J. Appl. Math. Comput. DOI 10.1007/s12190-014-0843-9 ORIGINAL RESEARCH

Positive solutions for singular Hadamard fractional differential system with four-point coupled boundary conditions Wengui Yang

Received: 16 June 2014 © Korean Society for Computational and Applied Mathematics 2014

Abstract In this paper, we investigate the four-point coupled boundary value problem of nonlinear semipositone Hadamard fractional differential equations

D α u(t) + λ f (t, u(t), v(t)) = 0,

D β v(t) + λg(t, u(t), v(t)) = 0, t ∈ (1, e), λ > 0,

u ( j) (1) = v ( j) (1) = 0, 0 ≤ j ≤ n − 2, u(e) = av(ξ ), v(e) = bu(η), ξ, η ∈ (1, e),

where λ, a, b are three parameters with 0 < ab(log η)α−1 (log ξ )β−1 < 1, α, β ∈ (n − 1, n] are two real numbers and n ≥ 3, D α , D β are the Hadamard fractional derivative of fractional order, and f, g are continuous and may be singular at t = 0 and t = 1. We firstly give the corresponding Green’s function for the boundary value problem and some of its properties. Moreover, by applying Guo-Krasnoselskii fixed point theorems, we derive an interval of λ such that any λ lying in this interval, the singular boundary value problem has at least one positive solution. As applications, two interesting examples are presented to illustrate the main results. Keywords Hadamard fractional differential equations · Four-point coupled boundary conditions · Positive solution · Green’s function · Fixed point theorems Mathematics Subject Classification

34A08 · 34A12 · 34B15

W. Yang (B) Ministry of Public Education, Sanmenxia Polytechnic, Sanmenxia 472000, Henan, China e-mail: [email protected]

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1 Introduction We consider the following four-point coupled boundary value problem of nonlinear semipositone Hadamard fractional differential equations

D α u(t) + λ f (t, u(t), v(t)) = 0, D β v(t) + λg(t, u(t), v(t)) = 0, t∈ (1, e), λ > 0, u ( j) (1) = v ( j) (1) = 0, 0 ≤ j ≤ n−2, u(e) = av(ξ ), v(e) = bu(η), ξ, η ∈ (1, e), (1.1) where λ, a, b are three parameters with 0 < ab(log η)α−1 (log ξ )β−1 < 1, α, β ∈ (n − 1, n] are two real numbers and n ≥ 3, D α , D β are the Hadamard fractional derivative of fractional order, and f, g are continuous and may be singular at t = 0 and t = 1. To the best of author’s knowledge, there are few results available in the literature to study the important problems for existence of positive solutions for nonlinear boundary value problems for coupled systems of Hadamard fractional differential equations. The recent studies on fractional differential equations has gained considerable popularity and importance mainly due to both the intensive development of the theory of fractional calculus itself and the applications of such constructions in numerous seemingly diverse and widespread fields of science and engineering, such as physics, mechanics, chemistry, engineering, etc. For the recent development of the theory and applications of fractional calculus and boundary value problems for nonlinear fractional differential equations, we refer the reader to a series of books and papers [1,3,5,9,13,19,21,22,25,26,29–31] and the references therein. Besides, many researchers [2,4,15,17,23,24,27,28,32,35] have established the existence and uniqueness for solutions or positive solutions of some coupled systems of nonlinear fractional differential equations. For example, Yuan [33] studied the multiple positive solutions to the (n − 1, 1)-type integral boundary value problems for systems of nonlinear semipositone fractional differential equations. Under different conditions, Yuan et al. [34] and Jiang et al. [20] considered the positive solutions to the four-point coupled boundary value problems for systems of nonlinear semipositone fractional differential equations, respectively. However, it has been noticed that most of the work on the topic is based on Riemann– Liouville and Caputo type fractional differential equations in the last few years. Except the Riemann–Liouville and Caputo derivatives in the literature, another kind of fractional derivatives appears due to Hadamard introduced in 1892, which differs from the preceding ones in the sense that the kernel of the integral and derivative contain logarithmic function of arbitrary exponent. Details and properties of Hadamard fractional derivative and integral can be found in [10–12,14,18,21]. Recently, few results have been obtained on Hadamard type fractional differential equations/inclusions. Ahmad and Ntouyas [6,7] studied the existence and uniqueness of solutions for fractional integral boundary value problem involving Hadamard type fractional differential equations/systems with integral boundary conditions by applying some standard fixed point theorems, respectively. Ahmad, Ntouyas and Alsaedi [8] investigated the existence of solutions for fractional boundary value problem involving Hadamard-type

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fractional differential inclusions and integral boundary conditions based on standard fixed point theorems for multivalued maps. Motivated by the results mentioned above, we consider the existence of positive solutions for singular Hadamard fractional differential equations boundary value problem (1.1). In Sect. 2, we present some preliminaries and lemmas that will be used to prove our main results. We also obtain the corresponding Green’s function and some of its properties. The main theorems are formulated and proved in Sect. 3. Furthermore, two examples are given to illustrate our main results in Sect. 4. 2 Preliminaries For the convenience of the reader, we firstly present some basic concepts of Hadamard type fractional calculus to facilitate analysis of problem (1.1). Definition 2.1 [21] The Hadamard derivative of fractional order q for a function g : [1, ∞) → R is defined as      t n−q−1 g(s) d n t 1 log t ds, n − 1 < q < n, D g(t) = (n − q) dt s s 1 q

where n = [q] + 1, [q] denotes the integer part of the real number q and log(·) = loge (·). Definition 2.2 [21] The Hadamard fractional integral of order q for a function g : [1, ∞) → R is defined as I q g(t) =

1 (q)

  t t q−1 g(s) log ds, q > 0, s s 1

provided the integral exists. Set q (t) = (log t)q−1 (1 − log t) and ρq (t) = (1 − log t)q−1 log t, for q > 2, t ∈ [1, e], and 1 G q (t, s) = (q)



(log t)q−1 (1 − log s)q−1 − (log(t/s))q−1 , 1 ≤ s ≤ t ≤ e, (2.1) 1 ≤ t ≤ s ≤ e. (log t)q−1 (1 − log s)q−1 ,

Lemma 2.3 The function G q (t, s) defined by (2.1) has the following properties: (I) G q (t, s) is continuous function on (t, s) ∈ [1, e]2 and G q (t, s) > 0, for t, s ∈ (1, e); (II) q (t)ρq (s) ≤ (q)G q (t, s) ≤ (q − 1)ρq (s), for t, s ∈ [1, e]; (III) q (t)ρq (s) ≤ (q)G q (t, s) ≤ (q − 1)q (t), for t, s ∈ [1, e].

