Positive Solutions for Second-Order Three-Point Eigenvalue Problems

Abstract

With the help of the fixed point index theorem in cones, we get an existence theorem concerning the existence of positive solution for a second-order three-point eigenvalue problem 𝑥^′′(𝑡)+𝜆𝑓(𝑡,𝑥(𝑡))=0,  0≤𝑡≤1,𝑥(0)=0,𝑥(1)=𝑥(𝜂), where 𝜆 is a parameter. An illustrative example is given to demonstrate the effectiveness of the obtained result.

1. Introduction

Motivated by the work of Bitsadze and Samarskii [1] and Ilyin and Moiseev [2], much attention has been paid to the study of certain nonlocal boundary value problems (BVPs) in recent years.

In the last twenty years, many mathematician, have considered the existence of positive solutions of nonlinear three-point boundary value problems; see, for example, Graef et al. [3] Webb [4], Gupta and Trofimchuk [5], Infante [6], Ehrke [7], Ma [8], Feng [9], He and Ge [10], Bai and Fang [11], and Guo [12]. Recently, by applying the Avery-Henderson [13] double fixed point theorem, Henderson [14] studied the existence of two positive solutions of the three-point boundary value problem for the second-order differential equation𝑦^′′+𝑓(𝑦)=0,0≤𝑡≤1,𝑦(0)=0,𝑦(𝑝)−𝑦(1)=0, (1.1) where 0<𝑝<1 and 𝑓∶𝑅→[0,+∞) is continuous.

In this paper, motivated and inspired by the above work and Wong [15], we apply a fixed point index theorem in cones to investigate the existence of positive solutions for nonlinear three-point eigenvalue problems𝑥^′′(𝑡)+𝜆𝑓(𝑡,𝑥(𝑡))=0,0≤𝑡≤1,𝑥(0)=0,𝑥(1)=𝑥(𝜂), (1.2) where 0<𝜂<1 and 𝑓∈𝐶([0,1]×[0,+∞),[0,∞)).

We need the following well-known lemma. See [16] for a proof and further discussion of the fixed point index 𝑖(𝐴,𝐾_𝑟,𝐾).

Lemma 1.1. Assume that 𝐸 is a Banach space, and 𝐾⊂𝐸 is a cone in 𝐸. Let 𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟}. Furthermore, assume that 𝐴∶𝐾→𝐾 is a completely continuous map, and 𝐴𝑥≠𝑥 for 𝑥∈𝜕𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖=𝑟}. Then, one has the following conclusions: (1)if ‖𝑥‖≤‖𝐴𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖(𝐴,𝐾_𝑟,𝐾)=0; (2)if ‖𝑥‖≥‖𝐴𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖(𝐴,𝐾_𝑟,𝐾)=1.

2. Main Results

In the following, we will denote by 𝐶[0,1] the space of all continuous functions 𝑥∶[0,1]→𝑅. This is a Banach space when it is furnished with usual sup-norm ‖𝑥‖∶=sup{|𝑥(𝑠)|∶𝑠∈[0,1]}.

By [14], the Green's function for the three-point boundary-value problem −𝑥^′′=0,𝑥(0)=0,𝑥(1)=𝑥(𝜂) (2.1) is given by𝐺(𝑡,𝑠)=𝑡,𝑡≤𝑠≤𝜂,𝑠,𝑠≤𝑡,𝑠≤𝜂,1−𝑠1−𝜂𝑡,𝑡≤𝑠,𝑠≥𝜂,𝑠+𝜂−𝑠1−𝜂𝑡,𝜂≤𝑠≤𝑡. (2.2)

From the Green's function 𝐺(𝑡,𝑠), we have that a function 𝑥 is a solution of the boundary value problem (1.2) if and only if it satisfies𝑥(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠,𝑡∈[0,1]. (2.3)

Lemma 2.1. Suppose that 𝐺(𝑡,𝑠) is defined as above. Then we have the following results: (1)0≤𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠),0≤𝑡,𝑠≤1,(2)𝐺(𝑡,𝑠)≥𝜂𝑡𝐺(𝑠,𝑠),0≤𝑡,𝑠≤1.

