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A coupled system of Hadamard type sequential fractional differential equations with coupled strip conditions
Chaos, Solitons and Fractals 91 (2016) 39–46
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Chaos, Solitons and Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos
A coupled system of Hadamard type sequential fractional differential equations with coupled strip conditions Shorog Aljoudi a, Bashir Ahmad a, Juan J. Nieto b,∗, Ahmed Alsaedi a 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 b Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela, Santiago de Compostela, 15782, Spain
a r t i c l e
i n f o
Article history: Received 24 March 2016 Revised 5 May 2016 Accepted 8 May 2016
MSC: 34A08 34B10 34B15
a b s t r a c t We study a nonlocal boundary value problem of Hadamard type coupled sequential fractional differential equations supplemented with coupled strip conditions (nonlocal Riemann-Liouville integral boundary conditions). The nonlinearities in the coupled system of equations depend on the unknown functions as well as their lower order fractional derivatives. We apply Leray-Schauder alternative and Banach’s contraction mapping principle to obtain the existence and uniqueness results for the given problem. An illustrative example is also discussed. © 2016 Elsevier Ltd. All rights reserved.
Keywords: Hadamard fractional derivative Coupled system Strip conditions Existence Fixed point
1. Introduction Fractional calculus has emerged as an interesting field of investigation in the last few decades. The subject has been extensively developed and the literature on the topic is much enriched now, covering theoretical as well as widespread applications of this branch of mathematical analysis. In fact, the tools of fractional calculus have led to much improved, realistic and practical mathematical modeling of many systems and processes, occurring in engineering and scientific disciplines such as control theory, signal and image processing, biophysics, blood flow phenomena, etc. [1–3]. The nonlocal characteristic of fractional order operators, which takes into account the hereditary properties of phenomena involved, has greatly contributed to the popularity of the subject. In particular, the topic of boundary value problems of RiemannLiouville or Liouville-Caputo type fractional order differential equations equipped with a variety of boundary conditions has
∗
Corresponding author. Fax: +34881813197. E-mail addresses: [email protected] (S. Aljoudi), bashirahmad_qau@ yahoo.com (B. Ahmad), [email protected] (J.J. Nieto), [email protected] (A. Alsaedi). http://dx.doi.org/10.1016/j.chaos.2016.05.005 0960-0779/© 2016 Elsevier Ltd. All rights reserved.
attracted significant attention, for example, see [4–14] and the references cited therein. In the classical text [15], it has been mentioned that Hadamard in 1892 [16] suggested a concept of fractional integrod q differentiation in terms of the fractional power of the type (t dt )
d q in contrast to its Riemann-Liouville counterpart of the form ( dt ) . The kind of derivative introduced by Hadamard contains logarithmic function of arbitrary exponent in the kernel of the integral appearing in its definition. Hadamard’s construction is invariant in relation to dilation and is well suited to the problems containing half axes. In [18], the authors have shown that the Lamb-Bateman integral equation can be expressed in terms of Hadamard fractional derivatives of order 1/2. Following an approach based on Hadamard derivatives and fractional Hyper-Bessel-type operators, the authors in [17] discussed a modified Lamb-Bateman equation. In fact, there are not many applications of Hadamard integrals and derivatives in view of possibly the intrinsic intractability of these operators. However, one can find some recent results on Hadamard type fractional differential equations, for instance, see [2,19–23]. Coupled systems of fractional order differential equations have also been investigated by many authors. Such systems appear naturally in many real world situations, for example, see [24]. Some recent results on the topic can be found in a series of
40
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
papers [25–30] and the references cited therein. More recently, in [27], the authors discussed a coupled system of Hadamard type fractional differential equations with Hadamard integral boundary conditions of the form:
⎧ Dα u(t ) = f (t , u(t ), v(t )), Dβ v(t ) = g(t , u(t ), v(t )), ⎪ ⎪ ⎪ 1 < α , β ≤ 2, 1 < t < e, ⎨ γ −1 u(s) σ u(1 )=0, u(e ) = (1γ ) 1 1 log σs1 ds, γ > 0, 1 < σ1 < e, ⎪ s ⎪ ⎪ ⎩v(1 ) = 0, v(e ) = 1 σ2 log σ2 γ −1 v(s) ds, 1 < σ < e, 2 s s (γ ) 1
(1.1) where f, g : [1, e] × R × R → R are given continuous functions. In this paper, motivated by aforementioned applications of Hadamard fractional integro-differentiation as well as by the recent interest in developing the existence theory for Hadamard type initial and boundary value problems, we study a coupled system of Hadamard type sequential fractional differential equations equipped with nonlocal coupled strip conditions given by
⎧ q (D + kDq−1 )u(t ) = f (t, u(t ), v(t ), Dα v(t )), k > 0, ⎪ ⎪ ⎪ 1 < q 2, 0 < α < 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (D p + kD p−1 )v(t ) = g(t, u(t ), Dδ u(t ), v(t )), ⎪ ⎪ ⎪ ⎨ 1 < p 2, 0 < δ < 1, γ −1 v(s) η u(1 ) = 0, u(e ) = Iγ v(η ) = (1γ ) 1 log ηs ds, γ > 0, ⎪ s ⎪ ⎪ ⎪ 1 < η < e, ⎪ ⎪ ⎪ ⎪ ⎪ ζ ζ β −1 u(s ) ⎪ 1 β ⎪ ds, β > 0, ⎪v(1 ) = 0, v(e ) = I u(ζ ) = (β ) 1 log s s ⎩ 1 < ζ < e, (1.2) D(.)
I(.)
where and respectively denote the Hadamard fractional derivative and Hadamard fractional integral (to be defined later), and f, g : [1, e] × R3 → R are given continuous functions to be chosen appropriately (according to the requirement). We show the existence of solutions for problem (1.1) by applying Leray-Schauder alternative criterion while uniqueness of solutions for (1.1) relies on Banach’s contraction mapping principle. Though we make use of the standard methodology to obtain the desired results, yet its exposition to the given problem is new. Section 2 contains some basic concepts and an auxiliary lemma, while the main results are presented in Section 3. Here we remark that problem (1.2) is a generalization of the problem (1.1) in the sense that the coupled system in (1.2) consists of sequential fractional differential equations with the nonlinearities depending on the unknown functions as well as their lower order fractional derivatives, and coupled strip boundary conditions, whereas problem (1.1) deals with fractional Hadamard type equations with nonlinearities only involving unknown functions and uncoupled integral boundary conditions.
First of all, we recall some important definitions [2] and then prove an auxiliary lemma for the linear variant of problem (1.1). Definition 2.1. The Hadamard fractional integral of order q for a function g ∈ Lp [y, x], 0 ≤ y ≤ t ≤ x ≤ ∞, is defined as
1 (q )
t
log y
D g(t ) = q
t s
q−1 g(s ) s
ds, q > 0.
d Definition 2.2. Let [y, x] ⊂ R, δ = t dt and ACδn [y, x] = {g : [y, x] → n −1 R:δ [g(t )] ∈ AC[y, x]}. The Hadamard derivative of fractional or-
δ (I n
n−q
1 d )(t ) = t (n − q ) dt
n t
log 1
t s
n−q−1 g(s ) s
ds,
n − 1 < q < n, n = [q] + 1, where [q] denotes the integer part of the real number q and log(· ) = loge (· ). Recall that the Hadamard fractional derivative is the left-inverse operator to the Hadamard fractional integral in the space Lp [y, x], 0 < y < x < ∞, 1 ≤ p ≤ ∞, that is, Dq Iq f (t ) = f (t ) (Theorem 4.8, [31]). In [32], Caputo-type modification of the Hadamard fractional derivatives was proposed as follows: c
Dq g(t ) = Dq g(s ) −
n−1
δ k g( y )
k!
k=0
log
s y
k
(t ), t ∈ (y, x ).
Further, it was shown in (Theorem 2.1, [32]) that In−q δ n g(t ). For 0 < q < 1, it follows from (2.1) that c
(2.1) c Dq g(t )
=
Dq g(t ) = Dq g(s ) − g(y ) (t ).
Furthermore, it was established in Lemma 2.4 and Lemma 2.5 of [32] respectively that c
Dq (Iq g)(t ) = g(t ), Iq (c Dq )g(t ) = g(t ) −
n−1 k=0
δ k g( y )
k!
log
t y
k
. (2.2)
From the second formula in (2.2), one can easily infer that the solution of Hadamard differential equation: c Dq u(t ) = σ (t ) can be written as n−1
u(t ) = Iq σ (t ) +
k=0
δ k u (y )
k!
log
t y
k
,
for appropriate function u(t) and σ (t) (as required in the above definitions). Note that the Hadamard integral and derivative defined above are left-sided. One can define the Hadamard right-sided integral and derivative in the same way, for instance, see [32]. Lemma 2.1. Let h1 , h2 ∈ AC ([1, e], R ). Then the solution for the linear system of sequential fractional differential equations:
(Dq + kDq−1 )u(t ) = h1 (t ), (D p + kD p−1 )v(t ) = h2 (t ),
(2.3)
supplemented with the boundary conditions in (1.2) is given by a coupled system of Hadamard integral equations
u(t ) =
2. Preliminaries
Iq g(t ) =
der q for a function g ∈ ACδn [y, x] is defined as
1
t −k
t 1
e
s
e−k s p−2 h2 (r ) sk−1 log dr ds ( p − 1 ) 1 r r 1 ζ
1 ζ β −1 −k−1 − log s (q − 1 )(β ) 1 s
s
r
r q−2 h1 (m ) × r k−1 log dm dr ds m m 1 1 η
1 η γ −1 −k−1 − B2 log s ( p − 1 )(γ ) 1 s
s
r
r p−2 h2 (m ) × r k−1 log dm dr ds m m 1 1 ×
− A2
sk−1 (log s )q−2 ds
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
e
s
e−k s q−2 h1 (r ) sk−1 log dr ds (q − 1 ) 1 r r 1 t +t −k sk−1 Iq−1 h1 (s )ds,
−
Now, using the coupled integral boundary conditions given by (1.2) in (2.11) and (2.12), we obtain (2.4)
1
and
1
v(t ) =
t −k
t
1
s
×
sk−1 (log s ) p−2 ds
log 1
− B1
s q−2 h1 (r ) r
r
e−k (q − 1 )
A1 c1 + A2 d1 = J1 , B1 c1 + B2 d1 = J2 ,
e
sk−1
J1 =
1
dr ds
−
J2 =
(γ )( p − 1 ) 1 s r
r k−1
1
e
(q − 1 )
(2.5)
1
where
1 e
sk−1
r k−1
e
( p − 1 )
1 e
sk−1
A1 = e
×
e
s
k−1
1 s r
k−1
1
(β )
B2 = e−k
e
1
q−2
log
1
ds, A2 =
(log r )
ζ
−1
(log s )
(2.6)
p−2
s
s
(γ )
r m
p−2
s
s
−k−1
s
r
k−1
q−2
(log r )
dr ds,
ds.
