Existence of Three Positive Solutions to Nonlinear Boundary Value Problems

Abstract

Criteria are established for existence of three solutions to the boundary value problems 𝐿𝑥=𝑓(𝑡,𝑥), 𝑤₁𝑥(0)−𝑤₂𝑥^′(0)=0=𝑤₃𝑥(1)+𝑤₄𝑥^′(1), where 𝐿𝑥∶=−(𝑝𝑥^′)^′+𝑞𝑥. Here, 𝑝∈𝐶¹[0,1], 𝑝>0, 𝑞∈𝐶[0,1], 𝑞≥0.

1. Introduction

In this paper, we are concerned with the existence of three positive solutions for the boundary value problem (BVP) 𝑤𝐿𝑥=𝑓(𝑡,𝑥),0<𝑡<1,₁𝑥(0)−𝑤₂𝑥^′𝑤(0)=0,₃𝑥(1)+𝑤₄𝑥^′(1)=0, (1.1)(1.2) where 𝑓∈𝐶([0,1]×[0,+∞)),[0,+∞), 𝑤_𝑖≥0(𝑖=1,…,4) with 𝜌∶=𝑤₂𝑤₃+𝑤₁𝑤₃+𝑤₁𝑤₄>0 and 𝐿𝑥∶=−(𝑝(𝑡)𝑥^′)^′+𝑞(𝑡)𝑥. Here 𝑝∈𝐶¹([0,1],(0,∞)),𝑞∈𝐶([0,1],[0,∞)). We shall also assume that 𝜆=0 is not an eigenvalue of 𝐿𝑥=𝜆𝑥 subject to conditions (1.2). As a consequence, it follows that the the smallest eigenvalue 𝜆₁ of the problem 𝐿𝑥=𝜆𝑥 subject to (1.2) satisfies 𝜆₁>0 and the corresponding eigenfunction 𝜑₁(𝑡) does not vanish on (0,1). Without loss of generality, we may assume 𝜑₁(𝑡)>0 on (0,1) and ‖𝜑₁‖=max_0≤𝑡≤1|𝜑₁(𝑡)|=1.

Let 𝐺(𝑡,𝑠) denote Green’s function for the problem 𝐿𝑥=0 subject to condition (1.2). It is well known that 𝐺(𝑡,𝑠) may be written as 1𝐺(𝑡,𝑠)=𝑑𝜙(𝑡)𝜓(𝑠),0≤𝑡≤𝑠≤1,𝜙(𝑠)𝜓(𝑡),0≤𝑠≤𝑡≤1, (1.3) where 𝜙(𝑡) and 𝜓(𝑡) satisfy 𝐿𝜙=0,𝜙(0)=𝑤₂,𝜙^′(0)=𝑤₁,𝐿𝜓=0,𝜓(1)=𝑤₄,𝜓^′(1)=−𝑤₃, (1.4) and where 𝑝(𝑡)(𝜙(𝑡)𝜓^′(𝑡)−𝜙^′(𝑡)𝜓(𝑡))≡−𝑑. (1.5) It may be shown that 𝜙(𝑡)≥0 and is increasing on [0,1] while 𝜓(𝑡)≥0 and is decreasing on [0,1]. As a consequence, it follows that 𝑑>0 and, furthermore, we have 0≤𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠),0≤𝑡,𝑠≤1. (1.6) We define the positive number 𝜂,𝜇 by 𝜂⁻¹∶=max_0≤𝑡≤1(¹₀𝐺(𝑡,𝑠)𝑑𝑠),𝜇⁻¹∶=^3/4_1/4𝐺(𝑠,𝑠)𝑑𝑠. (1.7)

For the case 𝐿𝑥=−𝑥^′′ (i.e, 𝑝(𝑡)≡1,𝑞(𝑡)≡0), the corresponding BVP −𝑥^′′=𝑓(𝑡,𝑥),0<𝑡<1 (1.8) subject to (1.2) has attracted considerable attention over the last number of years. Under certain condition, positive solutions of (1.8) and (1.2) are obtained in [1, 2]. In a recent paper, Erbe [3] investigated the existence of multiple positive solutions to (1.1)-(1.2) by applying the fixed point index.

