Applied Mathematics and Computation 217 (2011) 9278–9285

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate /amc

Positive solutions for boundary value problem of nonlinear fractionalfunctional differential equations q

Xiaoyan Li ⇑, Song Liu, Wei JiangSchool of Mathematical Science, Anhui University Hefei, Anhui 230039, PR China

a r t i c l e i n f o

Keywords:Captuo derivativefractional functional differential equationBoundary value problemPositive solution

0096-3003/$ - see front matter � 2011 Elsevier Incdoi:10.1016/j.amc.2011.04.006

q This research had been supported by the Nationa(No. 2009 3401110001), and Natural Science Found⇑ Corresponding author.

E-mail address: lxynnfjl@126.com (X. Li).

a b s t r a c t

In this paper, we investigate the existence of positive solutions for the nonlinear Captuofractional order functional differential equation

. All righ

l Natureation of

Da0þuðtÞ þ aðtÞf ðutÞ ¼ 0; 0 < t < 1; 1 < a 6 2;

where Da0þ is the Captuo fractional order derivative, subject to the boundary conditions

�auðtÞ þ bu0ðtÞ ¼ nðtÞ; �s 6 t 6 0;cuðtÞ þ du0ðtÞ ¼ gðtÞ; 1 6 t 6 1þ b;

we obtain the existence results of positive solutions by using some fixed point theorems.� 2011 Elsevier Inc. All rights reserved.

1. Introduction

Recently, fractional differential equations have been of great interest due to their varied applications in various fields ofapplied sciences and engineering, etc. for details, see [1–14]. It should be noted that most paper and books on fractional cal-culus are devoted to solvability for the initial value problems of fractional differential equations involving differential oper-ator. Moreover, there are some papers which deal with the existence and multiplicity of solutions for nonlinear fractionaldifferential equations’ boundary value problems.

In [12], Bai and Lu studied the following two point boundary problem of fractional differential equations

Da0þuðtÞ þ aðtÞf ðt; uðtÞÞ ¼ 0; 0 < t < 1; 1 < a 6 2;

uð0Þ ¼ uð1Þ ¼ 0;

where Da is Riemann–Liouville fractional derivative.In [8], Li, Luo, Zhou considered the following nonlinear fractional boundary value problem

Da0þ þ kf ðt;uðtÞÞ ¼ 0; 0 < t < 1; 1 < a 6 2;

uð0Þ ¼ 0; Db0þuð1Þ ¼ aDb

0þuðnÞ

where Da is Riemann–liouville fractional derivative. f : [0,1) ? [0,1) is a given continuous function and 0 6 b 6 1, 0 6 a 6 1,n 2 (0,1).

ts reserved.

Science Foundation of China (No. 11071001), Research Fund for the Doctoral Program of ChinaAnhui Province (NO. KJ2010ZD02).

X. Li et al. / Applied Mathematics and Computation 217 (2011) 9278–9285 9279

From above works, we can see that the fractional boundary value problem have been studied by some authors, howeverthe results dealing with the existence of positive solutions for boundary value problem of fractional functional differentialequations are relatively scare.

In [15], Peixuan Weng and Daqing Jiang studied the existence of positive solution for boundary value problem of second-order FDE. Motivated by the work above, in this paper, we investigate the existence of positive solutions for boundary valueproblem of fractional functional differential equation with the form

Da0uðtÞ þ aðtÞf ðutÞ ¼ 0; 0 < t < 1;1 < a 6 2;�auðtÞ þ bu0ðtÞ ¼ nðtÞ; �s 6 t 6 0;cuðtÞ þ du0ðtÞ ¼ gðtÞ; 1 6 t 6 1þ b;

ð1Þ

where ut = u(t + h) for h 2 [ � s,b], s, b P 0 are constants satisfying 0 6 s + b < 1. Let C = C([ � s,b],R) be a space with a normku(h)kC = sup�s6h6bju(h)j and

Cþ ¼ fu 2 C; uðhÞP 0; �s 6 h 6 bg:

Throughout this paper, we suppose the following are satisfied:

(H1) f is a nonnegative continuous functional defined on C+.(H2) a, b, c, d P 0, ac + bc + ad > 0.(H3) n(t) and g(t) are continuous functions defined, respectively, on [�s,0] and [1,c], where c = 1 + b, n(0) = g(1) = 0,

n(t) P 0 as b ¼ 0;R 0

t e�absnðsÞds P 0 as b > 0; g(t) P 0 as d ¼ 0;

R t1 e

cdsgðsÞds P 0 as d > 0.

