ChuanzhiBai Onthesolvabilityofanti-periodicboundaryvalueproblemswithimpulse

Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 11, 1-15;http://www.math.u-szeged.hu/ejqtde/

On the solvability of anti-periodic boundary value problems with impulse

Chuanzhi Bai ^∗

Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 223300, P R China

Abstract

In this paper, we are concerned with the existence of solutions for second order impulsive anti-periodic boundary value problem











u⁰⁰(t) +f(t, u(t), u⁰(t)) = 0, t6=t_k, t∈[0, T], 4u(t_k) =I_k(u(t_k)), k= 1,· · ·, m,

4u⁰(t_k) =I_k^∗(u(t_k)), k= 1,· · ·, m, u(0) +u(T) = 0, u⁰(0) +u⁰(T) = 0,

new criteria are established based on Schaefer’s fixed-point theorem.

Keywords: Anti-periodic boundary value problem; Impulsive; Schaefer’s fixed-point theorem 1. Introduction

In recent years, the solvability of the anti-periodic boundary value problems of first-order and second-order differential equations were studied by many authors, for example, we refer to [1-5] and the references therein. It should be noted that anti-periodic boundary value problems appear in physics in a variety of situations [6,7].

Impulsive differential equations, which arise in biology, physics, population dynamics, eco- nomics, etc., are a basic tool to study evolution processes that are subjected to abrupt in their states (see [8-12]). Recently, the existence results were extended to anti-periodic boundary value problems for first-order impulsive differential equations [13,14]. Very recently, Wang and Shen [15] investigated the anti-periodic boundary value problem for a class of second order differential equations by using Schauder’s fixed point theorem and the lower and upper solutions method.

Inspired by [13-15], in this paper, we investigate the anti-periodic boundary value problem for second order impulsive nonlinear differential equations of the form

∗E-mail address: czbai@hytc.edu.cn, czbai8@sohu.com











u⁰⁰(t) +f(t, u(t), u⁰(t)) = 0, t∈J₀ =J\ {t₁,· · ·, t_m}, 4u(t_k) =I_k(u(t_k)), k= 1,· · ·, m,

4u⁰(t_k) =I_k^∗(u(t_k)), k= 1,· · ·, m, u(0) +u(T) = 0, u⁰(0) +u⁰(T) = 0,

(1.1)

where J = [0, T], 0< t₁ < t₂ <· · · < t_m < T, f : [0, T]×R² → R is continuous on (t, x, y) ∈ J₀×R²,f(t⁺_k, x, y) := lim

t→t⁺k

f(t, x, y), f(t⁻_k, x, y) := lim

t→t⁻k

f(t, x, y) exist, f(t⁻_k, x, y) =f(t_k, x, y);

4u(t_k) =u(t⁺_k)−u(t⁻_k),4u⁰(t_k) =u⁰(t⁺_k)−u⁰(t⁻_k);I_k, I_k^∗∈C(R, R).

To the best of the authors knowledge, no one has studied the existence of solutions for impulsive anti-periodic boundary value problem (1.1). The aim of this paper is to fill the gap in the relevant literatures.

The following Schaefer’s fixed-point theorem is fundamental in the proof of our main results.

Lemma 1.1.[16] (Schaefer) Let E be a normed linear space with H : E → E a compact operator. If the set

S:={x∈E|x=λHx, f or some λ∈(0,1)} is bounded, thenH has at least one fixed-point.

The paper is formulated as follows. In section 2, some definitions and lemmas are given.

In section 3, we obtain a new existence theorem by using Schaefer’s fixed point theorem, and uniqueness result by using Banach’s fixed point theorem. In Section 4, an illustrative example is given to demonstrate the effectiveness of the obtained results.

2. Preliminaries

In order to define the concept of solution for (1.1), we introduce the following spaces of functions:

P C(J) = {u :J → R :u is continuous for any t ∈J₀, u(t⁺_k), u(t⁻_k) exist, and u(t⁻_k) = u(t_k), k= 1,· · ·, m},

P C¹(J) = {u : J → R : u is continuously differentiable for any t ∈ J₀, u⁰(t⁺_k), u⁰(t⁻_k) exist, andu⁰(t⁻_k) =u⁰(t_k), k= 1,· · ·, m}.

