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Multiple positive solutions for a second-order three-point boundary value problem
Applied Mathematics and Computation 182 (2006) 258–268 www.elsevier.com/locate/amc
Multiple positive solutions for a second-order three-point boundary value problem q Jianli Li a
a,*
, Jianhua Shen
a,b
Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China b Department of Mathematics, College of Huaihua, Huaihua, Hunan 418008, China
Abstract In this paper, we study the second-order three-point boundary value problem u00 ðtÞ þ aðtÞu0 ðtÞ þ bðtÞuðtÞ þ hðtÞf ðuÞ ¼ 0; uð0Þ ¼ 0;
t 2 ð0; 1Þ;
auðgÞ ¼ uð1Þ:
The existence criteria for multiple positive solutions is obtained by using Krasnoselskii fixed point theorem and Leggett– Williams fixed-point theorem. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Positive solution; Fixed point theorem; Three-point boundary value problem
1. Introduction The study of multi-point boundary value problems for linear second-order ordinary differential equations was initiated by Il’in and Moviseev [1,2]. Since then, nonlinear second-order multi-point boundary value problems have also been studied by several authors. We refer the reader to [3–7,10,11] and their references. Recently, Ma and Wang [3] used Krasnoselskii’s fixed point theorem in cones to prove the existence of at least one positive solutions for the three-point boundary value problem (BVP) u00 ðtÞ þ aðtÞu0 ðtÞ þ bðtÞuðtÞ þ hðtÞf ðuÞ ¼ 0; uð0Þ ¼ 0;
t 2 ð0; 1Þ;
auðgÞ ¼ uð1Þ;
ð1:1Þ
where 0 < g < 1, a is a positive constant, h, f, a and b satisfy q
This work is supported by the NNSF of China (No. 10571050), and supported by the Key Project of Chinese Ministry of Education, and supported by Hunan Provincial Natural Science Foundation of China (No. 005JJ4001). * Corresponding author. E-mail address: [email protected] (J. Li). 0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.01.095
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
259
(H1) f 2 C([0, 1), [0, 1)); (H2) h 2 C([0, 1], [0, 1)) and there exists x0 2 (0, 1) such that h(x0) > 0; (H3) a 2 C[0, 1], b 2 C([0, 1], (1, 0)). Motivated by the results [3], this paper is concerned with the existence of multiple positive solutions of the above boundary problems. Our tool in this paper will be well-known Krasnoselskii fixed point theorem [3,8,9] and Leggett–Williams fixed-point theorem [10–14]. 2. Preliminaries In order to obtain our result, in this section, we mention without proof several foundational results. Lemma 2.1 [3]. Assume that (H3) hold. Let /1 and /2 be the solutions of 00 /1 ðtÞ þ aðtÞ/01 ðtÞ þ bðtÞ/1 ðtÞ ¼ 0; t 2 ð0; 1Þ; /1 ð0Þ ¼ 0; /1 ð1Þ ¼ 1 and 00 /2 ðtÞ þ aðtÞ/02 ðtÞ þ bðtÞ/2 ðtÞ ¼ 0; t 2 ð0; 1Þ; /2 ð0Þ ¼ 1; /2 ð1Þ ¼ 0:
ð2:1Þ
ð2:2Þ
Then (i) /1 is strictly increasing on [0, 1], (ii) /2 is strictly decreasing on [0, 1].
Lemma 2.2 [3]. Assume that (H3) hold. Then (2.1) and (2.2) have unique solution, respectively. Now, we assume the following condition holds. (H4) 0 < a/1(g) < 1. Lemma 2.3 [3]. Assume that (H3) and (H4) hold. Let y 2 C[0, 1]. Then the problem 00 u ðtÞ þ aðtÞu0 ðtÞ þ bðtÞuðtÞ þ yðtÞ ¼ 0; t 2 ð0; 1Þ; uð0Þ ¼ 0; auðgÞ ¼ uð1Þ; is equivalent to the integral equation Z 1 Gðt; sÞpðsÞyðsÞ ds þ A/1 ðtÞ; uðtÞ ¼
ð2:3Þ
ð2:4Þ
0
where
Z 1 a Gðg; sÞpðsÞyðsÞ ds; 1 a/1 ðgÞ 0 Z t aðsÞ ds ; pðtÞ ¼ exp 0 ( 1 /1 ðtÞ/2 ðsÞ; if s P t; Gðt; sÞ ¼ q /1 ðsÞ/2 ðtÞ; if t P s;
A¼
and q :¼ /01 ð0Þ: Moreover, u(t) P 0 on [0, 1] provided y P 0.
ð2:5Þ ð2:6Þ ð2:7Þ
260
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
Set
/1 ðtÞ /2 ðtÞ qðtÞ ¼ min ; ; k/1 k k/2 k where kÆk denote the supremum norm. From (2.7) we have Gðs; sÞ P Gðt; sÞ P qðtÞGðs; sÞ P 0; for t 2 ½0; 1: Choose d 2 (0, 1/2) such that x0 2 (d,1 d), take c ¼ minfqðtÞjt 2 ½d; 1 dg; it follows from Lemma 2.1 that 0 < c < 1. Lemma 2.4 [3]. Assume that (H3) and (H4) hold. Let y 2 C[0, 1] and y P 0, then the unique solution u of the problem (2.3) satisfies t 2 ½d; 1 d:
uðtÞ P ckuk;
3. Existence of one or two positive solutions In this section, we establish the existence conditions of one or two positive solutions for the BVP (1.1). Let E = C[0, 1]. Then E is a Banach space with respect to the norm kuk ¼ sup juðtÞj: t2½0;1
Now BVP (1.1) has a solution u = u(t) if and only if u is a solution of the operator equation Z 1 Z 1 a uðtÞ ¼ Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ :¼ ðTuÞðtÞ 1 a/1 ðgÞ 0 0
ð3:1Þ
in E, where p and G are defined by (2.6) and (2.7), respectively. It is easy to check that T is completely continuous. We define a cone in E by P ¼ u 2 E : u P 0; min uðtÞ P ckuk : ð3:2Þ d6t61d
Lemma 3.1. T(P) P. Proof. Note that G(t, s) P 0 and 0 < a/1(g) < 1 implies (Tu)(t) P 0. For any y 2 P, we have from (3.1) and G(t, s) 6 G(s, s), /1(t) 6 1 that Z 1 Z 1 a TuðtÞ 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 1 a/1 ðgÞ 0 0 which implies kTuðtÞk 6
Z
1
Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
0
From (2.7) we have Gðt; sÞ P qðtÞ; Gðs; sÞ
t 2 ð0; 1; s 2 ð0; 1Þ
a 1 a/1 ðgÞ
Z 0
1
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds:
ð3:3Þ
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
and also q(t) 6 /1(t), so we have Z 1 Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ min ðTuÞðtÞ ¼ min d6t61d
d6t61d
0
Z
a 1 a/1 ðgÞ
Z
1
261
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ
0
Z 1 a P min qðtÞ Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds d6t61d 1 a/1 ðgÞ 0 0 Z 1 Z 1 a Pc Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 1 a/1 ðgÞ 0 0 1
which and (3.3) implies min ðTuÞðtÞ P ckTuk: d6t61d
Hence Tu 2 P. This proof is complete. h Set f0 ¼ limþ u!0
f ðuÞ ; u
f 1 ¼ lim
u!1
f ðuÞ : u
The following Theorem is main result of [3]. Theorem 3.1 [3]. Let (H1)(H4) hold. Then the problem (1.1) has at least one positive solution in the case (i) f0 = 0 and f1 = 1 (superlinear), or (ii) f0 = 1 and f1 = 0 (sublinear).
