Multiple Bounded Positive Solutions to Integral Type BVPs for Singular Second Order ODEs on the Whole Line

Abstract

This paper is concerned with the integral type boundary value problems of the second order differential equations with one-dimensional p-Laplacian on the whole line. By constructing a suitable Banach space and a operator equation, sufficient conditions to guarantee the existence of at least three positive solutions of the BVPs are established. An example is presented to illustrate the main results. The emphasis is put on the one-dimensional p-Laplacian term [𝜌(𝑡)Φ(𝑥′(𝑡))]′ involved with the function ρ, which makes the solutions un-concave.

1. Introduction

The multipoint boundary-value problems for linear second order ordinary differential equations (ODEs for short) was initiated by Il'in and Moiseev [1]. Since then, more general nonlinear multi-point boundary-value problems (BVPs for short) were studied by several authors, see the text books [2–4] and the references cited therein.

Differential equations governed by nonlinear differential operators have been widely studied. In this setting the most investigated operator is the classical one-dimensional 𝑝-Laplacian, that is, Φ_𝑝(𝑥)=|𝑥|^𝑝−2𝑥 with 𝑝>1. This operator is involved in some models, for example, in non-Newtonian fluid theory, diffusion of flows in porous media, nonlinear elasticity, and theory of capillary surfaces. The related nonlinear differential equation has the form Φ(𝑥′)^′=𝑓𝑡,𝑥,𝑥^′,𝑡∈(−∞,+∞), (1.1) where Φ(𝑥)=|𝑥|^𝑝−2𝑥 with 𝑝>1 is a one dimensional 𝑝-Laplacian. For a comprehensive bibliography on this subject, see, for example [5–9].

In this paper, we consider the more generalized BVP for second order differential equation on the whole line with 𝑝-Laplacian coupled with the integral type BCs, that is the BVP 𝜌(𝑡)Φ𝑥^′(𝑡)^′+𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)=0,𝑡∈𝑅,lim_{𝑡→−∞}𝑥(𝑡)=∶𝑥(−∞)=^+∞_−∞𝑔(𝑠)𝑥(𝑠)𝑑𝑠,lim_𝑡→+∞𝑥(𝑡)=∶𝑥(+∞)=^+∞_−∞ℎ(𝑠)𝑥(𝑠)𝑑𝑠, (1.2) where 𝑓∶𝑅³→𝑅 is a nonnegative Caratheodory function, 𝑔,ℎ∶𝑅→[0,∞) satisfy 𝑔,ℎ∈𝐿¹(𝑅),𝜌∈𝐶(𝑅,(0,∞)), the integrals in mentioned equations are meant in the sense of Riemann-Stieljes, Φ(𝑥)=|𝑥|^𝑝−2𝑥 with 𝑝>1 is called a one dimensional 𝑝-Laplacian, whose inverse function is denoted by Φ⁻¹.

The purpose is to establish sufficient conditions for the existence of at least three positive solutions of BVP(1.2). The result in this paper generalizes and improves some known ones since the one-dimensional 𝑝-Laplacian term [𝜌(𝑡)Φ(𝑥′(𝑡))]′ involved with the function 𝜌, which makes the solutions unconcave and there exists no paper concerned with the existence of at least three positive solutions of this kind of integral BVPs on the whole lines. This paper fills the gap.

The remainder of this paper is organized as follows: the main result (Theorem 2.8) is presented in Section 2, and the example to show the main result is given in Section 3.

2. Main Results

In this section, we first present some background definitions in Banach spaces and state an important three fixed point theorem. Then the main results are given and proved.

Definition 2.1. Let 𝑋 be a real Banach space. The nonempty convex closed subset 𝑃 of 𝑋 is called a cone in 𝑋 if 𝑎𝑥∈𝑃 for all 𝑥∈𝑃 and 𝑎≥0 and 𝑥∈𝑋 and −𝑥∈𝑋 imply 𝑥=0.

