Bull. Malays. Math. Sci. Soc. DOI 10.1007/s40840-017-0511-y

Existence of Nonoscillatory Solutions for Fractional Functional Differential Equations Yong Zhou1,2 · Bashir Ahmad2 · Ahmed Alsaedi2

Received: 10 March 2017 / Accepted: 11 May 2017 © Malaysian Mathematical Sciences Society and Penerbit Universiti Sains Malaysia 2017

Abstract In this paper, we develop sufficient criteria for the existence of a nonoscillatory solution to the fractional neutral functional differential equation of the form:

Dtα [x(t) + cx(t − τ )] +

m 

Pi (t)Fi (x(t − σi )) = 0, t ≥ t0 ,

i=1

where Dtα is Liouville fractional derivatives of order α ≥ 0 on the half-axis, c ∈ R, τ , σi ∈ R+ , Pi ∈ C([t0 , ∞), R), Fi ∈ C(R, R), i = 1, 2, . . . , m, m ≥ 1 is an integer. Our results are new and improve many known results on the integer-order functional differential equations. Keywords Fractional differential equations · Liouville derivative · Nonoscillatory solutions · Existence Mathematics Subject Classification 26A33 · 34K15 · 35K99

Communicated by Syakila Ahmad. Project supported by National Natural Science Foundation of China (11671339).

B

Yong Zhou [email protected]

1

Faculty of Mathematics and Computational Science, Xiangtan University, Xiangtan 411105, Hunan, People’s Republic of China

2

Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia

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1 Introduction Fractional differential equations have gained much popularity due to their extensive applications in a variety of fields such as physics, mechanics, chemistry and engineering. The recent development on fractional-order ordinary and partial differential equations can be found in the monographs by Podlubny [1], Kilbas et al. [2], Diethelm [3], and a series of recent research articles [4–8] and the references therein. Oscillation theory provides a useful platform to acquire much needed information about the qualitative properties of solutions of differential equations. This branch of mathematics has been developed for integer-order functional differential equations during the past three decades. One can find important works on the topic in the monographs by Ladde et al. [9], Györi and Ladas [10], Gopalsamy [11], Erbe et al. [12], Agarwal et al. [13]. Recently, Grace et al. [14], Bolat [15], Duan et al. [16], Harikrishnan et al. [17] investigated oscillation and forced oscillation characteristics for fractional-order delay differential equations. However, to the best of our knowledge, the nonoscillatory theory for fractional differential equations is yet to be explored. In this paper, we discuss the nonoscillatory characteristics of solutions for the following fractional neutral functional differential equation Dtα [x(t) + cx(t



− τ )] +

m 

Pi (t)Fi (x(t − σi )) = 0, t ≥ t0 ,

(1)

i=1

where Dtα is Liouville fractional derivatives of order α ≥ 0 on the half-axis, c ∈ R, τ , σi ∈ R+ , Pi ∈ C([t0 , ∞), R), Fi ∈ C(R, R), i = 1, 2, . . . , m, m ≥ 1 is an integer. Let r = max1≤i≤m {τ, σi , }. By a solution of Eq. (1), we mean a function x ∈ C([t1 − r, ∞), R) for some t1 ≥ t0 such that Dtα [x(t) + cx(t − τ )] exists on [t1 , ∞) and that Eq. (1) is satisfied for t ≥ t1 . A nontrivial solution x of Eq. (1) is said to be oscillatory if it has an arbitrarily large number of zeros. Otherwise, x is said to be nonoscillatory, that is, x is nonoscillatory if there exists a T > t1 such that x(t) = 0 for t ≥ T . In other words, a nonoscillatory solution must be eventually positive or eventually negative. For the case of α = n ∈ N, the existence of nonoscillatory solution to Eq. (1) has been studied extensively. The monographs [12,13] summarize some important works on nonoscillation theory of higher-order neutral differential equations of the form dn [x(t) + cx(t − τ )] + P(t)x(t − σ ) = 0, t ≥ t0 , dt n where τ , σ ∈ (0, ∞), P ∈ C([t0 , ∞), R). Theorem A [13] Assume that c = −1, P(t) ≥ 0, and  ∞ t n−1 P(t)dt < ∞, for some add n ∈ N. t0

Then Eq. (2) has a bounded positive solution.

