Impulsive fractional differential equations with nonlinear boundary conditions

Mathematical and Computer Modelling 55 (2012) 303–311

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Impulsive fractional differential equations with nonlinear boundary conditions Jianxin Cao a,b , Haibo Chen a,∗ a

Department of Mathematics, Central South University, Changsha, 410075 Hunan, People’s Republic of China

b

Faculty of Science, Hunan Institute of Engineering, Xiangtan, 411104 Hunan, People’s Republic of China

article

info

Article history: Received 11 February 2011 Accepted 27 July 2011 Keywords: Fractional differential equation Nonlinear boundary conditions Lower and upper solutions Monotone iterative technique

abstract In this paper, we study certain impulsive functional fractional differential equations with nonlinear boundary conditions. By means of monotone iterative method coupled with lower and upper solutions, some new sufficient conditions for the existence of solutions are established. © 2011 Published by Elsevier Ltd

1. Introduction Because of its wide applicability in biology, medicine and in more and more fields, the theory of fractional differential equations has recently been attracting increasing interest, see for instance [1–8] and references therein. The monotone iterative method coupled with upper and lower solutions is a useful tool to solve fractional differential equations. When the fractional order 0 < q ≤ 1, there are lots of progresses on this, such as [9–16] and the references therein. Due to technical difficulties of building a corresponding comparison result, the are few papers concerned with when the fractional order q > 1. As a result of unifying different techniques for initial or boundary conditions, nonlinear boundary conditions received more and more attention, see [17–20]. But, for fractional differential equations, to the best of our knowledge similar papers are rarely seen. In this paper, we consider the following fractional order impulsive problem:

C q D u(t ) = f (t , u(t ), u(ω(t ))), t ∈ J ′ ,   1u(t )|t =tk = Ik (u(tk )), k = 1, 2, . . . , m, 1u′ (t )|t =tk = ¯Ik (u(tk )), k = 1, 2, . . . , m,  u(t ) = u(0), t ∈ [−r , 0],

(1.1)

subject to the nonlinear boundary value conditions as follows g0 (u(0), u(T )) = 0,

g1 (u′ (0), u′ (T )) = 0,

(1.2)

where D is the Caputo fractional derivative, 1 < q ≤ 2. J = [0, T ], J = [−r , T ], J = J \ {t1 , t2 , . . . , tm }, ′ J+ = J + \ {t1 , t2 , . . . , tm }, 0 = t0 < t1 < · · · < tm < tm+1 = T . f ∈ C (J × R2 , R)Ik , ¯Ik ∈ C [R, R], g0 , g1 ∈ C (R2 , R), ω(t ) ∈ C (J , J + ), t − r < ω(t ) < t , r > 0, t ∈ J and tk < ω(t ) < t, t ∈ (tk , tk+1 ]. 1u(t )|t =tk = u(tk+ ) − u(tk− ), u(tk+ ) and u(tk− ) represent the right-hand limit and the left-hand limit of the function u(t ) at t = tk , respectively. C

∗

q

+

Corresponding author. E-mail addresses: [email protected] (J. Cao), [email protected], [email protected] (H. Chen).

0895-7177/$ – see front matter © 2011 Published by Elsevier Ltd doi:10.1016/j.mcm.2011.07.037

′

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Remark 1.1. If ω(t ) = t + θ , θ ∈ [−r , 0], then (1.1) can be regarded as a retarded differential equation. Similar problems with fractional derivatives have been studied in [21–23]. Taking advantage of a simple comparison result based on an initial value problem, we build two monotone sequences, which converge to a coupled quasisolution. Then we obtain some existence results for the initial and final value problem, anti-periodic boundary value problem, and general nonlinear boundary value problem. It is worth pointing out that our results, being new, unify the treatment of different problems and extend some previous works. 2. Preliminaries We now introduce some definitions, preliminary facts about the fractional calculus, notations, and some auxiliary results, which will be used later. Definition 2.1. The fractional (arbitrary) order integral of the function v(t ) ∈ L1 ([0, ∞], R) of µ ∈ R+ is defined by I µ v(t ) =

t

∫

1

Γ (µ)

(t − s)µ−1 v(s)ds,

t > 0.

