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Existence and unboundedness of positive solutions for singular boundary value problems on half-line
Applied Mathematics and Computation 144 (2003) 543–556 www.elsevier.com/locate/amc
Existence and unboundedness of positive solutions for singular boundary value problems on half-line q Yansheng Liu Department of Mathematics, Shandong Normal University, Jinan, 250014, PR China
Abstract By constructing a special cone and using cone compression and expansion fixed point theorem, this paper presents some existence results of positive solutions of singular boundary value problem (BVP, for short) on half-line for a class of second order differential equations. At the same time, the boundedness and unboundedness of positive solutions for BVP are considered also. Ó 2002 Elsevier Inc. All rights reserved. Keywords: Singularity; Cone; Positive solution; Boundedness; Half-line
1. Introduction This paper is concerned with the existence of single and multiple solutions of the following singular boundary value problems (BVPs) on half-line:
x00 ðtÞ þ f ðt; xðtÞÞ ¼ 0; t > 0; xð0Þ ¼ 0; x0 ðþ1Þ ¼ y1 P 0;
ð1:1Þ
where f 2 C½ð0; þ1Þ ð0; þ1Þ; Rþ , x0 ð1Þ ¼ limt!þ1 x0 ðtÞ, Rþ ¼ ½0; þ1Þ. q The project supported by the National Natural Science Foundation of PR China (10171057), Natural Science Foundation of Shandong Province (Z2000A02) and Rewarded Fund of Outstanding Middle and Young Scientists of Shandong Province. E-mail address: [email protected] (Y. Liu).
0096-3003/02/$ - see front matter Ó 2002 Elsevier Inc. All rights reserved. doi:10.1016/S0096-3003(02)00431-9
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
In recent years, singular BVPs on finite interval are extensively studied (see [1–5] and referees therein, for example). Only a small quantity of papers had discussed such question on half-line (see [6,7], for instance), in which it is considered simply the case that f ðt; xÞ is singular at t ¼ 0, not singular at x ¼ 0. To our knowledge, there is no paper to consider BVP (1.1) when f ðt; xÞ is singular at x ¼ 0 and also no result is available about the existence of multiple positive unbounded solutions for such system. As a result, the goal of present paper is to fill the gap in this area. The main features here are as follows. Firstly, f ðt; xÞ is singular both at t ¼ 0 and x ¼ 0. Secondly, the existence of multiple solutions of BVP (1.1) is considered. Thirdly, it is easy to see the solution of BVP (1.1) is unbounded as y1 > 0. But what about when y1 ¼ 0? We will show that the result depends upon f ðt; xÞ. Some sufficient conditions are given to guarantee the boundedness or unboundedness of solutions for BVP (1.1) when y1 ¼ 0. Generally speaking, the approximation method is often used to study singular problem. However, in present paper, the approximation problem is not needed. The main techniques used here are a new constructed cone and cone compression and expansion fixed point theorem. Let FC½Rþ ; R ¼:
x 2 C½Rþ ; R : sup t2Rþ
jxðtÞj < þ1 : 1þt
ð1:2Þ
Then FC½Rþ ; R is a Banach space with norm kxkF ¼: supt2Rþ ðjxðtÞj= 1 þ tÞ. In present paper, FC½Rþ ; R will be the basic space to study (1.1). The paper is organized as follows. Section 2 is devoted to the existence of positive solutions for BVP (1.1). Section 3 gives some sufficient conditions which guarantees the solution of BVP (1.1) to be bounded or unbounded as y1 ¼ 0. Finally in this section we state a result from the literature which will be used in Section 2. Lemma 2.4 ([8] fixed point theorem of cone expansion and compression of norm type). Let X1 and X2 be two bounded open sets in Banach space E such that h 2 X1 and X1 X2 . Let operator A : P \ ðX2 n X1 Þ ! P be completely continuous, where h denotes the zero element of E and P is a cone of E. Suppose that one of the two conditions (i) kAxk 6 kxk, 8x 2 P \ oX1 and kAxk P kxk, 8x 2 P \ oX2 and (ii) kAxk P kxk, 8x 2 P \ oX1 and kAxk 6 kxk, 8x 2 P \ oX2 is satisfied. Then A has at least one fixed point in P \ ðX2 n X1 Þ.
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
545
2. Existence results For convenience, we first list some conditions. (H1 ) There exist k; l such that 1 < k < 0 < l < þ1 and N ; M > 0 such that 0 < N 6 1 6 M satisfying cl f ðt; xÞ 6 f ðt; cxÞ 6 ck f ðt; xÞ; ck f ðt; xÞ 6 f ðt; cxÞ 6 cl f ðt; xÞ; R1 (H2 ) 0 < 0 s1þk f ðs; 1Þds < þ1;
0 < c 6 N; c P M: R þ1 l 0 < 1 ð1 þ sÞ f ðs; 1Þds < þ1:
Pn Remark. The typical function satisfying (H1 ) is f ðt; xÞ ¼ i¼1 ai ðtÞxki , where k ¼ k1 6 k2 6 6 kk < 0 < kkþ1 6 6 kn ¼ l, ai 2 C½ð0; þ1Þ; Rþ (i ¼ 1; 2; . . . ; n; k ¼ 1; 2; . . . ; n 1). For the sake of investigating the existence of positive solutions let P ¼ f x 2 FC½Rþ ; Rþ : xðtÞ 8 txðsÞ > < ; t 2 ½0; 1 ; 1 þs P > : xðsÞ ; t 2 ½1; þ1Þ; 1þs
s 2 Rþ ;
9 > =
xðtÞ is nondecreasing on Rþ : > ; s 2 Rþ ; ð2:1Þ
It is easy to see P is a nonempty, convex and closed subset of FC½Rþ ; Rþ and furthermore, P is a cone of Banach space FC½Rþ ; R . Notice that if x 2 P n fhg, then xðtÞ > 0 for t 2 ð0; þ1Þ, where h denotes the zero element of FC½Rþ ; R . In addition notice that tkxkF ; t 2 ½0; 1 ; xðtÞ P ð2:2Þ kxkF ; t 2 ½1; þ1Þ; and xðtÞ 6 ð1 þ tÞkxkF 6 2kxkF ;
t 2 ½0; 1 :
ð2:3Þ
Now we are in position to give the following results. Lemma 2.1. Assume that (H1 ) and (H2 ) hold. Then x 2 FC½Rþ ; Rþ \ C 2 ½ð0; þ1Þ; Rþ is a positive solution of BVP (1.1) if and only if x 2 FC½Rþ ; Rþ is a positive solution of the following integral equation Z þ1 xðtÞ ¼ ty1 þ Gðt; sÞf ðs; xðsÞÞds; ð2:4Þ 0
where Gðt; sÞ ¼ minft; sg.
