J. Appl. Math. Comput. DOI 10.1007/s12190-016-1025-8 ORIGINAL RESEARCH

Integral boundary value problems for nonlinear non-instantaneous impulsive differential equations Dan Yang1 · JinRong Wang1

Received: 21 April 2016 © Korean Society for Computational and Applied Mathematics 2016

Abstract In this paper, we study integral boundary value problems for two classes of nonlinear non-instantaneous impulsive ordinary differential equations. More precisely, one is integer order impulsive model, the other is fractional order impulsive model. By using standard fixed point approach, a series of existence results are presented under different conditions. Keywords Integral boundary value problems · Non-instantaneous impulsive differential equations · Fractional order · Existence Mathematics Subject Classification 26A33 · 34B37

1 Introduction Recently, models governed by differential equations with non-instantaneous impulses based on the background of injecting drugs for a person are firstly offered by Hernández and O’Regan [1]. It also can be widely used in mechanical engineer, medical science and any other fields. For instance, biological phenomena involving thresholds, bursting rhythm models in medicine and biology, and learning control model [2,3]. Concerning on this new type impulsive models, impulsive action start at an arbitrary fixed point and remains active on a finite time interval, which is much differ-

B

JinRong Wang [email protected] Dan Yang [email protected]

1

Department of Mathematics, Guizhou University, Guiyang, Guizhou 550025, People’s Republic of China

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D. Yang, J. Wang

ent from classical instantaneous impulsive that changes is relatively short compared to the overall duration of the whole process. For more recent development on integer or fractional order non-instantaneous impulsive differential equations, we refer to [4–18] and reference therein. In this paper, we study the following integral boundary value problems for integer order nonlinear differential equations with non-instantaneous impulses: ⎧  ⎨ u (t) = f (t, u(t)), u(t) = gi (t, u(t)), 1 ⎩ u(0) = 0 u(s)ds,

t ∈ (si , ti+1 ], t ∈ (ti , si ],

i = 0, 1, 2, . . . , m, i = 1, 2, . . . , m,

(1)

and fractional order nonlinear differential equations with non-instantaneous impulses: ⎧c q ⎨ D0,t u(t) = f (t, u(t)), u(t) = gi (t, u(t)), 1 ⎩ u(0) = 0 u(s)ds,

t ∈ (si , ti+1 ], t ∈ (ti , si ],

i = 0, 1, 2, . . . , m, q ∈ (0, 1), i = 1, 2, . . . , m, (2)

q

where c D0,t denotes the Caputo fractional derivative of the order q with the lower limit zero, 0 = s0 < t1 ≤ s1 ≤ t2 < · · · < tm ≤ sm < tm+1 = 1 are pre-fixed numbers, f : [0, 1] × R → R is continuous and gi : [ti , si ] × R → R is continuous for all i = 1, 2, . . . , m. We firstly derive the corresponding integral equations for linear problems for integer or fractional order non-instantaneous impulsive equations. Secondly, we present sufficient conditions for existence of the solutions for (1) and (2) by using fixed point method, which is a very powerful and important tool to the study of differential problems. Finally, examples are given to illustrate the theoretical results.

2 Preliminaries We set J := [0, 1] and C(J, R) = {v : v : J → R is continuous} with the standard norm vC := maxt∈J |v(t)|. Consider the piecewise continuous space PC(J, R) := {v : J → R : v ∈ C((tk , tk+1 ], R), k = 0, 1, . . . , m and there exist v(tk− ) and v(tk+ ), k = 1, . . . , m, with v(tk− ) = v(tk )} with the norm v PC := maxt∈J |v(t)|. Set PC 1 (J, R) := {v ∈ PC(J, R) : v  ∈ PC(J, R)} with v PC 1 := max{v PC , v   PC }. Then, (PC 1 (J, R),  ·  PC 1 ) is also a Banach space. Definition 2.1 The fractional integral of order γ with the lower limit zero for a func t f (s) γ 1 tion f : [0, ∞) → R is defined as It f (t) = (γ ) 0 (t−s)1−γ ds, t > 0, γ > 0, provided the right side is point-wise defined on [0, ∞), where (·) is the gamma function. Definition 2.2 The Riemann-Liouville derivative of order γ with the lower limit zero γ γ for a function f : [0, ∞) → R can be written as L D0,t f (t) := L Dt f (t) =  t f (s) dn 1 (n−γ ) dt n 0 (t−s)γ +1−n ds, t > 0, where n = [γ ] + 1 and [γ ] denotes the integer part of γ .

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Integral boundary value problems...

Definition 2.3 The generalized Caputo derivative of order γ with the lower limit zero γ γ γ for a function f : [0, ∞) → R can be written as c D0,t f (t) := c Dt f (t)= L Dt f (t)− n−1 t k (k)  (0) , t > 0, n = [γ ] + 1. k=0 k! f Definition 2.4 A function u ∈ PC 1 (J, R) is called a solution of the problem (1), if 1 u satisfies u(0) = 0 u(s)ds, u  (t) = f (t, u(t)), t ∈ (si , ti+1 ], i = 0, 1, 2, . . . , m, and u(t) = gi (t, u(t)), t ∈ (ti , si ], i = 1, 2, . . . , m. Definition 2.5 A function u ∈ PC 1 (J, R) is said to be a solution of the problem (2) if u(t) = u k (t) for t ∈ (sk , tk+1 ) and u k ∈ C([0, tk+1 ], R) satisfies c D q u (t) = f (t, u (t)) a.e. on (0, t k k+1 ), k = 0, 1, 2, . . . , m, the conditions t k 1 u(t) = gk (t, u(t)), t ∈ (ti , si ], k = 1, 2, . . . , m, and u(0) = 0 u(s)ds hold. Remark 2.6 In Definition 2.5, we establish our concept on solution to impulsive fractional differential equations based on Definition 2.3 keeping the lower limit with zero. Concerning on other concept on solution to impulsive fractional differential equations by changing the lower limit associated with impulsive points, the reader can refer to Agarwal et al. [19], where two mainstream viewpoints are collected. Lemma 2.7 Assume h : [0, 1] → R and G i : [ti , si ] → R are continuous functions. A function u ∈ PC 1 (J, R) is a solution of the problem ⎧  ⎨ u (t) = h(t), u(t) = G i (t), 1 ⎩ u(0) = 0 u(s)ds,

