J Appl Math Comput DOI 10.1007/s12190-013-0751-4

JAMC

O R I G I NA L R E S E A R C H

Periodic BVP for integer/fractional order nonlinear differential equations with non-instantaneous impulses JinRong Wang · Xuezhu Li

Received: 10 October 2013 © Korean Society for Computational and Applied Mathematics 2014

Abstract In this paper, we investigate periodic BVP for integer/fractional order nonlinear differential equations with non-instantaneous impulses. Several new existence results are obtained under different conditions via fixed point methods. Finally, two examples are given to illustrate our main results. Keywords Periodic BVP · Integer/fractional order · Nonlinear differential equations · Non-instantaneous impulses Mathematics Subject Classification 26A33 · 34B37

1 Introduction In this paper, we shall study the following periodic boundary value problems for integer/fractional order nonlinear differential equations with non-instantaneous impulses

This work is partially supported by the National Natural Science Foundation of China (11201091), Key Projects of Science and Technology Research in the Chinese Ministry of Education (211169), Key Support Subject (Applied Mathematics), Key Project on the Reforms of Teaching Contents and Course System of Guizhou Normal College and Doctor Project of Guizhou Normal College (13BS010).

B

J. Wang ( ) School of Mathematics and Computer Science, Guizhou Normal College, Guiyang, Guizhou 550018, P.R. China e-mail: [email protected] J. Wang e-mail: [email protected]

B

J. Wang · X. Li ( ) Department of Mathematics, Guizhou University, Guiyang, Guizhou 550025, P.R. China e-mail: [email protected]

J. Wang, X. Li

⎧  ⎪ ⎨ u (t) = f (t, u(t)), u(t) = gi (t, u(t)), ⎪ ⎩ u(0) = u(T ), and

⎧c q ⎪ ⎨ D0,t u(t) = f (t, u(t)), u(t) = gi (t, u(t)), ⎪ ⎩ u(0) = u(T ),

t ∈ (si , ti+1 ], i = 0, 1, 2, . . . , m, t ∈ (ti , si ], i = 1, 2, . . . , m,

(1)

t ∈ (si , ti+1 ], i = 0, 1, 2, . . . , m, q > 0, t ∈ (ti , si ], i = 1, 2, . . . , m,

(2)

where 0 = s0 < t1 ≤ s1 ≤ t2 < · · · < tm ≤ sm ≤ tm+1 = T are pre-fixed numbers, f : [0, T ] × R → R is continuous and gi : [ti , si ] × R → R is continuous for all i = 1, 2, . . . , m. q The symbol c D0,t denotes the Caputo fractional derivative of the order q with the lower limit 0. Here we use the following generalized definition from [1]. Definition 1.1 The Caputo derivative of order q for a function h : [a, b] → R can be  q q t k (k) written as c Da,t h(t) = L Da,t (h(t) − n−1 k=0 k! h (a)), t > 0, n − 1 < q < n. q

Here L Da,t is understood by the following definition: Definition 1.2 For a function h given on the interval [a, b], the qth Riemannq 1 d n ( dt ) × Liouville fractional order derivative of h is defined by L (Da,t h)(t) = Γ (n−q) t n−q−1 h(s)ds, here Γ is the Gamma function, n = [q] + 1 and [q] denotes (t − s) a the integer part of q. Remark 1.3 The function h in Definition 1.1 can be discontinuous. Of course, the fractional order integral of the function is understood by the following definition: Definition 1.4 The fractional order integral function h ∈ L1 ([a, b], R) with  t of the q−1 q 1 + order q ∈ R is defined by Ia,t h(t) = Γ (q) a (t − s) h(s)ds. The concept of fractional calculus appeared in 1695 in the letter between L’Hospital and Leibniz. Since then, further development in this area has been explored by many mathematicians and we recommend to read the study of Riemann, Liouville, Caputo, and other famous mathematicians. Fractional calculus plays an important role in various fields such as electricity, biology, economics and signal and image processing. Long-term perturbations are usually regarded as having acted non-instantaneous or appearing in the form of non-instantaneous impulses involving the corresponding differential equations. It is remarkable that Hernández and O’Regan [2] and Pierri et al. [3] introduced initial value problems for a new class of non-instantaneous impulsive differential equations to describe some certain dynamic change of evolution processes in the pharmacotherapy. Whether integer order or fractional order differential