(2.8)
Proof. As argued in [2], the general solution of the system (2.3) can be written as
u(t ) = c0 t −k + c1 t −k + t −k
t 1
t 1
t 1
1
t
sk−1 (log s )q−2 ds (2.9)
sk−1 (log s ) p−2 ds
sk−1 I p−1 h2 (s )ds,
(2.10)
where ci , di , (i = 0, 1 ) are unknown arbitrary constants. Making use of the conditions u(1 ) = 0 and v(1 ) = 0 in (2.9)–(2.10) yields c0 = 0 and d0 = 0, which leads to
u(t ) = c1 t −k
v(t ) = d1t −k
t 1
t 1
sk−1 (log s )q−2 ds + t −k
sk−1 (log s ) p−2 ds + t −k
c1 =
t
1
t 1
sk−1 Iq−1 h1 (s )ds,
(2.11)
sk−1 I p−1 h2 (s )ds.
(2.12)
1
×
r
r
dr ds,
(2.14)
s−k−1
s
q−2 h (m ) 1 m
log 1
1 (q − 1 )(β )
A2
s
r
k−1
1
dm dr ds
s p−2 h2 (r ) u
r
dr ds.
r
1
ζ
r log m s
log
ζ
(2.15)
1
β −1
s
q−2 h (m ) 1 m
s−k−1
dm dr ds
s q−2 h2 (r ) s log dr ds r r 1 1
e s
e−k s q−2 h1 (r ) k−1 + B2 s log dr ds (q − 1 ) 1 r r 1 η
1 η γ −1 −k−1 − log s ( p − 1 )(γ ) 1 s s r
r p−2 h2 (m ) × r k−1 log dm dr ds , m m 1 1 e−k − ( p − 1 )
e
k−1
(2.16)
η
1 η γ −1 −k−1 log s
( p − 1 )(γ ) 1 s s r
r p−2 h2 (m ) k−1 × r log dm dr ds m m 1 1
e s
e−k s q−2 h1 (r ) k−1 − s log dr ds (q − 1 ) 1 r r 1
e s
e−k s p−2 h2 (r ) k−1 + A1 s log dr ds ( p − 1 ) 1 r r 1
d1 =
sk−1 Iq−1 h1 (s )ds,
v(t ) = d0t −k + d1t −k + t −k
s q−2 h1 (r )
β −1
s
1
dm dr ds
Solving the system (2.13) for c1 and d1 and using the notation (2.6), we find that
−k−1
(2.7)
1
sk−1 (log s )
log
1
η γ −1
dr ds,
β −1
ζ
η
−1
ζ
log
= B1 A2 − A1 B2 = 0,
m
log
s−k−1
p−2 h (m ) 2 log
(β )(q − 1 ) 1 s r
s
1
ζ
1
η γ −1
s
1
1
−k
r m
log
−k
× −
η
log
1
×
η
η γ −1 −k−1 s k−1 1 log s r ( p − 1 )(γ ) 1 s 1
r
r p−2 h2 (m ) × log dm dr ds m m 1
β −1 ζ
s 1 ζ − A1 log s−k−1 r k−1 (q − 1 )(β ) 1 s 1
r
r q−2 h1 (m ) × log dm dr ds m m 1 e s
−k e s p−2 h2 (r ) − sk−1 log dr ds ( p − 1 ) 1 r r 1 t −k k−1 p−1 +t s I h2 (s )ds,
−k
(2.13)
where Ai and Bi (i = 1, 2 ) are respectively given by (2.7) and (2.8), and
−
B1 =
41
−
1
B1
β −1 ζ 1 ζ log s−k−1 (q − 1 )(β ) 1 s s r
r q−2 h1 (m ) × r k−1 log dm dr ds . m m 1 1
(2.17)
Substituting the values of c1 and d1 in (2.13), we get the desired solution (2.4)–(2.5). The converse of the lemma follows by direct computation. This completes the proof.
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S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
3. Existence and uniqueness results
Let X = u : u ∈ C ([1, e], R ) and Dδ u ∈ C ([1, e], R ) and Y = {v : v ∈ C ([1, e], R ) and Dα v ∈ C ([1, e], R )} denote the spaces respectively endowed with the norms u X = u + Dδ u = supt∈[1,e] |u(t )| + supt∈[1,e] |Dδ u(t )| and v Y = v + D α v = supt∈[1,e] |v(t )| + supt∈[1,e] |Dα v(t )|. Observe that ( X , . X ) and (Y, . Y ) are Banach spaces. In turn, the product space (X × Y, . X ×Y ) is a Banach space equipped with the norm (u, v ) X ×Y = u X + v Y for (u, v ) ∈ X × Y. In relation to problem (1.2), we introduce an operator T : X × Y → X × Y by Lemma 2.1 as follows:
T (u, v )(t ) := (T1 (u, v )(t ), T2 (u, v )(t )), where
T1 (u, v )(t ) =
+
A2
ζ
1
(β )
s 1
B2
1
log
1
η
1
r
k−1
log
I
1
+ B2 e + t −k 1
−k
e
1
t 1
s−k−1
η γ −1 s
s−k−1
sk−1 Iq−1 f (s, u(s ), v(s ), Dα v(s )) ds
(3.2)
t
sk−1 (log s ) p−2 ds
1 e
− B1 e−k sk−1 Iq−1 f (s, u(s ), v(s ), Dα v(s )) ds t −k
η
B1
(γ )
s 1
A1
1
1
+ A1 e−k + t −k
ζ
1
1
log
ζ
s−k−1
1
e
1
+
M2 =
Theorem 3.3. Let f, g : [1, e] × R3 → R be continuous functions satisfying the Lipschitz condition: (H1 ) | 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 | ), l, l1 > 0, ∀t ∈ [1, e], u j , v j ∈ R. Further,
(3.11)
2(r1 M1 + r2 M2 )(1 + (1 − δ )) , [(1 − δ ) − 2(1 + (1 − δ ))(lM1 + l1 M2 )]
Then we show that T Br ⊂ Br where (3.3)
|B2 | | | + , (q + 1 ) q(q − 1 ) (3.4)
(3.10)
Br = {(u, v ) ∈ X × Y : (u, v ) X ×Y ≤ r } For (u, v) ∈ Br , we have
q + β
| |(q − 1 ) (q + 1 )(β + 1 )
(3.9)
2(r1 N1 + r2 N2 )(1 + (1 − α )) , [(1 − α ) − 2(1 + (1 − α ))(lN1 + l1 N2 )]
For computational convenience, we set
M1 =
(3.8)
1 ω2 = μ1 M1 + max{λ1 , λ2 }M2 1 + (1 − δ )
1 + μ1 N1 + max{λ1 , λ2 }N2 1 + , (1 − α )
r ≥ max
sk−1 I p−1 g(s, u(s ), Dδ (s ), v(s ))ds.
|A2 | log ζ
1 μ0 M1 + λ0 M2 ω1 = 1 + (1 − δ )
1 μ0 N1 + λ0 N2 , + 1+ (1 − α )
s−k−1
sk−1 I p−1 g(s, u(s ), Dδ (s ), v(s )) ds
(3.7)
Proof. Let us set r1 = supt∈[1,e] f (t, 0, 0, 0 ) < ∞, r2 = supt∈[1,e] g(t, 0, 0, 0 ) < ∞ and define
β −1
1 |A1 | | | + + , | |( p − 1 ) (γ + 1 )( p + 1 ) ( p + 1 ) p( p − 1 )
where M1 , M2 , N1 , N2 are given by (3.7). Then the boundary value problem (1.2) has a unique solution.
s
p+γ |B1 | log η
lN1 + l1 N2 < (1 − α )/2(1 + (1 − α )),
r k−1 Iq−1 f (r, u(r ), v(r ), Dα (r )) dr ds
t
s
(3.6)
lM1 + l1 M2 < (1 − δ )/2(1 + (1 − δ )),
r k−1 I p−1 g(r, u(r ), Dδ u(r ), v(r )) dr ds
(β )
s
×
log
η γ −1
1 |B1 | + , | |( p − 1 ) (q + 1 ) (q + 1 )(β + 1 )
Now we are in a position to present our main results. The first result deals with the uniqueness of solutions for problem (1.2), which has been obtained by means of Banach’s contraction mapping principle.
g(r, u(r ), Dδ u(r ), v(r )) dr ds
N2 =
q + β |A1 | log ζ
1 ω3 = max{μ2 , μ3 }M1 + λ3 M2 1 + (1 − δ )
1 + max{μ2 , μ3 }N1 + λ3 N2 1 + . (1 − α )
sk−1 Iq−1 f (s, u(s ), v(s ), Dα v(s )) ds,
× −
s
p−1
1
+
ζ β −1
r k−1 Iq−1 f (r, u(r ), v(r ), Dα v(r )) dr ds
(γ )
s
×
T2 (u, v )(t ) =
t
sk−1 (log s )q−2 ds
1 e
× − A2 e−k sk−1 I p−1 g(s, u(s ), Dδ u(s ), v(s )) ds t −k
× −
(3.1)
N1 =
Similarly, we have
|g(t, u(t ), Dδ u(t ), v(t ))| ≤ l1 r + r2 .
p+γ
|A | |B2 | log η 1 2 + , | |(q − 1 ) ( p + 1 ) (γ + 1 )( p + 1 )
| f (t, u(t ), v(t ), Dα v(t ))| ≤ | f (t, u(t ), v(t ), Dα v(t )) − f (t, 0, 0, 0 )| + | f (t, 0, 0, 0 )| ≤ l[|u(t )| + |v(t )| + |Dα v(t )|] + r1 l[ u X + v Y ] + r1 l (u, v ) X ×Y + r1 lr + r1 .