P C(J) andP C¹(J) are Banach space with the norms : kukP C = sup_t∈J|u(t)|,

and

kukP C¹ = max{kukP C,ku⁰kP C}.

A solution to the impulsive BVP (1.1) is a function u∈P C¹(J)∩C²(J₀) that satisfies (1.1) for each t∈J.

Consider the following impulsive BVP withp≥0, q >0











−u⁰⁰(t) +pu⁰(t) +qu(t) =σ(t), t∈J0, 4u(t_k) =I_k(u(t_k)), k= 1,· · ·, m, 4u⁰(t_k) =I_k^∗(u(t_k)), k= 1,· · ·, m, u(0) +u(T) = 0, u⁰(0) +u⁰(T) = 0,

(2.1)

whereσ ∈P C(J).

For convenience, we setI_k=I_k(u(t_k)), I_k^∗ =I_k^∗(u(t_k)), r1 := p+^pp²+ 4q

2 >0, r2:= p−^pp²+ 4q

2 <0. (2.2)

Lemma 2.1. u∈P C¹(J)∩C²(J₀) is a solution of(2.1) if and only if u∈P C¹(J)is a solution of the impulsive integral equation

u(t) = Z _T

0

G(t, s)σ(s)ds+

m

X

k=1

[G(t, t_k)(−I_k^∗) +W(t, t_k)I_k], (2.3) where

G(t, s) = 1 r₁−r₂







e^r2(t−s)

1+e^r2T −^e_1+e^r1(tr−1^sT⁾, 0≤s < t≤T,

e^r1(T+t−s)

1+e^r1T − ^er_1+e^2(T+r2^t−T^s), 0≤t≤s≤T, (2.4) and

W(t, s) = 1 r₁−r₂







r1e^r2(t−s)

1+e^r2T −^r2_1+e^er1(rt1⁻T^s), 0≤s < t≤T,

r2e^r1(T+t−s)

1+e^r1T −^r1e_1+e^r2(Tr+2T^t−s), 0≤t≤s≤T. (2.5) Proof. Ifu∈P C¹(J)^TC²(J₀) is a solution of (2.1), setting

v(t) =u⁰(t)−r₂u(t), (2.6)

then by the first equation of (2.1) we have

v⁰(t)−r₁v(t) =−σ(t), t6=t_k. (2.7)

Multiplying (2.7) bye^−r1t and integrating on [0, t] and (t₁, t], respectively, we get e^−r1tv(t)−v(0) =−

Z t 0

σ(s)e^−r1sds, 0≤t≤t₁, e^−r1tv(t)−e^−r1t1v(t⁺₁) =−

Z t t1

σ(s)e^−r1sds, t₁ < t≤t₂.

So

v(t) =e^r1t[v(0)− Z _t

0

e^−r1sσ(s)ds+e^−r1t1(I₁^∗−r₂I₁), t₁< t≤t₂. In the same way, we can obtain that

v(t) =e^r1t



v(0)− Z t

0

e^−r1sσ(s)ds+ ^X

0<tk<t

e^−r1tk(I_k^∗−r₂I_k)



, t∈J, (2.8)

where v(0) = u⁰(0)−r₂u(0). Multiplying (2.6) by e^−r2t and integrating on [0, t_k] and (t_k, t]

(t_k< t≤t_k+1), respectively, similar to the proof of (2.8), we have u(t) =e^r2t



u(0) + Z _t

0

v(s)e^−r2sds+ ^X

0<t^k<t

e^−r2tkI_k



, t∈J. (2.9)

By some calculation, we get Z t

0

v(s)e^−r2sds = 1 r₁−r₂

v(0)(e^(r1−r2)t−1)− Z t

0

(e^(r1−r2)t−e^(r1−r2)s)σ(s)e^−r1sds

+ ^X

0<tk<t

e^(r1−r2)t−e^(r1−r2)tke^−r1tk(I_k^∗−r₂I_k)



.

