Positive solutions for singular semipositone boundary value problems ...
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Author's personal copy Applied Mathematics and Computation 227 (2014) 256–273
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Positive solutions for singular semipositone boundary value problems on infinite intervals Ying Wang a,b, Lishan Liu a,⇑, Yonghong Wu c a
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China School of Science, Linyi University, Linyi 276000, Shandong, People’s Republic of China c Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia b
a r t i c l e
i n f o
a b s t r a c t By using the fixed point theory on a cone with a special norm and space, we discuss the existence of positive solutions for a class of semipositone boundary value problems on infinite intervals. The work improves many known results including singular and non-singular cases. Ó 2013 Elsevier Inc. All rights reserved.
Keywords: Positive solutions Cone Semipositone boundary value problems Infinite intervals
1. Introduction The main purpose of this paper is to study the existence of positive solutions for the following nonlinear singular and semipositone boundary value problem (BVP) on infinite intervals:
8 ðpðtÞx0 ðtÞÞ0 þ kf ðt; xðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; t!0þ
i
i
i¼1 > > > m 2 X > > > > a2 lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ; : t!þ1 t!þ1
ð1:1Þ
i¼1
where k > 0 is a parameter, a1 ; a2 ; ci ; di P 0; b1 ; b2 > 0; gi 2 ð0; þ1Þ; ð1; 2; . . . ; m 2Þ are given constants, f : ð0; þ1Þ ð0; þ1Þ ! ð1; þ1Þ is a continuous function and f ðt; uÞ may be singular at t ¼ 0 and u ¼ 0, p 2 C½0; þ1Þ \ C 1 ð0; þ1Þ with R1 1 Rs p > 0 on ð0; þ1Þ; 0 pðsÞ ds < þ1, q ¼ a2 b1 þ a1 b2 þ a1 a2 Bð0; 1Þ > 0 in which Bðt; sÞ ¼ t pð1v Þ dv . Boundary value problems on infinite intervals arises from the study of many real world problems such as solution of nonlinear elliptic equations and modeling of gas pressure in a semi-infinite porous medium. A great deal of work has been done in these areas such as those in [1–22] and the references therein. Over the last couple of decades, a great deal of results have been developed for differential and integral boundary value problems. O’Regan et al. studied in [23] the BVP
(
x00 ðtÞÞ þ uðtÞf ðt; xðtÞÞ ¼ 0; xðaÞ ¼ 0;
t 2 ða; þ1Þ;
lim x0 ðtÞ ¼ 0;
t!þ1
where f : ½a; þ1Þ ð0; þ1Þ ! ½0; þ1Þ is a continuous function and u : ða; þ1Þ ! ½0; þ1Þ is continuous. The existence of multiple unbounded positive solutions is discussed using the theory of fixed point index. By using the fixed point theorem and the monotone iterative technique, Zhang [24] studied the problem ⇑ Corresponding author. E-mail addresses: [email protected] (Y. Wang), [email protected] (L. Liu), [email protected] (Y. Wu). 0096-3003/$ - see front matter Ó 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.11.009
Author's personal copy Y. Wang et al. / Applied Mathematics and Computation 227 (2014) 256–273
257
8 00 x ðtÞÞ þ qðtÞf ðt; xðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > < m2 X > ci xðgi Þ; lim x0 ðtÞ ¼ x1 P 0; : xð0Þ ¼ t!þ1
i¼1
Pm2 where ci P 0, 0 < g1 < g2 < < gm2 < þ1, i¼1 ci < 1; f : ½0; þ1Þ ½0; þ1Þ ! ½0; þ1Þ is a continuous function and q : ð0; þ1Þ ! ½0; þ1Þ is a Lebesgue integrable function. Liu et al. in [25] established the existence of positive solutions for the following equation on infinite intervals by applying the fixed point theorem of cone map
8 ðpðtÞx0 ðtÞÞ0 þ mðtÞf ðt; xðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > < a1 xð0Þ b1 limþ pðtÞx0 ðtÞ ¼ 0; t!0 > > : a2 lim xðtÞ þ b lim pðtÞx0 ðtÞ ¼ 0; 2 t!þ1
t!þ1
in which f : ½0; þ1Þ ½0; þ1Þ ! ½0; þ1Þ is a continuous function, m : ð0; þ1Þ ! ½0; þ1Þ is a Lebesgue integrable function and may be singular at t ¼ 0. Also, by the use of the Krasnosel-skii fixed point theorem, upper and lower solutions, Schauder point theorem and inequality technique, Xing et al. in [26], Lian et al. in [27,28], Li and Zhao in [29] studied many equations with infinite intervals. However, all of the above studies are limited to the cases in which the nonlinear term is positive. Inspired by the work of the above papers and many known results, in this paper, we study the existence of positive solutions to BVP (1.1), where x 2 C½0; þ1Þ is said to be a positive solution of BVP (1.1) if and only if x satisfies (1.1) and xðtÞ > 0 for any t 2 ½0; þ1Þ. By using the fixed point theory on a cone, some new existence results are obtained for the case where the nonlinearity is allowed to be sign changing and has singularity. We should address here that our work presented in this paper has various new features. Firstly, our study is on singular nonlinear differential boundary value problems, that is, f ðt; uÞ is allowed to be singular at t ¼ 0 and u ¼ 0, which leads to many difficulties in analysis. Secondly, the techniques used in this paper are the approximation method; and a special cone in a special space is established to overcome the difficulties caused by singularity and infinite interval. Also we find another substitute function to solve the problem associated with the semipositive property of the nonlinear function. Thirdly, we discuss the boundary value problem with multi-point boundary conditions, that is, BVP (1.1) includes two-point and three-point boundary value problems as special cases. To our knowledge, the theory of Sturm–Liouville multi-point boundary value problems on infinite interval is yet to be developed. Our model improves and generalizes the results of previous papers to some degree. The plan of the paper is as follows. In Section 2, we present the preliminaries and necessary lemmas that are to be used to prove our main results. The main results are given in Section 3, including results for a completely continuous operator, and the conditions for the existence of positive solutions for the BVP (1.1). In Section 4, an example is given to demonstrate the application of our theoretical results. 2. Preliminaries and lemmas For convenience, we let
aðtÞ ¼ b1 þ a1 Bð0; tÞ; bðtÞ ¼ b2 þ a2 Bðt; 1Þ; að1Þ ¼ lim aðtÞ ¼ b1 þ a1 Bð0; 1Þ < þ1; að0Þ ¼ lim aðtÞ ¼ b1 ; t!þ1
t!0
bð0Þ ¼ lim bðtÞ ¼ b2 þ a2 Bð0; 1Þ < þ1;
bð1Þ ¼ lim bðtÞ ¼ b2 ; t!þ1
t!0
Pm2 q Pm2 c bðg Þ i i¼1 i i¼1 ci aðgi Þ D ¼ Pm2 : P m2 d bð g Þ q d að g Þ i i i¼1 i i¼1 i It is obvious that aðtÞ is increasing and bðtÞ is decreasing on ½0; þ1Þ. Define
Gðt; sÞ ¼
1
aðsÞbðtÞ; 0 6 s 6 t < þ1;
ð2:1Þ
q aðtÞbðsÞ; 0 6 t 6 s < þ1:
Denote sðtÞ ¼ aðtÞbðtÞ, then for any 0 6 t; s < þ1, we get
0 6 Gðt; sÞ 6 Gðs; sÞ 6 GðsÞ ¼ lim Gðt; sÞ ¼ t!þ1
bð0ÞaðsÞ
q
b2 aðsÞ
q
;
0 6 Gðt; sÞ 6
sðtÞ ; q
6 Gðs; sÞ < þ1:
1 Lemma 2.1. Suppose h ¼ að1Þbð0Þ , then Gðt; sÞ P hsðtÞGðs; sÞ; 0 6 t; s < þ1.
ð2:2Þ
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Proof. From (2.2) and the property of aðtÞ; bðtÞ, for 0 st; s < þ1, we have
8 8 aðtÞbðtÞ bðtÞ Gðt; sÞ < bðsÞ ; s 6 t; < aðtÞbðsÞ ; s 6 t; ¼ ¼ Gðs; sÞ : aðtÞ ; t 6 s; : aðtÞbðtÞ ; t 6 s; aðsÞ aðsÞbðtÞ
P
sðtÞ : að1Þbð0Þ
Therefore, Gðt; sÞ P hsðtÞGðs; sÞ; 0 6 t; s < þ1. h In this paper, we always assume that the following conditions hold. P Pm2 ðH1 Þ : D > 0; q m2 i¼1 ci bðgi Þ > 0, q i¼1 di aðgi Þ > 0. ðH2 Þ : f : ð0; þ1Þ ð0; þ1Þ ! ð1; þ1Þ is a continuous function and
wðtÞ 6 f ðt; uÞ 6 /ðtÞðgðuÞ þ hðuÞÞ;
ðt; uÞ 2 ð0; þ1Þ ð0; þ1Þ;
where w; / : ð0; þ1Þ ! ½0; þ1Þ is continuous and singular at t ¼ 0; wðtÞ; /ðtÞ X 0 on ½0; þ1Þ, g : ð0; þ1Þ ! ½0; þ1Þ is continuous and nonincreasing, h : ½0; þ1Þ ! ½0; þ1Þ is continuous, g and h are bounded in any bounded set of ½0; þ1Þ. R1 R1 ðH3 Þ 0 wðsÞds < þ1; 0 Gðs; sÞðwðsÞ þ /ðsÞÞds < þ1. Let
P P m2 q m2 1 i¼1 ci sðgi Þ i¼1 ci bðgi Þ A ¼ Pm2 ; P m2 D di sðg Þ di bðg Þ i
i¼1
i
i¼1
P P m2 q m2 1 i¼1 di sðgi Þ i¼1 di aðgi Þ B ¼ Pm2 : P m2 D c sð g Þ c aðg Þ i¼1
i
i
i
i¼1
i
Choose a constant d, such that d P að1Þbð0Þ þ Aað1Þ þ Bbð0Þ, and denote
fðtÞ ¼
sðtÞ þ AaðtÞ þ BbðtÞ d
ð2:3Þ
;
P 1; 0 < fðtÞ 6 1. then dh Lemma 2.2. Suppose ðH1 Þ holds,
R1
1 0 pðsÞ ds
< þ1 and q > 0, then the BVP
8 ðpðtÞx0 ðtÞÞ0 þ wðtÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; i
t!0þ
i
i¼1 > > > m 2 X > > 0 > > a lim xðtÞ þ b lim pðtÞx ðtÞ ¼ di xðgi Þ; 2 2 : t!þ1 t!þ1 i¼1
has a unique solution for any w 2 Lð0; þ1Þ. Moreover, this unique solution can be expressed in the form
xðtÞ ¼
Z
1
Gðt; sÞwðsÞds þ AðwÞaðtÞ þ BðwÞbðtÞ;
ð2:4Þ
0
where Gðt; sÞ is defined as (2.1) and
Pm2 R 1 P q m2 1 i¼1 ci 0 Gðgi ; sÞwðsÞds i¼1 ci bðgi Þ AðwÞ ¼ Pm2 R 1 ; P m2 D di Gðg ; sÞwðsÞds di bðg Þ i¼1
0
i
i
i¼1
Pm2 R 1 P q m2 i¼1 di 0 Gðgi ; sÞwðsÞds i¼1 di aðgi Þ 1 BðwÞ ¼ : D Pm2 R 1 P m2 c Gðg ; sÞwðsÞds c aðg Þ i¼1
i
0
i
i¼1
i
i
The proof is similar to [30], so we omit it.
Lemma 2.3. The solution defined by (2.4) satisfies xðtÞ 6 dfðtÞ q
R1 0
wðsÞds.
Proof. Since xðtÞ is the unique solution of (2.4). By (2.2)–(2.4), we have
xðtÞ 6
Z 0
1
sðtÞwðsÞ ds þ AaðtÞ q
Z 0
1
wðsÞ
q
ds þ BbðtÞ
Z 0
1
wðsÞ
q
ds 6 ðsðtÞ þ AaðtÞ þ BbðtÞÞ
Z 0
1
wðsÞ
q
ds ¼
dfðtÞ
q
Z
1
wðsÞds: 0
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In this paper, the following space X will be used in the study of BVP(1.1).