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(2)

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Theorem B [12,13] Assume that c < 0, c = −1, and  ∞ t n−1 |P(t)|dt < ∞, for n ∈ N. t0

Then Eq. (2) has a bounded positive solution. In [18], the authors investigated the following second-order neutral functional differential equation with positive and negative coefficients d2 [x(t) + cx(t − τ )] + P1 (t)x(t − σ1 ) − P2 (t)x(t − σ2 ) = 0, t ≥ t0 . dt 2

(3)

Obviously, Eq. (3) is a special form of Eq. (1), and the following result has been proved for this equation in [18]. Theorem C [18] Assume that c = ±1, P1 (t) ≥ 0 and P2 (t) ≥ 0 and a P1 (t)−P2 (t) ≥ 0, for every t ≥ T and a > 0. Further, assume that  ∞  ∞ P1 (t)dt < ∞, P2 (t)dt < ∞. t0

t0

Then Eq. (3) has a nonoscillatory solution. In this paper, we obtain some sufficient conditions for the existence of a nonoscillatory solution of Eq. (1) by using fixed point theorems due to Krasnoselskii and Schauder, and some new techniques. Our results are new and more general as we relax the restrictive conditions and hypotheses assumed in proving Theorems A, B and C.

2 Preliminaries In this section, we introduce preliminary details which are used throughout this paper. Definition 1 [2] (Liouville fractional integrals on the half-axis) The Liouville fractional integral on the half-axis is defined by  ∞ 1 −α (s − t)α−1 f (s)ds, Dt f (t) = (α) t where t ∈ R and α ∈ [0, ∞). Definition 2 [2] (Liouville fractional derivatives on the half-axis) The Liouville fractional derivative on the half-axis is defined by dn −(n−α) D f (t)) dt n t   ∞ 1 dn (−1)n n = (s − t)n−α−1 f (s)ds , (n − α) dt t

Dtα f (t) = (−1)n

where n = [α] + 1, α ∈ (0, ∞), [α] denotes the integer part of α and t ∈ R.

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In particular, if α = n ∈ N, then Dtn f (t) = (−1)n f (n) (t), where f (n) (t) is the usual derivative of f (t) of order n. Property 1 [2] For α > 0, λ > 0,   Dtα Dt−α f (t) = f (t);

Dtα e−λt = λα e−λt .

Now we state fixed point theorems that we need to prove our main results. Lemma 1 (Krasnoselskii’s fixed point theorem) Let X be a Banach space, let  be a bounded closed convex subset of X and let A1 , A2 be maps of  into X such that A1 x + A2 y ∈  for every pair x, y ∈ . If A1 is a contraction and A2 is completely continuous, then the equation A1 x + A2 x = x has a solution in . Lemma 2 (Schauder’s fixed point theorem) Let  be a closed, convex and nonempty subset of a Banach space X . Let A :  →  be a continuous mapping such that A is a relatively compact subset of X . Then A has at least one fixed point in , that is, there exists an x ∈  such that Ax = x.

3 Main Results We will consider the two cases: c = ±1 and c = −1. Our main results are the following theorems. Theorem 1 Assume that c = ±1 and that  ∞ t α |Pi (t)|dt < ∞, i = 1, 2, . . . , m.

(4)

t0

Then (1) has a bounded nonoscillatory solution. Proof Case I (−1 < c ≤ 0). By (4), we choose a T > t0 sufficiently large so that  ∞  m 1 1+c α , s |Pi (s)|M1 ds ≤ (α + 1) T 3 i=1

where M1 =

max

{|Fi (x)| : 1 ≤ i ≤ m}.

