3.20) X×Y.
(21) 142. M. Houas and Z. Dahmani. Consequently,. φ x, y X×Y < ∞.. (3.21). . (2∗ :) : Now, we will prove that φ is equicontinuous on J : For x, y ∈ Bρ , and t1 , t2 ∈ J, such that t1 < t2 . We have: φ1 y (t2 ) − φ1 y (t1 ) ≤. 1 Γ (α). t1 0. . (t1 − s)α−1 − (t2 −s)α−1 f s, y (s) , Dδ y (s) ds. (t2 −s)α−1 f s, y (s) , Dδ y (s) ds t1 . η Γ p + 4 t32 − t31. η−s p+α−1 f s, y (s) , Dδy (s) ds. Γ (4) ηp+3− 6Γ p+4 Γ p+α 0 . x∗0 ηp Γ p + 4 t31 − t32 + . . . Γ (4) ηp+3 − 6Γ p+ 4 Γ p + 1 +. (3.22). 1 Γ (α). t2. Thus, φ1 y (t2 ) − φ1 y (t1 ). 2θ1 θ1 (t2 − t1 )α tα1 − tα2 + Γ (α + 1) Γ (α + 1). L1 Γ p + 4 ηp+α t32 − t31 + . Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1 . x∗0 ηp Γ p + 4 3 3 + . . t1 − t2 , p+3 Γ (4) η − 6Γ p + 4 Γ p + 1. ≤. (3.23). and. Dσ φ1 y (t2 ) − Dσ φ1 y (t1 ) ≤.
(22) t1 1 Γ(α−σ) 0. . (t1 − s)α−σ−1 − (t2 − s)α−σ−1 f s, y (s) , Dδ y (s) ds. 1 + Γ(α−σ). +.
(23) t2 t1. (t2 − s)α−σ−1 f s, y (s) , Dδ y (s) ds. Γ(p+4)(t3−σ −t3−σ 2 1 ). |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α)Γ(4−σ).
(24) η 0. . p+α−1 δ (s) (s) η−s y f s, y , D ds. −t3−σ |x∗0 |ηp Γ(p+4)(t3−σ 2 ) 1 , p+3 |Γ(4)η −6Γ(p+4)|Γ(p+1)Γ(4−σ).
(25) Fractional Differential Equations. 143. Using (H3), we obtain: Dσ φ1 y (t2 ) − Dσ φ1 y (t1 ) ≤. (3.24). θ1 2θ1 (t2 − t1 )α−σ + tα−σ − tα−σ 2 1 Γ (α − σ + 1) Γ (α − σ + 1). θ1 Γ p + 4 ηp+α t3−σ + −t3−σ. 2 1 Γ (4) ηp+3 − 6Γ p+4 Γ p+α+1 Γ (4−σ) . x∗0 ηp Γ p + 4 + . − t3−σ t3−σ. 2 1 Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1 Γ (4 − σ). Hence, by (3.23) and (3.24), we can write . φ1 y (t2 ) − φ1 y (t1 ) X ≤ (3.25). 2θ1 θ1 (t2 − t1 )α tα1 − tα2 + Γ (α + 1) Γ (α + 1). θ1 Γ p + 4 ηp+α t32 − t31 +. Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1 . x∗0 ηp Γ p + 4 t31 − t32 + . Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1 2θ1 θ1 α−σ (t2 − t1 )α−σ + tα−σ − t + 2 1 Γ (α − σ + 1) Γ (α − σ + 1). θ1 Γ p + 4 ηp+α 3−σ t3−σ + − t. 1 Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1 Γ (4 − σ) 2 . x∗0 ηp Γ p + 4 3−σ 3−σ + . t − t. 2 1 Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1 Γ (4 − σ). With the same arguments as before, we get. φ1 x (t2 ) − φ1 x (t1 ) Y ≤ (3.26). β β 2θ2 θ2. t1 − t2 + . (t2 − t1 )β Γ β+1 Γ β+1. θ2 Γ q + 4 ηq+β t32 − t31 +. Γ (4) ξq+3 − 6Γ q + 4 Γ q + β + 1 . y∗0 ξq Γ q + 4 + t31 − t32. Γ (4) ξq+3 − 6Γ q + 4 Γ q + 1 β−δ β−δ 2θ2 θ2. t1 − t2 + . (t2 − t1 )β−δ + Γ β−δ+1 Γ β−δ+1. θ2 Γ q + 4 ξq+β t3−δ + − t3−δ. 2 1 Γ (4) ξq+3 − 6Γ q + 4 Γ q + β + 1 Γ (4 − δ) . y∗0 ξq Γ q + 4 + . − t3−δ t3−δ. 2 1 Γ (4) ξq+3 − 6Γ q + 4 Γ q + 1 Γ (4 − δ).