X¼
x 2 C½0; þ1Þ : lim xðtÞ exists : t!þ1
Clearly ðX; k kÞ is a Banach space with the supremum norm kxk ¼ supt2½0;þ1Þ jxðtÞj, see [31]. Let
K¼
fðtÞ x 2 X : xðtÞ P kxk; t 2 ½0; þ1Þ : 2
It is easy to see that K is a cone in X. Next we consider the following singular nonlinear boundary value problem:
8 ðpðtÞx0 ðtÞÞ0 þ kðf ðt; ½xðtÞ kxðtÞ Þ þ wðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; i
t!0þ
i
ð2:5Þ
i¼1 > > > m2 X > > > > lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ; : a2 t!þ1 t!þ1 i¼1
where k > 0; xðtÞ is defined in Lemma 2.2, ½zðtÞ ¼ maxfzðtÞ; 0g. h Lemma 2.4. If x is a solution of BVP (2.5) with xðtÞ > kxðtÞ for any t 2 ½0; þ1Þ, then xðtÞ kxðtÞ is a positive solution of BVP (1.1). Proof. In fact, if x is a positive solution of BVP (2.5) such that xðtÞ > kxðtÞ for any t 2 ½0; þ1Þ, then from BVP (2.5) and the definition of ½zðtÞ , we have
8 ðpðtÞx0 ðtÞÞ0 þ kðf ðt; xðtÞ kxðtÞÞ þ wðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m 2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; i
t!0þ
i
ð2:6Þ
i¼1 > > > m2 X > > > > lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ: : a2 t!þ1 t!þ1 i¼1
Let uðtÞ ¼ xðtÞ kxðtÞ; t 2 ½0; þ1Þ, then ðpðtÞx0 ðtÞÞ0 ¼ ðpðtÞu0 ðtÞÞ0 þ kðpðtÞx0 ðtÞÞ0 . Thus, (2.6) becomes
8 ðpðtÞu0 ðtÞÞ0 þ kf ðt; uðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 uð0Þ b1 lim pðtÞu0 ðtÞ ¼ ci uðgi Þ; t!0þ
i¼1 > > > m2 X > > > > lim uðtÞ þ b2 lim pðtÞu0 ðtÞ ¼ di uðgi Þ: : a2 t!þ1 t!þ1 i¼1
Then uðtÞ ¼ xðtÞ kxðtÞ is a positive solution of BVP (1.1). h To overcome singularity, we consider the following approximate problem:
8 ðpðtÞx0 ðtÞÞ0 þ k f t; ½xðtÞ kxðtÞ þ 1n þ wðtÞ ¼ 0; t 2 ð0; þ1Þ; > > > > > m2 X > > < a1 xð0Þ b lim pðtÞx0 ðtÞ ¼ c xðg Þ; 1
i
t!0þ
i
ð2:7Þ
i¼1 > > > m2 > X > > > a2 lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ; : t!þ1 t!þ1 i¼1
in which n is a positive integer. Under the assumptions ðH1 Þ ðH3 Þ, for any n 2 N where N is a natural number set, we define a nonlinear integral operator T n : K ! X by
ðT n xÞðtÞ ¼ k
Z 0
1
1 þ wðsÞ ds þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ; Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n
where
Pm2 R 1 P 1 q m2 1 i¼1 ci 0 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ Aðkðfn þ wÞÞ ¼ Pm2 R 1 ; P m2 1 D di Gðg ; sÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ds di bðg Þ i¼1
0
i
n
i¼1
i
t 2 ½0; þ1Þ;
ð2:8Þ
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Pm2 R 1 P 1 q m2 1 i¼1 di 0 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ Bðkðfn þ wÞÞ ¼ Pm2 R 1 : P m2 D c Gðg ; sÞ f s; ½xðsÞ kxðsÞ þ 1 þ wðsÞ ds c aðg Þ i¼1
i
0
i
n
i¼1
i
i
Obviously, the existence of solutions to BVP (2.7) is equivalent to the existence of solutions in K for operator equation T n x ¼ x defined by (2.8). In this paper, the proof of the main theorem is based on the fixed point theory in cone. We list the following lemmas which are needed in our study. Lemma 2.5 [31]. Let X be defined by (2.1) and M X. Then M is relatively compact in X if the following conditions hold: (1) M is uniformly bounded in X; (2) the functions from M are equicontinuous on any compact interval of ½0; þ1Þ; (3) the functions from M are equiconvergent, that is, for any given e > 0, there exists a T ¼ TðeÞ > 0 such that jxðtÞ xðþ1Þj < e, for any t > T; x 2 M. Lemma 2.6 [32]. Let P be a positive cone in a real Banach space E. Denote Pr ¼ fx 2 P : kxk < rg, P r;R ¼ fx 2 P : r 6 kxk 6 Rg; 0 < r < R < þ1. Let A : P r;R ! Pbe a completely continuous operator. If the following conditions are satisfied: (1) kAxk 6 kxk; 8x 2 @PR . (2) there exists a x0 2 @P1 , such that x – Ax þ mx0 ; 8x 2 @Pr ; m > 0. Then A has fixed points in Pr;R . Remark 2.1. If (1) and (2) are satisfied for x 2 @Pr and x 2 @P R respectively. Then Lemma 2.6 is still true. Lemma 2.7 [33]. Let P be a positive cone in a Banach space E; X1 and X2 are bounded open sets in E; h 2 X1 ; X1 X2 ; A : P \ X2 n X1 ! P is a completely continuous operator. If the following conditions are satisfied: kAxk 6 kxk; 8x 2 P \ @ X1 , kAxk P kxk; 8x 2 P \ @ X2 , or kAxk P kxk; 8x 2 P \ @ X1 , kAxk 6 kxk; 8x 2 P \ @ X2 , then A has at least one fixed point in P \ ðX2 n X1 Þ. 3. Main results Lemma 3.1. Assume that ðH1 Þ–ðH3 Þ hold. Then T n : K ! K is a completely continuous operator for any fixed n 2 N. Proof. First we show that T n : K ! X is well defined and T n ðKÞ # K. For x 2 K, there exists r > 0 such that jxðtÞj 6 r, for t 2 ½0; þ1Þ, also j½xðtÞ kxðtÞ j 6 jxðtÞj 6 r; t 2 ½0; þ1Þ. From ðH2 Þ and the definition of g and h, we have
1 Sr;n :¼ sup gðuÞ þ hðuÞ : 6 u 6 r þ 1 < þ1: n Thus, by ðH2 Þ and ðH3 Þ, for any t 2 ½0; þ1Þ, we know
1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Z 1 1 Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 6k n 0 1 þ wðsÞ ds þh ½xðsÞ kxðsÞ þ n Z 1 Gðs; sÞð/ðsÞSr;n þ wðsÞÞds 6k 0 Z 1 6 kðSr;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds Z
1
k
0
< þ1: By (2.2), for i ¼ 1; 2; . . . ; m 2, we also have
Z k 0
1
Z 1 1 þ wðsÞ ds 6 kðSr;n þ 1Þ Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1: n 0
ð3:1Þ
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Therefore,
Aðkðfn þ wÞÞaðtÞ P R1 P m2 1 q m2 aðtÞ i¼1 ci k 0 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ ¼ Pm2 R 1 P m2 D i¼1 di k Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ 1n þ wðsÞ ds i¼1 di bðgi Þ 0 Pm2 R1 Pm2 i¼1 ci kðSr;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds q i¼1 ci bðgi Þ 0 að1Þ m2 6 R X Pm2 D di kðSr;n þ 1Þ 01 Gðs; sÞð/ðsÞ þ wðsÞÞds d bð g Þ i i i¼1 i¼1 P P Z m2 1 að1Þ ci q m2 i¼1 ci bðgi Þ ¼ Gðs; sÞð/ðsÞ þ wðsÞÞds kðSr;n þ 1Þ Pi¼1 Pm2 m2 D i¼1 di 0 i¼1 di bðgi Þ Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; ¼Aað1ÞkðSr;n þ 1Þ
ð3:2Þ
0
where
P m2 ci 1 A ¼ Pi¼1 D m2 di i¼1
q
Pm2
c g g
i¼1 i bð i Þ : Pm2 i¼1 di bð i Þ
In the same way, we get
Bðkðfn þ wÞÞbðtÞ 6 Bbð0ÞkðSr;n þ 1Þ
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1;
ð3:3Þ
0
where
P m2 di 1 B ¼ Pi¼1 D m2 c
i
i¼1
q
Pm2
g
i¼1 di að i Þ : Pm2 i¼1 i að i Þ
c g
Hence, by (3.1)–(3.3), we can see that
1 þ wðsÞ ds þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; t 2 ½0; þ1Þ: 6 ð1 þ Aað1Þ þ Bbð0ÞÞkðSr;n þ 1Þ
ðT n xÞðtÞ ¼ k
Z
1
0
Then, T n x is well defined for any x 2 K. On the other hand, for any t; t j 2 ½0; þ1Þ; tj ! t, by the continuity of Gðt; sÞ, we get
1 1 þ wðsÞ ! Gðt; sÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ; s 2 ½0; þ1Þ; j ! þ1: Gðtj ; sÞ f s; ½xðsÞ kxðsÞ þ n n
ð3:4Þ
By (2.2), we know
Z 1 1 þ wðsÞ ds 6 ðSr;n þ 1Þ Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; n Z0 1 Z 01 1 þ wðsÞ ds 6 ðSr;n þ 1Þ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1: n 0 0
Z
1
ð3:5Þ
So, by ðH3 Þ, (3.4) and (3.5) and the Lebesgue dominated convergence theorem, we have
1 þ wðsÞ ds lim Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ t j !t 0 n Z 1 1 þ wðsÞ ds lim Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ ¼ tj !t n Z0 1 1 þ wðsÞ ds: ¼ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Z
1
Consequently, together with the continuity of aðtÞ and bðtÞ, we have
jT n xðt j Þ T n xðtÞj Z 1 1 þ wðsÞ ds Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ ¼ k n 0
þAðkðfn þ wÞÞaðtj Þ þ Bðkðfn þ wÞÞbðt j Þ Z 1 1 þ wðsÞ ds k Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Aðkðfn þ wÞÞaðtÞ Bðkðfn þ wÞÞbðtÞj Z 1 1 ðGðt j ; sÞ Gðt; sÞÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ds 6 k n 0 þ Aðkðfn þ wÞÞjaðt j Þ aðtÞj þ Bðkðfn þ wÞÞjbðt j Þ bðtÞj ! 0; as j ! þ1:
ð3:6Þ
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Therefore, T n x 2 C½0; þ1Þ. In what follows, for any tj 2 ½0; þ1Þ; t j ! þ1, by (2.2), we have
1 1 þ wðsÞ ! GðsÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ; s 2 ½0; þ1Þ; j ! þ1: Gðtj ; sÞ f s; ½xðsÞ kxðsÞ þ n n Then by the Lebesgue dominated convergence theorem and the property of aðtÞ, bðtÞ, we also have
lim ðT n xÞðt j Þ Z 1 1 þ wðsÞ ds GðsÞ f s; ½xðsÞ kxðsÞ þ ¼k n 0
j!þ1
þ Aðkðfn þ wÞÞað1Þ þ Bðkðfn þ wÞÞb2 < þ1: So, for any x 2 K, we get T n x 2 X, which implies that T n maps K to X. In the following, we will prove T n ðKÞ # K. For any x 2 K, from the definition of k k and (2.2), we have
kT n xk 6 k
Z
1
0
1 þ wðsÞ ds þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ: Gðs; sÞ f s; ½xðsÞ kxðsÞ þ n
ð3:7Þ
By Lemma 2.1 and (2.3), we get
ðT n xÞðtÞ
1 P khsðtÞ þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z khsðtÞ 1 1 þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ P 2 n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 1 þ wðsÞ ds khsðtÞ Gðs; sÞ f s; ½xðsÞ kxðsÞ þ ¼ 2 n 0 þAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 1 þ ðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z 1 kh 1 þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ P ðsðtÞ þ AaðtÞ þ BbðtÞÞ 2 n 0 1 þ ðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z khdfðtÞ 1 1 þ wðsÞ ds ¼ Gðs; sÞ f s; ½xðsÞ kxðsÞ þ 2 n 0 1 þ ðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z kfðtÞ 1 1 þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ P 2 n 0 1 þ fðtÞðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z 1 fðtÞ 1 þ wðsÞ ds k Gðs; sÞ f s; ½xðsÞ kxðsÞ þ ¼ 2 n 0 þðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞÞ: Z
1
ð3:8Þ
Combining (3.7) with (3.8), we have ðT n xÞðtÞ P fðtÞ kT n xk, for any t 2 ½0; þ1Þ. Therefore, T n ðKÞ # K. 2 Next, for any positive integers n; k 2 N, we define an operator T n;k : K ! X by
ðT n;k xÞðtÞ ¼ k
Z 1 k
1
1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n
þ Ak ðkðfn þ wÞÞaðtÞ þ Bk ðkðfn þ wÞÞbðtÞ; t 2 ½0; þ1Þ; where
Ak ðkðfn þ wÞÞ Pm2 R 1 P 1 q m2 i¼1 ci k 1 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ 1 k ; ¼ Pm2 R 1 Pm2 1 D d k Gð g ; sÞ f s; ½xðsÞ k x ðsÞ þ d bð g Þ þ wðsÞ ds 1 i i i i i¼1 i¼1 n k
ð3:9Þ
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Bk ðkðfn þ wÞÞ P R1 P m2 1 q m2 i¼1 di k 1 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ k 1 : ¼ Pm2 D Pm2 R 1 1 i¼1 ci k 1 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci aðgi Þ k
As in the above discussion, we can prove that T n;k : K ! X is well defined and T n;k ðKÞ # K. In what follows, we will prove that T n;k : K ! K is completely continuous, for each k P 1. Firstly, we show that T n;k : K ! K is continuous. Let xt ; x 2 K are such that kxt xk ! 0 as t ! þ1. By (3.9) and ðH3 Þ, we know
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xt ðsÞ kxðsÞ þ k 1 n k Z 1 1 k þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 1 1 6k þ f s; ½xðsÞ kxðsÞ þ þ 2wðsÞ ds Gðs; sÞ f s; ½xt ðsÞ kxðsÞ þ 1 n n Zk 1 1 1 6k þ h ½xt ðsÞ kxðsÞ þ Gðs; sÞ /ðsÞ g ½xt ðsÞ kxðsÞ þ 1 n n k 1 1 þ/ðsÞ g ½xðsÞ kxðsÞ þ þ h ½xðsÞ kxðsÞ þ þ 2wðsÞ ds n n Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; 2kðSr0 ;n þ 1Þ
ð3:10Þ
0
where Sr0 ;n :¼ supfgðuÞ þ hðuÞ :
1 n
6 u 6 r0 þ 1g < þ1 (by ðH2 Þ), r0 is a real number such that r 0 P maxt2N fkxk; kxt kg. Denote
Pm2 R 1 P 1 q m2 i¼1 ci k 1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ 1 k ; Ak;t ðkðfn þ wÞÞ ¼ Pm2 R 1 Pm2 D di k 1 Gðg ; sÞ f s; ½xt ðsÞ kxðsÞ þ 1 þ wðsÞ ds di bðg Þ i¼1
i
k
i
i¼1
n
Pm2 R 1 P 1 q m2 i¼1 di k 1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ 1 k : Bk;t ðkðfn þ wÞÞ ¼ Pm2 R 1 Pm2 D c k 1 Gðg ; sÞ f s; ½xt ðsÞ kxðsÞ þ 1 þ wðsÞ ds c aðg Þ i¼1
i
i
k
i¼1
n
i
i
Though calculation, we obtain
Z P P m2 1 q m2 að1Þ 1 i¼1 ci i¼1 ci bðgi Þ jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjaðtÞ 6 Gðs; sÞ f s; ½xt ðsÞ kxðsÞ þ k Pm2 1 n D Pm2 di i¼1 i¼1 di bðgi Þ k 1 þ f s; ½xðsÞ kxðsÞ þ þ 2wðsÞ ds n Z 1 1 1 þ h ½xt ðsÞ kxðsÞ þ Gðs; sÞ /ðsÞ g ½xt ðsÞ kxðsÞ þ 6 Ak 1 n n k 1 1 þ h ½xðsÞ kxðsÞ þ þ wðsÞ ds þ wðsÞ þ /ðsÞ g ½xðsÞ kxðsÞ þ n n Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1: ð3:11Þ 6 2Aað1ÞkðSr0 ;n þ 1Þ 0
In the same way, we get
jBk;t ðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbðtÞ 6 2Bbð0ÞkðSr0 ;n þ 1Þ
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1:
ð3:12Þ
0
From (3.10), (3.12), (3.13), for any
e > 0, by ðH3 Þ, there exists a sufficiently large A0 ðA0 > 1=kÞ, such that
Z o max 1; að1ÞA; bð0ÞB kðSr0 ;n þ 1Þ n
1
A0
Gðs; sÞð/ðsÞ þ wðsÞÞds
0, there exists a d > 0 such that for any s 2 ½1=k; A0 and u; t 2 ½0; r 0 , when ju tj ¼ jðu þ 1nÞ ðt þ 1nÞj < d, we have
!1 n o Z A0 f s; u þ 1 f s; t þ 1 < e max 1; að1ÞA; bð0ÞB k Gðs; sÞds : 1 n n 6 k
ð3:14Þ
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From kxt xk ! 0 ðn ! þ1Þ and the definition of the norm k k in the space X, for the above d > 0, there exists a sufficiently large nature number V 0 such that when t > V 0 , for all s 2 ½1=k; A0 , we have
½xt ðsÞ kxðsÞ þ 1 ½xðsÞ kxðsÞ þ 1 n n jxt ðsÞ kxðsÞj þ xt ðsÞ kxðsÞ jxðsÞ kxðsÞj þ xðsÞ kxðsÞ 6 2 2 jxt ðsÞ kxðsÞj jxðsÞ kxðsÞj xt ðsÞ xðsÞ ¼ þ 2 2
ð3:15Þ
6 jxt ðsÞ xðsÞj 6 kxt xk < d: Hence, by (3.14)–(3.16), when
t > V 0 , we have the following inequality
jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjaðtÞ 6 jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjað1Þ Pm2 Pm2 að1Þ F 1 q i¼1 ci bðgi Þ H1 q i¼1 ci bðgi Þ ¼ Pm2 Pm2 H2 D F 2 i¼1 di bðgi Þ i¼1 di bðgi Þ Z A0 1 1 6 að1ÞAk f s; ½x kxðsÞ þ Gðs; sÞf s; ½xt ðsÞ kxðsÞ þ ds 1 n n k Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds þ 2að1ÞAðSr0 ;n þ 1Þk 6
e 3
ð3:16Þ
A0
;
where
!