Definition 2.2. A map 𝜓∶𝑃→[0,+∞) is a nonnegative continuous concave or convex functional map provided 𝜓 is nonnegative, continuous, and satisfies 𝜓(𝑡𝑥+(1−𝑡)𝑦)≥𝑡𝜓(𝑥)+(1−𝑡)𝜓(𝑦), or 𝜓(𝑡𝑥+(1−𝑡)𝑦)≤𝑡𝜓(𝑥)+(1−𝑡)𝜓(𝑦), for all 𝑥,𝑦∈𝑃 and 𝑡∈[0,1].

Definition 2.3. An operator 𝑇∶𝑋→𝑋 is completely continuous if it is continuous and maps bounded sets into precompact sets.

Definition 2.4. Let 𝑎,𝑏,𝑐,𝑑,ℎ>0 be positive constants, 𝛼,𝜓 be two nonnegative continuous concave functionals on the cone 𝑃, 𝛾,𝛽,𝜃 be three nonnegative continuous convex functionals on the cone 𝑃. Define the convex sets as follows: 𝑃_𝑐={‖𝑥‖},𝑃=(𝑥)(𝑥),𝑃=(𝑥)(𝑥)(𝑥),𝑄=(𝑥)(𝑥),𝑄=(𝑥)(𝑥)(𝑥).𝑥∈𝑃∶<𝑐𝛾,𝛼;𝑎,𝑐𝑥∈𝑃∶𝛼≥𝑎,𝛾≤𝑐𝛾,𝜃,𝛼;𝑎,𝑏,𝑐𝑥∈𝑃∶𝛼≥𝑎,𝜃≤𝑏,𝛾≤𝑐𝛾,𝛽;,𝑑,𝑐𝑥∈𝑃∶𝛽≤𝑑,𝛾≤𝑐𝛾,𝛽,𝜓;ℎ,𝑑,𝑐𝑥∈𝑃∶𝜓≥ℎ,𝛽≤𝑑,𝛾≤𝑐 (2.1)

Lemma 2.5 (see [10]). Let 𝑋 be a real Banach space, 𝑃 be a cone in 𝑋, 𝛼,𝜓 be two nonnegative continuous concave functionals on the cone 𝑃,𝛾,𝛽,𝜃 be three nonnegative continuous convex functionals on the cone 𝑃. Assume that there exists a constant 𝑀>0 such that 𝛼(𝑥)≤𝛽(𝑥),‖𝑥‖≤𝑀𝛾(𝑥),∀𝑥∈𝑃. (2.2) Furthermore, suppose that ℎ,𝑑,𝑎,𝑏,𝑐>0 are constants with 𝑑<𝑎. Let 𝑇∶𝑃_𝑐→𝑃_𝑐 be a completely continuous operator. If(C1){𝑥∈𝑃(𝛾,𝜃,𝛼;𝑎,𝑏,𝑐)∣𝛼(𝑥)>𝑎}≠∅ and 𝛼(𝑇𝑥)>𝑎forevery𝑥∈𝑃(𝛾,𝜃,𝛼;𝑎,𝑏,𝑐);(C2){𝑥∈𝑄(𝛾,𝜃,𝜓;ℎ,𝑑,𝑐)∣𝛽(𝑥)<𝑑}≠∅ and 𝛽(𝑇𝑥)<𝑑forevery𝑥∈𝑄(𝛾,𝜃,𝜓;ℎ,𝑑,𝑐);(C3)𝛼(𝑇𝑥)>𝑎 for 𝑥∈𝑃(𝛾,𝛼;𝑎,𝑐) with 𝜃(𝑇𝑥)>𝑏;(C4)𝛽(𝑇𝑥)<𝑑 for each 𝑥∈𝑄(𝛾,𝛽;𝑑,𝑐) with 𝜓(𝑇𝑥)<ℎ,then 𝑇 has at least three fixed points 𝑥₁, 𝑥₂ and 𝑥₃ such that 𝛽(𝑥₁)<𝑑,𝛼(𝑥₂)>𝑎,𝛽(𝑥₃)>𝑑,𝛼(𝑥₃)<𝑎.