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Proof. First of all, we show for any given t 2 ð0; þ1Þ, x 2 P n fhg, R þ1 Gðt; sÞf ðs; xðsÞÞds is convergent. 0 Choose c1 > 0 such that 2c1 kxkF 6 N , 1=c1 > M. Thus, by (2.3), c1 xðsÞ 6 N for s 2 ½0; 1 . This together with (H1 ), (H2 ) and (2.2) yields that Z 1 Z 1 Z 1 1 Gðt; sÞf ðs; xðsÞÞds 6 sf ðs; xðsÞÞds 6 sf s; c1 xðsÞ ds c1 0 0 0
Z 1 Z 1 1 6 sck1 jxðsÞjk f s; sjxðsÞjk f ðs; 1Þds ds 6 c1kl c 1 0 0 Z 1 k kl 1þk 6 c1 s f ðs; 1Þds kxkF < þ1: ð2:5Þ 0
On the other hand, choose c2 > 0 sufficiently large that c2 kxkF P M, 1=c2 < N . So c2 xðsÞ P c2 kxkF P M for s P 1. From (H1 ) and (H2 ) one can obtain Z þ1 Z þ1 Z þ1 1 l f ðs; xðsÞÞds 6 f s; c2 xðsÞ ds 6 clk 2 jxðsÞj f ðs; 1Þds c2 1 1 1 Z þ1 lk 6 c2 ð1 þ sÞl f ðs; 1Þds kxklF < þ1: ð2:6Þ 1
R þ1 This together with (2.5) guarantees that 0 Gðt; sÞf ðs; xðsÞÞds is convergent for t 2 ð0; þ1Þ, and x 2 P n fhg. Next, suppose x 2 FC½Rþ ; Rþ is a positive solution of (2.4). Now we show x 2 P n fhg. R þ1Firstly, we claim limt!0þ xðtÞ ¼ 0. By (2.4) we need only to prove limt!0þ t t f ðs; xðsÞÞds ¼ 0. In this situation, notice that R d (2.5) holds. As a result, for any given e > 0, there exists d1 > 0 such that 0 1 sf ðs; xðsÞÞds < e=2. Let ( Z 1 ) þ1 d ¼ min d1 ; 2 f ðs; xðsÞÞds e : d1
Then Z t
þ1
f ðs; xðsÞÞds ¼ t
t
Z
Z 6
d1
f ðs; xðsÞÞds þ t
þ1
f ðs; xðsÞÞds
d1
t d1
sf ðs; xðsÞÞds þ d
t
6
Z
e e þ ¼ e; 2 2
Z
þ1
f ðs; xðsÞÞds
d1
8t 2 ð0; dÞ:
ð2:7Þ
Immediately it follows that limt!0þ xðtÞ ¼ 0. Secondly, noting that x is a positive solution of BVP (1.1), so, Z þ1 f ðs; xðsÞÞds P 0: ð2:8Þ x0 ðtÞ ¼ y1 þ t
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
This means that x is nondecreasing. Thirdly, notice 8t t > P ; 0 < t; s < s; > > 1þs >s > > > >t P t ; > 0 < t < s < s; Gðt; sÞ < s 1þs ¼ s > P 1 P 1 ; 0 < s < s < t; Gðs; sÞ > > > > s 1þs > > > > : s ¼ 1 P 1 ; t; s > s > 0: s 1þs Therefore, Z 1 xðtÞ ¼ ty1 þ Gðt; sÞf ðs; xðsÞÞds 0 8 Z þ1 t t > > > Gðs; sÞf ðs; xðsÞÞds; 0 < t 6 1; < 1 þ s sy1 þ 1 þ s 0 P Z þ1 > 1 1 > > sy1 þ Gðs; sÞf ðs; xðsÞÞds; t P 1; : 1þs 1þs 0
547
s 2 Rþ ; ð2:9Þ s 2 Rþ :
This together with (2.8) yields x 2 P n fhg. In addition, from (2.6) and (2.8) it follows that x 2 FC½Rþ ; Rþ \ 2 C ½ð0; þ1Þ; Rþ is a positive solution of BVP (1.1). Conversely, if x 2 FC½Rþ ; Rþ \ C 2 ½ð0; þ1Þ; Rþ is a positive solution of BVP(1.1) it is easy to see that x 2 FC½Rþ ; Rþ is a positive solution of (2.4). By (2.5) and (2.6), we can define an operator A on P n fhg by Z þ1 ðAxÞðtÞ ¼ ty1 þ Gðt; sÞf ðs; xðsÞÞds;
ð2:10Þ
0
where Gðt; sÞ is the same as in (2.4). Also from the proof of Lemma 2.1 the existence of positive solutions belonging to FC½Rþ ; Rþ \ C 2 ½ð0; þ1Þ; Rþ of BVP (1.1) is equivalent to the existence of fixed point of A on P n fhg. To obtain the complete continuity of A, the following lemmas are still need. Lemma 2.2. Let W be a bounded subset of P. Then W is relatively compact in FC½Rþ ; R if fW ðtÞ=1 þ tg are equicontinuous on any finite subinterval of Rþ and for any e > 0, there exists N > 0 such that xðt1 Þ xðt2 Þ < e; 1 þ t 1 þ t2 1 uniformly with respect to x 2 W as t1 ; t2 P N , where W ðtÞ ¼ fxðtÞ : x 2 W g, t 2 Rþ . This lemma is a simple improvement of the Corduneanu theorem in [9].
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Lemma 2.3. Suppose (H1 ) and (H2 ) hold. Then A: P R n Pr ! P is completely continuous, where R > r > 0 and Pr ¼ fx 2 P : kxkF < rg, PR is very similar to Pr . Proof. First of all, from the proof of Lemma 2.1 and 0
ðAxÞ ðtÞ ¼ y1 þ
Z
þ1
f ðs; xðsÞÞds P 0
for x 2 P n fhg; t 2 ð0; þ1Þ;
t
similar to (2.9), it follows that A : P n fhg ! P . Of course, A : P R n Pr ! P . In addition, choose c1 ; c2 > 0 such that 2c1 r 6 N ;
1 PM c1
and
1 6 N: c2
c2 R P M;
Combining with (2.2) and (2.3), similar to (2.5) and (2.6), it follows that Z
1
sf ðs; xðsÞÞds 6 c1kl 0 k 6 ckl 1 r
Z
1 k
s1þk f ðs; 1Þds kxkF 0
Z
1
s1þk f ðs; 1Þds < þ 1;
8x 2 P R n Pr
ð2:11Þ
0
and Z
Z þ1 f ðs; xðsÞÞds 6 clk ð1 þ sÞl f ðs; 1Þds kxklF 2 1 1 Z þ1 l l 6 clk R ð1 þ sÞ f ðs; 1Þds < þ1; 8x 2 P R n Pr : 2 þ1
ð2:12Þ
1
It follows from (2.10)–(2.12) that AðP R n Pr Þ is bounded in FC½Rþ ; Rþ . Now we are ready to show AV is relatively compact for any bounded subsets V of P R n Pr . To see this, by (2.10) and Lemma 2.2, we first prove limt!0þ ððAxÞðtÞ=1 þ tÞ ¼ 0 uniformly with respect to x 2 V . Notice that Z t Z 1 ðAxÞðtÞ ty1 1 t ¼ þ sf ðs; xðsÞÞds þ f ðs; xðsÞÞds 1þt 1þt t 1þt 1þt 0 Z þ1 t f ðs; xðsÞÞds; 8t 2 ð0; 1Þ; x 2 V : þ 1þt 1 From this R 1 together with (2.11) and (2.12), similar to (2.7), one can see limt!0þ t t f ðs; xðsÞÞds ¼ 0 uniformly with respect to x 2 V . As a result, limt!0þ ððAxÞðtÞ=1 þ tÞ ¼ 0 uniformly with respect to x 2 V .