1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ þ wðsÞ ds; 1 n A0 i¼1 k ! Z Z m2 A0 1 X 1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ þ wðsÞ ds; F 2 ¼ di k þ 1 n A0 i¼1 k ! Z A0 Z 1 m2 X 1 H1 ¼ þ wðsÞ ds; ci k þ Gðgi ; sÞ f s; ½x kxðsÞ þ 1 n A0 i¼1 k ! Z Z m2 A0 1 X 1 H2 ¼ di k þ wðsÞ ds: þ Gðgi ; sÞ f s; ½x kxðsÞ þ 1 n A0 i¼1 k F1 ¼
m 2 X
ci k
Z
A0
þ
Z
1
Using the same method as (3.17), when
t > V 0 , we get
jBk;t ðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbðtÞ 6
e 3
ð3:17Þ
:
Then, by (3.17) and (3.18) and the above discussion, when
t > V 0 , we obtain
jðT n;k xt ÞðtÞ ðT n;k xÞðtÞj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xt ðsÞ kxðsÞ þ ¼ k 1 n k
þAk;t ðkðfn þ wÞÞaðtÞ þ Bk;t ðkðfn þ wÞÞbðtÞ Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k Ak ðkðfn þ wÞÞaðtÞ Bk ðkðfn þ wÞÞbðtÞj Z A0 1 1 f s; ½xðsÞ kxðsÞ þ 6k Gðs; sÞf s; ½xt ðsÞ kxðsÞ þ ds 1 n n k þ jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjað1Þ þ jBk;t ðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbð0Þ Z 1 1 1 þ f s; ½xðsÞ kxðsÞ þ þ 2wðsÞ ds Gðs; sÞ f s; ½xt ðsÞ kxðsÞ þ þk n n A0 Z 1 5e 6 þ 2kðSr0 ;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < e: 6 0 This implies that the operator T n;k : K ! K is continuous for any natural numbers n; k.
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It what follows, we need to prove that T n;k : K ! K is a compact operator for natural numbers n; k. Let M be any bounded subset of K. Then there exists a constant R > 0 such that kxk 6 R for any x 2 M. By (3.9), ðH2 Þ and ðH3 Þ, for any x 2 M, we have
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k Z 1 1 6k Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 1 n k 1 þ wðsÞ ds þh ½xðsÞ kxðsÞ þ n Z 1 Gðs; sÞð/ðsÞSR;n þ wðsÞÞds 6k 1 k
6 kðSR;n þ 1Þ
Z
ð3:18Þ
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1;
0
where SR;n :¼ supfgðuÞ þ hðuÞ :
1 n
6 u 6 R þ 1g. By the similar proof as (3.2) and (3.3), for any t 2 ½0; þ1Þ, we have
Z 1 Ak ðkðfn þ wÞÞaðtÞ 6 Aað1ÞkðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; Z 01 Bk ðkðfn þ wÞÞbðtÞ 6 Bbð0ÞkðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1:
ð3:19Þ
0
Therefore, from (3.18) and (3.19), T n;k M is bounded in K. Given a > 0, for any x 2 M and t; t0 2 ½0; a, by (3.9), we get
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k
1 k þ wðsÞ ds Gðt ; sÞ f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 1 þ wðsÞ ds 6k jGðt; sÞ þ Gðt 0 ; sÞj f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 2kðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1 Z
1
0
1 k
and so, for any
e0 > 0, we can find a sufficiently large H0 ðH0 > 1kÞ such that
kðSR;n þ 1Þ
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds
0 such that for any
!1
H0
ð/ðsÞ þ wðsÞÞds
:
1 k
Therefore, we get
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k
1 k þ wðsÞ ds Gðt ; sÞ f s; ½xðsÞ kxðsÞ þ 1 n k Z H0 1 0 6k þ wðsÞ ds jGðt; sÞ Gðt ; sÞj f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 1 þ wðsÞ ds jGðt; sÞ Gðt 0 ; sÞj f s; ½xðsÞ kxðsÞ þ þk n H0 Z 1 0 0 e e Gðs; sÞð/ðsÞ þ wðsÞÞds < : 6 þ 2kðSR;n þ 1Þ 1 6 3 k Z
1
0
ð3:20Þ
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Also by the uniformly continuity of aðtÞ; bðtÞ on ½0; a, for the above e0 > 0, there exists d00 > 0 such that for any t; t 0 2 ½0; a and jt t0 j < d00 , we have 0
jaðtÞ aðt Þj < 0
jbðtÞ bðt Þj
0, let d0 ¼ minfd0 ; d00 g, then for any t; t0 2 ½0; a with jt t0 j < d0 , and for any x 2 M, we have 0
jT n;k xðtÞ T n;k xðt0 Þj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ ¼ k 1 n k
þAk ðkðfn þ wÞÞaðtÞ þ Bk ðkðfn þ wÞÞbðtÞ Z 1 1 þ wðsÞ ds Gðt 0 ; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k Ak ðkðfn þ wÞÞaðt 0 Þ Bk ðkðfn þ wÞÞbðt0 Þj Z 1 1 þ wðsÞ ds 6k jGðt; sÞ Gðt 0 ; sÞj f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞdsjaðtÞ aðt 0 Þj þ AkðSR;n þ 1Þ 0 Z 1 þ BkðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞdsjbðtÞ bðt0 Þj 0
< e0 : So, fT n;k x : x 2 Mg are equicontinuous on ½0; a. Since a > 0 is arbitrary, fT n;k x : x 2 Mg are locally equicontinuous on ½0; þ1Þ. Let T n;k xðþ1Þ ¼ limt!þ1 T n;k xðtÞ, by a simple calculation, we can see that limt!þ1 T n;k xðtÞ < þ1, so we get
Z 1 1 þ wðsÞ ds þ Ak ðkðfn þ wÞÞjaðtÞ að1Þj jT n;k xðtÞ T n;k xðþ1Þj 6 k ðGðt; sÞ GðsÞÞ f s; ½xðsÞ kxðsÞ þ 1 n k
þ Bk ðkðfn þ wÞÞjbðtÞ bð1Þj: By the similar method as for (3.20), we get that, for any
e > 0, there exists N0 such that, when t > N0 , it is true that
Z 1 1 e þ wðsÞ ds < : k ðGðt; sÞ GðsÞÞ f s; ½xðsÞ kxðsÞ þ 3 1 n k
Together with the continuity of aðtÞ; bðtÞ on ½0; þ1Þ, we get that for the above e > 0, there exists N 0 such that, when t > N 0 , we have jT n;k xðtÞ T n;k xðþ1Þj < e. Hence, the functions fT n;k x : x 2 Mg are equiconvergent at þ1, which implies that fT n;k x : x 2 Mg is relatively compact (by Lemma 2.5). Therefore, we get that the operator T n;k : K ! K is completely continuous for natural numbers n; k. Finally, we show that T n : K ! K is a completely continuous operator. For any t 2 ½0; þ1Þ and x 2 S ¼ fx 2 K : kxk 6 1g, by (2.8) and (3.9), we have
Z k 0
1 k
Z 1 k 1 1 1 þ wðsÞ ds 6k þ h ½xðsÞ kxðsÞ þ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ n n n 0 Z 1 k Gðs; sÞð/ðsÞS1;n þ wðsÞÞds ! 0; k ! þ1; ð3:22Þ þwðsÞÞds 6 k 0
where S1;n :¼ supfgðuÞ þ hðuÞ :
1 n
6 u 6 2g < þ1. And then, we get
Z 1 P P m2 k q m2 að1Þ 1 i¼1 ci i¼1 ci bðgi Þ jAðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjaðtÞ 6 Gðs; sÞ f s; ½xðsÞ kxðsÞ þ k Pm2 0 n D Pm2 di i¼1 i¼1 di bðgi Þ 1 þ 2wðsÞ ds þ f s; ½xðsÞ kxðsÞ þ n Z 1 k Gðs; sÞð/ðsÞ þ wðsÞÞds ! 0; k ! þ1: 6 2AkðS1;n þ 1Það1Þ 0
ð3:23Þ
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Using the similar method, we also get
jBðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbðtÞ 6 2BkðS1;n þ 1Þbð0Þ
Z
1 k
Gðs; sÞð/ðsÞ þ wðsÞÞds ! 0;
k ! þ1:
ð3:24Þ
0
Together (3.22)–(3.24) imply that
kT n T n;k k ¼ supkT n x T n;k xk ! 0;
k ! þ1:
x2S
Therefore, by T n;k : K ! K is a completely continuous operator, we get that T n : K ! K is a completely continuous operator. h Theorem 3.1. Assume that ðH1 Þ-ðH3 Þ hold and f satisfies the following condition: ðH4 Þ There exists ½a; b ð0; þ1Þ such that
lim min
u!þ1 t2½a;b
f ðt; uÞ ¼ þ1: u
Then there exists k > 0 such that BVP (1.1) has at least one positive solution for any k 2 ð0; kÞ. n Proof. Choose r1 > max 4;
4d
q
R1 0
o wðsÞds , where d is defined as (2.3), let
8 9 < = r1
R
k ¼ min 1; ; 1 : 1 þ að1ÞA þ bð0ÞB 0 Gðs; sÞ /ðsÞ g ðfðsÞÞ þ hðr 1 Þ þ wðsÞ ds;
¼ sup06u6rþ1 hðuÞ; A and B are defined by (3.2) and (3.3). Let K r1 ¼ fx 2 K : kxk < r1 g. For any x 2 @K 1 ; t 2 ½0; þ1Þ, where hðrÞ by the definition of k k and Lemma 2.3, we have
½xðtÞ kxðtÞ 6 xðtÞ 6 kxk 6 r 1 ; xðtÞ kxðtÞ P xðtÞ
kdfðtÞ
q
Z
1
0
wðsÞds P xðtÞ
dfðtÞ
q
Z
1
wðsÞds P xðtÞ 0
2dxðtÞ qr1
Z
1
wðsÞds P 0
xðtÞ fðtÞr 1 P P fðtÞ: 2 4 ð3:25Þ
For any k 2 ð0; kÞ, we have
jðT n xÞðtÞj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ ¼k n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 1 þ h ½xðsÞ kxðsÞ þ þ wðsÞ ds Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 6k n n 0 þ Aðkðfn þ wÞÞað1Þ þ Bðkðfn þ wÞÞbð0Þ
Z 1
6 k 1 þ Aað1Þ þ Bbð0Þ Gðs; sÞ /ðsÞ g ðfðsÞÞ þ hðr 1 Þ þ wðsÞ ds 0
6 r1 : Thus,
kT n xk 6 kxk;
for any x 2 @K r1 :
On the other hand, by the inequality in ðH4 Þ, choose l such that khslf such that
f ðt; uÞ P lu; Let
r 2 > max
n
ð3:26Þ Rb a
Gðs; sÞds > 4; s ¼ mint2½a;b sðtÞ, then there exists N > 0
u P N ; t 2 ½a; b:
r1 ; 4Nf
o ; f ¼ mint2½a;b fðtÞ; K r2 ¼ fx 2 K : kxk < r 2 g.
Take
q1 1 2 @K 1 ; K 1 ¼ fx 2 K : kxk < 1g.