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Proof The continuity of G q is easily checked. For 1 ≤ s ≤ t ≤ e, then we have  (q)G q (t, s) = (log t)q−1 (1−log s)q−1 −(log(t/s))q−1 = (q −1)

(1−log s) log t

x q−2 d x

log t−log s

≤ (q − 1)((1 − log s) log t)q−2 ((1 − log s) log t − (log t − log s)) = (q − 1)(1 − log s)q−2 (log t)q−2 (1 − log t) log s ≤ (q − 1)(1 − log s)q−2 (1 − log s) log s = (q − 1)(1 − log s)q−1 log s = (q − 1)ρq (s), and (q)G q (t, s) = (log t)q−1 (1 − log s)q−1 − (log(t/s))q−1 = (log t −log t log s)q−2 (log t −log t log s) − (log t −log s)q−2 (log t −log s) ≥ (log t −log t log s)q−2 (log t −log t log s) − (log t −log t log s)q−2 (log t −log s) ≥ (log t)q−2 (1 − log s)q−2 (1 − log t) log s ≥ (log t)q−1 (1 − log t)(1 − log s)q−1 log s = q (t)ρq (s), For 1 ≤ t ≤ s ≤ e, then we have (q)G q (t, s) = (log t)q−1 (1 − log s)q−1 = (log t)q−2 (1 − log s)q−1 log t ≤ (q − 1)(log t)q−2 (1 − log s)q−1 log s ≤ (q − 1)(1 − log s)q−1 log s = (q − 1)ρq (s), and (q)G q (t, s) = (log t)q−1 (1 − log s)q−1 ≥ (log t)q−1 (1 − log t)(1 − log s)q−1 log s = q (t)ρq (s). Nextly, we prove the right side of (III). For 1 ≤ s ≤ t ≤ e, then we have (q)G q (t, s) ≤ (q − 1)(log t)q−2 (1 − log s)q−2 (1 − log t) log s ≤ (q − 1)(log t)q−2 (1 − log s)q−2 (1 − log t) log t = (q − 1)(log t)q−1 (1 − log t) = (q − 1)q (t).

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For 1 ≤ t ≤ s ≤ e, then we have (q)G q (t, s) = (log t)q−1 (1 − log s)q−1 = (log t)q−1 (1 − log s)(1 − log s)q−2 ≤ (q − 1)(log t)q−1 (1 − log t)(1 − log s)q−2 ≤ (q − 1)(log t)q−1 (1 − log t) = (q − 1)q (t), This completes the proof of the lemma.

 

Now we derive the corresponding Green’s function for boundary value problem (1.1), and obtain some properties of the Green’s function. Lemma 2.4 Let x, y ∈ C[0, 1] be given functions. Then the boundary value problem D α u(t) + x(t) = 0, D β v(t) + y(t) = 0, t ∈ (1, e), u ( j) (1) = v ( j) (1) = 0, 0 ≤ j ≤ n−2, u(e) = av(ξ ), v(e) = bu(η), ξ, η ∈ (1, e), (2.2) has an integral representation 

e e u(t) = 1 K 1 (t, s) x(s) ds + 1 H1 (t, s) y(s) s s ds, e e y(s) x(s) v(t) = 1 K 2 (t, s) s ds + 1 H2 (t, s) s ds,

(2.3)

where ab(log ξ )β−1 (log t)α−1 G α (η, s), 1 − ab(log η)α−1 (log ξ )β−1 ab(log η)α−1 (log t)β−1 K 2 (t, s) = G β (t, s) + G α (ξ, s), 1 − ab(log η)α−1 (log ξ )β−1 a(log t)α−1 H1 (t, s) = G β (ξ, s), 1 − ab(log η)α−1 (log ξ )β−1 b(log t)β−1 H2 (t, s) = G α (η, s). 1 − ab(log η)α−1 (log ξ )β−1 K 1 (t, s) = G α (t, s) +

(2.4) (2.5)

(2.6)

Proof As argued in [21], the solution of Hadamard differential system in (2.2) can be written the following equivalent integral equations u(t) = c11 (log t)α−1 + c12 (log t)α−2 + · · · + c1n (log t)α−n   t t α−1 x(s) 1 log ds, − (α) 1 s s v(t) = c21 (log t)β−1 + c22 (log t)β−2 + · · · + c2n (log t)β−n   t t β−1 y(s) 1 log ds. − (β) 1 s s

(2.7)

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From u ( j) (1) = v ( j) (1) = 0, 0 ≤ j ≤ n − 2, we have cin = ci(n−1) = · · · = ci2 = 0 (i = 1, 2). Thus, (2.7) reduces to   t t α−1 log u(t) = c11 (log t) s 1   t t β−1 1 log v(t) = c21 (log t)β−1 − (β) 1 s α−1

x(s) ds, s

1 − (α)

y(s) ds. s

(2.8)

Using the boundary conditions u(e) = av(ξ ) and v(e) = bu(η), from (2.8), we obtain c11 −a(log ξ )β−1 c21 =



e

1

c21 −b(log η)α−1 c11 =



e 1

a (1 − log s)α−1 x(s) ds − (α) s (β) (1 − log s)β−1 (β)

y(s) b ds − s (α)



ξ

 log

1



η 1

ξ s

β−1

y(s) ds, s

 η α−1 x(s) ds. log s s (2.9)

Solving for c11 and c21 in (2.9), we have c11

c21

 e (1 − log s)α−1 x(s) 1 ds = 1 − ab(log η)α−1 (log ξ )β−1 (α) s 1

  ξ ξ β−1 y(s) a ds log − (β) 1 s s  e a(log ξ )β−1 (1 − log s)β−1 y(s) + ds α−1 β−1 1 − ab(log η) (log ξ ) (β) s 1   η η α−1 x(s) b ds , log − (α) 1 s s  e (1 − log s)β−1 y(s) 1 ds = 1 − ab(log η)α−1 (log ξ )β−1 (β) s 1   η η α−1 x(s) b ds log − (α) 1 s s  e b(log η)α−1 (1 − log s)α−1 x(s) + ds 1 − ab(log η)α−1 (log ξ )β−1 (α) s 1

  ξ ξ β−1 y(s) a ds . (2.10) log − (β) 1 s s

Substituting (2.10) into (2.8), we get u(t) = −

1 (α) +

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  t t α−1 x(s) log ds s s 1

(log t)α−1 1 − ab(log η)α−1 (log ξ )β−1

 1

e

(1 − log s)α−1 x(s) ds (α) s

Solutions for singular Hadamard fractional differential system

a − (β)



ξ 1



ξ log s

y(s) ds s  e α−1

β−1

a(log ξ )β−1 (log t) (1 − log s)β−1 y(s) ds 1 − ab(log η)α−1 (log ξ )β−1 (β) s 1   η η α−1 x(s) b ds log − (α) 1 s s   e  t t α−1 x(s) (1 − log s)α−1 x(s) 1 log ds + (log t)α−1 ds =− (α) 1 s s (α) s 1  e (1 − log s)α−1 x(s) α−1 − (log t) ds (α) s 1  e (1 − log s)α−1 x(s) (log t)α−1 + ds 1 − ab(log η)α−1 (log ξ )β−1 1 (α) s  η ab(log ξ )β−1 (log t)α−1 log(η/s))α−1 x(s) − ds 1 − ab(log η)α−1 (log ξ )β−1 1 (α) s  e a(log ξ )β−1 (log t)α−1 (1 − log s)β−1 y(s) + ds α−1 β−1 1 − ab(log η) (log ξ ) (β) s 1  ξ a(log t)α−1 (log(ξ/s))β−1 y(s) − ds α−1 β−1 1 − ab(log η) (log ξ ) (β) s 1  e  e ab(log ξ )β−1 (log t)α−1 x(s) x(s) = ds + ds G α (t, s) G α (η, s) α−1 (log ξ )β−1 s 1 − ab(log η) s 1 1  e a(log t)α−1 y(s) + ds G β (ξ, s) α−1 β−1 1 − ab(log η) (log ξ ) s 1   e ab(log ξ )β−1 (log t)α−1 x(s) = G α (t, s) + ds G (η, s) α α−1 β−1 1 − ab(log η) (log ξ ) s 1  e a(log t)α−1 y(s) + ds. G (ξ, s) α−1 (log ξ )β−1 β 1 − ab(log η) s 1 +