Periodic BVP for integer/fractional order nonlinear differential

equations with impulses have been studied previously. One can see the monographs [4–6] and [7–15] and the references therein. In the present paper, we shall establish the corresponding integral equations for linear problems of integer/fractional order. Then, we establish sufficient conditions for existence of the solutions for both the linear periodic problems (1) and the problem (2). The topological method that we have chosen to study existence of the solutions is the theory of fixed point, which is a very powerful and important tool to the study of differential problems.

2 Linear problems of integer/fractional order Let J = [0, T ]. Denote C(J, R) by the Banach space of all continuous functions from J into R with the norm uC := sup{|u(t)| : t ∈ J } for u ∈ C(J, R). From the associate literature we consider the space PC(J, R) := {u : J → R : u ∈ C((tk , tk+1 ], R), k = 0, 1, . . . , m and there exist u(tk− ) and u(tk+ ), k = 1, . . . , m, with u(tk− ) = u(tk )} with the norm uPC := sup{|u(t)| : t ∈ J }. Set PC1 (J, R) := {u ∈ PC(J, R) : u ∈ PC(J, R)} with uPC1 := max{uPC , u PC }. Clearly, PC1 (J, R) endowed with the norm  · PC1 is a Banach space. By considering the developments in [2], next we say that a function u ∈ PC1 (J, R) is called a solution of the problem (1), if u satisfies u(0) = u(T ), and  u(t) = gi t, u(t) , t ∈ (ti , si ], i = 1, 2, . . . , m;

t  u(t) = u(0) + f s, u(s) ds, t ∈ [0, t1 ]; 0

 u(t) = gi si , u(si ) +



t

 f s, u(s) ds,

t ∈ [si , ti+1 ], i = 1, 2, . . . , m.

si

In order to study existence results for the problem (1), we need the following lemma. Lemma 2.1 Let h : J → R and Gi : [ti , si ] → R be continuous functions. A function u ∈ PC1 (J, R) is a solution of the problem ⎧  ⎪ ⎨ u (t) = h(t), t ∈ (si , ti+1 ], i = 0, 1, 2, . . . , m, u(t) = Gi (t), t ∈ (ti , si ], i = 1, 2, . . . , m, (3) ⎪ ⎩ u(0) = u(T ), if and only if ⎧ T t ⎪ ⎪ Gm (sm ) + sm h(s)ds + 0 h(s)ds, t ∈ [0, t1 ], ⎨ u(t) = Gi (t), t ∈ (ti , si ], i = 1, 2, . . . , m, ⎪ ⎪ ⎩ G (s ) +  t h(s)ds, t ∈ (s , t ], i = 1, 2, . . . , m. i

i

si

i

i+1

(4)

J. Wang, X. Li

Proof Assume that u satisfies (3). If t ∈ [0, t1 ]. By integrating the first equation of (3) from zero to t, we have

t u(t) = u(0) + h(s)ds. (5) 0

On the other hand, if t ∈ (si , ti+1 ], i = 1, 2, . . . , m. By integrating the first equation of (3) from si to t, we have

t u(t) = u(si ) + h(s)ds. (6) si

Applying the impulsive condition of (3), we find that u(si ) = Gi (si ).

(7)

Now combining (6) and (7), we obtain

u(t) = Gi (si ) +

t

h(s)ds.

(8)

si

So, we have

u(0) = u(T ) = Gm (sm ) +

T

h(s)ds.

