The existence of countably many positive solutions for one-dimensional p-Laplacian with infinitely many singularities on the half-line

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Applied Mathematics and Computation 201 (2008) 210–220 www.elsevier.com/locate/amc

The existence of countably many positive solutions for one-dimensional p-Laplacian with infinitely many singularities on the half-line q Sihua Liang a,b, Jihui Zhang a,* a

Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, 210097 Jiangsu, PR China b College of Mathematics, Changchun Normal College, Changchun 130032, Jilin, PR China

Abstract We study the existence of countably many positive solutions of boundary value problems on the half-line for differential equations of second-order. The fixed-point index theory and a new fixed-point theorem in cones are used. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Singularity; Multiple positive solutions; Multiple points boundary value problems; Fixed-point theorem; Half-line; Cone

1. Introduction In this paper, we study the existence of countable many positive solutions for m-point boundary value problem with a p-Laplacian operator on a half-line 0

ð/p ðu0 ÞÞ þ aðtÞf ðuðtÞÞ ¼ 0; uð0Þ ¼

m2 X

ai uðni Þ;

0 < t < þ1;

u0 ð1Þ ¼ 0;

ð1:1Þ ð1:2Þ

i¼1 p2

where /p ðsÞ ¼ jsj s, p > 1, ni 2 ð0; þ1Þ with 0 < n1 < n2 <    < nm2 < þ1 and ai satisfy ai 2 ½0; þ1Þ; Pm2 0 < i¼1 ai < 1; f 2 Cð½0; þ1Þ; ½0; þ1ÞÞ, aðtÞ : ½0; þ1Þ ! ½0; þ1Þ and has countably many singularities in ½1; þ1Þ: The existence and multiplicity of positive solutions for linear and nonlinear ordinary differential equations and difference equations have been studied extensively. To identify a few, we refer the reader to see [2,6–10,12,14]. Recently, in paper [7] study the existence of positive solutions of the following singular q

Project supported by Foundation of Major Project of Science and Technology of Chinese Education Ministry, SRFDP of Higher Education and NSF of Education Committee of Jiangsu Province. * Corresponding author. E-mail addresses: [email protected] (S. Liang), [email protected] (J. Zhang). 0096-3003/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2007.12.016

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211

three-point boundary value problems for one-dimensional under the condition that f 2 Cð½0; þ1Þ; ½0; þ1ÞÞ; aðtÞ : ½0; 1 ! ½0; þ1Þ and have countable many singularities on ð0; 12Þ p-Laplacian operator (

ð/p ðu0 ÞÞ0 þ aðtÞf ðuðtÞÞ ¼ 0; 0

u ð0Þ ¼ 0;

0 < t < 1;

uð1Þ ¼ buðgÞ:

In [6,8] they also obtain countable many positive solutions by assuming aðtÞ has infinitely many singularities in ð0; 12Þ. Seeing such a fact, we will think ‘‘whether or not we can obtain countable many positive solutions for the boundary value problems of differential equation on infinite intervals?” As a result, the goal of present paper is to fill the gap in this area. Obviously, using the similar method as [6–8] we can get the same conclusion when aðtÞ has infinitely many singularities in ð12 ; 1Þ if we change corresponding conditions. So we only assume aðtÞ has infinitely many singularities in ð1; þ1Þ in this paper. The motivation for the present work stems from both practical and theoretical aspects. In fact, boundary value problems on the half-line occur naturally in the study of radially symmetric solutions of nonlinear elliptic equations (see [4,13]) and various physical phenomena [3,11], such as unsteady flow of gas through a semiinfinite porous media, the theory of drain flows, plasma physics, in determining the electrical potential in an isolated neutral atom. In all these applications, it is frequent that only solutions that are positive are useful. Recently there have been many papers investigated the positive solutions of boundary value problem on the half-line, see [1,15–18], they discuss the existence and multiplicity (at least three) positive solutions to nonlinear differential equation. However, to the best knowledge of the authors, there is no paper concerned with the existence of countable many positive solutions to m-point boundary value problems of differential equation on infinite intervals so far. So in this paper, we use fixed-point index theory and a new fixed-point theorem in cones investigate the existence of countable solutions to boundary value problems (1.1) and (1.2). We will assume that the following conditions are satisfied throughout this paper: ðC 1 Þ f 2 Cð½0; þ1Þ; ½0; þ1ÞÞ; f ð0Þ 6 0 on any subinterval of ð0; þ1Þ and when u is bounded f ðð1 þ tÞuÞ is bounded on ½0; þ1Þ; 1 ðC 2 Þ There exists a sequence fti gi¼1 such that 1 < ti < tiþ1 , limi!1 ti ¼ t0 < þ1; limt!ti aðtÞ ¼ 1; i ¼ 1; 2; . . . ; and Z 0< 0