(H4) a(t) is nonnegative measurable function defined on (0,1), and satisfies

0 <Z

EHðtÞaðtÞdt 6

Z 1

0HðtÞaðtÞdt < þ1:

Where E = {t 2 [0,1]; 0 6 t + h 6 1, �s 6 h 6 b}, form the assumption that 0 6 s + b < 1, we conclude that E is not empty. H(t)is defined as

HðtÞ ¼ ðbþ atÞ cCðaÞ ð1� tÞa�1 þ d

Cða� 1Þ ð1� tÞa�2� �

:

We would mention that a(t) is allowed to be zero on some subset of E and have singularity at the the endpoints t = 0 and t = 1of [0,1]. For example, while b = d = 0, the function a(t) = t�m(1 � t)a�n, 0 < m < 2, 0 < n < 2 satisfies (H4).

2. Preliminaries

Let us start with some definitions and preliminaries which are used throughout this paper.

Definition 2.1 [1]. The left sided Riemann–Liouville fractional integral of order a > 0 of a function f is defined by

Iaaþ f ðtÞ ¼ 1CðaÞ

Z t

aðt � hÞa�1f ðhÞdh:

Definition 2.2 [1]. For a function f : [0,+1) ? R, the Caputo derivative of fractional order a is defined as

Daaþ f ðtÞ ¼ 1

Cðm� aÞ

Z t

a

f ðmÞðhÞðt � hÞa�mþ1 dh; ðx > aÞ;

where 0 6m � 1 6 a < m.

Lemma 2.1 [11]. Let a > 0. Then fractional differential equation CDaaþuðtÞ ¼ 0 has the solution u(t) = c0 + c1t +� � �+ cn�1tn�1,

ci 2 R,i = 0,1,2, � � � ,n � 1, n = [a] + 1.

Lemma 2.2 [11]. Let a > 0. Then

ðIaÞCDaaþuðtÞ ¼ uðtÞ þ c0 þ c1t þ � � � þ cn�1tn�1; ci 2 R; i ¼ 0;1;2; . . . ;n� 1; n ¼ ½a� þ 1:

Definition 2.3. If the function u(t) satisfies the following

1. u(t) is continuous and u(t) > 0, t 2 [�s,c].2. u(t) = u(�s, t) for t 2 [�s,0], where u(�s, t) is defined as

9280 X. Li et al. / Applied Mathematics and Computation 217 (2011) 9278–9285

uð�s; tÞ ¼e

abt 1

b

R 0t e�

absnðsÞdsþ uð0Þ

� �; b > 0;

1a nðtÞ; b ¼ 0:

8><>: ð2Þ

3. u(t) = u(c, t) for t 2 [1,c], where u(c, t) is defined as

uðc; tÞ ¼e�

cdt 1

d

R t1 e

cdsgðsÞdsþ uð1Þ

� �; d > 0;

1c gðtÞ; d ¼ 0:

8><>: ð3Þ

4. u(t) 2 AC2[0,1], where AC2[0,1] = {u : u0(t) 2 AC[0,1]}, and uðtÞ 2 AC½0;1� () uðtÞ ¼ c þR t

1 uðsÞds;uðtÞ 2 Lð0;1Þ is thespace of function u which are absolutely continuous on [0,1].

