Existence and nonexistence of positive solutions for a class of third order BVP with integral boundary conditions in Banach spaces

Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

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Existence and nonexistence of positive solutions for a class of third order BVP with integral boundary conditions in Banach spaces Junfang Zhao a,c,*,1, Peiguang Wang b,2, Weigao Ge c a

School of Information Engineering, China University of Geosciences, Beijing 100083, PR China College of Electronic and Information Engineering, Hebei University, Baoding 071002, PR China c Department of Mathematics, Beijing Institute of Technology, Beijing 100081, PR China b

a r t i c l e

i n f o

Article history: Received 23 April 2009 Received in revised form 23 May 2009 Accepted 12 October 2009 Available online 10 November 2009 Keywords: Third order Positive solutions Fixed-point index theory Boundary-value problem Integral boundary condition Measure of noncompactness

a b s t r a c t This paper investigates the existence, nonexistence, and multiplicity of positive solutions for a class of nonlinear boundary-value problems of third order differential equations with integral boundary conditions in ordered Banach spaces by means of fixed-point principle in cone and the fixed-point index theory for strict set contraction operator. In addition, an example is worked out to demonstrate the main results. Ó 2009 Elsevier B.V. All rights reserved.

1. Introduction Recently, the theory of ordinary differential equations in abstract spaces has become a new important branch (see [1–4]). On the other hand, boundary-value problems with integral boundary conditions arise in variety of different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics can be reduced to nonlocal problems with integral boundary conditions. They include two, three, and nonlocal boundary-value problems as special cases and have attracted the attention of Gallardo [5], Karakostas and Tsamatos [6], Lomtatidze and Malaguti [7] (also see the references therein). For more information about the general theory of integral equations and their relation to boundary-value problems we refer the reader to the book of Corduneanu [8] and Agarwal and O’Regan [9]. In addition, such kind of BVP in Banach Space has been studied by some researchers, we refer the readers to [10,11] and the references therein. For other kinds of BVPs in Banach Spaces, we refer the readers to Guo [12–14], Liu [15], Zhao [16] and the references therein. However, to the best knowledge of the authors, the corresponding theory for third-order such kind of BVP has not been investigated until now. So in this paper, we intend to fill this gap. Let E be a real Banach space with norm kxk and P  E be a cone of E. Motivated by the works mentioned above, the purpose of this paper is to investigate the existence and nonexistence of positive solutions of the following third order differential equations:

* Corresponding author. Address: Department of Mathematics, Beijing Institute of Technology, Beijing 100081, PR China. E-mail address: [email protected] (J. Zhao). 1 Supported by the NNSF of CHINA (10671012) and the DPFE of China (20050007011). 2 Supported by the NNSF of CHINA(10971045). 1007-5704/$ - see front matter Ó 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.cnsns.2009.10.011

J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

x000 þ f ðt; xðtÞÞ ¼ h;

t 2 J;

403

ð1:1Þ

subject to one of the following integral boundary conditions:

x00 ð0Þ ¼ h;

xð0Þ ¼ h; xð0Þ ¼

Z

xð1Þ ¼

Z

1

gðtÞxðtÞ dt;

ð1:2Þ

xð1Þ ¼ h;

ð1:3Þ

0 1

gðtÞxðtÞ dt;

x00 ð0Þ ¼ h;

0

where J ¼ ½0; 1; f 2 Cð½0; 1  P; PÞ, h is the zero element of E; and g 2 L½0; 1 is nonnegative. When E ¼ R, (1.1) subjects to (1.2) or (1.3) has been studied by Yao in [17]. Basic facts about an ordered Banach space E can be found in [1–4]. Here we just recall a few of them. The cone P in E induces a partial order on E, i.e., x 6 y if and only if y  x 2 P. P is said to be normal if there exists a positive constant N such that h 6 x 6 y implies kxk 6 NkykðN is called the normal constant of P). Without loss of generality, we suppose here that the _ we write y  x. For details on cone theory, see [3]. normal constant N ¼ 1. If P is solid and y  x 2 P, In the following, x 2 C 2 ðJ; EÞ is called a solution of problem (1.1) and (1.2) or (1.1) and (1.3) if it satisfies (1.1) and (1.2) or (1.1) and (1.3). x is a positive solution of (1.1) and (1.2) or (1.1) and (1.3) if, in addition, xðtÞ > h for t 2 ð0; 1Þ. Clearly, if f ðt; hÞ  h then xðtÞ  h is the trivial solution of problem (1.1) and (1.2) or (1.1) and (1.3). R1 Let xðtÞ : ð0; 1 ! E be continuous; the abstract generalized integral 0 xðtÞ dt is called convergent if the limit R1 lime!0þ e xðtÞ dt exists. The convergence or divergence of other kinds of generalized integrals can be defined similarly. For a bounded set V in Banach space E, we denote aðVÞ the Kuratowski’s measure of noncompactness (see [1–4] for further understanding). In the paper, we will denote aðÞ; aC ðÞ the Kuratowski’s measure of noncompactness of a bounded subset in E and in CðJ; EÞ, respectively. The operator A : D ! EðD  EÞ is said to be a k-set contraction if A : D ! E is continuous and bounded and there is a constant k P 0 such that aðAðSÞÞ 6 kaðSÞ for any bounded S  D; a k-set contraction with k < 1 is called a strict set contraction. For application, we first state some definitions and lemmas which can be found in [2], and some notations. Definition 1.1. Let S be a bounded set in a real Banach space E. Let aðSÞ ¼ inffd > 0g; S can be expressed as the union of a finite number of sets such that the diameter of each set does not exceed d, i.e., S ¼ [m i¼1 Si with diam ðSi Þ 6 d; i ¼ 1; 2; . . . ; mg. Clearly, 0 6 aðSÞ < 1. aðSÞ is called Kuratowski’s measure of noncompactness. Definition 1.2. Let P be a cone of real Banach space E. If P ¼ fw 2 E jwðxÞ P 0; 8x 2 Pg; then P is dual cone of cone P. Throughout this paper, for x 2 P we set