Similarly, we have 

e

v(t) = 1

 G β (t, s) +



+

e

1

ab(log η)α−1 (log t)β−1 G α (ξ, s) 1 − ab(log η)α−1 (log ξ )β−1



y(s) ds s

b(log t)β−1 x(s) ds. G α (η, s) 1 − ab(log η)α−1 (log ξ )β−1 s

Hence, we have 

u(t) = v(t) =

e 1e 1

K 1 (t, s) x(s) s ds + K 2 (t, s) y(s) s ds

This completes the proof of the lemma.

+

e 1e 1

H1 (t, s) y(s) s ds, H2 (t, s) x(s) s ds,  

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Employing Lemma 2.4, the system (1.1) can be expressed as 

e  e , u(t) = λ 1 K 1 (t, s) f (s, u(s), v(s)) ds + 1 H1 (t, s)g(s, u(s), v(s)) ds s s e  e ds ds v(t) = λ 1 K 2 (t, s)g(s, u(s), v(s)) s + 1 H2 (t, s) f (s, u(s), v(s)) s . (2.11)

Lemma 2.5 For t, s ∈ [1, e], the functions K 1 (t, s) and H1 (t, s) defined by (2.4) and (2.6) satisfy ab(log ξ )β−1 α (η) (log t)α−1 ρα (s) ≤ K 1 (t, s) (1 − ab(log η)α−1 (log ξ )β−1 )(α) ab(log ξ )β−1 (1 − (log η)α−1 ) ρα (s), ≤ (1 − ab(log η)α−1 (log ξ )β−1 )(α − 1) aβ (ξ ) (log t)α−1 ρβ (s) ≤ H1 (t, s) (1 − ab(log η)α−1 (log ξ )β−1 )(β) a ρβ (s), ≤ α−1 (1 − ab(log η) (log ξ )β−1 )(β − 1) ab(log ξ )β−1 (1 − (log η)α−1 ) (log t)α−1 , K 1 (t, s) ≤ (1 − ab(log η)α−1 (log ξ )β−1 )(α − 1) a (log t)α−1 . H1 (t, s) ≤ (1 − ab(log η)α−1 (log ξ )β−1 )(β − 1)

(2.12)

(2.13)

(2.14)

Proof First, we will show that (2.12) is true. By (2.4) and Lemma 2.3, we obtain ab(log ξ )β−1 (log t)α−1 G α (η, s) 1 − ab(log η)α−1 (log ξ )β−1 α−1 ab(log ξ )β−1 α−1 ρα (s) + ρα (s) ≤ · α−1 β−1 (α) 1 − ab(log η) (log ξ ) (α) ab(log ξ )β−1 (1 − (log η)α−1 ) ρα (s), t, s ∈ [1, e]. = (1 − ab(log η)α−1 (log ξ )β−1 )(α − 1)

K 1 (t, s) = G α (t, s) +

On the other hand, by (2.4) and Lemma 2.3, we also have ab(log ξ )β−1 (log t)α−1 G α (η, s) 1 − ab(log η)α−1 (log ξ )β−1 ab(log ξ )β−1 (log t)α−1 ≥ G α (η, s) 1 − ab(log η)α−1 (log ξ )β−1 ab(log ξ )β−1 (log t)α−1 α (η)ρα (s) ≥ · α−1 β−1 1 − ab(log η) (log ξ ) (α) β−1 α−1 α (η)(log t) ab(log ξ ) ρα (s), t, s ∈ [1, e]. = (1 − ab(log η)α−1 (log ξ )β−1 )(α)

K 1 (t, s) = G α (t, s) +

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Solutions for singular Hadamard fractional differential system

Next we show that (2.13) holds for t, s ∈ [1, e]. In fact, using (2.6) and Lemma 2.3, we get a(log t)α−1 a G β (ξ, s) ≤ G β (ξ, s) 1−ab(log η)α−1 (log ξ )β−1 1−ab(log η)α−1 (log ξ )β−1 a β −1 ≤ ρβ (s) · 1 − ab(log η)α−1 (log ξ )β−1 (β) a = ρβ (s), t, s ∈ [1, e]. (1 − ab(log η)α−1 (log ξ )β−1 )(β − 1)

H1 (t, s) =

On the other hand, by (2.6) and Lemma 2.3, we also have β (ξ )ρβ (s) a(log t)α−1 · α−1 β−1 1 − ab(log η) (log ξ ) (β) aβ (ξ )(log t)α−1 = ρβ (s), t, s ∈ [1, e]. (1 − ab(log η)α−1 (log ξ )β−1 )(β)

H1 (t, s) ≥

Finally, we will prove (2.14) is valid for any t, s ∈ [1, e]. Noticing α (t) ≤ (log t)α−1 , β (t) ≤ (log t)β−1 , α (η) ≤ 1, and β (ξ ) ≤ 1, it follows from (2.6) and Lemma 2.3 that ab(log ξ )β−1 (log t)α−1 G α (η, s) 1 − ab(log η)α−1 (log ξ )β−1 α−1 ab(log ξ )β−1 α−1 α (t) + α (t) ≤ · (α) 1 − ab(log η)α−1 (log ξ )β−1 (α) ab(log ξ )β−1 (1 − (log η)α−1 ) α (t) = (1 − ab(log η)α−1 (log ξ )β−1 )(α − 1) ab(log ξ )β−1 (1 − (log η)α−1 ) (log t)α−1 , t, s ∈ [1, e]. ≤ (1 − ab(log η)α−1 (log ξ )β−1 )(α − 1)

K 1 (t, s) = G α (t, s) +

and a(log t)α−1 β −1 ρβ (ξ ) · α−1 β−1 1 − ab(log η) (log ξ ) (β) a (log t)α−1 , t, s ∈ [1, e]. ≤ α−1 (1 − ab(log η) (log ξ )β−1 )(β − 1)

H1 (t, s) ≤

This completes the proof of the lemma.