Problem 1. Whether we can obtain a similar conclusion or not, if f0 = f1 = 0 or f0 = f1 = 1. Problem 2. Whether we can get a similar conclusion or not, if f0 = f1 62 {0, +1}. The aim of this section is to establish some simple criteria for the existence of positive solutions of BVP (1.1), which gives a positive answer to the questions stated above. The key tool in our approach is the following Krasnoselskii fixed point theorem, which is a useful method to prove the existence of positive solution for differential equations, for example [3,8]. Theorem 3.2 [9]. Let E be a Banach space, and let K E be a cone. Assume X1, X2 are open subsets of E with 0 2 X1, X1 X2 , and let A : K \ ðX2 n X1 Þ ! K be a completely continuous operator such that either (i) kAuk 6 kuk, u 2 K \ oX1 and kAuk P kuk, u 2 K \ oX2; or (ii) kAuj P kuk, u 2 K \ oX1 and kAuk 6 kuk, u 2 K \ oX2. Then A has a fixed point in K \ ðX2 n X1 Þ. For convenient, we denote Z 1 A1 :¼ Gðs; sÞpðsÞhðsÞ ds þ 0
a 1 a/1 ðgÞ
and Z
1d
Gðx0 ; sÞpðsÞhðsÞ ds:
A2 :¼ d
Z
1
Gðg; sÞpðsÞhðsÞ ds 0
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3.1. The existence results of the BVP (1.1) for the case: f0 = f1 = 1 or f0 = f1 = 0 Theorem 3.3. Assume that the following assumptions are satisfied. (B1) f0 = f1 = 1, (B2) There exists a constant q1 > 0 such that f ðuÞ 6 A1 1 q1 ;
f or
u 2 ½0; q1 :
Then BVP (1.1) has at least two solutions u1 and u2 such that 0 < ku1 k < q1 < ku2 k: Proof. At first, since f0 = 1, we may choose q0 2 (0, q1) such that f(u) P Mu for 0 < u < q0, where M satisfies McA2 P 1. Let Xq0 ¼ fu 2 C½0; 1 : kuk < q0 g, for u 2 P and kuk = q0 we have Z 1 Z 1 a Tuðx0 Þ ¼ Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞMuðsÞ ds P McA2 kuk P kuk 0
d
so kTuk P kuk;
ð3:4Þ
u 2 P \ oXq0 :
Further, in view of f1 = 1, there exists q10 > q1 such that f(u) P M1u for u P q10, where M1 > 0 is chosen so that M1cA2 P 1. Let q0 ¼ maxf2q1 ; q10 =cg and Xq0 ¼ fu 2 C½0; 1 : kuk < q0 g, then u 2 P and kuk ¼ q0 implies min uðtÞ P ckuk P q10 :
d6t61d
So we have Z 1 a Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ Tuðx0 Þ ¼ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞM 1 uðsÞ ds P M 1 cA2 kuk P kuk: Z
1
0
d
Hence kTuk P kuk
for u 2 P \ oXq0 :
ð3:5Þ
Finally, let Xq1 ¼ fu 2 C½0; 1 : kuk < q1 g. For u 2 P with kuk = q1, we have Z 1 Z 1 a TuðtÞ ¼ Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ 1 a/1 ðgÞ 0 0 Z 1 Z 1 a 1 6 A1 q1 Gðs; sÞpðsÞhðsÞ ds þ Gðg; sÞpðsÞhðsÞ ds ¼ q1 1 a/1 ðgÞ 0 0 which yields kTuk 6 kuk
for u 2 P \ oXq1 :
ð3:6Þ
Thus from (3.4)–(3.6) and Theorem 3.2, T has a fixed point u1 in P \ ðXq1 n Xq0 Þ, and a fixed point u2 in P \ ðXq0 n Xq1 Þ. Both are positive solutions of BVP (1.1) and 0 < ku1 k < q1 < ku2 k: The proof is complete.