2(1+c)/3≤x≤4/3

Let C([t0 , ∞), R) be the set of all continuous functions with the norm ||x|| = supt≥t0 |x(t)| < ∞. Then C([t0 , ∞), R) is a Banach space. We define a closed, bounded and convex subset  of C([t0 , ∞), R) by 

4 2(1 + c) ≤ x(t) ≤ , t ≥ t0 .  = x = x(t) ∈ C([t0 , ∞), R) : 3 3

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Define two maps A1 and A2 :  → C([t0 , ∞), R) as follows:

1 + c − cx(t − τ ), t ≥ T, t0 ≤ t ≤ T. (A1 x)(T ),  m ⎧  ∞  ⎪ 1 ⎨ α (s − t) Pi (s)Fi (x(s − σi )) ds, (A2 x)(t) = (α + 1) t i=1 ⎪ ⎩ (A2 x)(T ), (A1 x)(t) =

t ≥ T, t0 ≤ t ≤ T.

(i) We shall show that A1 x + A2 y ∈  for any x, y ∈ . Indeed, for every x, y ∈  and t ≥ T , we get (A1 x)(t) + (A2 y)(t) ≤ 1 + c − cx(t − τ )  m  ∞  1 α (s − t) |Pi (s)||Fi (y(s − σi ))| ds + (α + 1) t i=1  ∞  m 1 4 α ≤ 1+c− c+ s |Pi (s)|M1 ds 3 (α + 1) T i=1

4 1+c 4 ≤ 1+c− c+ = . 3 3 3 Furthermore, we have (A1 x)(t) + (A2 y)(t) ≥ 1 + c − cx(t − τ )  m  ∞  1 α − (s − t) |Pi (s)||Fi (y(s − σi ))| ds (α + 1) t i=1  ∞  m 1 ≥ 1+c− sα |Pi (s)|M1 ds (α + 1) T i=1

2(1 + c) 1+c = . ≥ 1+c− 3 3 From the above two inequalities, it follows that 2(1 + c) 4 ≤ (A1 x)(t) + (A2 y)(t) ≤ , for t ≥ t0 . 3 3 Thus, A1 x + A2 y ∈  for any x, y ∈ . (ii) We show that A1 is a contraction mapping on . In fact, for x, y ∈  and t ≥ T , we have |(A1 x)(t) − (A1 y)(t)| ≤ −c|x(t − τ ) − y(t − τ )| ≤ −c||x − y||,

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which implies that ||A1 x − A1 y|| ≤ −c||x − y||. Since 0 < −c < 1, we conclude that A1 is a contraction mapping on . (iii) Here we show that A2 is completely continuous. First, we will show that A2 is continuous. Let xk = xk (t) ∈  be such that xk (t) → x(t) as k → ∞. As  is closed, x = x(t) ∈ . For t ≥ T , we have |(A2 xk )(t) − (A2 x)(t)|  ∞  m 1 ≤ sα |Pi (s)||Fi (xk (s − σi )) − Fi (x(s − σi ))| ds (α + 1) t i=1  ∞  m 1 α ≤ s |Pi (s)||Fi (xk (s − σi )) − Fi (x(s − σi ))| ds. (α + 1) T i=1

Since |Fi (xk (t −σi ))− Fi (x(t −σi ))| → 0 as k → ∞ for i = 1, 2, . . . , m, by applying the Lebesgue-dominated convergence theorem, we deduce that limk→∞ ||(A2 xk )(t)− (A2 x)(t)|| = 0. This means that A2 is continuous. Next, we show A2  is relatively compact. It suffices to show that the family of functions {A2 x : x ∈ } is uniformly bounded and equicontinuous on [t0 , ∞). The uniform boundedness is obvious. For the equicontinuity, according to Levitan’s result, we only need to show that, for any given ε > 0, [T, ∞) can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have change of amplitude less than ε. By (4), for any ε > 0, take T ∗ ≥ T large enough so that  ∞  m  1 ε s α M1 |Pi (s)| ds < . (α + 1) T ∗ 2 i=1