For
any
x 2 @K r2 ; l > 0; n 2 N, we will show
x – T n x þ lq1 :
ð3:27Þ
Otherwise, there exist x0 2 @K r2 and l0 > 0, such that x0 ¼ T n x0 þ l0 q1 . From x0 2 @K r2 , we know that kx0 k ¼ r 2 . Then, for t 2 ½a; b, we have Z Z Z kdfðtÞ 1 dfðtÞ 1 2dx0 ðtÞ 1 x0 ðtÞ fðtÞr2 fr2 x0 ðtÞ kxðtÞ P x0 ðtÞ wðsÞds P x0 ðtÞ wðsÞds P x0 ðtÞ wðsÞds P P P P N : q 0 q 0 qr2 0 4 4 2
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Hence, we conclude that
x0 ðtÞ ¼ k
Z
1
0
1 þ wðsÞ ds Gðt; sÞ f s; ½x0 ðsÞ kxðsÞ þ n
þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ þ l0 Z 1 1 þ wðsÞ ds þ l0 Pk hsðtÞGðs; sÞ f s; ½x0 ðsÞ kxðsÞ þ n 0 Z b lfr2 Pk hsðtÞGðs; sÞ ds þ l0 4 a Z khslfr2 b P Gðs; sÞds þ l0 4 a P r 2 þ l0 > r 2 : This implies that r 2 > r2 , which is a contradiction. This yields that (3.27) holds. It follows from the above discussion, (3.26) and (3.27), Lemmas 2.6 and 3.1, that for any n 2 N; k 2 ð0; kÞ; T n has a fixed point xn 2 K r2 n K r1 . Let fxn g1 n¼1 be the sequence of solutions of BVP (2.7). It is easy to see that they are uniformly bounded. From xn 2 K r 2 n K r 1 , by the similar discussion as (3.25), we know that
½xn ðtÞ kxðtÞ 6 xn ðtÞ 6 kxn k 6 r2 ; t 2 ½0; þ1Þ; Z Z kdfðtÞ 1 dfðtÞ xn ðtÞ kxðtÞ P xn ðtÞ wðsÞds P xn ðtÞ
q
P fðtÞ;
q
0
1
wðsÞds P xn ðtÞ
0
2dxn ðtÞ qr1
Z
1
wðsÞds P 0
t 2 ½0; þ1Þ:
xn ðtÞ 2 ð3:28Þ
R1
0 0 Next, given a0 > 0, we will prove that fxn g1 n¼1 are equicontinuous on ½0; a . For any e > 0, by k 0 Gðs; sÞð/ðsÞðgðfðsÞÞþ hðr 2 ÞÞ þ wðsÞÞds < þ1, where hðr 2 Þ ¼ supfhðuÞ : 0 6 u 6 r 2 þ 1g, we can find a sufficiently large H0 > 0 such that
Z
1
k
Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞds
0 such that for any
:
0
Therefore, for any n 2 N; t; t0 2 ½0; a; s 2 ½0; H0 and jt t 0 j < d0 , we get
Z k
1 þ wðsÞ ds Gðt; sÞ f s; ½xn ðsÞ kxðsÞ þ n 0 Z 1 1 k þ wðsÞ ds Gðt 0 ; sÞ f s; ½xn ðsÞ kxðsÞ þ n 0 Z H0 1 6k þ wðsÞ ds jGðt; sÞ Gðt 0 ; sÞj f s; ½xn ðsÞ kxðsÞ þ n 0 Z 1 1 0 þk þ wðsÞ ds jGðt; sÞ Gðt ; sÞj f s; ½xn ðsÞ kxðsÞ þ n H0 Z H0
jGðt; sÞ Gðt 0 ; sÞj /ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞ ds 6k 0Z 1 jGðt; sÞ Gðt 0 ; sÞjð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞds þk H0 Z 1
e0 e0 6 þ 2k Gðs; sÞ /ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞ ds < : 6 3 0 1
ð3:29Þ
For
Pm2 R 1 P 1 q m2 1 i¼1 ci k 0 Gðgi ; sÞ f s; ½xn ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ An ðkðfn þ wÞÞ ¼ Pm2 R 1 ; P m2 D di k Gðg ; sÞ f s; ½xn ðsÞ kxðsÞ þ 1 þ wðsÞ ds di bðg Þ i¼1
0
i
n
i
i¼1
Pm2 R 1 P 1 q m2 1 i¼1 di k 0 Gðgi ; sÞ f s; ½xn ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ Bn ðkðfn þ wÞÞ ¼ Pm2 R 1 ; P m2 D c k Gðg ; sÞ f s; ½xn ðsÞ kxðsÞ þ 1 þ wðsÞ ds c aðg Þ i¼1
i
0
i
n
i¼1
i
i
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through a simple calculation, we get
An ðkðfn þ wÞÞ 6 Ak
Z
1
Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞds < þ1;
0
Bn ðkðfn þ wÞÞ 6 Bk
Z
1
Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr2 ÞÞ þ wðsÞÞds < þ1;
0
where A and B are defined as (3.2) and (3.3). So by the uniformly continuity of aðtÞ; bðtÞ on ½0; a0 , for the above exists d00 > 0 such that, for any t; t 0 2 ½0; a0 and jt t0 j < d00 , we have
jaðtÞ aðt 0 Þj < 0
jbðtÞ bðt Þj
0, there
ð3:30Þ
0
Then, by (3.29) and (3.30), for the above
e0 > 0, let d0 ¼ minfd0 ; d00 g, then for any n 2 N; t; t0 2 ½0; a0 and jt t0 j < d0 , we have
jxn ðtÞ xn ðt 0 Þj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xn ðsÞ kxðsÞ þ ¼ k n 0
þAn ðkðfn þ wÞÞaðtÞ þ Bn ðkðfn þ wÞÞbðtÞ Z 1 1 k þ wðsÞ ds Gðt 0 ; sÞ f s; ½xn ðsÞ kxðsÞ þ n 0 An ðkðfn þ wÞÞaðt 0 Þ Bn ðkðfn þ wÞÞbðt0 Þj Z 1 1 þ wðsÞ ds 6k jGðt; sÞ Gðt 0 ; sÞj f s; ½xn ðsÞ kxðsÞ þ n 0 Z 1 Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr2 ÞÞ þ wðsÞÞdsjaðtÞ aðt0 Þj þ Ak 0 Z 1 þ Bk Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞdsjbðtÞ bðt 0 Þj 0
< e0 : 1 0 0 So, fxn g1 n¼1 are equicontinuous on ½0; a . Since a > 0 is arbitrary, fxn gn¼1 are locally equicontinuous on ½0; þ1Þ. Let xn ðþ1Þ ¼ limt!þ1 xn ðtÞ, by a simple calculation, we can see that limt!þ1 xn ðtÞ < þ1, and so we get
jxn ðtÞ xn ðþ1Þj Z 1 1 þ wðsÞ ds ðGðt; sÞ GðsÞÞ f s; ½xn ðsÞ kxðsÞ þ 6 k n 0 þ An ðkðfn þ wÞÞjaðtÞ að1Þj þ Bn ðkðfn þ wÞÞjbðtÞ bð1Þj: By the similar method as for (3.29), we get that, for any
Z k
1
0
e0 > 0, there exists N0 such that, when t > N0 , it follows
e0 1 þ wðsÞ ds < : ðGðt; sÞ GðsÞÞ f s; ½xn ðsÞ kxðsÞ þ n 3
Together with the continuity of aðtÞ; bðtÞ on ½0; þ1Þ, we get that for the above e0 > 0, there exists N 0 such that, when t > N 0 , 1 we have jxn ðtÞ xn ðþ1Þj < e0 . Hence, the functions from fxn g1 n¼1 are equiconvergent at þ1, which implies that fxn gn¼1 is 1 relatively compact (by Lemma 2.5). Therefore, the sequence fxn gn¼1 has a subsequence being uniformly convergent on ½0; þ1Þ. Without loss of generality, we still assume that fxn g1 n¼1 itself uniformly converges to x on ½0; þ1Þ. Since fxn g1 2 K n K K, we have x P 0. By (2.7), we have r2 r1 n n¼1
Z t Z s 0 1 1 1 p ð1Þx0n ð1Þ þ x0n t ds d1 1 1 2 2 2 pð1Þ 2 2 Z t Z s f 1; ½xn ð1Þ kxð1Þ þ 1n þ wð1Þ k ds d1; 1 1 pð1Þ 2 2
xn ðtÞ ¼ xn
t 2 ð0; þ1Þ:
ð3:31Þ
1
As fx0n ð12Þgn¼1 is bounded, without loss of generality, we may assume x0n ð12Þ ! c0 as n ! þ1. Then, by (3.31) and the Lebesgue dominated convergence theorem, we have
Z t Z s 0 Z t Z s 1 1 p ð1Þx0 ð1Þ ðf ð1; ½xð1Þ kxð1Þ Þ þ wð1ÞÞ þ c0 t xðtÞ ¼ x ds d1 k ds d1; 1 1 1 1 2 2 pð1Þ pð1Þ 2 2 2 2
t 2 ð0; þ1Þ:
ð3:32Þ
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By (3.32), the direct computation shows that 0
ðpðtÞx0 ðtÞÞ þ kðf ðt; ½xðtÞ kxðtÞ Þ þ wðtÞÞ ¼ 0;
t 2 ð0; þ1Þ:
On the other hand, let n ! þ1 in the following boundary conditions:
a1 xn ð0Þ b1 limþ pðtÞx0n ðtÞ ¼
m2 X
t!0
ci xn ðgi Þ;
i¼1 m2 X
a2 t!þ1 lim xn ðtÞ þ b2 lim pðtÞx0n ðtÞ ¼ t!þ1
di xn ðgi Þ:
i¼1
Therefore, we deduce that x is a solution of BVP (2.7). Let xðtÞ ¼ xðtÞ kxðtÞ. By (3.28) and the convergence of the sequence fxn g1 n¼1 , we have xðtÞ P fðtÞ > 0; t 2 ½0; þ1Þ. It then follows from Lemma 2.4 that x is a positive solution of BVP (1.1). The proof is completed. h Remark 3.1. From the proof of Theorem 3.1, we get the main results under the condition that the function f ðt; uÞ not only has singularity on t but also has singularity on u. In addition, we obtain the positive solution of BVP (1.1) under the condition that the function f ðt; uÞ is semipositive, which are the great improvement of [23–25]. Also our method is different from those in [26–29]. Theorem 3.2. Assume that ðH1 Þ–ðH3 Þ hold, f and h satisfy the following condition: ðH5 Þ There exists ½c; d ð0; þ1Þ, such that
lim inf min f ðt; uÞ >
4d
0
wðsÞds
-¼
Rd c
lim
;
qhs0 -
u!þ1 t2½c;d
where s0 ¼ mint2½c;d sðtÞ; k 2 ðk; þ1Þ.
R1
u!þ1
hðuÞ ¼ 0; u
Gðs; sÞds. Then there exists k > 0 such that BVP (1.1) has at least one positive solution for any
Proof. By the first inequality of ðH5 Þ, we have that there exists N > 0 such that for any t 2 ½a; b; u > N , we have
f ðt; uÞ >
4d
R1
wðsÞds
0
qhs0 -
ð3:33Þ
:
Select
k¼
N q ; R1 df 0 wðsÞds
where f0 ¼ min fðtÞ:
0
t2½c;d
In the following proof, we suppose k > k. Let
R1 ¼
4kd
R1 0
wðsÞds
q
:
Assume K R1 ¼ fx 2 K : kxk < R1 g. For any x 2 @K R1 ; t 2 ½c; d,
xðtÞ kxðtÞ P P P
4kdfðtÞ 2q Z kdfðtÞ
q
Z
fðtÞR1 kdfðtÞ 2 q
1
wðsÞds
kdfðtÞ
q
0 1
wðsÞds P
0
Z
kdf0
q
Z
1
wðsÞds 0
Z
1
wðsÞds 0 1
wðsÞds P N :
0
Then, by (3.33), we have
jðT n xÞðtÞj ¼ k
Z
1
0
1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n
þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 þ wðsÞ ds hsðtÞGðs; sÞ f s; ½xðsÞ kxðsÞ þ Pk n 0 R Z d 1 4d 0 wðsÞds P khs0 Gðs; sÞ ds qhs0 c Z d R1 Gðs; sÞds: P
-
c
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Noticing that
-¼
Rd c
kT n xk P kxk;
271
Gðs; sÞds, we have
for any x 2 @K R1 :
ð3:34Þ
On the basis of the second inequality in ðH5 Þ and the continuity of hðuÞ on ½0; þ1Þ, we have
lim
u!þ1
For
hðuÞ ¼ 0: u (
Z c ¼ max 1; 4kð1 þ Aað1Þ þ Bbð0ÞÞ
1
1 ) ; Gðs; sÞð/ðsÞ þ wðsÞÞds
0 00
00
there exists N > 0, such that when x P N , for any 0 6 y 6 x, we have hðyÞ 6 cx. Select
Z 00 R2 P 2; R1 ; N ; 2kð1 þ Aað1Þ þ Bbð0ÞÞ
0
1
fðsÞ ds : Gðs; sÞ/ðsÞg 2
Then, for any x 2 @K R2 ; t 2 ½0; þ1Þ, we have
½xðtÞ kxðtÞ 6 xðtÞ 6 kxk 6 R2 ; Z Z kdfðtÞ 1 2kdxðtÞ 1 xðtÞ fðtÞR2 fðtÞ xðtÞ kxðtÞ P xðtÞ wðsÞds P xðtÞ wðsÞds P P P > 0: 2 2 q qR2 0 4 0 Hence, we gain
jðT n xÞðtÞj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ ¼k n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 1 þ h ½xðsÞ kxðsÞ þ þ wðsÞ ds Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 6k n n 0 þ Aðkðfn þ wÞÞað1Þ þ Bðkðfn þ wÞÞbð0Þ
Z 1 fðsÞ þ cðR2 þ 1Þ þ wðsÞ ds Gðs; sÞ /ðsÞ g 6 k 1 þ Aað1Þ þ Bbð0Þ 2 0 Z
1 fðsÞ ds Gðs; sÞ/ðsÞg 6 k 1 þ Aað1Þ þ Bbð0Þ 2 0 Z
1 Gðs; sÞð/ðsÞ þ wðsÞÞds 6 R2 : þ kc 1 þ Aað1Þ þ Bbð0Þ ðR2 þ 2Þ 0
Thus,
kT n xk 6 kxk;
for any x 2 @K R2 :
ð3:35Þ
It follows from the above discussion, (3.34) and (3.35), Lemmas 2.7 and 3.1, for n 2 N; k 2 ðk; þ1Þ; T n has a fixed point xn 2 K R2 n K R1 satisfying R1 6 kxn k 6 R2 . The rest of proof is similar to Theorem 3.1. That’s the proof of Theorem 3.2. h Corollary 3.1. The conclusion of Theorem 3.2 is valid if ðH5 Þ is replaced by: ðH5 Þ There exists ½c; d ð0; þ1Þ, such that
lim inf minf ðt; uÞ ¼ þ1; u!þ1 t2½c;d
lim
u!þ1
hðuÞ ¼ 0: u
Remark 3.2. From Theorems 3.1 and 3.2, we get the positive solution of BVP (1.1) when the parameter k is sufficiently large or small, and the solution x in BVP (1.1) satisfies xðtÞ > 0 for any t 2 ½0; þ1Þ. 4. Example Example 4.1. Consider the following BVP
8 1 1 0 2 0 > 2 1 > ¼ 0; t 2 ð0; þ1Þ; ðð1 þ tÞ x ðtÞÞ þ k þ x > 2 1 x 1þt > t 2 þt 2 > < xð0Þ limþ ð1 þ tÞ2 x0 ðtÞ ¼ 13 x 13 ; > t!0 > > > > : lim xðtÞ þ lim ð1 þ tÞ2 x0 ðtÞ ¼ 1 x 1 : 6 3 t!þ1
t!þ1
ð4:1Þ
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1 1 It is easy to see that a1 ¼ a2 ¼ b1 ¼ b2 ¼ 1, pðtÞ ¼ ð1 þ tÞ2 ; aðtÞ ¼ 2 1þt ; bðtÞ ¼ 1 þ 1þt , q ¼ 3; q 13 b 13 ¼ 29 , q 16 a 13 ¼ 67 12 24 and
29 q 1 b 1 1 a 13 3 3 3 ¼ 12 D ¼ 1 1 6b 3 q 16 a 13 247 As f ðt; uÞ ¼
1
wðtÞ ¼
x
þ x2
1 1 2
t þt
2
1 1þt 2
1
;
/ðtÞ ¼
1
t 2 þt 2
5 12 67 24
¼
53 > 0: 8
, we can suppose
1 ; 1 þ t2
gðuÞ ¼
1 ; u
hðuÞ ¼ u2 :
By direct calculation, we have
Z
1
wðsÞds < þ1;
0
lim inf min
u!þ1 t2½2;5
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1;
0
f ðt; uÞ ¼ þ1: u
So all conditions of Theorem 3.1 are satisfied. By Theorem 3.1, BVP (4.1) has at least one positive solution provided k is small enough. Acknowledgements The first and second authors were supported financially by the National Natural Science Foundation of China (11071141, 11371221), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001) and the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province. The third author was supported financially by the Australia Research Council through an ARC Discovery Project Grant. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26]
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[27] H. Lian, J. Zhao, Existence of unbounded solutions for a third-order boundary value problem on infinite intervals, Disc. Dyn. Nat. Soc. 2012 (2012) 14 (Article ID 357697). [28] H. Lian, Z. Cui, S. Zhang, The existence of unbounded solutions to three-point boundary value problem over a semi-infinite interval, Acta Math. Sci. 32(A) (3) (2012) 489–499. [29] X. Li, Z. Zhao, Existence of positive solutions for singular Sturm–Liouville boundary value problem on the half-line, J. Sys. Sci. Math. Sci. 32 (5) (2012) 562–570. [30] R. Ma, Multiple positive solutions for nonlinear m-point boundary value problems, Appl. Math. Comput. 148 (2004) 249–262. [31] R.P. Agarwal, D. O’egan, Infinite Interval Problems for Differential and Integral Equations, Kluwer Academic, 2001. [32] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach space, SIAM Rev. 18 (1976) 620–709. [33] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, New York, 1988.
Author's personal copy Applied Mathematics and Computation 227 (2014) 256–273
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Positive solutions for singular semipositone boundary value problems on infinite intervals Ying Wang a,b, Lishan Liu a,⇑, Yonghong Wu c a
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China School of Science, Linyi University, Linyi 276000, Shandong, People’s Republic of China c Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia b
a r t i c l e
i n f o
a b s t r a c t By using the fixed point theory on a cone with a special norm and space, we discuss the existence of positive solutions for a class of semipositone boundary value problems on infinite intervals. The work improves many known results including singular and non-singular cases. Ó 2013 Elsevier Inc. All rights reserved.
