# Nonlocal Boundary Value Problem for Impulsive Differential Equations of Fractional OrderAcademic research paper on "Mathematics"

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## Academic research paper on topic "Nonlocal Boundary Value Problem for Impulsive Differential Equations of Fractional Order"

﻿Hindawi Publishing Corporation Advances in Difference Equations Volume 2011, Article ID 404917,16 pages doi:10.1155/2011/404917

Research Article

Nonlocal Boundary Value Problem for Impulsive Differential Equations of Fractional Order

Liu Yang1,2 and Haibo Chen1

1 Department of Mathematics, Central South University, Changsha, Hunan 410075, China

2 Department of Mathematics and Computational Science, Hengyang Normal University, Hengyang, Hunan 421008, China

Correspondence should be addressed to Liu Yang, yangliu19731974@yahoo.com.cn

Received 18 September 2010; Accepted 4 January 2011

Copyright © 2011 L. Yang and H. Chen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We study a nonlocal boundary value problem of impulsive fractional differential equations. By means of a fixed point theorem due to O'Regan, we establish sufficient conditions for the existence of at least one solution of the problem. For the illustration of the main result, an example is given.

1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in various fields, such as physics, mechanics, aerodynamics, chemistry, and engineering and biological sciences, involves derivatives of fractional order. Fractional differential equations also provide an excellent tool for the description of memory and hereditary properties of many materials and processes. In consequence, fractional differential equations have emerged as a significant development in recent years, see [1-3].

As one of the important topics in the research differential equations, the boundary value problem has attained a great deal of attention from many researchers, see [4-11] and the references therein. As pointed out in [12], the nonlocal boundary condition can be more useful than the standard condition to describe some physical phenomena. There are three noteworthy papers dealing with the nonlocal boundary value problem of fractional differential equations. Benchohra et al. [12] investigated the following nonlocal boundary value problem

cDau{t) + f (t, u(t)) = 0, 0 <t<T, 1 <a < 2, ^^

u(0) = g (u), u(T) = uT,

where cDa denotes the Caputo's fractional derivative.

Zhong and Lin [13] studied the following nonlocal and multiple-point boundary value problem

cDau(t)+ f (t,u(t)) = 0, 0 <t< 1, 1 <a < 2,

m-2 (1.2)

u(0) = uo + g (u), u'(1) = u + ^ biu!(li).

Ahmad and Sivasundaram [14] studied a class of four-point nonlocal boundary value problem of nonlinear integrodifferential equations of fractional order by applying some fixed point theorems.

On the other hand, impulsive differential equations of fractional order play an important role in theory and applications, see the references [15-21] and references therein. However, as pointed out in [15, 16], the theory of boundary value problems for nonlinear impulsive fractional differential equations is still in the initial stages. Ahmad and Sivasundaram [15,16] studied the following impulsive hybrid boundary value problems for fractional differential equations, respectively,

cDqu(t) + f (t,u(t))= 0, 1 <q < 2, t e J1 = [0,1] \ {tx,t2,...,tp},

Au(tk) = Ik(u(t-)), Au'(tk) = Jk(u(t-)), tk e (0,1), k = 1,2.....p, (1.3)

u(0)+ u'(0) = 0, u(1)+ u'(1)= 0, cDqu(t)+ f (t,u(t))= 0, 1 <q < 2, t e J1 = [0,1] \ {t1,t2,...,tp}, Au(tk) = Ik(u(t-)), Au'(tk) = Jk(u(t-)), tk e (0,1), k = 1,2,...,p,

au(0) + (u'(0)=\ q\(u(s))ds, au(1) + ftu'(1)=\ q2(u(s))ds.