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

(3)

if and only if u satisfies the following integral equation ⎧ t J +J +J ⎪ ⎨ 1 1−t2 1 3 + 0 h(s)ds, t ∈ [0, t1 ], t ∈ (ti , si ], i = 1, 2, . . . , m, u(t) = G i (t),  ⎪ ⎩ G i (si ) + t h(s)ds, t ∈ (si , ti+1 ], i = 1, 2, . . . , m, si

(4)

where

J1 = J2 = J3 =

t1



s

0 0 m si 

h(τ )dτ ds, G i (s)ds,

i=1 ti m ti+1  i=1





s

G i (si ) +

si

h(τ )dτ ds.

si

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

t

u(t) = u(0) +

h(s)ds. 0

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In general, for t ∈ (si , ti+1 ], i = 1, 2, . . . , m, and one can apply the impulsive condition of (3), we obtain

t

u(t) = G i (si ) +

h(s)ds. si

Now we remain to determine the initial value u(0). Note that

u(0) =

1

0

=

t1



u(s)ds =



s

u(0) +

s1

+

0



0

+

t1

t1

i=1

ti+1

+··· +



i=1



s

G i (si ) +

sm

+

tm

m

 h(τ )dτ ds +

si

= u(0)t1 +

+

t2 s1

0 m





si

tm+1

u(s)ds

sm

G i (s)ds

ti

h(τ )dτ ds

si 3 

Ji .

i=1

Thus, one can solve u(0) =

J1 + J2 + J3 . 1 − t1

As a result, sufficiency is verified. Conversely, assume that u satisfies (4). One can verify the results immediately. The proof is completed.  Lemma 2.8 Define

t1 s 1 V1 = (s − τ )q−1 h(τ )dτ ds, (q) 0 0 m si  V2 = G i (s)ds, V3 =

i=1 ti m ti+1  i=1 si

si

s 1 1 q−1 q−1 G i (si ) − (si − τ ) h(τ )dτ + (s − τ ) h(τ )dτ ds. (q) 0 (q) 0

A function u given by ⎧

t V1 + V2 + V3 1 ⎪ ⎪ + (t − s)q−1 h(s)ds, t ∈ [0, t1 ], ⎪ ⎪ ⎪ 1 − t1 (q) 0 ⎪ ⎪ ⎪ ⎨ G i (t), t ∈ (ti , si ], i = 1, 2, . . . , m, u(t) = (5)

t

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

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Integral boundary value problems...

is a solution of the following problem ⎧c q ⎨ D0,t u(t) = h(t), t ∈ (si , ti+1 ], u(t) = G i (t), t ∈ (ti , si ], 1 ⎩ u(0) = 0 u(s)ds.

i = 0, 1, 2, . . . , m, i = 1, 2, . . . , m,

(6)

Proof Assume that u satisfies (6). For t ∈ [0, t1 ], integrating the first equation of (6) from zero to t, we have

t 1 (t − s)q−1 h(s)ds. u(t) = u(0) + (q) 0 In addition, for t ∈ (si , ti+1 ], i = 1, 2, . . . , m integrating the first equation of (6) from si to t, we have

si

t 1 1 q−1 u(t) = u(si ) − (si − s) h(s)ds + (t − s)q−1 h(s)ds. (q) 0 (q) 0 For t ∈ (ti , si ], apply the impulsive condition of (6) we have u(si ) = G i (si ). So 1 u(t) = G i (si ) − (q)



si

(si − s)

q−1

0

1 h(s)ds + (q)



t

(t − s)q−1 h(s)ds.

0

Next

u(0) =

1

0

=

t1



t1

u(s)ds = 0

u(0) +

+



s

+

1 (q)

= u(0)t1 +

s

t2

+

+··· +

s1

sm tm



tm+1

+

1 (q)

i=1



si

u(s)ds

sm

m

 q−1 (s − τ ) h(τ )dτ ds +

si





t1

1 (q) 0 0

m  ti+1 G i (si ) − + i=1

s1

si

G i (s)ds

ti

(si − τ )q−1 h(τ )dτ

0

(s − τ )q−1 h(τ )dτ ds

0 3 

Vi .

i=1

Thus, we obtain u(0) = The proof is completed.