(2.10) Substituting (2.10) into (2.9), we obtain

u(t) = 1

r₁−r₂

h(r₁u(0)−u⁰(0))e^r2t+ (u⁰(0)−r₂u(0))e^r1t +

Z t

0 (e^r2(t−s)−e^r1(t−s))σ(s)ds+ ^X

0<tk<t

e^r1(t−tk)(I_k^∗−r₂I_k)

− ^X

0<tk<t

e^r2(t−tk)(I_k^∗−r₁I_k)



, t∈[0, T],

(2.11) u⁰(t) = 1

r₁−r₂

hr₂(r₁u(0)−u⁰(0))e^r2t+r₁(u⁰(0)−r₂u(0))e^r1t +

Z t 0

(r₂e^r2(t−s)−r₁e^r1(t−s))σ(s)ds+ ^X

0<tk<t

r₁e^r1(t−tk)(I_k^∗−r₂I_k)

− ^X

0<tk<t

r2e^r2(t−tk)(I_k^∗−r1I_k)



, t∈[0, T].

(2.12)

In view of u(0) +u(T) = 0, u⁰(0) +u⁰(T) = 0, we have u⁰(0)−r₂u(0) = 1

1 +e^r1T



 Z T

0

e^r1(T−s)σ(s)ds− ^X

0<tk<T

e^r1(T−tk)(I_k^∗−r₂I_k)



, (2.13)

r₁u(0)−u⁰(0) = 1 1 +e^r2T



− Z T

0 e^r2(T−s)σ(s)ds+ ^X

0<t^k<T

e^r2(T−tk)(I_k^∗−r₁I_k)



. (2.14) Substituting (2.13) and (2.14) into (2.11), by routine calculation, we can get (2.3).

Conversely, if u is a solution of (2.3), then direct differentiation of (2.3) gives −u⁰⁰(t) = σ(t) −pu⁰(t)−qu(t), t 6= t_k. Moreover, we obtain 4u|t=tk = I_k(u(t_k)), 4u⁰|t=tk = I_k^∗(u(t_k)), u(0) +u(T) = 0 andu⁰(0) +u⁰(T) = 0. Hence,u∈P C¹(J)∩C²(J0) is a solution of (2.1) 2

Define a mapping A:P C¹(J)→P C¹(J) by Au(t) =

Z _T

0

G(t, s)[f(s, u(s), u⁰(s)) +pu⁰(s) +qu(s)]ds +

m

X

k=1

[G(t, t_k)(−I_k^∗) +W(t, t_k)I_k], t∈[0, T]. (2.15) In view of Lemma 2.1, we easily know that u is a fixed point of operatorA if and only if u is a solution to the impulsive boundary value problem (1.1).

Lemma 2.2Ifu∈P C¹(J) and u(0) +u(T) = 0, then kukP C ≤ 1

2 Z T

0 |u⁰(s)|ds+

m

X

k=1

|4u(t_k)|

! . Proof. Since u∈P C¹(J), we have

u(t) =u(0) + ^X

0<tk<t

4u(t_k) + Z _t

0 u⁰(s)ds. (2.16)

Sett=T, we obtain from u(0) +u(T) = 0 that u(0) =−1

2

m

X

k=1

4u(t_k) + Z _T

0 u⁰(s)ds

!

. (2.17)

Substituting (2.17) into (2.16), we get

|u(t)|= 1 2

Z _t

0 u⁰(s)ds− Z _T

t u⁰(s)ds

! +1

2



 X

0<tk<t

4u(t_k)−^X

t≤tk

4u(t_k)





≤ 1 2

Z _t

0 |u⁰(s)|ds+ Z _T

t |u⁰(s)|ds

! +1

2



 X

0<tk<t

|4u(t_k)|+ ^X

t≤tk

|4u(t_k)|





= 1 2

Z T

0 |u⁰(s)|ds+

m

X

k=1

|4u(t_k)|

! . The proof is complete. 2

It is easy to check that

|G(t, s)| ≤ 1 r₁−r₂

1

1 +e^r2T − 1 1 +e^r1T

:=G₁. (2.18)

Since

∂

∂tG(t, s) = 1 r₁−r₂







−r1e^r1(t−s)

1+e^r1T +^r2_1+e^er2(r^t2⁻T^s), 0≤s < t≤T

r1e^r1(T+t−s)

1+e^r1T +^−r2_1+e^er2(rT₂^+Tt−s), 0≤t≤s≤T, and

∂

∂tW(t, s) = 1 r1−r2







r1r2e^r2(t−s)

1+e^r2T − ^r2r_1+e^1er1(r1^tT^−s), 0≤s < t≤T,

r1r2e^r1(T+t−s)