Motivated by the facts mentioned above, in this paper, we consider the following problem:

cDqu(t)= f (t,u(t),u'(t)), 1 < q < 2, t ej1 = [0,1] \ {t1,t2,...,tp},

Au(tk) = Ik(u(tk)), Au'(tk) = Jk(u(tk)), tk e (0,1), k = 1,2.....p, (1.5)

au(0) + (u'(0) = gi(u), au(1) + (u'(1) = g2(u),

where J = [0,1], f : J x R x R ^ R is a continuous function, and Ik,Jk : R ^ R are continuous functions, Au(tk) = u(t+) - u(tk) with u(t+) = lim^^0+ u(tk + h), u(tk) = limh^0-u(tk + h), k = 1,2,...,p, 0 = t0 < ti < t2 < ■■■ < tp < tp+\ = 1, a > 0, ( > 0, and gi, g2 : PC(J^) ^ R are two continuous functions. We will define PC(J, R) in Section 2.

To the best of our knowledge, this is the first time in the literatures that a nonlocal boundary value problem of impulsive differential equations of fractional order is considered.

In addition, the nonlinear term f (t,u(t),u'(t)) involves u!(t). Evidently, problem (1.5) not only includes boundary value problems mentioned above but also extends them to a much wider case. Our main tools are the fixed point theorem of O'Regan. Some recent results in the literatures are generalized and significantly improved (see Remark 3.6)

The organization of this paper is as follows. In Section 2, we will give some lemmas which are essential to prove our main results. In Section 3, main results are given, and an example is presented to illustrate our main results.

2. Preliminaries

At first, we present here the necessary definitions for fractional calculus theory. These definitions and properties can be found in recent literature.

Definition 2.1 (see [1-3]). The Riemann-Liouville fractional integral of order a > 0 of a function y : (0, to) ^ R is given by

«0 = f^) { (i - (2-1)

where the right side is pointwise defined on (0, to).

Definition 2.2 (see [1-3]). The Caputo fractional derivative of order a > 0 of a function y : (0, to) ^ R is given by

CD*u(t) = fi^aj Jo(i" s)'"a~1y"(s)ds>

where n = [a] +1, [a] denotes the integer part of the number a, and the right side is pointwise defined on (0, to).

Lemma 2.3 (see [1-3]). Let a > 0, then the fractional differential equation cDqu(t) = 0 has solutions

u(t) = co + cit + c2t2 + ••• + cn_i tn-1, (2.3)

where ci e R, i = 0,1,...,n- 1, n = [q] + 1. Lemma 2.4 (see [1-3]). Let a> 0, then one has

I0a+cDau(t) = u(t) + c0 + c1t + c212 + ••• + cn-1tn-1, (2.4)

where ci e R, i = 0,1,...,n- 1, n = [q] + 1. Second, we define

PC(/,R) = {x : J ^ R x e C((tfc,tfc+1],R), k = 0,1,...,p + 1 and x(t+),x(tk) exist with x(t-) = x(tk), k = 1,...,p}.

PC1 (J,W) = {x e PC(J,R); x'(t) e C((tk/tk+1],W), k = 0,1,...,p k 1, x'(t+k), x'(tk) exist, and x' is left continuous at tk, k = 1,...,p}. Let C = PC!(J,R); it is a Banach space with the norm ||x|| = suptej{||x(t)||PC, ||x'(t)||PC}, where ||x||PC = sUPteJ lx(t)|.

Like Definition 2.1 in [16], we give the following definition.

Definition 2.5. A function u e C with its Caputo derivative of order q existing on J1 is a solution of (1.5) if it satisfies (1.5).

To deal with problem (1.5), we first consider the associated linear problem and give its solution.

Lemma 2.6. Assume that

([to ,h], i = 0,

J' = 1

l(tirtik1], i = 1, 2,..

f0, t e Jo,

[l, t e Jo.

For any a e C[0,1], the solution of the problem

cDqu(t)= a(t), 1 <q < 2, t eJ = [0,1] \ {tx,t2,...,tp}, Au(tk) = Ik(u(tk)), Au'(tk) = Jk(u(tk)), tk e (0,1), k = 1,2,...,p, au(0) k pu'(0) = gi(u), au(1) k pu'(1) = g2(u)

is given by

u(t) =

(t - s)'?-1g(s)

(1- s)q-'a(sK , P (1 (1 - s)q-2a(s)

. . -ds + - ----—ds

(tk - s)q-1a(s)

0<tk<Aa / \Jt

ds k h{u(tk))

tk (tk - s)q-2a(s)

r( q k 1

ds k Jk u tk

..,(ox(i" tew)

-mZv-tJf {tk: f2°{s)ds+h(u(ti))

0<tk<t \Jtk-1 H q -1; 1

- -^[agi(u) + £(t)(at - p)(g2(u) - gi(u))], forte Jit i = 0,1,..., p.