0

Definition 2.2. The Caputo fractional derivative of order µ > 0 for a function v(t ) given in the interval [0, ∞) is defined by C

t

∫

1

Dµ v(t ) =

Γ (n − µ)

(t − s)n−µ−1 v (n) (s)ds

0

provided that the right hand side is point-wise defined. Here n = [µ] + 1 and [µ] means the integral part of the number µ, and Γ is the Euler gamma function. This definition is of course more restrictive than the Riemann–Liouville definition, in that it requires the absolute integrability of the derivative of order n. Whenever we use the operator C Dµ we (tacitly) assume that this condition is met. The following properties of the fractional calculus theory are well known. (i) C Dβ I β v(t ) = v(t ) for a.e. t ∈ J, where v(t ) ∈ L1 [0, 1], β > 0. ∑ n −1 (ii) I β C Dβ v(t ) = v(t ) − j=0 cj t j for a.e. t ∈ J, where v(t ) ∈ L1 [0, 1], β > 0, cj (j = 0, 1, . . . , n − 1) are some constants, n = [β] + 1. (iii) I β : C [0, 1] → C [0, 1], I β : L1 [0, 1] → L1 [0, 1], β > 0. (iv) C Dβ I α = I α−β and C Dβ 1 = 0 for t ∈ J, α > β > 0. More details on fractional derivatives and their properties can be found in [7,8,24]. For the sake of convenience, we introduce the following notations. ′ Let J0 = [0, t1 ], Jk = (tk , tk+1 ], k = 1, 2, . . . , m, PC (J + ) = {u : [−r , T ] → R : u is continuous on J+ , u(tk+ ) and − − 1 + + u(tk ) exist, and u(tk ) = u(tk ), k = 1, 2, . . . , m}, and PC (J ) = {u ∈ PC (J ) : u is continuously differentiable on J ′ , u′ (0+ ), u′ (T − ), u′ (tk− ), u′ (tk+ ) exist; k = 1, 2, . . . , m}. Obviously, PC (J + ) and PC 1 (J + ) are Banach spaces with the norms ‖u‖PC = sup{|u| : t ∈ J + } and ‖u‖PC 1 = max{‖u‖PC , ‖u′ ‖PC }, respectively. Lemma 2.1 ([3, Lemma 4.8]). Let v(t ) ∈ L1 (J , R). Then u is the unique solution of the initial value problem:

C q D u(t ) = v(t ), t ∈ J ′ ,   1u(t )|t =tk = Ik (u(tk )), k = 1, 2, . . . , m, ′ ¯ 1   u (t )|t =tk = Ik ′(u(tk )), k = 1, 2, . . . , m, u(0) = a, u (0) = b,

(2.1)

if and only if u(t ) =

1

Γ (q)

+

+

t

∫

(t − s)q−1 v(s)ds + tk

1

Γ (q − 1) k − i=1

∫ k − (t − ti ) i =1

ti

1

Γ (q)

k ∫ − i =1

ti

(ti − s)q−1 v(s)ds

ti−1

(ti − s)q−2 v(s)ds +

ti−1

(t − ti )¯Ii (u(ti )) + a + bt ,

k −

Ii (u(ti ))

i =1

t ∈ Jk , k = 0, 1, . . . , m.

(2.2)

J. Cao, H. Chen / Mathematical and Computer Modelling 55 (2012) 303–311

305

In what follows, by that the u′ (t ) has some properties on J + , we mean that the u′ (t ) has some properties on J ′ and at tk , tk+ k = 1, 2, . . . , m. −

Lemma 2.2 (Comparison Result). Suppose that u ∈ PC 1 (J + ) is a function with

C q D u(t ) ≥ 0, t ∈ J ′ ,    1u(t )|t =tk = Ik (u(tk )) ≥ 0, k = 1, 2, . . . , m, 1u′ (t )|t =tk = ¯Ik (u(tk )) ≥ 0, k = 1, 2, . . . , m,    u(t ) = u(0), t′ ∈ [−r , 0], u(0) ≥ 0, u (0) ≥ 0.