 

Similarly, we have Lemma 2.6 For t, s ∈ [1, e], the functions K 2 (t, s) and H2 (t, s) defined by (2.5) and (2.6) satisfy

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ab(log η)α−1 β (ξ ) (log t)β−1 ρβ (s) ≤ K 2 (t, s) (1 − ab(log η)α−1 (log ξ )β−1 )(β) ab(log η)α−1 (1 − (log ξ )β−1 ) ρβ (s), ≤ (1 − ab(log η)α−1 (log ξ )β−1 )(β − 1) bα (η) (log t)β−1 ρα (s) ≤ H2 (t, s) (1 − ab(log η)α−1 (log ξ )β−1 )(α) b ≤ ρα (s), α−1 (1 − ab(log η) (log ξ )β−1 )(α − 1) ab(log η)α−1 (1 − (log ξ )β−1 ) (log t)β−1 , K 2 (t, s) ≤ (1 − ab(log η)α−1 (log ξ )β−1 )(α − 1) b (log t)β−1 . H2 (t, s) ≤ α−1 (1 − ab(log η) (log ξ )β−1 )(α − 1) Remark 2.7 From Lemmas 2.5 and 2.6, for t, s ∈ [1, e], we have ν(log t)α−1 ρα (s) ≤ K 1 (t, s) ≤ μρα (s), α−1

K 1 (t, s) ≤ μ(log t)α−1 ,

ρβ (s) ≤ H1 (t, s) ≤ μρβ (s),

H1 (t, s) ≤ μ(log t)α−1 ,

ν(log t)β−1 ρβ (s) ≤ K 2 (t, s) ≤ μρβ (s),

K 2 (t, s) ≤ μ(log t)β−1 ,

ν(log t)β−1 ρα (s) ≤ H2 (t, s) ≤ μρα (s),

H2 (t, s) ≤ μ(log t)β−1 ,

ν(log t)

where 

min{ab(log ξ )β−1 α (η), bα (η)} , (1 − ab(log η)α−1 (log ξ )β−1 )(α)

min{ab(log η)α−1 β (ξ ), aβ (ξ )} , (1 − ab(log η)α−1 (log ξ )β−1 )(β)  max{b, ab(log ξ )β−1 (1 − (log η)α−1 )} μ = max , (1 − ab(log η)α−1 (log ξ )β−1 )(α − 1)

max{a, ab(log η)α−1 (1 − (log ξ )β−1 )} . (1 − ab(log η)α−1 (log ξ )β−1 )(β − 1) ν = min

In the rest of the paper, we always suppose the following assumption holds: (A) f, g : (1, e) × [0, +∞) × [0, +∞) → (−∞, +∞) are continuous and satisfy −q1 (t) ≤ f (t, u, v) ≤ p1 (t)h 1 (t, u, v), −q2 (t) ≤ g(t, u, v) ≤ p2 (t)h 2 (t, u, v), t ∈ (1, e), u, v ≥ 0, where h i ∈ C([1, e]×[0,+∞)×[0,+∞), [0,+∞)), qi , pi ∈ C((1, e), [0, +∞)), and 

e

0< 1

123

pi (s)

ds < +∞, 0 < s



e 1

qi (s)

ds < +∞, i = 1, 2. s

Solutions for singular Hadamard fractional differential system

Lemma 2.8 Assume the condition (A) holds, then the boundary value problem −D α ω1 (t) = λq1 (t), −D β ω2 (t) = λq2 (t), t ∈ (1, e), λ > 0, ( j)

( j)

ω1 (1) = ω2 (1) = 0, 0 ≤ j ≤ n − 2, ω1 (e) = aω2 (ξ ), ω2 (e) = bω1 (η), ξ, η ∈ (1, e), has an unique solution ⎧   ⎨ ω1 (t) = λ e K 1 (t, s) q1 (s) ds + e H1 (t, s) q2 (s) ds , 1 s s 1  ⎩ ω2 (t) = λ e K 2 (t, s) q2 (s) ds + e H2 (t, s) q1 (s) ds , 1 1 s s

(2.15)

which satisfies 

ω1 (t) ≤ λμ(log t)α−1 ω2 (t) ≤ λμ(log t)

e

1 β−1 e

(q1 (s) + q2 (s)) ds s , t ∈ [1, e],

ds 1 (q1 (s) + q2 (s)) s ,

(2.16)

t ∈ [1, e].

Proof It follows from Lemma 2.4, Remark 2.7 and the condition (A) that (2.15) and (2.16) hold.   Let E = [1, e] × [1, e], then E is a Banach space with the norm (u, v) 1 = u + v , u = max |u(t)|, v = max |v(t)| t∈[1,e]

t∈[1,e]

for any (u, v) ∈ E. Let   P = (u, v) ∈ E : u(t) ≥ γ (log t)α−1 u , v(t) ≥ γ (log t)β−1 v for t ∈ [1, e] , where 0 < γ = ν/μ < 1. Then P is a cone of E. Next we only consider the following singular boundary value problem D α x(t)+λ( f (t, [x(t)−ω1 (t)]∗ , [y(t)−ω2 (t)]∗ )+q1 (t)) = 0, t ∈ (1, e), λ > 0, D β y(t)+λ(g(t, [x(t)−ω1 (t)]∗ , [y(t)−ω2 (t)]∗ )+q2 (t)) = 0, t ∈ (1, e), λ > 0, x ( j) (1) = y ( j) (1) = 0, 0 ≤ j ≤ n−2, x(e) = ay(ξ ),

y(e) = bx(η), ξ, η ∈ (1, e), (2.17)

where a modified function [z(t)]∗ for any z ∈ C[1, e] by [z(t)]∗ = z(t), if z(t) ≥ 0, and [z(t)]∗ = 0, if z(t) < 0. Lemma 2.9 If (x, y) ∈ C[1, e] × C[1, e] with x(t) > ω1 (t) and y(t) > ω2 (t) for any t ∈ (1, e) is a positive solution of the singular system (2.17), then (x − ω1 , y − ω2 ) is a positive solution of the singular system (1.1).

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Proof In fact, if (x, y) ∈ C[1, e] × C[1, e] is a positive solution of the singular system (2.17) such that x(t) > ω1 (t) and y(t) > ω2 (t) for any t ∈ (1, e], then from (2.17) and the definition of [·]∗ , we have D α x(t) + λ( f (t, x(t) − ω1 (t), y(t) − ω2 (t)) + q1 (t)) = 0, t ∈ (1, e), λ > 0, D β y(t) + λ(g(t, x(t) − ω1 (t), y(t) − ω2 (t)) + q2 (t)) = 0, t ∈ (1, e), λ > 0, x ( j) (1) = y ( j) (1) = 0, 0 ≤ j ≤ n−2, x(e) = ay(ξ ), y(e) = bx(η), ξ, η ∈ (1, e). (2.18) Let u = x − ω1 and v = y − ω2 , then D α u(t) = D α x(t) − D α ω1 (t) and D β v(t) = D β y(t) − D β ω2 (t) for t ∈ (1, e), which imply that −D α u(t) = −D α x(t) + D α ω1 (t) = −D α x(t) − λq1 (t), t ∈ (1, e), −D β v(t) = −D β y(t) + D β ω2 (t) = −D β y(t) − λq2 (t), t ∈ (1, e). Thus (2.17) becomes D α u(t)+λ f (t, u(t), v(t)) = 0, D β v(t)+λg(t, u(t), v(t)) = 0, t ∈ (1, e), λ > 0, u ( j) (1) = v ( j) (1) = 0, 0 ≤ j ≤ n−2, u(e) = av(ξ ), v(e) = bu(η), ξ, η ∈ (1, e), i.e., (x − ω1 , y − ω2 ) is a positive solution of the singular system (1.1). This proves Lemma 2.9.   Employing Lemma 2.4, the singular system (2.17) can be expressed as ⎧ e u(t) = λ 1 K 1 (t, s)( f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) ds ⎪ ⎪ s ⎪ e ⎪ ⎨ + λ 1 H1 (t, s)(g(s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q2 (s)) ds s , t ∈ [1, e], e ds ∗ ∗ ⎪ ⎪ ⎪ v(t) = λ 1 K 2 (t, s)(g(s, [x(s) − ω1 (s)] , [y(s) − ω2 (s)] ) + q2 (s)) s ⎪ ⎩ e + λ 1 H2 (t, s)( f (s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q1 (s)) ds s , t ∈ [1, e].