Proof. It is easy to see that (1) holds. To show that (2) holds, we distinguish four cases.(i)If 𝑡≤𝑠≤𝜂, then 𝐺(𝑡,𝑠)=𝑡≥𝜂𝑡𝑠=𝜂𝑡𝐺(𝑠,𝑠). (2.4)(ii)If 𝑠≤𝑡 and 𝑠≤𝜂, then 𝐺(𝑡,𝑠)=𝑠≥𝜂𝑡𝑠=𝜂𝑡𝐺(𝑠,𝑠). (2.5)(iii)If 𝑡≤𝑠 and 𝑠≥𝜂, then 𝐺(𝑡,𝑠)=1−𝑠1−𝜂𝑡≥𝜂𝑠𝑡1−𝑠1−𝜂=𝜂𝑡𝐺(𝑠,𝑠). (2.6)(iv)Finally, if 𝜂≤𝑠≤𝑡, then 𝐺(𝑡,𝑠)=𝑠−𝑠−𝜂1−𝜂𝑡≥𝑠−𝑠−𝜂=1−𝜂𝜂(1−𝑠)1−𝜂≥𝑡𝑠𝜂(1−𝑠)1−𝜂=𝜂𝑡𝑠(1−𝑠)1−𝜂=𝜂𝑡𝐺(𝑠,𝑠). (2.7)

Remark 2.2. If 𝑠≤𝜂 and 𝑠≥𝜂, then 𝐺(𝑠,𝑠)=𝑠 and 𝐺(𝑠,𝑠)=𝑠(1−𝑠)/(1−𝜂), respectively.
Define𝑃={𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0,min_{𝑡∈[𝜂/2,1]}𝜂𝑢(𝑡)≥²2‖𝑢‖}. (2.8) Obviously, 𝑃 is a cone in the Banach space 𝐶[0,1].
Define an operator 𝐴∶𝑃→𝐶[0,1] as follows:(𝐴𝑥)(𝑡)∶=𝜆¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠,𝑡∈[0,1]. (2.9)
It is easy to know that fixed points of 𝐴 are solutions of the BVP (1.2).
Now, we can state and prove our main results.

Lemma 2.3. 𝐴∶𝑃→𝑃 is completely continuous.

Proof. For any 𝑥∈𝑃, by Lemma 2.1 (1), we have (𝐴𝑥)(𝑡)≥0, for each 𝑡∈[0,1]. It follows from Lemma 2.1 that min_{𝑡∈[𝜂/2,1]}(𝐴𝑥)(𝑡)=𝜆min_{𝑡∈[𝜂/2,1]}¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥𝜆𝜂²2¹₀≥𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠𝜆𝜂²2¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (2.10) Hence, min_{𝑡∈[𝜂/2,1]}(𝐴𝑥)(𝑡)≥(𝜂²/2)‖𝐴𝑥‖, which implies 𝐴𝑃⊂𝑃. Moreover, it is easy to check that 𝐴∶𝑃→𝑃 is completely continuous.

By simple calculation, we obtain that¹_𝜂/2𝜂𝐺(2,𝑠)𝑑𝑠=^𝜂_𝜂/2𝜂2𝑑𝑠+¹_𝜂1−𝑠𝜂1−𝜂2𝜂𝑑𝑠=²4+𝜂(1−𝜂)4=𝜂4. (2.11)

Lemma 2.4. Suppose that there exists a positive constant 𝑟>0 such that (H1)𝑓(𝑡,𝑥)≥4𝑟𝜂𝜂o𝑛[2𝜂,1]×[²2𝑟,𝑟] (2.12) holds. If 𝜆>1, then 𝑖(𝐴,𝑃_𝑟,𝑃)=0. (2.13)