(9)

sm

Now it is clear that (5), (8) and (9) imply (4). Conversely, assume that u satisfies (4). One can verify the results immediately. The proof is completed.  By using Definitions 1.1, 1.2 and 1.4, Lemma 2.5 and Remark 2.6 in [13] and proceeding the steps in Lemma 4.1 again in [13]/or just like the proof in Lemma 2.1, we can derive the following result. Lemma 2.2 A function u given by T  sm ⎧ 1 1 q−1 q−1 ⎪ ⎪ Gm (sm ) + Γ (q) 0 (T − s) h(s)ds − Γ (q) 0 (sm − s) h(s)ds ⎪  ⎪ t ⎪ 1 q−1 ⎪ ⎪ ⎨ + Γ (q) 0 (t − s) h(s)ds, t ∈ [0, t1 ], u(t) =

Gi (t), t ∈ (ti , si ], i = 1, 2, . . . , m, ⎪ t ⎪ ⎪ ⎪ Gi (si ) + Γ 1(q) 0 (t − s)q−1 h(s)ds ⎪ ⎪ ⎪ s ⎩ − Γ 1(q) 0 i (si − s)q−1 h(s)ds, t ∈ (si , ti+1 ], i = 1, 2, . . . , m,

(10)

is a solution of the following problem ⎧ q c ⎪ ⎨ D0,t u(t) = h(t), t ∈ (si , ti+1 ], i = 0, 1, 2, . . . , m, q > 0, u(t) = Gi (t), ⎪ ⎩ u(0) = u(T ).

t ∈ (ti , si ], i = 1, 2, . . . , m,

(11)

Periodic BVP for integer/fractional order nonlinear differential

Proof Assume that u satisfies (11). If t ∈ [0, t1 ]. By integrating the first equation of (11) from zero to t via Definition 1.4, we have u(t) = u(0) +

1 Γ (q)



t

(t − s)q−1 h(s)ds.

(12)

0

On the other hand, if t ∈ (si , ti+1 ], i = 1, 2, . . . , m and one can apply the impulsive condition of (11) to derive 1 u(t) = Gi (si ) − Γ (q)



si

(si − s)

q−1

0

1 h(s)ds + Γ (q)



t

(t − s)q−1 h(s)ds. (13)

0

So, we have u(0) = u(T ) = Gm (sm ) − +

1 Γ (q)



T

1 Γ (q)



sm

(sm − s)q−1 h(s)ds

0

(T − s)q−1 h(s)ds.

(14)

0

Now it is clear that (12), (13) and (14) imply (10).



3 Main results We introduce the following conditions: [A1] The function f : J × R → R jointly continuous and gi ∈ C([ti , si ] × R, R), i = 1, 2, . . . , m. [A2] There is a positive constant Lf such that   f (t, u1 ) − f (t, u2 ) ≤ Lf |u1 − u2 |, for each t ∈ [si , ti+1 ], and all u1 , u2 ∈ R. [A2 ] There is a constant L > 0 such that    f (t, u) ≤ L 1 + |u| for each t ∈ [si , ti+1 ] and all u ∈ R. [A3] There is a positive constant Lgi , i = 1, 2, . . . , m such that   gi (t, u1 ) − gi (t, u2 ) ≤ Lg |u1 − u2 |, for each t ∈ [ti , si ], and all u1 , u2 ∈ R. i [A3 ] There is a function ϕi (t), i = 1, 2, . . . , m such that   gi (t, u) ≤ ϕi (t) for each t ∈ [ti , si ] and all u ∈ R. In the remainder of this work, Mi = supt∈[ti ,si ] ϕi (t) < ∞, i = 1, 2, . . . , m.