(2.3)

Then u(t ) ≥ 0 and u′ (t ) ≥ 0 on J + . Proof. By (2.2) in Lemma 2.1, we have u′ (t ) =

∫

1

Γ (q − 1)

+

k −

t

(t − s)q−2 v(s)ds + tk

¯Ii (u(ti )) + b,

1

Γ (q − 1)

k ∫ − i =1

ti

(ti − s)q−2 v(s)ds

ti−1

t ∈ Jk , k = 0, 1, . . . , m.

(2.4)

i=1

By Lemma 2.1, we need only to set v(t ) ≥ 0, t ∈ [0, T ] and a ≥ 0, b ≥ 0. Then we can accomplish the purpose desired, which complete the proof.  Now, we are in position to define the lower and upper solution and coupled quasisolutions of the problem (1.1)–(1.2). Definition 2.3. Let α0 , β0 ∈ PC 1 (J + ). α0 , β0 are called coupled lower and upper solutions, respectively, of problem (1.1)–(1.2) if the following conditions hold: (a)

C q D α0 (t ) ≤ f (t , α0 (t ), α0 (ω(t ))), t ∈ J ′ ,    1α0 (t )|t =tk ≤ Ik (α0 (tk )), k = 1, 2, . . . , m, 1α0′ (t )|t =tk ≤ ¯Ik (α0 (tk )), k = 1, 2, . . . , m,   α  0 (t ) ≤ α0 (0), t ∈ [−r , 0],  g0 (α0 (0), β0 (T )) ≤ 0, g1 (α0′ (0), β0′ (T )) ≤ 0. (b)

C q D β0 (t ) ≥ f (t , β0 (t ), β0 (ω(t ))), t ∈ J ′ ,   1β0 (t )|t =t ≥ Ik (β0 (tk )), k = 1, 2, . . . , m,  k 1β0′ (t )|t =tk ≥ ¯Ik (β0 (tk )), k = 1, 2, . . . , m,    β0 (t ) ≥ β0 (0), t ∈ [−r , 0], ′ g0 (β0 (0), α0 (T )) ≥ 0, g1 (β0 (0), α0′ (T )) ≥ 0. Definition 2.4. We say that α, β ∈ PC 1 (J + ) are coupled quasisolutions for the problem (1.1)–(1.2) if α and β are solutions of the Eq. (1.1) such that g0 (α(0), β(T )) = 0 = g0 (β(0), α(T )) and g1 (α ′ (0), β ′ (T )) = 0 = g1 (β ′ (0), α ′ (T )) hold. ∞ Then we build a basic result about the monotone iterative sequences {αn }∞ n=0 , {βn }n=0 defined as below:

C q D αn = f (t , αn−1 , αn−1 (ω(t ))), t ∈ J ′ ,   1αn |t =tk = Ik (αn−1 (tk )), k = 1, 2, . . . , m,     1α ′ |t =t = ¯Ik (αn−1 (tk )), k = 1, 2, . . . , m,    n k αn (t ) = αn (0), t ∈ [−r , 0], g0 (φn−1 (0), ϕn−1 (T ))    αn (0) = φn−1 (0) − ,   m0   ′ ′  g (φ ( 0 ), ϕ ( T ))  n −1 α ′ (0) = φ ′ (0) − 1 n−1 , n

n−1

m1

(2.5)

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J. Cao, H. Chen / Mathematical and Computer Modelling 55 (2012) 303–311

and

C q D βn = f (t , βn−1 , βn−1 (ω(t ))), t ∈ J ′ ,    1βn |t =tk = Ik (βn−1 (tk )), k = 1, 2, . . . , m,     1β ′ | = ¯I (β (t )), k = 1, 2, . . . , m,   β (nt )t ==tk β (0k), n−t1 ∈k[−r , 0], n

n

(2.6)

g0 (ϕn−1 (0), φn−1 (T ))   , βn (0) = ϕn−1 (0) −    m0   ′ ′  g (ϕ (0), φn−1 (T ))  β ′ (0) = ϕ ′ (0) − 1 n−1 , n n −1 m1

where ϕn−1 = max{αn−1 , βn−1 }, ϕn′ −1 = max{αn′ −1 , βn′ −1 }, φn−1 = min{αn−1 , βn−1 }, φn′ −1 = min{αn′ −1 , βn′ −1 }, n = 1, 2, . . . and m0 , m1 will be defined later. This is an adequate definition. Since by general results on the initial value problem of impulsive differential equations [3], the existence and uniqueness of solution for Eqs. (2.5) and (2.6) are guaranteed. We will make use of the following assumptions:

(H 1) α0 , β0 are coupled lower and upper solutions, respectively, of problem (1.1)–(1.2) such that α0(i) ≤ β0(i) (i = 0, 1) on J + .

(H 2) For all α0 ≤ x1 ≤ x2 ≤ β0 f (t , x1 , x1 (ω(t ))) ≤ f (t , x2 , x2 (ω(t ))), Ik (x1 ) ≤ Ik (x2 ),

I¯k (x1 ) ≤ I¯k (x2 ).

(H 3) There exist mi > 0 (i = 0, 1) such that g0 (x2 , y1 ) − g0 (x1 , y1 ) ≤ m0 (x2 − x1 ), g0 (x1 , y1 ) ≤ g0 (x1 , y2 ), and

for all α0 ≤ x1 ≤ x2 ≤ β0 , α0 ≤ y1 ≤ y2 ≤ β0 ,

g1 (x2 , y1 ) − g1 (x1 , y1 ) ≤ m1 (x2 − x1 ),

g1 (x1 , y1 ) ≤ g1 (x1 , y2 ),

for all α0′ ≤ x1 ≤ x2 ≤ β0′ , α0′ ≤ y1 ≤ y2 ≤ β0′ .

Our basic result about the monotone iterative sequences is the following theorem. ∞ Theorem 2.1. Suppose that (H 1) –(H 3) hold. Then the sequences {αn }∞ n=0 , {βn }n=0 , defined as in (2.5) and (2.6), are converging to α and β , respectively, uniformly on J + . Moreover, α and β are coupled quasisolutions for the problem (1.1)–(1.2).

Proof. We prove firstly that these sequences satisfy the property

αn(i−) 1 ≤ αn(i) ≤ βn(i) ≤ βn(i−) 1 (i = 0, 1), t ∈ J + , n ≥ 1.

(2.7)

Indeed, let p = α1 − α0 , then we have

C q D p = f (t , α0 , α0 (ω(t ))) − C Dq α0 ≥ 0, t ∈ J ′    1p|t =tk = Ik (α0 (tk )) − 1α0 |t =tk ≥ 0, k = 1, 2, . . . , m,     1p′ |t =tk = ¯Ik (α0 (tk )) − 1α0′ |t =tk ≥ 0, k = 1, 2, . . . , m,   p(t ) = p(0), t ∈ [−r , 0], g0 (φ0 (0), ϕ0 (T )) g0 (φ0 (0), ϕ0 (T ))   − α0 (0) = − ≥ 0, p(0) = φ0 (0) −    m m0 0  ′ ′ ′ ′   p′ (0) = φ ′ (0) − g1 (φ0 (0), ϕ0 (T )) − α ′ (0) = − g1 (φ0 (0), ϕ0 (T )) ≥ 0. 0

m0

0

(2.8)

m1

(i)

(i)

Thus, by Lemma 2.2, we conclude that p(t ) ≥ 0 and p′ (t ) ≥ 0 on J + . That is, α1 ≥ α0 (i = 0, 1). Analogously, one can (i)

(i)

show that β1 ≤ β0 (i = 0, 1). Now, let p = β1 − α1 . Using (H 1)–(H 3), we have

C q D p = f (t , β0 , β0′ ) − f (t , α0 , α0 (ω(t ))) ≥ 0, t ∈ J ′   1p|t =tk = Ik (β0 (tk )) − Ik (α0 (tk )) ≥ 0, k = 1, 2, . . . , m, ′ ¯ ¯ 1   p |t =tk = Ik (β0 (tk )) − Ik (α0 (tk )) ≥ 0, k = 1, 2, . . . , m, p(t ) = p(0), t ∈ [−r , 0],