V1 + V2 + V3 . 1 − t1 

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Theorem 2.9 (Krasnoselskii’s fixed point theorem) Let M be a nonempty and closed convex, a mapping P z = Az + Bz, such that (i) Ax + By ∈ M for each x, y ∈ M, (ii) A is compact and continuous, (iii) B is a contraction mapping. Then there exist z ∈ M such that z = Az + Bz. Theorem 2.10 (Hölder inequality) If f (x) and g(x) are continuous, nonnegab tive functions on the closed interval a ≤ x ≤ b, then a f (x)g(x)d x ≤ 1  1 b b ( a f p1 (x)d x) p1 ( a g p2 (x)d x) p2 , where p11 + p12 = 1, p1 and p2 are real numbers greater than 1.

3 Main results To consider the existence results, we list the following assumptions: [A1] The function f : J × R → R is jointly continuous and gi ∈ C([ti , si ] × R, R), i = 1, 2, . . . , m. [A2] There exists a positive constant L f such that | f (t, u 1 ) − f (t, u 2 )| ≤ L f |u 1 − u 2 |, for each t ∈ [si , ti+1 ], and all u 1 , u 2 ∈ R. [A2 ] There exists an integrable function ψ : [si , ti+1 ] → R+ , and an integrable, monotonous function ν : R+ → R+ such that | f (t, u)| ≤ ψ(t)ν(u PC ), for each t ∈ [si , ti+1 ], and all u ∈ R. [A3] There exists a positive constant L gi , i = 1, 2, . . . , m such that |gi (t, u 1 ) − gi (t, u 2 )| ≤ L gi |u 1 − u 2 |, for each t ∈ [ti , si ], and all u 1 , u 2 ∈ R. [A3 ] There exists an integrable function ϕi : [ti , si ] → R+ , i = 1, 2, . . . , m and an integrable, monotonous function μ : R+ → R+ such that |gi (t, u)|  s ≤ ϕi (t)μ(u PC ), for each t ∈ [ti , si ], and all u ∈ R, where Mi = ti i ϕi (t)dt = ϕi  L 1 . [ti ,si ]

At beginning, we study the existence of solutions to problem (1). Let L g := maxi=1,2,...,m L gi . Now we are ready to state our first result. Theorem 3.1 Suppose that the conditions [A1], [A2] and [A3] are satisfied. If 

m  1 1 2 Lf  := max (ti+1 − si ) + L g (1 − t1 ) + L f t1 , 1 − t1 2 i=0  L g + L f (ti+1 − si ) < 1, 

then the problem (1) has a unique solution.

123

(7)

Integral boundary value problems...

Proof We turn the problem (1) into a fixed point problem. Define an operator F : PC(J, R) → PC(J, R) by

(Fu)(t) =

⎧ ⎪ ⎨

t I1 +I2 +I3 + 0 1−t1 gi (t, u(t)),

f (s, u(s))ds,

t ∈ [0, t1 ], t ∈ (ti , si ], i = 1, 2, . . . , m, t ∈ (si , ti+1 ], i = 1, 2, . . . , m, (8)

⎪ ⎩ gi (si , u(si )) +  t f (s, u(s))ds, si

where

I1 = I2 = I3 =

t1



s

0 0 m si 

f (τ, u(τ ))dτ ds, gi (s, u(s))ds,

i=1 ti m ti+1  i=1



gi (si , u(si )) +

si

s

f (τ, u(τ ))dτ ds.

si

Clearly, 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 u 1 , u 2 ∈ PC(J, R) and for t ∈ [0, t1 ], we have |(Fu 1 )(t) − (Fu 2 )(t)| ≤

 1   |I1 − I1 | + |I2 − I2 | + |I3 − I3 | 1 − t1

t |u 1 (s) − u 2 (s)|ds, +Lf 0

where  t1 s 

t1 s   |I1 − I1 | =  f (τ, u 1 (τ ))dτ ds − f (τ, u 2 (τ ))dτ ds  0 0 0 0

t1 s ≤ | f (τ, u 1 (τ )) − f (τ, u 2 (τ ))|dτ ds 0 0

t1 s ≤ Lf |u 1 (τ ) − u 2 (τ )|dτ ds 0

0

1 ≤ L f t12 u 1 − u 2  PC , 2 and |I2

−

I2 |

 m

 =  ≤

i=1 ti m  si i=1

si

gi (s, u 1 (s))ds −

m

 i=1

si ti

  gi (s, u 2 (s))ds 

|gi (s, u 1 (s)) − gi (s, u 2 (s))|ds

ti

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D. Yang, J. Wang m



≤

ti

i=1

≤ Lg

si

L gi |u 1 (s) − u 2 (s)|ds

m  (si − ti )u 1 − u 2  PC , i=1

and |I3

−

I3 |



 m  =

ti+1



i=1 si m ti+1 

gi (si , u 1 (si )) +

si



≤

s

si

si

m ti+1 

  f (τ, u 2 (τ ))dτ ds 

|gi (si , u 1 (si ) − gi (si , u 2 (si )|ds

si

i=1

+

gi (si , u 2 (si )) +

−

i=1

f (τ, u 1 (τ ))dτ ds

s

m



ti+1



s

| f (τ, u 1 (τ )) − f (τ, u 2 (τ ))|dτ ds

si i=1 si

m ti+1 

|u 1 (si ) − u 2 (si )|ds + L f

≤ Lg ≤ Lg =

i=1 si m 

(ti+1 i=1 m 

Lg

i=1

− si )u 1 − u 2  PC +

m



ti+1

i=1 si m 

1 Lf 2



s

|u 1 (τ ) − u 2 (τ )|dτ ds

si

(ti+1 − si ) u 1 − u 2  PC 2

i=1

m  1 (ti+1 − si ) + L f (ti+1 − si )2 u 1 − u 2  PC . 2 i=1

Thus |(Fu 1 )(t) − (Fu 2 )(t)|  m m   1 1 2 L f t1 + L g ≤ (si − ti ) + L g (ti+1 − si ) 1 − t1 2 i=1 i=1