The aim of this paper is to establish criteria for the existence of three positive solutions to (1.1) and (1.2), which improve the corresponding result of [3]. Our tool in this paper will be well-known Five Functionals Fixed Point Theorem [4–7].

2. Preliminaries

Definition 2.1. Suppose 𝑃 is a cone in a Banach. The map 𝛼 is a nonnegative continuous concave functional on 𝑃 provided 𝛼∶𝑃→[0,∞) is continuous and 𝛼(𝑡𝑥+(1−𝑡)𝑦)≥𝑡𝛼(𝑥)+(1−𝑡)𝛼(𝑦) (2.1) for all 𝑥,𝑦∈𝑃 and 𝑡∈[0,1]. Similarly, the map 𝛽 is a nonnegative continuous convex functional on 𝑃 provided 𝛽∶𝑃→[0,∞) is continuous and 𝛽(𝑡𝑥+(1−𝑡)𝑦)≤𝑡𝛽(𝑥)+(1−𝑡)𝛽(𝑦) (2.2) for all 𝑥,𝑦∈𝑃 and 𝑡∈[0,1].
Let 𝛾,𝛽,𝜃 be nonnegative, continuous, convex functionals on 𝑃 and 𝛼,𝜓 be nonnegative, continuous, concave functionals on 𝑃. Then, for nonnegative real numbers ℎ,𝑎,𝑏,𝑑 and 𝑐, we define the convex sets 𝑃(𝛾,𝑐)={𝑥∈𝑃∶𝛾(𝑥)<𝑐},𝑃(𝛾,𝛼,𝑎,𝑐)={𝑥∈𝑃∶𝑎≤𝛼(𝑥),𝛾(𝑥)≤𝑐},𝑄(𝛾,𝛽,𝑑,𝑐)={𝑥∈𝑃∶𝛽(𝑥)≤𝑑,𝛾(𝑥)≤𝑐},𝑃(𝛾,𝜃,𝛼,𝑎,𝑏,𝑐)={𝑥∈𝑃∶𝑎≤𝛼(𝑥),𝜃(𝑥)≤𝑏,𝛾(𝑥)≤𝑐},𝑄(𝛾,𝛽,𝜓,ℎ,𝑑,𝑐)={𝑥∈𝑃∶ℎ≤𝜓(𝑥),𝛽(𝑥)≤𝑑,𝛾(𝑥)≤𝑐}. (2.3) To prove our main results, we need the following theorem, which is the Five Functionals Fixed Point Theorem [4].

Theorem 2.2. Let 𝑃 be a cone in a real Banach space 𝐸. Suppose there exist positive numbers 𝑐 and 𝑀, nonnegative, continuous, concave functionals 𝛼 and 𝜓 on 𝑃, and nonnegative, continuous, convex functionals 𝛾,𝛽 and 𝜃 on 𝑃, with 𝛼(𝑥)≤𝛽(𝑥),‖𝑥‖≤𝑀𝛾(𝑥) (2.4) for all 𝑥∈𝑃(𝛾,𝑐). Suppose Φ∶𝑃(𝛾,𝑐)→𝑃(𝛾,𝑐) (2.5) is completely continuous and there exist nonnegative numbers ℎ,𝑎,𝑘,𝑏, with 0<𝑎<𝑏 such that(i){𝑥∈𝑃(𝛾,𝜃,𝛼,𝑏,𝑘,𝑐)∶𝛼(𝑥)>𝑏}≠∅ and 𝛼(Φ𝑥)>𝑏 for 𝑥∈𝑃(𝛾,𝜃,𝛼,𝑏,𝑘,𝑐);(ii){𝑥∈𝑄(𝛾,𝛽,𝜓,ℎ,𝑎,𝑐)∶𝛽(𝑥)<𝑎}≠∅ and 𝛽(Φ𝑥)<𝑎 for 𝑥∈𝑄(𝛾,𝛽,𝜓,ℎ,𝑎,𝑐);(iii)𝛼(Φ𝑥)>𝑏 for 𝑥∈𝑃(𝛾,𝛼,𝑏,𝑐) with 𝜃(Φ𝑥)>𝑘;(iv)𝛽(Φ𝑥)<𝑎 for 𝑥∈𝑄(𝛾,𝛽,𝑎,𝑐) with 𝜓(Φ𝑥)<ℎ.Then Φ has at least three fixed points 𝑥₁,𝑥₂,𝑥₃∈𝑃(𝛾,𝑐) such that 𝛽(𝑥₁)<𝑎,𝑏<𝛼(𝑥₂),𝑎<𝛽(𝑥₃)with𝛼(𝑥₃)<𝑏. (2.6)