Proof. For 𝑥∈𝜕𝑃_𝑟, it follows from the definition of the cone that 𝜂²2𝜂𝑟=²2‖𝑥‖≤min_{𝑡∈[𝜂/2,1]}𝜂𝑥(𝑡)≤𝑥(𝑡)≤‖𝑥‖=𝑟,𝑡∈[2,1], (2.14) which implies 𝜂²2𝜂𝑟≤𝑥≤𝑟,𝑡∈[2,1]. (2.15) Thus, we have by (H1) and (2.11) that 𝜂(𝐴𝑥)(2)=𝜆¹₀𝜂𝐺(2,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠>¹_𝜂/2𝜂𝐺(2≥,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠4𝑟𝜂¹_𝜂/2𝜂𝐺(2,𝑠)𝑑𝑠=𝑟=‖𝑥‖ (2.16) since 𝑥∈𝜕𝑃_𝑟. This shows that ‖𝐴𝑥‖>‖𝑥‖,∀𝑥∈𝜕𝑃_𝑟. (2.17) It is obvious that 𝐴𝑥≠𝑥 for 𝑥∈𝜕𝑃_𝑟. Therefore, by Lemma 1.1 (1), we conclude that 𝑖(𝐴,𝑃_𝑟,𝑃)=0.

Lemma 2.5. Suppose that there exists a positive constant 𝑚>0 such that (H2)𝑓(𝑡,𝑥)≤𝑝(𝑡)𝑔(𝑥)on[0,1]×[0,𝑚], (2.18) where 𝑝∈𝐶([0,1],[0,+∞)) and 𝑔∈𝐶([0,+∞),[0,+∞)). If 1𝜆<2𝑝₁(^𝑚₀𝑑𝑠)𝐻(𝑚)−𝐻(𝑠)², (2.19) where 𝐻(𝑢)∶=∫^𝑢₀𝑔(𝑠)𝑑𝑠 and 𝑝₁=max_𝑡∈[0,1]𝑝(𝑡)>0, then 𝑖(𝐴,𝑃_𝑚,𝑃)=1. (2.20)

Proof. First, we claim that 𝐴𝑥≠𝜇𝑥,for𝑥∈𝜕𝑃_𝑚,𝜇≥1. (2.21) Suppose to the contrary that there exist 𝑥∈𝜕𝑃_𝑚 and 𝜇₀≥1 such that 𝐴𝑥=𝜇₀𝑥,for𝑥∈𝜕𝑃_𝑚. (2.22) It is clear that (2.22) is equivalent to 𝑥^′′𝜆(𝑡)+𝜇₀𝑓(𝑡,𝑥)=0. (2.23)
Since 𝑥∈𝐶[0,1] and 𝑥(𝜂)=𝑥(1), it follows that there exists a 𝜉∈(𝜂,1) such that 𝑥^′(𝜉)=0. From 𝑥^′′≤0 on (0,1), we see that 𝑥(𝜉)=‖𝑥‖∶=𝑚>0, 𝑥^′(𝑡)≥0 on (0,𝜉), and 𝑥^′(𝑡)≤0 on (𝜉,1). By (2.18) and (2.23), we have𝑥^′′𝜆(𝑡)=−𝜇₀𝜆𝑓(𝑡,𝑥(𝑡))≥−𝜇₀𝑝(𝑡)𝑔(𝑥(𝑡)),𝑡∈[0,1]. (2.24) Multiplying (2.24) by 𝑥^′ and then integrating from 𝑡 to 𝜉 (𝑡∈[0,𝜉)), we get from 𝑥^′(𝜉)=0 that −12(𝑥^′(𝑡))²𝜆≥−𝜇₀^𝜉_𝑡𝑝(𝑠)𝑔(𝑥(𝑠))𝑥^′(𝑠)𝑑𝑠,𝑡∈[0,𝜉), (2.25) that is (𝑥^′(𝑡))²≤2𝜆𝜇₀^𝜉_𝑡𝑝(𝑠)𝑔(𝑥(𝑠))𝑥^′(𝑠)𝑑𝑠≤2𝜆𝑝₁𝜇₀^𝑥(𝜉)_𝑥(𝑡)𝑔(𝑢)𝑑𝑢,𝑡∈[0,𝜉), (2.26) which implies that 0≤𝑥^′(𝑡)≤2𝜆𝑝₁𝜇₀(𝐻(𝑚)−𝐻(𝑥(𝑡))),𝑡∈[0,𝜉). (2.27) Thus, ^𝑚₀𝑑𝑠(2𝜆𝑝₁/𝜇₀)(𝐻(𝑚)−𝐻(𝑠))≤^𝜉₀𝑑𝑡=𝜉≤1. (2.28) Hence, we obtain from (2.19) and (2.28) that 1≥𝜇₀2𝜆𝑝₁^𝑚₀𝑑𝑠≥𝐻(𝑚)−𝐻(𝑠)12𝜆𝑝₁^𝑚₀𝑑𝑠𝐻(𝑚)−𝐻(𝑠)>1. (2.29) This contradiction implies that (2.21) holds. By (2.21), we have ‖𝐴𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝑃_𝑚, and 𝐴𝑥≠𝑥 for 𝑥∈𝜕𝑃_𝑚. Thus, by Lemma 1.1 (2), we obtain 𝑖(𝐴,𝑃_𝑚,𝑃)=1. (2.30)