(2.19)

By a solution of the singular system (2.17), we mean a solution of the corresponding system of integral equation (2.19). Defined an operator T : P → P by T (x, y) = (T1 (x, y), T2 (x, y)),

123

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where operators Ti : P → C[1, e] (i = 1, 2) are defined by T1 (x, y)(t)  e ds =λ K 1 (t, s)( f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) s 1  e ds + λ H1 (t, s)(g(s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q2 (s)) , t ∈ [1, e], s 1 (2.20) T2 (x, y)(t)  e ds =λ K 2 (t, s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) s 1  e ds + λ H2 (t, s)( f (s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q1 (s)) , t ∈ [1, e]. s 1 Clearly, if (x, y) ∈ P is a fixed point of T , then (x, y) is a solution of the singular system (2.17). Lemma 2.10 Assume the condition (A) holds, then T : P → P is a completely continuous operator. Proof For any fixed (x, y) ∈ P, there exists a constant L > 0 such that (x, y) 1 ≤ L. And then, [x(s) − ω1 (s)]∗ ≤ x(s) ≤ x ≤ (x, y) 1 ≤ L , [y(s) − ω2 (s)]∗ ≤ y(s) ≤ y ≤ (x, y) 1 ≤ L , s ∈ [1, e]. For any t ∈ [1, e], it follows from (2.20) and Remark 2.7 that  |T1 (x, y)(t)| = λ 1

e

ds s ds ∗ ∗ H1 (t, s)(g(s, [x(s)−ω1 (s)] , [y(s)−ω2 (s)] )+q2 (s)) s

K 1 (t, s)( f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))

 e +λ  1e ds ρα (s)( p1 (s)h 1 (s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q1 (s)) ≤ λμ s 1  e ds + λμ ρβ (s)( p2 (s)h 2 (s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q2 (s)) s  e 1 e ds ds ≤ λMμ ρα (s)( p1 (s)+q1 (s)) +λMμ ρβ (s)( p2 (s)+q2 (s)) s s 1 1  e  e ds ds + λMμ < +∞, ( p1 (s) + q1 (s)) ( p2 (s) + q2 (s)) ≤ λMμ s s 1 1

where  M = max

max

t∈[1,e],u,v∈[0,L]

h 1 (t, u, v),

max

t∈[1,e],u,v∈[0,L]

h 2 (t, u, v) + 1.

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Similarly, we have 

e

|T2 (x, y)(t)| ≤ λMμ 1

ds ( p1 (s) + q1 (s)) + λMμ s



e

( p2 (s) + q2 (s))

1

ds < +∞, s

Thus T : P → E is well defined. Next, we show that T : P → P. For any fixed (x, y) ∈ P, t ∈ [1, e], by (2.20) and Remark 2.7, we have 

e



e

ds s 1  e ds + λμ ρβ (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) , s  e1 ds ρβ (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) T2 (x, y)(t) ≤ λμ s 1  e ds + λμ ρα (s)( f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) , s 1 T1 (x, y)(t) ≤ λμ

ρα (s)( f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))

which implies that ds s 1  e ds ρβ (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) , + λμ s  e1 ds T2 (x, y) ≤ λμ ρβ (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) s 1  e ds ρα (s)( f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) . + λμ s 1 T1 (x, y) ≤ λμ

ρα (s)( f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))

On the other hand, from (2.20) and Remark 2.7, we also obtain T1 (x, y)(t)



e

ds s 1  e ds ρβ (s)(g(s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q2 (s)) , + λν(log t)β−1 s 1 T2 (x, y)(t)  e ds β−1 ≥ λν(log t) ρβ (s)(g(s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q2 (s)) s 1  e ds α−1 ∗ ∗ ρα (s)( f (s, [x(s)−ω1 (s)] , [y(s)−ω2 (s)] )+q1 (s)) . + λν(log t) s 1 ≥ λν(log t)α−1

123

ρα (s)( f (s, [x(s)−ω1 (s)]∗ , [y(s)−ω2 (s)]∗ )+q1 (s))

Solutions for singular Hadamard fractional differential system

So we have T1 (x, y) ≥ γ (log t)α−1 T1 (x, y) , T2 (x, y) ≥ γ (log t)β−1 T2 (x, y) , t ∈ [1, e]. This implies that T (P) ⊂ P. According to the Ascoli-Arzela theorem, we can easily get that T : P → P is completely continuous. The proof is completed.   Lemma 2.11 (Krasnoselskii’s fixed point theorem, see [16]) Let X be a Banach space, and let P ⊂ X be a cone in X . Assume 1 , 2 are open subsets of X with 0 ∈ 1 ⊂

1 ⊂ 2 , and let S : P → P be a completely continuous operator such that, either (a) Sw ≤ w , w ∈ P ∩ ∂ 1 , Sw ≥ w , w ∈ P ∩ ∂ 2 , or (b) Sw ≥ w , w ∈ P ∩ ∂ 1 , Sw ≤ w , w ∈ P ∩ ∂ 2 . Then S has a fixed point in P ∩ ( 2 \ 1 ). 3 Main results For convenience, for fixed [δ, σ ] ⊂ (1, e), we denote  α = μ  f (∞) =

e 1

ds , β = μ s min f (t, u, v),  g (∞) =

ρα (s)( p1 (s) + q1 (s))

lim inf

u+v↑+∞,u,v≥0 t∈[δ,σ ]



e 1

ds , s min g(t, u, v),

ρβ (s)( p2 (s) + q2 (s)) lim inf

u+v↑+∞,u,v≥0 t∈[δ,σ ]

f (t, u, v) g(t, u, v) , g(∞) = , lim inf min u+v↑+∞,u,v≥0 t∈[δ,σ ] u + v u+v h i (t, u, v) , i = 1, 2 h i (∞) = lim sup max u+v u+v↑+∞,u,v≥0 t∈[1,e] f (∞) =

lim inf

min

u+v↑+∞,u,v≥0 t∈[δ,σ ]

For any R > 0, set G iR = max{h i (t, u, v) : t ∈ [1, e], u, v ≥ 0, u + v ≤ R}, i = 1, 2. Theorem 3.1 Let the condition (A) holds. In addition, suppose that the following condition is satisfied: (B1) f (∞) = +∞ or g(∞) = +∞. Then there exists λ∗ such that the singular system (1.1) has at least one positive u , v ) satisfying  u (t) ≥ m (log t)α−1 solution ( u , v ) for any λ ∈ (0, λ∗ ). Moreover, ( β−1 and  v (t) ≥ m (log t) , t ∈ [1, e] for some positive constant m .    e ds 2μ 2μ Proof Choose R1 > max 2μ · , , (q1 (s) + q2 (s)) . Let β−1 α−1 γ aγ (log ξ ) bγ (log η) s 1  λ∗ = min 1,

R1 2[(G 1R1 + 1)α + (G 2R1 + 1)β ]

 .