 m  1 + Lf (ti+1 − si )2 + L f t1 u 1 − u 2  PC 2 i=1



m  1 1 2 Lf = (ti+1 − si ) + L g (1 − t1 ) + L f t1 u 1 − u 2  PC . 1 − t1 2 i=0

Case 2. For u 1 , u 2 ∈ PC(J, R) and for t ∈ (ti , si ], i = 1, 2, . . . , m, we have |(Fu 1 )(t) − (Fu 2 )(t)| = |gi (t, u 1 (t)) − gi (t, u 2 (t))| ≤ L gi u 1 − u 2  PC .

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Integral boundary value problems...

Case 3. For u 1 , u 2 ∈ PC(J, R) and for t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we have

t

|(Fu 1 )(t) − (Fu 2 )(t)| ≤ L gi |u 1 (si ) − u 2 (si )| + L f |u 1 (s) − u 2 (s)|ds si

≤ L gi + L f (ti+1 − si ) u 1 − u 2  PC . From above, we obtain Fu 1 − Fu 2  PC ≤ u 1 − u 2  PC , which implies that F is a contractive mapping and there exists a unique solution u ∈ PC(J, R) of the problem(1).  Let M = maxi=1,2,...,m Mi , N = maxi=1,2,...,m ϕi (si ) and K = ψ L 1 . Now we are ready to state below theorem.

1 0

ψ(t)dt =

[0,1]

Theorem 3.2 Assume that [A1], [A2 ], [A3] and [A3 ] hold. If L g < 1 and 2 1 max{ 1−t m Mμ(r ) + 2−t 1−t1 K ν(r ), N μ(r ) + K ν(r )} ≤ r holds for some r > 0, then 1 the problem (1) has at least one solution. Proof Setting Br = {u ∈ PC(J, R) : u PC ≤ r } where 

 2 2 − t1 r ≥ max m Mμ(r ) + K ν(r ), N μ(r ) + K ν(r ) . 1 − t1 1 − t1 We define the operators P and Q on Br as follows:

(Pu)(t) =

⎧ ⎪ ⎪ ⎨

1 1−t1

m  si i=1 ti

gi (s, u(s))ds +

m  ti+1 i=1 si

gi (si , u(si ))ds ,

gi (t, u(t)), t ∈ (ti , si ], ⎪ ⎪ ⎩ gi (si , u(si )), t ∈ (si , ti+1 ],

t ∈ [0, t1 ], i = 1, 2, . . . , m, i = 1, 2, . . . , m,

(9) and ⎧ t1 s

m ti+1 s  ⎪ 1 ⎪ ⎪ f (τ, u(τ ))dτ ds + f (τ, u(τ ))dτ ds ⎪ ⎪ ⎪ 0 0 si ⎪ 1 − t1 i=1 si ⎪ ⎪

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

⎪ t ⎪ ⎪ ⎪ f (s, u(s))ds, t ∈ (si , ti+1 ], i = 1, 2, . . . , m. ⎩ si

For the sake of convenience, we divide the proof into several steps. Step 1. For any u ∈ Br ,we prove that Pu + Qu ∈ Br .

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Case 1. For each t ∈ [0, t1 ], we get |(Pu + Qu)(t)| ⎛ m

m ti+1  1 ⎝ si ≤ |gi (s, u(s))|ds + |gi (si , u(si ))|ds 1 − t1 ti si i=1

+

t1 s 0

0

i=1

| f (τ, u(τ ))|dτ ds +

m ti+1 s  i=1 si

si

| f (τ, u(τ ))|dτ ds ⎠ +

⎛ m

m ti+1  1 ⎝ si ≤ ϕi (s)μ(u)ds + ϕi (si )μ(u)ds 1 − t1 ti si i=1

+

t1 s 0

0

i=1

ψ(τ )ν(u)dτ ds +

m ti+1 s  i=1 si

si

⎞

⎞

ψ(τ )ν(u)dτ ds ⎠ +

t 0

t 0

| f (s, u(s))|ds

ψ(s)ν(u)ds

⎞ ⎛ m m m    1 ⎝ Mi + μ(r ) Mi + K t1 ν(r ) + K ν(r ) (ti+1 − si )⎠ + K ν(r ) ≤ μ(r ) 1 − t1 i=1 i=1 i=1 ⎞ ⎛ m  1 ⎝ ≤ (ti+1 − si )⎠ + K ν(r ) 2m Mμ(r ) + K ν(r ) 1 − t1 i=0

 1  ≤ 2m Mμ(r ) + K ν(r ) + K ν(r ) 1 − t1 2 2 − t1 = m Mμ(r ) + K ν(r ) ≤ r. 1 − t1 1 − t1

Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we get |(Pu + Qu)(t)| ≤ |gi (si , u(si ))| ≤ ϕi (si )μ(r ) ≤ r. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get

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

≤ ϕi (si )μ(r ) + K ν(r ) ≤ r. From above, we infer that Pu + Qu ∈ Br . Step 2. P is contractive mapping on Br . Case 1. For u 1 , u 2 ∈ Br and for t ∈ [0, t1 ], we have 

m m  1 Lg (si − ti ) + L g (ti+1 − si ) u 1 − u 2  PC |(Pu 1 )(t) − (Pu 2 )(t)| ≤ 1 − t1 i=1

i=1

1 = L g (1 − t1 )u 1 − u 2  PC 1 − t1 = L g u 1 − u 2  PC .