For convenience, let(H3)2𝑝₁<(^𝑚₀𝑑𝑠)𝐻(𝑚)−𝐻(𝑠)². (2.31)

Theorem 2.6. Assume that there exist two distinct positive constants 𝑟,𝑚 such that (H1)–(H3) hold. If 11<𝜆<2𝑝₁(^𝑚₀𝑑𝑠)𝐻(𝑚)−𝐻(𝑠)², (2.32) then BVP (1.2) has at least one positive solution.

Proof. From Lemmas 2.4 and 2.5, and the property of the fixed point index, we can easily get that the operator 𝐴 has a fixed point in 𝑃_𝑚⧵𝑃_𝑟 (𝑚>𝑟) or in 𝑃_𝑟⧵𝑃_𝑚 (𝑟>𝑚). Therefore, BVP (1.2) has at least one positive solution.

3. An Example

To illustrate our results we present the following example.

Example 3.1. Consider the following boundary value problem 𝑥^′′𝜆(𝑡)+2𝑒^𝑡(1+|sin(𝑥𝑡)|)𝑥²1(𝑡)=0,0<𝑡<1,𝑥(0)=0,𝑥(1)=𝑥(2). (3.1) Let 𝑓(𝑡,𝑥)=(1/2)𝑒^𝑡(1+|sin(𝑥𝑡)|)𝑥² and 𝜂=1/2. Choosing 𝑚=1/3, 𝑟=800, we have 1𝑓(𝑡,𝑥)≤𝑝(𝑡)𝑔(𝑥),on[0,1]×[0,3], (3.2) where 𝑝(𝑡)=𝑒^𝑡 and 𝑔(𝑥)=𝑥². Thus, 𝑝₁=𝑒, 𝐻(𝑢)=(1/3)𝑢³, and (^𝑚₀𝑑𝑠)𝐻(𝑚)−𝐻(𝑠)²=(₀^1/3𝑑𝑠1/81−𝑠³)/3²=6.3701>5.4366=2𝑝₁. (3.3) Hence, (H2) and (H3) hold. Moreover, we get 1𝑓(𝑡,𝑥)≥2𝑒^𝑡𝑥²𝜂(on[2𝜂,1]×[²2≥1𝑟,𝑟])2𝑒^1/4100²1(on[4,1]×[100,800])=6420>6400=8𝑟=4𝑟𝜂1(on[4,1]×[100,800]), (3.4) which implies that (H1) holds. Therefore, it follows from Theorem 2.6 that BVP (3.1) has at least one positive solution if 11<𝜆<2𝑝₁(^𝑚₀𝑑𝑠)𝐻(𝑚)−𝐻(𝑠)²=6.37015.4366=1.1717. (3.5)

Acknowledgments

The author would like to thank the referees for their valuable suggestions and comments. This paper was supported by the National Natural Science Foundation of China (10771212) and the Natural Science Foundation of Jiangsu Education Office (06KJB110010).