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In order to prove the theorem, we assume 0 < λ < λ∗ . Let 1 = {(x, y) ∈ E : (x, y) 1 < R1 }. For any (x, y) ∈ P ∩ ∂ 1 , s ∈ [1, e], we have [x(s) − ω1 (s)]∗ ≤ x(s) ≤ x , [y(s) − ω2 (s)]∗ ≤ y(s) ≤ y . Then for any (x, y) ∈ P ∩ ∂ 1 , using (2.20) and Remark 2.7, we have 

e

ds ρα (s)( p1 (s)h 1 (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) s  e ds ∗ ∗ ρβ (s)( p2 (s)h 2 (s, [x(s) − ω1 (s)] , [y(s) − ω2 (s)] ) + q2 (s)) + λμ s 1  e  e ds ds R1 R1 ρα (s)( p1 (s)G 1 + q1 (s)) ρβ (s)( p2 (s)G 2 + q2 (s)) λμ + λμ s s 1 1  e  e ds ds ρα (s)( p1 (s) + q1 (s)) ρβ (s)( p2 (s) + q2 (s)) λ(G 1R1 + 1)μ + λ(G 2R1 + 1)μ s s 1 1 (x, y) 1 , λ(G 1R1 + 1)α + λ(G 2R1 + 1)β ≤ 2 (x, y) 1 λ(G 1R1 + 1)α + λ(G 2R1 + 1)β ≤ . 2

T1 (x, y) ≤ λμ

1

≤ ≤ = T2 (x, y) ≤

Thus we get T (x, y) 1 = T1 (x, y) + T2 (x, y) ≤ (x, y) 1 , ∀(x, y) ∈ P ∩ ∂ 1 . (3.1) On the other hand, choose N1 > 0 big enough such that 1 λνγ (log δ)2(α−1) N1 2



σ

δ

α (s)

ds ≥ 1. s

(3.2)

It follows from f (∞) = +∞ that there exists a constant L 1 > 0 such that f (t, u, v) ≥ N1 (u + v), t ∈ [δ, σ ], u, v ≥ 0, u + v ≥ L 1 .

(3.3)

Let R2 = max{2R1 , 4γ −1 (log δ)1−α L 1 } and 2 = {(x, y) ∈ E : (x, y) 1 < R2 }. For any (x, y) ∈ P ∩ ∂ 2 , we have (x, y) 1 = R2 , so there exists a component x or y such that x ≥ R22 or y ≥ R22 . Without loss of generality, we may suppose that x ≥ R22 . So for any (x, y) ∈ P ∩ ∂ 2 , notice (2.16), we have 

e ds x(t) − ω1 (t) ≥ x(t) − λμ(log t) (q1 (s) + q2 (s)) s 1  ds μx(t) e (q1 (s) + q2 (s)) ≥ x(t) − γ x 1 s  e 2μx(t) 1 ds ≥ x(t) − ≥ x(t) (q1 (s) + q2 (s)) γ R2 1 s 2 1 1 ≥ γ (log t)α−1 x ≥ γ (log δ)α−1 R2 ≥ L 1 , 2 4 α−1

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and then [x(t)−ω1 (t)]∗ +[y(t)−ω2 (t)]∗ ≥ [x(t)−ω1 (t)]∗ 1 = x(t)−ω1 (t) ≥ γ (log δ)α−1 R2 ≥ L 1 . 4

(3.4)

Thus for any (x, y) ∈ P ∩ ∂ 2 , t ∈ [δ, σ ], by (3.3) and (3.4), we have f (t, [x(t) − ω1 (t)]∗ , [y(t) − ω2 (t)]∗ ) ≥ N1 ([x(t) − ω1 (t)]∗ + [y(t) − ω2 (t)]∗ ). (3.5) Then for any (x, y) ∈ P ∩ ∂ 2 , t ∈ [δ, σ ], by (3.2), (3.4) and (3.5), we have T1 (x, y)(t) + T2 (x, y)(t)  e ds ≥λ K 1 (t, s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) s 1  e ds H2 (t, s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) +λ s 1  e ds α (s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) ≥ 2λν(log t)α−1 s 1 σ ds ≥ 2λν(log t)α−1 α (s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) s δ σ ds ≥ 2λν(log t)α−1 α (s)N1 ([x(s) − ω1 (s)]∗ + [y(s) − ω2 (s)]∗ ) s δ  σ 1 ds ≥ λνγ (log δ)2(α−1) N1 R2 ≥ R2 = (x, y) 1 . α (s) 2 s δ Consequently, T (x, y) 1 ≥ (x, y) 1 , ∀(x, y) ∈ P ∩ ∂ 1 .

(3.6)

Obviously, if g(∞) = +∞ holds, (3.6) is still valid. By (3.1), (3.6) and Lemma 2.11, T has a fixed point ( x,  y) ∈ P, and R1 ≤ x, y) 1 ≥ R1 , so there exists a component  x or  y such ( x, y) 1 ≤ R2 . Since ( y ≥ R21 . Without loss of generality, we may suppose  x ≥ R21 , that  x ≥ R21 or  by (2.16), we have 

e ds  x (t) − ω1 (t) ≥  x (t) − λμ(log t) (q1 (s) + q2 (s)) s 1  ds μ x (t) e (q1 (s) + q2 (s)) ≥ x (t) − γ  x 1 s    e ds 2μ  x (t) (q1 (s) + q2 (s)) ≥ 1− γ R1 1 s α−1

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W. Yang

   e 2μ ds γ (log t)α−1  ≥ 1− (q1 (s) + q2 (s)) x γ R1 1 s    e ds 1 2μ γ R1 (log t)α−1 (q1 (s) + q2 (s)) ≥ 1− γ R1 1 s 2 = m 1 (log t)α−1 ,

t ∈ [1, e],

 e ds > 0. From boundary conditions (q (s) + q (s)) where m 1 = 21 γ R1 1 − γ2μ 1 2 R1 1 s of singular system (2.17), we have  y(e) = b x (η) > 0, and so we have  y ≥  y(e) = b x (η) ≥ bγ (log η)α−1  x ≥

1 bγ (log η)α−1 R1 , 2

this together with (2.17), we have  e ds ≥ y(t) y(t) − λμ(log t)β−1 (q1 (s) + q2 (s))  y(t) − ω2 (t) ≥  s 1  e μ y(t) ds − (q1 (s) + q2 (s)) γ  y 1 s    e 2μ ds  y(t) ≥ 1− (q1 (s) + q2 (s)) s bγ 2 (log η)α−1 R1 1    e 2μ ds ≥ 1− γ (log t)β−1  (q1 (s) + q2 (s)) y 2 α−1 s bγ (log η) R1 1    e 2μ ds 1 2 ≥ 1− 2 (q1 (s)+q2 (s)) bγ (log η)α−1 R1 (log t)β−1 s 2 bγ (log η)α−1 R1 1 = m 2 (log t)β−1 ,

t ∈ [1, e],

 e ds > 0. where m 2 = 21 bγ 2 (log η)α−1 R1 1 − bγ 2 (log2μ (q (s) + q (s)) 1 2 α−1 s η) R1 1 Let  u (t) =  x (t) − ω1 (t),  v (t) =  y(t) − ω2 (t), and m  = min{m 1 , m 2 }, then we have v (t) ≥ m (log t)β−1 > 0, t ∈ (1, e].  u (t) ≥ m (log t)α−1 > 0,  By Lemma 2.11, we know the system (1.1) has at least one positive solution ( u , v) v (t) ≥ m (log t)β−1 for any t ∈ [1, e]. The proof is satisfying  u (t) ≥ m (log t)α−1 ,  completed.   Theorem 3.2 Let the condition (A) holds. In addition, suppose that the following condition is satisfied: (B2) h i (∞) = 0 (i = 1, 2) and  f (∞) >  or  g (∞) > ,    e max 2, a γ(log2 ξ )β−1 , b γ(log2η)α−1 · 1 (q1 (s) + q2 (s)) ds s =  σ .  ds β−1 γ2 · min (log δ)α−1 δσ α (s) ds δ β (s) s s , (log δ)