(26) 144. M. Houas and Z. Dahmani. . . Thanks to (3.25) and (3.26), we can state that φ x, y (t2 ) − φ x, y (t1 ) X×Y → 0 as t2 → t1 . Combining (1∗) and (2∗) and using Arzela-Ascoli theorem, we conclude that φ is completely continuous operator. (3∗ :) : Finally, we shall show that the set Ω defined by . . . Ω = { x, y ∈ X × Y, x, y = μφ x, y , 0 < μ < 1},. (3.27). is bounded: . . . Let x, y ∈ Ω, then x, y = μφ x, y , for some 0 < μ < 1. Thus, for each t ∈ J, we have: (3.28). y (t) = μφ1 y (t) , x (t) = μφ2 x (t) .. Then. 1 y (t) μ
(27) t α−1 1 δ ds + x∗ (t (s) (s) − s) f s, y , D ≤ Γ(α) 0 0 +. Γ(p+4)t3. |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α) +. |.
(28) η 0. . p+α−1 δ (s) (s) η−s y f s, y , D ds. | |. x∗0 ηp Γ p+3 Γ(4)η −6Γ. (p+4)t3 . (p+4)|Γ(p+1). Thanks to (H3), we can write 1 y (t) ≤ μ. (3.29). Therefore, (3.30). Hence, (3.31). y (t). sup l1 (t) t∈J. sup l1 (t) Γ p + 4 ηp+α t∈J. + . . Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1 . x∗0 ηp Γ p + 4 ∗ + x0 + . . . Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1 Γ (α + 1). ⎡ ⎤. ⎢⎢ ⎥⎥ Γ p + 4 ηp+α 1 ⎢ + ≤ μθ1 ⎢⎣. . ⎥⎥ Γ (α + 1) Γ (4) ηp+3 − 6Γ p + 4 Γ p + α + 1 ⎦ ⎞ ⎛. x∗0 ηp Γ p + 4 ⎟⎟ ⎜⎜ ∗ ⎟⎟ . +μ ⎜⎜⎝x0 + . ⎠ Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1. ⎞ ⎛. x∗0 ηp Γ p + 4 ⎟⎟ ⎜⎜ ∗ ⎟⎟ , ⎜ y (t) ≤ μ ⎜⎝θ1 N1 + x0 + . ⎠ Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1.
(29) 145. Fractional Differential Equations. On the other hand, 1 σ D y (t) μ
(30) t α−σ−1 1 δ (s) (t (s) − s) ≤ Γ(α−δ) f s, y , D ds 0 +. Γ(p+4)t3−σ. |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α)Γ(4−σ) +.
(31) η 0. | |. . p+α−1 δ (s) (s) y η−s f s, y , D ds. x∗0 ηp Γ p+4 t3−σ p+3 Γ(4)η −6Γ p+4 Γ p+1. |. (. (. ) )| (. )Γ(4−σ). .. By (H3), we have 1 σ D y (t) μ ≤ sup l1 (t) t∈J. . 1 Γ(α−σ+1). +. |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α+1)Γ(4−σ). | |ηp Γ(p+4) . (p+4)|Γ(p+1)Γ(4−σ). |. Dσ y (t) ≤ μθ1. . 1 Γ(α−σ+1). +μ. (3.32). . Γ(p+4)ηp+α. x∗0 p+3 Γ(4)η −6Γ. Therefore,. Thus,. +. +. Γ(p+4)ηp+α. . |Γ(4)ηp+3 −6Γ(p+4)|Γ(p+α+1)Γ(4−σ) | |ηp Γ(p+4) . (p+4)|Γ(p+1)Γ(4−σ). x∗0 p+3 Γ(4)η −6Γ. |. ⎞ ⎛. x∗0 ηp Γ p + 4 ⎟⎟ ⎜⎜ β ⎟⎟ . D y (t) ≤ μ ⎜⎜⎝θ1 N2 + . ⎠ Γ (4) ηp+3 − 6Γ p + 4 Γ p + 1 Γ (4 − σ). From (3.31) and (3.32) , we get ⎡ |x∗0 |ηp Γ(p+4) ⎢⎢ θ1 (N1 + N2 ) + x∗ + ⎢ p+3 −6Γ p+4 Γ p+1 0 Γ(4)η ⎢ ( )| ( ) | y ≤ μ ⎢⎢⎢ (3.33) X |x∗0 |ηp Γ(p+4) ⎢⎣ + Γ(4)ηp+3 −6Γ p+4 Γ p+1 Γ(4−σ) ( )| ( ) |. ⎤ ⎥⎥ ⎥⎥ ⎥⎥ . ⎥⎥ ⎦. Analogously, we can obtain ⎡ |y∗0 |ηq Γ(q+4) ⎢⎢ θ2 (N3 + N4 ) + y∗ + ⎢⎢ q+3 −6Γ q+4 Γ q+1 0 Γ(4)ξ ( )| ( ) | (3.34) xY ≤ μ ⎢⎢⎢ |y∗ |ηq Γ(q+4) ⎢⎣ + Γ(4)ξq+3 −6Γ0 q+4 Γ q+1 Γ(4−δ) ( )| ( ) |. ⎤ ⎥⎥ ⎥⎥ ⎥⎥ . ⎥⎥ ⎦.