Proof. By Lemmas 2.3 and 2.4, the solution of (2.6) can be written as

u(t) = Ig+cr(f) - bo - bit = f ^ s)? a(s)ds - bo - bit, t e [0,ii], (2.8)

J0 M q

where b0,b1 e R. Taking into account that cDqIq+ u(t) = u(t), Iq+ Ip+ u(t) = Ip++qu(t) for p,q > 0, we obtain

(i_^£)ds_(„. (2.9)

J0 r(q -1)

Using au(0) + ¡u'(0) = g1(u), we get

u(t)= (^p^a(s)ds + bi(^-t) + -gi(u), te[0,ti]. (2.10)

J0 r(q) \a f a

If t e (t1,t2], then we have

"(0= f {t Sj![a{S)ds-co-ci(t-ti), (2.11)

K r( q)

where c0,c1 e R. In view of the impulse conditions Au(t1) = u(t+)-u(t-) = I1(u(t-)), Au'(t1) = u'(t+) - u'(t-) = J1(u(t-)), we have

+ (t -11)

, t e (t1,t2].

(2.12)

Repeating the process in this way, the solution u(t) for t e (tk, tk+1] can be written as

z (f ) (in,

0<tk<t\Jtk-i Hq)

^wl' ds + Hu(rk)) )> te(tk,tk+i],

<tk<t \'tk

(tk - s)q-2a(s)

0<tk<t \J tk-i n q -1

Applying the boundary condition au(1) + pu'(1) = g2(u), we find that

. f 0z^lds

p f1 (1 - s)q-2a(s)

r(q - I) ^k (t,_ - s)q-i

£ (f S^^** *№») ("4)

o<tk<A Jtk-i H q) /

0<tk<Aa / \Jtk

a /\Jtk-i r(q - 1) g(u) - g2(u)).

Substituting the value of into (2.10) and (2.13), we obtain (2.7). □

Now, we introduce the fixed point theorem which was established by O'Regan in [22]. This theorem will be applied to prove our main results in the next section.

Lemma 2.7 (see [13, 22]). Denote by U an open set in a closed, convex set Y of a Banach space E. Assume that 0 e %. Also assume that F(K) is bounded and that F \'U Y is given by F = Fi+ F2, in which Fi : % -H- E is continuous and completely continuous and F2 : % -h- E is a nonlinear contraction (i.e., there exists a nonnegative nondecreasing function \$ : [0, <) ^ [0, <) satisfying <j>(z) < zfor z>0, such that ||F2(x) - (y) 11 < <j>(\\x - y\\) for all x,y e %), then either

(Ci) F has a fixed point u e U, or

(C2) there exists a point u e &K and A e (0,1) with u = XF(u), where K,dK represent the closure and boundary of U, respectively.

3. Main Results

In order to apply Lemma 2.7 to prove our main results, we first give F, F1, F2 as follows. Let

Qr = {u£C: ||«||<r}, r > 0,

(t - s)q-1 f (s,x(s),x'(s))

[Fium=\ti-m-ds

1 (1 -sr1f(s/x(s)/x'(s))^ _ /3 f1 (1 -sr2f(s,x(s),x'(s))_ds

+ //3_ \iy (l-sr1f(sfx(s)fx'(s)) | /3 r1 (l-s)^2

V« /[Jfp r(q) S a)tp T(q-1)

+ 2j ( -wT\-ds+Ik(u(tk))

0<tk<A •,tk-1 H q)

o<?<A«+1 - "XL —?(FT)—^w*»/

0<tk<A-/tk-1 Hq; /

0<tk<t \Jtk-1 Hq -1) /

for t e Ji, i = 0,1,...,p,

[.F2u] (0 = (m) + X(t)(at - /3) (£2 («) - gi (m) )], for f G /,, i = 0,1,..., p,

F = F1 + F2.