Then, for x ∈ , t2 > t1 ≥ T ∗ , we have 1 |(A2 x)(t2 ) − (A2 x)(t1 )| ≤ (α + 1)



∞

t2

sα

 m 

|Pi (s)||Fi (x(s − σi ))| ds

i=1  m 

 ∞ 1 α + s |Pi (s)||Fi (x(s − σi ))| ds (α + 1) t1 i=1  ∞  m  1 α ≤ s M1 |Pi (s)| ds (α + 1) t2 i=1  ∞  m  1 α + s M1 |Pi (s)| ds (α + 1) t1 i=1 ε ε < + = ε. 2 2

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Existence of Nonoscillatory Solutions for Fractional…

For x ∈  and T ≤ t1 < t2 ≤ T ∗ , we obtain 1 |(A2 x)(t2 ) − (A2 x)(t1 )| ≤ (α + 1) 1 ≤ (α + 1)



t2

s

α

t1



t2

 m  

sα

|Pi (s)||Fi (x(s − σi ))| ds

i=1

M1

t1

m 

|Pi (s)| ds

i=1

 1 max ∗ s α ≤ (α + 1) T ≤s≤T



M1

m 

|Pi (s)|

 (t2 − t1 ).

i=1

Thus, there exists a δ > 0 such that |(A2 x)(t2 ) − (A2 x)(t1 )| < ε, if 0 < t2 − t1 < δ. For any x ∈ , t0 ≤ t1 < t2 ≤ T , it is easy to see that |(A2 x)(t2 ) − (A2 x)(t1 )| = 0 < ε. Therefore, {A2 x : x ∈ } is uniformly bounded and equicontinuous on [t0 , ∞), and hence A2  is relatively compact. In consequence, the conclusion of Lemma 1 (Krasnoselskii’s fixed point theorem) applies and there exits x0 ∈  such that A1 x0 + A2 x0 = x0 , that is,  m  ∞  1 α x0 (t) = 1+c − cx0 (t −τ ) + (s − t) Pi (s)Fi (x0 (s − σi )) ds, (α + 1) t i=1

which implies that  ∞  s 1 x0 (t) = 1 + c − cx0 (t − τ ) + ds (s − u)α−1 (α) t t  m  × Pi (s)Fi (x0 (s − σi )) du. i=1

Hence 1 [x0 (t) + cx0 (t − τ )] = (α) 



∞

(s − t)

t

α−1

 m 

Pi (s)Fi (x0 (s − σi )) ds.

i=1

By Property 1, it is easy to see that x0 (t) is a nonoscillatory solution of Eq. (1). Case II (−∞ < c < −1). By (4), we choose a T > t0 sufficiently large such that  m  ∞  1 c+1 α − , s |Pi (s)|M2 ds ≤ − c(α + 1) T +τ 2 i=1

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where M2 =

max

−(c+1)/2≤x≤−2c

{|Fi (x)| : 1 ≤ i ≤ m}.

Let C([t0 , ∞), R) be the set as in the proof of Theorem 1. We define a closed, bounded and convex subset  of C([t0 , ∞), R) as follows:

 c+1  = x = x(t) ∈ C([t0 , ∞), R) : − ≤ x(t) ≤ −2c, t ≥ t0 . 2 Define two maps A1 and A2 :  → C([t0 , ∞), R) by 

1 x(t + τ ), t ≥ T, c t0 ≤ t ≤ T. (A1 x)(T ),  m ⎧  ∞  ⎪ 1 ⎨ α (s − t − τ ) Pi (s)Fi (x(s − σi )) ds, (A2 x)(t) = c(α + 1) t+τ i=1 ⎪ ⎩ (A2 x)(T ), (A1 x)(t) =

−c − 1 −

t ≥ T, t0 ≤ t ≤ T.