þ1

aðtÞdt < þ1;

Z

þ1

/1 p

Z

0

þ1

 aðsÞds ds < þ1:

s

Moreover aðtÞ does not vanish identically on any subinterval of ½0; þ1Þ: The plan of the paper is as follows. In Section 2, for the convenience of the reader we give some definitions. In Section 3, we present some lemmas in order to prove our main results. Section 4 is developed to presenting and proving our main results. 2. Some definitions and fixed-point theorems In this section, we provide some background definitions cited from cone theory in Banach spaces. Definition 2.1. Let ðE; k  kÞ be a real Banach space. A nonempty, closed, convex set P  E is said to be a cone provided the following are satisfied (a) if y 2 P and k P 0, then ky 2 P ; (b) if y 2 P and y 2 P , then y ¼ 0. If P  E is a cone, we denote the order induced by P on E by 6, that is, x 6 y if and only if y  x 2 P .

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Definition 2.2. A map a is said to be a nonnegative, continuous, concave functional on a cone P of a real Banach space E, if a : P ! ½0; 1Þ; is continuous, and aðtx þ ð1  tÞyÞ P taðxÞ þ ð1  tÞaðyÞ; for all x, y 2 P and t 2 ½0; 1. Definition 2.3. Given a nonnegative continuous functional c on a cone P of E, we define for each d > 0 the set P ðc; dÞ ¼ fx 2 P : cðxÞ < d g: The following fixed-point theorems are fundamental and important to the proofs of our main results. Theorem 2.1 [5]. Let T E be a Banach space and P  E be a cone in E. Let r > 0 define Xr ¼ fx 2 P : kxk < rg. Assume that T : P Xr ! P is completely continuous operator such that Tx 6¼ x for x 2 oXr . (i) If kTuk 6 kuk for u 2 oXr , then iðA; Xr ; P Þ ¼ 1: (ii) If kTuk P kuk for u 2 oXr , then iðA; Xr ; P Þ ¼ 0: Theorem 2.2 [19]. Let P be a cone in a Banach space E. Let a, b and c be three increasing, nonnegative and continuous functionals on P, satisfying for some c > 0 and M > 0 such that cðxÞ 6 bðxÞ 6 aðxÞ;

kxk 6 McðxÞ

for all x 2 P ðc; cÞ. Suppose there exists a completely continuous operator T : P ðc; cÞ ! P and 0 < a < b < c such that (i) cðTxÞ < c, for all x 2 oP ðc; cÞ; (ii) bðTxÞ > b, for all x 2 oP ðb; bÞ; (iii) P ða; aÞ 6¼ ;; and aðTxÞ < a; for all x 2 oP ða; aÞ. Then T has at least three fixed-points x1 , x2 , x3 2 P ðc; cÞ such that 0 6 aðx1 Þ < a < aðx2 Þ;

bðx2 Þ < b < bðx3 Þ;

cðx3 Þ < c:

3. Preliminaries and lemmas In this paper, the following space E will be the basic space to the study (1.1) and (1.2), which is denoted by   j uðtÞ j < þ1 : E ¼ u 2 C½0; þ1Þ : sup 06t<þ1 1 þ t Then E is a Banach space with norm kuk ¼ sup06t<þ1 juðtÞj < þ1. 1þt Define cone K  E by   0 K ¼ u 2 E : uðtÞ is a nonnegative concave function on ½0; þ1Þ and lim u ðtÞ ¼ 0 : t!þ1

0

0

It is easily to see that if uðtÞ satisfies (1.1), then ð/p ðu ðtÞÞÞ ¼ aðtÞf ðuðtÞÞ 6 0 on ½0; þ1Þ, which implies that uðtÞ is concave on ½0; þ1Þ: Moreover, if u0 ð1Þ ¼ 0; then u0 ðtÞ P 0 on ½0; þ1Þ. Therefore, uðtÞ is monotone increasing on ½0; þ1Þ.