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Solutions for singular Hadamard fractional differential system

Then there exists λ∗ such that the singular system (1.1) has at least one positive u , v ) satisfying  u (t) ≥ m (log t)α−1 solution ( u , v ) for any λ ∈ (λ∗ , +∞). Moreover, ( β−1 and  v (t) ≥ m (log t) , t ∈ [1, e] for some positive constant m . Proof It follows from  f (∞) >  that there exists a constant L 1 > 0 such that, for any t ∈ [δ, σ ],    e max 2, a γ(log2 ξ )β−1 , b γ(log2η)α−1 · 1 (q1 (s) + q2 (s)) ds s , f (t, u, v) ≥  =   β−1 σ  (s) ds } γ2 · min{(log δ)α−1 δσ α (s) ds , (log δ) δ β s s u, v ≥ 0, u + v ≥ L 1 . (3.7) Select λ∗ =

max{L 1 , L 1 γ2 (log ξ )β−1 , L 1 γ2 (log η)α−1 } . e μ · min{(log δ)α−1 , (log δ)β−1 } · 1 (q1 (s) + q2 (s)) ds s

In proving the theorem, we assume λ > λ∗ . Let  R1 = max

4λμ 4λμ 4λμ , , 2 2 β−1 γ a γ (log ξ ) b γ (log η)α−1

 e ds · (q1 (s) + q2 (s)) . s 1

and 1 = {(x, y) ∈ E : (x, y) 1 < R1 }. For any (x, y) ∈ P ∩ ∂ 1 , similar to Theorem 3.1, we may suppose x ≥ R21 , then  e

ds s 1  e ds 1 (q1 (s) + q2 (s)) ≥ γ(log t)α−1 R1 − λμ(log t)α−1 2 s 1  e  e 4λμ ds ds 1 (q1 (s)+q2 (s)) − λμ(log t)α−1 (q1 (s)+q2 (s)) ≥ γ(log t)α−1 γ 2 s s 1 1  e L1 ds α−1 α−1 ≥ (q1 (s) + q2 (s)) (log t) , t ∈ [1, e]. = λμ(log t) s (log δ)α−1 1

x(t) − ω1 (t) ≥ γ(log t)α−1 x − λμ(log t)α−1

(q1 (s) + q2 (s))

So for any (x, y) ∈ P ∩ ∂ 1 , t ∈ [δ, σ ], we have [x(t) − ω1 (t)]∗ + [y(t) − ω2 (t)]∗ ≥ [x(t) − ω1 (t)]∗ = x(t) − ω1 (t) L1 (log t)α−1 ≥ L 1 . ≥ (log δ)α−1

(3.8)

Then for any (x, y) ∈ P ∩ ∂ 2 , t ∈ [δ, σ ], by (3.7), and (3.8), we have T1 (x, y)(t) + T2 (x, y)(t)  e ds ≥λ K 1 (t, s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) s 1  e ds H2 (t, s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) +λ s 1

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W. Yang

 e

ds s 1 σ ds ≥ 2λν(log t)α−1 α (s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) s δ    e  σ max 2, γ 2 β−1 , γ 2 α−1 · 1 (q1 (s) + q2 (s)) ds s ds a (log ξ ) b (log η) α (s) ≥ 2λν(log δ)α−1 σ σ ds ds 2 α−1 β−1 s γ · min{(log δ) δ δ α (s) s , (log δ) δ β (s) s }

 e  4λμ 4λμ ds 4λμ · (q1 (s)+q2 (s)) = R2 = (x, y) 1 . , , ≥ max γ a γ2 (log ξ )β−1 b γ2 (log η)α−1 s 1 ≥ 2λν(log t)α−1

α (s) f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ )

Consequently, T (x, y) 1 ≥ (x, y) 1 , ∀(x, y) ∈ P ∩ ∂ 1 .

(3.9)

On the other hand, for  ε = min

8λμ

e 1

1

1

, e ds α (s) p1 (s) ds s 8λμ 1 β (s) p2 (s) s

 ,

(3.10)

by h i (∞) = 0 (i = 1, 2), there exists L 2 > 0 such that, for any t ∈ [1, e], we have h i (t, u, v) ≤ ε(u + v), u, v ≥ 0, u + v ≥ L 2 , i = 1, 2. Then h i (t, u, v) ≤ G iL 2 + ε(u + v), t ∈ [1, e], u, v ≥ 0, i = 1, 2.

(3.11)

Choose R2 = max{2R1 , 8λ(G 1L 2 + 1)α , 8λ(G 2L 2 + 1)β },

(3.12)

and let 2 = {(x, y) ∈ E : (x, y) 1 < R2 }. for any (x, y) ∈ P ∩ ∂ 1 , by (2.10), (2.11) and (3.12), we have 

e

ds ρα (s)( p1 (s)h 1 (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) s  e ds ρβ (s)( p2 (s)h 2 (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) + λμ s 1  e ds ρα (s)( p1 (s)(G 1L 2 + ε[x(s) − ω1 (s)]∗ + [y(s) − ω2 (s)]∗ ) + q1 (s)) ≤ λμ s 1  e ds L2 ∗ ∗ ρβ (s)( p2 (s)(G 2 + ε[x(s) − ω1 (s)] + [y(s) − ω2 (s)] ) + q2 (s)) +λμ s 1  e  e ds ds l2 ρα (s)( p1 (s) + q1 (s)) ρα (s) p1 (s) ≤ λ(G 1 + 1)μ + λμε( x + y ) s s 1 1  e  e ds ds L2 ρβ (s)( p2 (s) + q2 (s)) ρβ (s) p2 (s) +λ(G 2 + 1)μ + λμε( x + y ) s s 1 1

T1 (x, y) ≤ λμ

1

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Solutions for singular Hadamard fractional differential system



e

ds +λ(G 2L 2 +1)β +λμε R2 s 1 R2 R2 R2 R2 R2 (x, y) 1 = + + + = = . 8 8 8 8 2 2 ≤ λ(G l12 +1)α +λμε R2

Similarly, we have T2 (x, y) ≤



ρα (s) p1 (s)