In the first step, let us show that A1 x + A2 y ∈  for any x, y ∈ . Indeed, for every x, y ∈  and t ≥ T , we get 1 (A1 x)(t) + (A2 y)(t) ≤ −c − 1 − x(t + τ ) c  ∞ 1 1 − (s − t − τ )α c (α + 1) t+τ  m  × |Pi (s)||Fi (y(s − σi ))| ds i=1

1 1 ≤ −c − 1 + 2 − c (α + 1) ≤ −c + 1 −



∞ T +τ

 sα

m 

|Pi (s)|M2 ds

i=1

c+1 ≤ −2c 2

and 1 (A1 x)(t) + (A2 y)(t) ≥ −c − 1 − x(t + τ ) c  m  ∞  1 1 α + (s − t) |Pi (s)||Fi (y(s − σi ))| ds c (α + 1) t+τ i=1  ∞  m 1 1 α ≥ −c − 1 + s |Pi (s)|M2 ds c (α + 1) T i=1

c+1 c+1 ≥ −c − 1 + =− , 2 2

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Existence of Nonoscillatory Solutions for Fractional…

which imply that −

c+1 ≤ (A1 x)(t) + (A2 y)(t) ≤ −2c, for t ≥ t0 . 2

Thus, A1 x + A2 y ∈  for any x, y ∈ . Next we show that A1 is a contraction mapping on . For x, y ∈  and t ≥ T , we have 1 1 |(A1 x)(t) − (A1 y)(t)| ≤ − |x(t + τ ) − y(t + τ )| ≤ − ||x − y||, c c which implies that 1 ||A1 x − A1 y|| ≤ − ||x − y||. c In view of the condition 0 < −1/c < 1, it follows that A1 is a contraction mapping on . As in the proof of Case I, we can obtain that the mapping A2 is completely continuous. Therefore, all the conditions of Lemma 1 are satisfied. Hence there exists x0 ∈  such that A1 x0 + A2 x0 = x0 . Clearly, x0 = x0 (t) is a bounded positive solution of Eq. (1). Case III (0 ≤ c < 1). By (4), we choose a T > t0 sufficiently large so that 1 (α + 1)



∞

s

α

 m 

T

|Pi (s)|M3 ds ≤ 1 − c,

i=1

where M3 =

max

2(1−c)≤x≤4

{Fi (x) : 1 ≤ i ≤ m}.

Let C([t0 , ∞), R) be the set defined in the proof of Theorem 1. We define a closed, bounded and convex subset  of C([t0 , ∞), R) as follows:  = {x = x(t) ∈ C([t0 , ∞), R) : 2(1 − c) ≤ x(t) ≤ 4, t ≥ t0 }, and consider two maps A1 and A2 :  → C([t0 , ∞), R) defined by

(A1 x)(t) =

3 + c − cx(t − τ ), (A1 x)(T ),

t ≥ T, t0 ≤ t ≤ T,

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and

(A2 x)(t) =

⎧ ⎪ ⎪ ⎨

1 (α + 1)



⎪ ⎪ ⎩ (A x)(T ), 2

∞

(s − t)α

t

 m 

Pi (s)Fi (x(s − σi )) ds,

t ≥ T,

i=1

t0 ≤ t ≤ T.

As before, for any x, y ∈  and t ≥ T , we have (A1 x)(t) + (A2 y)(t) ≤ 3 + c − cx(t − τ )  m  ∞  1 (s − t)α |Pi (s)||Fi (y(s − σi ))| ds + (α + 1) t i=1  ∞  m 1 α ≤ 3+c+ s |Pi (s)|M3 ds (α + 1) T i=1

≤ 3 + c + 1 − c = 4, and (A1 x)(t) + (A2 y)(t) ≥ 3 + c − cx(t − τ )  m  ∞  1 α − (s − t) |Pi (s)||Fi (y(s − σi ))| ds (α + 1) t i=1  ∞  m 1 α ≥ 3 + c − 4c − s |Pi (s)|M3 ds (α + 1) T i=1