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Lemma 3.1. Suppose ðC 2 Þ holds. Then (i) for any a constant h 2 ð1; þ1Þ which satisfies Z h 0< aðtÞdt < þ1; 1 h

(ii) the function H ðtÞ ¼

Z

t1

/1 p

Z

t

t1

Pm2  i¼1 ai aðsÞds ds þ

s

R t  /1 aðsÞds ds p s Pm2 1  i¼1 ai

Rt

1 t1

is continuous and positive on ½t11 ; t1 : Furthermore, L ¼ min H ðtÞ > 0: t2½t1 ;t1  1

Proof. Firstly, we can easy to obtain (i) from the condition ðC 2 Þ. Next, we prove the conclusion (ii) also true. It is easily seen that H ðtÞ is continuous on ½t11 ; t1 . Let Z t1  Z t1 1 /p aðsÞds ds; H 1 ðtÞ ¼ t s  Pm2 R t 1 R t aðsÞds ds i¼1 ai t1 /p s 1 H 2 ðtÞ ¼ : Pm2 1  i¼1 ai Then, from condition ðC 2 Þ, we know that H 1 ðtÞ strictly monotone decreasing on ½t11 ; t1  and H 1 ðt1 Þ ¼ 0. Similarly, function H 2 ðtÞ is strictly monotone increasing on ½t11 ; t1  and H 2 ðt11 Þ ¼ 0. Since, H 1 ðtÞ and H 2 ðtÞ are not equal to zero at the same time. So, the function H ðtÞ ¼ H 1 ðtÞ þ H 2 ðtÞ is positive on ½t11 ; t1 , which implies L ¼ mint2½t1 ;t1  H ðtÞ > 0. h 1

Lemma 3.2. Let u 2 K and h of Lemma 3.1, then  1 1 uðtÞ P kuk; t 2 ; h : h h Proof. From the definition of K, we can get uðtÞ is increasing on ½0; þ1Þ. Moreover, u0 ð1Þ ¼ 0 implies that uðtÞ the function 1þt achieves its maximum at l 2 ½0; þ1Þ. So by the concavity of uðtÞ, we have     1 h  1 þ hl 1 1 þ l ¼u 16t6h h h þ hl h  1 þ hl h þ hl h   h  1 þ hl 1 1 uðlÞ 1 uðlÞ 1 P u uðlÞ P ¼ ¼ kuk: þ h þ hl h  1 þ hl h þ hl hð1 þ lÞ h 1 þ l h

uðtÞ P min uðtÞ ¼ u



Now we define an operator T : K ! C½0; þ1Þ by  Pm2 R ni 1 R þ1 aðsÞf ðuðsÞÞds ds i¼1 ai 0 /p s ðTuÞðtÞ ¼ aðsÞf ðuðsÞÞds ds þ : ð3:1Þ Pm2 1  i¼1 ai 0 s R þ1 Obviously ðTuÞðtÞ P 0, for t 2 ð0; þ1Þ and ðTuÞ0 ðtÞ ¼ /1 aðsÞf ðuðsÞÞdsÞ, furthermore ð/p ðTuÞ0 ðtÞÞ0 ¼ p ð t aðtÞf ðuðtÞÞ 6 0. This shows ðTKÞ  K. To obtain the complete continuity of T, the following lemma is till need. n o ðtÞ Lemma 3.3 [18]. Let W be a bounded subset of K. The W is relatively compact in E if W1þt are equicontinuous on any finite subinterval of ½0; þ1Þ and for any e > 0; there exists N > 0 such that Z

t

/1 p

Z

þ1



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xðt1 Þ xðt2 Þ



1 þ t  1 þ t < e; 1 2 uniformly with respect to x 2 W as t1 ; t2 P N ; where W ðtÞ ¼ fxðtÞ : x 2 W g; t 2 ½0; þ1Þ. Lemma 3.4. Let ðC 1 Þ and ðC 2 Þ hold. Then T : K ! K is completely continuous. Proof. Firstly, it is easy to check that T : K ! K is well defined. Let Br0 as defined in [17] and X be any bounded subset of K. Then, there existence r > 0 such that kuk 6 r; for all u 2 X. Therefore, we have