1+e^r1T −^r1r2_1+e^er2(rT2⁺T^t−s), 0≤t≤s≤T, we obtain by r₁≥ −r₂ >0 that

∂

∂tG(t, s)

≤ r₁ r₁−r₂







e^r1(t−s)

1+e^r1T +^e_1+e^r2(tr⁻2^sT⁾, 0≤s < t≤T

e^r1(T+t−s)

1+e^r1T +^er2(T+t−s)

1+e^r2T , 0≤t≤s≤T,

|W(t, s)| ≤ r₁ r₁−r₂







e^r1(t−s)

1+e^r1T +^er2(t−s)

1+e^r2T, 0≤s < t≤T

e^r1(T+t−s)

1+e^r1T +^er_1+e^2(T+r2^t−T^s), 0≤t≤s≤T, and

∂

∂tW(t, s)

≤ −r1r2

r₁−r₂







e^r2(t−s)

1+e^r2T +^e_1+e^r1(tr−₁^sT), 0≤s < t≤T,

e^r1(T+t−s)

1+e^r1T +^er_1+e^2(T+r2^t−T^s), 0≤t≤s≤T.

Let h(x) =Be^r1x+Ce^r2x, forx ∈[0, T], where B, C are two nonnegative constants. Obvi- ously,h⁰⁰(x)≥0, that is, h is a convex function on [0, T]. Thus,

h(x)≤max{h(0), h(T)}= max{B+C, Be^r1T +Ce^r2T}. (2.19) By (2.19), we easily obtain that

∂

∂tG(t, s)

≤ r₁ r₁−r₂

e^r1T

1 +e^r1T + e^r2T 1 +e^r2T

!

:=G2, |W(t, s)| ≤G2, (2.20) and

∂

∂tW(t, s)

≤ −r₁r₂ r₁−r₂

e^r1T

1 +e^r1T + e^r2T 1 +e^r2T

!

:=G₃. (2.21)

3.Main results

Throughout this section, we assume that

(H1) 3 Z _T

0 c(t)dt+ 6pT <2;

(H₂) There exist constant 0< η <1 and functionsa, b, c, h∈C(J,[0,+∞)) such that

|f(t, u, v)| ≤a(t)|u|+b(t)|u|^η+c(t)|v|+h(t);

(H₃) There exist nonnegative constantsα_k, β_k, γ_k, δ_k (k= 1,2,· · ·, m) such that

|I_k(u)| ≤α_k|u|+β_k, |I_k^∗(u)| ≤γ_k|u|+δ_k, k= 1,· · ·m.

For convenience, let

a₁ = 3^R₀^Ta(t)dt+ 2qT +^Pm_i=1γ_i

2−3^R₀^T c(t)dt−6pT , a₂ = 3^R₀^T b(t)dt 2−3^R₀^T c(t)dt−6pT,

(3.1) a₃ = 3^R₀^Th(t)dt+^Pm_i=1δ_i

2−3^R₀^T c(t)dt−6pT .

Theorem 3.1. Suppose that (H1)−(H3)hold. Further assume that b₁T

2(2−c^∗) +

s b₁T 2(2−c^∗)

m

X

i=1

α_i+m 4

m

X

i=1

α²_i <1, (3.2)

where b₁=

Z _T

0

a(t)dt+1 2

Z _T

0

c(t)dt+

m

X

i=1

[(p+a₁)α_i+γ_i],

a1 as in(3.1) and c^∗ = maxt∈Jc(t)<2. Then BVP(1.1) has at least one solution.

Proof. It is easy to check by Arzela-Ascoli theorem that the operatorAis completely continuous.

Assume thatu is a solution of the equation u=λAu, λ∈(0,1).