J. Wang, X. Li

3.1 Integer order case Theorem 3.1 Suppose that the conditions [A1], [A2] and [A3] are satisfied. If

Λ := max max Lgi + Lf (ti+1 − si ), Lgm + Lf (T − sm + t1 ) < 1, (15) i=1,2,...,m

then the problem (1) has an unique solution on J . Proof We turn the problem (1) into a fixed point problem. Define an operator F : PC(J, R) → PC(J, R) by ⎧ T t ⎪ ⎪ ⎨ gm (sm , u(sm )) + sm f (s, u(s))ds + 0 f (s, u(s))ds, t ∈ [0, t1 ]; (16) (Fu)(t) = gi (t, u(t)), t ∈ (ti , si ], i = 1, 2, . . . , m; ⎪ ⎪ ⎩ g (s , u(s )) +  t f (s, u(s))ds, t ∈ (s , t ], i = 1, 2, . . . , m. i

i

i

i

si

i+1

It is easy to see that F is well defined and Fu ∈ PC(J, R) for all u ∈ PC(J, R). Now we only need to show that F is a contraction mapping. To this end, we consider three cases. Case 1. For u1 , u2 ∈ PC(J, R) and for t ∈ [0, t1 ], we have   (Fu1 )(t) − (Fu2 )(t)

T

t       u1 (s) − u2 (s)ds + Lf u1 (s) − u2 (s)ds ≤ Lgm u1 (sm ) − u2 (sm ) + Lf 0

sm

  ≤ Lgm + Lf (T − sm + t1 ) u1 − u2 PC .

Case 2. For u1 , u2 ∈ PC(J, R) and for t ∈ (ti , si ], i = 1, 2, . . . , m, we have   (Fu1 )(t) − (Fu2 )(t) ≤ Lg u1 − u2 PC . i Case 3. For u1 , u2 ∈ PC(J, R) and for t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get     (Fu1 )(t) − (Fu2 )(t) ≤ Lg u1 (si ) − u2 (si ) + Lf i 



 u1 (s) − u2 (s)ds

t

si

 ≤ Lgi + Lf (ti+1 − si ) u1 − u2 PC . From above, we obtain Fu1 − Fu2 PC ≤ Λu1 − u2 PC , which implies that F is a contraction and there exists a unique solution u ∈ PC(J, R) of the problem (1).  Theorem 3.2 Assume that [A1], [A2 ], [A3] and [A3 ] hold. If LT < 1 and L¯ := max{Lgi , i = 1, 2, . . . , m} < 1 Then the problem (1) has at least one solution.

Periodic BVP for integer/fractional order nonlinear differential

Proof Let M = max{M1 , M2 , M3 , . . . , Mm }. Setting   Br = u ∈ PC(J, R) : uPC ≤ r where M + LT . 1 − LT We define the operators P and Q on Br as ⎧ ⎪ ⎨ gm (sm , u(sm )), t ∈ [0, t1 ]; (Pu)(t) = gi (t, u(t)), t ∈ (ti , si ], i = 1, 2, . . . , m; ⎪ ⎩ gi (si , u(si )), t ∈ (si , ti+1 ], i = 1, 2, . . . , m. r≥

(17)

⎧ t T ⎪ ⎪ ⎨ sm f (s, u(s))ds + 0 f (s, u(s))ds, t ∈ [0, t1 ]; (Qu)(t) = 0, t ∈ (ti , si ], i = 1, 2, . . . , m; ⎪ ⎪ ⎩  t f (s, u(s))ds, t ∈ (s , t ], i = 1, 2, . . . , m. i

si

i+1

For the sake of convenience, we divide the proof into several steps. Step 1. For any u ∈ Br , we prove that Pu + Qu ∈ Br . Case 1. For each t ∈ [0, t1 ], we get      (Pu + Qu)(t) ≤ gm sm , u(sm )  +