Z 

Pm2 R ni 1 R þ1 Z þ1  aðsÞf ðuðsÞÞds ds

1

t 1 i¼1 ai 0 /p s kTuk ¼ sup aðsÞf ðuðsÞÞds ds þ

/p

Pm2

1  i¼1 ai t2½0;þ1Þ 1 þ t 0 s Z þ1  Z t 1 6 sup /1 aðsÞf ðuðsÞÞds ds p t2½0;þ1Þ 1 þ t 0 s  Pm2 R nm2 1 R þ1 /p s aðsÞf ðuðsÞÞds ds 1 i¼1 ai 0 þ sup Pm2 ai 1  i¼1 t2½0;þ1Þ 1 þ t ! Pm2 Z þ1  1 i¼1 ai nm2 6 /p aðsÞf ðuðsÞÞds 1þ 8u 2 X: 6 C/1 Pm2 p ðBr Þ 1  i¼1 ai 0 So T X is bounded. By the similar estimate in [17], we can get T X is equicontinuous on any finite subinterval of ½0; þ1Þ. Next we prove for any e > 0, there exists sufficiently large N > 0 such that



ðTuÞðt1 Þ ðTuÞðt2 Þ



ð3:2Þ

1 þ t  1 þ t < e for all t1 ; t2 P N 8u 2 X: 1 2 Since

R þ1

aðsÞf ðuðsÞÞds < þ1. Therefore, we can choose N 1 > 0 such that 0  Pm2 R ni 1 R þ1 aðsÞf ðuðsÞÞds ds e i¼1 ai 0 /p s  < : Pm2 5 N 1 1  i¼1 ai

We can also select N 2 ; N 3 > 0 are satisfied, respectively R þ1  R þ1 Z þ1  5 0 /1 aðsÞf ðuðsÞÞds ds e p s 1 ; /p N2 > aðsÞf ðuðsÞÞds < : 5 e N3 Then let N ¼ maxfN 1 ; N 2 ; N 3 g. It is not lose generalization, we assume t2 > t1 P N . So it follow that



Z þ1  Z þ1 

Z t1 Z t2

ðTuÞðt1 Þ ðTuÞðt2 Þ 1

1 1 1



/p aðsÞf ðuðsÞÞds ds  /p aðsÞf ðuðsÞÞds ds

1 þ t  1 þ t 6 1 þ t 1 þ t2 0 1 2 1 0 s s   Pm2 R ni 1 R þ1 Pm2 R ni 1 R þ1 aðsÞf ðun ðsÞÞds ds aðsÞf ðuðsÞÞds ds i¼1 ai 0 /p i¼1 ai 0 /p s s   þ þ Pm2 Pm2 ð1 þ t1 Þ 1  i¼1 ai ð1 þ t2 Þ 1  i¼1 ai

Z þ1 

Z þ1

1 1

1

/p aðsÞf ðuðsÞÞds ds

 < 1 þ t1 1 þ t2

0 s Z þ1  Z t2 1 2e 2e e 2e þ /1 aðsÞf ðuðsÞÞds ds þ 6 þ þ ¼ e: 1 þ t 2 t1 p 5 5 5 5 t1 That is, (3.2) holds. By Lemma 3.3, T X is relatively compact. Therefore, we know that T is a compact operator.

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215

Thirdly, we prove T is continuous. Let un ! u as n ! þ1 in K. Then by the Lebesgue dominated convergence theorem and continuity of f, we can get

Z þ1

Z þ1 Z þ1



aðsÞf ðun ðsÞÞds  aðsÞf ðuðsÞÞds

6 aðsÞjf ðun ðsÞÞ  f ðuðsÞÞjds ! 0; as n ! þ1;

t

t

t

i.e., Z

þ1

aðsÞf ðun ðsÞÞds !

t

Z

þ1

aðsÞf ðuðsÞÞds;