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Integral boundary value problems...

Case 2. For u 1 , u 2 ∈ Br and for t ∈ (ti , si ], i = 1, 2, . . . , m, we have |(Pu 1 )(t) − (Pu 2 )(t)| ≤ L gi u 1 − u 2  PC . Case 3. For u 1 , u 2 ∈ Br and for t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we have |(Pu 1 )(t) − (Pu 2 )(t)| ≤ L gi u 1 − u 2  PC . From above, we obtain Pu 1 − Pu 2  PC ≤ L g u 1 − u 2  PC , which implies that P is a contractive. Step 3. We show that Q is continuous. Let {u n } be a sequence such that u n → u in PC(J, R). Case 1. For each t ∈ [0, t1 ], we have |(Qu n )(t) − (Qu)(t)| ≤

m  1 (ti+1 − si )2 + t1  f (·, u n (·)) − f (·, u(·)) PC . 2 − 2t1 i=0

Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we have |(Qu n )(t) − (Qu)(t)| = 0. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we have |(Qu n )(t) − (Qu)(t)| ≤ (ti+1 − si ) f (·, u n (·)) − f (·, u(·)) PC . Since the function f is defined on finite dimensional spaces, from the continuity of the function f and the last inequality, we infer that Qu n − Qu PC → 0 as n → ∞. Step 4. We prove Q is compact. First, Q is uniformly bounded on Br . Next, we show that Q maps bounded set into equicontinuous set of Br . Case 1. For a1 , a2 ∈ [0, t1 ], and a1 < a2 , u ∈ Br , one has

|(Qu)(a2 ) − (Qu)(a1 )| ≤

a2

| f (s, u(s))|ds

a

1a2

ψ(s)ν(u)ds

a2 ψ(s)ds. ≤ ν(r )

≤

a1

a1

Case 2. For a1 , a2 ∈ (ti , si ], i = 1, 2, . . . , m, and a1 < a2 , u ∈ Br , one has |(Qu)(a2 ) − (Qu)(a1 )| = 0. Case 3. For a1 , a2 ∈ (si , ti+1 ], i = 1, 2, . . . , m, and a1 < a2 , u ∈ Br , one has

|(Qu)(a2 ) − (Qu)(a1 )| ≤ ν(r )

a2

ψ(s)ds.

a1

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D. Yang, J. Wang

From above, we get |(Qu)(a2 ) − (Qu)(a1 )| → 0 as a2 → a1 . So Q is equicontinuous. Now we recall the Steps 3 − 4 via PC−type Arzela-Ascoli theorem [20, Theorem 2.1] to find Q : Br → Br is continuous and compact. Then, one can utilize Theorem 2.9, F = P + Q to obtain existence of solution of problem (1). The proof is completed.  Next, we study the existence of solutions to problem (2). Theorem 3.3 Suppose that the conditions [A1], [A2] and [A3] are satisfied. If  q := max



m    Lf q+1 q+1 ti+1 − si + L g (1 − t1 ) (q + 2) i=0   m  q Lf Lf Lf q q q + < 1, si (ti+1 −si ) + t , Lg + t +s (q +1) (q +1) 1 (q + 1) i+1 i

1 1 − t1

i=1

(10) then the problem (2) has a unique solution. Proof Define an operator Fq : PC(J, R) → PC(J, R) by ⎧

t I1 (q) + I2 + I3 (q) 1 ⎪ ⎪ ⎪ + (t − s)q−1 f (s, u(s))ds, t ∈ [0, t1 ], ⎪ ⎪ 1 − t (q) 1 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎨ gi (t, u(t)), t ∈ (ti , si ], i = 1, 2, . . . , m,

t (Fq u)(t) = 1 ⎪ g (s , u(s )) + (t − s)q−1 f (s, u(s))ds ⎪ i i i ⎪ ⎪ (q) 0 ⎪ ⎪

si ⎪ ⎪ 1 ⎪ ⎪ (si − s)q−1 f (s, u(s))ds, t ∈ (si , ti+1 ], i = 1, 2, . . . , m. ⎩− (q) 0 (11) where I2 is defined in Theorem 3.1 and I1 (q) = I3 (q) =

1 (q) m

 i=1

t1



0 ti+1





si

1 + (q)



s

(s − τ )q−1 f (τ, u(τ ))dτ ds,

0

gi (si , u(si )) − s

(s − τ )

q−1

1 (q)

0

si

(si − τ )q−1 f (τ, u(τ ))dτ

f (τ, u(τ ))dτ ds.