T

  f s, u(s) ds +



t

  f s, u(s) ds

0

sm

≤ Mm + L(1 + r)(T − sm + t1 ) ≤ M + L(1 + r)T ≤ r. Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we have    (Pu + Qu)(t) ≤ |gi si , u(si ) | ≤ Mi ≤ r. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get

t         (Pu + Qu)(t) ≤ gi si , u(si )  + f s, u(s) ds si

≤ Mi + L(1 + r)(ti+1 − si ) ≤ M + L(1 + r)T ≤ r. From above, we infer that Pu + Qu ∈ Br . Step 2. P is contraction on Br . Case 1. For u1 , u2 ∈ Br and for t ∈ [0, t1 ], we have     (Pu1 )(t) − (Pu2 )(t) ≤ Lg u1 (sm ) − u2 (sm ) ≤ Lg u1 − u2 PC . m m Case 2. For u1 , u2 ∈ Br and for t ∈ (ti , si ], i = 1, 2, . . . , m, we have   (Pu1 )(t) − (Pu2 )(t) ≤ Lg u1 − u2 PC . i

J. Wang, X. Li

Case 3. For u1 , u2 ∈ Br and for t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get   (Pu1 )(t) − (Pu2 )(t) ≤ Lg u1 − u2 PC . i From above, we obtain ¯ 1 − u2 PC , Pu1 − Pu2 PC ≤ Lu which implies that P is a contraction Step 3. We show that Q is continuous. Let {un } be a sequence such that un → u in PC(J, R). Case 1. For each t ∈ [0, t1 ], we have       (Qun )(t) − (Qu)(t) ≤ (T − sm + t1 )f ·, un (·) − f ·, u(·)  . PC Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we have   (Qun )(t) − (Qu)(t) = 0. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get       (Qun )(t) − (Qu)(t) ≤ (ti+1 − si )f ·, un (·) − f ·, u(·)  . PC Since the functions f are defined on finite dimensional spaces, from the continuity of the function f and the last inequality, we infer that Qun − QuPC → 0 as n → ∞. Step 4. We prove Q is compact. First, Q is uniformly bounded on Br since Qu ≤ L(1 + r)T < r. Next, we show that Q maps bounded set into equicontinuous set of Br . Case 1. For interval [0, t1 ], 0 ≤ e1 < e2 ≤ t1 , u ∈ Br , one has   (Qu)(e2 ) − (Qu)(e1 ) ≤ L(1 + r)(e2 − e1 ). Case 2. For interval (ti , si ], i = 1, 2, . . . , m, ti ≤ e1 < e2 ≤ si , u ∈ Br , one has   (Qu)(e2 ) − (Qu)(e1 ) = 0. Case 3. For interval (si , ti+1 ], i = 1, 2, . . . , m, si ≤ e1 < e2 ≤ ti+1 , u ∈ Br , one has   (Qu)(e2 ) − (Qu)(e1 ) ≤ L(1 + r)(e2 − e1 ). From above, we get |(Qu)(e2 ) − (Qu)(e1 )| → 0 as e2 → e1 . So Q is equicontinuous. As a consequence of Steps 3–4 together with the PC-type Arzela-Ascoli theorem (Theorem 2.1 of [16] when X = R), we can conclude that Q : Br → Br is continuous and completely continuous. By using Krasnoselskii’s fixed point theorem, F = P + Q has a fixed point which is a solution of the problem (1). The proof is completed. 