Moreover, Z /1 p



þ1

aðsÞf ðun ðsÞÞds

! /1 p

Z

t

So

as n ! þ1:

t

þ1

 aðsÞf ðuðsÞÞds ;

as n ! þ1:

t

Z  Z þ1  Z þ1 

1

t 1 1 kTun  Tuk 6 sup /p aðsÞf ðun ðsÞÞds  /p aðsÞf ðuðsÞÞds ds

þ sup

t2½0;þ1Þ 1 þ t t2½0;þ1Þ 0 s s

Pm2 R n 1 R þ1   

R R P n þ1 m2 1 i i aðsÞf ðuðsÞÞds ds

1

i¼1 ai 0 /p s aðsÞf ðun ðsÞÞds ds i¼1 ai 0 /p s  



Pm2 Pm2

1þt

1  i¼1 ai 1  i¼1 ai Z þ1  Z þ1 

Z t

1

1

/ 6 sup aðsÞf ðun ðsÞÞds  /1 aðsÞf ðuðsÞÞds

ds p p

t2½0;þ1Þ 1 þ t 0 s s

Pm2 Z þ1  Z þ1 



1 i¼1 ai nm2 1 aðsÞf ðun ðsÞÞds  /p aðsÞf ðuðsÞÞds

þ Pm2 /p 1  i¼1 ai s s ! 0;

as n ! þ1:

Therefore, T is continuous. In sum, T : K ! K is completely continuous. h 4. Main results For notational convenience, we denote by 1 þ t1 k1 ¼ ; L

Pm2 1  i¼1 ai : k2 ¼ R þ1  Pm2 Pm2 /1 aðsÞds 1  i¼1 ai þ i¼1 ai nm2 p 0

The main results of this paper are the following. Theorem 4.1. Suppose that conditions ðC 1 Þ, ðC 2 Þ hold. Let fhk g1 k¼1 be such that hk 2 ðtk ; tkþ1 Þ ðk ¼ 1; 2; . . .Þ: Let 1 frk g1 and fR g be such that k k¼1 k¼1 rk Rk1 < 2 < rk < mrk < Rk ; k ¼ 2; 3; . . . : 2hk Furthermore for each natural number k we assume that f satisfy: ðC 3 Þ f ðð1 þ tÞuÞ P /p ðmrk Þ; for all 2hrk2 6 uðtÞ 6 rk ; k ðC 4 Þ f ðð1 þ tÞuÞ 6 /p ðMRk Þ; for all 0 6 uðtÞ 6 Rk ; where m 2 ðk1 ; þ1Þ; M 2 ð0; k2 Þ: Then the boundary value problem (1.1) and (1.2) has infinitely many solutions 1 fuk gk¼1 such that rk 6 kuk k 6 Rk ;

k ¼ 1; 2; . . . :

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Proof. Since 1 < tk < hk < tkþ1 6 t0 < þ1; k ¼ 1; 2; . . . ; then, for any k 2 N and u 2 K; by the Lemma 3.2, we have  1 1 uðtÞ P kuk; t 2 ; hk : ð4:1Þ hk hk 1

1

Consider the sequence fX1;k gk¼1 and fX2;k gk¼1 of open subsets of E defined by X1;k ¼ fu 2 K : kuk < rk g;

k ¼ 1; 2; . . . ;

X2;k ¼ fu 2 K : kuk < Rk g;

k ¼ 1; 2; . . . :

For a fixed k and u 2 oX1;k . From (4.1) we have uðh1k Þ juðtÞj juðtÞj uðtÞ uðtÞ 1 P sup P P inf P kuk P 1 1 þ hk 1þt hk ð1 þ hk Þ 1 6t6h 1 þ t 6t6hk 1 þ t 06t<þ1 1 þ t hk k hk  1 1 P 2 rk for all t 2 ; hk : hk 2hk By ðC 3 Þ, we have  1 f ðuÞ P /p ðmrk Þ for all t 2 ; hk : hk rk ¼ kuk ¼ sup

Since ðt11 ; t1 Þ  ½h1k ; hk ; if ðC 2 Þ holds, in the following, we consider three cases: (i) If n1 2 ½t11 ; t1 . In this case, from (3.1), condition ðC 3 Þ and Lemma 3.1, we have