Clearly, for any t e Ji, i = 0,1,...,p,

r( q -1)

s «Ji, s

p nq) a )tp r(q -1)

+ 2j ( -wTx-ds + Ik(u(tk))

0<tk<A •,tk-1 H

fifc MTM^fs)) o<t*<Aa "VW, r(q-i)

x ym V /Y* (t^^fjsMsU^ \ + *(f) Jj ( --ds + Jk(u(tk)) ),

0<tk<^\^tk-1 1

[F2u]\t) = -[X(t)(g2(u) - gl(u))].

Now, we make the following hypotheses.

(A1) f : [0,1] x R x R ^ R is continuous. There exists a nonnegative function p(t) E C[0,1] with p(t) > 0 on a subinterval of [0,1]. Also there exists a nondecreasing function f : [0,to) ^ [0,to) such that \f (t,u,v)\ < p(t)f (\u\) for any (t,u,v) E [0,1] xRxR

(A2) There exist two positive constants l1,l2 such that ((a + ß)/a2)(l1 + l2) = L < 1. Moreover, g^0) = 0, g2(0) = 0, and

\g1(u) - g1 (v) \ < l1^u - v\\, \g2(u) - g2(V \ < h\\u - v\\, Vu,v EC. (3.3)

(A3) Ik,Jk : R ^ R are continuous. There exists a positive constant M such that

\Ik(u)\< M, J(u)\< M, k = 1,2.....p. (3.4)

Hi = 0 + l^Mp + 0 + i) Mp + 2pM, H2 = Mp+ 0 + Mp + pM,

Kl=f<Iz£Äs

Jü T(q)

(1 - s)q-1p(s) ß f1 (1 - s)q-2p(s)

r^srm pr

Jt„ T(q) a )tr

r (q) a),p r (q - 1)

V f* <!izfj!i<£)ds+ y <" »"-<»*

<tk<utk-1 r\q) 0<tk<Aa ' Jtk-1

0<tk<1J tk-1 1 \q) 0<tk<1

]<ik<1 ' tk-1 L\q) 0<tk<1 tk-1

a >!,,_, r(q -1) p

0<tk<1J tk-1 r( q) 0<tk<1J tk-1 r(q - 1

)t„ T(q) a)ip Y(q-1)

+ y (tk (h-s^pjs) y /ß+ \ (tk-syi-2p(s)d'

0<tk<1 tk-1

q) 0<tk<Aa ' Jtk-1 r(q 1)

0<tk<1 ^ tk-1

r(q -1

* (tk-syi-2p(s)_ds

where P = maxse[0/1]p(s).

Now, we state our main results.

Theorem 3.1. Assume that (A1), (A2), and (A3) are satisfied; moreover, supre(0(X))(r/(H + Ky(r))) > 1/(1 - L), where H = max{H1 ,H2}, K = max{K1 ,K2}, then the problem (1.5) has at least one solution.

Proof. The proof will be given in several steps.

Step 1. The operator F\ : Q,r —» C is completely continuous.

LetMr = maxSG[o/i]{|/(s,x(s),x/(s))|, x e Qr}. In fact, by (Ai), Mr can be replaced by P(p(r). For any u e Q,r, we have

\[FlU](t)\ < f - - '^v;---------->-dS

f1 (l-s)q~1\f(s,x(s),x'(s))\ --as

/3 r1 (1 -s)*-2\f{s,x{s),x'(s))\j + — -;---as

aJtP r(q-l)

+ 21 --Lrfs+IM"(ffc))l

o<tk<1\ Jtk-1 H q)