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549
Secondly, notice Z ðAxÞðtÞ ðAxÞðt0 Þ t t0 þ1 0 f ðs; xðsÞÞds 1 þ t 1 þ t0 6 jt t jy1 þ 1 þ t 1 þ t0 t Z 0 Z 1 t 1 t f ðs; xðsÞÞds þ sf ðs; xðsÞÞds þ t 1 þ t 1 þ t0 0 Z 0 t sf ðs; xðsÞÞds; 8t; t0 2 ð0; þ1Þ; x 2 V : þ t Combining this with (2.11) and (2.12), it follows that fððAxÞðtÞ=1 þ tÞ : x 2 V g are equicontinuous on any finite closed interval of ð0; þ1Þ. Thirdly, it remains to show for any e > 0, there exists sufficiently large N > 0 such that ðAxÞðtÞ ðAxÞðt0 Þ 0 ð2:13Þ 1 þ t 1 þ t0 < e; 8t; t P N ; 8x 2 V : To see this, combining with (2.10) we need only to prove that for any given e > 0, there exists N > 0 such that Z Z t0 t s s f ðs; xðsÞÞds f ðs; xðsÞÞds < e; 8t; t0 P N ; 8x 2 V : 0 0 1þt 1 þ t 0 R þ1 Notice (2.12) guarantees that there exists L > 0 satisfying L f ðs; xðsÞÞds < e=4 uniformly with respect to x 2 V . By (2.11) and (2.12) we know Z L M ¼: sup sf ðs; xðsÞÞds < þ1: x2P R nPr
0
Then let N ¼ maxfL; ð4M=eÞg. It follows that Z Z t0 t s s f ðs; xðsÞÞds f ðs; xðsÞÞds 0 0 1þt 1 þ t 0
Z L Z t 1 1 s f ðs; xðsÞÞds 6 sf ðs; xðsÞÞds þ þ 1 þ t 1 þ t0 1 þ t 0 L Z t0 s þ f ðs; xðsÞÞds 1 þ t0 L e e e e 6 þ þ þ ¼ e; 8t; t0 P N ; 8x 2 V ; 4 4 4 4 that is, (2.13) holds. In conclusion, Lemma 2.2 guarantees that AV is a relatively compact subset of P.
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Finally, we are in position to show A : P R n Pr ! P is continuous. Suppose xn , x 2 P R n Pr and kxn xkF ! 0. Then, for any t > 0, limn!þ1 xn ðtÞ ¼ xðtÞ. This together with (2.11) and (2.12) and Lebesgue dominated theorem yields that ðAxn ÞðtÞ ðAxÞðtÞ ! ðn ! þ1Þ; 1þt 1þt
8t 2 ð0; þ1Þ:
ð2:14Þ
Notice from above that fAxn g is relatively compact also. Therefore, (2.14) guarantees that kAxn AxkF ! 0 ðn ! þ1Þ, that is, A: P R n Pr ! P is continuous. To sum up, the conclusion of Lemma 2.3 follows. The following theorem gives sufficient conditions to guarantee the existence of positive solutions for BVP (1.1). Theorem 2.1. Assume that (H1 ) and (H2 ) hold and l 2 ð0; 1Þ. Then BVP (1.1) has at least one positive solution. Proof. We first prove there exists a sufficiently small r > 0 such that kAxkF P kxkF ;
8x 2 oPr :
ð2:15Þ
In fact, if y1 > 0, it is obvious because we can chooseR r ¼ y1 =2. Now 1 consider the case y1 ¼ 0. Notice that (H2 ) guarantees that 0 s1þl f ðs; 1Þds is convergent. Let c ¼ 1=M and (
1=1l ) Z N 1 kl 1 1þl ; M r ¼ min s f ðs; 1Þds : 2c 2 0 Then sr ¼ skxkF 6 xðsÞ 6 2r;
8x 2 oPr ;
s 2 ½0; 1 ;
that is, cxðsÞ 6 2rc 6 N for x 2 oPr and s 2 ½0; 1 . By 1=c ¼ M together with (H1 ) and (H2 ), we know that Z ðAxÞðtÞ ðAxÞð1Þ 1 1 P P sf ðs; xðsÞÞds kAxkF ¼ sup 1þ1 2 0 t2Rþ 1 þ t
Z Z 1 1 1 1 1 lk l sf s; cxðsÞ ds P sc jxðsÞj f ðs; 1Þds P 2 0 c 2 0 Z 1 1 P M kl s1þl f ðs; 1Þds kxklF P r ¼ kxkF ; 8x 2 oPr 2 0 which implies that (2.15) holds.
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551
Next we show there exists a sufficiently large R > r such that kAxkF 6 kxkF ;
8x 2 oPR :
ð2:16Þ
Indeed, choose R1 > r satisfying 2R1 =N P M. Let c1 ¼ c1 ðR1 Þ ¼: ðN =2R1 Þ. Then 2c1 R1 ¼ N , 1=c1 P M. It follows from (2.11) that Z 1 Z 1 sf ðs; xðsÞÞds 6 c1kl Rk1 s1þk f ðs; 1Þds 0
0
kl Z 1 N ¼ Rk1 s1þk f ðs; 1Þds 2R1 0
kl Z 1 N Rl1 s1þk f ðs; 1Þds; 8x 2 oPR1 : ¼ 2 0
ð2:17Þ
On the other hand, let c2 ¼ 1=N and 8
kl Z 1 < N R ¼ max MN ; R1 ; 2y1 ; 2 s1þk f ðs; 1Þds : 2 0 9 1 ! Z þ1 1l = l þ 2N kl : ð1 þ sÞ f ðs; 1Þds ; 1 Then c2 R P M, 1=c2 6 N . It follows from (2.12) that Z þ1 Z þ1 l f ðs; xðsÞÞds 6 N kl Rl ð1 þ sÞ f ðs; 1Þds; 1
8x 2 oPR :
ð2:18Þ
1
Letting R1 ¼ R and c2 ¼ c2 ðRÞ in (2.17) and (2.11) together with (2.17) and (2.18) guarantees that Z 1 Z þ1 ðAxÞðtÞ 6 y1 þ kAxkF ¼ sup sf ðs; xðsÞÞds þ f ðs; xðsÞÞds 1þt t2Rþ 0 1 " Z kl 1 N l 6 y1 þ R s1þk f ðs; 1Þds 2 0 # Z þ1 l kl þN ð1 þ sÞ f ðs; 1Þds 6 R; 8x 2 oPR : 1
This implies that (2.16) holds. Consequently, (2.15) and (2.16) and Lemma 1.1, Lemma 2.1 yield our result. The next theorem generates a multiplicity result for BVP (1.1).
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Theorem 2.2. Assume that (H1 ) and (H2 ) hold. In addition suppose the following two conditions are satisfied. (H3 ) There exist k0 < 0 and l0 > 1 such that 0
f ðt; cxÞ P ck f ðt; xÞ; 0
f ðt; cxÞ P cl f ðt; xÞ;
0 < c 6 N; c P M:
(H4 ) y1 þ akl
Z
1
s1þk f ðs; 1Þds þ blk
0
Z
þ1
ð1 þ sÞl f ðs; 1Þds < 1;
1
where a ¼ minfðN =2Þ; ð1=MÞg, b ¼ maxfM; ð1=N Þg. Then BVP (1.1) has at least two positive solutions x, y 2 P such that 0 < kxkF < 1 < kykF < þ1: Proof. Firstly, similar to the proof of Theorem 2.1, there exists a sufficiently small r 2 ð0; 1Þ to guarantee (2.15) holds. Secondly, we show kAxkF < kxkF ;
8x 2 oP1 :
ð2:19Þ
Notice that s 6 xðsÞ 6 2kxkF ¼ 2 for s 2 ½0; 1 and x 2 oP1 . Therefore, axðsÞ 6 N for s 2 ½0; 1 . This together with (H1 ) and (H2 ) guarantees that Z 1 Z 1 1 sf ðs; xðsÞÞds ¼ sf s; axðsÞ ds a 0 0 Z 1 Z 1 6 akl sjxðsÞjk f ðs; 1Þds 6 akl s1þk f ðs; 1Þds; 8x 2 oP1 : ð2:20Þ 0
0
On the other hand, notice 1 ¼ kxkF 6 xðsÞ 6 ð1 þ sÞkxkF ¼ 1 þ s for s 2 ½1; þ1Þ and x 2 oP1 . This implies bxðsÞ P M for s P 1. Therefore, it follows from (H1 ) and (H2 ) that Z þ1 Z þ1 Z þ1 1 l f ðs; xðsÞÞds ¼ f s; bxðsÞ ds 6 blk jxðsÞj f ðs; 1Þds b 1 1 1 Z þ1 l 6 blk ð1 þ sÞ f ðs; 1Þds: 1
This together with (H4 ) and (2.20) means (2.19) holds, which also guarantees that there is no positive solution of BVP (1.1) on oP1 .