h
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263
Theorem 3.4. Assume that the following assumptions are satisfied. (B3) f0 = f1 = 0, (B4) There exists q2 > 0 such that f ðuÞ P A1 2 q2
for u 2 ½cq2 ; q2 :
Then the BVP (1.1) has at least two positive solutions u1 and u2 such that 0 < ku1 k < q2 < ku2 k: Proof. Since f0 = 0, we may choose 0 < q < q2 such that f(u) 6 eu for 0 < u < q, where e > 0 satisfies eA1 6 1. Thus for u 2 P and kuk = q, we get Z 1 a Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ 1 a/1 ðgÞ 0 0 Z 1 Z 1 a 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 6 eA1 kuk 6 kuk: 1 a/1 ðgÞ 0 0
TuðtÞ ¼
Z
1
Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
Let Xq = {u 2 C[0, 1] : kuk < q}, it follows that: kTuk 6 kuk
for u 2 P \ oXq :
ð3:7Þ
In view of f1 = 0, there exists r0 > q2 such that f(u) 6 e1u for u P r0, where e1 > 0 satisfies e1A1 6 1. We consider two cases: Case (i). Suppose that f is unbounded, there exists q* > r0 such that f(u) 6 f(q*) for 0 < u 6 q*. Then for u 2 P and kuk = q*, we have Z 1 Z 1 a TuðtÞ ¼ Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ 1 a/1 ðgÞ 0 0 Z 1 Z 1 a 6 e1 q Gðs; sÞpðsÞhðsÞ ds þ Gðg; sÞpðsÞhðsÞ ds 6 q ¼ kuk: 1 a/1 ðgÞ 0 0 Case (ii) . If f is bounded, say f(u) 6 N for all u 2 [0, 1). Taking q* P max{2q2, NA1}, for u 2 P with kuk = q*, we have Z 1 Z 1 a TuðtÞ 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 6 NA1 6 q ¼ kuk: 1 a/ ðgÞ 0 0 1 Hence, in either case, we always may set Xq ¼ fu 2 C½0; 1 : kuk < q g such that kTuk 6 kuk
ð3:8Þ
for u 2 P \ oXq :
Finally, set Xq2 ¼ fu 2 C½0; 1 : kuk < q2 g, then u 2 P and kuk = q2 implies min uðtÞ P ckuk ¼ cq2
d6t61d
and so Z 1 a Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ Tuðx0 Þ ¼ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞA1 2 q2 ds ¼ q2 ¼ kuk: Z
1
0
d
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Hence for u 2 P \ oXq2 , we have kTuk P kuk: ð3:9Þ From (3.7)–(3.9) and Theorem 3.2, T has a fixed point u1 in P \ ðXq2 n Xq Þ, and a fixed point u2 2 P \ ðXq n Xq2 Þ. Both are positive solutions of BVP (1.1) and 0 < ku1 k < q2 < ku2 k; which complete the proof.
h
3.2. The existence results of the BVP (1.1) for the case: f0, f1 62 {0, 1} Now, we discuss the existence for the positive solutions of the BVP (1.1) under f0, f1 62 {0, 1}. We state and prove the following main result. Theorem 3.5. Assume that there exist two positive constants q1 5 q2 such that (C1) f ðuÞ 6 A1 1 q1 for u 2 [0, q1], (C2) f ðuÞ P A1 2 q2 for u 2 [cq2, q2]. Then the BVP (1.1) has at least one positive solution u such that kuk between q1 and q2. Proof. With loss of generality, we may assume that q1 < q2. Let Xq1 ¼ fu 2 C½0; 1 : kuk < q1 g, for u 2 P and kuk = q1, one has Z 1 Z 1 a TuðtÞ 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 6 A1 1 q1 A1 ¼ q1 ¼ kuk; 1 a/1 ðgÞ 0 0 which shows that kTuk 6 kuk
ð3:10Þ
for u 2 P \ oXq1 :
Now let Xq2 ¼ fu 2 C½0; 1 : kuk < q2 g for u 2 P with kuk = q2, we have min uðtÞ P ckuk ¼ cq2 ;
d6t61d
thus, we get Z 1 a Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P A1 2 q2 A2 ¼ q2 ¼ kuk
Tuðx0 Þ ¼
Z
1
Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
0
d
and so kTuk P kuk
for u 2 P \ oXq2 :
Hence, from (3.10) and (3.11) and Theorem 3.2, we complete the proof.
Corollary 3.1. Assume that the following assumptions hold. (C3) f0 ¼ a1 2 ½0; A11 Þ; (C4) f1 ¼ b1 2 cA12 ; þ1 . Then BVP (1.1) has at least one positive solution.
ð3:11Þ h
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265
Proof. By (C3), for e ¼ A1 1 a1 > 0, there exists a sufficiently small q1 > 0 such that f ðuÞ 1 6 a1 þ e ¼ u A1
for 0 < u 6 q1 ;
which implies 1 f ðuÞ 6 A1 1 u 6 A1 q1
for 0 < u 6 q1 :
This shows that the condition (C1) of Theorem 3.5 is satisfied. By (C4), for e1 ¼ b1 cA1 2 > 0, there exists a sufficiently large q2 such that f ðuÞ 1 P b1 e1 ¼ u cA2
for u P cq2 ;
thus f ðuÞ P ðcA2 Þ1 u P A1 2 q2 ; which implies the condition (C2) of Theorem 3.5 hold. Hence, by Theorem 3.5, BVP (1.1) has at least one positive solution. The proof is complete. h Similarly, we can prove the following conclusion. Corollary 3.2. Assume that the following hypothesis hold. (C5) f0 ¼ a2 2 ðcA12 ; þ1Þ; (C6) f1 ¼ b2 2 ½0; A11 Þ. Then the BVP (1.1) has at least one positive solution. Corollary 3.3. Assume that the previous hypotheses (C1), (C4) and (C5) hold. Then the BVP (1.1) has at least two positive solutions u1 and u2 such that 0 < ku1 k < q1 < ku2 k: Corollary 3.4. Assume that the previous hypotheses (C2), (C3) and (C6) hold. Then the BVP (1.1) has at least two positive solutions u1 and u2 such that 0 < ku1 k < q2 < ku2 k: 4. Existence of three positive solutions Let E be a real Banach space with cone P. A map b : P ! [0, +1) is said to be a nonnegative continuous concave functional on P if b is continuous and bðtx þ ð1 tÞyÞ P tbðxÞ þ ð1 tÞbðyÞ for all x, y 2 P and t 2 [0, 1]. Let a, b be two numbers such that 0 < a < b and b a nonnegative continuous concave functional on P. We define the following convex sets: P a ¼ fx 2 P : kxk < ag; P ðb; a; bÞ ¼ fx 2 P : a 6 bðxÞ; kxk 6 bg: Theorem 4.1 (Leggett–Williams fixed point theorem [14]). Let T: P c ! P c be completely continuous and b be a nonnegative continuous concave functional on P such that b(x) 6 kxk for all x 2 P c . Suppose there exist 0 < d < a < b 6 c such that (i) {x 2 P(b, a, b):b(x) > a} 5 ; and b(Tx) > a for x 2 P(b, a, b), (ii) kTxk < d for kxk 6 d, (iii) b(Tx) > a for x 2 P(b, a, c) with kTxk > b.