+ 1,(1+1-* 0<tk<1X a

(;tk-sr2\f(s,x(s),x'(s))\

-ds+ \Jk(u(tk))\

(-L rfa-i)

ftk (tk - s)q-1 \f (s,x(s),x'(s))|

+ Mt) 2j ( -TTTn-1^0(0)1 )

o<tk<t\Jtk-1 H qJ /

V- fftk (tk - S)q-2\f (S,X(S),X' (S))\ , , ,

o<tk<t \Jtk~1 uq -/

+ l) \MT C ^—r^ds + P-Mr f

V« JI )tP T(q) a r)tpT(q-l)

T(q - ^

+ £ (Mr f

0<tk<1 \ ^k-L

oiiA" 'Vfc-. r(9-l) /

+ 2 (f (i^t^.M)

o<T<1\.' tk-1 r( q) J

n^AJ.. r(i-i) /

1 +f£+l

r( q + 1) \a

1 ¡ 1 / 1

-Mr—— +p[ Mr-;-- + M

' r(q + 1) a ' r(q) ^ ' + 1)

+ p( Mr—-—- + M ) + p( Mr—+ M V for t 6 Ji, i = 0,l,...,p, \ r(i + l) ) H\ T(q) )

|[F1 u]'(t)| <|(

t (t - s)q-2|f (s,x(s),x'(s))|

r(q - 1)

(l-sr1\f(s,x(s)rx'(s))[ds + ^ f1 (l-s)q_2|/(s,x(s),x'(s))|

r( q -1)

0<tk<1\ Jtk-1 U q)

\a / wt^

0<tk<1

tk (tk - s)q-2If(s/x(s)/x'(s))|

r(q -1)

ds + | I

+ X(t) £

0<tk<^^tk-1

•tk {tk - s)q-2|f (s,x(s),x'(s))|

r( q -1)

ds + I I

'(l-sy-M^I" (1-«)*>,,,

H q -1)

+ ^ (t + M

0<tk<1\ ^tk-1

o£<A* T(q-1)

0<tk<1 \ Jtk-1

(tfc-s)q-2Mr Tfa-l)

ds + M

1 0 1 / 1

+ p( +M ), for t eji, i = 0,1,... ,p.

These imply that y[F1 u](t)|| < B, where B is a positive constant, that is, F1 is uniformly

bounded. In addition, for any u 6 Qr, for all Ti, t2 6 /„ T\ < t2, we can obtain

|[F1 u](n) - [F1 u](T2 )|

< p (T2-s)'i~1\f(s/x(s),x'(s))\^

r ((T2 - s)q-1 - (T1 - s)q-1) |f (s,x(s),x'(s))| + --—-ds

Jo T(q)

Jf1 (1-s)^1 1/(5^(5)^(5))^

+(T2~TlU-m-

+ l f1 (l-s)q-2|/(s,x(s),x'(s))|^

r(q -1)

+ 2j ( ---ds + \h(u(tk))\

0<tk<1\J tk-1 uq)

X-. / P , . \ I I k (tk - s)q-2

0<tk <1 a tk-1

V (P 1 Ait (tk-srz\f(s,x(s),x'(s))\j \

U*+1~tk) (L —wd—+IMu(0) 9.

+ 2j (t2-ti)( -^--ds + \jk(u(tk))\

0<tk<T1 x^tk^ h q -1;

- (T2 " Tl)" + Mr[-(T2 - n)q + (T2)q - (n)q] ; 1

r(q + 1) L > v > v J^q +1)

+ (T2 - T1)

Ji„ r(<?) r(,-l)

+ 2 (f (AzJpkds + M) 0<k<1\} tk-1 r( q) /

0^ V^ T(q - 1) /

|[F1 u](T1) - [F1 u]'(T2)|

< p (T2 -s)q"2|/(s/x(s)/x'(s))|^ J:

T1 r( q - 1)

T1 ((T1 - s)q-2 - (T2 - s)q-2) |f (s,x(s),x'(s))|

H q -1)

Taking into account the uniform continuity of the function tq, tq-1 on [0,1], we get that F1 is equicontinuity on Q,r. By the Lemma 5.4.1 in [23], we have Fi(Qr) as relatively compact. Due to the continuity of f, Ik, Jk, it is clear that F1 is continuous. Hence, we complete the proof of Step 1.