Then,

u⁰⁰(t) =λ(Au)⁰⁰(t) =λ[−f(t, u(t), u⁰(t))−pu⁰(t)−qu(t) +p(Au)⁰(t) +q(Au)(t)]

=−λf(t, u(t), u⁰(t))−p(λ−1)u⁰(t)−q(λ−1)u(t), (3.3)

−u(t)u⁰⁰(t) =λu(t)f(t, u(t), u⁰(t)) +p(λ−1)u(t)u⁰(t) +q(λ−1)u²(t)

≤λu(t)f(t, u(t), u⁰(t)) +p(λ−1)u(t)u⁰(t). (3.4) Integrating (3.3) from 0 toT, we get that

u⁰(T)−u⁰(0) = Z T

0

u⁰⁰(t)dt+

m

X

i=1

I_i^∗

=−λ Z T

0 f(t, u(t), u⁰(t))dt−p(λ−1) Z T

0 u⁰(t)dt−q(λ−1) Z T

0 u(t)dt+

m

X

i=1

I_i^∗. (3.5) In view of u⁰(0) +u⁰(T) = 0, we obtain by (3.5) that

|u⁰(0)| ≤ 1 2

Z _T

0 |f(t, u(t), u⁰(t))|dt+p Z _T

0 |u⁰(t)|dt+q Z _T

0 |u(t)|dt+1 2

m

X

i=1

|I_i^∗|. (3.6) Integrating (3.3) from 0 tot, we obtain that

u⁰(t)−u⁰(0) = Z _t

0

u⁰⁰(s)ds+ ^X

0<tⁱ<t

I_i^∗

=−λ Z _t

0

f(s, u(s), u⁰(s))ds−p(λ−1) Z _t

0

u⁰(s)ds−q(λ−1) Z _t

0

u(s)ds+ ^X

0<tⁱ<t

I_i^∗. (3.7) From (3.6), (3.7) and assumptions (H₂), (H₃), we have

|u⁰(t)| ≤ |u⁰(0)|+ Z T

0 |f(s, u(s), u⁰(s))|ds+ 2p Z T

0 |u⁰(s)|ds+ 2q Z T

0 |u(s)|ds+

m

X

i=1

|I_i^∗|

≤ 3 2

Z T 0

(a(t)|u(t)|+b(t)|u(t)|^η +c(t)|u⁰(t)|+h(t))dt+ 3p Z T

0 |u⁰(t)|dt +3q

Z T

0 |u(t)|dt+3 2

m

X

i=1

(γ_ikukP C+δ_i)

≤ 3

2 kuk^{P C} Z _T

0 a(t)dt+kuk^ηP C

Z _T

0 b(t)dt+ku⁰k^{P C} Z _T

0 c(t)dt+ Z _T

0 h(t)dt

!

+3pTku⁰kP C+ 3qTkukP C +3 2

m

X

i=1

(γ_ikukP C+δ_i), that is,

ku⁰kP C ≤ 3 2

Z _T

0

a(t)dt+ 3qT +3 2

m

X

i=1

γ_i

! kukP C

+kuk^ηP C

Z T

0 b(t)dt+ 3 2

Z T

0 c(t)dt+ 3pT

!

ku⁰kP C+3 2

Z T

0 h(t)dt+ 3 2

m

X

i=1

δ_i. Thus, in view of assumption (H₁), we have

ku⁰kP C ≤a₁kukP C+a₂kuk^ηP C+a₃, (3.8)

wherea₁, a₂, a₃ are as in (3.1). Integrating (3.4) from 0 toT, we get that

− Z _T

0

u(t)u⁰⁰(t)dt≤λ Z _T

0

u(t)f(t, u(t), u⁰(t))dt+p(λ−1) Z _T

0

u(t)u⁰(t)dt. (3.9) In view of u(0) +u(T) = 0, u⁰(0) +u⁰(T) = 0, we have

Z T 0

u(t)u⁰(t)dt= 1 2

Z T 0

d(u²(t))

= 1 2

"

Z t1

0

d((u⁰(t))²) + Z t2

t1

d(u²(t)) +· · ·+ Z T

t^m

d(u²(t))

#

= 1 2

h(u²(t₁−0)−u²(0)) + (u²(t₂−0)−u²(t₁+ 0)) +· · ·+ (u²(T)−u²(t_m+ 0))ⁱ

= 1 2

h(u²(t₁−0)−u²(t₁+ 0)) + (u²(t₂−0)−u²(t₂+ 0)) +· · ·+ (u²(t_m−0)−u²(t_m+ 0))ⁱ

= 1

2[−(u(t₁−0) +u(t₁+ 0))I₁−(u(t₂−0) +u(t₂+ 0))I₂− · · · −(u(t_m−0) +u(t_m+ 0))I_m], (3.10) and