(3.5)
Then
|T1 (u, v )(t )| (lM1 + l1 M2 )r + (r1 M1 + r2 M2 ),
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
and
|Dδ T1 (u, v )(t )|
d t
1
t
(1 − δ ) dt 1 (lM1 + l1 M2 )r + (r1 M1 + r2 M2 ) . (1 − δ )
log
t −δ |T1 (u, v )(s )| s
s
≤ ds
Therefore,
1 T1 (u, v ) X = T1 (u, v ) + Dδ T1 (u, v ) 1 + (1 − δ )
r × (lM1 + l1 M2 )r + (r1 M1 + r2 M2 ) ≤ (3.12) 2
In similar manner, we obtain
|T2 (u, v )(t )| ≤ (lN1 + l1 N2 )r + (r1 N1 + r2 N2 ), |Dα T2 (u, v )(t )| (lN1 + l1 N2 )r + (r1 N1 + r2 N2 ) . (1 − α ) In consequence, we get
T2 (u, v ) Y = T2 (u, v ) + Dα T2 (u, v )
r 1 1+ (lN1 + l1 N2 )r + (r1 N1 + r2 N2 ) ≤ (1 − α ) 2
t 1 |T1 (u1 , v1 )(t ) − T1 (u2 , v2 )(t )| ≤ t −k sk−1 (log s )q−2 ds | | 1
s
|A |e−k e p−2 s sk−1
+
log
1
r
×
1 g(r, u1 (r ), Dδ u1 (r ), v1 (r ))−g(r, u2 (r ), Dδ u2 (r ), v2 (r )) r dr ds ζ
s
r
|A | ζ γ −1 r q−2 2
log
s−k−1
r k−1
log
(q − 1 )(β ) 1 s m 1 1 α × f (m, u1 (m ), v1 (m ), D v1 (m ) 1 − f (m, u2 (m ), v2 (m ), Dα v2 (m ) dm dr ds m s
−k e q−2 |B2 |e s + sk−1 log f (r, u1 (r ), v1 (r ), Dα v1 (r ) (q − 1 ) 1 r 1 1 − f (r, u2 (r ), v2 (r ), Dα v2 (r ) dr ds r η
s
r
|B | η γ −1 r p−2 2
+
log
s−k−1
r k−1
r ≤ (lM1 + l1 M2 )[ u1 − u2 X + v1 − v2 Y ].
d t 1 |Dδ T1 (u1 , v1 )(t ) − Dδ T1 (u2 , v2 )(t )| ≤ t (1 − δ ) dt 1
t −δ 1 × log |T1 (u1 , v1 )(s ) − T1 (u2 , v2 )(s )|ds s
s
(3.14) Similarly, we can find that
T 2 ( u 1 , v 1 ) − T 2 ( u 2 , v 2 ) Y (1 + (1 − α ))(lN1 + l1 N2 ) ≤ [ u1 − u2 X + v1 − v2 Y ]. (1 − α ) (3.15)
= T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) X + T 2 ( u 1 , v 1 ) − T 2 ( u 2 , v 2 ) Y
(1 + (1 − δ ))(lM1 + l1 M2 ) (1 + (1 − α ))(lN1 + l1 N2 ) + (1 − δ ) (1 − α )
[ u1 − u2 X + v1 − v2 Y ]. In view of the assumption (3.11), there exists a positive number such that
(1 + (1 − δ ))(lM1 + l1 M2 ) (1 − δ ) (1 + (1 − α ))(lN1 + l1 N2 ) + ≤ < 1. (1 − α ) Therefore,
T (u1 , v1 ) − T (u2 , v2 ) X ×Y ≤ ( u1 − u2 X + v1 − v2 Y ). This shows that T is a contraction. Hence, by Banach’s fixed point theorem, the operator T has a unique fixed point which corresponds to a unique solution of problem (1.2). This completes the proof. Our next existence result is based on Leray-Schauder alternative. Lemma 3.2. (Leray-Schauder alternative [33]) Let F: E → E be a completely continuous operator. Let ε (F ) = {x ∈ E : x = λF (x ) for some 0 < λ < 1}. Then either the set ε(F) is unbounded or F has at least one fixed point.
log
( p − 1 )(γ ) 1 s m 1 1 δ ×g(m, u1 (m ), D u1 (m ), v1 (m )) 1 − g(m, u2 (m ), Dδ u2 (m ), v2 (m )) dm dr ds m t t
−k q−2 t s + sk−1 log f (r, u1 (r ), v1 (r ), Dα v1 (r )) (q − 1 ) 1 r 1 1 − f (r, u2 (r ), v2 (r ), Dα v2 (r )) dr ds
Also we have
T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) X = T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) + Dδ T 1 ( u 1 , v1 ) − Dδ T 1 ( u 2 , v2 ) (1 + (1 − δ ))(lM1 + l1 M2 ) ≤ [ u1 − u2 X + v1 − v2 Y ]. (1 − δ )
≤
which implies T Br ⊂ Br . Now we prove that the operator T is a contraction. For ui , vi ∈ Br ; i = 1, 2 and for each t ∈ [1, e] we have
1
From the above inequalities, we get
T (u1 , v1 ) − T (u2 , v2 ) X ×Y (3.13)
T (u, v ) X ×Y = T1 (u, v ) X + T2 (u, v ) Y r,
2
(lM1 + l1 M2 ) [ u1 − u2 X + v1 − v2 Y ]. (1 − δ )
Consequently, we obtain from (3.14 ) and (3.15 ) that
Thus, it follows from (3.12) and (3.13) that
( p − 1 )
43
Theorem 3.4. Let f, g : [1, e] × R3 → R be continuous functions and there exist real constants μ j , λ j ≥ 0 ( j = 1, 2, 3 ) and μ0 > 0, λ0 > 0 such that (H2 ) | f (t, x1 , x2 , x3 )| ≤ μ0 + μ1 |x1 | + μ2 |x2 | + μ3 |x3 |, |g(t, x1 , x2 , x3 )| ≤ λ0 + λ1 |x1 | + λ2 |x2 | + λ3 |x3 |, ∀x j ∈ R, j = 1, 2, 3. In addition it is assumed that max{ω2 , ω3 } < 1, where ω2 and ω3 are given by (3.19) and (3.20) respectively. Then the boundary value problem (1.2) has at least one solution [1, e]. Proof. Let us first show that the operator T : X × Y → X × Y is completely continuous. Clearly continuity of the functions f and g implies that the operators T1 and T2 are continuous. Hence the operator T is continuous. In order to establish that the operator T is uniformly bounded, we consider a bounded set ⊂ X × Y such that |f(t, u(t), v(t), Dα v(t))| ≤ L1 , |g(t, u(t), Dδ u(t), v(t))| ≤ L2 , ∀(u, v) ∈ , where L1 and L2 are positive constants. Then, for any (u, v) ∈ , we have
44
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
t L |T1 (u, v )(t )| ≤ 1 t −k sk−1 (log s )q−2 ds | | 1 |B |e−k e
s
s q−2 1 2
×
sk−1
log
From (3.16) and (3.17), we infer that T1 and T2 are uniformly bounded, which implies that the operator T is uniformly bounded. Next, we show that T is equicontinuous. Let t1 , t2 ∈ [1, e] with t1 < t2 . Then we have
dr ds
(q − 1 ) 1 r r 1 ζ
|A2 | ζ β −1 −k−1 + log s (q − 1 )(β ) 1 s
s
r
r q−2 1 r k−1
×
log
1
1
m
m
|T1 (u, v )(t2 ) − T1 (u, v )(t1 )| ≤ ×
dm dr ds
t L2 + t −k sk−1 (log s )q−2 ds | | 1 |A |e−k e
s
s p−2 1 2 × sk−1 log dr ds ( p − 1 ) 1 r r 1 η
γ −1 |B2 | η + log s−k−1 ( p − 1 )(γ ) 1 s
s
r
r p−2 1 r k−1
×
log
1
+ L1 t −k
1
t
sk−1
s
1
1
m
s log r
m
+
+
dr ds
×
which, on taking the norm for t ∈ [1, e] and using (3.4), (3.5) yields
T1 (u, v ) ≤ L1 M1 + L2 M2 . By definition of Hadamard fractional derivative with 0 < δ < 1, we get
d t
1 t −δ |T1 (u, v )(s )| t log ds (1 − δ ) dt 1 s s 1 ≤ (L M + L2 M2 ). (1 − δ ) 1 1
|Dδ T1 (u, v )(t )| ≤
Hence
(3.16)
Similarly, we can get
L1
| | ( p − 1 ) q + β
|A1 | log ζ
L2 |B1 | + + (q + 1 ) (β + 1 )(q + 1 ) | | ( p − 1 )
p+γ |B1 | log η |A1 | | | × + + (γ + 1 )( p + 1 ) ( p + 1 ) p( p − 1 ) ×
|Dα T2 (u, v )(t )| ≤
+
e
sk−1
log
t1
2
t1k t2k
1