(32) 146. M. Houas and Z. Dahmani. It follows from (3.33) and (3.34) that (3.35). x, y . X×Y. Hence,. ≤ μ [θ1 (N1 + N2 ) + θ2 (N3 + N4 ) + M1 + M2 ] . φ x, y . (3.36). X×Y. < ∞.. This shows that the set Ω is bounded. Thanks to (1∗) , (2∗) and (3∗) , we deduce that φ has at least one fixed point, which is a solution of the problem (1.1). 6Γ(p+4) 6Γ(q+4) Corollary 3.1. Assume that (H1) holds and ηp+3 Γ(4) , ξq+3 Γ(4) . If there exist L1 > 0, L2 > 0, such that f1 ≤ L1 , f2 ≤ L2 on J × R2 , then the coupled system (1.1) has at least a solution on J.. 4. Examples To illustrate our results, we will present three examples: Example 4.1. Let us consider the following coupled system: ⎧ 1 ⎪ |y(t)|+D 2 y(t) 7 ⎪ ⎪ + arctan (1 + t) = 0, t ∈ [0, 1] , ⎪ D 2 x (t) + ⎪ ⎪ 1 ⎪ (32+t2 ) e−t +| y(t)|+D 2 y(t) ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎨ D 72 y (t) + 1 2 sin |x (t)| + sin D 13 x (t) + ln 2 + t2 = 0, t ∈ [0, 1] , 20π+t (4.1) ⎪ ⎪ √ √ ⎪ ⎪ ⎪ x (0) = 2, y (0) = 3, ⎪ ⎪ ⎪ ⎪ ⎪ |x (0)| + |x (0)| + y (0) + y (0) = ⎪ ⎪ 0, ⎪ 3 ⎩ 3 3 2 2 x (0) = J x , y (0) = J y . 4. 5. For this example, we have f1 (t, x, y). =. f2 (t, x, y). =. |x| + y + arctan (1 + t) , t ∈ [0, 1] , x, y ∈ R, (32 + t2 ) e−t + |x| + y 1 sin |x| + sin y + ln 2 + t2 , , t ∈ [0, 1] , x, y ∈ R. 2 20π + t. Taking x, y, x1 , y1 ∈ R, t ∈ [0, 1] , then: . f1 t, x, y − f1 t, x1 , y1 . ≤. . f2 t, x, y − f2 t, x1 , y1 . ≤. 1 1 y − y1 , |x − x1 | + 2 2 32 + t 32 + t 1 1 y − y1 . |x − x1 | + 2 2 20π + t 20π + t. So, we can take (4.2). a1 (t) = b1 (t) =. 1 1 , a2 (t) = b2 (t) = . 32 + t2 20π + t2.