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553
Now it remains to show there exists a sufficiently large R > 1 such that kAxkF P kxkF ; 8x 2 oPR :
ð2:21Þ R 10
Without loss of generality, by (H2 ), suppose 1 f ðs; 1Þds > 0. Let c ¼ 1=N , and
1=1l0 Z 1 k0 l0 10 N R ¼ 1 þ NM þ f ðs; 1Þds : 2 1 Then 1=c ¼ N , cR ¼ R=N P M. Note that xðsÞ P kxkF ¼ R for s 2 ½1; þ1Þ and x 2 oPR . Combining this with (H3 ), it follows that Z 10 ðAxÞðtÞ P ðAxÞð1Þ P 1 f ðs; xðsÞÞds kAxkF ¼ sup 1þt 1þ1 2 1 t2Rþ
Z Z 0 1 10 1 1 k0 l0 10 ¼ f s; N xðsÞ ds P N jxðsÞjl f ðs; 1Þds 2 1 N 2 1 Z 1 k0 l0 l0 10 R f ðs; 1Þds P R ¼ kAxkF ; 8x 2 oPR ; P N 2 1 that is, (2.21) holds. Consequently, from (2.15), (2.19), (2.20) and Lemmas 1.1 and 2.1, our conclusion follows. To illustrate how the Theorems 2.1 and 2.2 can be used in practice we consider the following two examples. Example 2.1. Consider BVP (1.1) with f ðt; xÞ ¼ a1 ðtÞx5 þ a2 ðtÞx2=3 ; where ai 2 C½ð0; þ1Þ; Rþ (i ¼ 1; 2) and Z 1 t4 ½a1 ðtÞ þ a2 ðtÞ dt < þ1; 0< 0 Z þ1 2=3 ð1 þ tÞ ½a1 ðtÞ þ a2 ðtÞ dt < þ1: 0< 1
It is not difficult to see, by Theorem 2.1, that BVP (1.1) has at least one positive solution. Example 2.2. Consider BVP (1.1) with f ðt; xÞ ¼ b1 ðtÞx2 þ b2 ðtÞx2 ; where bi 2 C½ð0; þ1Þ; Rþ (i ¼ 1; 2),
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Z
1
t1 ½b1 ðtÞ þ b2 ðtÞ dt < þ1;
0< 0
Z
þ1
0<
2
ð1 þ tÞ ½b1 ðtÞ þ b2 ðtÞ dt < þ1
1
and y1 þ
1 16
Z
1
t1 ½b1 ðtÞ þ b2 ðtÞ dt þ 0
Z
þ1
2
ð1 þ tÞ ½b1 ðtÞ þ b2 ðtÞ dt < 1: 1
By Theorem 2.2, one can see that BVP (1.1) has at least two positive solutions. 3. The boundedness and unboundedness of positive solutions for BVP (1.1) In what follows sufficient conditions are given to guarantee that the positive solution of BVP (1.1) is bounded or unbounded. Throughout this section always suppose y1 ¼ 0 because every positive solution of BVP (1.1) is unbounded when y1 > 0. Theorem 3.1. Suppose (H1 ) holds and in addition, Z þ1 1þl ð1 þ sÞ f ðs; 1Þds < þ1: 1
Then every positive solution of BVP (1.1) is bounded. Proof. Assume that x is an any given positive solution of BVP (1.1). To prove x R þ1 is bounded on ½0; þ1Þ, by (2.10), we need only to show 1 sf ðs; xðsÞÞds < þ1. In fact, choose c > 0 such that ckxkF P M and 1=c 6 N . Notice that ð1 þ sÞkxkF P xðsÞ P kxkF for s 2 ½1; þ1Þ. This yields that cxðsÞ P M for s 2 ½1; þ1Þ. Therefore, by (H1 ), we have Z þ1 Z þ1 1 sf ðs; xðsÞÞds ¼ sf s; cxðsÞ ds c 1 1 Z þ1 l 6 clk sjxðsÞj f ðs; 1Þds 1 Z þ1 l sð1 þ sÞ f ðs; 1Þds kxkF 6 clk 1 Z þ1 lk 6c ð1 þ sÞ1þl f ðs; 1Þds kxkF < þ1; 1
that is, our conclusion follows.
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
555
Theorem 3.2. Suppose that (H1 ) holds. Then every positive solution of BVP (1.1) is unbounded if one of the following two conditions is fulfilled: (C1 ) Z
þ1
sf ðs; 1Þds ¼ þ1: 1
(C2 ) lim t2 f ðt; 1Þ ¼ þ1:
t!þ1
Proof. In what follows, let x be an arbitrary positive solution of BVP (1.1). First, suppose that (C1 ) is satisfied. Choose c sufficiently large that ckxkF P M and 1=c 6 N . Notice that xðtÞ is nondecreasing and xðtÞ P kxkF on ½1; þ1Þ. Then for t > 1, Z t Z t 1 sf ðs; xðsÞÞds ¼ sf s; cxðsÞ ds xðtÞ P c 1 1 Z t Z t k k P ckl sjxðsÞj f ðs; 1Þds P ckl jxðtÞj sf ðs; 1Þds: 1 1k
kl
Rt
1
Consequently, xðtÞ P c sf ðs; 1Þds, that is, our conclusion follows. 1 Second, assume that (C2 ) holds. In this situation, suppose, on the contrary, x is bounded on Rþ . Choose c sufficiently small such that cxðtÞ 6 N for t 2 Rþ and 1=c P M. Then by (2.10) we know Z þ1 Z þ1 1 f ðs; xðsÞÞds P t f s; cxðsÞ ds xðtÞ P t c t t Z þ1 Z þ1 l l P clk t jxðsÞj f ðs; 1Þds P clk jxðtÞj t f ðs; 1Þds: t 1l
lk
R þ1
t
Consequently, xðtÞ P c t t f ðs; 1Þds. In addition, it follows from LÕHospital rule that R þ1 Z þ1 f ðs; 1Þds ¼ lim t2 f ðt; 1Þ ¼ þ1 f ðs; 1Þds ¼ lim t lim t t!þ1 t!þ1 t!þ1 1=t t which is a contradiction with the boundedness of x on Rþ . Therefore, our conclusion follows.
References [1] R.P. Agarwal, D. OÕregan, Second-order boundary value problems of singular type, J. Math. Anal. Appl. 226 (1998) 414–430.