Keywords: Positive solutions Cone Semipositone boundary value problems Infinite intervals
1. Introduction The main purpose of this paper is to study the existence of positive solutions for the following nonlinear singular and semipositone boundary value problem (BVP) on infinite intervals:
8 ðpðtÞx0 ðtÞÞ0 þ kf ðt; xðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; t!0þ
i
i
i¼1 > > > m 2 X > > > > a2 lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ; : t!þ1 t!þ1
ð1:1Þ
i¼1
where k > 0 is a parameter, a1 ; a2 ; ci ; di P 0; b1 ; b2 > 0; gi 2 ð0; þ1Þ; ð1; 2; . . . ; m 2Þ are given constants, f : ð0; þ1Þ ð0; þ1Þ ! ð1; þ1Þ is a continuous function and f ðt; uÞ may be singular at t ¼ 0 and u ¼ 0, p 2 C½0; þ1Þ \ C 1 ð0; þ1Þ with R1 1 Rs p > 0 on ð0; þ1Þ; 0 pðsÞ ds < þ1, q ¼ a2 b1 þ a1 b2 þ a1 a2 Bð0; 1Þ > 0 in which Bðt; sÞ ¼ t pð1v Þ dv . Boundary value problems on infinite intervals arises from the study of many real world problems such as solution of nonlinear elliptic equations and modeling of gas pressure in a semi-infinite porous medium. A great deal of work has been done in these areas such as those in [1–22] and the references therein. Over the last couple of decades, a great deal of results have been developed for differential and integral boundary value problems. O’Regan et al. studied in [23] the BVP
(
x00 ðtÞÞ þ uðtÞf ðt; xðtÞÞ ¼ 0; xðaÞ ¼ 0;
t 2 ða; þ1Þ;
lim x0 ðtÞ ¼ 0;
t!þ1
where f : ½a; þ1Þ ð0; þ1Þ ! ½0; þ1Þ is a continuous function and u : ða; þ1Þ ! ½0; þ1Þ is continuous. The existence of multiple unbounded positive solutions is discussed using the theory of fixed point index. By using the fixed point theorem and the monotone iterative technique, Zhang [24] studied the problem ⇑ Corresponding author. E-mail addresses: [email protected] (Y. Wang), [email protected] (L. Liu), [email protected] (Y. Wu). 0096-3003/$ - see front matter Ó 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.11.009
Author's personal copy Y. Wang et al. / Applied Mathematics and Computation 227 (2014) 256–273
257
8 00 x ðtÞÞ þ qðtÞf ðt; xðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > < m2 X > ci xðgi Þ; lim x0 ðtÞ ¼ x1 P 0; : xð0Þ ¼ t!þ1
i¼1
Pm2 where ci P 0, 0 < g1 < g2 < < gm2 < þ1, i¼1 ci < 1; f : ½0; þ1Þ ½0; þ1Þ ! ½0; þ1Þ is a continuous function and q : ð0; þ1Þ ! ½0; þ1Þ is a Lebesgue integrable function. Liu et al. in [25] established the existence of positive solutions for the following equation on infinite intervals by applying the fixed point theorem of cone map
8 ðpðtÞx0 ðtÞÞ0 þ mðtÞf ðt; xðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > < a1 xð0Þ b1 limþ pðtÞx0 ðtÞ ¼ 0; t!0 > > : a2 lim xðtÞ þ b lim pðtÞx0 ðtÞ ¼ 0; 2 t!þ1
t!þ1
in which f : ½0; þ1Þ ½0; þ1Þ ! ½0; þ1Þ is a continuous function, m : ð0; þ1Þ ! ½0; þ1Þ is a Lebesgue integrable function and may be singular at t ¼ 0. Also, by the use of the Krasnosel-skii fixed point theorem, upper and lower solutions, Schauder point theorem and inequality technique, Xing et al. in [26], Lian et al. in [27,28], Li and Zhao in [29] studied many equations with infinite intervals. However, all of the above studies are limited to the cases in which the nonlinear term is positive. Inspired by the work of the above papers and many known results, in this paper, we study the existence of positive solutions to BVP (1.1), where x 2 C½0; þ1Þ is said to be a positive solution of BVP (1.1) if and only if x satisfies (1.1) and xðtÞ > 0 for any t 2 ½0; þ1Þ. By using the fixed point theory on a cone, some new existence results are obtained for the case where the nonlinearity is allowed to be sign changing and has singularity. We should address here that our work presented in this paper has various new features. Firstly, our study is on singular nonlinear differential boundary value problems, that is, f ðt; uÞ is allowed to be singular at t ¼ 0 and u ¼ 0, which leads to many difficulties in analysis. Secondly, the techniques used in this paper are the approximation method; and a special cone in a special space is established to overcome the difficulties caused by singularity and infinite interval. Also we find another substitute function to solve the problem associated with the semipositive property of the nonlinear function. Thirdly, we discuss the boundary value problem with multi-point boundary conditions, that is, BVP (1.1) includes two-point and three-point boundary value problems as special cases. To our knowledge, the theory of Sturm–Liouville multi-point boundary value problems on infinite interval is yet to be developed. Our model improves and generalizes the results of previous papers to some degree. The plan of the paper is as follows. In Section 2, we present the preliminaries and necessary lemmas that are to be used to prove our main results. The main results are given in Section 3, including results for a completely continuous operator, and the conditions for the existence of positive solutions for the BVP (1.1). In Section 4, an example is given to demonstrate the application of our theoretical results. 2. Preliminaries and lemmas For convenience, we let
aðtÞ ¼ b1 þ a1 Bð0; tÞ; bðtÞ ¼ b2 þ a2 Bðt; 1Þ; að1Þ ¼ lim aðtÞ ¼ b1 þ a1 Bð0; 1Þ < þ1; að0Þ ¼ lim aðtÞ ¼ b1 ; t!þ1
t!0
bð0Þ ¼ lim bðtÞ ¼ b2 þ a2 Bð0; 1Þ < þ1;
bð1Þ ¼ lim bðtÞ ¼ b2 ; t!þ1
t!0
Pm2 q Pm2 c bðg Þ i i¼1 i i¼1 ci aðgi Þ D ¼ Pm2 : P m2 d bð g Þ q d að g Þ i i i¼1 i i¼1 i It is obvious that aðtÞ is increasing and bðtÞ is decreasing on ½0; þ1Þ. Define
Gðt; sÞ ¼
1
aðsÞbðtÞ; 0 6 s 6 t < þ1;
ð2:1Þ
q aðtÞbðsÞ; 0 6 t 6 s < þ1:
Denote sðtÞ ¼ aðtÞbðtÞ, then for any 0 6 t; s < þ1, we get
0 6 Gðt; sÞ 6 Gðs; sÞ 6 GðsÞ ¼ lim Gðt; sÞ ¼ t!þ1
bð0ÞaðsÞ
q
b2 aðsÞ
q
;
0 6 Gðt; sÞ 6
sðtÞ ; q
6 Gðs; sÞ < þ1:
1 Lemma 2.1. Suppose h ¼ að1Þbð0Þ , then Gðt; sÞ P hsðtÞGðs; sÞ; 0 6 t; s < þ1.
ð2:2Þ
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Proof. From (2.2) and the property of aðtÞ; bðtÞ, for 0 st; s < þ1, we have
8 8 aðtÞbðtÞ bðtÞ Gðt; sÞ < bðsÞ ; s 6 t; < aðtÞbðsÞ ; s 6 t; ¼ ¼ Gðs; sÞ : aðtÞ ; t 6 s; : aðtÞbðtÞ ; t 6 s; aðsÞ aðsÞbðtÞ
P
sðtÞ : að1Þbð0Þ
Therefore, Gðt; sÞ P hsðtÞGðs; sÞ; 0 6 t; s < þ1. h In this paper, we always assume that the following conditions hold. P Pm2 ðH1 Þ : D > 0; q m2 i¼1 ci bðgi Þ > 0, q i¼1 di aðgi Þ > 0. ðH2 Þ : f : ð0; þ1Þ ð0; þ1Þ ! ð1; þ1Þ is a continuous function and
wðtÞ 6 f ðt; uÞ 6 /ðtÞðgðuÞ þ hðuÞÞ;
ðt; uÞ 2 ð0; þ1Þ ð0; þ1Þ;
where w; / : ð0; þ1Þ ! ½0; þ1Þ is continuous and singular at t ¼ 0; wðtÞ; /ðtÞ X 0 on ½0; þ1Þ, g : ð0; þ1Þ ! ½0; þ1Þ is continuous and nonincreasing, h : ½0; þ1Þ ! ½0; þ1Þ is continuous, g and h are bounded in any bounded set of ½0; þ1Þ. R1 R1 ðH3 Þ 0 wðsÞds < þ1; 0 Gðs; sÞðwðsÞ þ /ðsÞÞds < þ1. Let
P P m2 q m2 1 i¼1 ci sðgi Þ i¼1 ci bðgi Þ A ¼ Pm2 ; P m2 D di sðg Þ di bðg Þ i
i¼1
i
i¼1
P P m2 q m2 1 i¼1 di sðgi Þ i¼1 di aðgi Þ B ¼ Pm2 : P m2 D c sð g Þ c aðg Þ i¼1
i
i
i
i¼1
i
Choose a constant d, such that d P að1Þbð0Þ þ Aað1Þ þ Bbð0Þ, and denote
fðtÞ ¼
sðtÞ þ AaðtÞ þ BbðtÞ d
ð2:3Þ
;
P 1; 0 < fðtÞ 6 1. then dh Lemma 2.2. Suppose ðH1 Þ holds,
R1
1 0 pðsÞ ds
< þ1 and q > 0, then the BVP
8 ðpðtÞx0 ðtÞÞ0 þ wðtÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; i
t!0þ
i
i¼1 > > > m 2 X > > 0 > > a lim xðtÞ þ b lim pðtÞx ðtÞ ¼ di xðgi Þ; 2 2 : t!þ1 t!þ1 i¼1
has a unique solution for any w 2 Lð0; þ1Þ. Moreover, this unique solution can be expressed in the form
xðtÞ ¼
Z
1
Gðt; sÞwðsÞds þ AðwÞaðtÞ þ BðwÞbðtÞ;
ð2:4Þ
0
where Gðt; sÞ is defined as (2.1) and
Pm2 R 1 P q m2 1 i¼1 ci 0 Gðgi ; sÞwðsÞds i¼1 ci bðgi Þ AðwÞ ¼ Pm2 R 1 ; P m2 D di Gðg ; sÞwðsÞds di bðg Þ i¼1
0
i
i
i¼1
Pm2 R 1 P q m2 i¼1 di 0 Gðgi ; sÞwðsÞds i¼1 di aðgi Þ 1 BðwÞ ¼ : D Pm2 R 1 P m2 c Gðg ; sÞwðsÞds c aðg Þ i¼1
i
0
i
i¼1
i
i
The proof is similar to [30], so we omit it.
Lemma 2.3. The solution defined by (2.4) satisfies xðtÞ 6 dfðtÞ q
R1 0
wðsÞds.
Proof. Since xðtÞ is the unique solution of (2.4). By (2.2)–(2.4), we have
xðtÞ 6
Z 0
1
sðtÞwðsÞ ds þ AaðtÞ q
Z 0
1
wðsÞ
q
ds þ BbðtÞ
Z 0
1
wðsÞ
q
ds 6 ðsðtÞ þ AaðtÞ þ BbðtÞÞ
Z 0
1
wðsÞ
q
ds ¼
dfðtÞ
q
Z
1
wðsÞds: 0
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In this paper, the following space X will be used in the study of BVP(1.1).