f b ¼ lim sup max t2J

kxk!b

kf ðt; xÞk ; kxk

f b ¼ lim inf min kxk!b

t2J

kf ðt; xÞk ; kxk

ðwf Þb ¼ lim inf min kxk!b

t2J

wðf ðt; xÞÞ ; kxk

where b denotes 0 or 1; w 2 P ; and kwk ¼ 1: Lemma 1.1. If H 2 CðJ; EÞ is bounded and equicontinuous, then aC ðHÞ ¼ aðHðJÞÞ ¼ maxt2J aðHðtÞÞ; where HðJÞ ¼ fxðtÞ : t 2 J; x 2 Hg; HðtÞ ¼ fxðtÞ : x 2 Hg: Lemma 1.2. Let D  E and D be a bounded set; if f is uniformly continuous and bounded from J  S into E; then

aðf ðJ; SÞÞ ¼ max aðf ðt; SÞÞ 6 gl aðSÞ; for S  D: t2J

ð1:4Þ

Lemma 1.3. Let K be a cone of Banach space E and K r ¼ fx 2 K : kxk 6 rg; K r;R ¼ fx 2 K; r 6 kxk 6 Rg with R > r > 0: Suppose that A : K R ! K is a strict set acontraction such that one of the following two conditions is satisfied: (a) kAxk P kxk; 8x 2 K; kxk ¼ r; kAxk 6 kxk; 8x 2 K; kxk ¼ R: (b) kAxk 6 kxk; 8x 2 K; kxk ¼ r; kAxk P kxk; 8x 2 K; kxk ¼ R: Then, A has a fixed point x 2 K r;R : Lemma 1.4. Let K be a cone in real Banach space E and X be a nonempty bounded open convex subset of K: Suppose that A : X ! Kis a strict set contraction and AðXÞ  X, where X denotes the closure of X in K. Then the fixed-point index

iðA; X; KÞ ¼ 1:

ð1:5Þ

2. Preliminary To establish the existence and nonexistence of positive solutions in C 2 ðJ; PÞ of (1.1) and (1.2), let us list the following assumptions:

404

J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

(H0) f 2 CðJ  P; PÞ, and for any l > 0; f is uniformly continuous on J  Pl . Further suppose that g 2 L½0; 1 is nonnegative, r 2 ½0; 1Þ and there exists nonnegative constants gl with

cgl < 1 such that

aðf ðt; SÞÞ 6 gl aðSÞ; t 2 J; SR 2 Pl ;

where P l ¼ fx 2 P; kxk 6 lg; c ¼

1þ

1

0

ð1sÞgðsÞ ds 1r

ð2:1Þ ;

r¼

R1 0

sgðsÞ ds:

In our main results, we will make use of the following lemmas. Evidently, ðCðJ; EÞ; k  kc Þ is a Banach space, and the norm is defined as kxkc ¼ maxt2J kxðtÞk. We shall reduce BVP (1.1) to an integral equation in E. To this end, we first consider operator A defined by

ðAxÞðtÞ ¼

Z

1

Hðt; sÞf ðs; xðsÞÞ ds;

ð2:2Þ

0

where

Z 1 t Hðt; sÞ ¼ Gðt; sÞ þ Gðs; sÞgðsÞ ds; 1r 0 ( 1 tð1  sÞ2  12 ðt  sÞ2 ; 0 6 s 6 t 6 1; Gðt; sÞ ¼ 2 1 tð1  sÞ2 ; 0 6 t 6 s 6 1: 2

ð2:3Þ ð2:4Þ

Lemma 2.1. If condition (H0) is satisfied, then operator A defined by (2.2) is a continuous operator. Proof. It can be verified easily by the definition of Hðt; sÞ; we omit it here. h By (2.3) and (2.4), we can prove that Hðt; sÞ and Gðt; sÞ have the following properties. Lemma 2.2. For t; s 2 ½0; 1; 0 6 Gðt; sÞ 6 max06t;s61 Gðt; sÞ 6 18. Proof. If 0 6 t 6 s 6 1; easily, Gðt; sÞ P 0. If 0 6 s 6 t 6 1; Gðt; sÞ is concave with respect to t, clearly, Gðs; sÞ P 0, Gð1; sÞ ¼ 0, thus, Gðt; sÞ P 0 for t; s 2 ½0; 1. In addition,

 max Gðt; sÞ 6 max

06t;s61

06s61

1 þ s2 1 1 þ s2 s ð1  sÞ2  ð  sÞ2 ; ð1  sÞ2 2 2 4 2

 6

1 ; 8

which implies the proof is complete. h 2

Remark 2.1. It is easy to check that Gðt; sÞ > 0 for t; s 2 ð0; 1Þ. In addition, the maximum of Gðt; sÞ occurs at t ¼ 1þs 2 if 0 6 s 6 t 6 1. Lemma 2.3. Choose d 2 ð0; 12Þ and let J d ¼ ½d; 1  d, then for all t 2 J d , v ; s 2 ½0; 1, we have

Gðt; sÞ P qGðv ; sÞ;

ð2:5Þ

2

where q ¼ 4d ð1  dÞ. Proof. Obviously, for t 2 J d ;

v;

s 2 f0; 1g; Gðt; sÞ P qGðv ; sÞ holds. And for

v; s

2 ð0; 1Þ, we have four cases:

Case I: maxfv ; tg 6 s, then 1 tð1  sÞ2 Gðt; sÞ t ¼ P d P q: ¼ 2 Gðv ; sÞ 1 v ð1  sÞ2 v

ð2:6Þ

2

Case II: s 6 minfv ; tg, then 1 tð1  sÞ2  12 ðt  sÞ2 Gðt; sÞ d2 ð1  dÞ P ¼ 4d2 ð1  dÞ ¼ q: ¼ 2 1 Gðv ; sÞ 1 v ð1  sÞ2  1 ðv  sÞ2 4 2

Case III: t 6 s 6 v , then

2

ð2:7Þ

405

J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413 1 tð1  sÞ2 Gðt; sÞ tð1  sÞ2 t 2 P ¼ P d P q: ¼1 2 2 2 1 Gðv ; sÞ v v ð1  sÞ  ð v  sÞ v ð1  sÞ 2 2

Case IV:

ð2:8Þ

v 6 s 6 t, then

2 2 Gðt; sÞ 12 tð1  sÞ  12 ðt  sÞ d2 ð1  dÞ P ¼ 4d2 ð1  dÞ ¼ q: ¼ 2 1 1 Gðv ; sÞ v ð1  sÞ 4

ð2:9Þ

2

The proof is complete. h Remark 2.2. For 0 6 s 6 t 6 1, denote Gðt; sÞ ¼ G1 ðt; sÞ. Notice that G1 ðt; sÞ is concave with respect to ‘‘t”, we have

min G1 ðt; sÞ ¼ minfG1 ðd; sÞ; G1 ð1  d; sÞg ¼

t2J d ;06s6t

1 2 d ð1  dÞ; 2

therefore, (2.7) and (2.9) are established. Lemma 2.4. Assume that (H0) holds, then

1 c; for t 2 ½0; 1; 2 Hðt; sÞ P qHðv ; sÞ; for t 2 J d ;

Hðt; sÞ 6

ð2:10Þ

v;

s 2 ½0; 1;

ð2:11Þ

where c is as defined in (H0). Proof. For t 2 ½0; 1,

Hðt; sÞ ¼ Gðt; sÞ þ

t 1r

Z

1

Gðs; sÞgðsÞ ds 6

0

1 t ð1  sÞ2 ð1 þ 2 1r 1 6 c; t 2 ½0; 1: 2 ¼

Z

1

gðsÞ dsÞ ¼

0

1 t ð1  sÞ2 þ 2 1r

Z

1

0

1 1 ð1  sÞ2 ð1 þ 2 1r

1 ð1  sÞ2 gðsÞ ds 2

Z 0

1

gðsÞ dsÞ ¼

1þ 1 ð1  sÞ2 2

R1 0

ð1  sÞgðsÞ ds 1r ð2:12Þ

For t 2 J d , in view of Lemma 2.3, we have

Hðt; sÞ ¼ Gðt; sÞ þ

t 1r

P qGðv ; sÞ þ

Z 0

qv 1r

1

Gðs; sÞgðsÞ ds P qGðv ; sÞ þ Z

d 1r

Z

1

Gðs; sÞgðsÞ ds

0

1

Gðs; sÞgðsÞ ds ¼ qHðv ; sÞ:

ð2:13Þ

0

The proof is complete. h In the following, we construct a cone K by

K ¼ fx 2 Q : xðtÞ P qxðv Þ; t 2 J d ;

v 2 ½0; 1g;

ð2:14Þ

where

Q ¼ fx 2 C 2 ðJ; PÞ : xðtÞ P h; t 2 Jg;

ð2:15Þ

and let

Bl ¼ fx 2 CðJ; PÞ : kxkc 6 lg;

l > 0:

ð2:16Þ

2

It is easy to see that K is a cone of C ðJ; EÞ and K r;R ¼ fx 2 K : r 6 kxk 6 Rg  K; K  Q . Lemma 2.5. Let (H0) be satisfied, then xðtÞ is a solution of BVP (1.1) and (1.2) if and only if x 2 K is a solution of the integral equation

xðtÞ ¼

Z

1

Hðt; sÞf ðs; xðsÞÞ ds;

0

i.e., x is a fixed point of operator A defined by (2.2) in K. Proof. First we suppose that x is a solution of Eq. (1.1). Obviously, x 2 C 2 ð½0; 1; PÞ, and xðtÞ can be expressed as

ð2:17Þ

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J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

xðtÞ ¼

Z 2 X tj ðjÞ 1 t ðt  sÞ2 f ðs; xðsÞÞ ds; x ð0Þ  2 j! 0 j¼0

ð2:18Þ

considering boundary condition (1.2), one gets

xðtÞ ¼ tx0 ð0Þ 

1 2

Z

t

ðt  sÞ2 f ðs; xðsÞÞ ds:

ð2:19Þ

0

Letting t ¼ 1, we find that

xð1Þ ¼ x0 ð0Þ 

1 2

Z

1

ð1  sÞ2 f ðs; xðsÞÞ ds;

ð2:20Þ

0

thus

x0 ð0Þ ¼ xð1Þ þ

1 2

Z

1

Z

ð1  sÞ2 f ðs; xðsÞÞ ds ¼

0

1

gðsÞxðsÞ ds þ

0

1 2

Z

1

ð1  sÞ2 f ðs; xðsÞÞ ds:

ð2:21Þ

0

Substituting (2.21) into (2.19), we obtain

xðtÞ ¼ tð

Z

1

gðsÞxðsÞ ds þ

0

1 2

Z

1

ð1  sÞ2 f ðs; xðsÞÞ dsÞ 

0

Z

1 2

t

ðt  sÞ2 f ðs; xðsÞÞ ds ¼

0

Z

1

Gðt; sÞf ðs; xðsÞÞ ds þ t

0

Z

1

gðsÞxðsÞ ds:

0

ð2:22Þ Thus

Z

1

gðsÞxðsÞ ds ¼

0

Z

1

gðsÞ

0

Z

1

Gðs; sÞf ðs; xðsÞÞ ds ds þ

Z

0

1

sgðsÞ ds

0

Z

1

gðsÞxðsÞ ds:

ð2:23Þ

0

Then we obtain that

Z

1

gðsÞxðsÞ ds ¼

0

1

R1 0

Z

1 sgðsÞ ds

1

gðsÞ

Z

0

1

Gðs; sÞf ðs; xðsÞÞ ds ds;

ð2:24Þ

0

therefore,

xðtÞ ¼

Z

1

Gðt; sÞf ðs; xðsÞÞ ds þ

0

¼

Z

1

R1 0

Z

t sgðsÞ ds

1

Gðt; sÞf ðs; xðsÞÞ ds þ

0

1

R1 0

1

gðsÞ

Z

0

t sgðsÞ ds

1

Gðs; sÞf ðs; xðsÞÞ ds ds

0

Z 0

1

ð

Z

1

gðsÞGðs; sÞ dsÞf ðs; xðsÞÞ ds ¼

0

Z

1

Hðt; sÞf ðs; xðsÞÞ ds:

ð2:25Þ

0

By Lemma 2.2, there holds xðtÞ P h, that is x 2 Q . Together with Lemma 2.4, we have

xðtÞ ¼

Z

1

Hðt; sÞyðsÞ ds P q

0

Z

1

Hðv ; sÞyðsÞ ds ¼ qxðv Þ;

ð2:26Þ

0

which implies x 2 K. To sum up, we know that x is a solution of the integral equation (2.17) in K. Conversely, if x 2 K is a solution of the integral equation (2.17). Evidently,

@Gðt; sÞ ¼ @t

(

1 ð1 2

 sÞ2  ðt  sÞ; 0 6 s 6 t 6 1;

1 ð1 2

 sÞ2 ;

0 6 t 6 s 6 1;

ð2:27Þ

and

@ 2 Gðt; sÞ ¼ @t 2



1; 0 6 s 6 t 6 1; 0; 0 6 t 6 s 6 1;

ð2:28Þ

thus

Z 1 Z 1 @Gðt; sÞ 1 f ðs; xðsÞÞ ds þ gðsÞ Gðs; sÞf ðs; xðsÞÞ ds ds; R1 @t 0 0 1  0 sgðsÞ ds 0 Z 1 2 Z t @ Gðt; sÞ f ðs; xðsÞÞ ds ¼  f ðs; xðsÞÞ ds: x00 ðtÞ ¼ @t 2 0 0 000 x ðtÞ ¼ f ðt; xðtÞÞ; Z 1 xð0Þ ¼ h; x00 ð0Þ ¼ h; xð1Þ ¼ xðsÞgðsÞ ds;

x0 ðtÞ ¼

Z

1

0

ð2:29Þ ð2:30Þ ð2:31Þ ð2:32Þ

407

J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

which implies xðtÞ is a solution of BVP (1.1) and (1.2). The proof is complete. h Lemma 2.6. Suppose that (H0) holds. Then for each l > 0; A is a strict set contraction on Q \ Bl , i.e., there exists a constant 0 6 kl < 1 such that aC ðAðSÞÞ 6 kl aC ðSÞ for any S  Q \ Bl . Proof. By (H0), Lemmas 2.2 and 2.5, we know that A : Q ! Q is continuous and bounded. Now let S  Q be bounded. Then again by (H0), we can get that

aðAðSÞÞ 6 aðcofHðt; sÞf ðs; xðsÞÞ : s 2 ½0; t; t 2 J; x 2 SgÞ

ð2:33Þ

1 6 cgl aðSðJÞÞ: 2

ð2:34Þ

Thus,

1 2

aC ðAðSÞÞ 6 cgl aðSðJÞÞ:

ð2:35Þ

On the other hand, using a similar method as in the proof of Lemma 2 in [12], we can get that

aðSðJÞÞ 6 2aC ðSÞ:

ð2:36Þ

Therefore, it follows from (2.35) and (2.36) that

aC ðAðSÞÞ 6 kl aðSÞ; S  Q \ Bl ;

ð2:37Þ

where kl ¼ cgl ; 0 6 kl < 1. The proof is complete.

h

Lemma 2.7. Suppose that (H0) holds. Then AðKÞ  K and A : K r;R ! K is a strict set contraction. Proof. From Lemma 2.4 we have

minðAxÞðtÞ ¼ min t2Jd

t2J d

Z

1

Hðt; sÞf ðs; xðsÞÞ ds P q

0

Z

1

Hðv ; sÞf ðs; xðsÞÞ ds P qðAxÞðv Þ;

v 2 J:

ð2:38Þ

0

Therefore, Ax 2 K, i.e., AðKÞ  K. Also we have AðK r;R Þ  K by K r;R  K. Hence we have A : K r;R ! K. Next by Lemma 2.6, one can prove that A : K r;R ! K is a strict set contraction. The proof is complete. h 3. Existence of positive solutions to BVP (1.1) and (1.2) In this section, we impose growth conditions on f which allow us to apply Lemmas 1.3 and 1.4 to establish the existence of positive solutions of (1.1) and (1.2). At the beginning, we introduce a notation

K¼q

Z

1d

d

1 Hð ; sÞ ds: 2

ð3:1Þ

Theorem 3.1. Assume that (H0) holds and P is normal. If positive solution.