|B2 | ( p − 1 )(γ )
s
r
r k−1
η
1
1
|A2 |e−k ( p − 1 )
e
sk−1
1
log
r m
1
s−k−1
dm dr ds
r
η γ −1 s
m log
sk−1 (log s )q−2 ds
q−2 1
s r
p−2 1
s
t2
t1
dr ds
sk−1 (log s )q−2 ds + t2−k
log
1
s
m
1
|t k − t k |
ζ β −1
q−2 1
r m
s
1
1
log
1
1
s r
t2
t1
sk−1 (log s )q−2 ds
s−k−1
dm dr ds
p−2 1 r
dr ds
|t k − t k | t1
s
L1 s q−2 1 k−1 1 2 s log dr ds k k (q − 1 ) r r t1 t2 1 1 t2
s
q−2 s 1 + t2−k sk−1 log d r d s → 0, r r t1 1
independent of (u, v) ∈ as t2 − t1 → 0. Also
1
|Dδ T1 (u, v )(t2 ) − Dδ T1 (u, v )(t1 )| ≤ (1 − δ ) t2
−δ d t2 T1 (u, v )(s ) log ds t2 dt2 s s 1
d t1
t1 −δ T1 (u, v )(s ) − t1 log ds dt1 s s 1
d t1
− δ t 2 ≤ L1 M1 + L2 M2 t2 log dt2 1 s
−δ t1 d t1 t1 −δ ds ds d − log + t2 − t1 log s s dt2 dt1 1 s s t2
d t −δ ds dt2
log
t1
2
s
s
→ 0 as t2 − t1 → 0,
independent of (u, v) ∈ . Analogously we can have
|T2 (u, v )(t2 ) − T2 (u, v )(t1 )| → 0, |Dα T2 (u, v )(t2 ) − Dα T2 (u, v )(t1 )| → 0,
1 L N + L2 N2 , (1 − α ) 1 1
where N1 and N2 are respectively given by (3.6) and (3.7). In consequence, we get
T2 (u, v ) Y = T2 (u, v ) + Dα T2 (u, v )
1 ≤ 1+ (L1 N1 + L2 N2 ). (2 − α )
+
+ t2
≤ L1 N1 + L2 N2 , which leads to
| |
ζ
log
1
L2
×
p+γ
|A | |B2 | log η L2 2 + , | |(q − 1 ) ( p + 1 ) (γ + 1 )( p + 1 )
|T2 (u, v )(t )| ≤
1
sk−1 (log s )q−2 ds + t2−k
r k−1
|B2 |e−k (q − 1 )
q + β
T1 (u, v ) X = T1 (u, v ) + Dδ T1 (u, v )
1 ≤ 1+ (L1 M1 + L2 M2 ). (1 − δ )
t1k t2k
×
dm dr ds
t1
2
|A2 | × (q − 1 )(β )
s
r
|A2 | log ζ L1 |B2 | | | ≤ + + | |(q − 1 ) (q + 1 )(β + 1 ) (q + 1 ) q(q − 1 ) +
1
q−2 1 r
|t k − t k |
L1
| |
(3.17)
as t2 − t1 → 0, independent of (u, v) ∈ . Thus it follows the equicontinuity of T1 and T2 that operator T is equicontinuous. Therefore, by Arzela-Ascoli’s theorem, we deduce that the operator T is completely continuous. In the last step, we define a set E (T ) = {(u, v ) ∈ X × Y : (u, v ) = λT (u, v ); 0 ≤ λ ≤ 1} and show that it is bounded. Let (u, v ) ∈ E. Then (u, v ) = λT (u, v ). For any t ∈ [1, e], we have u(t ) = λT1 (u, v )(t ), v(t ) = λT2 (u, v )(t ). Thus
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
t 1 |u(t )| = |λT1 (u, v )(t )| ≤ t −k sk−1 (log s )q−2 ds | | 1 s p−2 |A2 |e−k e k−1 s
× s log ( p − 1 ) 1 r 1 1 × λ0 + λ1 |u(r )| + λ2 |Dδ u(r )| + λ3 |v(r )| dr ds r ζ
|A | ζ γ −1 2
+
(q − 1 )(β )
s
r
log
1
+
×
+
( p − 1 )(γ )
s
r
r k−1
log
1
log
1
1
r p−2
m
s
s
−k−1
(u, v ) X ×Y ≤
×
1
cos(x(t )) + |y(t )| + sin(D 2 y(t )) 1
5
1
sin(x(t )) + cos(D 3 x(t )) + |y(t )| 3t 2 + 25 + cos(t + 1 ), 1
=
(3.21)
equipped with nonlocal strip conditions:
u ( 1 ) = 0 , u ( e ) = I 1/2 v ( 2 ) (3.22)
k = 2, q = 3/2, p = 5/4, α = 1/2, δ = 1 /3, ζ = 3/2, η = 1 α 2, γ = 1/2, β = 1/4, f (t, x(t ), y(t ), D y(t )) = 30et cos(x(t )) +
1 (μ0 + μ1 u X + max{μ2 , μ3 } v Y )M1 (1 − δ )
+ (λ0 + max{λ1 , λ2 } u X + λ3 v Y )M2 .
1 |y(t )| + sin(D 2 y(t )) + 12 + t 1 sin(x(t ))+cos(D 3 x(t ))+|y (t )| 3t 2 +25
ties, we get l =
1 30
and
g(t, x(t ), Dδ x(t ), y(t )) =
+ cos(t + 1 ). From the following inequali-
and l1 =
1 25
:
| f (t, x1 (t ), y1 (t ), Dα y1 (t )) − f (t, x2 (t ), y2 (t ), Dα y2 (t ))|
1 = u + Dδ u ≤ 1 + (1 − δ )
≤
+ (λ0 + max{λ1 , λ2 } u X + λ3 v Y )M2 .
1 1 |x1 (t ) − x2 (t )| + |y1 (t ) − y2 (t )| + |D 2 y1 (t ) − D 2 y2 (t )| ,
25
(3.18)
1 (μ0 + μ1 u X + max{μ2 , μ3 } v Y )N1 (1 − α ) + (λ0 + max{λ1 , λ2 } u X + λ3 v Y )N2 . (3.19) Taking into account the estimates (3.18) and (3.19) together with (3.8)–(3.10), we have
1 u X + v Y = 1 + (1 − δ )
1 × μ0 M1 + λ0 M2 + 1 + μ0 N1 + λ0 N2 (1 − α )
1 + u X μ1 M1 + max{λ1 , λ2 }M2 1 + (1 − δ )
1 + μ1 N1 + max{λ1 , λ2 }N2 1 + (1 − α )
1 30
|g(t, x1 (t ), Dδ x1 (t ), y1 (t )) − g(t, x2 (t ), Dδ x2 (t ), y2 (t ))
1 1 1 |x1 (t ) − x2 (t )| + |D 3 x1 (t ) − D 3 x2 (t )| + |y1 (t ) − y2 (t )| . ≤
× (μ0 + μ1 u X + max{μ2 , μ3 } v Y )M1
v Y ≤ 1 +
Here,
Likewise, we can obtain
1 30et
v ( 1 ) = 0 , v ( e ) = I 1/4 u ( 3 /2 ) .
u ≤ (μ0 + μ1 u X + max{μ2 , μ3 } v Y )M1 + (λ0 + max{λ1 , λ2 } u X + λ3 v Y )M2 .
In consequence, we get
.
+ 12 + t, t ∈ [1, e], (D 4 + 2D 4 )y(t )
log
which, on taking the norm for t ∈ [1, e], yields
Dδ u ≤
(3.20)
Consider the following coupled system of Hadamard fractional differential equations
(q − 1 ) 1 r 1 1 × μ0 + μ1 |u(r )| + μ2 |v(r )| + μ3 |Dα v(r )| dr ds,
Also
3.1. Example
3
1
1+
This shows that E (T ) is bounded. Thus, by Lemma 3.2, the operator T has at least one fixed point. Consequently, problem (1.2 )has at least one solution on [1, e]. This completes the proof.
λ0 + λ1 |u(m )| + λ2 |Dδ u(m )| + λ3 |v(m )| dm dr ds m t
t
t −k s q−2 sk−1
ω1
1 − max{ω2 , ω3 }
(D 2 + 2D 2 )x(t ) =
r
u X
max{μ2 , μ3 }M1 + λ3 M2
which, in view of (u, v ) X ×Y = u X + v Y , yields
r q−2 × r k−1 log μ0 + μ1 | u ( m ) | m 1 1 1 + μ2 | v ( m ) | + μ 3 | D α v ( m ) | dm dr ds m
s |B2 |e−k e k−1 s q−2 + s log μ0 + μ1 | u ( r ) | (q − 1 ) 1 r 1 1 + μ2 |v(r )| + μ3 |Dα v(r )| dr ds r η
|B | η γ −1 2
1 (1 − δ )
1 + max{μ2 , μ3 }N1 + λ3 N2 1 + (1 − α ) ≤ ω1 + max{ω2 , ω3 } (u, v ) X ×Y ,
v Y
+
s−k−1
s
45
Using the given data, we find that A1 ≤ 2, |A2 | ≤ 3.42874, |B1 | ≤ 1.1212, B2 ≤ 4, | | ≤ 4.15578, M1 ≤ 2.6434, M2 ≤ 2.4659, N1 ≤ 1.1409, N2 ≤ 3.1476. Further
(lM1 + l1 M2 )((1 + (2/3 )) ≈ 0.32466 < 1/2, (2/3 ) (lN1 + l1 N2 )((1 + (1/2 ))) ≈ 0.25642 < 1/2. (1/2 ) Thus all the conditions of Theorem 3.3 are satisfied. Hence it follows by the conclusion of Theorem 3.3 that there exists a unique solution for problem (3.21)–(3.22) on [1, e].
Acknowledgements The authors thank the reviewers for their useful and constructive remarks that led to the improvement of the manuscript.