Step 2. F(U) is bounded.

From suprG(0^(r/(H + Ky(r))) > 1/(1 - L), it follows that there exists a positive constant r0, such that

r° > 7~~T- (3.8)

H + Ky(ro ) 1 - L

Now, we verify the validity of all the conditions in Lemma 2.6 with respect to the operator

F1, F2, and F. Let Qro = U. From (A2), we have

I[Fiu№\ < -2 [(a(l - t) + /3)\gl(u) - gl(0)| + ¡at - /31 - #(0)|] 1

<—(a + fi) (hr0 + hro), for t e Ju (3.9)

|[F2m]'(î)| = -[/2ro + /iro], for tefr, i = 0,1,

Combining with the property that F1 (U) is bounded (Step 1), we have F bounded on U.

Hence, we can assume that ||F(2i)|| < G, G > 0 is a constant.

Step 3. F2 is a nonlinear contraction.

Let Y = Qri, n = max{G,r0}, E = C. By (A2), we obtain |[F2w](i) - [F2r?](i)| < (1/a2)[\(a(1 - t)+fi)(g1 (u) - g1 (v))\ + \(at - ftg(u) - g2(v))\] < ((a + ft/a2)(h + l2)\\u - v\\, and \[F2u](t) - [F2v]f(t)\ < (1/a)[h + I2]\\u - v\\ < L\\u - v\\, for t e Ji. Since L< 1, we have \\F2(u) - F2(v)\\ < \$(\\u - v\\), that is, F2 is a nonlinear contraction (\$(z) = Lz).

Step 4. (C2) in Lemma 2.7 does not occur.

To this end, we perform the argument by contradiction. Suppose that (C2) holds, then there exist X e (0,1), u e dQro, such that u = XFu. Hence, we can obtain \\u\\ = r0 and

(P+1\\(1 (l-srV(s)y(ro) | p C1 (l-sr2p(s)tp(ro) U / ltP r(i) S ajt T(q-l)

+ 2 /p 0<k<i\J tk-1 r( q)

otiA« AJ.;-,

6<k<t\Jtkki T( q) /

0<k<t VA^ r( q-1)

+ -p(a + p)(hr0 + l2r0)

< —(a + p)(!1+l2)r0 +

- + 1 )Mp + ( - + 1 ) Mp + 2pM

+ f (r0)

[f1 (1 - s)q-1 p(s)

1 (1 - s)q-1 p(s) P r1 (1 - s)q-2p(s)

-ds + — -;-r—ds

T(q-l)

0<tk<1 tk-1

tk (tk - s)q-1 p(s)

<tk<1\ a / Jtk-1

tk (tk - s)q-2p(s)

0<tk<1J tk-1

0<tk<1 q1

tk (tk - s)q-1 p(s)

0<tk<1J tk-1

r(q -1)

(tk - s)q-2P(s) Tfa-l)

< Lr0 + H1 + K1 f (r0),

21 < I

(t - s)q-2p(s)f (r0)

'"'H, r(q -1)

(1 - s)q-1 p(s)f (r0) , P f 1 (1 - s)q-2p(s)f (r0)

+ I I --' _ ; ,......ds + — -—---ds

ftk (tk - s)q-1 p(s)f (r0)

0<tk<1 \ ^-1

r( q -1)

-ds + M

0<tk<1^ a

tk {tk - s)q-2p(s)f (r0)

r( q -1)

ds + M

+X(t) 2

0<tk<t\J tk-1

•tk (tk - s)q p(s)f (r0)

Mp + pM

■ <p(ro)

fi (1 sr^s) ttsYMs)ds

Jt„ r (q) cx)tr

Hq) a hp T(q - 1)

0<tk<iJ tk-1 r(q)

x^f^i) f {tk:sr2p{s)*s

]<tk<A a ' j tk-i

o<t;J<A a / Jtk-i r(q -1

tk (tk - s)q-2p(s)

x f r-'ds

<t <1 t

o<tk<1 J tk-1 r(q -1) < Lro + H2 + K2f (ro).