Z _T

0

u(t)u⁰⁰(t)dt= Z _T

0

u(t)d(u⁰(t))

= Z t1

0

u(t)d(u⁰(t)) + Z t2

t1

u(t)d(u⁰(t)) +· · ·+ Z T

t^m

u(t)d(u⁰(t))

=u(t)u⁰(t)|^t0¹ − Z _t1

0

(u⁰(t))²dt+u(t)u⁰(t)|^tt²1 − Z _t2

t1

(u⁰(t))²dt+· · ·+u(t)u⁰(t)|^Tt^m− Z _T

t^m

(u⁰(t))²dt

=u(t₁−0)u⁰(t₁−0)−u(0)u⁰(0) +u(t₂−0)u⁰(t₂−0)−u(t₁+ 0)u⁰(t₁+ 0) +· · ·+u(T)u⁰(T)−u(tn+ 0)u⁰(tn+ 0)−

Z _T

0 (u⁰(t))²dt

=u(t₁−0)u⁰(t₁−0)−u(t₁+ 0)u⁰(t₁+ 0) +· · · +u(t_n−0)u⁰(t_m−0)−u(t_m+ 0)u⁰(t_m+ 0)−

Z T 0

(u⁰(t))²dt

=u(t₁−0)u⁰(t₁−0)−u(t₁−0)u⁰(t₁+ 0) +u(t₁−0)u⁰(t₁+ 0)−u(t₁+ 0)u⁰(t₁+ 0) +· · ·+u(t_m−0)u⁰(t_m−0)−u(t_m−0)u⁰(t_m+ 0)

+u(tm−0)u⁰(tm+ 0)−u(tm+ 0)u⁰(tm+ 0)− Z _T

0 (u⁰(t))²dt

=−u(t1−0)I₁^∗−u⁰(t1+ 0)I1− · · · −u(tm−0)I_n^∗−u⁰(tm+ 0)Im− Z _T

0 (u⁰(t))²dt.(3.11) Substituting (3.10) and (3.11) into (3.9), we obtain by (H₂),(H₃) and (3.8) that

Z T

0 (u⁰(t))²dt≤λ Z T

0 u(t)f(t, u(t), u⁰(t))dt−u(t₁−0)I₁^∗−u⁰(t₁+ 0)I₁− · · · −u(t_m−0)I_m^∗

−u⁰(t_m+ 0)I_m+p(1−λ)

2 [(u(t₁−0) +u(t₁+ 0))I₁+· · ·+ (u(t_m−0) +u(t_m+ 0))I_m]

≤λ Z _T

0

u(t)f(t, u(t), u⁰(t))dt+kukP C m

X

i=1

|I_i^∗|+ku⁰kP C

X

i=1

|I_i|+p(1−λ)kukP C m

X

i=1

|I_i|

≤λ Z T

0 u(t)f(t, u(t), u⁰(t))dt+kukP C m

X

i=1

(p|I_i|+|I_i^∗|) +ku⁰kP C m

X

i=1

|I_i|

≤ Z T

0 (a(t)u²(t) +b(t)|u(t)|^1+η +1

2c(t)(u²(t) + (u⁰(t))²) +h(t))dt +

m

X

i=1

[p(α_ikukP C+β_i) +γ_ikukP C +δ_i]kukP C +

m

X

i=1

(α_ikukP C+β_i)ku⁰kP C

≤ kuk²P C

Z _T

0

a(t)dt+kuk^1+η_{P C} Z _T

0

b(t)dt+1 2kuk²P C

Z _T

0

c(t)dt+c^∗ 2

Z _T

0

(u⁰(t))²dt +

Z _T

0 h(t)dt+

m

X

i=1

(pαi+γi)kuk²P C +

m

X

i=1

((pβi+δi)kuk^{P C}

+a₁

m

X

i=1

α_ikuk²P C +a₂

m

X

i=1

α_ikuk^1+ηP C +a₂

m

X

i=1

β_ikuk^ηP C+

m

X

i=1

(a₁β_i+a₃α_i)kukP C +a₃

m

X

i=1

β_i.