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S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
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Contents lists available at ScienceDirect
Chaos, Solitons and Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos
A coupled system of Hadamard type sequential fractional differential equations with coupled strip conditions Shorog Aljoudi a, Bashir Ahmad a, Juan J. Nieto b,∗, Ahmed Alsaedi a 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 b Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela, Santiago de Compostela, 15782, Spain
a r t i c l e
i n f o
Article history: Received 24 March 2016 Revised 5 May 2016 Accepted 8 May 2016
MSC: 34A08 34B10 34B15
a b s t r a c t We study a nonlocal boundary value problem of Hadamard type coupled sequential fractional differential equations supplemented with coupled strip conditions (nonlocal Riemann-Liouville integral boundary conditions). The nonlinearities in the coupled system of equations depend on the unknown functions as well as their lower order fractional derivatives. We apply Leray-Schauder alternative and Banach’s contraction mapping principle to obtain the existence and uniqueness results for the given problem. An illustrative example is also discussed. © 2016 Elsevier Ltd. All rights reserved.
Keywords: Hadamard fractional derivative Coupled system Strip conditions Existence Fixed point
1. Introduction Fractional calculus has emerged as an interesting field of investigation in the last few decades. The subject has been extensively developed and the literature on the topic is much enriched now, covering theoretical as well as widespread applications of this branch of mathematical analysis. In fact, the tools of fractional calculus have led to much improved, realistic and practical mathematical modeling of many systems and processes, occurring in engineering and scientific disciplines such as control theory, signal and image processing, biophysics, blood flow phenomena, etc. [1–3]. The nonlocal characteristic of fractional order operators, which takes into account the hereditary properties of phenomena involved, has greatly contributed to the popularity of the subject. In particular, the topic of boundary value problems of RiemannLiouville or Liouville-Caputo type fractional order differential equations equipped with a variety of boundary conditions has
∗
Corresponding author. Fax: +34881813197. E-mail addresses: [email protected] (S. Aljoudi), bashirahmad_qau@ yahoo.com (B. Ahmad), [email protected] (J.J. Nieto), [email protected] (A. Alsaedi). http://dx.doi.org/10.1016/j.chaos.2016.05.005 0960-0779/© 2016 Elsevier Ltd. All rights reserved.
attracted significant attention, for example, see [4–14] and the references cited therein. In the classical text [15], it has been mentioned that Hadamard in 1892 [16] suggested a concept of fractional integrod q differentiation in terms of the fractional power of the type (t dt )
d q in contrast to its Riemann-Liouville counterpart of the form ( dt ) . The kind of derivative introduced by Hadamard contains logarithmic function of arbitrary exponent in the kernel of the integral appearing in its definition. Hadamard’s construction is invariant in relation to dilation and is well suited to the problems containing half axes. In [18], the authors have shown that the Lamb-Bateman integral equation can be expressed in terms of Hadamard fractional derivatives of order 1/2. Following an approach based on Hadamard derivatives and fractional Hyper-Bessel-type operators, the authors in [17] discussed a modified Lamb-Bateman equation. In fact, there are not many applications of Hadamard integrals and derivatives in view of possibly the intrinsic intractability of these operators. However, one can find some recent results on Hadamard type fractional differential equations, for instance, see [2,19–23]. Coupled systems of fractional order differential equations have also been investigated by many authors. Such systems appear naturally in many real world situations, for example, see [24]. Some recent results on the topic can be found in a series of
40
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
papers [25–30] and the references cited therein. More recently, in [27], the authors discussed a coupled system of Hadamard type fractional differential equations with Hadamard integral boundary conditions of the form:
⎧ Dα u(t ) = f (t , u(t ), v(t )), Dβ v(t ) = g(t , u(t ), v(t )), ⎪ ⎪ ⎪ 1 < α , β ≤ 2, 1 < t < e, ⎨ γ −1 u(s) σ u(1 )=0, u(e ) = (1γ ) 1 1 log σs1 ds, γ > 0, 1 < σ1 < e, ⎪ s ⎪ ⎪ ⎩v(1 ) = 0, v(e ) = 1 σ2 log σ2 γ −1 v(s) ds, 1 < σ < e, 2 s s (γ ) 1
(1.1) where f, g : [1, e] × R × R → R are given continuous functions. In this paper, motivated by aforementioned applications of Hadamard fractional integro-differentiation as well as by the recent interest in developing the existence theory for Hadamard type initial and boundary value problems, we study a coupled system of Hadamard type sequential fractional differential equations equipped with nonlocal coupled strip conditions given by
⎧ q (D + kDq−1 )u(t ) = f (t, u(t ), v(t ), Dα v(t )), k > 0, ⎪ ⎪ ⎪ 1 < q 2, 0 < α < 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (D p + kD p−1 )v(t ) = g(t, u(t ), Dδ u(t ), v(t )), ⎪ ⎪ ⎪ ⎨ 1 < p 2, 0 < δ < 1, γ −1 v(s) η u(1 ) = 0, u(e ) = Iγ v(η ) = (1γ ) 1 log ηs ds, γ > 0, ⎪ s ⎪ ⎪ ⎪ 1 < η < e, ⎪ ⎪ ⎪ ⎪ ⎪ ζ ζ β −1 u(s ) ⎪ 1 β ⎪ ds, β > 0, ⎪v(1 ) = 0, v(e ) = I u(ζ ) = (β ) 1 log s s ⎩ 1 < ζ < e, (1.2) D(.)
I(.)
where and respectively denote the Hadamard fractional derivative and Hadamard fractional integral (to be defined later), and f, g : [1, e] × R3 → R are given continuous functions to be chosen appropriately (according to the requirement). We show the existence of solutions for problem (1.1) by applying Leray-Schauder alternative criterion while uniqueness of solutions for (1.1) relies on Banach’s contraction mapping principle. Though we make use of the standard methodology to obtain the desired results, yet its exposition to the given problem is new. Section 2 contains some basic concepts and an auxiliary lemma, while the main results are presented in Section 3. Here we remark that problem (1.2) is a generalization of the problem (1.1) in the sense that the coupled system in (1.2) consists of sequential fractional differential equations with the nonlinearities depending on the unknown functions as well as their lower order fractional derivatives, and coupled strip boundary conditions, whereas problem (1.1) deals with fractional Hadamard type equations with nonlinearities only involving unknown functions and uncoupled integral boundary conditions.
First of all, we recall some important definitions [2] and then prove an auxiliary lemma for the linear variant of problem (1.1). Definition 2.1. The Hadamard fractional integral of order q for a function g ∈ Lp [y, x], 0 ≤ y ≤ t ≤ x ≤ ∞, is defined as
1 (q )
t
log y
D g(t ) = q
t s
q−1 g(s ) s
ds, q > 0.
d Definition 2.2. Let [y, x] ⊂ R, δ = t dt and ACδn [y, x] = {g : [y, x] → n −1 R:δ [g(t )] ∈ AC[y, x]}. The Hadamard derivative of fractional or-
δ (I n
n−q
1 d )(t ) = t (n − q ) dt
n t
log 1
t s
n−q−1 g(s ) s
ds,
n − 1 < q < n, n = [q] + 1, where [q] denotes the integer part of the real number q and log(· ) = loge (· ). Recall that the Hadamard fractional derivative is the left-inverse operator to the Hadamard fractional integral in the space Lp [y, x], 0 < y < x < ∞, 1 ≤ p ≤ ∞, that is, Dq Iq f (t ) = f (t ) (Theorem 4.8, [31]). In [32], Caputo-type modification of the Hadamard fractional derivatives was proposed as follows: c
Dq g(t ) = Dq g(s ) −
n−1
δ k g( y )
k!
k=0
log
s y
k
(t ), t ∈ (y, x ).
Further, it was shown in (Theorem 2.1, [32]) that In−q δ n g(t ). For 0 < q < 1, it follows from (2.1) that c
(2.1) c Dq g(t )
=
Dq g(t ) = Dq g(s ) − g(y ) (t ).
Furthermore, it was established in Lemma 2.4 and Lemma 2.5 of [32] respectively that c
Dq (Iq g)(t ) = g(t ), Iq (c Dq )g(t ) = g(t ) −
n−1 k=0
δ k g( y )
k!
log
t y
k
. (2.2)
From the second formula in (2.2), one can easily infer that the solution of Hadamard differential equation: c Dq u(t ) = σ (t ) can be written as n−1
u(t ) = Iq σ (t ) +
k=0
δ k u (y )
k!
log
t y
k
,
for appropriate function u(t) and σ (t) (as required in the above definitions). Note that the Hadamard integral and derivative defined above are left-sided. One can define the Hadamard right-sided integral and derivative in the same way, for instance, see [32]. Lemma 2.1. Let h1 , h2 ∈ AC ([1, e], R ). Then the solution for the linear system of sequential fractional differential equations:
(Dq + kDq−1 )u(t ) = h1 (t ), (D p + kD p−1 )v(t ) = h2 (t ),
(2.3)
supplemented with the boundary conditions in (1.2) is given by a coupled system of Hadamard integral equations
u(t ) =
2. Preliminaries
Iq g(t ) =
der q for a function g ∈ ACδn [y, x] is defined as
1
t −k
t 1
e
s
e−k s p−2 h2 (r ) sk−1 log dr ds ( p − 1 ) 1 r r 1 ζ
1 ζ β −1 −k−1 − log s (q − 1 )(β ) 1 s
s
r
r q−2 h1 (m ) × r k−1 log dm dr ds m m 1 1 η
1 η γ −1 −k−1 − B2 log s ( p − 1 )(γ ) 1 s
s
r
r p−2 h2 (m ) × r k−1 log dm dr ds m m 1 1 ×
− A2
sk−1 (log s )q−2 ds
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
e
s
e−k s q−2 h1 (r ) sk−1 log dr ds (q − 1 ) 1 r r 1 t +t −k sk−1 Iq−1 h1 (s )ds,
−
Now, using the coupled integral boundary conditions given by (1.2) in (2.11) and (2.12), we obtain (2.4)
1
and
1
v(t ) =
t −k
t
1
s
×
sk−1 (log s ) p−2 ds
log 1
− B1
s q−2 h1 (r ) r
r
e−k (q − 1 )
A1 c1 + A2 d1 = J1 , B1 c1 + B2 d1 = J2 ,
e
sk−1
J1 =
1
dr ds
−
J2 =
(γ )( p − 1 ) 1 s r
r k−1
1
e
(q − 1 )
(2.5)
1
where
1 e
sk−1
r k−1
e
( p − 1 )
1 e
sk−1
A1 = e
×
e
s
k−1
1 s r
k−1
1
(β )
B2 = e−k
e
1
q−2
log
1
ds, A2 =
(log r )
ζ
−1
(log s )
(2.6)
p−2
s
s
(γ )
r m
p−2
s
s
−k−1
s
r
k−1
q−2
(log r )
dr ds,
ds.