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[2] R.P. Agarwal, D. OÕRegan, Nonlinear superlinear singular and nonsingular second order boundary value problems, J. Different. Eqs. 143 (1998) 60–95. [3] K. Lan, J.L. Webb, Positive solutions of semilinear differential equations with singularities, J. Different. Eqs. 148 (1998) 407–421. [4] X. Liu, B. Yan, Boundary-irregular solutions to singular boundary value problems, Nonlinear Anal. (TMA) 32 (1998) 633–646. [5] Z. Wei, Positive solutions of singular sublinear second order boundary value problems, Syst. Sci. Math. Sci. 11 (1) (1998) 82–88. [6] S.Z. Chen, Y. Zhang, Singular boundary value problems on a half-line, J. Math. Anal. Appl. 195 (1995) 449–468. [7] B. Yan, Boundary value problems on the half-line with impulses and infinite delay, J. Math. Anal. Appl. 259 (2001) 94–114. [8] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Inc., New York, 1988. [9] C. Corduneanu, Integral Equations and Stability of Feedback Systems, Academic Press, New York, 1973.
Existence and unboundedness of positive solutions for singular boundary value problems on half-line q Yansheng Liu Department of Mathematics, Shandong Normal University, Jinan, 250014, PR China
Abstract By constructing a special cone and using cone compression and expansion fixed point theorem, this paper presents some existence results of positive solutions of singular boundary value problem (BVP, for short) on half-line for a class of second order differential equations. At the same time, the boundedness and unboundedness of positive solutions for BVP are considered also. Ó 2002 Elsevier Inc. All rights reserved. Keywords: Singularity; Cone; Positive solution; Boundedness; Half-line
1. Introduction This paper is concerned with the existence of single and multiple solutions of the following singular boundary value problems (BVPs) on half-line:
x00 ðtÞ þ f ðt; xðtÞÞ ¼ 0; t > 0; xð0Þ ¼ 0; x0 ðþ1Þ ¼ y1 P 0;
ð1:1Þ
where f 2 C½ð0; þ1Þ ð0; þ1Þ; Rþ , x0 ð1Þ ¼ limt!þ1 x0 ðtÞ, Rþ ¼ ½0; þ1Þ. q The project supported by the National Natural Science Foundation of PR China (10171057), Natural Science Foundation of Shandong Province (Z2000A02) and Rewarded Fund of Outstanding Middle and Young Scientists of Shandong Province. E-mail address: [email protected] (Y. Liu).
0096-3003/02/$ - see front matter Ó 2002 Elsevier Inc. All rights reserved. doi:10.1016/S0096-3003(02)00431-9
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
In recent years, singular BVPs on finite interval are extensively studied (see [1–5] and referees therein, for example). Only a small quantity of papers had discussed such question on half-line (see [6,7], for instance), in which it is considered simply the case that f ðt; xÞ is singular at t ¼ 0, not singular at x ¼ 0. To our knowledge, there is no paper to consider BVP (1.1) when f ðt; xÞ is singular at x ¼ 0 and also no result is available about the existence of multiple positive unbounded solutions for such system. As a result, the goal of present paper is to fill the gap in this area. The main features here are as follows. Firstly, f ðt; xÞ is singular both at t ¼ 0 and x ¼ 0. Secondly, the existence of multiple solutions of BVP (1.1) is considered. Thirdly, it is easy to see the solution of BVP (1.1) is unbounded as y1 > 0. But what about when y1 ¼ 0? We will show that the result depends upon f ðt; xÞ. Some sufficient conditions are given to guarantee the boundedness or unboundedness of solutions for BVP (1.1) when y1 ¼ 0. Generally speaking, the approximation method is often used to study singular problem. However, in present paper, the approximation problem is not needed. The main techniques used here are a new constructed cone and cone compression and expansion fixed point theorem. Let FC½Rþ ; R ¼:
x 2 C½Rþ ; R : sup t2Rþ
jxðtÞj < þ1 : 1þt
ð1:2Þ
Then FC½Rþ ; R is a Banach space with norm kxkF ¼: supt2Rþ ðjxðtÞj= 1 þ tÞ. In present paper, FC½Rþ ; R will be the basic space to study (1.1). The paper is organized as follows. Section 2 is devoted to the existence of positive solutions for BVP (1.1). Section 3 gives some sufficient conditions which guarantees the solution of BVP (1.1) to be bounded or unbounded as y1 ¼ 0. Finally in this section we state a result from the literature which will be used in Section 2. Lemma 2.4 ([8] fixed point theorem of cone expansion and compression of norm type). Let X1 and X2 be two bounded open sets in Banach space E such that h 2 X1 and X1 X2 . Let operator A : P \ ðX2 n X1 Þ ! P be completely continuous, where h denotes the zero element of E and P is a cone of E. Suppose that one of the two conditions (i) kAxk 6 kxk, 8x 2 P \ oX1 and kAxk P kxk, 8x 2 P \ oX2 and (ii) kAxk P kxk, 8x 2 P \ oX1 and kAxk 6 kxk, 8x 2 P \ oX2 is satisfied. Then A has at least one fixed point in P \ ðX2 n X1 Þ.
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
545
2. Existence results For convenience, we first list some conditions. (H1 ) There exist k; l such that 1 < k < 0 < l < þ1 and N ; M > 0 such that 0 < N 6 1 6 M satisfying cl f ðt; xÞ 6 f ðt; cxÞ 6 ck f ðt; xÞ; ck f ðt; xÞ 6 f ðt; cxÞ 6 cl f ðt; xÞ; R1 (H2 ) 0 < 0 s1þk f ðs; 1Þds < þ1;
0 < c 6 N; c P M: R þ1 l 0 < 1 ð1 þ sÞ f ðs; 1Þds < þ1:
Pn Remark. The typical function satisfying (H1 ) is f ðt; xÞ ¼ i¼1 ai ðtÞxki , where k ¼ k1 6 k2 6 6 kk < 0 < kkþ1 6 6 kn ¼ l, ai 2 C½ð0; þ1Þ; Rþ (i ¼ 1; 2; . . . ; n; k ¼ 1; 2; . . . ; n 1). For the sake of investigating the existence of positive solutions let P ¼ f x 2 FC½Rþ ; Rþ : xðtÞ 8 txðsÞ > < ; t 2 ½0; 1 ; 1 þs P > : xðsÞ ; t 2 ½1; þ1Þ; 1þs
s 2 Rþ ;
9 > =
xðtÞ is nondecreasing on Rþ : > ; s 2 Rþ ; ð2:1Þ
It is easy to see P is a nonempty, convex and closed subset of FC½Rþ ; Rþ and furthermore, P is a cone of Banach space FC½Rþ ; R . Notice that if x 2 P n fhg, then xðtÞ > 0 for t 2 ð0; þ1Þ, where h denotes the zero element of FC½Rþ ; R . In addition notice that tkxkF ; t 2 ½0; 1 ; xðtÞ P ð2:2Þ kxkF ; t 2 ½1; þ1Þ; and xðtÞ 6 ð1 þ tÞkxkF 6 2kxkF ;
t 2 ½0; 1 :
ð2:3Þ
Now we are in position to give the following results. Lemma 2.1. Assume that (H1 ) and (H2 ) hold. Then x 2 FC½Rþ ; Rþ \ C 2 ½ð0; þ1Þ; Rþ is a positive solution of BVP (1.1) if and only if x 2 FC½Rþ ; Rþ is a positive solution of the following integral equation Z þ1 xðtÞ ¼ ty1 þ Gðt; sÞf ðs; xðsÞÞds; ð2:4Þ 0
where Gðt; sÞ ¼ minft; sg.