(x,y) 1 . 2

1

e

ρβ (s) p2 (s)

ds s

Thus we get

T (x, y) 1 = T1 (x, y) + T2 (x, y) ≤ (x, y) 1 , ∀(x, y) ∈ P ∩ ∂ 2 . (3.13) By (3.11), (3.13) and Lemma 2.11, T has a fixed point ( x, y) ∈ P, and R1 ≤ x, y) 1 ≥ R1 , so there exists a component  x or  y such ( x, y) 1 ≤ R2 . Since ( y ≥ R21 . Without loss of generality, we may suppose  x ≥ R21 , that  x ≥ R21 or  we have  x (t) − ω1 (t) ≥

L1 (log t)α−1 = m (log t)α−1 , t ∈ [1, e]. (log δ)α−1

where m  = (logLδ)1 α−1 > 0. From boundary conditions of singular system (2.17), we have  y(e) = b x (η) > 0, and so we have  y ≥  y(e) = b x (η) ≥ bγ (log η)α−1  x ≥

1 bγ (log η)α−1 R1 , 2

this together with (2.17), we have  y(t) − ω2 (t) ≥  y(t) − λμ(log t)β−1



e

(q1 (s) + q2 (s))

ds s

1  e ds 1 2 α−1 β−1 β−1 R1 − λμ(log t) (q1 (s) + q2 (s)) ≥ bγ (log η) (log t) 2 s  e1 4λμ ds 1 2 (q1 (s) + q2 (s)) ≥ bγ (log η)α−1 (log t)β−1 · 2 bγ 2 (log η)α−1 1 s  e ds (q1 (s) + q2 (s)) − λμ(log t)β−1 s  e1 L1 ds ≥ (q1 (s) + q2 (s)) (log t)α−1 ≥ λμ(log t)β−1 s (log δ)α−1 1

=m (log t)α−1 , t ∈ [1, e]. Let  u (t) =  x (t) − ω1 (t) and  v (t) =  y(t) − ω2 (t), then we have  u (t) ≥ m (log t)α−1 > 0,  v (t) ≥ m (log t)β−1 > 0, t ∈ (1, e]. By Lemma 2.11, we know the system (1.1) has at least one positive solution ( u , v) v (t) ≥ m (log t)β−1 for any t ∈ [1, e]. The proof is satisfying  u (t) ≥ m (log t)α−1 ,  completed.   From the proof of Theorem 3.2, we have the following conclusion:

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Corollary 3.3 The conclusion of Theorem 3.2 is valid if (B2) is replaced by (B’2): h i (∞) = 0 (i = 1, 2) and  f (∞) = +∞ or  g (∞) = +∞.

4 Two examples Example 4.1 Consider the following singular Hadamard fractional differential semipositone system 

5 2(u + v)2 + log log t = 0, t ∈ (1, e), D 2 u(t) + λ  log t (1 − log t) 

5 2 log(1 + u + v) D 2 v(t) + λ  + log(1 − log t) = 0, t ∈ (1, e), log t (1 − log t) √ 1 3 2 u ( j) (1) = v ( j) (1) = 0, j = 1, 2, u(e) = v(e 4 ), v(e) = 4 3u(e 4 ), (4.1) 3 √ 1 3 2 where λ is a positive parameter. We take a = 3 , ξ = e 4 , b = 4 3, η = e 4 , then ab(log η)α−1 (log ξ )β−1 = 38 ∈ (0, 1). Let f (t, u, v) = 

(u + v)2

+ log log t, log t (1 − log t) log(1 + u + v) + log(1 − log t), t ∈ (1, e), u, v ≥ 0. g(t, u, v) =  log t (1 − log t) Take p1 (t) = p2 (t) = √

2 , log t (1−log t)

q1 (t) = − log log t, q2 (t) = − log(1 − log t),

h 1 (t, u, v) = (u + v)2 , h 2 (t, u, v) = log(1 + u + v), then −q1 (t) ≤ f (t, u, v) ≤ p1 (t)h 1 (t, u, v), −q2 (t) ≤ g(t, u, v) ≤ p2 (t)h 2 (t, u, v), t ∈ (1, e), u, v ≥ 0, By direct calculation, we have 

e

1

ds = p1 (s) s

 1

e

ds = 2π, p2 (s) s



e 1

ds = q1 (s) s

 1

e

q2 (s)

ds = 1. s

So the condition (A) holds. In addition, choose [δ, σ ] ∈ (1, e), we can easily to check that f (∞) = +∞, and so the condition (B1) of Theorem 3.1 is satisfied. Then by Theorem 3.1, the system (4.1) has at least one positive solution provided λ > 0 is small enough. Example 4.2 Consider the following singular Hadamard fractional differential semipositone system

123

Solutions for singular Hadamard fractional differential system

 5 2

D u(t)+λ

√  4

1

2(u + v) 2 e2(u+v)

2 −√ 2(u+v) log t +e ]



(log t)3 (1 − log t)[1 + log t (1 − log t)eu+v = 0, t ∈ (1, e), 

1 5 2(u + v) 2 [(log t − 1)2 + tanh(u + v)] 3  D 2 v(t) + λ −√ 1 − log t log t (1 − log t) = 0, t ∈ (1, e),

u

( j)

(1) = v ( j) (1) = 0,

√ 4

1

j = 1, 2, u(e) = 2v(e 4 ), v(e) = 1

where λ is a positive parameter. We take a = 2, ξ = e 4 , b = 9 ∈ (0, 1). Let ab(log η)α−1 (log ξ )β−1 = 16

3

27u(e 4 ),

(4.2)

√ 3 4 27, η = e 4 , then

√ 1 2(u + v) 2 e2(u+v) f (t, u, v) =    4 (log t)3 (1 − log t) 1 + log t (1 − log t)eu+v + e2(u+v) 2 −√ , t ∈ (1, e), u, v ≥ 0 log t  1  (u + v) 2 (1 − log t)2 + tanh(u + v) 3  −√ , g(t, u, v) = 1 − log t log t (1 − log t) t ∈ (1, e), u, v ≥ 0. √

Take p1 (t) = √ 4

2 , (log t)3 (1−log t)

p2 (t) = √

2 , log t (1−log t)

1 (u+v) 2 e2(u+v) 1+log t (1−log t)eu+v +e2(u+v)

√ 3 , h 1 (t, u, v) = 1−log t 2 log t) + tanh(u + v)], then

,

q1 (t) =

√2 , log t

q2 (t) = 1

h 2 (t, u, v) = (u + v) 2 [(1 −

−q1 (t) ≤ f (t, u, v) ≤ p1 (t)h 1 (t, u, v), −q2 (t) ≤ g(t, u, v) ≤ p2 (t)h 2 (t, u, v), t ∈ (1, e), u, v ≥ 0, By direct calculation, we have 

e 1

p1 (s)

ds = s



e 1

p2 (s)

ds = 2π, s

 1

e

q1 (s)

ds = 4, s

 1

e

q2 (s)

ds = 6. s

So the condition (A) holds. In addition, choose [δ, σ ] ∈ (1, e), we can easily to check that h i (∞) = 0 (i = 1, 2),  f (∞) = +∞ or  g (∞) = +∞, and so the condition (B’2) of Corollary 3.3 is satisfied. Then by Corollary 3.3, the system (4.2) has at least one positive solution provided λ > 0 is large enough. Acknowledgments The author sincerely thanks the editor and reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original manuscript.

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