Periodic BVP for integer/fractional order nonlinear differential

3.2 Fractional order case In this section, we study the existence of problem (2). Theorem 3.3 Suppose that the conditions [A1], [A2] and [A3] are satisfied. If  q Lf q ti+1 + si , Λ := max max Lg i + i=1,2,...,m−1 Γ (q + 1)   Lf q q T q + sm + t1 < 1, (18) Lg m + Γ (q + 1) then the problem (2) has an unique solution on J . Proof Define an operator Fq : PC(J, R) → PC(J, R) by ⎧ T gm (sm , u(sm )) + Γ 1(q) 0 (T − s)q−1 h(s)ds ⎪ ⎪ ⎪ s ⎪ ⎪ ⎪ − Γ 1(q) 0 m (sm − s)q−1 h(s)ds ⎪ ⎪ ⎪ ⎪ ⎨ + 1  t (t − s)q−1 f (s, u(s))ds, t ∈ [0, t ]; 1 Γ (q) 0 (Fq u)(t) = (19) ⎪ ⎪ gi (t, u(t)), t ∈ (ti , si ], i = 1, 2, . . . , m; ⎪ ⎪ ⎪ ⎪ g (s , u(s )) + 1  t (t − s)q−1 h(s)ds ⎪ i i i ⎪ Γ (q) 0 ⎪ ⎪  si ⎩ 1 − Γ (q) 0 (si − s)q−1 h(s)ds, t ∈ (si , ti+1 ], i = 1, 2, . . . , m. It is easy to see that Fq is well defined. Next, we show that Fq is a contraction mapping. Case 1. For u1 , u2 ∈ PC(J, R) and for each t ∈ [0, t1 ], we have   (Fq u1 )(t) − (Fq u2 )(t)     ≤ gm sm , u1 (sm ) − gm sm , u2 (sm ) 

T     1 + (T − s)q−1 f s, u1 (s) − f s, u2 (s) ds Γ (q) 0

sm     1 + (sm − s)q−1 f s, u1 (s) − f s, u2 (s) ds Γ (q) 0

t     1 + (t − s)q−1 f s, u1 (s) − f s, u2 (s) ds Γ (q) 0    q Lf q q ≤ Lg m + T + sm + t1 u1 − u2 PC . Γ (q + 1) Case 2. For u1 , u2 ∈ PC(J, R) and for each t ∈ (ti , si ], i = 1, 2, . . . , m, we have   (Fq u1 )(t) − (Fq u2 )(t) ≤ Lg u1 − u2 PC . i

J. Wang, X. Li

Case 3. For u1 , u2 ∈ PC(J, R) and for each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get   (Fq u1 )(t) − (Fq u2 )(t)     ≤ gi si , u1 (si ) − gi si , u2 (si ) 

t     1 + (t − s)q−1 f s, u1 (s) − f s, u2 (s) ds Γ (q) 0

si     1 + (si − s)q−1 f s, u1 (s) − f s, u2 (s) ds Γ (q) 0   q Lf q ≤ Lg i + ti+1 + si u1 − u2 PC . Γ (q + 1) From above, we obtain Fq u1 − Fq u2 PC ≤ Λu1 − u2 PC , which implies that Fq is a contraction and there exists a unique solution u ∈ PC(J, R)of the problem (2).  Theorem 3.4 Assume that [A1], [A2 ], [A3] and [A3 ] hold. If  q q  q q L(ti+1 + si ) L(T q + sm + t1 ) , max <1 γ := max i=1,2,...,m Γ (q + 1) Γ (q + 1)

and L¯ < 1,

then the problem (2) has at least one solution. Proof Setting

  Bq,r = u ∈ PC(J, R) : uPC ≤ r

where r≥

M +γ . 1−γ

Consider the operator P defined in (17) and the operator Qq on Bq,r as ⎧ T s 1 ⎪ (T − s)q−1 f (s, u(s))ds − Γ 1(q) 0 m (sm − s)q−1 f (s, u(s))ds ⎪ 0 Γ (q) ⎪ ⎪ t ⎪ 1 q−1 ⎪ ⎪ ⎨ + Γ (q) 0 (t − s) f (s, u(s))ds, t ∈ [0, t1 ]; (Qq u)(t) = 0, t ∈ (ti , si ], i = 1, 2, . . . , m; ⎪ t s ⎪ ⎪ 1 ⎪ (t − s)q−1 f (s, u(s))ds − Γ 1(q) 0 i (si − s)q−1 f (s, u(s))ds, ⎪ 0 Γ (q) ⎪ ⎪ ⎩ t ∈ (si , ti+1 ], i = 1, 2, . . . , m. For the sake of convenience, we divide the proof into several steps. Step 1. For any u ∈ Bq,r , we prove that Pu + Qq u ∈ Bq,r . Case 1. For each t ∈ [0, t1 ], we get