J. Cao, H. Chen / Mathematical and Computer Modelling 55 (2012) 303–311

307

and p(0) = ϕ0 (0) −

g0 (φ0 (0), ϕ0 (T )) − φ0 (0) + m0 m0 g0 (φ0 (0), ϕ0 (T )) − g0 (φ0 (0), φ0 (T )) + g0 (φ0 (0), φ0 (T )) − g0 (ϕ0 (0), φ0 (T ))

g0 (ϕ0 (0), φ0 (T ))

= ϕ0 (0) − φ0 (0) + ≥ ϕ0 (0) − φ0 (0) +

m0 g0 (φ0 (0), ϕ0 (T )) − g0 (φ0 (0), φ0 (T )) + m0 (φ0 (0) − ϕ0 (0)) m0

≥ 0, p′ (0) = ϕ0′ (0) −

g1 (ϕ0′ (0), φ0′ (T )) m1

= ϕ0′ (0) − φ0′ (0) + ≥ ϕ0′ (0) − φ0′ (0) +

− φ0′ (0) +

(2.9)

g1 (φ0′ (0), ϕ0′ (T )) m1

g1 (φ0′ (0), ϕ0′ (T )) − g1 (φ0′ (0), φ0′ (T )) + g1 (φ0′ (0), φ0′ (T )) − g1 (ϕ0′ (0), φ0′ (T )) m1 g1 (φ0′ (0), ϕ0′ (T )) − g1 (φ0′ (0), φ0′ (T )) + m1 (φ0′ (0) − ϕ0′ (0)) m1

≥ 0. (i)

(i)

This allows us to conclude that β1 ≥ α1 (i = 0, 1) on J + . Thus, (2.7) is valid for n = 1. Assuming that it is valid for m ≥ 1 we show that it is valid for m + 1. Let αm+1 − αm . By (H 1)–(H 3), αm(i)−1 ≤ αm(i) ≤ βm(i) ≤ βm(i)−1 (i = 0, 1), we have

C q D p = f (t , αm , αm (ω(t ))) − f (t , αm−1 , αm−1 (ω(t ))) ≥ 0, t ∈ J ′   1p|t =tk = Ik (αm (tk )) − Ik (αm−1 (tk )) ≥ 0, k = 1, 2, . . . , m, ′ ¯ ¯  1p |t =tk = Ik (αm (tk )) − Ik (αm−1 (tk )) ≥ 0, k = 1, 2, . . . , m, p(t ) = p(0), t ∈ [−r , 0], and, similarly to (2.9), p(0) ≥ p′ (0) ≥

g0 (φm (0), ϕm−1 (T )) − g0 (φm (0), ϕm (T )) + m0 (φm−1 (0) − φm (0)) m0 ′ g1 (φm (0), ϕm′ −1 (T )) − g1 (φm′ (0), ϕm′ (T )) + m1 (φm′ −1 (0) − φm′ (0))

m1 (i)

+ φm (0) − φm−1 (0) ≥ 0, + φm′ (0) − φm′ −1 (0) ≥ 0.

(i)

(i)

(i)

Hence, we have p(t ) ≥ 0 and p′ (t ) ≥ 0 on J + , which show that αm+1 ≥ αm (i = 0, 1). Analogously, both βm ≥ βm+1 and βm(i)+1 ≥ αm(i)+1 (i = 0, 1) are also true. As a consequence, by mathematical induction, we have proved the validity of Eq. (2.7). We have two monotone sequences in PC 1 (J + ) that are bounded. By standard arguments, there exist α, β ∈ PC 1 (J + ) with {αn } ↑ α , {βn } ↓ β and α ≤ β . Moreover the convergence is uniformly on J + . Now, there exist α, β ∈ PC 1 (J + ) as before such that

C q D α = f (t , α, α(ω(t ))), t ∈ J ′ ,     1 α| = Ik (α(tk )), k = 1, 2, . . . , m,   ′ t = tk 1α |t =tk = ¯Ik (α(tk )), k = 1, 2, . . . , m, α(t ) = α(0), t ∈ [−r , 0],     g0 (α(0), β(T )) = 0,  g1 (α ′ (0), β ′ (T )) = 0,

(2.10)

C q D β = f (t , β, β(ω(t ))), t ∈ J ′ ,     1β| = Ik (β(tk )), k = 1, 2, . . . , m,   ′ t = tk 1β |t =tk = ¯Ik (β(tk )), k = 1, 2, . . . , m, β(t ) = β(0), t ∈ [−r , 0],      g0 (β(′ 0), α(′T )) = 0, g1 (β (0), α (T )) = 0.