5. CDa0þuðtÞ ¼ �aðtÞf ðutÞ for almost everywhere.We say the function u(t) is a positive solution of Eq. (1).

Lemma 2.3 (Krasnoselskii fixed point Theorem). Assume that X is a Banach space and K � X is a cone in X; X1, X2 are opensubsets of X, and 0 2 X1;X1 � X2. Furthermore, let T : K \ ðX2 nX1Þ ! K be a completely continuous operator satisfying one of thefollowing conditions:

(i) kTuk 6 kuk, u 2 K \ oX1 and kTukP kuk, u 2 K \ oX2,(ii) kTukP kuk, u 2 K \ oX1 and kTukP kuk, u 2 K \ oX2.

Then there is a fixed point of T in K \ ðX2 nX1Þ.

3. Existence of positive solutions of Eq. (1)

In this section, we will give our main results on existence of positive solutions of functional fractional differential Eq. (1).

Lemma 3.1. Let 1 < a 6 2, the unique solution of Eq. (1) is

uðtÞ ¼

uð�s; tÞ; �s 6 t 6 0;R 10 Gðt; sÞaðsÞf ðusÞds; 0 6 t 6 1;

uðc; tÞ; 1 6 t 6 c;

8>>><>>>:

ð4Þ

where

Gðt; sÞ ¼g1ðt; sÞ; 0 6 s 6 t 6 1;

g2ðt; sÞ; 0 6 t 6 s 6 1;

(

and g1ðt; sÞ ¼ � ðt�sÞa�1

CðaÞ þ bþatbcþacþad

cCðaÞ ð1� sÞa�1 þ d

Cða�1Þ ð1� sÞa�2h i

; g2ðt; sÞ ¼ bþatbcþacþad

cCðaÞ ð1� sÞa�1 þ d

Cða�1Þ ð1� sÞa�2h i

respectively.

Proof. In the view of Lemma 2.2, the Eq. (1) is equivalent to the integral equation

uðtÞ ¼ �Ia0þaðtÞf ðutÞ þ c0 þ c1t; 0 < t < 1;

for some c0, c1 2 R. Then we have

uðtÞ ¼ � 1CðaÞ

R t0ðt � sÞa�1aðsÞf ðusÞdsþ c0 þ c1t;

u0ðtÞ ¼ � 1Cða�1Þ

R t0ðt � sÞa�2aðsÞf ðusÞdsþ c1:

The boundary condition �au(t) + bu0(t) = n(t), �s 6 t 6 0, n(0) = 0 and cu(t) + du0(t) = g(t), 1 6 t 6 1 + b, g(1) = 0 imply�au(0) + bu0(0) = 0, cu(1) + du0(1) = 0. Then we get

uðtÞ ¼ � 1CðaÞ

Z t

0ðt � sÞa�1aðsÞf ðusÞds

þ bþ atbc þ ac þ ad

cCðaÞ

Z 1

0ð1� sÞa�1aðsÞf ðusÞdsþ d

Cða� 1Þ

Z 1

0ð1� sÞa�2aðsÞf ðusÞds

� �; ð5Þ

Then we get the result of the Lemma. h

X. Li et al. / Applied Mathematics and Computation 217 (2011) 9278–9285 9281

Lemma 3.2. The Green function G(t, s) in Lemma 3.1 satisfies the following conditions

(i) G(t,s) is continuous on [0,1] � [0,1],(i) for b > 2�a

a�1 a,we have G(t,s) > 0, for any s, t 2 [0,1].

Proof. It is easy to check that (i) hold. So we prove that (ii) is true. When 0 6 s 6 t 6 1, we have