0

Clearly, Fq is well defined. Next, we show that Fq is a contractive mapping. Case 1. For u 1 , u 2 ∈ PC(J, R) and for each t ∈ [0, t1 ], we have |(Fq u 1 )(t) − (Fq u 2 )(t)| Lf 1 q t u 1 − u 2  PC , (|I  (q) − I1 (q)| + |I2 − I2 | + |I3 (q) − I3 (q)|) + ≤ 1 − t1 1 (q + 1) 1

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where |I1 (q) − I1 (q)| 

t1 s  1 =  (s − τ )q−1 f (τ, u 1 (τ ))dτ ds (q) 0 0 

t1 s  1 − (s − τ )q−1 f (τ, u 2 (τ ))dτ ds  (q) 0 0

t1 s 1 ≤ (s − τ )q−1 | f (τ, u 1 (τ )) − f (τ, u 2 (τ ))|dτ ds (q) 0 0 Lf q+1 t u 1 − u 2  PC , ≤ (q + 2) 1 and |I2

−

I2 |

m  ≤ Lg (si − ti )u 1 − u 2  PC , i=1

and |I3 (q) − I3 (q)| m ti+1  ≤ |gi (si , u 1 (si )) − gi (si , u 2 (si ))|ds i=1

si

1  (q) m

+



ti+1 si

i=1



si

(si − τ )q−1 | f (τ, u 1 (τ )) − f (τ, u 2 (τ ))|dτ ds

0

m

1  ti+1 s + (s − τ )q−1 | f (τ, u 1 (τ )) − f (τ, u 2 (τ ))|dτ ds (q) 0 s i i=1 m ti+1  ≤ Lg |u 1 (si ) − u 2 (si )|ds i=1

+ +

Lf (q) Lf (q)

si m ti+1  i=1

si

m

 i=1

ti+1 si



si

(si − τ )q−1 |u 1 (τ ) − u 2 (τ )|dτ ds

0



s

(s − τ )q−1 |u 1 (τ ) − u 2 (τ )|dτ ds

0

m  ≤ Lg (ti+1 − si )u 1 − u 2  PC + i=1

 q Lf si (ti+1 − si )u 1 − u 2  PC (q + 1) m

i=1

m  



Lf q+1 q+1 ti+1 − si u 1 − u 2  PC (q + 2) i=1  m m  m    Lf Lf q q+1 q+1 = Lg (ti+1 − si ) + si (ti+1 − si ) + ti+1 − si (q + 1) (q + 2) +

i=1

i=1

i=1

×u 1 − u 2  PC .

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Thus |(Fq u 1 )(t) − (Fq u 2 )(t)|   m m   Lf 1 q+1 t1 + L g ≤ (si − ti ) + L g (ti+1 − si ) 1 − t1 (q + 2) i=1 i=1   m m     Lf Lf Lf q q+1 q+1 q t si (ti+1 − si ) + + + ti+1 − si (q + 1) (q + 2) (q + 1) 1 i=1

i=1

× u 1 − u 2  PC 

m  m    Lf Lf 1 q+1 q+1 q = si (ti+1 − si ) ti+1 − si + L g (1 − t1 ) + 1 − t1 (q + 2) (q + 1) i=0 i=1  Lf q + t u 1 − u 2  PC . (q + 1) 1

Case 2. For u 1 , u 2 ∈ PC(J, R) and for t ∈ (ti , si ], i = 1, 2, . . . , m, we have |(Fq u 1 )(t) − (Fq u 2 )(t)| = |gi (t, u 1 (t)) − gi (t, u 2 (t))| ≤ L gi u 1 − u 2  PC . Case 3. For u 1 , u 2 ∈ PC(J, R) and for t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we have |(Fq u 1 )(t) − (Fq u 2 )(t)|

t 1 ≤ L gi |u 1 (si ) − u 2 (si )| + (t − s)q−1 | f (s, u 1 (s)) − f (s, u 2 (s))|ds (q) 0

si 1 (si − s)q−1 | f (s, u 1 (s)) − f (s, u 2 (s))|ds + (q) 0

q Lf q t ≤ L gi + + si u 1 − u 2  PC . (q + 1) i+1

From above, we obtain Fq u 1 − Fq u 2  PC ≤ q u 1 − u 2  PC , which implies that Fq is a contractive mapping. Thus there exists a unique solution u ∈ PC(J, R) of the problem (2).  Let (q − 1) p1 + 1 > 0 for some p1 > 1 and K q = ψ L p2 = ( [0,1] q−1+ p1 1

si

1 0

1

ψ p2 (t)dt) p2 . Denote γ :=

ψ

p2 L [0,1]

1

, where

((q−1) p1 +1) p1 (q) q−1+ p1 q−1+ p1 1 1 K q max{( 1+m + t ), (t 1 i+1 1−t1

+

)}. Now we are ready to give the following result.

Theorem 3.4 Assume that [A1], [A2 ], [A3] and [A3 ] hold. If (q − 1) p1 + 1 > 0 for 2 some p1 > 1, L g < 1 and r ≥ max{ 1−t m Mμ(r ) + γ ν(r ), N μ(r ) + γ ν(r )} holds 1 for some r > 0, then the problem (2) has at least one solution.

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Integral boundary value problems... 2 Proof Setting Bq,r = {u ∈ PC(J, R) : u PC ≤ r } where r ≥ max{ 1−t m Mμ(r ) + 1 γ ν(r ), N μ(r ) + γ ν(r )}. Consider the operator P defined in (9) and the operator Q q on Bq,r as



t1 s ⎧ 1 1 ⎪ ⎪ (s − τ )q−1 f (τ, u(τ ))dτ ds ⎪ ⎪ 1 − t (q) ⎪ 1 0 0 ⎪ ⎪ ⎪

m

⎪ ⎪ 1  ti+1 si ⎪ ⎪ − (si − τ )q−1 f (τ, u(τ ))dτ ds ⎪ ⎪ ⎪ (q) 0 ⎪ i=1 si ⎪ ⎪ ⎪

⎪ m ti+1 s ⎪  ⎪ 1 q−1 ⎪ ⎪ (s − τ ) f (τ, u(τ ))dτ ds ⎨ + (q) 0 i=1 si (Q q u)(t) =