Thus, Z _T

0 (u⁰(t))²dt≤ 1 1−c^∗/2

hb1kuk²P C+b2kuk^1+ηP C +b3kuk^{P C}+b4kuk^ηP C +b5

i, (3.12)

where b1=

Z _T

0 a(t)dt+1 2

Z _T

0 c(t)dt+

m

X

i=1

((p+a1)αi+γi), b2 = Z _T

0 b(t)dt+a2 m

X

i=1

αi,

(3.13) b₃=

m

X

i=1

((p+a₁)β_i+a₃α_i+δ_i), b₄ =a₂

m

X

i=1

β_i, b₅ = Z _T

0

b(t)dt+a₃

m

X

i=1

β_i. By Lemma 2.2 and (3.12), we have

kuk²P C ≤ 1 4

Z _T

0 |u⁰(t)|dt

!2

+ 1 2

Z _T

0 |u⁰(t)|dt

m

X

i=1

|Ii|+ 1 4

m

X

i=1

|Ii|

!2

≤ T 4

Z _T

0

(u⁰(t))²dt+

√T 2

Z _T

0

(u⁰(t))²dt

!1/2 m

X

i=1

|I_i|+1 4m

m

X

i=1

|I_i|²

≤ T

2(2−c^∗)

hb1kuk²P C +b2kuk^1+ηP C +b3kuk^{P C} +b4kuk^ηP C +b5

i

+

s T 2(2−c^∗)

hb₁kuk²P C+b₂kuk^1+ηP C +b₃kukP C+b₄kuk^ηP C+b₅^i1/2

m

X

i=1

(α_ikukP C +β_i)

+m 4

m

X

i=1

(α²_ikuk²P C+ 2α_iβ_ikukP C +β_i²).

It follows from the above inequality and condition (3.2) that there exists M₁ > 0 such that kuk^{P C} ≤M1. Thus, we get by (3.8) that

ku⁰kP C ≤a₁M₁+a₂M₁^η+a₃ :=M₂. (3.14)

Thus, kukP C¹ ≤ max{M₁, M₂}. It follows from Lemma 1.1 that BVP (1.1) has at least one solution. The proof is complete. 2

Corollary 3.2. Assume that (H₁),(H₂) hold. Suppose that there exist nonnegative constants β_k, δ_k (k= 1,2,· · ·, m) such that

(H4) |I_k(u)| ≤β_k, |I_k^∗(u)| ≤δ_k, k= 1,· · ·m, holds. Further assume that

Z T 0

a(t)dt+1 2

Z T 0

c(t)dt <2(2−c^∗), (3.15)

wherec^∗ = max_t∈Jc(t)<2. Then BVP (1.1) has at least one solution.

Proof. Set α_k = γ_k = 0, k = 1,2,· · ·, m. Then (H₃) reduces to (H₄), and (3.1) reduces to (3.15). So, by Theorem 3.1, we know that Corollary 3.2 holds.

Theorem 3.3. Suppose that there exist constants K₁, K₂, and L_k, l_k (k = 1,2,· · ·, m) such that

|f(t, u, v)−f(t, x, y)| ≤K₁|u−x|+K₂|v−y|, ∀u, v, x, y∈R, and

|I_k(u)−I_k(v)| ≤L_k|u−v|, |I_k^∗(u)−I_k^∗(v)| ≤l_k|u−v|, ∀u, v∈R.

Moreover suppose that

(K₁+K₂+p+q)Tmax{G₁, G₂}+

m

X

k=1

(max{G₁, G₂}L_k+ max{G₂, G₃}l_k)<1, (3.16) where G₁, G₂ and G₃ are as in (2.18), (2.20) and (2.21), respectively, then BVP (1.1) has a unique solution.

Proof. From (2.15), we have

|Au(t)−Av(t)|=

Z _T

0 G(t, s)[(f(s, u(s), u⁰(s))−f(s, v(s), v⁰(s)) +p(u⁰(s)−v⁰(s)) +q(u(s)−v(s))]ds+

m

X

k=1

[G(t, t_k)(I_k^∗(v(t_k))−I_k^∗(u(t_k))) +W(t, t_k)(I_k(u(t_k))−I_k(v(t_k)))]

≤ Z _T

0 |G(t, s)|[|f(s, u(s), u⁰(s))−f(s, v(s), v⁰(s)|+p|u⁰(s)−v⁰(s)| +q|u(s)−v(s)|]ds+

m

X

k=1

[|G(t, t_k)||I_k^∗(v(t_k))−I_k^∗(u(t_k))|+|W(t, t_k)||I_k(u(t_k))−I_k(v(t_k))|]