@g1ðt; sÞ@t

¼ �ðt � sÞa�2

Cða� 1Þ þac

ðac þ adþ bcÞCðaÞ ð1� sÞa�1 þ adðac þ adþ bcÞCða� 1Þ ð1� sÞa�2

;

and

@2g1ðt; sÞ@t2 ¼ ð2� aÞðt � sÞa�3

Cða� 1Þ P 0;

so we get

@g1ðt; sÞ@t

6@g1ð1; sÞ

@t¼ �ð1� sÞa�2

Cða� 1Þ þac

ðac þ adþ bcÞCðaÞ ð1� sÞa�1 þ adðac þ adþ bcÞCða� 1Þ ð1� sÞa�2

6�ðac þ bcÞð1� aÞ þ acðac þ adþ bcÞCðaÞ ð1� sÞa�2

;

from the condition b > 2�aa�1 a, we have @g1ðt;sÞ

@t 6 0, then g1(t,s) is decreasing with respect to t on [s,1], we obtain0 < g1(1,s) 6 g1(t,s) 6 g1(s,s). And

g1ð1; sÞ ¼ �ð1� sÞa�1

CðaÞ þ bþ abc þ ac þ ad

cCðaÞ ð1� sÞa�1 þ d

Cða� 1Þ ð1� sÞa�2� �

¼ ðða� 1Þb� ð2� aÞaÞdþ ads

ðbc þ ac þ adÞCðaÞð1� sÞ2�a ;

when b > 2�aa�1 a, we get g1(1,s) > 0. When 0 6 s 6 t 6 1. Because

@g2ðt; sÞ@t

¼ abc þ ac þ ad

cCðaÞ ð1� sÞa�1 þ d

Cða� 1Þ ð1� sÞa�2� �

P 0;

so 0 < g2(0,s) 6 g2(t,s) 6 g2(s,s). It is easy to see g2(0,s) > 0, s 2 (0,1). Therefore we get G(t,s) > 0. h

Lemma 3.3. Assume b > 2�aa�1 a, the Green function G(t, s) in Lemma 3.1 satisfies the following conditions

(i) G(t,s) 6 G(s,s) for s, t 2 (0,1),(ii) there exist positive numbers k, k⁄, such that kH(s) 6 G(t,s) 6 k⁄H(s), where HðsÞ ¼ ðbþ asÞ

cCðaÞ ð1� sÞa�1 þ d

Cða�1Þ ð1� sÞa�2h i

.

Proof. From Lemma 3.2 we can get G(t,s) 6 G(s,s) easily. We also have

h1ðsÞ 6 Gðt; sÞ 6 h2ðsÞ; ð6Þ�

where h1ðsÞ ¼

g1ð1; sÞ; 0 < s 6 rg2ð0; sÞ; r 6 s < 1 where r is the unique solution of the equation

�ð1� sÞa�1

CðaÞ þ abc þ ac þ ad

cCðaÞ ð1� sÞa�1 þ d

Cða� 1Þ ð1� sÞa�2� �

¼ 0;

and h2(s) = g2(s,s) = g1(s,s). From (6) we have kðsÞHðsÞ 6 Gðt; sÞ 6 1bcþacþad HðsÞ; for any s, t 2 (0,1), where kðsÞ ¼ h1ðsÞ

HðsÞ . Since

g1ð1; sÞHðsÞ ¼

�adð1� sÞ þ ða� 1Þðbdþ adÞðbþ asÞðcð1� sÞ þ ða� 1ÞdÞÞðac þ adþ bcÞ ;

and

g2ð0; sÞHðsÞ ¼

bcð1� sÞ þ ða� 1Þbdðbþ asÞðcð1� sÞ þ ða� 1ÞdÞÞðac þ adþ bcÞ ;

we get

inf0<s<1

g1ð1; sÞHðsÞ P

4acdðða� 2Þaþ ða� 1ÞbÞðac þ adþ bcÞ½ða� 1Þadþ ac � bc�2 þ 4ac½ða� 1Þbdþ bc�

¼ k1;