t ⎪ ⎪ 1 ⎪ ⎪ ⎪ + (t − s)q−1 f (s, u(s))ds, t ∈ [0, t1 ], ⎪ ⎪ (q) ⎪ 0 ⎪ ⎪ ⎪ ⎪ , s ], i = 1, 2, . . . , m, 0, t ∈ (t ⎪ i i ⎪ ⎪

t

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

Step 1. For any u ∈ Bq,r ,we prove that Pu + Q q u ∈ Bq,r . Case 1. For each t ∈ [0, t1 ], we get |(Pu + Q q u)(t)|  m

m ti+1  si  1 |gi (s, u(s))|ds + |gi (si , u(si ))|ds ≤ 1 − t1 i=1 ti i=1 si

t1 s 1 + (s − τ )q−1 | f (τ, u(τ ))|dτ ds (q) 0 0

m

1  ti+1 si + (si − τ )q−1 | f (τ, u(τ ))|dτ ds (q) 0 i=1 si 



m 1  ti+1 s q−1 + (s − τ ) | f (τ, u(τ ))|dτ ds (q) 0 i=1 si

t 1 + (t − s)q−1 | f (s, u(s))|ds (q) 0 By using Theorem 2.10, we obtain |(Pu + Q q u)(t)| m  q−1+ p1 p1 K q ν(r ) q+ p11 1 1 ≤ t1 2m Mμ(r ) + + K q ν(r ) si (ti+1 − si ) 1 − t1 qp1 + 1 i=1

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D. Yang, J. Wang

m q+ p1 q−1+ p1 p1 K q ν(r )  q+ p11 1 1 ti+1 − si + K q ν(r )t1 + qp1 + 1 i=1

m q+ p1 p1 K q ν(r )  q+ p11 1 2m Mμ(r ) + ti+1 − si 1 ≤ 1 − t1 qp1 + 1 i=0

m  q−1+ p1 q−1+ p1 1 1 +K q ν(r ) si (ti+1 − si ) + K q ν(r )t1 i=1



m  q+ p1 q+ p1 1 2 1 1 ti+1 − si ≤ m Mμ(r ) + K q ν(r ) 1 − t1 (qp1 + 1) − (qp1 + 1)t1 i=0  m q−1+ p1 1  q−1+ p11 1 si (ti+1 − si ) + t1 + 1 − t1 i=1 ⎛  m   2(q−1+ p1 ) 1 2 1 1 ≤ m Mμ(r ) + K q ν(r ) ⎝ + si 1 − t1 1 − t1 1 − t1 i=1 ⎞  1  m 2  q−1+ p1 1⎠ (ti+1 − si )2 + t1 i=1

q−1+ p1 2 1+m 1 m Mμ(r ) + K q ν(r ) + t1 1 − t1 1 − t1 2 = m Mμ(r ) + γ ν(r ) ≤ r. 1 − t1 =

Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we get |(Pu + Q q u)(t)| ≤ |gi (si , u(si ))| ≤ ϕi (si )μ(r ) ≤ r. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m, we get

t 1 (t − s)q−1 | f (s, u(s))|ds |(Pu + Q q u)(t)| ≤ |gi (si , u(si ))| + (q) 0

si 1 (si − s)q−1 | f (s, u(s))|ds + (q) 0

q−1+ p1 q−1+ p1 1 ≤ ϕi (si )μ(r ) + K q ν(r ) ti+1 1 + si ≤ ϕi (si )μ(r ) + γ ν(r ) ≤ r. So, we infer that Pu + Q q u ∈ Bq,r . Step 2. From Step 2 of Theorem 3.2, we have known that P is contractive on Bq,r . Step 3. We show that Q q is continuous. Let {u n } be a sequence such that u n → u in PC(J, R).

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Integral boundary value problems...

Case 1. For each t ∈ [0, t1 ], we have |(Q q u n )(t) − (Q q u)(t)| 

m  m    1 1 1 q+1 q+1 q ≤ ti+1 − si + si (ti+1 − si ) 1 − t1 (q + 2) (q + 1) i=0 i=1  1 q + t  f (·, u n (·)) − f (·, u(·)) PC . (q + 1) 1 Case 2. For each t ∈ (ti , si ], i = 1, 2, . . . , m, we have |(Q q u n )(t) − (Q q u)(t)| = 0. Case 3. For each t ∈ (si , ti+1 ], i = 1, 2, . . . , m,we have |(Q q u n )(t) − (Q q u)(t)| ≤

q q ti+1 + si (q + 1)

 f (·, u n (·)) − f (·, u(·)) PC .

Thus, we infer that Q q u n − Q q u PC → 0 as n → ∞. Step 4. We show that Q q is compact. First, Q q is uniformly bounded on Br . Next, we show that Q q maps bounded set into equicontinuous set of Bq,r . Case 1. For a1 , a2 ∈ [0, t1 ], and a1 < a2 , u ∈ Bq,r . We have |(Q q u)(a2 ) − (Q q u)(a1 )|

a1 1 ≤ |(a2 − s)q−1 − (a1 − s)q−1 || f (s, u(s))|ds (q) 0

a2 1 (a2 − s)q−1 | f (s, u(s))|ds + (q) a1

q−1+ p1 q−1+ p1 q−1+ p1 1 1 1 − a2 ) + 2(a2 − a1 ) ≤ K q ν(r ) (a1 q−1+ p1

≤ 3K q ν(r )(a2 − a1 )

1

.