(33) 147. Fractional Differential Equations It follows then that ω1. =. 1. =. N1. =. 1 = ω2 , 32 1 = 2 , 20π 0, 0860791, N2 = 0, 4264507, N3 = 0, 0860755, N4 = 0, 1666720.. And then, (N1 + N2 ) (ω1 + ω2 ) + (N3 + N4 ) ( 1 + 2 ) = 0, 0360577 < 1.. (4.3). Hence by Theorem 3.1, the system (4.1) has a unique solution on [0, 1]. Example 4.2. The second example is the following: ⎛ ⎞ ⎧ 1 ⎜⎜ y(t) ⎟⎟ ⎪ tD 5 y(t) ⎪. 7 | | ⎜ ⎟ ⎪ 1 ⎜ ⎪ D 2 x (t) + 8(t+2)2 ⎜⎜ 1+ y(t) + 1 ⎟⎟⎟ + cos 1 + t + t2 = 0, t ∈ [0, 1] , ⎪ ⎪ | | ⎝ ⎠ ⎪ 2π 1+D 5 y(t) ⎪ ⎪ ⎪ ⎪. 1 7 ⎪ 1 t+t2 ⎪ 2 4 x (t) ⎨ D y (t) + (t) sin 2πD + cosh 2 + t2 = 0, t ∈ [0, 1] , sin x + 2 8π 32(t +1) ⎪ ⎪ √ ⎪ ⎪ ⎪ x (0) = 5, y (0) = 10, ⎪ ⎪ ⎪ ⎪ |x (0)| + |x (0)| ⎪ ⎪ + y (0) + y 3(0) = 0, ⎪ 2 ⎪ ⎩ x (0) = J 5 x 2 , y (0) = J 5 y 3 ,. (4.4). 13. 11. where f1 (t, x, y). =. f2 (t, x, y). =. ⎛ ⎞ ⎜⎜ |x| ⎟⎟ t y ⎜⎜ ⎟ + ⎟⎟⎟ + cos 1 + t + t2 , t ∈ [0, 1] , x, y ∈ R, ⎜ 2 ⎜ 8 (t + 2) ⎝ 1 + |x| 2π 1 + y ⎠ . t + t2 1 sin x + sin 2πy + cosh 2 + t2 , t ∈ [0, 1] , x, y ∈ R. 2 32 (t + 1) 8π 1. For t ∈ [0, 1] and x, y, x1 , y1 ∈ R, we have: . f1 t, x, y − f1 t, x1 , y1 . ≤. . f2 t, x, y − f2 t, x1 , y1 . ≤. 1 2. t. |x − x1 | +. 2. 8 (t + 2) 16π (t + 2) 1 1 |x − x1 | + 32 (t2 + 1) 256π (t2 + 1). with (4.5). a1 (t) =. 1 8 (t + 2). 2. , b1 (t) =. t 16π (t + 2)2. ,. and (4.6). a2 (t) =. y − y1 ,. 1 1 , b2 (t) = . 32 (t2 + 1) 256π (t2 + 1). y − y1 ,.
(34) 148. M. Houas and Z. Dahmani. It follows then that ω1. =. 1. =. N1. =. 1 1 , ω2 = , 32 400π 1 1 , 2 = , 32 256 0, 0859751, N2 = 0, 2413984, N3 = 0, 0860209, N4 = 0, 1129376,. and (4.7). (N1 + N2 ) (ω1 + ω2 ) + (N3 + N4 ) ( 1 + 2 ) = 0, 0176382 < 1.. Hence by Theorem 3.1, we conclude that the system (4.4) has a unique solution on [0, 1]. Example 4.3. Taking. (4.8). ⎧ 1 ⎪ cos y(t)+D 2 y ⎪ 11 ⎪ ⎪ D 3 x (t) + 16+t+t2 = 0, t ∈ [0, 1] , ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ sin x(t)+D 4 x(t) ⎪ ⎪ ⎨ D 103 y (t) + = 0, t ∈ [0, 1] , 18+t+t ⎪ √ 2 √ ⎪ ⎪ ⎪ (0) (0) 2, y x = ⎪ = 3, ⎪ ⎪ ⎪ ⎪ |x (0)| + |x (0)| + y (0) + y (0) = 0, ⎪ ⎪ ⎪ ⎩ x (0) = J 52 x 3 , y (0) = J 34 y 1 . 7 3. Then, we have f1 (t, x, y). =. f2 (t, x, y). =. Let x, y ∈ R and t ∈ [0, 1] . Then . f1 t, x, y ≤ (4.9). cos x + y , t ∈ [0, 1] , x, y ∈ R, 16 + t + t2. sin x + y , t ∈ [0, 1] , x, y ∈ R. 18 + t + t2. . 1 1 , f2 t, x, y ≤ . 16 + t + t2 18 + t + t2. So we take (4.10). l1 (t) =. 1 1 , l2 (t) = . 16 + t + t2 18 + t + t2. Then, (4.11). θ1 =. 1 1 , θ2 = . 16 18. Thanks to Theorem 3.2, the system (4.8) has at least one solution on [0, 1] .. 5. Conclusion We have presented some existence and uniqueness results for a nonlinear coupled system of fractional differential equations involving Caputo derivative. The proof of the existence results is based on Schaefer fixed point theorem, and the uniqueness result is proved by applying Banach contraction principle. This work can be extended to coupled systems of nonlinear fractional differential equations involving Riemann-Liouville fractional derivative..