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J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
Then T has at least three fixed points x1, x2, x3 in P c such that kx1 k < d;
a < bðx2 Þ and
kx3 k > d
with bðx3 Þ < a:
Now, we establish the existence conditions of three positive solutions for the BVP (1.1). Theorem 4.2. Suppose that (H1)(H4) hold and that there exist numbers a and d with 0 < d < a such that f ðuÞ <
d ; D
u 2 ½0; d
ð4:1Þ
f ðuÞ >
a ; C
u 2 ½a; a=c;
ð4:2Þ
and
where D ¼ max
Z
t2½0;1
C ¼ min
d6t61d
Gðt; sÞpðsÞhðsÞ ds 1 þ
1
0
Z
a ; 1 a/1 ðgÞ
1d
Gðt; sÞpðsÞhðsÞ ds: d
Suppose further that one of the following conditions holds: (P1) limu!1 f ðuÞ < D1 , u (P2) there exists a number c such that c > a/c and if u 2 [0,c]. Then f ðuÞ < Dc . Then the boundary value problem (1.1) has at least three positive solutions. Proof. Let P be defined by (3.2) and T be defined by (3.1). For u 2 P, let bðuÞ ¼ min uðtÞ: d6t61d
Then it is easy to check that b is a nonnegative continuous concave functional on P with b(x) 6 kxk for x 2 P and T : P ! P is completely continuous. First, we prove that if (P1) holds, then there exists a number c such that c > a/c and T : P c ! P c . To do this, by (P1), there exists M > 0 and r < 1/D such that f ðuÞ 6 ru;
for u > M:
Set e ¼ max f ðuÞ; u2½0;M
it follows that f(u) 6 ru + e for all u 2 [0, +1). Take eD a ; : c > max 1 rD c If u 2 P c , then Z
Z 1 a Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 1 a/1 ðgÞ 0 0 Z 1 a 6 max Gðt; sÞpðsÞhðsÞ ds 1 þ ðrkuk þ eÞ 6 ðrc þ eÞD < c; t2½0;1 0 1 a/1 ðgÞ
TuðtÞ 6 maxt2½0;1
1
Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
that is kTuk < c:
ð4:3Þ
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
267
Next, we assert that if there exists a positive number r such that f(u) < r/D for u 2 [0, r], then T : P r ! P r . Indeed, if u 2 P r ; then Z 1 r a max kTuk < Gðt; sÞpðsÞhðsÞ ds 1 þ ¼ r: ð4:4Þ D t2½0;1 0 1 a/1 ðgÞ Hence (4.3) and (4.4) show that if either (P1) or (P2) holds, then there exists a number c > a/c such that T maps P c into Pc. Note that if r = d, then we may assert further that T maps P d into Pd by (4.1). Now we show that {u 2 P(b, a, a/c) : b (u) > a} 5 ; and b(Tu) > a for all u 2 P(b, a, a/c). In fact, take xðtÞ aþa=c > a; so x 2 {u 2 P(b, a, a/c) : b(u) > a}. Moreover, for u 2 P(b, a, a/c), then b(u) P a, and we have 2 a=c P kuk P min uðtÞ ¼ bðuÞ P a: d6t61d
Thus, in view of (4.2) we obtain Z 1 Gðt; sÞpðsÞhðsÞf ðuðsÞÞ dsþ bðTuÞ ¼ min d6t61d
Z
0
Z
1
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ
0
1d
Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds ¼ a:
P min
d6t61d
a 1 a/1 ðgÞ
d
Finally, we assert that if u 2 P(b, a, c) and kTuk > a/c, then b(Tu) > a. To see this, if u 2 P(b, a, c) and kTuk > a/c, then we have from G(t, s) P q(t)G(s, s) and /1(t) P q(t) that Z 1 Z 1 a bðTuÞ ¼ min Gðt; sÞpðsÞhðsÞf ðuðsÞÞ dsþ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ d6t61d 1 a/1 ðgÞ 0 0 Z 1 Z 1 a P min qðtÞ Gðs; sÞpðsÞhðsÞf ðuðsÞÞ dsþ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds d6t61d 1 a/1 ðgÞ 0 0 Z 1 Z 1 a ¼c Gðs; sÞpðsÞhðsÞf ðuðsÞÞ dsþ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds : ð4:5Þ 1 a/1 ðgÞ 0 0 On the other hand Z 1 kTuk 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ 0
a 1 a/1 ðgÞ
Z
1
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds:
ð4:6Þ
0
So we get from (4.5) and (4.6) that a bðTuÞ P ckTuk > c ¼ a: c To sum up, all the hypotheses of Theorem 4.1 are satisfied by taking b = a/c. Hence, T has at least three fixed points, that is, BVP (1.1) has at least three positive solutions u1, u2 and u3 such that ku1 k < d;
a < aðu2 Þ
The proof is complete.
and
ku3 k > d
with aðu3 Þ < a:
h
References [1] V.A. Il’in, E.I. Moviseev, Nonlocal boundary value problem of the second kind for a Sturm–Liouville operator, Differen. Equat. 23 (1987) 979–987. [2] V.A. Il’in, E.I. Moviseev, Nonlocal boundary value problem of the first kind for a Sturm–Liouville operator in its differential and finite difference aspects, Differen. Equat. 23 (1987) 803–810. [3] R.Y. Ma, H.Y. Wang, Positive solutions of nonlinear three-point boundary value problems, J. Math. Anal. Appl. 279 (2003) 216–227. [4] R.Y. Ma, Positive solutions of a nonlinear three-point boundary value problem, Electron J. Differen. Equat. 34 (1999) 1–8.