Case 2. For a1 , a2 ∈ (ti , si ], i = 1, 2, . . . , m, and a1 < a2 , u ∈ Bq,r . We have |(Q q u)(a2 ) − (Q q u)(a1 )| = 0. Case 3. For a1 , a2 ∈ (si , ti+1 ], i = 1, 2, . . . , m, and a1 < a2 , u ∈ Bq,r . We have q−1+ p1

|(Q q u)(a2 ) − (Q q u)(a1 )| ≤ 3K q ν(r )(a2 − a1 )

1

.

From above, we get |(Q q u)(a2 ) − (Q q u)(a1 )| → 0 as a2 → a1 . So Q q is equicontinuous.

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D. Yang, J. Wang

Now we know that Q q : Bq,r → Bq,r is continuous and compact. One can apply Theorem 2.9, Fq = P + Q q , which derive existence of 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  c 2 ⎪ ⎪ ⎪ u (t)[or D0,t u(t)] = ⎨

|u(t)| 1+4et ,

u(t) = |u(t)| 5+4t , ⎪ ⎪ ⎪ 1 ⎩ u(0) = 0 u(s)ds.

t ∈ (0, 41 ] ∪ ( 21 , 43 ], t ∈ ( 41 , 21 ],

(12)

Set J = [0, 1] and 0 = s0 < t1 = 41 < s1 = 21 < t2 = 43 = s2 < t3 = 1, m = 2. |u(t)| |u(t)| 1 1 3 Set f (t, u(t)) = 1+4e t , g1 (t, u(t)) = 5+4t . Let u 1 , u 2 ∈ R and t ∈ [0, 4 ] ∪ ( 2 , 4 ], then we have | f (t, u 1 ) − f (t, u 2 )| ≤ 15 |u 1 − u 2 |. Let u 1 , u 2 ∈ R and t ∈ ( 41 , 21 ], then we have |g1 (t, u 1 ) − g1 (t, u 2 )| ≤ 16 |u 1 − u 2 |. 13 Set L f = 15 , L g = 16 , one can deduce  = L g + L f (t2 − s1 ) = 60 < 1, or √ √ 1 1 . Lf  1 = L g + (q+1) (t22 + s12 ) = 16 + 5√2 π ( 23 + 22 ) = 0.5217 < 1. 2 Thus all the assumptions in Theorem 3.1 or Theorem 3.3 are satisfied, our results can be applied to the problem (12). Example 4.2 Consider ⎧ 2 3 ⎪ ⎪ u  (t)[or c D0,t u(t)] = ⎪ ⎨ u(t) = |u(t)| 9 , ⎪ ⎪ ⎪ 1 ⎩ u(0) = 0 u(s)ds.

|u(t)| 9 ,

t ∈ (0, 41 ] ∪ ( 21 , 43 ], t ∈ ( 41 , 21 ],

(13)

Set J = [0, 1] and 0 = s0 < t1 = 41 < s1 = 21 < t2 = 43 = s2 < t3 = 1, m = 2. 1 1 3 Set f (t, u(t)) = g1 (t, u(t)) = |u(t)| 9 . Let u 1 , u 2 ∈ R and t ∈ [0, 4 ] ∪ ( 2 , 4 ], then we have | f (t, u 1 ) − f (t, u 2 )| ≤ 19 |u 1 − u 2 |. Let u 1 , u 2 ∈ R and t ∈ ( 41 , 21 ], then we have |g(t, u 1 ) − g(t, u 2 )| ≤ 19 |u 1 − u 2 |, L g1 = L g = 19 < 1. For all u ∈ R and each t ∈ [0, 41 ] ∪ ( 21 , 43 ], we have | f (t, u)| ≤ 19 u PC , ψ(t) = 1 1 1 1 9 , ν(t) = t. For all u ∈ R and each t ∈ ( 4 , 2 ], we have |g1 (t, u)| ≤ 9 u PC , 1  1 ϕ1 (t) = 19 , μ(t) = t. Moreover, 12 ϕ1 (t)dt = 36 = M1 := M. 4 ! " 2−t1 2 Obviously, the inequality r ≥ max 1−t m Mμ(r ) + K ν(r ), N μ(r ) + K ν(r ) 1−t 1 1 $ #4 7 r + 27 r, 29 r = 11r reduces to r ≥ max 27 27 , which holds for every r > 0.

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Integral boundary value problems...

. p1 = 2 then (q − 1) p1 + 1 = 13 > 0, K q = % 1 1 2 =0.1421, 9( ) 3 3 % % % % . γ = K q max{4 + 6 41 , 6 43 + 6 21 , 1 + 6 43 }=0.6812. Moreover, the inequality r ≥ #4 2 1 max{ 1−t m Mμ(r )+γ ν(r ), N μ(r )+γ ν(r )} reduces to r ≥ max 27 r + 0.6812r, 9 r 1 . + 0.6812r } = 0.8293r , which holds for every r > 0. Thus, all the assumptions in Theorem 3.2 or Theorem 3.4 are satisfied, our results can be applied to the problem (13). Next, q =

2 3,

Acknowledgments We thank to the referee for valuable comments and suggestions which improved our paper. The authors acknowledge Training Object of High Level and Innovative Talents of Guizhou Province ((2016)4006), Unite Foundation of Guizhou Province ([2015]7640) and Outstanding Scientific and Technological Innovation Talent Award of Education Department of Guizhou Province ([2014]240).

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