(2.11)

and

These show that α and β are coupled quasisolutions for the problem (1.1)–(1.2). The proof is complete.



The result above is important to establish the existence for the initial and final value problem as well as the nonlinear boundary value problem including anti-periodic conditions.

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3. Initial and final value problems In this section we establish some corollaries of Theorem 2.1, which answer the existence of problem (1.1) with the condition of initial type h00 (u(0)) = 0,

h01 (u′ (0)) = 0

(3.1)

hT 1 (u′ (T )) = 0,

(3.2)

and final type hT 0 (u(T )) = 0, respectively. Corollary 3.1. Suppose that (H 1), (H 2) and the following (H0 3) hold.

(H0 3) There exist mi > 0, (i = 0, 1) such that h00 (x2 ) − h00 (x1 ) ≤ m0 (x2 − x1 ), and

for all α0 ≤ x1 ≤ x2 ≤ β0

h01 (x2 ) − h01 (x1 ) ≤ m1 (x2 − x1 ),

for all α0′ ≤ x1 ≤ x2 ≤ β0′ .

Then there exists at least one solution of problem (1.1)–(3.1) between the lower and the upper solution. Proof. Applying Theorem 2.1 with gi (x, y) = h0i (x), i = 0, 1, we have that there exist α, β coupled quasisolutions of the problem (1.1)–(3.1). That is, α, β satisfy the problem (1.1) and h00 (α(0)) = h01 (α ′ (0)) = 0,

h00 (β(0)) = h01 (β ′ (0)) = 0,



Remark 3.1. When h0i (x) = x − ci , ci are constants (i = 0, 1), we obtain naturally the existence of the problem (1.1) with simple initial condition. For the final type condition (3.2), we can get the corresponding corollary as below. Corollary 3.2. Suppose that (H 1), (H 2) and the following (HT 3) hold. (HT 3) There exist mi > 0 (i = 0, 1) such that hT 0 (y1 ) ≤ hT 0 (y2 ), and

for all α0 ≤ y1 ≤ y2 ≤ β0

hT 1 (y1 ) ≤ hT 1 (y2 ),

for all α0′ ≤ y1 ≤ y2 ≤ β0′ .

Then there exists at least one solution of problem (1.1)–(3.2) between the lower and the upper solution. Proof. Similarly to the proof of the previous corollary, there exist α, β coupled quasisolutions of the problem (1.1)–(3.2). That is, α, β satisfy the problem (1.1) and hT 0 (α(T )) = hT 1 (α ′ (T )) = 0,

hT 0 (β(T )) = hT 1 (β ′ (T )) = 0. 

4. Antiperiodic boundary value problem For the antiperiodic boundary value condition gi (x, y) = x + y(i = 0, 1) and general nonlinear boundary condition (1.2), the situation is different. We should construct some suitable conditions under which the coupled quasisolutions α, β satisfy α(T ) = β(T ) and α ′ (T ) = β ′ (T ). In this section, we consider the antiperiodic boundary value problem, that is, the problem (1.1) with the boundary value condition u(0) + u(T ) = 0,

u′ (0) + u′ (T ) = 0.

We have the following theorem. Theorem 4.1. Suppose (H 1) and the following (HA 2) hold.

(HA 2) The function f ∈ C [J × R2 , R] and for all α0 ≤ x1 ≤ x2 ≤ β0 f (t , x1 , x1 (ω(t ))) < f (t , x2 , x2 (ω(t ))), when x1 (τ ) < x2 (τ ) and

Ik (x1 ) < Ik (x2 ),

when x1 (τ ) < x2 (τ )

for some τ ∈ J ′ , I¯k (x1 ) < I¯k (x2 ), for some τ ∈ {t1 , t2 , . . . , tm }.