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[5] C.P. Gupta, A sharper condition for the solvability of a three-point second order boundary value problem, J. Math. Anal. Appl. 205 (1997) 579–586. [6] J.P. Sun, W.T. Li, Multiple positive solutions to second order Neumann boundary value problems, Appl. Math. Comput. 146 (2003) 187–194. [7] C.P. Gupta, A generalized multi-point boundary value problem for second order ordinary differential equations, Appl. Math. Comput. 89 (1998) 133–146. [8] B. Liu, Positive solutions of fourth-order two point boundary value problems, Appl. Math. Comput. 148 (2004) 407–420. [9] M.A. Krasnoselskii, Positive Solution of Operator Equations, Noordhoof, Gronignen, 1964. [10] T.P. Sun, W.T. Li, S.S. Cheng, Three positive solutions for second order Neumann boundary value problems, Appl. Math. Lett. 17 (2004) 1079–1084. [11] R.I. Avery, J. Henderson, Three symmetric positive solutions for second order boundary value problem, Appl. Math. Lett. 13 (2000) 1–7. [12] Y.P. Guo, X.J. Liu, J.Q. Qiu, Three positive solutions for higher order m-point boundary value problems, J. Math. Anal. Appl. 289 (2004) 545–553. [13] D.R. Anderson, Multiple periodic solutions for second-order problem on periodic time scales, Nonlinear Anal. 60 (2005) 101–115. [14] R.W. Leggett, L.R. Williams, Multiple positive fixed points of nonlinear operator on ordered Banach spaces, Indiana Univ. Math. J. 28 (1979) 673–688.
Multiple positive solutions for a second-order three-point boundary value problem q Jianli Li a
a,*
, Jianhua Shen
a,b
Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China b Department of Mathematics, College of Huaihua, Huaihua, Hunan 418008, China
Abstract In this paper, we study the second-order three-point boundary value problem u00 ðtÞ þ aðtÞu0 ðtÞ þ bðtÞuðtÞ þ hðtÞf ðuÞ ¼ 0; uð0Þ ¼ 0;
t 2 ð0; 1Þ;
auðgÞ ¼ uð1Þ:
The existence criteria for multiple positive solutions is obtained by using Krasnoselskii fixed point theorem and Leggett– Williams fixed-point theorem. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Positive solution; Fixed point theorem; Three-point boundary value problem
1. Introduction The study of multi-point boundary value problems for linear second-order ordinary differential equations was initiated by Il’in and Moviseev [1,2]. Since then, nonlinear second-order multi-point boundary value problems have also been studied by several authors. We refer the reader to [3–7,10,11] and their references. Recently, Ma and Wang [3] used Krasnoselskii’s fixed point theorem in cones to prove the existence of at least one positive solutions for the three-point boundary value problem (BVP) u00 ðtÞ þ aðtÞu0 ðtÞ þ bðtÞuðtÞ þ hðtÞf ðuÞ ¼ 0; uð0Þ ¼ 0;
t 2 ð0; 1Þ;
auðgÞ ¼ uð1Þ;
ð1:1Þ
where 0 < g < 1, a is a positive constant, h, f, a and b satisfy q
This work is supported by the NNSF of China (No. 10571050), and supported by the Key Project of Chinese Ministry of Education, and supported by Hunan Provincial Natural Science Foundation of China (No. 005JJ4001). * Corresponding author. E-mail address: [email protected] (J. Li). 0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.01.095
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
259
(H1) f 2 C([0, 1), [0, 1)); (H2) h 2 C([0, 1], [0, 1)) and there exists x0 2 (0, 1) such that h(x0) > 0; (H3) a 2 C[0, 1], b 2 C([0, 1], (1, 0)). Motivated by the results [3], this paper is concerned with the existence of multiple positive solutions of the above boundary problems. Our tool in this paper will be well-known Krasnoselskii fixed point theorem [3,8,9] and Leggett–Williams fixed-point theorem [10–14]. 2. Preliminaries In order to obtain our result, in this section, we mention without proof several foundational results. Lemma 2.1 [3]. Assume that (H3) hold. Let /1 and /2 be the solutions of 00 /1 ðtÞ þ aðtÞ/01 ðtÞ þ bðtÞ/1 ðtÞ ¼ 0; t 2 ð0; 1Þ; /1 ð0Þ ¼ 0; /1 ð1Þ ¼ 1 and 00 /2 ðtÞ þ aðtÞ/02 ðtÞ þ bðtÞ/2 ðtÞ ¼ 0; t 2 ð0; 1Þ; /2 ð0Þ ¼ 1; /2 ð1Þ ¼ 0:
ð2:1Þ
ð2:2Þ
Then (i) /1 is strictly increasing on [0, 1], (ii) /2 is strictly decreasing on [0, 1].
Lemma 2.2 [3]. Assume that (H3) hold. Then (2.1) and (2.2) have unique solution, respectively. Now, we assume the following condition holds. (H4) 0 < a/1(g) < 1. Lemma 2.3 [3]. Assume that (H3) and (H4) hold. Let y 2 C[0, 1]. Then the problem 00 u ðtÞ þ aðtÞu0 ðtÞ þ bðtÞuðtÞ þ yðtÞ ¼ 0; t 2 ð0; 1Þ; uð0Þ ¼ 0; auðgÞ ¼ uð1Þ; is equivalent to the integral equation Z 1 Gðt; sÞpðsÞyðsÞ ds þ A/1 ðtÞ; uðtÞ ¼
ð2:3Þ
ð2:4Þ
0
where
Z 1 a Gðg; sÞpðsÞyðsÞ ds; 1 a/1 ðgÞ 0 Z t aðsÞ ds ; pðtÞ ¼ exp 0 ( 1 /1 ðtÞ/2 ðsÞ; if s P t; Gðt; sÞ ¼ q /1 ðsÞ/2 ðtÞ; if t P s;
A¼
and q :¼ /01 ð0Þ: Moreover, u(t) P 0 on [0, 1] provided y P 0.