3. Main Result

In this section, we shall obtain existence results for BVP (1.1) and (1.2) by using the Five Functional Fixed Point Theorem.

By [3], it is well known that BVP associated with (1.1), (1.2) is equivalent to the operator equation 𝑥=𝐴𝑥,𝑥∈𝐶[0,1], (3.1) where (𝐴𝑥)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (3.2) Now with 𝑋=𝐶[0,1],‖𝑥‖=max_0≤𝑡≤1|𝑥(𝑡)|, it is easy to see that 𝐴∶𝑋→𝑋 is completely continuous. We define a cone 𝑃⊂𝑋 by 𝑃∶={𝑥∈𝑋∶𝑥(𝑡)≥0,min_{1/4≤𝑡≤3/4}𝑥(𝑡)≥𝜎‖𝑥‖}, (3.3) where 𝜎 is defined by 𝜎∶=min{𝐺(𝑡,𝑠)∶1𝐺(𝑠,𝑠)43≤𝑡≤4,0≤𝑠≤1}. (3.4) By (1.3) and the properties of 𝜑(𝑡),𝜓(𝑡), we have 𝜎∶=min{𝜙(1/4),𝜙(1)𝜓(3/4)𝜓(0)}. (3.5) Clearly, 0<𝜎<1 and 𝐺(𝑡,𝑠)≥𝜎𝐺(𝑠,𝑠) for 1/4≤𝑡≤3/4,0≤𝑠≤1.

Lemma 3.1. The operator 𝐴 maps 𝑃 into 𝑃.

Proof. Let 𝑥∈𝑃. From (1.6) and the condition of 𝑓, we see that 𝐴𝑥≥0. Next, for 𝑥∈𝑃, we have |(𝐴𝑥)(𝑡)|=(𝐴𝑥)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≤¹₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (3.6) Hence, ‖𝐴𝑥‖≤¹₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (3.7) Now, from 𝐺(𝑡,𝑠)≥𝜎𝐺(𝑠,𝑠) for 1/4≤𝑡≤3/4,0≤𝑠≤1, we have min_{𝑡∈[1/4,3/4]}(𝐴𝑥)(𝑡)=min_{𝑡∈[1/4,3/4]}¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥𝜎¹₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥𝜎‖𝐴𝑥‖. (3.8) This show that 𝐴𝑥∈𝑃, which completes this proof.

Theorem 3.2. Let 0<𝑎<𝑏 and 𝜇𝑏<𝜂𝜎𝑐, and suppose 𝑓(𝑡,𝑥) satisfies the following conditions:(H1)𝑓(𝑡,𝑥)<𝜂𝑎 for 0≤𝑡≤1 and 0≤𝑥≤𝑎, (H2)𝑓(𝑡,𝑥)≥(𝜇𝑏)/𝜎 for 1/4≤𝑡≤3/4 and 𝑏≤𝑥≤𝑏/𝜎,(H3)𝑓(𝑡,𝑥)≤𝜂𝑐 for 0≤𝑡≤1 and 0≤𝑥≤𝑐.Then the BVP (1.1)-(1.2) has at least three positive solutions.