Then there exists at least one solution of problem (1.1)–(4.1) between the lower and the upper solution.

(4.1)

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309

Proof. Obviously, gi (x, y) = x + y (i = 0, 1) satisfy (H 3). By Theorem 2.1, we know that the problem (1.1)–(1.1) has a coupled quasisolutions α, β . That is, α, β satisfy

C q D α = f (t , α, α(ω(t ))), t ∈ J ′ ,    1α|t =tk = Ik (α(tk )), k = 1, 2, . . . , m, 1α ′ |t =tk = ¯Ik (α(tk )), k = 1, 2, . . . , m,    α(t ) = α(0), t ∈ [−′r , 0], α(0) = −β(T ), α (0) = −β ′ (T ),

(4.2)

C q D β = f (t , β, β(ω(t ))), t ∈ J ′ ,   1β|t =t = Ik (β(tk )), k = 1, 2, . . . , m,  k 1β ′ |t =tk = ¯Ik (β(tk )), k = 1, 2, . . . , m,    β(t ) = β(0), t ∈ [−′ r , 0], β(0) = −α(T ), β (0) = −α ′ (T ).

(4.3)

and

Regarding (4.2) and (4.3) as an initial value problem, by Lemma 2.1, we have m+1

ti

i=1

ti−1

−∫

1

α(T ) =

Γ (q)

+

+

1

Γ (q − 1) m −

(ti − s)q−1 f (s, α, α(ω(s)))ds

∫ m − (T − ti )

i=1

(ti − s)q−2 f (s, α, α(ω(s)))ds

ti−1

i =1

Ii (α(ti )) +

ti

m − (T − ti )¯Ii (α(ti )) − β(T ) − β ′ (T )T ,

(4.4)

i=1

and

β(T ) =

m+1

ti

i=1

ti−1

−∫

1

Γ (q)

+

+

1

Γ (q − 1) m −

(ti − s)q−1 f (s, β, β(ω(s)))ds

∫ m − (T − ti )

i=1

(ti − s)q−2 f (s, β, β(ω(s)))ds

ti−1

i =1

Ii (β(ti )) +

ti

m − (T − ti )¯Ii (β(ti )) − α(T ) − α ′ (T )T .

(4.5)

i=1

Let p(t ) = β(t )−α(t ), then, by (2.7), (4.2) and (4.3), p(t ) ≥ 0, p(0) = p(T ) and p′ (0) = p′ (T ). Suppose, by contradiction, that there exists some τ ∈ J such that β(τ ) > α(τ ). By (4.4), (4.5) and (HA 2), we can get

β(T ) − α(T ) =

m+1

ti

i=1

ti−1

−∫

1

Γ (q)

+ +

1

Γ (q − 1) m −

(ti − s)q−1 (f (s, β, β(ω(s))) − f (s, α, α(ω(s))))ds

∫ m − (T − ti ) i =1

ti

(ti − s)q−2 (f (s, β, β(ω(s))) − f (s, α, α(ω(s))))ds

ti−1

(Ii (β(ti )) − Ii (α(ti ))) +

i=1

m − (T − ti )(¯Ii (β(ti )) − ¯Ii (α(ti ))) + β(T ) − α(T ) + p′ (T )T i=1

> β(T ) − α(T ) + p′ (0)T . Immediately, we have 0 > lim

t →0+

p(t ) − p(0) t

T ≥ lim (p(t ) − p(0)) = 0,

(4.6)

t →0+

which is a contradiction. Then α(t ) = β(t ) on J and obviously on J + , which complete the proof. 5. General nonlinear boundary condition Now, we consider the general nonlinear boundary condition (1.2).



310

J. Cao, H. Chen / Mathematical and Computer Modelling 55 (2012) 303–311

Theorem 5.1. Suppose that (H 1), (H 2) and the following (HG 3) hold.