≥ 3 + c − 4c − (1 − c) = 2(1 − c). In consequence, we get 2(1 − c) ≤ (A1 x)(t) + (A2 y)(t) ≤ 4, for t ≥ t0 . This shows that A1 x + A2 y ∈  for any x, y ∈ . Proceeding as in the proof of Case I, we can establish that the mapping A1 is a contraction mapping on  and the mapping A2 is completely continuous. By Lemma 1, there exists x0 ∈  such that A1 x0 + A2 x0 = x0 . Clearly, x0 = x0 (t) is a bounded positive solution of (1). Case IV (1 < c < ∞). Again, by (4), we can choose a T > t0 sufficiently large so that 1 c(α + 1)

123



∞ T +τ

s

α

 m  i=1

|Pi (s)|M4 ds ≤ c − 1,

Existence of Nonoscillatory Solutions for Fractional…

where M4 =

max

2(c−1)≤x≤4c

{Fi (x) : i = 1, 2, . . . , m}.

Let C([t0 , ∞), R) be the set considered in the proof of Theorem 1. Let  be a closed, bounded and convex subset of C([t0 , ∞), R) defined by  = {x = x(t) ∈ C([t0 , ∞), R) : 2(c − 1) ≤ x(t) ≤ 4c, t ≥ t0 }. Define two maps A1 and A2 :  → C([t0 , ∞), R) as follows: ⎧ ⎨ 3c + 1 − 1 x(t + τ ), c (A1 x)(t) = ⎩ (A1 x)(T ),

t ≥ T, t0 ≤ t ≤ T,

and

(A2 x)(t) =

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

1 c(α+1)



∞

(s −t −τ )

t+τ

α

 m 

Pi (s)Fi (x(s −σi )) ds, t ≥ T,

i=1

(A2 x)(T ),

t0 ≤ t ≤ T.

In order to show that A1 x + A2 y ∈  for any x, y ∈  and t ≥ T , we consider 1 (A1 x)(t) + (A2 y)(t) ≤ 3c + 1 − x(t + τ ) c  ∞ 1 1 (s − t − τ )α + c (α + 1) t+τ  m  × |Pi (s)||Fi (y(s − σi ))| ds i=1

1 1 ≤ 3c + 1 + c (α + 1)



∞ T +τ

s

α

 m 

(|Pi (s)|M4 ds

i=1

≤ 3c + 1 + (c − 1) = 4c, (A1 x)(t) + (A2 y)(t) 1 ≥ 3c + 1 − x(t + τ ) c  m  ∞  1 1 − (s − t)α |Pi (s)||Fi (y(s − σi ))| ds c (α + 1) t+τ i=1  ∞  m 1 1 α ≥ 3c + 1 − 4 − s |Pi (s)|M4 ds c (α + 1) T i=1

≥ 3c − 3 − (c − 1) = 2(c − 1).

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Hence, we obtain 2(c − 1) ≤ A1 x(t) + A2 y(t) ≤ 4c, for t ≥ t0 , which shows that A1 x + A2 y ∈  for any x, y ∈ . As in the proof of Case I, it can be shown that the mapping A1 is a contraction mapping on  and the mapping A2 is completely continuous. In consequence, the conclusion of Lemma 1 applies, and there exists x0 ∈  such that A1 x0 + A2 x0 = x0 . By Property 1, it is easy to see that x0 = x0 (t) is a bounded positive solution of Eq. (1). The proof is complete. 

Remark 1 We emphasize that Theorem 1 is a new result in the context of fractional functional differential equations. In particular, for α = n ∈ N, Theorem 1 improves essentially Theorems A, B and C by removing the restrictive conditions: P(t) ≥ 0 in Theorem A, a P1 (t) − P2 (t) ≥ 0 in Theorem C, and relaxing the hypothesis that c < 0 in Theorem B. Remark 2 Minor adjustments are only necessary to discuss the neutral functional differential equation of the form Dtα [x(t) + C(t)x(t − τ )] + F(t, x(σ1 (t)), . . . , x(σm (t))) = f (t), t ≥ t0 , where τ ∈ R+ = [0, ∞), σi (t) → ∞ (i = 1, 2, . . . , m) as t → ∞, m ≥ 1 is an integer, and F : [t0 , ∞) × R × · · · × R → R is continuous and bounded, C, f ∈ C([t0 , ∞), R). So we omit the details. Theorem 2 Assume that c = −1 and that ∞  

∞

j=0 t0 + jτ

t α |Pi (t)|dt < ∞, i = 1, 2, . . . , m.