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Proof. First of all, we show for any given t 2 ð0; þ1Þ, x 2 P n fhg, R þ1 Gðt; sÞf ðs; xðsÞÞds is convergent. 0 Choose c1 > 0 such that 2c1 kxkF 6 N , 1=c1 > M. Thus, by (2.3), c1 xðsÞ 6 N for s 2 ½0; 1 . This together with (H1 ), (H2 ) and (2.2) yields that Z 1 Z 1 Z 1 1 Gðt; sÞf ðs; xðsÞÞds 6 sf ðs; xðsÞÞds 6 sf s; c1 xðsÞ ds c1 0 0 0
Z 1 Z 1 1 6 sck1 jxðsÞjk f s; sjxðsÞjk f ðs; 1Þds ds 6 c1kl c 1 0 0 Z 1 k kl 1þk 6 c1 s f ðs; 1Þds kxkF < þ1: ð2:5Þ 0
On the other hand, choose c2 > 0 sufficiently large that c2 kxkF P M, 1=c2 < N . So c2 xðsÞ P c2 kxkF P M for s P 1. From (H1 ) and (H2 ) one can obtain Z þ1 Z þ1 Z þ1 1 l f ðs; xðsÞÞds 6 f s; c2 xðsÞ ds 6 clk 2 jxðsÞj f ðs; 1Þds c2 1 1 1 Z þ1 lk 6 c2 ð1 þ sÞl f ðs; 1Þds kxklF < þ1: ð2:6Þ 1
R þ1 This together with (2.5) guarantees that 0 Gðt; sÞf ðs; xðsÞÞds is convergent for t 2 ð0; þ1Þ, and x 2 P n fhg. Next, suppose x 2 FC½Rþ ; Rþ is a positive solution of (2.4). Now we show x 2 P n fhg. R þ1Firstly, we claim limt!0þ xðtÞ ¼ 0. By (2.4) we need only to prove limt!0þ t t f ðs; xðsÞÞds ¼ 0. In this situation, notice that R d (2.5) holds. As a result, for any given e > 0, there exists d1 > 0 such that 0 1 sf ðs; xðsÞÞds < e=2. Let ( Z 1 ) þ1 d ¼ min d1 ; 2 f ðs; xðsÞÞds e : d1
Then Z t
þ1
f ðs; xðsÞÞds ¼ t
t
Z
Z 6
d1
f ðs; xðsÞÞds þ t
þ1
f ðs; xðsÞÞds
d1
t d1
sf ðs; xðsÞÞds þ d
t
6
Z
e e þ ¼ e; 2 2
Z
þ1
f ðs; xðsÞÞds
d1
8t 2 ð0; dÞ:
ð2:7Þ
Immediately it follows that limt!0þ xðtÞ ¼ 0. Secondly, noting that x is a positive solution of BVP (1.1), so, Z þ1 f ðs; xðsÞÞds P 0: ð2:8Þ x0 ðtÞ ¼ y1 þ t
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
This means that x is nondecreasing. Thirdly, notice 8t t > P ; 0 < t; s < s; > > 1þs >s > > > >t P t ; > 0 < t < s < s; Gðt; sÞ < s 1þs ¼ s > P 1 P 1 ; 0 < s < s < t; Gðs; sÞ > > > > s 1þs > > > > : s ¼ 1 P 1 ; t; s > s > 0: s 1þs Therefore, Z 1 xðtÞ ¼ ty1 þ Gðt; sÞf ðs; xðsÞÞds 0 8 Z þ1 t t > > > Gðs; sÞf ðs; xðsÞÞds; 0 < t 6 1; < 1 þ s sy1 þ 1 þ s 0 P Z þ1 > 1 1 > > sy1 þ Gðs; sÞf ðs; xðsÞÞds; t P 1; : 1þs 1þs 0
547
s 2 Rþ ; ð2:9Þ s 2 Rþ :
This together with (2.8) yields x 2 P n fhg. In addition, from (2.6) and (2.8) it follows that x 2 FC½Rþ ; Rþ \ 2 C ½ð0; þ1Þ; Rþ is a positive solution of BVP (1.1). Conversely, if x 2 FC½Rþ ; Rþ \ C 2 ½ð0; þ1Þ; Rþ is a positive solution of BVP(1.1) it is easy to see that x 2 FC½Rþ ; Rþ is a positive solution of (2.4). By (2.5) and (2.6), we can define an operator A on P n fhg by Z þ1 ðAxÞðtÞ ¼ ty1 þ Gðt; sÞf ðs; xðsÞÞds;
ð2:10Þ
0
where Gðt; sÞ is the same as in (2.4). Also from the proof of Lemma 2.1 the existence of positive solutions belonging to FC½Rþ ; Rþ \ C 2 ½ð0; þ1Þ; Rþ of BVP (1.1) is equivalent to the existence of fixed point of A on P n fhg. To obtain the complete continuity of A, the following lemmas are still need. Lemma 2.2. Let W be a bounded subset of P. Then W is relatively compact in FC½Rþ ; R if fW ðtÞ=1 þ tg are equicontinuous on any finite subinterval of Rþ and for any e > 0, there exists N > 0 such that xðt1 Þ xðt2 Þ < e; 1 þ t 1 þ t2 1 uniformly with respect to x 2 W as t1 ; t2 P N , where W ðtÞ ¼ fxðtÞ : x 2 W g, t 2 Rþ . This lemma is a simple improvement of the Corduneanu theorem in [9].
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Lemma 2.3. Suppose (H1 ) and (H2 ) hold. Then A: P R n Pr ! P is completely continuous, where R > r > 0 and Pr ¼ fx 2 P : kxkF < rg, PR is very similar to Pr . Proof. First of all, from the proof of Lemma 2.1 and 0
ðAxÞ ðtÞ ¼ y1 þ
Z
þ1
f ðs; xðsÞÞds P 0
for x 2 P n fhg; t 2 ð0; þ1Þ;
t
similar to (2.9), it follows that A : P n fhg ! P . Of course, A : P R n Pr ! P . In addition, choose c1 ; c2 > 0 such that 2c1 r 6 N ;
1 PM c1
and
1 6 N: c2
c2 R P M;
Combining with (2.2) and (2.3), similar to (2.5) and (2.6), it follows that Z
1
sf ðs; xðsÞÞds 6 c1kl 0 k 6 ckl 1 r
Z
1 k
s1þk f ðs; 1Þds kxkF 0
Z
1
s1þk f ðs; 1Þds < þ 1;
8x 2 P R n Pr
ð2:11Þ
0
and Z
Z þ1 f ðs; xðsÞÞds 6 clk ð1 þ sÞl f ðs; 1Þds kxklF 2 1 1 Z þ1 l l 6 clk R ð1 þ sÞ f ðs; 1Þds < þ1; 8x 2 P R n Pr : 2 þ1
ð2:12Þ
1
It follows from (2.10)–(2.12) that AðP R n Pr Þ is bounded in FC½Rþ ; Rþ . Now we are ready to show AV is relatively compact for any bounded subsets V of P R n Pr . To see this, by (2.10) and Lemma 2.2, we first prove limt!0þ ððAxÞðtÞ=1 þ tÞ ¼ 0 uniformly with respect to x 2 V . Notice that Z t Z 1 ðAxÞðtÞ ty1 1 t ¼ þ sf ðs; xðsÞÞds þ f ðs; xðsÞÞds 1þt 1þt t 1þt 1þt 0 Z þ1 t f ðs; xðsÞÞds; 8t 2 ð0; 1Þ; x 2 V : þ 1þt 1 From this R 1 together with (2.11) and (2.12), similar to (2.7), one can see limt!0þ t t f ðs; xðsÞÞds ¼ 0 uniformly with respect to x 2 V . As a result, limt!0þ ððAxÞðtÞ=1 þ tÞ ¼ 0 uniformly with respect to x 2 V .