(3.10)

Therefore, r0 < Lr0 + H + Kf(r0). However, it contradicts with (3.8).

Hence, by using Steps 1-4, Lemmas 2.6 and 2.7, F has at least one fixed point u e Q,-0/ which is the solution of problem (1.5). □

Next, we will give some corollaries.

Corollary 3.2. Assume that (A\), (A2), and (A3) are satisfied; moreover, limsupre(0^)(r/(H + Ky(r))) = +<x>, where H = max{H1,H2}, K = max{K1,K2}; then the problem (1.5) has at least one solution.

Assume that,

(A1) (sublinear growth), f : [0,1] xRxR ^ R is continuous. There exists a nonnegative function p(t) e C[0,1] with p(t) > 0 on a subinterval of [0,1]. Also there exists a constant y e [0,1), such that \f (t,u,v)\ < p(t)\u\Y for any (t,u,v) e [0,1] x R x R.

Corollary 3.3. Assume that (A[), (A2), and (A3) are satisfied, then the problem (1.5) has at least one solution.

Assume that

(B1) f : [0,1] x R ^ R is continuous. There exists a nonnegative function p(t) e C[0,1] with p(t) > 0 on a subinterval of [0,1]. Also there exists a constant y e [0,1) such that\f (t,u)\ <p(t)\u\Y for any (t,u) e [0,1] x R,

(B2) there exist two positive constants l1,l2 such that ((a + ¡5)/a2)(l1 + l2) = L < 1. Moreover, q1(0) = 0, q2(0) = 0, and

|q1(u) - q^v)| < l1 yu - v\\, |q2(u) - q2(v)\< l2\\u - v\\, Vu,v eC, (3.11)

(B3) Ik,Jk : R ^ R are continuous. There exists a positive constant M, such that

\Ik(u)\< M, \Jk(u)\< M, k = 1,2.....p.

(3.12)

Corollary 3.4. Assume that (B1), (B2), and (B3) are satisfied, then the problem (1.4) has at least one solution.

Assume that

(B1) f : [0,1] xR ^ R is continuous. There exists a nonnegative function p(t) e C[0,1] with p(t) > 0 on a subinterval of [0,1]. \f (t, u)\<p(t) for any (t, u) e [0,1] x ffi.

Corollary 3.5. Assume that (B[), (B2), and (B3) are satisfied, then the problem (1.4) has at least one solution.

Remark 3.6. Compared with Theorem 3.2 in [16], Corollary 3.5 does not need conditions \\f (t,u) - f (t,v)\\ < L1 \\u - v\\, \\Ik(u) - Ik(V)\\ < L2\\u - v\\, and \\Jk(u) - Jk(V)\\ < L2\\u - v\\. Moreover, we only need ((a + ¡)/a2)(l1 + l2) = L < 1.

Example 3.7. Consider the following problem:

cD3/2u(t) = 0u2sin2(z/(O), 0 < t < 1, t e = [0,1] \ J,

u(0) + u'(0) = f QlUiS),l..ds, u( 1) + u'( 1) = f

J0 8 + \u(s)\ J0 8 + \u(s)\

where 0 > 0. Here, a = ¡5 = 1, p = 1, q = 3/2. Let p(s) = 9 = P and y (u) = u2, then we can see that (Ai) holds. Choosing l\ - l2 = 1/8, L = 1/2, we can easily obtain that (A2) holds. Let M = 1, then we have that (A3) also holds. Moreover, H = 8, K= ((4 + 26V2)/3Hence,

we get suprG(0 oo)(r/(H + K(p(r))) = 1/2^/(80(4 + 26^2))/3Vx > 1/(1 -L) = 2 for any given 0 < 6 < 3^57(128(4 + 26V2)). Therefore, By Theorem 3.1, the above problem (3.13) has at least one solution for 0 < 0 < 3^57(128(4 + 26V5)).

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