1 2

cf 0 < 1 < Kðwf Þ1 , then problem (1.1) and (1.2) has at least one

Proof. A is defined as (2.2). Considering 12 cf 0 < 1 < Kðwf Þ1 , there exists r 1 such that kf ðt; xÞk 6 ðf 0 þ e1 Þkxk, for t 2 J; x 2 K; kxk 6 r 1 , where e1 > 0 satisfies 12 cðf 0 þ e1 Þ 6 1. Let r1 2 ð0; r1 Þ. Then, for t 2 J; x 2 K; kxkc ¼ r 1 , by Lemma 2.4, we have

kðAxÞðtÞk 6

1 c 2

Z

1

kf ðs; xðsÞÞk ds 6

0

1 cðf 0 þ e1 Þ 2

Z

1

kxðsÞk ds 6

0

1 cðf 0 þ e1 Þ 2

Z

1

0

kxkc ds 6 kxkc ;

ð3:2Þ

i.e., for x 2 K; kxkc ¼ r1 , there holds

kAxkc 6 kxkc :

ð3:3Þ

Next, turning to 1 < Kðwf Þ1 , there exists r 2 > 0 such that

wðf ðt; xðtÞÞÞ P ððwf Þ1  e2 Þkxk;

for

t 2 J;

x 2 P;

kxk P r 2 ;

where e2 > 0 satisfies ððwf Þ1  e2 ÞkxkK P 1. Let r 2 ¼ maxf2r1 ; rq2 g. Then for t 2 J d ; x 2 K; kxkc ¼ r2 , wððAxÞð12ÞÞ 6 kwðAxÞð12Þk 6 kwkkðAxÞð12Þk ¼ kðAxÞð12Þk, thus

we

ð3:4Þ have

by

Lemma

2.4,

kxk P qkxkc P r 2 ,

and

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J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

Z

Z 1d Z 1d 1 1 1 Hð ; sÞwðf ðs; xðsÞÞÞ ds P Hð ; sÞwðf ðs; xðsÞÞÞ ds P ððwf Þ1  e2 Þ Hð ; sÞkxðsÞk ds 2 2 2 0 d d Z 1d 1 Hð ; sÞ ds P kxkc ; ð3:5Þ P ððwf Þ1  e2 Þqkxkc 2 d

1 1 kðAxÞð Þk P wððAxÞð ÞÞ ¼ 2 2

1

i.e., for x 2 K; kxkc ¼ r 2 , there holds

kAxkc P kxkc :

ð3:6Þ

Lemma 1.3 yields that A has at least one fixed point x 2 K r1 ;r2 ; r 1 6 kx k 6 r 2 and x ðtÞ P qkx k > 0; t 2 J d . Thus BVP (1.1) and (1.2) has at least one positive solution x . The proof is complete. h Similarly, we have the following results. Corollary 3.2. Assume that (H0) holds and P is normal. If 12 cf 0 < 1 < Kf1 , then problem (1.1) and (1.2) has at least one positive solution. Theorem 3.3. Assume that (H0) holds and P is normal. If 12 cf 1 < 1 < Kðwf Þ0 , then problem (1.1) and (1.2) has at least one positive solution. Proof. Let A be defined as (2.2). Considering Kðwf Þ0 > 1, there exists r3 > 0 such that wðf ðt; xÞÞ P ððwf Þ0  e3 Þkxk, for t 2 J; x 2 K; kxk 6 r 3 , where e3 > 0 satisfies ððwf Þ0  e3 ÞK P 1. Let r3 2 ð0; r3 Þ. Then, for t 2 J; x 2 K; kxkc ¼ r 3 , we have

1 1 kðAxÞð Þk P wððAxÞð ÞÞ ¼ 2 2 Z P ððwf Þ0  e3 Þ

Z

1

1 Hð ; sÞwðf ðs; xðsÞÞÞ ds P 2

Z

1d

1 Hð ; sÞwðf ðs; xðsÞÞÞ ds 2 0 d Z 1d 1d 1 1 Hð ; sÞkxðsÞk ds P ððwf Þ0  e3 Þqkxkc Hð ; sÞ ds P kxkc ; 2 2 d d

ð3:7Þ

i.e., for x 2 K; kxkc ¼ r 3 , there holds

kAxkc P kxkc : 1 2

Next, turning to cf

kf ðt; xÞk 6 ðf

ð3:8Þ 1

1

< 1, there exists r4 > 0 such that

þ e4 Þkxk;

for t 2 J;

x 2 P;

kxk P r4 ;

ð3:9Þ

1 ðf 1 2

where e4 > 0 satisfies þ e4 Þc < 1. Denote M ¼ maxx2K;kxkc 6r4 ;t2J kf ðt; xðtÞÞk, then kf ðt; xÞk 6 M þ ðf 1 þ e4 Þkxk for x 2 K; t 2 J. Choose r 4 P maxfr3 ; r4 ; 12 cMð1  12 cðf 1 þ e4 ÞÞ1 g. Thus for x 2 K; kxkc ¼ r4 , we have

Z 1 1 1 1 c ½M þ ðf 1 þ e4 ÞkxðsÞk ds 6 cM þ cðf 1 þ e4 Þkxkc 2 0 2 2 0 1 1 1 1 6 ð1  cðf þ e4 ÞÞr 4 þ cðf þ e4 Þkxkc ¼ r4 ¼ kxkc ; 2 2

kðAxÞðtÞk 6

1 c 2

Z

1

kf ðs; xðsÞÞk ds 6

ð3:10Þ

i.e., for x 2 K; kxkc ¼ r 4 , there holds

kAxkc 6 kxkc :