Proof. Theorem 2.2 will be applied. We begin by defining the nonnegative continuous concave functional 𝛼,𝜓 and the nonnegative continuous convex functional 𝛽,𝜃,𝛾 on 𝑃𝜓(𝑥)=min_𝑡∈[0,1]𝑥(𝑡),𝛽(𝑥)=𝜃(𝑥)=max_𝑡∈[0,1]𝑥(𝑡),𝛼(𝑥)=min_{𝑡∈[1/4,3/4]}𝑥(𝑡),𝛾(𝑥)=‖𝑥‖. (3.9) It is clear that 𝛼(𝑥)≤𝛽(𝑥) for all 𝑥∈𝑃.
First, we shall show that the operator 𝐴 maps 𝑃(𝛾,𝑐) into 𝑃(𝛾,𝑐). Let 𝑥∈𝑃(𝛾,𝑐). Thus we have 0≤𝑥(𝑡)≤𝑐 for 0≤𝑡≤1. Using (H3), we have |(𝐴𝑥)(𝑡)|=(𝐴𝑥)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≤𝜂𝑐¹₀𝐺(𝑡,𝑠)𝑑𝑠≤𝑐. (3.10) Hence 𝛾(𝐴𝑥)=‖𝐴𝑥‖≤𝑐. (3.11) Therefore, we have shown that 𝐴∶𝑃(𝛾,𝑐)→𝑃(𝛾,𝑐).
We next prove that Condition (i) of Theorem 2.2 holds. Let 𝑥≡(1/2)(𝑏+𝑘),𝑘=𝑏/𝜎. Then 1𝛼(𝑥)=21(𝑏+𝑘)>𝑏,𝜃(𝑥)=21(𝑏+𝑘)<𝑘,𝛾(𝑥)=2(𝑏+𝑘)<𝑐, (3.12) which shows that {𝑥∈𝑃(𝛾,𝜃,𝛼,𝑏,𝑘,𝑐),𝛼(𝑥)>𝑏}≠∅. Let 𝑥∈𝑃(𝛾,𝜃,𝛼,𝑏,𝑘,𝑐). Then 𝛼(𝑥)>𝑏,𝜃(𝑥)<𝑘=𝑏/𝜎 imply that 𝑏𝑏<𝑥(𝑡)<𝜎1,for𝑡∈[4,34]. (3.13) By (H2) we can obtain 𝛼(𝐴𝑥)=min_{𝑡∈[1/4,3/4]}(𝐴𝑥)(𝑡)=min_{𝑡∈[1/4,3/4]}¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥𝜎¹₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠>𝜎^3/4_1/4𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥𝜇𝑏^3/4_1/4𝐺(𝑠,𝑠)𝑑𝑠=𝑏. (3.14) Hence, 𝛼(𝐴𝑥)>𝑏 for all 𝑥∈𝑃(𝛾,𝜃,𝛼,𝑏,𝑘,𝑐) and so Condition (i) of Theorem 2.2 holds.
Next, we verify that Condition (ii) of Theorem 2.2 is satisfied. Take 𝑥≡𝜎𝑎,ℎ=𝜎𝑎, then 𝛾(𝑥)=𝜎𝑎<𝑐,𝜓(𝑥)=𝜎𝑎=ℎ,𝛽(𝑥)=𝜎𝑎<𝑎. (3.15) From this we know that {𝑥∈𝑄(𝛾,𝛽,𝜓,ℎ,𝑎,𝑐),𝛽(𝑥)<𝑎}≠∅. Let 𝑥∈𝑄(𝛾,𝛽,𝜓,ℎ,𝑎,𝑐). Then we have 𝛽(𝑥)≤𝑎, which lead to 0≤𝑥(𝑡)≤𝑎, for 𝑡∈[0,1]. In view of (H1), we have 𝛽(𝐴𝑥)=max_𝑡∈[0,1](𝐴𝑥)(𝑡)=max_𝑡∈[0,1]¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≤𝜂𝑎⋅max_𝑡∈[0,1]¹₀𝐺(𝑡,𝑠)𝑑𝑠=𝑎. (3.16) Hence, 𝛽(𝐴𝑥)<𝑎 for all 𝑥∈𝑄(𝛾,𝛽,𝜓,ℎ,a,𝑐). Thus, Condition (ii) of Theorem 2.2 is fulfilled.
We shall next show that Condition (iii) of Theorem 2.2 is met. Observe that for 𝑥∈𝑃𝜃(𝐴𝑥)=max_𝑡∈[0,1](𝐴𝑥)(𝑡)=max_𝑡∈[0,1]¹₀≤𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠₁₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (3.17) On the other hand, 𝛼(𝐴𝑥)=min_{𝑡∈[1/4,3/4]}(𝐴𝑥)(𝑡)=min_{𝑡∈[1/4,3/4]}¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥𝜎¹₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (3.18)(3.17) together (3.18) implies that 𝛼(𝐴𝑥)≥𝜎𝜃(𝐴𝑥),𝑥∈𝑃. (3.19)
Let 𝑥∈𝑃(𝛾,𝛼,𝑏,𝑐) with 𝜃(𝐴𝑥)>𝑘=𝑏/𝜎. Then, it follows from (3.19) that 𝛼(𝐴𝑥)≥𝜎𝜃(𝐴𝑥)>𝑏. (3.20) Thus, 𝛼(𝐴𝑥)>𝑏 for all 𝑥∈𝑃(𝛾,𝛼,𝑏,𝑐) with 𝜃(𝐴𝑥)>𝑏/𝜎. Hence, Condition (iii) of Theorem 2.2 holds.
Finally, we shall prove that Condition (iv) of Theorem 2.2 is fulfilled. Let 𝑥∈𝑄(𝛾,𝛽,𝑎,𝑐) and 𝜓(𝐴𝑥)<ℎ=𝜎𝑎. Then 0≤𝑥(𝑡)≤𝑎,𝑡∈[0,1]. By (H1), we have 𝛽(𝐴𝑥)=max_𝑡∈[0,1](𝐴𝑥)(𝑡)=max_𝑡∈[0,1]¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠<𝜂𝑎⋅max_𝑡∈[0,1]¹₀𝐺(𝑡,𝑠)𝑑𝑠=𝑎. (3.21) Thus, Condition (iv) of Theorem 2.2 is satisfied.
Now, an application of Theorem 2.2 ensures that the BVP (1.1) and (1.2) has at least three positive solutions 𝑥₁,𝑥₂,𝑥₃ such that 𝛽(𝑥₁)<𝑎,𝑏<𝛼(𝑥₂),𝑎<𝛽(𝑥₃)with𝛼(𝑥₃)<𝑏. (3.22) This proof is complete.