(5)

Then Eq. (1) has a bounded positive solution. Proof By the condition (5), we can choose a sufficiently large T > t0 so that ∞

 1 (α + 1)



∞

j=1 T + jτ

s

α

 m 

|Pi (s)|M5 ds ≤ 1,

i=1

where M5 = max0≤x≤1 {Fi (x) : 1 ≤ i ≤ m}. We consider a closed, bounded and convex subset  of C([t0 , ∞), R) given by  = {x = x(t) ∈ C([t0 , ∞), R) : 2 ≤ x(t) ≤ 4, t ≥ t0 }

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Existence of Nonoscillatory Solutions for Fractional…

and define a mapping A :  → C([t0 , ∞), R) as follows: ⎧ ∞  ∞  1 ⎪ ⎪ ⎪ 3 − (s − t − jτ )α ⎪ ⎪ (α + 1) ⎪ t+ jτ ⎨ j=1  m (Ax)(t) =  ⎪ ⎪ × Pi (s)Fi (x(s − σi )) ds, ⎪ ⎪ ⎪ ⎪ i=1 ⎩ (Ax)(T ),

t ≥ T, t0 ≤ t ≤ T.

We first show that A ⊂ . Indeed, for every x ∈  and t ≥ T , we get ∞

 1 (Ax)(t) ≤ 3 + (α + 1)



∞

α

(s − t − jτ )

j=1 t+ jτ ∞

 1 ≤ 3+ (α + 1)



∞

j=1 T + jτ

s

α

 m 

 m  i=1

|Pi (s)||Fi (x(s − σi ))| ds

|Pi (s)|M5 ds

i=1

≤ 4, and ∞

 1 (Ax)(t) ≥ 3 − (α + 1)



∞

α

(s − t − jτ )

j=1 t+ jτ ∞

 1 ≥ 3− (α + 1)



∞

j=1 T + jτ

s

α

 m 

 m  i=1

|Pi (s)||Fi (x(s − σi ))| ds

|Pi (s)|M5 ds

i=1

≥ 2. Hence A ⊂ . We now show that A is continuous. Let xk = xk (t) ∈  be such that xk (t) → x(t) as k → ∞. Since  is closed, x = x(t) ∈ . For t ≥ T , we have |(Axk )(t) − (Ax)(t)|  m ∞  ∞   1 α ≤ s |Pi (s)||Fi (xk (s − σi )) − Fi (x(s − σi ))| ds. (α + 1) T + jτ j=1

i=1

Noting that |Fi (xk (t − σi )) − Fi (x(t − σi ))| → 0 as k → ∞ for i = 1, 2, . . . , m, and applying the Lebesgue-dominated convergence theorem, we conclude that limk→∞ ||(Axk )(t) − (Ax)(t)|| = 0. This means that A is continuous. In what follows, we show that A is relatively compact. By (5), for any ε > 0, take T ∗ ≥ T large enough so that ∞

 1 (α + 1)





∞

∗ j=1 T + jτ

s

α

M5

m  i=1

|Pi (s)| ds <

ε . 2

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Then, for x ∈ , t2 > t1 ≥ T ∗ , we get |(Ax)(t2 ) − (Ax)(t1 )|  m ∞  ∞   1 α ≤ s |Pi (s)||Fi (x(s − σi ))| ds (α + 1) j=1 t2 + jτ i=1  m ∞  ∞   1 α + s |Pi (s)||Fi (x(s − σi ))| ds (α + 1) j=1 t1 + jτ i=1  m ∞  ∞   1 α ≤ s M5 |Pi (s)| ds (α + 1) j=1 t2 + jτ i=1  m ∞  ∞   1 α + s M5 |Pi (s)| ds (α + 1) t1 + jτ ε ε < + = ε. 2 2 T∗