ð3:11Þ

Lemma 1.3 yields that A has at least one fixed point x 2 K r3 ;r4 ; r3 6 kx k 6 r 4 and x ðtÞ P qkx k > 0; t 2 J d . Thus BVP (1.1) and (1.2) has at least one positive solution x . The proof is complete. h







Similarly, we have the following results. Corollary 3.4. Assume that (H0) holds and P is normal. If 12 cf 1 < 1 < Kf0 , then problem (1.1) and (1.2) has at least one positive solution. Theorem 3.5. Assume (H0) holds, P is normal, and the following two conditions hold: (i) Kðwf Þ0 > 1 and Kðwf Þ1 > 1; (ii) There exists b > 0 such that supt2J;x2Pb kf ðt; xÞk < 2c1 b. Then BVP (1.1) and (1.2) has at least two positive solutions x1 ðtÞ; x2 ðtÞ, which satisfy

0 < kx1 kc < b < kx2 kc :

ð3:12Þ

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J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

Proof. Choose r; R such that 0 < r < b < R. If Kðwf Þ0 > 1, then by the proof of (3.8), we have

kAxkc P kxkc ;

for x 2 K;

kxk ¼ r:

ð3:13Þ

If Kðwf Þ1 > 1, then by the proof of (3.6), we have

kAxkc P kxkc ;

for x 2 K;

kxk ¼ R:

ð3:14Þ

On the other hand, by Lemma 2.4, for x 2 K; kxkc ¼ b, we have

kðAxÞðtÞk 6

1 c 2

Z

1

kf ðs; xðsÞÞk ds 6

0

1 cM; 2

ð3:15Þ

where

M ¼ fkf ðt; xÞk : t 2 J; x 2 Pb g < 2c1 b:

ð3:16Þ

By (3.15) and (3.16), we have

kAxkc < b ¼ kxkc :

ð3:17Þ

Lemma 1.3 implies that A has at least one fixed point x1 2 K r;b , and has at least one fixed point x2 2 K b;R . Thus BVP (1.1) and (1.2) has at least two positive solutions x1 ðtÞ and x2 ðtÞ. Noticing (3.17), we have kx1 kc – b and kx2 kc – b, therefore (3.12) holds. The proof is complete. h Similarly, we have the following results. Corollary 3.6. Assume (H0) holds, P is normal, and the following two conditions hold: (i) Kf0 > 1 and Kðwf Þ1 > 1; (ii) There exists b > 0 such that supt2J;x2Pb kf ðt; xÞk < 2c1 b. Then BVP (1.1) and (1.2) has at least two positive solutions x1 ðtÞ; x2 ðtÞ which satisfy (3.12). Corollary 3.7. Assume (H0) holds, P is normal, and the following two conditions hold: (i) Kðwf Þ0 > 1 and Kf1 > 1; (ii) There exists b > 0 such that supt2J;x2Pb kf ðt; xÞk < 2c1 b. Then BVP (1.1) and (1.2) has at least two positive solutions x1 ðtÞ; x2 ðtÞ, which satisfy (3.12). Corollary 3.8. Assume (H0) holds, P is normal, and the following two conditions hold: (i) Kf0 > 1 and Kf1 > 1; (ii) There exists b > 0 such that supt2J;x2Pb kf ðt; xÞk < 2c1 b. Then BVP (1.1) and (1.2) has at least two positive solutions x1 ðtÞ; x2 ðtÞ, which satisfy (3.12). Theorem 3.9. Assume (H0), P is normal and solid, and the following two conditions are satisfied: (i) 12 cf 0 < 1 and 12 cf 1 < 1; (ii) There exist u  h, t 2 J d and m > 0 such that

f ðt; xÞ P mu;

t 2 J;

x P u;

ð3:18Þ

and

mq > 1;

ð3:19Þ

where

q ¼ q

Z

1d

Hðu; sÞ ds:

ð3:20Þ

d

Then BVP (1.1) and (1.2) has at least two solutions. Proof. Let A be defined as (2.2). Considering (i), by Theorems 3.1 and 3.3, we know that (3.3) and (3.11) hold. Choose

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 R > max

2

q

 kuk; r 4 ;

ð3:21Þ

where r 4 is defined as in Theorem 3.3. Let U 1 ¼ fx 2 Q ; kxkc < Rg, then U 1 ¼ fx 2 Q ; kxkc 6 Rg, and by (3.11), we obtain

8x 2 U 1 ;

kAxkc < kxkc ;

ð3:22Þ

i.e., 8x 2 U 1 ,

AðU 1 Þ  U 1 :

ð3:23Þ q

Choose 0 < r < minf2 kuk; r1 g, where r1 is as defined in Theorem 3.1. Let U 2 ¼ fx 2 Q ; kxkc < rg. Then U 2 ¼ fx 2 Q ; kxkc 6 rg, and by (3.3), we have

8x 2 U 2 ;

kAxkc 6 kxkc ;

ð3:24Þ

i.e., 8x 2 U 2 ,

AðU 2 Þ  U 2 :

ð3:25Þ

Let U 3 ¼ fx 2 Q ; kxkc < R; x  u; 8t 2 J d g. As in the proof of Theorem 1 in [2], we can show that U 3 is open in Q. Let hðtÞ ¼ q2 tu, it is easy to see that h 2 Q ; khkc 6 q2 kuk < R and hðtÞ P 2u  u for t 2 J d . Hence, h 2 U 3 and thus U 3 – ;. Let x 2 U 3 . By (3.21), we have kAxkc < R. On the other hand, (3.19) and (3.20) imply that