(2.8)
Proof. As argued in [2], the general solution of the system (2.3) can be written as
u(t ) = c0 t −k + c1 t −k + t −k
t 1
t 1
t 1
1
t
sk−1 (log s )q−2 ds (2.9)
sk−1 (log s ) p−2 ds
sk−1 I p−1 h2 (s )ds,
(2.10)
where ci , di , (i = 0, 1 ) are unknown arbitrary constants. Making use of the conditions u(1 ) = 0 and v(1 ) = 0 in (2.9)–(2.10) yields c0 = 0 and d0 = 0, which leads to
u(t ) = c1 t −k
v(t ) = d1t −k
t 1
t 1
sk−1 (log s )q−2 ds + t −k
sk−1 (log s ) p−2 ds + t −k
c1 =
t
1
t 1
sk−1 Iq−1 h1 (s )ds,
(2.11)
sk−1 I p−1 h2 (s )ds.
(2.12)
1
×
r
r
dr ds,
(2.14)
s−k−1
s
q−2 h (m ) 1 m
log 1
1 (q − 1 )(β )
A2
s
r
k−1
1
dm dr ds
s p−2 h2 (r ) u
r
dr ds.
r
1
ζ
r log m s
log
ζ
(2.15)
1
β −1
s
q−2 h (m ) 1 m
s−k−1
dm dr ds
s q−2 h2 (r ) s log dr ds r r 1 1
e s
e−k s q−2 h1 (r ) k−1 + B2 s log dr ds (q − 1 ) 1 r r 1 η
1 η γ −1 −k−1 − log s ( p − 1 )(γ ) 1 s s r
r p−2 h2 (m ) × r k−1 log dm dr ds , m m 1 1 e−k − ( p − 1 )
e
k−1
(2.16)
η
1 η γ −1 −k−1 log s
( p − 1 )(γ ) 1 s s r
r p−2 h2 (m ) k−1 × r log dm dr ds m m 1 1
e s
e−k s q−2 h1 (r ) k−1 − s log dr ds (q − 1 ) 1 r r 1
e s
e−k s p−2 h2 (r ) k−1 + A1 s log dr ds ( p − 1 ) 1 r r 1
d1 =
sk−1 Iq−1 h1 (s )ds,
v(t ) = d0t −k + d1t −k + t −k
s q−2 h1 (r )
β −1
s
1
dm dr ds
Solving the system (2.13) for c1 and d1 and using the notation (2.6), we find that
−k−1
(2.7)
1
sk−1 (log s )
log
1
η γ −1
dr ds,
β −1
ζ
η
−1
ζ
log
= B1 A2 − A1 B2 = 0,
m
log
s−k−1
p−2 h (m ) 2 log
(β )(q − 1 ) 1 s r
s
1
ζ
1
η γ −1
s
1
1
−k
r m
log
−k
× −
η
log
1
×
η
η γ −1 −k−1 s k−1 1 log s r ( p − 1 )(γ ) 1 s 1
r
r p−2 h2 (m ) × log dm dr ds m m 1
β −1 ζ
s 1 ζ − A1 log s−k−1 r k−1 (q − 1 )(β ) 1 s 1
r
r q−2 h1 (m ) × log dm dr ds m m 1 e s
−k e s p−2 h2 (r ) − sk−1 log dr ds ( p − 1 ) 1 r r 1 t −k k−1 p−1 +t s I h2 (s )ds,
−k
(2.13)
where Ai and Bi (i = 1, 2 ) are respectively given by (2.7) and (2.8), and
−
B1 =
41
−
1
B1
β −1 ζ 1 ζ log s−k−1 (q − 1 )(β ) 1 s s r
r q−2 h1 (m ) × r k−1 log dm dr ds . m m 1 1
(2.17)
Substituting the values of c1 and d1 in (2.13), we get the desired solution (2.4)–(2.5). The converse of the lemma follows by direct computation. This completes the proof.
42
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
3. Existence and uniqueness results
Let X = u : u ∈ C ([1, e], R ) and Dδ u ∈ C ([1, e], R ) and Y = {v : v ∈ C ([1, e], R ) and Dα v ∈ C ([1, e], R )} denote the spaces respectively endowed with the norms u X = u + Dδ u = supt∈[1,e] |u(t )| + supt∈[1,e] |Dδ u(t )| and v Y = v + D α v = supt∈[1,e] |v(t )| + supt∈[1,e] |Dα v(t )|. Observe that ( X , . X ) and (Y, . Y ) are Banach spaces. In turn, the product space (X × Y, . X ×Y ) is a Banach space equipped with the norm (u, v ) X ×Y = u X + v Y for (u, v ) ∈ X × Y. In relation to problem (1.2), we introduce an operator T : X × Y → X × Y by Lemma 2.1 as follows:
T (u, v )(t ) := (T1 (u, v )(t ), T2 (u, v )(t )), where
T1 (u, v )(t ) =
+
A2
ζ
1
(β )
s 1
B2
1
log
1
η
1
r
k−1
log
I
1
+ B2 e + t −k 1
−k
e
1
t 1
s−k−1
η γ −1 s
s−k−1
sk−1 Iq−1 f (s, u(s ), v(s ), Dα v(s )) ds
(3.2)
t
sk−1 (log s ) p−2 ds
1 e
− B1 e−k sk−1 Iq−1 f (s, u(s ), v(s ), Dα v(s )) ds t −k
η
B1
(γ )
s 1
A1
1
1
+ A1 e−k + t −k
ζ
1
1
log
ζ
s−k−1
1
e
1
+
M2 =
Theorem 3.3. Let f, g : [1, e] × R3 → R be continuous functions satisfying the Lipschitz condition: (H1 ) | 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 | ), l, l1 > 0, ∀t ∈ [1, e], u j , v j ∈ R. Further,
(3.11)
2(r1 M1 + r2 M2 )(1 + (1 − δ )) , [(1 − δ ) − 2(1 + (1 − δ ))(lM1 + l1 M2 )]
Then we show that T Br ⊂ Br where (3.3)
|B2 | | | + , (q + 1 ) q(q − 1 ) (3.4)
(3.10)
Br = {(u, v ) ∈ X × Y : (u, v ) X ×Y ≤ r } For (u, v) ∈ Br , we have
q + β
| |(q − 1 ) (q + 1 )(β + 1 )
(3.9)
2(r1 N1 + r2 N2 )(1 + (1 − α )) , [(1 − α ) − 2(1 + (1 − α ))(lN1 + l1 N2 )]
For computational convenience, we set
M1 =
(3.8)
1 ω2 = μ1 M1 + max{λ1 , λ2 }M2 1 + (1 − δ )
1 + μ1 N1 + max{λ1 , λ2 }N2 1 + , (1 − α )
r ≥ max
sk−1 I p−1 g(s, u(s ), Dδ (s ), v(s ))ds.
|A2 | log ζ
1 μ0 M1 + λ0 M2 ω1 = 1 + (1 − δ )
1 μ0 N1 + λ0 N2 , + 1+ (1 − α )
s−k−1
sk−1 I p−1 g(s, u(s ), Dδ (s ), v(s )) ds
(3.7)
Proof. Let us set r1 = supt∈[1,e] f (t, 0, 0, 0 ) < ∞, r2 = supt∈[1,e] g(t, 0, 0, 0 ) < ∞ and define
β −1
1 |A1 | | | + + , | |( p − 1 ) (γ + 1 )( p + 1 ) ( p + 1 ) p( p − 1 )
where M1 , M2 , N1 , N2 are given by (3.7). Then the boundary value problem (1.2) has a unique solution.
s
p+γ |B1 | log η
lN1 + l1 N2 < (1 − α )/2(1 + (1 − α )),
r k−1 Iq−1 f (r, u(r ), v(r ), Dα (r )) dr ds
t
s
(3.6)
lM1 + l1 M2 < (1 − δ )/2(1 + (1 − δ )),
r k−1 I p−1 g(r, u(r ), Dδ u(r ), v(r )) dr ds
(β )
s
×
log
η γ −1
1 |B1 | + , | |( p − 1 ) (q + 1 ) (q + 1 )(β + 1 )
Now we are in a position to present our main results. The first result deals with the uniqueness of solutions for problem (1.2), which has been obtained by means of Banach’s contraction mapping principle.
g(r, u(r ), Dδ u(r ), v(r )) dr ds
N2 =
q + β |A1 | log ζ
1 ω3 = max{μ2 , μ3 }M1 + λ3 M2 1 + (1 − δ )
1 + max{μ2 , μ3 }N1 + λ3 N2 1 + . (1 − α )
sk−1 Iq−1 f (s, u(s ), v(s ), Dα v(s )) ds,
× −
s
p−1
1
+
ζ β −1
r k−1 Iq−1 f (r, u(r ), v(r ), Dα v(r )) dr ds
(γ )
s
×
T2 (u, v )(t ) =
t
sk−1 (log s )q−2 ds
1 e
× − A2 e−k sk−1 I p−1 g(s, u(s ), Dδ u(s ), v(s )) ds t −k
× −
(3.1)
N1 =
Similarly, we have
|g(t, u(t ), Dδ u(t ), v(t ))| ≤ l1 r + r2 .
p+γ
|A | |B2 | log η 1 2 + , | |(q − 1 ) ( p + 1 ) (γ + 1 )( p + 1 )
| f (t, u(t ), v(t ), Dα v(t ))| ≤ | f (t, u(t ), v(t ), Dα v(t )) − f (t, 0, 0, 0 )| + | f (t, 0, 0, 0 )| ≤ l[|u(t )| + |v(t )| + |Dα v(t )|] + r1 l[ u X + v Y ] + r1 l (u, v ) X ×Y + r1 lr + r1 .