ð2:5Þ ð2:6Þ ð2:7Þ
260
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
Set
/1 ðtÞ /2 ðtÞ qðtÞ ¼ min ; ; k/1 k k/2 k where kÆk denote the supremum norm. From (2.7) we have Gðs; sÞ P Gðt; sÞ P qðtÞGðs; sÞ P 0; for t 2 ½0; 1: Choose d 2 (0, 1/2) such that x0 2 (d,1 d), take c ¼ minfqðtÞjt 2 ½d; 1 dg; it follows from Lemma 2.1 that 0 < c < 1. Lemma 2.4 [3]. Assume that (H3) and (H4) hold. Let y 2 C[0, 1] and y P 0, then the unique solution u of the problem (2.3) satisfies t 2 ½d; 1 d:
uðtÞ P ckuk;
3. Existence of one or two positive solutions In this section, we establish the existence conditions of one or two positive solutions for the BVP (1.1). Let E = C[0, 1]. Then E is a Banach space with respect to the norm kuk ¼ sup juðtÞj: t2½0;1
Now BVP (1.1) has a solution u = u(t) if and only if u is a solution of the operator equation Z 1 Z 1 a uðtÞ ¼ Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ :¼ ðTuÞðtÞ 1 a/1 ðgÞ 0 0
ð3:1Þ
in E, where p and G are defined by (2.6) and (2.7), respectively. It is easy to check that T is completely continuous. We define a cone in E by P ¼ u 2 E : u P 0; min uðtÞ P ckuk : ð3:2Þ d6t61d
Lemma 3.1. T(P) P. Proof. Note that G(t, s) P 0 and 0 < a/1(g) < 1 implies (Tu)(t) P 0. For any y 2 P, we have from (3.1) and G(t, s) 6 G(s, s), /1(t) 6 1 that Z 1 Z 1 a TuðtÞ 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 1 a/1 ðgÞ 0 0 which implies kTuðtÞk 6
Z
1
Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
0
From (2.7) we have Gðt; sÞ P qðtÞ; Gðs; sÞ
t 2 ð0; 1; s 2 ð0; 1Þ
a 1 a/1 ðgÞ
Z 0
1
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds:
ð3:3Þ
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
and also q(t) 6 /1(t), so we have Z 1 Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ min ðTuÞðtÞ ¼ min d6t61d
d6t61d
0
Z
a 1 a/1 ðgÞ
Z
1
261
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ
0
Z 1 a P min qðtÞ Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds d6t61d 1 a/1 ðgÞ 0 0 Z 1 Z 1 a Pc Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 1 a/1 ðgÞ 0 0 1
which and (3.3) implies min ðTuÞðtÞ P ckTuk: d6t61d
Hence Tu 2 P. This proof is complete. h Set f0 ¼ limþ u!0
f ðuÞ ; u
f 1 ¼ lim
u!1
f ðuÞ : u
The following Theorem is main result of [3]. Theorem 3.1 [3]. Let (H1)(H4) hold. Then the problem (1.1) has at least one positive solution in the case (i) f0 = 0 and f1 = 1 (superlinear), or (ii) f0 = 1 and f1 = 0 (sublinear).
Problem 1. Whether we can obtain a similar conclusion or not, if f0 = f1 = 0 or f0 = f1 = 1. Problem 2. Whether we can get a similar conclusion or not, if f0 = f1 62 {0, +1}. The aim of this section is to establish some simple criteria for the existence of positive solutions of BVP (1.1), which gives a positive answer to the questions stated above. The key tool in our approach is the following Krasnoselskii fixed point theorem, which is a useful method to prove the existence of positive solution for differential equations, for example [3,8]. Theorem 3.2 [9]. Let E be a Banach space, and let K E be a cone. Assume X1, X2 are open subsets of E with 0 2 X1, X1 X2 , and let A : K \ ðX2 n X1 Þ ! K be a completely continuous operator such that either (i) kAuk 6 kuk, u 2 K \ oX1 and kAuk P kuk, u 2 K \ oX2; or (ii) kAuj P kuk, u 2 K \ oX1 and kAuk 6 kuk, u 2 K \ oX2. Then A has a fixed point in K \ ðX2 n X1 Þ. For convenient, we denote Z 1 A1 :¼ Gðs; sÞpðsÞhðsÞ ds þ 0
a 1 a/1 ðgÞ
and Z
1d
Gðx0 ; sÞpðsÞhðsÞ ds:
A2 :¼ d
Z
1
Gðg; sÞpðsÞhðsÞ ds 0
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3.1. The existence results of the BVP (1.1) for the case: f0 = f1 = 1 or f0 = f1 = 0 Theorem 3.3. Assume that the following assumptions are satisfied. (B1) f0 = f1 = 1, (B2) There exists a constant q1 > 0 such that f ðuÞ 6 A1 1 q1 ;
f or
u 2 ½0; q1 :
Then BVP (1.1) has at least two solutions u1 and u2 such that 0 < ku1 k < q1 < ku2 k: Proof. At first, since f0 = 1, we may choose q0 2 (0, q1) such that f(u) P Mu for 0 < u < q0, where M satisfies McA2 P 1. Let Xq0 ¼ fu 2 C½0; 1 : kuk < q0 g, for u 2 P and kuk = q0 we have Z 1 Z 1 a Tuðx0 Þ ¼ Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞMuðsÞ ds P McA2 kuk P kuk 0
d
so kTuk P kuk;
ð3:4Þ
u 2 P \ oXq0 :
Further, in view of f1 = 1, there exists q10 > q1 such that f(u) P M1u for u P q10, where M1 > 0 is chosen so that M1cA2 P 1. Let q0 ¼ maxf2q1 ; q10 =cg and Xq0 ¼ fu 2 C½0; 1 : kuk < q0 g, then u 2 P and kuk ¼ q0 implies min uðtÞ P ckuk P q10 :
d6t61d
So we have Z 1 a Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ Tuðx0 Þ ¼ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞM 1 uðsÞ ds P M 1 cA2 kuk P kuk: Z
1
0
d
Hence kTuk P kuk
for u 2 P \ oXq0 :
ð3:5Þ
Finally, let Xq1 ¼ fu 2 C½0; 1 : kuk < q1 g. For u 2 P with kuk = q1, we have Z 1 Z 1 a TuðtÞ ¼ Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ 1 a/1 ðgÞ 0 0 Z 1 Z 1 a 1 6 A1 q1 Gðs; sÞpðsÞhðsÞ ds þ Gðg; sÞpðsÞhðsÞ ds ¼ q1 1 a/1 ðgÞ 0 0 which yields kTuk 6 kuk
for u 2 P \ oXq1 :
ð3:6Þ
Thus from (3.4)–(3.6) and Theorem 3.2, T has a fixed point u1 in P \ ðXq1 n Xq0 Þ, and a fixed point u2 in P \ ðXq0 n Xq1 Þ. Both are positive solutions of BVP (1.1) and 0 < ku1 k < q1 < ku2 k: The proof is complete.