ðAxÞðtÞ ¼

Z

1

Hðt; sÞf ðs; xðsÞÞ ds P q

0

Z

1d

Hðv ; sÞmu ds  u;

v 2 J;

t 2 Jd :

ð3:26Þ

d

Therefore,

AðU 3 Þ  U 3 :

ð3:27Þ

Since U 1 ; U 2 and U 3 are nonempty bounded convex open sets of Q, we see that (3.23), (3.25), (3.27) and [2, Corollary 0.1.2], imply, the fixed-point indices

iðA; U i ; Q Þ ¼ 1;

i ¼ 1; 2; 3:

ð3:28Þ

On the other hand, for x 2 U 3 , we have xðqÞ  u, and so

kxkc P kxðqÞk P kuk:

ð3:29Þ

Consequently, the definition of r implies

U2  U1 ;

U3  U1 ;

U 2 \ U 3 ¼ ;:

ð3:30Þ

It follows from (3.28) and (3.30) that

iðA; U 1 n ðU 2 [ U 3 Þ; Q Þ ¼ iðA; U 1 ; Q Þ  iðA; U 2 ; QÞ  iðA; U 3 ; QÞ ¼ 1:

ð3:31Þ

Finally, (3.28) and (3.31) imply that A has two fixed points x1 2 U 3 and x2 2 U 1 n ðU 2 [ U 3 Þ. We have, by (3.26), x1 ðtÞ  u for t 2 J d and, it is clear kx2 kc > r, hence x1 ðtÞ – h and x2 ðtÞ – h. This and Lemma 1.4 complete the proof. h Our next result corresponds to the case when BVP (1.1) and (1.2) has no positive solution. Theorem 3.10. Assume (H0), P is normal, and Kwðf ðt; xÞÞ > kxk for all x 2 P; kxk > 0, then problem (1.1) and (1.2) has no positive solution. Proof. Assume that xðtÞ is a solution of BVP (1.1) and (1.2). Then x 2 K; kxkc > 0 for t 2 J, and

kxkc P

Z

1

0

1 Hð ; sÞwðf ðs; xðsÞÞÞ ds > K1 2

Z 0

1

1 Hð ; sÞkxðsÞk ds P K1 qkxkc 2

which is a contradiction, and the proof is complete.

Z d

1d

1 Hð ; sÞ ds ¼ kxkc ; 2

ð3:32Þ

h

Theorem 3.11. Assume that (H0) holds, P is normal, and 12 ckf ðt; xÞk < kxk, for all x 2 P; kxk > 0, then BVP (1.1) and (1.2) has no positive solution. Proof. Assume that xðtÞ is a solution of BVP (1.1) and (1.2). Then x 2 K; kxkc > 0 for t 2 J, and

kxkc ¼ max kxðtÞk 6 t2J

1 c 2

Z 0

1

kf ðs; xðsÞÞk ds <

1 c2c1 kxkc 6 kxkc ; 2

which is a contradiction, and the proof is complete.

h

ð3:33Þ

J. Zhao et al. / Commun Nonlinear Sci Numer Simulat 16 (2011) 402–413

411

Corollary 3.12. Assume (H0), P is normal, and Kkf ðt; xÞk > kxk, for all x 2 P; kxk > 0, then problem (1.1) and (1.2) has no positive solution.

4. Existence of positive solutions to BVP (1.1) and (1.3) Now we deal with problem (1.1) and (1.3). The method is just similar to what we have done in Section 3, so we omit the proof of the main results of this section. For convenience, we list the following assumptions: f f 2 CðJ  P; PÞ, and for any l > 0, f is uniformly continuous on J  P1 . Further suppose that g 2 L½0; 1 is nonnegative, ð H0Þ e 2 ½0; 1Þ and there exists nonnegative constants gl with r

e c ge l < 1 such that

e l aðSÞ; t 2 J; S 2 P1 ; aðf ðt; SÞÞ 6g R

where e c¼

1þ

1

0

sgðsÞ ds

r 1e

;

e ¼ r

R1 0

ð4:1Þ

ð1  sÞgðsÞ ds.

e defined by We first consider operator A

e ð AxÞðtÞ ¼

Z

1

e sÞf ðs; xðsÞÞ ds; Hðt;

ð4:2Þ

0

where

e sÞ ¼ Gðt; sÞ þ 1  t Hðt; e 1r

Z

1

Gðs; sÞgðsÞ ds;

ð4:3Þ

0

and Gðt; sÞ is as defined in (2.4). It is clear that xðtÞ is a positive solution of problem (1.1) and (1.3) if and only if x is a solution of operator equation (4.2). By analogous analysis, we have the following results. f holds, then for all we have Lemma 4.1. Assume that ð H0Þ

e sÞ 6 1 e Hðt; c ; for t 2 ½0; 1; 2 e v ; sÞ; for t 2 J ; e sÞ P q Hð Hðt; d

ð4:4Þ

v;

s 2 ½0; 1;

ð4:5Þ

f where e c is as defined in ð H0Þ. e is a strict set contraction on Q \ Bl , i.e., there exists a constant f holds. Then for each l > 0, A Lemma 4.2. Suppose that ð H0Þ e e 0 6 k l < 1 such that aðAðSÞÞ 6 k l aðSÞ for any S  Q \ Bl . e e : K r;R ! K is a strict set contraction. f holds. Then AðKÞ Lemma 4.3. Suppose that ð H0Þ  K and A R 1d 1 e ; sÞ ds. e ¼q In the following, we write K Hð d 2 f holds and P is normal. If Theorem 4.4. Assume that ð H0Þ positive solution.