Z 

Pm2 R ni 1 R þ1 Z þ1  aðsÞf ðuðsÞÞds ds

1

t 1 i¼1 ai 0 /p s kTuk ¼ sup aðsÞf ðuðsÞÞds ds þ

/p

Pm2

1  i¼1 ai t2½0;þ1Þ 1 þ t 0 s Pm2 Z t1  Z n1  Z t1 Z n1 1 1 1 i¼1 ai P /1 aðsÞf ðuðsÞÞds ds þ / aðsÞf ðuðsÞÞds ds P p m2 1 þ t1 n1 p 1 þ t1 1  i¼1 ai t11 s s "Z Pm2 Z t 1  Z n1  # Z n1 t1 1 a i i¼1 P ðmrk Þ /1 aðsÞds ds þ /1 aðsÞds ds P p p 1 1 þ t1 1  m2 n1 s s i¼1 ai t1 ¼

mrk Lmrk H ðn1 Þ > > rk ¼ kuk: 1 þ t1 1 þ t1

(ii) If n1 2 ð0; t11 Þ: In this case, from (3.1), condition ðC 3 Þ and Lemma 3.1, we have

Z 

Pm2 R ni 1 R þ1 Z þ1  a / aðsÞf ðuðsÞÞds ds

1

t 1 i p i¼1 s 0 kTuk ¼ sup aðsÞf ðuðsÞÞds ds þ

/p Pm2



1  i¼1 ai t2½0;þ1Þ 1 þ t 0 s Z þ1  Z t1  Z t Z t1 1 1 P sup /1 aðsÞf ðuðsÞÞds ds P /1 aðsÞf ðuðsÞÞds ds p p 1 1 þ t 1 þ t 1 t t2½0;þ1Þ s 0 s 1 Z t1    Z t1 mrk mrk 1 Lmrk P /1 aðsÞds ds ¼ H > rk ¼ kuk: > p t 1 þ t1 t1 1 þ t 1 þ t1 1 1 s 1

(iii) If n1 2 ðt1 ; þ1Þ: In this case, from (3.1), condition ðC 3 Þ and Lemma 3.1, we have

Z 

Pm2 R ni 1 R þ1 Z þ1  aðsÞf ðuðsÞÞds ds

1

t 1 i¼1 ai 0 /p s kTuk ¼ sup aðsÞf ðuðsÞÞds ds þ

/p Pm2

1  i¼1 ai t2½0;þ1Þ 1 þ t 0 s  Pm2 R t1 1 R t1 aðsÞf ðuðsÞÞds ds i¼1 ai t1 /p s 1 mrk Lmrk 1 P P H ðt1 Þ > > rk ¼ kuk: Pm2 1 þ t1 1 þ t 1 þ t1 1 1  i¼1 ai

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Thus in all cases, an application of Theorem 2.1 implies that iðT ; X1;k ; KÞ ¼ 0: On the another hand, let u 2 oX2;k ; we have

ð4:2Þ uðtÞ 1þt

6

sup06t<þ1 juðtÞj 1þt

6 kuk ¼ Rk , by ðC 4 Þ we have

f ðuðtÞÞ 6 /p ðMRk Þ for all t 2 ½0; þ1Þ: So

Z 

Pm2 R ni 1 R þ1 Z þ1  aðsÞf ðuðsÞÞds ds

1

t 1 i¼1 ai 0 /p s kTuk ¼ sup aðsÞf ðuðsÞÞds ds þ

/p Pm2

ai 1  i¼1 t2½0;þ1Þ 1 þ t 0 s  Pm2 R nm2 1 R þ1 Z þ1  Z t /p s aðsÞf ðuðsÞÞds ds 1 i¼1 ai 0 1 6 sup /p aðsÞf ðuðsÞÞds ds þ Pm2 1  i¼1 ai t2½0;þ1Þ 1 þ t 0 s Z þ1  Pm2 Z þ1  1 i¼1 ai nm2 1 6 /p aðsÞf ðuðsÞÞds þ aðsÞf ðuðsÞÞds Pm2 /p 1  i¼1 ai 0 0 " !# Pm2 Z þ1  1 i¼1 ai nm2 6 MRk /p aðsÞds 1þ 6 Rk ¼ kuk: P 1  m2 0 i¼1 ai Thus Theorem 2.1 implies that iðT ; X2;k ; KÞ ¼ 1:

ð4:3Þ

Hence, since rk < Rk , for k 2 N , (4.2) and (4.3), it follows from additivity of the fixed-point index that iðT ; X2;k n X1;k ; KÞ ¼ 1 for k 2 N : Thus T has a fixed-point in X2;k n X1;k such that rk 6 kuk k 6 Rk : Since k 2 N was arbitrary, the proof is complete. h In order to use Theorem 2.2, let h1k < rk < hk and hk of Theorem 4.1, we define the nonnegative, increasing, continuous functional ck , bk , and ak by ck ðuÞ ¼ max uðtÞ ¼ uðrk Þ; 1 6t6r k hk

bk ðuÞ ¼ min uðtÞ ¼ uðrk Þ; rk 6t6hk

ak ðuÞ ¼ max uðtÞ ¼ uðhk Þ: 1 6t6h k hk

It is obvious that for each u 2 K; ck ðuÞ 6 bk ðuÞ 6 ak ðuÞ: In addition, by Lemma 4.2, for each u 2 K; ck ðuÞ ¼ uðrk Þ P

1 kuk: hk

kuk 6 hk ck ðuÞ

for all u 2 K:

Thus In the following, we denote by Pm2 ! Z þ1  nm2 i¼1 ai 1 qk ¼ /p aðsÞds hk þ ; Pm2 1  i¼1 ai 0

gk ¼

1 1 / hk p

Z

hk rk

 aðsÞds :

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Theorem 4.2. Suppose that conditions ðC 1 Þ, ðC 2 Þ hold. Let fhk g1 k¼1 be such that hk 2 ðtk ; tkþ1 Þ ðk ¼ 1; 2; . . .Þ. Let 1 1 fak g1 , fb g and fc g be such that k k k¼1 k¼1 k¼1 ck1 < ak <

bk bk < < ð1 þ hk Þck ; and qk bk < gk ck ; hk ðhk þ 1Þ hk

for k ¼ 2; 3; . . .

Furthermore, for each natural number k we assume that f satisfy: ðC 5 Þ f ðð1 þ tÞuÞ < /p ðqckk Þ, for all 0 6 uðtÞ 6 hk ck , ðC 6 Þ f ðð1 þ tÞuÞ > /p ðbgkk Þ, for all ðC 7 Þ f ðð1 þ tÞuÞ <

/p ðqakk Þ,

bk 1þhk

6 uðtÞ 6 hk bk ,

for all 0 6 uðtÞ 6 hk ak . 1

1

Then the boundary value problem (1.1) and (1.2) has three infinite families of solutions fu1k gk¼1 fu2k gk¼1 and fu3k g1 k¼1 satisfying 0 6 ak ðu1k Þ < ak < ak ðu2k Þ;

bk ðu2k Þ < bk < bk ðu3k Þ;

cðu3k Þ < ck ;

for k 2 N :

Proof. We define the completely continuous operator T by (3.1). So, it is easy to check that T : Kðck ; ck Þ ! K; for k 2 N : We now show that all the conditions of Theorem 2.2 are satisfied. To make use of property (i) of Theorem 2.2, we choose u 2 oKðck ; ck Þ. Then ck ðuÞ ¼ max 1 6t6rk uðtÞ ¼ uðrk Þ ¼ ck ; this implies that 0 6 uðtÞ 6 ck for hk

t 2 ½0; rk , i.e., 0 6 uðtÞ 1þt 6 ck : If we recall that kuk 6 hk ck ¼ hk ck : So, we have 06

uðtÞ 6 hk c k ; 1þt

0 6 t < þ1:

Then assumption ðC 5 Þ implies   ck f ðuÞ < /p ; 0 6 t < þ1: qk Therefore, ck ðTuÞ ¼ max ðTuÞðtÞ ¼ ðTuÞðrk Þ 1 6t6r k hk

 Pm2 R ni 1 R þ1 aðsÞf ðuðsÞÞds ds i¼1 ai 0 /p s aðsÞf ðuðsÞÞds ds þ ¼ Pm2 1  i¼1 ai 0 s  Pm2 R nm2 1 R þ1 Z þ1  Z hk /p 0 aðsÞf ðuðsÞÞds ds i¼1 ai 0 1 6 /p aðsÞf ðuðsÞÞds ds þ Pm2 1  i¼1 ai 0 0 Pm2 ! Z þ1  nm2 i¼1 ai 1 ¼ /p aðsÞf ðuðsÞÞds hk þ Pm2 1  i¼1 ai 0 Pm2 ! Z þ1  ck 1 nm2 i¼1 ai < /p aðsÞds hk þ ¼ ck : Pm2 qk 1  i¼1 ai 0 Z