h
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263
Theorem 3.4. Assume that the following assumptions are satisfied. (B3) f0 = f1 = 0, (B4) There exists q2 > 0 such that f ðuÞ P A1 2 q2
for u 2 ½cq2 ; q2 :
Then the BVP (1.1) has at least two positive solutions u1 and u2 such that 0 < ku1 k < q2 < ku2 k: Proof. Since f0 = 0, we may choose 0 < q < q2 such that f(u) 6 eu for 0 < u < q, where e > 0 satisfies eA1 6 1. Thus for u 2 P and kuk = q, we get Z 1 a Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ 1 a/1 ðgÞ 0 0 Z 1 Z 1 a 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 6 eA1 kuk 6 kuk: 1 a/1 ðgÞ 0 0
TuðtÞ ¼
Z
1
Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
Let Xq = {u 2 C[0, 1] : kuk < q}, it follows that: kTuk 6 kuk
for u 2 P \ oXq :
ð3:7Þ
In view of f1 = 0, there exists r0 > q2 such that f(u) 6 e1u for u P r0, where e1 > 0 satisfies e1A1 6 1. We consider two cases: Case (i). Suppose that f is unbounded, there exists q* > r0 such that f(u) 6 f(q*) for 0 < u 6 q*. Then for u 2 P and kuk = q*, we have Z 1 Z 1 a TuðtÞ ¼ Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ 1 a/1 ðgÞ 0 0 Z 1 Z 1 a 6 e1 q Gðs; sÞpðsÞhðsÞ ds þ Gðg; sÞpðsÞhðsÞ ds 6 q ¼ kuk: 1 a/1 ðgÞ 0 0 Case (ii) . If f is bounded, say f(u) 6 N for all u 2 [0, 1). Taking q* P max{2q2, NA1}, for u 2 P with kuk = q*, we have Z 1 Z 1 a TuðtÞ 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 6 NA1 6 q ¼ kuk: 1 a/ ðgÞ 0 0 1 Hence, in either case, we always may set Xq ¼ fu 2 C½0; 1 : kuk < q g such that kTuk 6 kuk
ð3:8Þ
for u 2 P \ oXq :
Finally, set Xq2 ¼ fu 2 C½0; 1 : kuk < q2 g, then u 2 P and kuk = q2 implies min uðtÞ P ckuk ¼ cq2
d6t61d
and so Z 1 a Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ Tuðx0 Þ ¼ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞA1 2 q2 ds ¼ q2 ¼ kuk: Z
1
0
d
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J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
Hence for u 2 P \ oXq2 , we have kTuk P kuk: ð3:9Þ From (3.7)–(3.9) and Theorem 3.2, T has a fixed point u1 in P \ ðXq2 n Xq Þ, and a fixed point u2 2 P \ ðXq n Xq2 Þ. Both are positive solutions of BVP (1.1) and 0 < ku1 k < q2 < ku2 k; which complete the proof.
h
3.2. The existence results of the BVP (1.1) for the case: f0, f1 62 {0, 1} Now, we discuss the existence for the positive solutions of the BVP (1.1) under f0, f1 62 {0, 1}. We state and prove the following main result. Theorem 3.5. Assume that there exist two positive constants q1 5 q2 such that (C1) f ðuÞ 6 A1 1 q1 for u 2 [0, q1], (C2) f ðuÞ P A1 2 q2 for u 2 [cq2, q2]. Then the BVP (1.1) has at least one positive solution u such that kuk between q1 and q2. Proof. With loss of generality, we may assume that q1 < q2. Let Xq1 ¼ fu 2 C½0; 1 : kuk < q1 g, for u 2 P and kuk = q1, one has Z 1 Z 1 a TuðtÞ 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 6 A1 1 q1 A1 ¼ q1 ¼ kuk; 1 a/1 ðgÞ 0 0 which shows that kTuk 6 kuk
ð3:10Þ
for u 2 P \ oXq1 :
Now let Xq2 ¼ fu 2 C½0; 1 : kuk < q2 g for u 2 P with kuk = q2, we have min uðtÞ P ckuk ¼ cq2 ;
d6t61d
thus, we get Z 1 a Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðx0 Þ 1 a/1 ðgÞ 0 0 Z 1 Z 1d P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds P A1 2 q2 A2 ¼ q2 ¼ kuk
Tuðx0 Þ ¼
Z
1
Gðx0 ; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
0
d
and so kTuk P kuk
for u 2 P \ oXq2 :
Hence, from (3.10) and (3.11) and Theorem 3.2, we complete the proof.
Corollary 3.1. Assume that the following assumptions hold. (C3) f0 ¼ a1 2 ½0; A11 Þ; (C4) f1 ¼ b1 2 cA12 ; þ1 . Then BVP (1.1) has at least one positive solution.
ð3:11Þ h
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
265
Proof. By (C3), for e ¼ A1 1 a1 > 0, there exists a sufficiently small q1 > 0 such that f ðuÞ 1 6 a1 þ e ¼ u A1
for 0 < u 6 q1 ;
which implies 1 f ðuÞ 6 A1 1 u 6 A1 q1
for 0 < u 6 q1 :
This shows that the condition (C1) of Theorem 3.5 is satisfied. By (C4), for e1 ¼ b1 cA1 2 > 0, there exists a sufficiently large q2 such that f ðuÞ 1 P b1 e1 ¼ u cA2
for u P cq2 ;
thus f ðuÞ P ðcA2 Þ1 u P A1 2 q2 ; which implies the condition (C2) of Theorem 3.5 hold. Hence, by Theorem 3.5, BVP (1.1) has at least one positive solution. The proof is complete. h Similarly, we can prove the following conclusion. Corollary 3.2. Assume that the following hypothesis hold. (C5) f0 ¼ a2 2 ðcA12 ; þ1Þ; (C6) f1 ¼ b2 2 ½0; A11 Þ. Then the BVP (1.1) has at least one positive solution. Corollary 3.3. Assume that the previous hypotheses (C1), (C4) and (C5) hold. Then the BVP (1.1) has at least two positive solutions u1 and u2 such that 0 < ku1 k < q1 < ku2 k: Corollary 3.4. Assume that the previous hypotheses (C2), (C3) and (C6) hold. Then the BVP (1.1) has at least two positive solutions u1 and u2 such that 0 < ku1 k < q2 < ku2 k: 4. Existence of three positive solutions Let E be a real Banach space with cone P. A map b : P ! [0, +1) is said to be a nonnegative continuous concave functional on P if b is continuous and bðtx þ ð1 tÞyÞ P tbðxÞ þ ð1 tÞbðyÞ for all x, y 2 P and t 2 [0, 1]. Let a, b be two numbers such that 0 < a < b and b a nonnegative continuous concave functional on P. We define the following convex sets: P a ¼ fx 2 P : kxk < ag; P ðb; a; bÞ ¼ fx 2 P : a 6 bðxÞ; kxk 6 bg: Theorem 4.1 (Leggett–Williams fixed point theorem [14]). Let T: P c ! P c be completely continuous and b be a nonnegative continuous concave functional on P such that b(x) 6 kxk for all x 2 P c . Suppose there exist 0 < d < a < b 6 c such that (i) {x 2 P(b, a, b):b(x) > a} 5 ; and b(Tx) > a for x 2 P(b, a, b), (ii) kTxk < d for kxk 6 d, (iii) b(Tx) > a for x 2 P(b, a, c) with kTxk > b.