9282 X. Li et al. / Applied Mathematics and Computation 217 (2011) 9278–9285

and

inf0<s<1

g2ð0; sÞHðsÞ P

4acða� 1Þbd

ðac þ adþ bcÞ½ða� 1Þadþ ac � bc�2 þ 4ac½ða� 1Þbdþ bc�¼ k2;

let k = min{k1,k2} and k� ¼ 1bcþacþad. So we have

kHðsÞ 6 Gðt; sÞ 6 k�HðsÞ: ð7Þ

For every solution u(t) of Eq. (1), we have from (4) and (7) that

kuk½0;1� 6 k�Z 1

0HðsÞaðsÞf ðusÞds; ð8Þ

uðtÞP kk�kuk½0;1�; ð9Þ

where kuk[0,1] = sup{ju(t)j:0 6 t 6 1}. In what follows, we shall show the conclusion of our theorem only for case b > 0, d > 0,since the other situation could be discussed similarly. Suppose that u0(t) is the solution of Eq. (1) with f � 0, then it can beexpressed as

u0ðtÞ ¼

eab

t

b

R 0t e�

absnðsÞds; �s 6 t 6 0;

0;0 6 t 6 1;e�dc t

d

R t1 e

dcsfðsÞds; 1 6 t 6 c:

8>><>>: ð10Þ

Let y(t) = u(t) � u0(t), then we have from (6) that

yðtÞ ¼e

abtuð0Þ; �s 6 t 6 0;R 10 Gðt; sÞaðsÞf ðys þ u0sÞds; 0 6 t 6 1;

e�dctuð1Þ; 1 6 t 6 c:

8><>: ð11Þ

Let K be a cone in the Banach space X = C[�s,c] defined by K = {y 2 X ; y(t) P g(t)kyk}, where kyk = sup{jy(t)j : �s 6 t 6 c}, and

gðtÞ ¼e�

asb ; �s 6 t 6 0;

kk� ; 0 6 t 6 1;

e�dcc ; 1 6 t 6 c:

8><>:

Define T : K ? K

ðTyÞðtÞ ¼e

abtR 1

0 Gð0; sÞaðsÞf ðys þ u0sÞds; �s 6 t 6 0;R 10 Gðt; sÞaðsÞf ðys þ u0sÞds; 0 6 t 6 1;

e�dctR 1

0 Gð1; sÞaðsÞf ðys þ u0sÞds; 1 6 t 6 c:

8>><>>: ð12Þ

Then we have the following Lemmas. h

Lemma 3.4. The operator T : K ? K is completely continuous.

Proof. For any yn, y 2 K, n = 1,2, . . . with limn?1kyn � yk = 0. Thus for t 2 [0,1], we have that jðTynÞðtÞ � ðTyÞðtÞj6 sup06t61jf ðynt þ u0tÞ � f ðyt þ u0tÞjk�

R 10 HðsÞaðsÞds: Hence kTyn � Tyk? 0 as n ?1. This means that T is continuous. Let

X � K is bounded and M1 is the constant such that kyk 6M1. Suppose that ku0k = M2, then ky + u0k 6M1 + M2 = M fory 2X. Define a set S � C+ as S = {u 2 C+; kukC 6M}. Let L = max06t61,u2Sjf(u)j + 1, for y 2X then from (8) we havekTyðtÞk 6 Lk�

R 10 HðsÞaðsÞds, we get T(X) is bounded.

For each y 2X, we have

ðTyÞ0ðtÞ ¼ � 1Cða�1Þ

R t0ðt � sÞa�2aðsÞf ðys þ u0sÞds

þ aðbcþacþadÞ

R 10

cCðaÞ ð1� sÞa�1h

þ dCða�1Þ ð1� sÞa�2

iaðsÞf ðys þ u0sÞds; 0 6 t < 1;

ðTyÞ0ðtÞ ¼ ab eð

abÞtR 1

0 Gð0; sÞaðsÞf ðys þ u0sÞds; �s 6 t 6 0:

ðTyÞ0ðtÞ ¼ � dc eð�

dcÞtR 1

0 Gð1; sÞaðsÞf ðys þ u0sÞds; 1 6 t 6 c:

ð13Þ

X. Li et al. / Applied Mathematics and Computation 217 (2011) 9278–9285 9283

From (13) and (H4), we have

jðTyÞ0ðtÞj 6 LCða�1Þ

R 10 ð1� sÞa�2aðsÞds

þ aLðbcþacþadÞ

R 10

cCðaÞ ð1� sÞa�1h

þ dCða�1Þ ð1� sÞa�2

iaðsÞds 6 L1; 0 < t < 1� d1

jðTyÞ0ðtÞj 6 ab Lk�

R 10 HðsÞaðsÞds ¼ L2; �s 6 t 6 0:

jðTyÞ0ðtÞj 6 dc Lk�

R 10 HðsÞaðsÞds ¼ L3; 1 6 t 6 c:

ð14Þ

Let d1 ¼ �maxfL1 ;L2 ;L3g

, then for t1, t2 2 [�s,c], jt1 � t2j < d1, we get

jðTyÞðt1Þ � ðTyÞðt2Þj 6maxfL1; L2; L3gjt1 � t2j < �: ð15Þ

That is to say that T is completely continuous. h

Theorem 3.1. Assume that (H1)–(H4) hold and b > 2�aa�1 a. Then Eq. (1) has got at least a positive solution if one of the following

conditions is satisfied:

1. f0 ¼ 0;�f1 ¼ þ1; nðtÞ � 0;gðtÞ � 0;2. �f 0 ¼ þ1; f1 ¼ 0, where f0; f1;�f 0;

�f1 are defined as

f0 ¼ limkukC!0

f ðuÞkukC

; f1 ¼ limkukC!þ1

f ðuÞkukC

;

and

�f 0 ¼ limu2C� ;kukC!0

f ðuÞkukC

; �f1 ¼ limu2C� ;kukC!þ1

f ðuÞkukC

;

where C⁄ = {u 2 C+ ; 0 < ckukC 6 u(h), h 2 [�s,b]}, 0 < c < 1 is some constant depending on u.

Proof. If condition 1 is satisfied, we have u0 � 0. From f0 = 0, there is a q1 > 0 such that

f ðuÞ 6 �kukC ;u 2 Cþ; 0 6 kukC 6 q1; ð16Þ

where � > 0 satisfies

0 < �k�Z 1

0HðsÞaðsÞds <

12: ð17Þ

For any y 2 K satisfies kyk = q1, we deduce that kysk 6 q1 for s 2 [0,1] and thus

0 6 ðTyÞðtÞ 6 k�Z 1

0HðsÞaðsÞf ðysÞds 6 k��

Z 1

0HðsÞaðsÞkyskCds 6 k��q1

Z 1

0HðsÞaðsÞds < kyk;

for t 2 [0,1], which lead to kTyk < kyk, "y 2 K \ oX1, where X1 = {y 2 C[�s,c];kyk < q1}. On the other hand, since f1 = +1, forany M > 0, there is a q2 > q1 such that

f ðuÞP MkukC ;u 2 C�; kukC > rq2: ð18Þ

By (H4) we can choose r 2 0;minfe�asb ; e�

dcc ; k

k�g� �

such that Q ¼R

ErHðsÞaðsÞds > 0 where Er = {t 2 E ; r 6 t + h 6 1 � r,

�s 6 h 6 b}. Now we choose M > 0 such that kQMrq2k� > 1. Define X2 = {y 2 C[�s,c] ; kyk < q2}. For any y 2 K satisfies kyk = q2,

we deduce that

rkyk 6 gðt þ hÞkyk 6 yðt þ hÞ; t 2 Er; ð19Þ

which implies that y 2 C⁄, and kytkC P rkyk. Thus, we have

ðTyÞðtÞP kMrq2

k�

ZEr

HðsÞaðsÞds > q2 ¼ kyk; ð20Þ

which leads to kTyk > kyk, "y 2 K \ oX2. According the first part of Lemma 2.3, it follows that T has a fixed point y inK \X2 nX1 such that 0 < q1 6 kyk 6 q2.