≤ Z _T

0 |G(t, s)|[K₁|u(s)−v(s)|+K₂|u⁰(s)−v⁰(s)|+p|u⁰(s)−v⁰(s)| +q|u(s)−v(s)|]ds+

m

X

k=1

[|G(t, t_k)|L_k|u(t_k))−v(t_k))|+|W(t, t_k)|l_k|u(t_k))−v(t_k))|]

≤(K1+K2+p+q)ku−vkP C¹

Z _T

0 |G(t, s)|ds+

m

X

k=1

[|G(t, t_k)|L_k+|W(t, t_k)|l_k]ku−vkP C¹

≤[(K₁+K₂+p+q)T G₁+

m

X

k=1

(G₁L_k+G₂l_k)]ku−vkP C¹.

|(Au)⁰(t)−(Av)⁰(t)|=

Z T 0

∂

∂tG(t, s)[(f(s, u(s), u⁰(s))−f(s, v(s), v⁰(s)) +p(u⁰(s)−v⁰(s)) +q(u(s)−v(s))]ds+

m

X

k=1

[∂

∂tG(t, t_k)(I_k^∗(v(t_k))−I_k^∗(u(t_k))) + ∂

∂tW(t, t_k)(I_k(u(t_k))−I_k(v(t_k)))]

≤ Z T

0

∂

∂tG(t, s)

[K₁|u(s)−v(s)|+K₂|u⁰(s)−v⁰(s)|+p|u⁰(s)−v⁰(s)| +q|u(s)−v(s)|]ds+

m

X

k=1

∂

∂tG(t, t_k)

L_k|u(t_k))−v(t_k))|+

∂

∂tW(t, t_k)

l_k|u(t_k))−v(t_k))|

≤(K₁+K₂+p+q)ku−vkP C¹

Z T 0

∂

∂tG(t, s) ds+

m

X

k=1

∂

∂tG(t, t_k) L_k

+

∂

∂tW(t, t_k) l_k

ku−vkP C¹

≤[(K₁+K₂+p+q)T G₂+

m

X

k=1

(G₂L_k+G₃l_k)]ku−vkP C¹. Hence,

kAu−AvkP C ≤[(K₁+K₂+p+q)T G₁+

m

X

k=1

(G₁L_k+G₂l_k)]ku−vkP C¹,

k(Au)⁰−(Av)⁰kP C ≤[(K₁+K₂+p+q)T G₂+

m

X

k=1

(G₂L_k+G₃l_k)]ku−vkP C¹. Thus, we obtain

kAu−AvkP C¹ ≤[(K1+K2+p+q)Tmax{G1, G2} +

m

X

k=1

(max{G1, G2}L_k+ max{G2, G3}l_k)]ku−vkP C¹.

In view of (3.16) and Banach fixed point theorem, A has a unique fixed point. The proof is complete. 2

4. Example

In this section, we give an example to illustrate the effectiveness of our results.

Example 4.1. Consider the problem















u⁰⁰(t) +u(t) sin²t−e^tu^1/2(t) + ¹₂u⁰(t) cos²t+t²+ sin 2t= 0, t∈0,^π₂\ {¹₃,²₃}, 4u¹₃= sinu¹₃, 4u²₃= cosu²₃,

4u⁰¹₃= 1 +¹₂cos²u¹₃, 4u⁰²₃= ¹₃sinu²₃, u(0) +u ^π₂= 0, u⁰(0) +u^{0 π}₂= 0,

(4.1)

Letf(t, u, v) =u(t) sin²t−e^tu^1/2(t) +^v₂cos²t+t²+ sin 2t,I₁(u) = sinu,I₂(u) = cosu,I₁^∗(u) = 1 +¹₂cos²u,I₂^∗(u) = ¹₃sinu,T = ^π₂,J =0,^π₂.

It is easy to show that

|I₁(u)| ≤1, |I₂(u)| ≤1, |I₁^∗(u)| ≤ ³₂, |I₂^∗(u)| ≤ ¹₃, and

|f(t, u, v)| ≤a(t)|u|+b(t)|u|^1/2+c(t)|v|+h(t), where