1e 0 2 f

c

e ðwf Þ , then problem (1.1) and (1.3) has at least one <1
f holds and P is normal. If 1 e e ðwf Þ , then problem (1.1) and (1.3) has at least one posicf 1 < 1 < K Theorem 4.5. Assume that ð H0Þ 0 2 tive solution. f holds, P is normal, and the following two conditions hold: Theorem 4.6. Assume ð H0Þ e ðwf Þ > 1 and K e ðwf Þ > 1; (i) K 0 1 (ii) There exists b > 0 such that supt2J;x2Pb kf ðt; xÞk < 2 e c 1 b. Then BVP (1.1) and (1.3) has at least two positive solutions x1 ðtÞ; x2 ðtÞ, which satisfy

0 < kx1 kc < b < kx2 kc :

ð4:6Þ

f holds, P is normal, and K e wðf ðt; xÞÞ > kxk, for all x 2 P; kxk > 0, then problem (1.1) and (1.3) has no Theorem 4.7. Assume ð H0Þ positive solution.

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f holds, P is normal, and 1 e Theorem 4.8. Assume ð H0Þ c kf ðt; xÞk < kxk, for all x 2 P; kxk > 0, then BVP (1.1) and (1.3) has no posi2 tive solution.

5. Example

Example 5.1. Consider the finite system of scalar third order differential equations

(

t x000 i ðtÞ þ 3 xi þ 300 sin xi ¼ 0;

x00i ð0Þ

xi ð0Þ ¼ 0;

¼ 0;

0 < t < 1; R1 xi ð1Þ ¼ 0 xi ðtÞ dt

ð5:1Þ

5.1. Conclusion System (5.1) has at least one positive solution xi ðtÞ for t 2 J. Proof. Let E ¼ Rn ¼ fx ¼ ðx1 ; x2 ; . . . ; xn Þ : xi 2 R; i ¼ 1; 2; . . . ; ng with the norm kxk ¼ max16i6n jxi j, and P ¼ fx ¼ ðx1 ; x2 ; . . . ; xn Þ : xi P 0; i ¼ 1; 2; . . . ; ng. Then P is a normal cone in E and system (5.1) can be translated into the form of (1.1) with boundary condition (1.2) in E. In this situation, x ¼ ðx1 ; x2 ; . . . ; xn Þ; g ¼ 1 and f ¼ ðf1 ; f2 ; . . . ; fn Þ and fi ði ¼ 1; 2; . . . ; nÞ is defined by

fi ðt; xi Þ ¼

t xi þ 300 sin xi : 3

On the other hand, we can get that P ¼ P. If we choose w ¼ ð1; 1;    ; 1Þ, then for any x 2 P we have

wðf ðt; uÞÞ ¼

n X

fi ðt; uÞ:

i¼1

It is clear that (H0) is satisfied automatically when E is finite dimensional. R 1 we try to prove that all the conditions in Theorem 3.3 are satisfied. It is easy that Next, 1þ ð1sÞgðsÞ ds c ¼ 0 1r ¼ 3. And

  (1 ð1  sÞ2  12 ð12  sÞ2 ; 0 6 s 6 12 6 1; 1 G ;s ¼ 4 1 2 ð1  sÞ2 ; 0 6 12 6 s 6 1: 4     Z 1 1 1 1=2 1 1 1 Gðs; sÞgðsÞ ds; ¼ G ; s þ ð1  sÞ2  ð1  sÞ3 : H ; s ¼ Gð ; sÞ þ 2 2 1r 0 2 4 6 7 Choose d ¼ 18 2 ð0; 12Þ, then q ¼ 4d2 ð1  dÞ ¼ 128 , and thus K ¼ q Since

f 1 ¼ lim sup max t2J

kuk 1 2

kf ðt; xÞk ¼ lim kxk!1 kxk

R 78 1 8

maxf3t xi þ 300 sin xi g 16i6n

¼

max xi 16i6n

r¼

R1 0

sgðsÞ ds ¼ 12 ;

ð5:2Þ

35 Hð12 ; sÞ ds ¼ 8192 .

1 ; 3

1 2

thus we have cf ¼ < 1. On the other hand, noticing wðxÞ P kxk; 8x 2 P, we have 1

wðf ðt; xÞÞ kf ðt; xÞk P ¼ kxk kxk

maxf3t xi þ 300 sin xi g 16i6n

max xi 16i6n

! 300

1 ðas kxk ! 0Þ; 3

and so Kðwf Þ0 > 1. To sum up, all the conditions in Theorem 3.3 are satisfied, then the conclusion follows, and the proof is complete. h References [1] Demling K. Ordinary differential equations in banach spaces. Springer-Verlag; 1977. [2] Guo DJ. Multiple positive solutions for first order nonlinear impulsive integro-differential equations in a Banach space. Appl Math Comput 2003;143:233–49. [3] Guo DJ, Lakshmikantham V, Liu XZ. Nonlinear integral equations in abstract spaces. Dordrecht: Kluwer Academic Publishers; 1996. [4] Lakshmikantham V, Leela S. Nonlinear differential equations in abstract spaces. Oxford: Pergamon; 1981. [5] Gallardo JM. Second order differential operators with integral boundary conditions and generation of semigroups. Rocky Mountain J Math 2000;30:1265–92. [6] Karakostas GL, Tsamatos PCh. Multiple positive solutions of some Fredholm integral equations arisen from nonlocal boundary-value problems. Electron. J. Differen. Eqn 2002;30:1–17.

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