T         (Pu + Qq u)(t) ≤ gm sm , u(sm )  + 1 (T − s)q−1 f s, u(s) ds Γ (q) 0

sm    1 + (sm − s)q−1 f s, u(s) ds Γ (q) 0

Periodic BVP for integer/fractional order nonlinear differential

1 Γ (q)

+ ≤ Mm +



t

   (t − s)q−1 f s, u(s) ds

0 q

q

+ sm + t1 ) (1 + r) < M + γ (1 + r) < r. Γ (q + 1)

L(T q

Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we have      (Pu + Qq u)(t) ≤ gi si , u(si )  ≤ Mi ≤ M < r. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get   (Pu + Qq u)(t)    ≤ gi si , u(si )  + +

1 Γ (q)

≤ Mi +



si

1 Γ (q)



t

   (t − s)q−1 f s, u(s) ds

0

   (si − s)q−1 f s, u(s) ds

0 q L(ti+1

q

+ si )

Γ (q + 1)

(1 + r) < M + γ (1 + r) < r.

So, we infer that Pu + Qq u ∈ Bq,r . Step 2. From Step 2 of Theorem 3.2, we have known that P is contraction on Bq,r . Step 3. We show that Qq is continuous. Let {un } be a sequence such that un → u in PC(J, R). Case 1. For each t ∈ [0, t1 ], we have q q q       (Qq un )(t) − (Qq u)(t) ≤ (T + sm + t1 ) f ·, un (·) − f ·, u(·)  . PC Γ (q + 1)

Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we have   (Qq un )(t) − (Qq u)(t) = 0. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get  (t q + siq )      (Qq un )(t) − (Qq u)(t) ≤ i+1 f ·, un (·) − f ·, u(·)  . PC Γ (q + 1) Thus, we infer that Qq un − Qq uPC → 0 as n → ∞. Step 4. We prove Qq is compact. First, Qq is uniformly bounded on Bq,r since Qq uPC ≤ r. Next, we show that Qq maps bounded set into equicontinuous set of Br . Case 1. For interval [0, t1 ], 0 ≤ e1 < e2 ≤ t1 , u ∈ Bq,r . We have

J. Wang, X. Li

  (Qq u)(e2 ) − (Qq u)(e1 )

e1     1 (e2 − s)q−1 − (e1 − s)q−1 f s, u(s) ds ≤ Γ (q) 0

e2    1 + (e2 − s)q−1 f s, u(s) ds Γ (q) e1 ⎧ q ⎨ L(1+r) [(e1q − e2q ) + 2(e2 − e1 )q ] ≤ 3L(1+r)(e2 −e1 ) , 0 < q < 1, Γ (q+1) Γ (q+1) ≤ ⎩ L(1+r) (eq − eq ) = L(1+r)qξ q−1 (e2 −e1 ) , ξ ∈ (e1 , e2 ), q ≥ 1. 1 Γ (q+1) 2 Γ (q+1) Case 2. For interval (ti , si ], i = 1, 2, . . . , m, ti ≤ e1 < e2 ≤ si , u ∈ Bq,r . We have   (Qq u)(e2 ) − (Qq u)(e1 ) = 0. Case 3. For interval (si , ti+1 ], i = 1, 2, . . . , m, si ≤ e1 < e2 ≤ ti+1 , u ∈ Bq,r . We have ⎧ 3L(1+r)(e2 −e1 )q , 0 < q < 1,  ⎨  Γ (q+1) (Qq u)(e2 ) − (Qq u)(e1 ) ≤ q−1 ⎩ L(1+r)qξ (e2 −e1 ) , ξ ∈ (e1 , e2 ), q ≥ 1. Γ (q+1)