X¼
x 2 C½0; þ1Þ : lim xðtÞ exists : t!þ1
Clearly ðX; k kÞ is a Banach space with the supremum norm kxk ¼ supt2½0;þ1Þ jxðtÞj, see [31]. Let
K¼
fðtÞ x 2 X : xðtÞ P kxk; t 2 ½0; þ1Þ : 2
It is easy to see that K is a cone in X. Next we consider the following singular nonlinear boundary value problem:
8 ðpðtÞx0 ðtÞÞ0 þ kðf ðt; ½xðtÞ kxðtÞ Þ þ wðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; i
t!0þ
i
ð2:5Þ
i¼1 > > > m2 X > > > > lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ; : a2 t!þ1 t!þ1 i¼1
where k > 0; xðtÞ is defined in Lemma 2.2, ½zðtÞ ¼ maxfzðtÞ; 0g. h Lemma 2.4. If x is a solution of BVP (2.5) with xðtÞ > kxðtÞ for any t 2 ½0; þ1Þ, then xðtÞ kxðtÞ is a positive solution of BVP (1.1). Proof. In fact, if x is a positive solution of BVP (2.5) such that xðtÞ > kxðtÞ for any t 2 ½0; þ1Þ, then from BVP (2.5) and the definition of ½zðtÞ , we have
8 ðpðtÞx0 ðtÞÞ0 þ kðf ðt; xðtÞ kxðtÞÞ þ wðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m 2 > X > > < a1 xð0Þ b1 lim pðtÞx0 ðtÞ ¼ c xðg Þ; i
t!0þ
i
ð2:6Þ
i¼1 > > > m2 X > > > > lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ: : a2 t!þ1 t!þ1 i¼1
Let uðtÞ ¼ xðtÞ kxðtÞ; t 2 ½0; þ1Þ, then ðpðtÞx0 ðtÞÞ0 ¼ ðpðtÞu0 ðtÞÞ0 þ kðpðtÞx0 ðtÞÞ0 . Thus, (2.6) becomes
8 ðpðtÞu0 ðtÞÞ0 þ kf ðt; uðtÞÞ ¼ 0; t 2 ð0; þ1Þ; > > > > m2 > X > > < a1 uð0Þ b1 lim pðtÞu0 ðtÞ ¼ ci uðgi Þ; t!0þ
i¼1 > > > m2 X > > > > lim uðtÞ þ b2 lim pðtÞu0 ðtÞ ¼ di uðgi Þ: : a2 t!þ1 t!þ1 i¼1
Then uðtÞ ¼ xðtÞ kxðtÞ is a positive solution of BVP (1.1). h To overcome singularity, we consider the following approximate problem:
8 ðpðtÞx0 ðtÞÞ0 þ k f t; ½xðtÞ kxðtÞ þ 1n þ wðtÞ ¼ 0; t 2 ð0; þ1Þ; > > > > > m2 X > > < a1 xð0Þ b lim pðtÞx0 ðtÞ ¼ c xðg Þ; 1
i
t!0þ
i
ð2:7Þ
i¼1 > > > m2 > X > > > a2 lim xðtÞ þ b2 lim pðtÞx0 ðtÞ ¼ di xðgi Þ; : t!þ1 t!þ1 i¼1
in which n is a positive integer. Under the assumptions ðH1 Þ ðH3 Þ, for any n 2 N where N is a natural number set, we define a nonlinear integral operator T n : K ! X by
ðT n xÞðtÞ ¼ k
Z 0
1
1 þ wðsÞ ds þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ; Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n
where
Pm2 R 1 P 1 q m2 1 i¼1 ci 0 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ Aðkðfn þ wÞÞ ¼ Pm2 R 1 ; P m2 1 D di Gðg ; sÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ds di bðg Þ i¼1
0
i
n
i¼1
i
t 2 ½0; þ1Þ;
ð2:8Þ
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Pm2 R 1 P 1 q m2 1 i¼1 di 0 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ Bðkðfn þ wÞÞ ¼ Pm2 R 1 : P m2 D c Gðg ; sÞ f s; ½xðsÞ kxðsÞ þ 1 þ wðsÞ ds c aðg Þ i¼1
i
0
i
n
i¼1
i
i
Obviously, the existence of solutions to BVP (2.7) is equivalent to the existence of solutions in K for operator equation T n x ¼ x defined by (2.8). In this paper, the proof of the main theorem is based on the fixed point theory in cone. We list the following lemmas which are needed in our study. Lemma 2.5 [31]. Let X be defined by (2.1) and M X. Then M is relatively compact in X if the following conditions hold: (1) M is uniformly bounded in X; (2) the functions from M are equicontinuous on any compact interval of ½0; þ1Þ; (3) the functions from M are equiconvergent, that is, for any given e > 0, there exists a T ¼ TðeÞ > 0 such that jxðtÞ xðþ1Þj < e, for any t > T; x 2 M. Lemma 2.6 [32]. Let P be a positive cone in a real Banach space E. Denote Pr ¼ fx 2 P : kxk < rg, P r;R ¼ fx 2 P : r 6 kxk 6 Rg; 0 < r < R < þ1. Let A : P r;R ! Pbe a completely continuous operator. If the following conditions are satisfied: (1) kAxk 6 kxk; 8x 2 @PR . (2) there exists a x0 2 @P1 , such that x – Ax þ mx0 ; 8x 2 @Pr ; m > 0. Then A has fixed points in Pr;R . Remark 2.1. If (1) and (2) are satisfied for x 2 @Pr and x 2 @P R respectively. Then Lemma 2.6 is still true. Lemma 2.7 [33]. Let P be a positive cone in a Banach space E; X1 and X2 are bounded open sets in E; h 2 X1 ; X1 X2 ; A : P \ X2 n X1 ! P is a completely continuous operator. If the following conditions are satisfied: kAxk 6 kxk; 8x 2 P \ @ X1 , kAxk P kxk; 8x 2 P \ @ X2 , or kAxk P kxk; 8x 2 P \ @ X1 , kAxk 6 kxk; 8x 2 P \ @ X2 , then A has at least one fixed point in P \ ðX2 n X1 Þ. 3. Main results Lemma 3.1. Assume that ðH1 Þ–ðH3 Þ hold. Then T n : K ! K is a completely continuous operator for any fixed n 2 N. Proof. First we show that T n : K ! X is well defined and T n ðKÞ # K. For x 2 K, there exists r > 0 such that jxðtÞj 6 r, for t 2 ½0; þ1Þ, also j½xðtÞ kxðtÞ j 6 jxðtÞj 6 r; t 2 ½0; þ1Þ. From ðH2 Þ and the definition of g and h, we have
1 Sr;n :¼ sup gðuÞ þ hðuÞ : 6 u 6 r þ 1 < þ1: n Thus, by ðH2 Þ and ðH3 Þ, for any t 2 ½0; þ1Þ, we know
1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Z 1 1 Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 6k n 0 1 þ wðsÞ ds þh ½xðsÞ kxðsÞ þ n Z 1 Gðs; sÞð/ðsÞSr;n þ wðsÞÞds 6k 0 Z 1 6 kðSr;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds Z
1
k
0
< þ1: By (2.2), for i ¼ 1; 2; . . . ; m 2, we also have
Z k 0
1
Z 1 1 þ wðsÞ ds 6 kðSr;n þ 1Þ Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1: n 0
ð3:1Þ
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Therefore,
Aðkðfn þ wÞÞaðtÞ P R1 P m2 1 q m2 aðtÞ i¼1 ci k 0 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ ¼ Pm2 R 1 P m2 D i¼1 di k Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ 1n þ wðsÞ ds i¼1 di bðgi Þ 0 Pm2 R1 Pm2 i¼1 ci kðSr;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds q i¼1 ci bðgi Þ 0 að1Þ m2 6 R X Pm2 D di kðSr;n þ 1Þ 01 Gðs; sÞð/ðsÞ þ wðsÞÞds d bð g Þ i i i¼1 i¼1 P P Z m2 1 að1Þ ci q m2 i¼1 ci bðgi Þ ¼ Gðs; sÞð/ðsÞ þ wðsÞÞds kðSr;n þ 1Þ Pi¼1 Pm2 m2 D i¼1 di 0 i¼1 di bðgi Þ Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; ¼Aað1ÞkðSr;n þ 1Þ
ð3:2Þ
0
where
P m2 ci 1 A ¼ Pi¼1 D m2 di i¼1
q
Pm2
c g g
i¼1 i bð i Þ : Pm2 i¼1 di bð i Þ
In the same way, we get
Bðkðfn þ wÞÞbðtÞ 6 Bbð0ÞkðSr;n þ 1Þ
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1;
ð3:3Þ
0
where
P m2 di 1 B ¼ Pi¼1 D m2 c
i
i¼1
q
Pm2
g
i¼1 di að i Þ : Pm2 i¼1 i að i Þ
c g
Hence, by (3.1)–(3.3), we can see that
1 þ wðsÞ ds þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; t 2 ½0; þ1Þ: 6 ð1 þ Aað1Þ þ Bbð0ÞÞkðSr;n þ 1Þ
ðT n xÞðtÞ ¼ k
Z
1
0
Then, T n x is well defined for any x 2 K. On the other hand, for any t; t j 2 ½0; þ1Þ; tj ! t, by the continuity of Gðt; sÞ, we get
1 1 þ wðsÞ ! Gðt; sÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ; s 2 ½0; þ1Þ; j ! þ1: Gðtj ; sÞ f s; ½xðsÞ kxðsÞ þ n n
ð3:4Þ
By (2.2), we know
Z 1 1 þ wðsÞ ds 6 ðSr;n þ 1Þ Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; n Z0 1 Z 01 1 þ wðsÞ ds 6 ðSr;n þ 1Þ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1: n 0 0
Z
1
ð3:5Þ
So, by ðH3 Þ, (3.4) and (3.5) and the Lebesgue dominated convergence theorem, we have
1 þ wðsÞ ds lim Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ t j !t 0 n Z 1 1 þ wðsÞ ds lim Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ ¼ tj !t n Z0 1 1 þ wðsÞ ds: ¼ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Z
1
Consequently, together with the continuity of aðtÞ and bðtÞ, we have
jT n xðt j Þ T n xðtÞj Z 1 1 þ wðsÞ ds Gðt j ; sÞ f s; ½xðsÞ kxðsÞ þ ¼ k n 0
þAðkðfn þ wÞÞaðtj Þ þ Bðkðfn þ wÞÞbðt j Þ Z 1 1 þ wðsÞ ds k Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n 0 Aðkðfn þ wÞÞaðtÞ Bðkðfn þ wÞÞbðtÞj Z 1 1 ðGðt j ; sÞ Gðt; sÞÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ds 6 k n 0 þ Aðkðfn þ wÞÞjaðt j Þ aðtÞj þ Bðkðfn þ wÞÞjbðt j Þ bðtÞj ! 0; as j ! þ1:
ð3:6Þ
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Therefore, T n x 2 C½0; þ1Þ. In what follows, for any tj 2 ½0; þ1Þ; t j ! þ1, by (2.2), we have
1 1 þ wðsÞ ! GðsÞ f s; ½xðsÞ kxðsÞ þ þ wðsÞ ; s 2 ½0; þ1Þ; j ! þ1: Gðtj ; sÞ f s; ½xðsÞ kxðsÞ þ n n Then by the Lebesgue dominated convergence theorem and the property of aðtÞ, bðtÞ, we also have
lim ðT n xÞðt j Þ Z 1 1 þ wðsÞ ds GðsÞ f s; ½xðsÞ kxðsÞ þ ¼k n 0
j!þ1
þ Aðkðfn þ wÞÞað1Þ þ Bðkðfn þ wÞÞb2 < þ1: So, for any x 2 K, we get T n x 2 X, which implies that T n maps K to X. In the following, we will prove T n ðKÞ # K. For any x 2 K, from the definition of k k and (2.2), we have
kT n xk 6 k
Z
1
0
1 þ wðsÞ ds þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ: Gðs; sÞ f s; ½xðsÞ kxðsÞ þ n
ð3:7Þ
By Lemma 2.1 and (2.3), we get
ðT n xÞðtÞ
1 P khsðtÞ þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z khsðtÞ 1 1 þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ P 2 n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 1 þ wðsÞ ds khsðtÞ Gðs; sÞ f s; ½xðsÞ kxðsÞ þ ¼ 2 n 0 þAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 1 þ ðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z 1 kh 1 þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ P ðsðtÞ þ AaðtÞ þ BbðtÞÞ 2 n 0 1 þ ðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z khdfðtÞ 1 1 þ wðsÞ ds ¼ Gðs; sÞ f s; ½xðsÞ kxðsÞ þ 2 n 0 1 þ ðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z kfðtÞ 1 1 þ wðsÞ ds Gðs; sÞ f s; ½xðsÞ kxðsÞ þ P 2 n 0 1 þ fðtÞðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞ 2 Z 1 fðtÞ 1 þ wðsÞ ds k Gðs; sÞ f s; ½xðsÞ kxðsÞ þ ¼ 2 n 0 þðAðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞÞÞ: Z
1
ð3:8Þ
Combining (3.7) with (3.8), we have ðT n xÞðtÞ P fðtÞ kT n xk, for any t 2 ½0; þ1Þ. Therefore, T n ðKÞ # K. 2 Next, for any positive integers n; k 2 N, we define an operator T n;k : K ! X by
ðT n;k xÞðtÞ ¼ k
Z 1 k
1
1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n
þ Ak ðkðfn þ wÞÞaðtÞ þ Bk ðkðfn þ wÞÞbðtÞ; t 2 ½0; þ1Þ; where
Ak ðkðfn þ wÞÞ Pm2 R 1 P 1 q m2 i¼1 ci k 1 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ 1 k ; ¼ Pm2 R 1 Pm2 1 D d k Gð g ; sÞ f s; ½xðsÞ k x ðsÞ þ d bð g Þ þ wðsÞ ds 1 i i i i i¼1 i¼1 n k
ð3:9Þ
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Bk ðkðfn þ wÞÞ P R1 P m2 1 q m2 i¼1 di k 1 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ k 1 : ¼ Pm2 D Pm2 R 1 1 i¼1 ci k 1 Gðgi ; sÞ f s; ½xðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci aðgi Þ k
As in the above discussion, we can prove that T n;k : K ! X is well defined and T n;k ðKÞ # K. In what follows, we will prove that T n;k : K ! K is completely continuous, for each k P 1. Firstly, we show that T n;k : K ! K is continuous. Let xt ; x 2 K are such that kxt xk ! 0 as t ! þ1. By (3.9) and ðH3 Þ, we know
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xt ðsÞ kxðsÞ þ k 1 n k Z 1 1 k þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 1 1 6k þ f s; ½xðsÞ kxðsÞ þ þ 2wðsÞ ds Gðs; sÞ f s; ½xt ðsÞ kxðsÞ þ 1 n n Zk 1 1 1 6k þ h ½xt ðsÞ kxðsÞ þ Gðs; sÞ /ðsÞ g ½xt ðsÞ kxðsÞ þ 1 n n k 1 1 þ/ðsÞ g ½xðsÞ kxðsÞ þ þ h ½xðsÞ kxðsÞ þ þ 2wðsÞ ds n n Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; 2kðSr0 ;n þ 1Þ
ð3:10Þ
0
where Sr0 ;n :¼ supfgðuÞ þ hðuÞ :
1 n
6 u 6 r0 þ 1g < þ1 (by ðH2 Þ), r0 is a real number such that r 0 P maxt2N fkxk; kxt kg. Denote
Pm2 R 1 P 1 q m2 i¼1 ci k 1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ 1 k ; Ak;t ðkðfn þ wÞÞ ¼ Pm2 R 1 Pm2 D di k 1 Gðg ; sÞ f s; ½xt ðsÞ kxðsÞ þ 1 þ wðsÞ ds di bðg Þ i¼1
i
k
i
i¼1
n
Pm2 R 1 P 1 q m2 i¼1 di k 1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ 1 k : Bk;t ðkðfn þ wÞÞ ¼ Pm2 R 1 Pm2 D c k 1 Gðg ; sÞ f s; ½xt ðsÞ kxðsÞ þ 1 þ wðsÞ ds c aðg Þ i¼1
i
i
k
i¼1
n
i
i
Though calculation, we obtain
Z P P m2 1 q m2 að1Þ 1 i¼1 ci i¼1 ci bðgi Þ jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjaðtÞ 6 Gðs; sÞ f s; ½xt ðsÞ kxðsÞ þ k Pm2 1 n D Pm2 di i¼1 i¼1 di bðgi Þ k 1 þ f s; ½xðsÞ kxðsÞ þ þ 2wðsÞ ds n Z 1 1 1 þ h ½xt ðsÞ kxðsÞ þ Gðs; sÞ /ðsÞ g ½xt ðsÞ kxðsÞ þ 6 Ak 1 n n k 1 1 þ h ½xðsÞ kxðsÞ þ þ wðsÞ ds þ wðsÞ þ /ðsÞ g ½xðsÞ kxðsÞ þ n n Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1: ð3:11Þ 6 2Aað1ÞkðSr0 ;n þ 1Þ 0
In the same way, we get
jBk;t ðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbðtÞ 6 2Bbð0ÞkðSr0 ;n þ 1Þ
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1:
ð3:12Þ
0
From (3.10), (3.12), (3.13), for any
e > 0, by ðH3 Þ, there exists a sufficiently large A0 ðA0 > 1=kÞ, such that
Z o max 1; að1ÞA; bð0ÞB kðSr0 ;n þ 1Þ n
1
A0
Gðs; sÞð/ðsÞ þ wðsÞÞds
0, there exists a d > 0 such that for any s 2 ½1=k; A0 and u; t 2 ½0; r 0 , when ju tj ¼ jðu þ 1nÞ ðt þ 1nÞj < d, we have
!1 n o Z A0 f s; u þ 1 f s; t þ 1 < e max 1; að1ÞA; bð0ÞB k Gðs; sÞds : 1 n n 6 k
ð3:14Þ
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From kxt xk ! 0 ðn ! þ1Þ and the definition of the norm k k in the space X, for the above d > 0, there exists a sufficiently large nature number V 0 such that when t > V 0 , for all s 2 ½1=k; A0 , we have
½xt ðsÞ kxðsÞ þ 1 ½xðsÞ kxðsÞ þ 1 n n jxt ðsÞ kxðsÞj þ xt ðsÞ kxðsÞ jxðsÞ kxðsÞj þ xðsÞ kxðsÞ 6 2 2 jxt ðsÞ kxðsÞj jxðsÞ kxðsÞj xt ðsÞ xðsÞ ¼ þ 2 2
ð3:15Þ
6 jxt ðsÞ xðsÞj 6 kxt xk < d: Hence, by (3.14)–(3.16), when
t > V 0 , we have the following inequality
jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjaðtÞ 6 jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjað1Þ Pm2 Pm2 að1Þ F 1 q i¼1 ci bðgi Þ H1 q i¼1 ci bðgi Þ ¼ Pm2 Pm2 H2 D F 2 i¼1 di bðgi Þ i¼1 di bðgi Þ Z A0 1 1 6 að1ÞAk f s; ½x kxðsÞ þ Gðs; sÞf s; ½xt ðsÞ kxðsÞ þ ds 1 n n k Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞds þ 2að1ÞAðSr0 ;n þ 1Þk 6
e 3
ð3:16Þ
A0
;
where
!