(HG 3) There exist some constants mi0 > 0, miT > 0 (i = 0, 1) satisfying β0 , α0 ≤ y1 < y2 ≤ β0 0<

g0 (x2 , y1 ) − g0 (x1 , y1 ) x2 − x1

g0 (x1 , y2 ) − g0 (x1 , y1 )

≤ m00 ,

y2 − y1

miT mi0

> 1 (i = 0, 1) such that for all α0 ≤ x1 < x2 ≤

≥ m0T ,

and for all α0′ ≤ x1 < x2 ≤ β0′ , α0′ ≤ y1 < y2 ≤ β0′ 0<

g1 (x2 , y1 ) − g1 (x1 , y1 ) x2 − x1

g1 (x1 , y2 ) − g1 (x1 , y1 )

≤ m10 ,

y2 − y1

≥ m1T .

Then there exists at least one solution of problem (1.1)–(1.2) between the lower and the upper solution. Proof. By Theorem 2.1, we have that there exist α, β coupled quasisolutions of the problem (1.1)–(1.2). That is, α, β satisfy (2.10) and (2.11). Let p(t ) = β(t ) − α(t ), by (2.7), p(t ) ≥ 0, p′ (t ) ≥ 0. And we have, 0 = g0 (α(0), β(T )) = g0 (β(0), α(T )), 0 = g1 (α ′ (0), β ′ (T )) = g1 (β ′ (0), α ′ (T )). Immediately, g0 (α(0), β(T )) − g0 (α(0), α(T )) = g0 (β(0), α(T )) − g0 (α(0), α(T )), g1 (α ′ (0), β ′ (T )) − g1 (α ′ (0), α ′ (T )) = g1 (β ′ (0), α ′ (T )) − g1 (α ′ (0), α ′ (T )). But now, using (HG 3), we can write, c0T p(T ) = c00 p(0),

(5.1)

c1T p′ (T ) = c10 p′ (0), where 0 < ci0 ≤ mi0 , ciT ≥ miT (i = 0, 1), and, naturally, On the other hand, we have C

Dq p(t ) = f (t , β, β(ω(t ))) − f (t , α, α(ω(t ))),

ciT ci0

> 1 (i = 0, 1).

t ∈ J ′,

1p|t =tk = Ik (β(tk )) − Ik (α(tk )), k = 1, 2, . . . , m, 1p′ |t =tk = ¯Ik (β(tk )) − ¯Ik (α(tk )), k = 1, 2, . . . , m p(t ) = p(0), t ∈ [−r , 0]. Applying Lemma 2.1 with v(t ) = f (t , β, β(ω(t )))− f (t , α, α(ω(t ))), a = p(0), b = p′ (0), and the new impulsive conditions as above, we have that p(T ) =

m+1

ti

i=1

ti−1

−∫

1

Γ (q)

+

1

Γ (q − 1)

(ti − s)q−1 (f (s, β, β(ω(s))) − f (s, α, α(ω(s))))ds

∫ m − (T − ti )

ti

(ti − s)q−2 (f (s, β, β(ω(s))) − f (s, α, α(ω(s))))ds

ti−1

i=1

m m − − + (Ii (β(ti )) − Ii (α(ti ))) + (T − ti )(¯Ii (β(ti )) − ¯Ii (α(ti ))) + p(0) + p′ (0)T i =1

i=1

≥ p(0) + p (0)T ≥ p(0) (similarly to 4.6) , ′

(5.2)

and analogously p′ (T ) =

1

Γ (q − 1)

m+1

ti

i =1

ti−1

−∫

(ti − s)q−2 (f (s, β, β(ω(s))) − f (s, α, α(ω(s))))ds

m − + (¯Ii (β(ti )) − ¯Ii (α(ti ))) + p′ (0) i=1

≥ p′ (0).

(5.3)

Hence, (5.1)–(5.3) imply p(T ) ≥

c0T c00

p(T ),

p′ (T ) ≥

c1T ′ p (T ). c10

J. Cao, H. Chen / Mathematical and Computer Modelling 55 (2012) 303–311

Therefore, p(T ) ≥ 0, p′ (T ) ≥ 0 and

ciT ci0

> 1 (i = 0, 1) showed before imply p(T ) = 0 and p′ (T ) = 0, that is, α(T ) = β(T )

and α (T ) = β (T ). The proof is complete. ′

′

311



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