Let us list the assumptions(H1)𝑔,ℎ∶𝑅→[0,∞) satisfy ∫^+∞_−∞𝑔(𝑠)𝑑𝑠<1, ∫^+∞_−∞ℎ(𝑠)𝑑𝑠<1 and 𝑔(𝑡)1−^+∞_−∞ℎ(𝑠)𝑑𝑠−ℎ(𝑡)1−^+∞_−∞𝑔(𝑠)𝑑𝑠≥0,𝑡∈𝑅. (2.3)(H2)𝜌∈𝐶(𝑅),𝜌(𝑡)>0 for 𝑡∈𝑅 with ∫⁰_−∞Φ⁻¹(1/𝜌(𝑡))𝑑𝑡=∫₀^+∞Φ⁻¹(1/𝜌(𝑡))𝑑𝑡<+∞.(H3)𝑓(𝑡,𝑐,0)≢0 on any finite subinterval of 𝑅 for each 𝑐∈𝑅, 𝑓∶𝑅³→𝑅 is a Carathédory function, that is,(i)𝑡→𝑓(𝑡,𝑥,(1/Φ⁻¹(𝜌(𝑡)))𝑦) is measurable for any (𝑥,𝑦)∈𝑅²,(ii)(𝑥,𝑦)→𝑓(𝑡,𝑥,(1/Φ⁻¹(𝜌(𝑡)))𝑦) is continuous for a.e. 𝑡∈𝑅,(iii) for each 𝑟>0, there exists nonnegative function 𝜙_𝑟∈𝐿¹(𝑅) such that max{|𝑢|,|𝑣|}≤𝑟 implies 𝑓1𝑡,𝑢,Φ⁻¹𝜌(𝑡)𝑣≤𝜙_𝑟(𝑡),𝑎.𝑒.𝑡∈𝑅. (2.4)

Choose 𝑋=andthereexistthelimitslim_{𝑡→−∞}𝑥(𝑡),𝑥∈𝐶¹(𝑅)∶lim_𝑡→+∞𝑥(𝑡),lim_{𝑡→−∞}Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)andlim_𝑡→+∞Φ⁻¹𝜌(𝑡)𝑥^′(𝑡). (2.5) For 𝑥∈𝑋, define the norm of 𝑥 by ‖𝑥‖=maxsup_𝑡∈𝑅|𝑥(𝑡)|,sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑥^′(𝑡). (2.6) One can prove that 𝑋 is a Banach space with the norm ||𝑥|| for 𝑥∈𝑋.

Let 𝑥∈𝑋. Consider the following auxiliary BVP 𝜌(𝑡)Φ𝑦^′(𝑡)^′+𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑦()==0,𝑡∈𝑅,−∞^+∞_−∞𝑔(𝑠)𝑦(𝑠)𝑦()=𝑑𝑠,+∞^+∞_−∞ℎ(𝑠)𝑦(𝑠)𝑑𝑠. (2.7)

Lemma 2.6. Suppose that (H1)–(H3) hold. If 𝑦∈𝐶¹(𝑅) such that [𝜌Φ(𝑦′)]′∈𝐿¹(𝑅) is a solution of BVP(2.7), then(i)𝑦 is bounded and nonnegative on 𝑅;(ii)𝑦(𝑡) is concave with respect to 𝜏=∫^𝑡_−∞Φ⁻¹(1/𝜌(𝑠))𝑑𝑠/∫^+∞_−∞Φ⁻¹(1/𝜌(𝑠))𝑑𝑠;(iii)for 𝑘>0, it holds that min_{𝑡∈[−𝑘,𝑘]}𝑦(𝑡)≥𝜇sup_𝑡∈𝑅𝑦(𝑡) with 𝜇=∫^−𝑘_−∞Φ⁻¹(1/𝜌(𝑠))𝑑𝑠/2∫^+∞_−∞Φ⁻¹(1/𝜌(𝑠))𝑑𝑠;(iv)there exists a unique constant 𝐴_𝑥∈[−∫^+∞_−∞𝑓(𝑠,𝑥(𝑠),𝑥′(𝑠))𝑑𝑠,0] such that ^+∞_−∞𝑔(𝑡)1−∫^+∞_−∞ℎ(𝑠)𝑑𝑠−ℎ(𝑡)1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠×^𝑡_−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)+𝑑𝑢𝑑𝑠𝑑𝑡_+∞−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥′(𝑢)𝑑𝑢𝑑𝑠=0. (2.8)