1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ þ wðsÞ ds; 1 n A0 i¼1 k ! Z Z m2 A0 1 X 1 Gðgi ; sÞ f s; ½xt ðsÞ kxðsÞ þ þ wðsÞ ds; F 2 ¼ di k þ 1 n A0 i¼1 k ! Z A0 Z 1 m2 X 1 H1 ¼ þ wðsÞ ds; ci k þ Gðgi ; sÞ f s; ½x kxðsÞ þ 1 n A0 i¼1 k ! Z Z m2 A0 1 X 1 H2 ¼ di k þ wðsÞ ds: þ Gðgi ; sÞ f s; ½x kxðsÞ þ 1 n A0 i¼1 k F1 ¼
m 2 X
ci k
Z
A0
þ
Z
1
Using the same method as (3.17), when
t > V 0 , we get
jBk;t ðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbðtÞ 6
e 3
ð3:17Þ
:
Then, by (3.17) and (3.18) and the above discussion, when
t > V 0 , we obtain
jðT n;k xt ÞðtÞ ðT n;k xÞðtÞj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xt ðsÞ kxðsÞ þ ¼ k 1 n k
þAk;t ðkðfn þ wÞÞaðtÞ þ Bk;t ðkðfn þ wÞÞbðtÞ Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k Ak ðkðfn þ wÞÞaðtÞ Bk ðkðfn þ wÞÞbðtÞj Z A0 1 1 f s; ½xðsÞ kxðsÞ þ 6k Gðs; sÞf s; ½xt ðsÞ kxðsÞ þ ds 1 n n k þ jAk;t ðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjað1Þ þ jBk;t ðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbð0Þ Z 1 1 1 þ f s; ½xðsÞ kxðsÞ þ þ 2wðsÞ ds Gðs; sÞ f s; ½xt ðsÞ kxðsÞ þ þk n n A0 Z 1 5e 6 þ 2kðSr0 ;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < e: 6 0 This implies that the operator T n;k : K ! K is continuous for any natural numbers n; k.
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It what follows, we need to prove that T n;k : K ! K is a compact operator for natural numbers n; k. Let M be any bounded subset of K. Then there exists a constant R > 0 such that kxk 6 R for any x 2 M. By (3.9), ðH2 Þ and ðH3 Þ, for any x 2 M, we have
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k Z 1 1 6k Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 1 n k 1 þ wðsÞ ds þh ½xðsÞ kxðsÞ þ n Z 1 Gðs; sÞð/ðsÞSR;n þ wðsÞÞds 6k 1 k
6 kðSR;n þ 1Þ
Z
ð3:18Þ
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1;
0
where SR;n :¼ supfgðuÞ þ hðuÞ :
1 n
6 u 6 R þ 1g. By the similar proof as (3.2) and (3.3), for any t 2 ½0; þ1Þ, we have
Z 1 Ak ðkðfn þ wÞÞaðtÞ 6 Aað1ÞkðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1; Z 01 Bk ðkðfn þ wÞÞbðtÞ 6 Bbð0ÞkðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1:
ð3:19Þ
0
Therefore, from (3.18) and (3.19), T n;k M is bounded in K. Given a > 0, for any x 2 M and t; t0 2 ½0; a, by (3.9), we get
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k
1 k þ wðsÞ ds Gðt ; sÞ f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 1 þ wðsÞ ds 6k jGðt; sÞ þ Gðt 0 ; sÞj f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 2kðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1 Z
1
0
1 k
and so, for any
e0 > 0, we can find a sufficiently large H0 ðH0 > 1kÞ such that
kðSR;n þ 1Þ
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds
0 such that for any
!1
H0
ð/ðsÞ þ wðsÞÞds
:
1 k
Therefore, we get
Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k
1 k þ wðsÞ ds Gðt ; sÞ f s; ½xðsÞ kxðsÞ þ 1 n k Z H0 1 0 6k þ wðsÞ ds jGðt; sÞ Gðt ; sÞj f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 1 þ wðsÞ ds jGðt; sÞ Gðt 0 ; sÞj f s; ½xðsÞ kxðsÞ þ þk n H0 Z 1 0 0 e e Gðs; sÞð/ðsÞ þ wðsÞÞds < : 6 þ 2kðSR;n þ 1Þ 1 6 3 k Z
1
0
ð3:20Þ
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Also by the uniformly continuity of aðtÞ; bðtÞ on ½0; a, for the above e0 > 0, there exists d00 > 0 such that for any t; t 0 2 ½0; a and jt t0 j < d00 , we have 0
jaðtÞ aðt Þj < 0
jbðtÞ bðt Þj
0, let d0 ¼ minfd0 ; d00 g, then for any t; t0 2 ½0; a with jt t0 j < d0 , and for any x 2 M, we have 0
jT n;k xðtÞ T n;k xðt0 Þj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ ¼ k 1 n k
þAk ðkðfn þ wÞÞaðtÞ þ Bk ðkðfn þ wÞÞbðtÞ Z 1 1 þ wðsÞ ds Gðt 0 ; sÞ f s; ½xðsÞ kxðsÞ þ k 1 n k Ak ðkðfn þ wÞÞaðt 0 Þ Bk ðkðfn þ wÞÞbðt0 Þj Z 1 1 þ wðsÞ ds 6k jGðt; sÞ Gðt 0 ; sÞj f s; ½xðsÞ kxðsÞ þ 1 n k Z 1 Gðs; sÞð/ðsÞ þ wðsÞÞdsjaðtÞ aðt 0 Þj þ AkðSR;n þ 1Þ 0 Z 1 þ BkðSR;n þ 1Þ Gðs; sÞð/ðsÞ þ wðsÞÞdsjbðtÞ bðt0 Þj 0
< e0 : So, fT n;k x : x 2 Mg are equicontinuous on ½0; a. Since a > 0 is arbitrary, fT n;k x : x 2 Mg are locally equicontinuous on ½0; þ1Þ. Let T n;k xðþ1Þ ¼ limt!þ1 T n;k xðtÞ, by a simple calculation, we can see that limt!þ1 T n;k xðtÞ < þ1, so we get
Z 1 1 þ wðsÞ ds þ Ak ðkðfn þ wÞÞjaðtÞ að1Þj jT n;k xðtÞ T n;k xðþ1Þj 6 k ðGðt; sÞ GðsÞÞ f s; ½xðsÞ kxðsÞ þ 1 n k
þ Bk ðkðfn þ wÞÞjbðtÞ bð1Þj: By the similar method as for (3.20), we get that, for any
e > 0, there exists N0 such that, when t > N0 , it is true that
Z 1 1 e þ wðsÞ ds < : k ðGðt; sÞ GðsÞÞ f s; ½xðsÞ kxðsÞ þ 3 1 n k
Together with the continuity of aðtÞ; bðtÞ on ½0; þ1Þ, we get that for the above e > 0, there exists N 0 such that, when t > N 0 , we have jT n;k xðtÞ T n;k xðþ1Þj < e. Hence, the functions fT n;k x : x 2 Mg are equiconvergent at þ1, which implies that fT n;k x : x 2 Mg is relatively compact (by Lemma 2.5). Therefore, we get that the operator T n;k : K ! K is completely continuous for natural numbers n; k. Finally, we show that T n : K ! K is a completely continuous operator. For any t 2 ½0; þ1Þ and x 2 S ¼ fx 2 K : kxk 6 1g, by (2.8) and (3.9), we have
Z k 0
1 k
Z 1 k 1 1 1 þ wðsÞ ds 6k þ h ½xðsÞ kxðsÞ þ Gðt; sÞ f s; ½xðsÞ kxðsÞ þ Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ n n n 0 Z 1 k Gðs; sÞð/ðsÞS1;n þ wðsÞÞds ! 0; k ! þ1; ð3:22Þ þwðsÞÞds 6 k 0
where S1;n :¼ supfgðuÞ þ hðuÞ :
1 n
6 u 6 2g < þ1. And then, we get
Z 1 P P m2 k q m2 að1Þ 1 i¼1 ci i¼1 ci bðgi Þ jAðkðfn þ wÞÞ Ak ðkðfn þ wÞÞjaðtÞ 6 Gðs; sÞ f s; ½xðsÞ kxðsÞ þ k Pm2 0 n D Pm2 di i¼1 i¼1 di bðgi Þ 1 þ 2wðsÞ ds þ f s; ½xðsÞ kxðsÞ þ n Z 1 k Gðs; sÞð/ðsÞ þ wðsÞÞds ! 0; k ! þ1: 6 2AkðS1;n þ 1Það1Þ 0
ð3:23Þ
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Using the similar method, we also get
jBðkðfn þ wÞÞ Bk ðkðfn þ wÞÞjbðtÞ 6 2BkðS1;n þ 1Þbð0Þ
Z
1 k
Gðs; sÞð/ðsÞ þ wðsÞÞds ! 0;
k ! þ1:
ð3:24Þ
0
Together (3.22)–(3.24) imply that
kT n T n;k k ¼ supkT n x T n;k xk ! 0;
k ! þ1:
x2S
Therefore, by T n;k : K ! K is a completely continuous operator, we get that T n : K ! K is a completely continuous operator. h Theorem 3.1. Assume that ðH1 Þ-ðH3 Þ hold and f satisfies the following condition: ðH4 Þ There exists ½a; b ð0; þ1Þ such that
lim min
u!þ1 t2½a;b
f ðt; uÞ ¼ þ1: u
Then there exists k > 0 such that BVP (1.1) has at least one positive solution for any k 2 ð0; kÞ. n Proof. Choose r1 > max 4;
4d
q
R1 0
o wðsÞds , where d is defined as (2.3), let
8 9 < = r1
R
k ¼ min 1; ; 1 : 1 þ að1ÞA þ bð0ÞB 0 Gðs; sÞ /ðsÞ g ðfðsÞÞ þ hðr 1 Þ þ wðsÞ ds;
¼ sup06u6rþ1 hðuÞ; A and B are defined by (3.2) and (3.3). Let K r1 ¼ fx 2 K : kxk < r1 g. For any x 2 @K 1 ; t 2 ½0; þ1Þ, where hðrÞ by the definition of k k and Lemma 2.3, we have
½xðtÞ kxðtÞ 6 xðtÞ 6 kxk 6 r 1 ; xðtÞ kxðtÞ P xðtÞ
kdfðtÞ
q
Z
1
0
wðsÞds P xðtÞ
dfðtÞ
q
Z
1
wðsÞds P xðtÞ 0
2dxðtÞ qr1
Z
1
wðsÞds P 0
xðtÞ fðtÞr 1 P P fðtÞ: 2 4 ð3:25Þ
For any k 2 ð0; kÞ, we have
jðT n xÞðtÞj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ ¼k n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 1 þ h ½xðsÞ kxðsÞ þ þ wðsÞ ds Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 6k n n 0 þ Aðkðfn þ wÞÞað1Þ þ Bðkðfn þ wÞÞbð0Þ
Z 1
6 k 1 þ Aað1Þ þ Bbð0Þ Gðs; sÞ /ðsÞ g ðfðsÞÞ þ hðr 1 Þ þ wðsÞ ds 0
6 r1 : Thus,
kT n xk 6 kxk;
for any x 2 @K r1 :
On the other hand, by the inequality in ðH4 Þ, choose l such that khslf such that
f ðt; uÞ P lu; Let
r 2 > max
n
ð3:26Þ Rb a
Gðs; sÞds > 4; s ¼ mint2½a;b sðtÞ, then there exists N > 0
u P N ; t 2 ½a; b:
r1 ; 4Nf
o ; f ¼ mint2½a;b fðtÞ; K r2 ¼ fx 2 K : kxk < r 2 g.
Take
q1 1 2 @K 1 ; K 1 ¼ fx 2 K : kxk < 1g.