(3.5)
Then
|T1 (u, v )(t )| (lM1 + l1 M2 )r + (r1 M1 + r2 M2 ),
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
and
|Dδ T1 (u, v )(t )|
d t
1
t
(1 − δ ) dt 1 (lM1 + l1 M2 )r + (r1 M1 + r2 M2 ) . (1 − δ )
log
t −δ |T1 (u, v )(s )| s
s
≤ ds
Therefore,
1 T1 (u, v ) X = T1 (u, v ) + Dδ T1 (u, v ) 1 + (1 − δ )
r × (lM1 + l1 M2 )r + (r1 M1 + r2 M2 ) ≤ (3.12) 2
In similar manner, we obtain
|T2 (u, v )(t )| ≤ (lN1 + l1 N2 )r + (r1 N1 + r2 N2 ), |Dα T2 (u, v )(t )| (lN1 + l1 N2 )r + (r1 N1 + r2 N2 ) . (1 − α ) In consequence, we get
T2 (u, v ) Y = T2 (u, v ) + Dα T2 (u, v )
r 1 1+ (lN1 + l1 N2 )r + (r1 N1 + r2 N2 ) ≤ (1 − α ) 2
t 1 |T1 (u1 , v1 )(t ) − T1 (u2 , v2 )(t )| ≤ t −k sk−1 (log s )q−2 ds | | 1
s
|A |e−k e p−2 s sk−1
+
log
1
r
×
1 g(r, u1 (r ), Dδ u1 (r ), v1 (r ))−g(r, u2 (r ), Dδ u2 (r ), v2 (r )) r dr ds ζ
s
r
|A | ζ γ −1 r q−2 2
log
s−k−1
r k−1
log
(q − 1 )(β ) 1 s m 1 1 α × f (m, u1 (m ), v1 (m ), D v1 (m ) 1 − f (m, u2 (m ), v2 (m ), Dα v2 (m ) dm dr ds m s
−k e q−2 |B2 |e s + sk−1 log f (r, u1 (r ), v1 (r ), Dα v1 (r ) (q − 1 ) 1 r 1 1 − f (r, u2 (r ), v2 (r ), Dα v2 (r ) dr ds r η
s
r
|B | η γ −1 r p−2 2
+
log
s−k−1
r k−1
r ≤ (lM1 + l1 M2 )[ u1 − u2 X + v1 − v2 Y ].
d t 1 |Dδ T1 (u1 , v1 )(t ) − Dδ T1 (u2 , v2 )(t )| ≤ t (1 − δ ) dt 1
t −δ 1 × log |T1 (u1 , v1 )(s ) − T1 (u2 , v2 )(s )|ds s
s
(3.14) Similarly, we can find that
T 2 ( u 1 , v 1 ) − T 2 ( u 2 , v 2 ) Y (1 + (1 − α ))(lN1 + l1 N2 ) ≤ [ u1 − u2 X + v1 − v2 Y ]. (1 − α ) (3.15)
= T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) X + T 2 ( u 1 , v 1 ) − T 2 ( u 2 , v 2 ) Y
(1 + (1 − δ ))(lM1 + l1 M2 ) (1 + (1 − α ))(lN1 + l1 N2 ) + (1 − δ ) (1 − α )
[ u1 − u2 X + v1 − v2 Y ]. In view of the assumption (3.11), there exists a positive number such that
(1 + (1 − δ ))(lM1 + l1 M2 ) (1 − δ ) (1 + (1 − α ))(lN1 + l1 N2 ) + ≤ < 1. (1 − α ) Therefore,
T (u1 , v1 ) − T (u2 , v2 ) X ×Y ≤ ( u1 − u2 X + v1 − v2 Y ). This shows that T is a contraction. Hence, by Banach’s fixed point theorem, the operator T has a unique fixed point which corresponds to a unique solution of problem (1.2). This completes the proof. Our next existence result is based on Leray-Schauder alternative. Lemma 3.2. (Leray-Schauder alternative [33]) Let F: E → E be a completely continuous operator. Let ε (F ) = {x ∈ E : x = λF (x ) for some 0 < λ < 1}. Then either the set ε(F) is unbounded or F has at least one fixed point.
log
( p − 1 )(γ ) 1 s m 1 1 δ ×g(m, u1 (m ), D u1 (m ), v1 (m )) 1 − g(m, u2 (m ), Dδ u2 (m ), v2 (m )) dm dr ds m t t
−k q−2 t s + sk−1 log f (r, u1 (r ), v1 (r ), Dα v1 (r )) (q − 1 ) 1 r 1 1 − f (r, u2 (r ), v2 (r ), Dα v2 (r )) dr ds
Also we have
T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) X = T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) + Dδ T 1 ( u 1 , v1 ) − Dδ T 1 ( u 2 , v2 ) (1 + (1 − δ ))(lM1 + l1 M2 ) ≤ [ u1 − u2 X + v1 − v2 Y ]. (1 − δ )
≤
which implies T Br ⊂ Br . Now we prove that the operator T is a contraction. For ui , vi ∈ Br ; i = 1, 2 and for each t ∈ [1, e] we have
1
From the above inequalities, we get
T (u1 , v1 ) − T (u2 , v2 ) X ×Y (3.13)
T (u, v ) X ×Y = T1 (u, v ) X + T2 (u, v ) Y r,
2
(lM1 + l1 M2 ) [ u1 − u2 X + v1 − v2 Y ]. (1 − δ )
Consequently, we obtain from (3.14 ) and (3.15 ) that
Thus, it follows from (3.12) and (3.13) that
( p − 1 )
43
Theorem 3.4. Let f, g : [1, e] × R3 → R be continuous functions and there exist real constants μ j , λ j ≥ 0 ( j = 1, 2, 3 ) and μ0 > 0, λ0 > 0 such that (H2 ) | f (t, x1 , x2 , x3 )| ≤ μ0 + μ1 |x1 | + μ2 |x2 | + μ3 |x3 |, |g(t, x1 , x2 , x3 )| ≤ λ0 + λ1 |x1 | + λ2 |x2 | + λ3 |x3 |, ∀x j ∈ R, j = 1, 2, 3. In addition it is assumed that max{ω2 , ω3 } < 1, where ω2 and ω3 are given by (3.19) and (3.20) respectively. Then the boundary value problem (1.2) has at least one solution [1, e]. Proof. Let us first show that the operator T : X × Y → X × Y is completely continuous. Clearly continuity of the functions f and g implies that the operators T1 and T2 are continuous. Hence the operator T is continuous. In order to establish that the operator T is uniformly bounded, we consider a bounded set ⊂ X × Y such that |f(t, u(t), v(t), Dα v(t))| ≤ L1 , |g(t, u(t), Dδ u(t), v(t))| ≤ L2 , ∀(u, v) ∈ , where L1 and L2 are positive constants. Then, for any (u, v) ∈ , we have
44
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
t L |T1 (u, v )(t )| ≤ 1 t −k sk−1 (log s )q−2 ds | | 1 |B |e−k e
s
s q−2 1 2
×
sk−1
log
From (3.16) and (3.17), we infer that T1 and T2 are uniformly bounded, which implies that the operator T is uniformly bounded. Next, we show that T is equicontinuous. Let t1 , t2 ∈ [1, e] with t1 < t2 . Then we have
dr ds
(q − 1 ) 1 r r 1 ζ
|A2 | ζ β −1 −k−1 + log s (q − 1 )(β ) 1 s
s
r
r q−2 1 r k−1
×
log
1
1
m
m
|T1 (u, v )(t2 ) − T1 (u, v )(t1 )| ≤ ×
dm dr ds
t L2 + t −k sk−1 (log s )q−2 ds | | 1 |A |e−k e
s
s p−2 1 2 × sk−1 log dr ds ( p − 1 ) 1 r r 1 η
γ −1 |B2 | η + log s−k−1 ( p − 1 )(γ ) 1 s
s
r
r p−2 1 r k−1
×
log
1
+ L1 t −k
1
t
sk−1
s
1
1
m
s log r
m
+
+
dr ds
×
which, on taking the norm for t ∈ [1, e] and using (3.4), (3.5) yields
T1 (u, v ) ≤ L1 M1 + L2 M2 . By definition of Hadamard fractional derivative with 0 < δ < 1, we get
d t
1 t −δ |T1 (u, v )(s )| t log ds (1 − δ ) dt 1 s s 1 ≤ (L M + L2 M2 ). (1 − δ ) 1 1
|Dδ T1 (u, v )(t )| ≤
Hence
(3.16)
Similarly, we can get
L1
| | ( p − 1 ) q + β
|A1 | log ζ
L2 |B1 | + + (q + 1 ) (β + 1 )(q + 1 ) | | ( p − 1 )
p+γ |B1 | log η |A1 | | | × + + (γ + 1 )( p + 1 ) ( p + 1 ) p( p − 1 ) ×
|Dα T2 (u, v )(t )| ≤
+
e
sk−1
log
t1
2
t1k t2k
1
|B2 | ( p − 1 )(γ )
s
r
r k−1
η
1
1
|A2 |e−k ( p − 1 )
e
sk−1