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Then T has at least three fixed points x1, x2, x3 in P c such that kx1 k < d;
a < bðx2 Þ and
kx3 k > d
with bðx3 Þ < a:
Now, we establish the existence conditions of three positive solutions for the BVP (1.1). Theorem 4.2. Suppose that (H1)(H4) hold and that there exist numbers a and d with 0 < d < a such that f ðuÞ <
d ; D
u 2 ½0; d
ð4:1Þ
f ðuÞ >
a ; C
u 2 ½a; a=c;
ð4:2Þ
and
where D ¼ max
Z
t2½0;1
C ¼ min
d6t61d
Gðt; sÞpðsÞhðsÞ ds 1 þ
1
0
Z
a ; 1 a/1 ðgÞ
1d
Gðt; sÞpðsÞhðsÞ ds: d
Suppose further that one of the following conditions holds: (P1) limu!1 f ðuÞ < D1 , u (P2) there exists a number c such that c > a/c and if u 2 [0,c]. Then f ðuÞ < Dc . Then the boundary value problem (1.1) has at least three positive solutions. Proof. Let P be defined by (3.2) and T be defined by (3.1). For u 2 P, let bðuÞ ¼ min uðtÞ: d6t61d
Then it is easy to check that b is a nonnegative continuous concave functional on P with b(x) 6 kxk for x 2 P and T : P ! P is completely continuous. First, we prove that if (P1) holds, then there exists a number c such that c > a/c and T : P c ! P c . To do this, by (P1), there exists M > 0 and r < 1/D such that f ðuÞ 6 ru;
for u > M:
Set e ¼ max f ðuÞ; u2½0;M
it follows that f(u) 6 ru + e for all u 2 [0, +1). Take eD a ; : c > max 1 rD c If u 2 P c , then Z
Z 1 a Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds 1 a/1 ðgÞ 0 0 Z 1 a 6 max Gðt; sÞpðsÞhðsÞ ds 1 þ ðrkuk þ eÞ 6 ðrc þ eÞD < c; t2½0;1 0 1 a/1 ðgÞ
TuðtÞ 6 maxt2½0;1
1
Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds þ
that is kTuk < c:
ð4:3Þ
J. Li, J. Shen / Applied Mathematics and Computation 182 (2006) 258–268
267
Next, we assert that if there exists a positive number r such that f(u) < r/D for u 2 [0, r], then T : P r ! P r . Indeed, if u 2 P r ; then Z 1 r a max kTuk < Gðt; sÞpðsÞhðsÞ ds 1 þ ¼ r: ð4:4Þ D t2½0;1 0 1 a/1 ðgÞ Hence (4.3) and (4.4) show that if either (P1) or (P2) holds, then there exists a number c > a/c such that T maps P c into Pc. Note that if r = d, then we may assert further that T maps P d into Pd by (4.1). Now we show that {u 2 P(b, a, a/c) : b (u) > a} 5 ; and b(Tu) > a for all u 2 P(b, a, a/c). In fact, take xðtÞ aþa=c > a; so x 2 {u 2 P(b, a, a/c) : b(u) > a}. Moreover, for u 2 P(b, a, a/c), then b(u) P a, and we have 2 a=c P kuk P min uðtÞ ¼ bðuÞ P a: d6t61d
Thus, in view of (4.2) we obtain Z 1 Gðt; sÞpðsÞhðsÞf ðuðsÞÞ dsþ bðTuÞ ¼ min d6t61d
Z
0
Z
1
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ
0
1d
Gðt; sÞpðsÞhðsÞf ðuðsÞÞ ds ¼ a:
P min
d6t61d
a 1 a/1 ðgÞ
d
Finally, we assert that if u 2 P(b, a, c) and kTuk > a/c, then b(Tu) > a. To see this, if u 2 P(b, a, c) and kTuk > a/c, then we have from G(t, s) P q(t)G(s, s) and /1(t) P q(t) that Z 1 Z 1 a bðTuÞ ¼ min Gðt; sÞpðsÞhðsÞf ðuðsÞÞ dsþ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds /1 ðtÞ d6t61d 1 a/1 ðgÞ 0 0 Z 1 Z 1 a P min qðtÞ Gðs; sÞpðsÞhðsÞf ðuðsÞÞ dsþ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds d6t61d 1 a/1 ðgÞ 0 0 Z 1 Z 1 a ¼c Gðs; sÞpðsÞhðsÞf ðuðsÞÞ dsþ Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds : ð4:5Þ 1 a/1 ðgÞ 0 0 On the other hand Z 1 kTuk 6 Gðs; sÞpðsÞhðsÞf ðuðsÞÞ ds þ 0
a 1 a/1 ðgÞ
Z
1
Gðg; sÞpðsÞhðsÞf ðuðsÞÞ ds:
ð4:6Þ
0
So we get from (4.5) and (4.6) that a bðTuÞ P ckTuk > c ¼ a: c To sum up, all the hypotheses of Theorem 4.1 are satisfied by taking b = a/c. Hence, T has at least three fixed points, that is, BVP (1.1) has at least three positive solutions u1, u2 and u3 such that ku1 k < d;
a < aðu2 Þ
The proof is complete.
and
ku3 k > d
with aðu3 Þ < a:
h
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