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
549
Secondly, notice Z ðAxÞðtÞ ðAxÞðt0 Þ t t0 þ1 0 f ðs; xðsÞÞds 1 þ t 1 þ t0 6 jt t jy1 þ 1 þ t 1 þ t0 t Z 0 Z 1 t 1 t f ðs; xðsÞÞds þ sf ðs; xðsÞÞds þ t 1 þ t 1 þ t0 0 Z 0 t sf ðs; xðsÞÞds; 8t; t0 2 ð0; þ1Þ; x 2 V : þ t Combining this with (2.11) and (2.12), it follows that fððAxÞðtÞ=1 þ tÞ : x 2 V g are equicontinuous on any finite closed interval of ð0; þ1Þ. Thirdly, it remains to show for any e > 0, there exists sufficiently large N > 0 such that ðAxÞðtÞ ðAxÞðt0 Þ 0 ð2:13Þ 1 þ t 1 þ t0 < e; 8t; t P N ; 8x 2 V : To see this, combining with (2.10) we need only to prove that for any given e > 0, there exists N > 0 such that Z Z t0 t s s f ðs; xðsÞÞds f ðs; xðsÞÞds < e; 8t; t0 P N ; 8x 2 V : 0 0 1þt 1 þ t 0 R þ1 Notice (2.12) guarantees that there exists L > 0 satisfying L f ðs; xðsÞÞds < e=4 uniformly with respect to x 2 V . By (2.11) and (2.12) we know Z L M ¼: sup sf ðs; xðsÞÞds < þ1: x2P R nPr
0
Then let N ¼ maxfL; ð4M=eÞg. It follows that Z Z t0 t s s f ðs; xðsÞÞds f ðs; xðsÞÞds 0 0 1þt 1 þ t 0
Z L Z t 1 1 s f ðs; xðsÞÞds 6 sf ðs; xðsÞÞds þ þ 1 þ t 1 þ t0 1 þ t 0 L Z t0 s þ f ðs; xðsÞÞds 1 þ t0 L e e e e 6 þ þ þ ¼ e; 8t; t0 P N ; 8x 2 V ; 4 4 4 4 that is, (2.13) holds. In conclusion, Lemma 2.2 guarantees that AV is a relatively compact subset of P.
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Finally, we are in position to show A : P R n Pr ! P is continuous. Suppose xn , x 2 P R n Pr and kxn xkF ! 0. Then, for any t > 0, limn!þ1 xn ðtÞ ¼ xðtÞ. This together with (2.11) and (2.12) and Lebesgue dominated theorem yields that ðAxn ÞðtÞ ðAxÞðtÞ ! ðn ! þ1Þ; 1þt 1þt
8t 2 ð0; þ1Þ:
ð2:14Þ
Notice from above that fAxn g is relatively compact also. Therefore, (2.14) guarantees that kAxn AxkF ! 0 ðn ! þ1Þ, that is, A: P R n Pr ! P is continuous. To sum up, the conclusion of Lemma 2.3 follows. The following theorem gives sufficient conditions to guarantee the existence of positive solutions for BVP (1.1). Theorem 2.1. Assume that (H1 ) and (H2 ) hold and l 2 ð0; 1Þ. Then BVP (1.1) has at least one positive solution. Proof. We first prove there exists a sufficiently small r > 0 such that kAxkF P kxkF ;
8x 2 oPr :
ð2:15Þ
In fact, if y1 > 0, it is obvious because we can chooseR r ¼ y1 =2. Now 1 consider the case y1 ¼ 0. Notice that (H2 ) guarantees that 0 s1þl f ðs; 1Þds is convergent. Let c ¼ 1=M and (
1=1l ) Z N 1 kl 1 1þl ; M r ¼ min s f ðs; 1Þds : 2c 2 0 Then sr ¼ skxkF 6 xðsÞ 6 2r;
8x 2 oPr ;
s 2 ½0; 1 ;
that is, cxðsÞ 6 2rc 6 N for x 2 oPr and s 2 ½0; 1 . By 1=c ¼ M together with (H1 ) and (H2 ), we know that Z ðAxÞðtÞ ðAxÞð1Þ 1 1 P P sf ðs; xðsÞÞds kAxkF ¼ sup 1þ1 2 0 t2Rþ 1 þ t
Z Z 1 1 1 1 1 lk l sf s; cxðsÞ ds P sc jxðsÞj f ðs; 1Þds P 2 0 c 2 0 Z 1 1 P M kl s1þl f ðs; 1Þds kxklF P r ¼ kxkF ; 8x 2 oPr 2 0 which implies that (2.15) holds.
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
551
Next we show there exists a sufficiently large R > r such that kAxkF 6 kxkF ;
8x 2 oPR :
ð2:16Þ
Indeed, choose R1 > r satisfying 2R1 =N P M. Let c1 ¼ c1 ðR1 Þ ¼: ðN =2R1 Þ. Then 2c1 R1 ¼ N , 1=c1 P M. It follows from (2.11) that Z 1 Z 1 sf ðs; xðsÞÞds 6 c1kl Rk1 s1þk f ðs; 1Þds 0
0
kl Z 1 N ¼ Rk1 s1þk f ðs; 1Þds 2R1 0
kl Z 1 N Rl1 s1þk f ðs; 1Þds; 8x 2 oPR1 : ¼ 2 0
ð2:17Þ
On the other hand, let c2 ¼ 1=N and 8
kl Z 1 < N R ¼ max MN ; R1 ; 2y1 ; 2 s1þk f ðs; 1Þds : 2 0 9 1 ! Z þ1 1l = l þ 2N kl : ð1 þ sÞ f ðs; 1Þds ; 1 Then c2 R P M, 1=c2 6 N . It follows from (2.12) that Z þ1 Z þ1 l f ðs; xðsÞÞds 6 N kl Rl ð1 þ sÞ f ðs; 1Þds; 1
8x 2 oPR :
ð2:18Þ
1
Letting R1 ¼ R and c2 ¼ c2 ðRÞ in (2.17) and (2.11) together with (2.17) and (2.18) guarantees that Z 1 Z þ1 ðAxÞðtÞ 6 y1 þ kAxkF ¼ sup sf ðs; xðsÞÞds þ f ðs; xðsÞÞds 1þt t2Rþ 0 1 " Z kl 1 N l 6 y1 þ R s1þk f ðs; 1Þds 2 0 # Z þ1 l kl þN ð1 þ sÞ f ðs; 1Þds 6 R; 8x 2 oPR : 1
This implies that (2.16) holds. Consequently, (2.15) and (2.16) and Lemma 1.1, Lemma 2.1 yield our result. The next theorem generates a multiplicity result for BVP (1.1).