1
log
r m
1
s−k−1
dm dr ds
r
η γ −1 s
m log
sk−1 (log s )q−2 ds
q−2 1
s r
p−2 1
s
t2
t1
dr ds
sk−1 (log s )q−2 ds + t2−k
log
1
s
m
1
|t k − t k |
ζ β −1
q−2 1
r m
s
1
1
log
1
1
s r
t2
t1
sk−1 (log s )q−2 ds
s−k−1
dm dr ds
p−2 1 r
dr ds
|t k − t k | t1
s
L1 s q−2 1 k−1 1 2 s log dr ds k k (q − 1 ) r r t1 t2 1 1 t2
s
q−2 s 1 + t2−k sk−1 log d r d s → 0, r r t1 1
independent of (u, v) ∈ as t2 − t1 → 0. Also
1
|Dδ T1 (u, v )(t2 ) − Dδ T1 (u, v )(t1 )| ≤ (1 − δ ) t2
−δ d t2 T1 (u, v )(s ) log ds t2 dt2 s s 1
d t1
t1 −δ T1 (u, v )(s ) − t1 log ds dt1 s s 1
d t1
− δ t 2 ≤ L1 M1 + L2 M2 t2 log dt2 1 s
−δ t1 d t1 t1 −δ ds ds d − log + t2 − t1 log s s dt2 dt1 1 s s t2
d t −δ ds dt2
log
t1
2
s
s
→ 0 as t2 − t1 → 0,
independent of (u, v) ∈ . Analogously we can have
|T2 (u, v )(t2 ) − T2 (u, v )(t1 )| → 0, |Dα T2 (u, v )(t2 ) − Dα T2 (u, v )(t1 )| → 0,
1 L N + L2 N2 , (1 − α ) 1 1
where N1 and N2 are respectively given by (3.6) and (3.7). In consequence, we get
T2 (u, v ) Y = T2 (u, v ) + Dα T2 (u, v )
1 ≤ 1+ (L1 N1 + L2 N2 ). (2 − α )
+
+ t2
≤ L1 N1 + L2 N2 , which leads to
| |
ζ
log
1
L2
×
p+γ
|A | |B2 | log η L2 2 + , | |(q − 1 ) ( p + 1 ) (γ + 1 )( p + 1 )
|T2 (u, v )(t )| ≤
1
sk−1 (log s )q−2 ds + t2−k
r k−1
|B2 |e−k (q − 1 )
q + β
T1 (u, v ) X = T1 (u, v ) + Dδ T1 (u, v )
1 ≤ 1+ (L1 M1 + L2 M2 ). (1 − δ )
t1k t2k
×
dm dr ds
t1
2
|A2 | × (q − 1 )(β )
s
r
|A2 | log ζ L1 |B2 | | | ≤ + + | |(q − 1 ) (q + 1 )(β + 1 ) (q + 1 ) q(q − 1 ) +
1
q−2 1 r
|t k − t k |
L1
| |
(3.17)
as t2 − t1 → 0, independent of (u, v) ∈ . Thus it follows the equicontinuity of T1 and T2 that operator T is equicontinuous. Therefore, by Arzela-Ascoli’s theorem, we deduce that the operator T is completely continuous. In the last step, we define a set E (T ) = {(u, v ) ∈ X × Y : (u, v ) = λT (u, v ); 0 ≤ λ ≤ 1} and show that it is bounded. Let (u, v ) ∈ E. Then (u, v ) = λT (u, v ). For any t ∈ [1, e], we have u(t ) = λT1 (u, v )(t ), v(t ) = λT2 (u, v )(t ). Thus
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
t 1 |u(t )| = |λT1 (u, v )(t )| ≤ t −k sk−1 (log s )q−2 ds | | 1 s p−2 |A2 |e−k e k−1 s
× s log ( p − 1 ) 1 r 1 1 × λ0 + λ1 |u(r )| + λ2 |Dδ u(r )| + λ3 |v(r )| dr ds r ζ
|A | ζ γ −1 2
+
(q − 1 )(β )
s
r
log
1
+
×
+
( p − 1 )(γ )
s
r
r k−1
log
1
log
1
1
r p−2
m
s
s
−k−1
(u, v ) X ×Y ≤
×
1
cos(x(t )) + |y(t )| + sin(D 2 y(t )) 1
5
1
sin(x(t )) + cos(D 3 x(t )) + |y(t )| 3t 2 + 25 + cos(t + 1 ), 1
=
(3.21)
equipped with nonlocal strip conditions:
u ( 1 ) = 0 , u ( e ) = I 1/2 v ( 2 ) (3.22)
k = 2, q = 3/2, p = 5/4, α = 1/2, δ = 1 /3, ζ = 3/2, η = 1 α 2, γ = 1/2, β = 1/4, f (t, x(t ), y(t ), D y(t )) = 30et cos(x(t )) +
1 (μ0 + μ1 u X + max{μ2 , μ3 } v Y )M1 (1 − δ )
+ (λ0 + max{λ1 , λ2 } u X + λ3 v Y )M2 .
1 |y(t )| + sin(D 2 y(t )) + 12 + t 1 sin(x(t ))+cos(D 3 x(t ))+|y (t )| 3t 2 +25
ties, we get l =
1 30
and
g(t, x(t ), Dδ x(t ), y(t )) =
+ cos(t + 1 ). From the following inequali-
and l1 =
1 25
:
| f (t, x1 (t ), y1 (t ), Dα y1 (t )) − f (t, x2 (t ), y2 (t ), Dα y2 (t ))|
1 = u + Dδ u ≤ 1 + (1 − δ )
≤
+ (λ0 + max{λ1 , λ2 } u X + λ3 v Y )M2 .
1 1 |x1 (t ) − x2 (t )| + |y1 (t ) − y2 (t )| + |D 2 y1 (t ) − D 2 y2 (t )| ,
25
(3.18)
1 (μ0 + μ1 u X + max{μ2 , μ3 } v Y )N1 (1 − α ) + (λ0 + max{λ1 , λ2 } u X + λ3 v Y )N2 . (3.19) Taking into account the estimates (3.18) and (3.19) together with (3.8)–(3.10), we have
1 u X + v Y = 1 + (1 − δ )
1 × μ0 M1 + λ0 M2 + 1 + μ0 N1 + λ0 N2 (1 − α )
1 + u X μ1 M1 + max{λ1 , λ2 }M2 1 + (1 − δ )
1 + μ1 N1 + max{λ1 , λ2 }N2 1 + (1 − α )
1 30
|g(t, x1 (t ), Dδ x1 (t ), y1 (t )) − g(t, x2 (t ), Dδ x2 (t ), y2 (t ))
1 1 1 |x1 (t ) − x2 (t )| + |D 3 x1 (t ) − D 3 x2 (t )| + |y1 (t ) − y2 (t )| . ≤
× (μ0 + μ1 u X + max{μ2 , μ3 } v Y )M1
v Y ≤ 1 +
Here,
Likewise, we can obtain
1 30et
v ( 1 ) = 0 , v ( e ) = I 1/4 u ( 3 /2 ) .
u ≤ (μ0 + μ1 u X + max{μ2 , μ3 } v Y )M1 + (λ0 + max{λ1 , λ2 } u X + λ3 v Y )M2 .
In consequence, we get
.
+ 12 + t, t ∈ [1, e], (D 4 + 2D 4 )y(t )
log
which, on taking the norm for t ∈ [1, e], yields
Dδ u ≤
(3.20)
Consider the following coupled system of Hadamard fractional differential equations
(q − 1 ) 1 r 1 1 × μ0 + μ1 |u(r )| + μ2 |v(r )| + μ3 |Dα v(r )| dr ds,
Also
3.1. Example
3
1
1+
This shows that E (T ) is bounded. Thus, by Lemma 3.2, the operator T has at least one fixed point. Consequently, problem (1.2 )has at least one solution on [1, e]. This completes the proof.
λ0 + λ1 |u(m )| + λ2 |Dδ u(m )| + λ3 |v(m )| dm dr ds m t
t
t −k s q−2 sk−1
ω1
1 − max{ω2 , ω3 }
(D 2 + 2D 2 )x(t ) =
r
u X
max{μ2 , μ3 }M1 + λ3 M2
which, in view of (u, v ) X ×Y = u X + v Y , yields
r q−2 × r k−1 log μ0 + μ1 | u ( m ) | m 1 1 1 + μ2 | v ( m ) | + μ 3 | D α v ( m ) | dm dr ds m
s |B2 |e−k e k−1 s q−2 + s log μ0 + μ1 | u ( r ) | (q − 1 ) 1 r 1 1 + μ2 |v(r )| + μ3 |Dα v(r )| dr ds r η
|B | η γ −1 2
1 (1 − δ )
1 + max{μ2 , μ3 }N1 + λ3 N2 1 + (1 − α ) ≤ ω1 + max{ω2 , ω3 } (u, v ) X ×Y ,
v Y
+
s−k−1
s
45
Using the given data, we find that A1 ≤ 2, |A2 | ≤ 3.42874, |B1 | ≤ 1.1212, B2 ≤ 4, | | ≤ 4.15578, M1 ≤ 2.6434, M2 ≤ 2.4659, N1 ≤ 1.1409, N2 ≤ 3.1476. Further
(lM1 + l1 M2 )((1 + (2/3 )) ≈ 0.32466 < 1/2, (2/3 ) (lN1 + l1 N2 )((1 + (1/2 ))) ≈ 0.25642 < 1/2. (1/2 ) Thus all the conditions of Theorem 3.3 are satisfied. Hence it follows by the conclusion of Theorem 3.3 that there exists a unique solution for problem (3.21)–(3.22) on [1, e].
Acknowledgements The authors thank the reviewers for their useful and constructive remarks that led to the improvement of the manuscript.
46
S. Aljoudi et al. / Chaos, Solitons and Fractals 91 (2016) 39–46
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