Remark 3.3. This Theorem improves the Corollary 2.5 in [3].

Example 3.4. For simplicity, we consider the boundary value problem −𝑥^′′=𝑓(𝑡,𝑥),0<𝑡<1,𝑥(0)=𝑥^′(1)=0, (3.23) where 1𝑓(𝑡,𝑥)=10|sin𝑡|+20𝑥⁵1,𝑥≤1,10|sin𝑡|+20,𝑥≥1. (3.24) By direct calculation we can obtain that 𝜂=2,𝜇=4,𝜎=1/4. Set 𝑎=1/2,𝑏=1,𝑐=12, so the nonlinear term 𝑓 satisfies 1𝑓(𝑡,𝑥)≤0.1+20×(2)⁵1<1=𝜂𝑎,(𝑡,𝑥)∈[0,1]×[0,2],𝑓(𝑡,𝑥)>20>16=(𝜇𝑏)𝜎1,(𝑡,𝑥)∈[4,34]×[1,4],𝑓(𝑡,𝑥)<21<24=𝜂𝑐,(𝑡,𝑥)∈[0,1]×[0,12]. (3.25) Then the conditions in Theorem 3.2 are all satisfied, so the boundary value problem (3.23) has at least three positive solutions 𝑥₁,𝑥₂,𝑥₃ such that max_0≤𝑡≤1𝑥₁1(𝑡)<2,1<min_{1/4≤𝑡≤3/4}𝑥₂(𝑡),1<max_0≤𝑡≤1𝑥₃(𝑡)withmin_{1/4≤𝑡≤3/4}𝑥₃(𝑡)<1. (3.26)

Acknowledgment

This work is supported by the NNSF of China (no. 10871062), A project supported by Scientifc Research Fund of Hunan Provincial Education Department (07B041) and Program for Young Excellent Talents in Hunan Normal University, and Program by Hunan Provincial Natural Science Foundation of China (No. 07JJ6010).