rk

/1 p

Z

þ1



Hence, condition (i) is satisfied. Secondly, we show that (ii) of Theorem 2.2 is fulled. For this, we select u 2 oKðbk ; bk Þ. Then bk ðuÞ ¼ bk bk minrk 6t6hk uðtÞ ¼ uðrk Þ ¼ bk , this means uðtÞ P bk , for rk 6 t 6 hk . So, we have kuk P uðtÞ 1þt P 1þt P 1þhk , for rk 6 t 6 hk . Noticing that kuk 6 hk ck 6 hk bk ¼ hk bk , we have bk uðtÞ 6 hk bk 6 1 þ hk 1 þ t

for rk 6 t 6 hk :

S. Liang, J. Zhang / Applied Mathematics and Computation 201 (2008) 210–220

By ðC 6 Þ, we have   bk f ðuÞ > /p gk

219

for rk 6 t 6 hk :

bk ðTuÞ ¼ min ðTuÞðtÞ ¼ ðTuÞðrk Þ rk 6t6hk

Pm2 R ni 1 R hk  a / aðsÞf ðuðsÞÞds ds i p i¼1 0 s ¼ /1 aðsÞf ðuðsÞÞds ds þ Pm2 p 1  i¼1 ai 0 s Z hk  Z hk  Z hk  Z rk Z rk bk bk 1 1 1 > /p aðsÞf ðuðsÞÞds ds P /p aðsÞds ds ¼ rk /p aðsÞds P bk : gk 0 gk 0 s rk rk Z

Z

rk

þ1

Hence, condition (ii) is satisfied. Finally, we verify that (iii) of Theorem 2.2 is also satisfied. We note that uðtÞ  a2k ; 0 6 t < þ1 is a member of Kðak ; ak Þ and ak ðuÞ ¼ a2k < ak . So Kðak ; ak Þ 6¼ ;. Now let u 2 oKðak ; ak Þ. Then ak ðuÞ ¼ max 1 6t6hk uðtÞ ¼ hk

ak uðhk Þ ¼ ak . This implies that 0 6 uðtÞ 6 ak , for 0 6 t 6 hk , we have 0 6 uðtÞ 1þt 6 1þt < ak , for 0 6 t 6 hk . Together with kuk 6 hk ck 6 hk ak ¼ hk ak . Then we get

06

uðtÞ 6 hk ak ; 1þt

By ðC 7 Þ we have   ak f ðuÞ < /p ; qk As before, we get

0 6 t < þ1:

0 6 t < þ1:

ak ðTuÞ ¼ max ðTuÞðtÞ ¼ ðTuÞðhk Þ 1 6t6h k hk

 Pm2 R ni 1 R þ1  aðsÞf ðuðsÞÞds ds i¼1 ai 0 /p s aðsÞf ðuðsÞÞds ds þ ¼ P 1  m2 0 s i¼1 ai  Pm2 R nm2 1 R þ1 Z þ1  Z hk /p 0 aðsÞf ðuðsÞÞds ds i¼1 ai 0 1 6 /p aðsÞf ðuðsÞÞds ds þ Pm2 1  i¼1 ai 0 0 Pm2 ! Z þ1  nm2 i¼1 ai 1 6 /p aðsÞf ðuðsÞÞds hk þ Pm2 1  i¼1 ai 0 Pm2 ! Z þ1  ak n i¼1 ai < /1 aðsÞds hk þ m2Pm2 ¼ ak : p qk 1  i¼1 ai 0 Z

hk

/1 p

Z

þ1

Thus (iii) of Theorem 2.2 is satisfied. Since all hypotheses of Theorem 2.2 are satisfied, the assertion follows. h Remark. If we add the condition of aðtÞf ðuðtÞÞ 6 0; t 2 ½0; þ1Þ, to Theorem 4.2, we can get three infinite 1 1 1 families of positive solutions fu1k gk¼1 fu2k gk¼1 and fu3k gk¼1 satisfying 0 < ak ðu1k Þ < ak < ak ðu2k Þ;

bk ðu2k Þ < bk < bk ðu3k Þ;

cðu3k Þ < ck

for n 2 N :

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