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Theorem 2.2. Assume that (H1 ) and (H2 ) hold. In addition suppose the following two conditions are satisfied. (H3 ) There exist k0 < 0 and l0 > 1 such that 0
f ðt; cxÞ P ck f ðt; xÞ; 0
f ðt; cxÞ P cl f ðt; xÞ;
0 < c 6 N; c P M:
(H4 ) y1 þ akl
Z
1
s1þk f ðs; 1Þds þ blk
0
Z
þ1
ð1 þ sÞl f ðs; 1Þds < 1;
1
where a ¼ minfðN =2Þ; ð1=MÞg, b ¼ maxfM; ð1=N Þg. Then BVP (1.1) has at least two positive solutions x, y 2 P such that 0 < kxkF < 1 < kykF < þ1: Proof. Firstly, similar to the proof of Theorem 2.1, there exists a sufficiently small r 2 ð0; 1Þ to guarantee (2.15) holds. Secondly, we show kAxkF < kxkF ;
8x 2 oP1 :
ð2:19Þ
Notice that s 6 xðsÞ 6 2kxkF ¼ 2 for s 2 ½0; 1 and x 2 oP1 . Therefore, axðsÞ 6 N for s 2 ½0; 1 . This together with (H1 ) and (H2 ) guarantees that Z 1 Z 1 1 sf ðs; xðsÞÞds ¼ sf s; axðsÞ ds a 0 0 Z 1 Z 1 6 akl sjxðsÞjk f ðs; 1Þds 6 akl s1þk f ðs; 1Þds; 8x 2 oP1 : ð2:20Þ 0
0
On the other hand, notice 1 ¼ kxkF 6 xðsÞ 6 ð1 þ sÞkxkF ¼ 1 þ s for s 2 ½1; þ1Þ and x 2 oP1 . This implies bxðsÞ P M for s P 1. Therefore, it follows from (H1 ) and (H2 ) that Z þ1 Z þ1 Z þ1 1 l f ðs; xðsÞÞds ¼ f s; bxðsÞ ds 6 blk jxðsÞj f ðs; 1Þds b 1 1 1 Z þ1 l 6 blk ð1 þ sÞ f ðs; 1Þds: 1
This together with (H4 ) and (2.20) means (2.19) holds, which also guarantees that there is no positive solution of BVP (1.1) on oP1 .
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
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Now it remains to show there exists a sufficiently large R > 1 such that kAxkF P kxkF ; 8x 2 oPR :
ð2:21Þ R 10
Without loss of generality, by (H2 ), suppose 1 f ðs; 1Þds > 0. Let c ¼ 1=N , and
1=1l0 Z 1 k0 l0 10 N R ¼ 1 þ NM þ f ðs; 1Þds : 2 1 Then 1=c ¼ N , cR ¼ R=N P M. Note that xðsÞ P kxkF ¼ R for s 2 ½1; þ1Þ and x 2 oPR . Combining this with (H3 ), it follows that Z 10 ðAxÞðtÞ P ðAxÞð1Þ P 1 f ðs; xðsÞÞds kAxkF ¼ sup 1þt 1þ1 2 1 t2Rþ
Z Z 0 1 10 1 1 k0 l0 10 ¼ f s; N xðsÞ ds P N jxðsÞjl f ðs; 1Þds 2 1 N 2 1 Z 1 k0 l0 l0 10 R f ðs; 1Þds P R ¼ kAxkF ; 8x 2 oPR ; P N 2 1 that is, (2.21) holds. Consequently, from (2.15), (2.19), (2.20) and Lemmas 1.1 and 2.1, our conclusion follows. To illustrate how the Theorems 2.1 and 2.2 can be used in practice we consider the following two examples. Example 2.1. Consider BVP (1.1) with f ðt; xÞ ¼ a1 ðtÞx5 þ a2 ðtÞx2=3 ; where ai 2 C½ð0; þ1Þ; Rþ (i ¼ 1; 2) and Z 1 t4 ½a1 ðtÞ þ a2 ðtÞ dt < þ1; 0< 0 Z þ1 2=3 ð1 þ tÞ ½a1 ðtÞ þ a2 ðtÞ dt < þ1: 0< 1
It is not difficult to see, by Theorem 2.1, that BVP (1.1) has at least one positive solution. Example 2.2. Consider BVP (1.1) with f ðt; xÞ ¼ b1 ðtÞx2 þ b2 ðtÞx2 ; where bi 2 C½ð0; þ1Þ; Rþ (i ¼ 1; 2),
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Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
Z
1
t1 ½b1 ðtÞ þ b2 ðtÞ dt < þ1;
0< 0
Z
þ1
0<
2
ð1 þ tÞ ½b1 ðtÞ þ b2 ðtÞ dt < þ1
1
and y1 þ
1 16
Z
1
t1 ½b1 ðtÞ þ b2 ðtÞ dt þ 0
Z
þ1
2
ð1 þ tÞ ½b1 ðtÞ þ b2 ðtÞ dt < 1: 1
By Theorem 2.2, one can see that BVP (1.1) has at least two positive solutions. 3. The boundedness and unboundedness of positive solutions for BVP (1.1) In what follows sufficient conditions are given to guarantee that the positive solution of BVP (1.1) is bounded or unbounded. Throughout this section always suppose y1 ¼ 0 because every positive solution of BVP (1.1) is unbounded when y1 > 0. Theorem 3.1. Suppose (H1 ) holds and in addition, Z þ1 1þl ð1 þ sÞ f ðs; 1Þds < þ1: 1
Then every positive solution of BVP (1.1) is bounded. Proof. Assume that x is an any given positive solution of BVP (1.1). To prove x R þ1 is bounded on ½0; þ1Þ, by (2.10), we need only to show 1 sf ðs; xðsÞÞds < þ1. In fact, choose c > 0 such that ckxkF P M and 1=c 6 N . Notice that ð1 þ sÞkxkF P xðsÞ P kxkF for s 2 ½1; þ1Þ. This yields that cxðsÞ P M for s 2 ½1; þ1Þ. Therefore, by (H1 ), we have Z þ1 Z þ1 1 sf ðs; xðsÞÞds ¼ sf s; cxðsÞ ds c 1 1 Z þ1 l 6 clk sjxðsÞj f ðs; 1Þds 1 Z þ1 l sð1 þ sÞ f ðs; 1Þds kxkF 6 clk 1 Z þ1 lk 6c ð1 þ sÞ1þl f ðs; 1Þds kxkF < þ1; 1
that is, our conclusion follows.
Y. Liu / Appl. Math. Comput. 144 (2003) 543–556
555
Theorem 3.2. Suppose that (H1 ) holds. Then every positive solution of BVP (1.1) is unbounded if one of the following two conditions is fulfilled: (C1 ) Z
þ1
sf ðs; 1Þds ¼ þ1: 1
(C2 ) lim t2 f ðt; 1Þ ¼ þ1:
t!þ1
Proof. In what follows, let x be an arbitrary positive solution of BVP (1.1). First, suppose that (C1 ) is satisfied. Choose c sufficiently large that ckxkF P M and 1=c 6 N . Notice that xðtÞ is nondecreasing and xðtÞ P kxkF on ½1; þ1Þ. Then for t > 1, Z t Z t 1 sf ðs; xðsÞÞds ¼ sf s; cxðsÞ ds xðtÞ P c 1 1 Z t Z t k k P ckl sjxðsÞj f ðs; 1Þds P ckl jxðtÞj sf ðs; 1Þds: 1 1k
kl
Rt
1
Consequently, xðtÞ P c sf ðs; 1Þds, that is, our conclusion follows. 1 Second, assume that (C2 ) holds. In this situation, suppose, on the contrary, x is bounded on Rþ . Choose c sufficiently small such that cxðtÞ 6 N for t 2 Rþ and 1=c P M. Then by (2.10) we know Z þ1 Z þ1 1 f ðs; xðsÞÞds P t f s; cxðsÞ ds xðtÞ P t c t t Z þ1 Z þ1 l l P clk t jxðsÞj f ðs; 1Þds P clk jxðtÞj t f ðs; 1Þds: t 1l
lk
R þ1
t
Consequently, xðtÞ P c t t f ðs; 1Þds. In addition, it follows from LÕHospital rule that R þ1 Z þ1 f ðs; 1Þds ¼ lim t2 f ðt; 1Þ ¼ þ1 f ðs; 1Þds ¼ lim t lim t t!þ1 t!þ1 t!þ1 1=t t which is a contradiction with the boundedness of x on Rþ . Therefore, our conclusion follows.
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