When condition 2 is satisfied, and r, M, Q are chosen as above. Since �f 0 ¼ þ1, there is a q1 > 0 such that

f ðuÞP MkukC ; u 2 C�; kukC < q1: ð21Þ

9284 X. Li et al. / Applied Mathematics and Computation 217 (2011) 9278–9285

For any y 2 K satisfies kyk = q1, we have that kytkC P rkyk = rq1, for t 2 E. Thus

ðTyÞðtÞP kMkyt þ u0tkC

k�

ZEr

HðsÞaðsÞds PkMrq1

k�

ZEr

HðsÞaðsÞds > kyk;

which implies that kTyk > kyk, "y 2 K \ oX1, where X1 = {y 2 C[�s,c] ; kyk < q1}.On the other hand, since �f1 ¼ 0, there is an N > q1 + ku0k such that

f ðuÞ 6 �kukC ; u 2 Cþ; kukC > N; ð22Þ

where � is chosen to satisfy

k��ð1þ ku0kCÞZ 1

0HðsÞaðsÞds <

12: ð23Þ

Choose a positive constant q2 such that

q2 > 1þ 2k� maxff ðuÞ : 0 6 kukC 6 N þ ku0kgZ 1

0HðsÞaðsÞds: ð24Þ

For any y 2 K and kyk = q2, we have from the facts: u0(t) P 0 for t 2 [�s,0] that for t 2 [0,1],

kyt þ u0tkC P kytkC > N; kytkC > N;

kyt þ u0tkC 6 kytkC þ ku0tkC 6 N þ ku0k; kytkC 6 N:

It follows from (23) and (24) we have that

ðTyÞðtÞ 6 k�Z 1

0hðsÞaðsÞf ðys þ u0sÞds

¼ k�ZkyskC>N

hðsÞaðsÞf ðys þ u0sÞdsþ k�Z

06kyskC6Nþku0khðsÞaðsÞf ðys þ u0sÞds

6 k��ðkytkC þ ku0tkCÞZ 1

0HðsÞaðsÞds

þ k� maxff ðuÞ : 0 6 kukC 6 N þ ku0kgZ 1

0HðsÞaðsÞds

6 k��ðkyk þ ku0kÞZ 1

0HðsÞaðsÞds

þ k� maxff ðuÞ : 0 6 kukC 6 N þ ku0kgZ 1

0HðsÞaðsÞds

612kyk þ 1

2þ k� maxff ðuÞ : 0 6 kukC 6 N þ ku0kg

Z 1

0HðsÞaðsÞds

<12kyk þ 1

2q2 ¼ q2;

which implies that kTyk < kyk, "y 2 K \ oX2, where X2 = {y 2 C[ � s,c] ; kyk < q2}. Therefore, by Lemma 2.3, it follows that Thas a fixed point y in y 2 K \ ð �X2 nX1Þ such that 0 < q1 6 kyk 6 q2. Suppose that y is the fixed point of T in K \ ðX2 nX1Þ. Letu(t) = y(t) + u0(t), by the facts 0 < q1 6 kyk 6 q2 and u0(t) P 0, we conclude that u(t) is a positive solution of Eq. (1). We com-plete the proof. h

Remark 3.1. The above theorem generalizes and refines the results in [15].

Example 1. Consider the equation

D320uðtÞ þ ð1þ t2Þu1

3 t � 15

� �¼ 0; 0 < t < 1;

� 13 uðtÞ þ u0ðtÞ ¼ �sint; � 1

3 6 t 6 0;

u0ðtÞ ¼ ðt � 1Þ2 1 6 t 6 65 :

ð25Þ

A simple computation showed that b > 2�aa�1 a and

f ðuÞkukC

¼u1

3 t � 15

� �kukC

6kuk

13C

kukC¼ kuk�

23

C ! 0;

as kukC ? +1. We have f+1 = 0. On the other hand, if u 2 C⁄, we have

X. Li et al. / Applied Mathematics and Computation 217 (2011) 9278–9285 9285

f ðuÞkukC

¼u1

3 t � 15

� �kukC

Pckuk

13C

kukC¼ kuk�

23

C ! þ1;

as kukC ? 0 thus �f 0 ¼ þ1. So from the Theorem 3.1, the equation has got at least a positive solution.

References

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