(35) Fractional Differential Equations. 149. REFERENCES 1. B. Ahmad, J. Nieto: Existence results for a coupled system of nonlinear fractional differential equations with three-point boundary conditions, Comput. Math. Appl., 58(9) (2009), 1838-1843. 2. B. Ahmad, J. Nieto: Riemann-Liouville fractional differential equations with fractional boundary conditions, Fixed Point Theory, 13 (2012), 329-336. 3. B. Ahmad, J. Nieto: Anti-periodic fractional boundary value problems with nonlinear term depending on lower order derivative, Fractional Calculus and Applied Analysis, 15(3) (2012), 451-462. 4. C.Z. Bai, J.X. Fang: The existence of a positive solution for a singular coupled system of nonlinear fractional differential equations, Appl. Math. Comput., 150(3) (2004), 611-621. 5. M.E. Bengrine, Z. Dahmani: Boundary value problems for fractional differential equations, Int. J. Open Problems Compt. Math., 5(4) (2012), 7-15. 6. D. Delbosco, L. Rodino: Existence and uniqueness for a nonlinear fractional differential equation. J. Math. Anal. Appl., 204(3-4) (1996), 429-440. 7. K. Diethelm, N.J. Ford: Analysis of fractiona differential equations. J. Math. Anal. Appl., 265(2) (2002), 229-248. 8. K. Diethelm, G. Walz: Numerical solution of fraction order differential equations by extrapolation, Numer. Algorithms, 16(3) (1998), 231-253. 9. A.M.A. El-Sayed: Nonlinear functional differential equations of arbitrary orders, Nonlinear Anal., 33(2) (1998), 181-186. 10. I.B. Grozdanovi´c, N.I. Buri´c, K.D. Todorovi´c, N.T. Vasovi´c: Coupled stochastic oscilators with delays in coupling , Facta Universitatis (Niˇs) Series Math. Inform., 27(1) (2012), 27-40. 11. M. Houas, Z. Dahmani: New fractional results for a boundary value problem with caputo derivative, Int. J. Open Problems Compt. Math., 6(2) (2013), 30-42. 12. A.A. Kilbas, S.A. Marzan: Nonlinear differential equation with the Caputo fraction derivative in the space of continuously differentiable functions, Differ. Equ., 41(1) (2005), 84-89. 13. V. Lakshmikantham, A.S. Vatsala: Basic theory of fractional differential equations, Nonlinear Anal., 69(8) (2008), 2677-2682. 14. J. Liang, Z. Liu, X. Wang: Solvability for a couple system of nonlinear fractional differential equations in a Banach space, Fractional Calculus and Applied Analysis, 16(1) (2013), 51-63. 15. F. Mainardi: Fractional calculus: Some basic problem in continuum and statistical mechanics. Fractals and fractional calculus in continuum mechanics, Springer, Vienna. 1997. 16. J. Nieto, J. Pimente: Positive solutions of a fractional thermostat model, Boundary Value Problems, 1(5) (2013), 1-11. 17. S.K. Ntouyas: Existence results for first order boundary value problems for fractional differential equations and inclusions with fractional integral boundary conditions, Journal of Fractional Calculus and Applications, 3(9) (2012), 1-14. 18. I. Podlubny, I. Petras, B.M. Vinagre, P. O’leary, L. Dorcak: Analogue realizations of fractional-order controllers, Fractional order calculus and its applications. Nonlinear Dynam., 29(4) (2002), 281-296..
(36) 150. M. Houas and Z. Dahmani 19. M.U. Rehman, R.A. Khan and N.A. Asif: Three point boundary value problems for nonlinear fractional differential equations, Acta Mathematica Scientia, 31B(4) (2011), 1337-1346. 20. X. Su: Boundary value problem for a coupled system of nonlinear fractional differential equations, Applied Mathematics Letters, 22(1) (2009), 64-69. 21. J. Wang, H. Xiang, Z. Liu: Positive solution to nonzero boundary values problem for a coupled system of nonlinear fractional differential equations, International Journal of Differential Equations (2010), Article ID 186928, 12 pages. 22. W. Yang: Positive solutions for a coupled system of nonlinear fractional differential equations with integral boundary conditions, Comput. Math. Appl., 63 (2012), 288-297. 23. Y. Zhang, Z. Bai, T. Feng: Existence results for a coupled system of nonlinear fractional three-point boundary value problems at resonance, Comput. Math. Appl., 61 (2011), 1032-1047.. Mohamed Houas Faculty of Exact Scicences Department of Mathematics and Informatics University of Khemis Miliana houasmed@yahoo.fr Zoubir Dahmani Laboratory LPAM Faculty SEI UMAB, University of Mostaganem zzdahmani@yahoo.fr.
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