For
any
x 2 @K r2 ; l > 0; n 2 N, we will show
x – T n x þ lq1 :
ð3:27Þ
Otherwise, there exist x0 2 @K r2 and l0 > 0, such that x0 ¼ T n x0 þ l0 q1 . From x0 2 @K r2 , we know that kx0 k ¼ r 2 . Then, for t 2 ½a; b, we have Z Z Z kdfðtÞ 1 dfðtÞ 1 2dx0 ðtÞ 1 x0 ðtÞ fðtÞr2 fr2 x0 ðtÞ kxðtÞ P x0 ðtÞ wðsÞds P x0 ðtÞ wðsÞds P x0 ðtÞ wðsÞds P P P P N : q 0 q 0 qr2 0 4 4 2
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Hence, we conclude that
x0 ðtÞ ¼ k
Z
1
0
1 þ wðsÞ ds Gðt; sÞ f s; ½x0 ðsÞ kxðsÞ þ n
þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ þ l0 Z 1 1 þ wðsÞ ds þ l0 Pk hsðtÞGðs; sÞ f s; ½x0 ðsÞ kxðsÞ þ n 0 Z b lfr2 Pk hsðtÞGðs; sÞ ds þ l0 4 a Z khslfr2 b P Gðs; sÞds þ l0 4 a P r 2 þ l0 > r 2 : This implies that r 2 > r2 , which is a contradiction. This yields that (3.27) holds. It follows from the above discussion, (3.26) and (3.27), Lemmas 2.6 and 3.1, that for any n 2 N; k 2 ð0; kÞ; T n has a fixed point xn 2 K r2 n K r1 . Let fxn g1 n¼1 be the sequence of solutions of BVP (2.7). It is easy to see that they are uniformly bounded. From xn 2 K r 2 n K r 1 , by the similar discussion as (3.25), we know that
½xn ðtÞ kxðtÞ 6 xn ðtÞ 6 kxn k 6 r2 ; t 2 ½0; þ1Þ; Z Z kdfðtÞ 1 dfðtÞ xn ðtÞ kxðtÞ P xn ðtÞ wðsÞds P xn ðtÞ
q
P fðtÞ;
q
0
1
wðsÞds P xn ðtÞ
0
2dxn ðtÞ qr1
Z
1
wðsÞds P 0
t 2 ½0; þ1Þ:
xn ðtÞ 2 ð3:28Þ
R1
0 0 Next, given a0 > 0, we will prove that fxn g1 n¼1 are equicontinuous on ½0; a . For any e > 0, by k 0 Gðs; sÞð/ðsÞðgðfðsÞÞþ hðr 2 ÞÞ þ wðsÞÞds < þ1, where hðr 2 Þ ¼ supfhðuÞ : 0 6 u 6 r 2 þ 1g, we can find a sufficiently large H0 > 0 such that
Z
1
k
Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞds
0 such that for any
:
0
Therefore, for any n 2 N; t; t0 2 ½0; a; s 2 ½0; H0 and jt t 0 j < d0 , we get
Z k
1 þ wðsÞ ds Gðt; sÞ f s; ½xn ðsÞ kxðsÞ þ n 0 Z 1 1 k þ wðsÞ ds Gðt 0 ; sÞ f s; ½xn ðsÞ kxðsÞ þ n 0 Z H0 1 6k þ wðsÞ ds jGðt; sÞ Gðt 0 ; sÞj f s; ½xn ðsÞ kxðsÞ þ n 0 Z 1 1 0 þk þ wðsÞ ds jGðt; sÞ Gðt ; sÞj f s; ½xn ðsÞ kxðsÞ þ n H0 Z H0
jGðt; sÞ Gðt 0 ; sÞj /ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞ ds 6k 0Z 1 jGðt; sÞ Gðt 0 ; sÞjð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞds þk H0 Z 1
e0 e0 6 þ 2k Gðs; sÞ /ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞ ds < : 6 3 0 1
ð3:29Þ
For
Pm2 R 1 P 1 q m2 1 i¼1 ci k 0 Gðgi ; sÞ f s; ½xn ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 ci bðgi Þ An ðkðfn þ wÞÞ ¼ Pm2 R 1 ; P m2 D di k Gðg ; sÞ f s; ½xn ðsÞ kxðsÞ þ 1 þ wðsÞ ds di bðg Þ i¼1
0
i
n
i
i¼1
Pm2 R 1 P 1 q m2 1 i¼1 di k 0 Gðgi ; sÞ f s; ½xn ðsÞ kxðsÞ þ n þ wðsÞ ds i¼1 di aðgi Þ Bn ðkðfn þ wÞÞ ¼ Pm2 R 1 ; P m2 D c k Gðg ; sÞ f s; ½xn ðsÞ kxðsÞ þ 1 þ wðsÞ ds c aðg Þ i¼1
i
0
i
n
i¼1
i
i
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through a simple calculation, we get
An ðkðfn þ wÞÞ 6 Ak
Z
1
Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞds < þ1;
0
Bn ðkðfn þ wÞÞ 6 Bk
Z
1
Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr2 ÞÞ þ wðsÞÞds < þ1;
0
where A and B are defined as (3.2) and (3.3). So by the uniformly continuity of aðtÞ; bðtÞ on ½0; a0 , for the above exists d00 > 0 such that, for any t; t 0 2 ½0; a0 and jt t0 j < d00 , we have
jaðtÞ aðt 0 Þj < 0
jbðtÞ bðt Þj
0, there
ð3:30Þ
0
Then, by (3.29) and (3.30), for the above
e0 > 0, let d0 ¼ minfd0 ; d00 g, then for any n 2 N; t; t0 2 ½0; a0 and jt t0 j < d0 , we have
jxn ðtÞ xn ðt 0 Þj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xn ðsÞ kxðsÞ þ ¼ k n 0
þAn ðkðfn þ wÞÞaðtÞ þ Bn ðkðfn þ wÞÞbðtÞ Z 1 1 k þ wðsÞ ds Gðt 0 ; sÞ f s; ½xn ðsÞ kxðsÞ þ n 0 An ðkðfn þ wÞÞaðt 0 Þ Bn ðkðfn þ wÞÞbðt0 Þj Z 1 1 þ wðsÞ ds 6k jGðt; sÞ Gðt 0 ; sÞj f s; ½xn ðsÞ kxðsÞ þ n 0 Z 1 Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr2 ÞÞ þ wðsÞÞdsjaðtÞ aðt0 Þj þ Ak 0 Z 1 þ Bk Gðs; sÞð/ðsÞðgðfðsÞÞ þ hðr 2 ÞÞ þ wðsÞÞdsjbðtÞ bðt 0 Þj 0
< e0 : 1 0 0 So, fxn g1 n¼1 are equicontinuous on ½0; a . Since a > 0 is arbitrary, fxn gn¼1 are locally equicontinuous on ½0; þ1Þ. Let xn ðþ1Þ ¼ limt!þ1 xn ðtÞ, by a simple calculation, we can see that limt!þ1 xn ðtÞ < þ1, and so we get
jxn ðtÞ xn ðþ1Þj Z 1 1 þ wðsÞ ds ðGðt; sÞ GðsÞÞ f s; ½xn ðsÞ kxðsÞ þ 6 k n 0 þ An ðkðfn þ wÞÞjaðtÞ að1Þj þ Bn ðkðfn þ wÞÞjbðtÞ bð1Þj: By the similar method as for (3.29), we get that, for any
Z k
1
0
e0 > 0, there exists N0 such that, when t > N0 , it follows
e0 1 þ wðsÞ ds < : ðGðt; sÞ GðsÞÞ f s; ½xn ðsÞ kxðsÞ þ n 3
Together with the continuity of aðtÞ; bðtÞ on ½0; þ1Þ, we get that for the above e0 > 0, there exists N 0 such that, when t > N 0 , 1 we have jxn ðtÞ xn ðþ1Þj < e0 . Hence, the functions from fxn g1 n¼1 are equiconvergent at þ1, which implies that fxn gn¼1 is 1 relatively compact (by Lemma 2.5). Therefore, the sequence fxn gn¼1 has a subsequence being uniformly convergent on ½0; þ1Þ. Without loss of generality, we still assume that fxn g1 n¼1 itself uniformly converges to x on ½0; þ1Þ. Since fxn g1 2 K n K K, we have x P 0. By (2.7), we have r2 r1 n n¼1
Z t Z s 0 1 1 1 p ð1Þx0n ð1Þ þ x0n t ds d1 1 1 2 2 2 pð1Þ 2 2 Z t Z s f 1; ½xn ð1Þ kxð1Þ þ 1n þ wð1Þ k ds d1; 1 1 pð1Þ 2 2
xn ðtÞ ¼ xn
t 2 ð0; þ1Þ:
ð3:31Þ
1
As fx0n ð12Þgn¼1 is bounded, without loss of generality, we may assume x0n ð12Þ ! c0 as n ! þ1. Then, by (3.31) and the Lebesgue dominated convergence theorem, we have
Z t Z s 0 Z t Z s 1 1 p ð1Þx0 ð1Þ ðf ð1; ½xð1Þ kxð1Þ Þ þ wð1ÞÞ þ c0 t xðtÞ ¼ x ds d1 k ds d1; 1 1 1 1 2 2 pð1Þ pð1Þ 2 2 2 2
t 2 ð0; þ1Þ:
ð3:32Þ
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By (3.32), the direct computation shows that 0
ðpðtÞx0 ðtÞÞ þ kðf ðt; ½xðtÞ kxðtÞ Þ þ wðtÞÞ ¼ 0;
t 2 ð0; þ1Þ:
On the other hand, let n ! þ1 in the following boundary conditions:
a1 xn ð0Þ b1 limþ pðtÞx0n ðtÞ ¼
m2 X
t!0
ci xn ðgi Þ;
i¼1 m2 X
a2 t!þ1 lim xn ðtÞ þ b2 lim pðtÞx0n ðtÞ ¼ t!þ1
di xn ðgi Þ:
i¼1
Therefore, we deduce that x is a solution of BVP (2.7). Let xðtÞ ¼ xðtÞ kxðtÞ. By (3.28) and the convergence of the sequence fxn g1 n¼1 , we have xðtÞ P fðtÞ > 0; t 2 ½0; þ1Þ. It then follows from Lemma 2.4 that x is a positive solution of BVP (1.1). The proof is completed. h Remark 3.1. From the proof of Theorem 3.1, we get the main results under the condition that the function f ðt; uÞ not only has singularity on t but also has singularity on u. In addition, we obtain the positive solution of BVP (1.1) under the condition that the function f ðt; uÞ is semipositive, which are the great improvement of [23–25]. Also our method is different from those in [26–29]. Theorem 3.2. Assume that ðH1 Þ–ðH3 Þ hold, f and h satisfy the following condition: ðH5 Þ There exists ½c; d ð0; þ1Þ, such that
lim inf min f ðt; uÞ >
4d
0
wðsÞds
-¼
Rd c
lim
;
qhs0 -
u!þ1 t2½c;d
where s0 ¼ mint2½c;d sðtÞ; k 2 ðk; þ1Þ.
R1
u!þ1
hðuÞ ¼ 0; u
Gðs; sÞds. Then there exists k > 0 such that BVP (1.1) has at least one positive solution for any
Proof. By the first inequality of ðH5 Þ, we have that there exists N > 0 such that for any t 2 ½a; b; u > N , we have
f ðt; uÞ >
4d
R1
wðsÞds
0
qhs0 -
ð3:33Þ
:
Select
k¼
N q ; R1 df 0 wðsÞds
where f0 ¼ min fðtÞ:
0
t2½c;d
In the following proof, we suppose k > k. Let
R1 ¼
4kd
R1 0
wðsÞds
q
:
Assume K R1 ¼ fx 2 K : kxk < R1 g. For any x 2 @K R1 ; t 2 ½c; d,
xðtÞ kxðtÞ P P P
4kdfðtÞ 2q Z kdfðtÞ
q
Z
fðtÞR1 kdfðtÞ 2 q
1
wðsÞds
kdfðtÞ
q
0 1
wðsÞds P
0
Z
kdf0
q
Z
1
wðsÞds 0
Z
1
wðsÞds 0 1
wðsÞds P N :
0
Then, by (3.33), we have
jðT n xÞðtÞj ¼ k
Z
1
0
1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ n
þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 þ wðsÞ ds hsðtÞGðs; sÞ f s; ½xðsÞ kxðsÞ þ Pk n 0 R Z d 1 4d 0 wðsÞds P khs0 Gðs; sÞ ds qhs0 c Z d R1 Gðs; sÞds: P
-
c
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Noticing that
-¼
Rd c
kT n xk P kxk;
271
Gðs; sÞds, we have
for any x 2 @K R1 :
ð3:34Þ
On the basis of the second inequality in ðH5 Þ and the continuity of hðuÞ on ½0; þ1Þ, we have
lim
u!þ1
For
hðuÞ ¼ 0: u (
Z c ¼ max 1; 4kð1 þ Aað1Þ þ Bbð0ÞÞ
1
1 ) ; Gðs; sÞð/ðsÞ þ wðsÞÞds
0 00
00
there exists N > 0, such that when x P N , for any 0 6 y 6 x, we have hðyÞ 6 cx. Select
Z 00 R2 P 2; R1 ; N ; 2kð1 þ Aað1Þ þ Bbð0ÞÞ
0
1
fðsÞ ds : Gðs; sÞ/ðsÞg 2
Then, for any x 2 @K R2 ; t 2 ½0; þ1Þ, we have
½xðtÞ kxðtÞ 6 xðtÞ 6 kxk 6 R2 ; Z Z kdfðtÞ 1 2kdxðtÞ 1 xðtÞ fðtÞR2 fðtÞ xðtÞ kxðtÞ P xðtÞ wðsÞds P xðtÞ wðsÞds P P P > 0: 2 2 q qR2 0 4 0 Hence, we gain
jðT n xÞðtÞj Z 1 1 þ wðsÞ ds Gðt; sÞ f s; ½xðsÞ kxðsÞ þ ¼k n 0 þ Aðkðfn þ wÞÞaðtÞ þ Bðkðfn þ wÞÞbðtÞ Z 1 1 1 þ h ½xðsÞ kxðsÞ þ þ wðsÞ ds Gðs; sÞ /ðsÞ g ½xðsÞ kxðsÞ þ 6k n n 0 þ Aðkðfn þ wÞÞað1Þ þ Bðkðfn þ wÞÞbð0Þ
Z 1 fðsÞ þ cðR2 þ 1Þ þ wðsÞ ds Gðs; sÞ /ðsÞ g 6 k 1 þ Aað1Þ þ Bbð0Þ 2 0 Z
1 fðsÞ ds Gðs; sÞ/ðsÞg 6 k 1 þ Aað1Þ þ Bbð0Þ 2 0 Z
1 Gðs; sÞð/ðsÞ þ wðsÞÞds 6 R2 : þ kc 1 þ Aað1Þ þ Bbð0Þ ðR2 þ 2Þ 0
Thus,
kT n xk 6 kxk;
for any x 2 @K R2 :
ð3:35Þ
It follows from the above discussion, (3.34) and (3.35), Lemmas 2.7 and 3.1, for n 2 N; k 2 ðk; þ1Þ; T n has a fixed point xn 2 K R2 n K R1 satisfying R1 6 kxn k 6 R2 . The rest of proof is similar to Theorem 3.1. That’s the proof of Theorem 3.2. h Corollary 3.1. The conclusion of Theorem 3.2 is valid if ðH5 Þ is replaced by: ðH5 Þ There exists ½c; d ð0; þ1Þ, such that
lim inf minf ðt; uÞ ¼ þ1; u!þ1 t2½c;d
lim
u!þ1
hðuÞ ¼ 0: u
Remark 3.2. From Theorems 3.1 and 3.2, we get the positive solution of BVP (1.1) when the parameter k is sufficiently large or small, and the solution x in BVP (1.1) satisfies xðtÞ > 0 for any t 2 ½0; þ1Þ. 4. Example Example 4.1. Consider the following BVP
8 1 1 0 2 0 > 2 1 > ¼ 0; t 2 ð0; þ1Þ; ðð1 þ tÞ x ðtÞÞ þ k þ x > 2 1 x 1þt > t 2 þt 2 > < xð0Þ limþ ð1 þ tÞ2 x0 ðtÞ ¼ 13 x 13 ; > t!0 > > > > : lim xðtÞ þ lim ð1 þ tÞ2 x0 ðtÞ ¼ 1 x 1 : 6 3 t!þ1
t!þ1
ð4:1Þ
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1 1 It is easy to see that a1 ¼ a2 ¼ b1 ¼ b2 ¼ 1, pðtÞ ¼ ð1 þ tÞ2 ; aðtÞ ¼ 2 1þt ; bðtÞ ¼ 1 þ 1þt , q ¼ 3; q 13 b 13 ¼ 29 , q 16 a 13 ¼ 67 12 24 and
29 q 1 b 1 1 a 13 3 3 3 ¼ 12 D ¼ 1 1 6b 3 q 16 a 13 247 As f ðt; uÞ ¼
1
wðtÞ ¼
x
þ x2
1 1 2
t þt
2
1 1þt 2
1
;
/ðtÞ ¼
1
t 2 þt 2
5 12 67 24
¼
53 > 0: 8
, we can suppose
1 ; 1 þ t2
gðuÞ ¼
1 ; u
hðuÞ ¼ u2 :
By direct calculation, we have
Z
1
wðsÞds < þ1;
0
lim inf min
u!þ1 t2½2;5
Z
1
Gðs; sÞð/ðsÞ þ wðsÞÞds < þ1;
0
f ðt; uÞ ¼ þ1: u
So all conditions of Theorem 3.1 are satisfied. By Theorem 3.1, BVP (4.1) has at least one positive solution provided k is small enough. Acknowledgements The first and second authors were supported financially by the National Natural Science Foundation of China (11071141, 11371221), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001) and the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province. The third author was supported financially by the Australia Research Council through an ARC Discovery Project Grant. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26]
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