From above, we get |(Qq u)(e2 ) − (Qq u)(e1 )| → 0 as e2 → e1 . So Qq is equicontinuous. As a result, we can conclude that Qq : Bq,r → Bq,r is continuous and completely continuous. By using Krasnoselskii’s fixed point theorem again, Fq = P + Qq has a fixed point which is a solution of the problem (2). The proof is completed.  4 Applications In this section we present some applications of the results in Sect. 3. Example 4.1 Consider ⎧ 1  (t) (or c D 2 u(t)) = ⎪ u ⎪ 0,t ⎨

|u(t)| (1+9et )(1+|u(t)|) ,

|u(t)| (5+et )(1+|u(t)|) ,

u(t) = ⎪ ⎪ ⎩ u(0) = u(3),

t ∈ (0, 1] ∪ (2, 3], (20)

t ∈ (1, 2],

where q = 12 , J = [0, 3] and 0 = s0 < t1 = 1 < s1 = 2 < t2 = 3. Set f (t, u(t)) =

|u(t)| (1+9et )(1+|u(t)|) , g1 (t, u(t))

|u(t)| (5+et )(1+|u(t)|) . Let 1 |f (t, u1 ) − f (t, u2 )| ≤ 10 |u1 − u2 |.

=

u1 , u2 ∈ R and

t ∈ [0, 1] ∪ (2, 3]. Then we have Let u1 , u2 ∈ R and t ∈ (1, 2]. Then we have |g1 (t, u1 ) − g1 (t, u2 )| ≤ 16 |u1 − u2 |. 1 , Lg1 = 16 , one can deduce Λ = Lg1 + Lf (T − s1 + t1 ) = 11 Set Lf = 10 30 < 1, or Λ = Lg 1 +

Lf q Γ (q+1) (T

q

q

+ s1 + t1 ) =

1 6

+

√

√ 3+√ 2+1 5 π

< 0.6345 < 1.

Periodic BVP for integer/fractional order nonlinear differential

Thus all the assumptions in Theorem 3.1/3.3 are satisfied, our results can be applied to the problem (20). Example 4.2 Consider ⎧ 1 |u(t)|  c 2 ⎪ ⎪ ⎨ u (t)(or D0,t u(t)) = (9+et )(1+|u(t)|) , |u(t)| u(t) = (9t+1)(1+|u(t)|) , t ∈ (1, 2], ⎪ ⎪ ⎩ u(0) = u(3),

t ∈ (0, 1] ∪ (2, 3], (21)

where q = 12 , J = [0, 3] and 0 = s0 < t1 = 1 < s1 = 2 < t2 = 3. |u(t)| |u(t)| (9+et )(1+|u(t)|) , g1 (t, u(t)) = (9t+1)(1+|u(t)|) . For u1 , u2 ∈ R and 1 t ∈ (1, 2], |g1 (t, u1 ) − g1 (t, u2 )| ≤ 10 |u1 − u2 |. For all u ∈ R and each t ∈ [0, 1] ∪ 1 1 := (2, 3], |f (t, u)| ≤ 10 (1 + |u|). For all u ∈ R and each t ∈ (1, 2], |g1 (t, u)| ≤ 9t+1 1 1 ϕ1 (t) with supt∈[1,2] ϕ1 (t) = 10 = M1 := M. Put L = Lg1 = L¯ = 10 < 1. Moreover, LT = 13 < 1,

Set f (t, u(t)) =

√ √  q q q q  L(T q + s1 + t1 ) L(t2 + s1 ) 3+ 2+1 , = γ := max = 0.4679 < 1. √ Γ (q + 1) Γ (q + 1) 5 π Thus, all the assumptions in Theorem 3.2/3.4 are satisfied, our results can be applied to the problem (21). Acknowledgements The authors thank the referees for their careful reading of the manuscript and insightful comments, which helped to improve the quality of the paper. We would also like to acknowledge the valuable comments and suggestions from the editors, which vastly contributed to improve the presentation of the paper.

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