j=1

i=1

For T ≤ t1 < t2 ≤ T ∗ , we choose a sufficiently large J ∈ N+ such that T + jτ ≥ as j ≥ J . For x ∈ , we obtain |(Ax)(t2 ) − (Ax)(t1 )|  m ∞  t2 + jτ   1 α ≤ s |Pi (s)||Fi (x(s − σi ))| ds (α + 1) j=1 t1 + jτ i=1 ⎡  m J  t2 + jτ   1 ⎣ s α M5 |Pi (s)| ds ≤ (α + 1) t + jτ 1 j=1 i=1 ⎤  m ∞  t2 + jτ   + s α M5 |Pi (s)| ds ⎦ j=J +1 t1 + jτ

1 ≤ (α + 1) +

∞   j=1

i=1



max ∗

T +τ ≤s≤T + jτ



∞ T ∗ + jτ

sα

M5

m 

 sα

 M5

m 

|Pi (s)|



J (t2 − t1 )

i=1

⎤

|Pi (s)| ds ⎦ .

i=1

Then there exists a δ > 0 such that |(Ax)(t2 ) − (Ax)(t1 )| < ε, if 0 < t2 − t1 < δ. For any x ∈ , t0 ≤ t1 < t2 ≤ T , it is easy to see that |(Ax)(t2 ) − (Ax)(t1 )| = 0 < ε.

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Therefore, {Ax : x ∈ } is uniformly bounded and equicontinuous on [t0 , ∞), and hence, A is relatively compact. By Lemma 2 (Schauder’s fixed point theorem), there exists x0 ∈  such that Ax0 = x0 , that is, ⎧ ∞  ∞  1 ⎪ ⎪ ⎪ 3− (s − t − jτ )α ⎪ ⎪ (α + 1) ⎪ t+ jτ ⎨ j=1  m x0 (t) =  ⎪ ⎪ × Pi (s)Fi (x0 (s − σi )) ds, ⎪ ⎪ ⎪ ⎪ i=1 ⎩ x0 (T ),

t ≥ T, t0 ≤ t ≤ T.

Then, we have 

1 x0 (t) − x0 (t − τ ) = (α + 1)

∞

(s − t)

α

 m 

t

Pi (s)Fi (x0 (s − σi )) ds, t ≥ T,

i=1

which implies that 1 [x0 (t) − x0 (t − τ )] = (α) 



∞

(s − t)

α−1

 m 

t

Pi (s)Fi (x0 (s −σi )) ds, t ≥ T.

i=1

By Property 1, it follows that x0 = x0 (t) is a bounded positive solution of Eq. (1). This completes the proof. 

Remark 3 When Fi (x) ≡ x, Pi (t) ≡ pi ∈ R, i = 1, 2, . . . , m, Eq. (1) reduces to Dtα [x(t) + cx(t − τ )] +

m 

pi x(t − σi ) = 0, t ≥ t0 .

(6)

i=1

In this case, (4) and (5) cannot be satisfied. So, we provide an alternative sufficient condition for existence of nonoscillatory solutions of Eq. (1). Theorem 3 Assume that α > 0, c, τ , pi , σi ∈ R, i = 1, 2, . . . , m. If the characteristic equation of (6): λα+1 + cλα+1 e−λτ =

m 

pi e−λσi

(7)

i=1

has a positive real root, then Eq. (6) has a bounded positive solution. Proof Let λ0 > 0 be a real root of (7). Set y(t) = e−λ0 t . By using Property 1 and Eq. (7), we get

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Dtα [y(t) + cy(t − τ )] = −(λα+1 + cλα+1 e−λ0 τ )e−λ0 t 0 0   m =− pi e−λ0 σi e−λ0 t i=1

=−

m 

pi y(t − σi ).

i=1

Clearly, y(t) is a bounded positive solution of Eq. (6). The proof is complete.



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