Proof. Since 𝑥∈𝑋, we get 𝑟=maxsup_𝑡∈𝑅|𝑥(𝑡)|,sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)<+∞. (2.9) Then there exists a nonnegative function 𝜙_𝑟∈𝐿¹(−∞,+∞) such that0≤𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)(𝑡),1=𝑓𝑡,𝑥Φ⁻¹𝜌(𝑡)Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)≤𝜙_𝑟(𝑡),𝑡∈𝑅. (2.10) Then ^+∞_−∞𝑓𝑠,𝑥(𝑠),𝑥^′(𝑠)𝑑𝑠isconvergent. (2.11)
(i) We know there exist the limits lim_{𝑡→−∞}𝜌(𝑡)Φ(𝑦′(𝑡)) and lim_𝑡→+∞𝜌(𝑡)Φ(𝑦′(𝑡)). We claim that there exists 𝜏₀∈𝑅 such that 𝑦′(𝜏₀)=0. In fact, if 𝑦′(𝑡)>0 for all 𝑡∈𝑅, we get 𝑦(−∞)<𝑦(𝑡)<𝑦(+∞) for all 𝑡∈𝑅. It follows that 𝑦(−∞)≥^+∞_−∞𝑔(𝑠)𝑑𝑠𝑦(−∞),𝑦(+∞)≤^+∞_−∞ℎ(𝑠)𝑑𝑠𝑦(+∞). (2.12) Then (H1) implies 𝑦(−∞)≥0≥𝑦(+∞), which contradicts to 𝑦′(𝑡)>0 for all 𝑡∈𝑅. Similarly we can prove that 𝑦′(𝑡)<0 does not hold. Then there exists 𝜏₀∈𝑅 such that 𝑦′(𝜏₀)=0.
Since [𝜌(𝑡)Φ(𝑦′(𝑡))]′=−𝑓(𝑡,𝑥(𝑡),𝑥′(𝑡))≤0, we know that 𝜌(𝑡)Φ(𝑦′(𝑡)) is decreasing on 𝑅. Then 𝜌(𝑡)Φ(𝑦′(𝑡))≥0 for all 𝑡≤𝜏₀ and 𝜌(𝑡)Φ(𝑦′(𝑡))≤0 for all 𝑡≥𝜏₀. Hence 𝑦isincreasingfor𝑡∈−∞,𝜏₀anddecreasingfor𝑡∈𝜏₀,+∞. (2.13) One sees that 𝑦^′(𝑡)=−Φ⁻¹1𝜌(𝑡)∫^𝑡_𝜏0𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)𝑑𝑢=∶−𝐺(𝑡),𝑡≥𝜏₀,Φ⁻¹1𝜌(𝑡)∫^𝜏0_𝑡𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)𝑑𝑢=∶𝐻(𝑡),𝑡≤𝜏₀. (2.14) Since ∫^+∞_−∞Φ⁻¹(1/𝜌(𝑡))𝑑𝑡<+∞ and ∫^+∞_−∞𝑓(𝑡,𝑥(𝑡),𝑥′(𝑡))𝑑𝑡<+∞, we see that both_𝑡^+∞𝐺(𝑠)𝑑𝑠,^𝑡_−∞𝐻(𝑠)𝑑𝑠areconvergent. (2.15) Then we get that 𝑦(𝑡)=𝑦(+∞)+∫_𝑡^+∞𝐺(𝑠)𝑑𝑠,𝑡≥𝜏₀,𝑦(−∞)+∫^𝑡_−∞𝐻(𝑠)𝑑𝑠,𝑡≤𝜏₀. (2.16) This tells us that 𝑦 is bounded on 𝑅.
It follows from (2.7) and (2.16) that 𝑦(−∞)=^𝜏0_−∞𝑔(𝑠)𝑑𝑠𝑦(−∞)+_𝜏^+∞₀𝑔(𝑠)𝑑𝑠𝑦(+∞)+^𝜏0_−∞𝑔(𝑡)^𝑡_−∞𝐻(𝑠)+𝑑𝑠𝑑𝑡_𝜏+∞0𝑔(𝑡)_𝑡^+∞𝐺(𝑠)𝑦()=𝑑𝑠𝑑𝑡,+∞^𝜏0_−∞ℎ(𝑠)𝑑𝑠𝑦(−∞)+_𝜏^+∞₀ℎ(𝑠)𝑑𝑠𝑦(+∞)+^𝜏0_−∞ℎ(𝑡)^𝑡_−∞𝐻(𝑠)+𝑑𝑠𝑑𝑡_𝜏+∞0ℎ(𝑡)_𝑡^+∞𝐺(𝑠)𝑑𝑠𝑑𝑡. (2.17) Then 1−^𝜏0_−∞𝑔(𝑠)𝑑𝑠𝑦(−∞)−_𝜏^+∞₀𝑔(𝑠)()=𝑑𝑠𝑦+∞_𝜏0−∞𝑔(𝑡)^𝑡_−∞𝑡𝐻(𝑠)𝑑𝑠𝑑𝑡+_𝜏^+∞₀𝑔(𝑡)_𝑡^+∞𝐺(𝑠)−𝑑𝑠𝑑𝑡,_𝜏0−∞ℎ(𝑠)𝑑𝑠𝑦(−∞)+1−_𝜏^+∞₀ℎ(𝑠)𝑦()=𝑑𝑠+∞_𝜏0−∞ℎ(𝑡)^𝑡_−∞𝐻(𝑠)𝑑𝑠𝑑𝑡+_𝜏^+∞₀ℎ(𝑡)_𝑡^+∞𝐺(𝑠)𝑑𝑠𝑑𝑡. (2.18) Hence 𝑦(−∞)=^{∫𝜏0−∞𝑔(𝑡)∫𝑡−∞𝐻(𝑠)𝑑𝑠𝑑𝑡+∫𝜏0+∞𝑔(𝑡)∫𝑡+∞𝐺(𝑠)𝑑𝑠𝑑𝑡−∫𝜏0+∞𝑔(𝑠)∫𝑑𝑠𝜏0−∞ℎ(𝑡)∫𝑡−∞𝐻(𝑠)𝑑𝑠𝑑𝑡+∫𝜏0+∞ℎ(𝑡)∫𝑡+∞𝐺(𝑠)𝑑𝑠𝑑𝑡1−∫𝜏0+∞ℎ(𝑠)𝑑𝑠}_{1−∫𝜏0−∞𝑔(𝑠)𝑑𝑠−∫𝜏0+∞𝑔(𝑠)𝑑𝑠−∫𝜏0−∞ℎ(𝑠)𝑑𝑠1−∫𝜏0+∞ℎ(𝑠)𝑑𝑠},𝑦(+∞)=^{1−∫𝜏0−∞𝑔(𝑠)𝑑𝑠∫𝜏0−∞𝑔(𝑡)∫𝑡−∞𝐻(𝑠)𝑑𝑠𝑑𝑡+∫𝜏0+∞𝑔(𝑡)∫𝑡+∞𝐺(𝑠)𝑑𝑠𝑑𝑡−∫𝜏0−∞ℎ(𝑠)𝑑𝑠∫𝜏0−∞ℎ(𝑡)∫𝑡−∞𝐻(𝑠)𝑑𝑠𝑑𝑡+∫𝜏0+∞ℎ(𝑡)∫𝑡+∞𝐺(𝑠)𝑑𝑠𝑑𝑡}_{1−∫𝜏0−∞𝑔(𝑠)𝑑𝑠−∫𝜏0+∞𝑔(𝑠)𝑑𝑠−∫𝜏0−∞ℎ(𝑠)𝑑𝑠1−∫𝜏0+∞ℎ(𝑠)𝑑𝑠}. (2.19)
Since 1−∫^𝜏0_−∞𝑔(𝑠)𝑑𝑠−∫_𝜏^+∞₀𝑔(𝑠)𝑑𝑠−∫^𝜏0_−∞ℎ(𝑠)𝑑𝑠1−∫_𝜏^+∞₀ℎ(𝑠)=𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠−∫_𝜏^+∞₀𝑔(𝑠)𝑑𝑠1−∫^+∞_−∞ℎ(𝑠)𝑑𝑠1−∫_𝜏^+∞₀ℎ(𝑠)𝑑𝑠>0, (2.20) we get from (2.19) that 𝑦(−∞)≥0 and 𝑦(+∞)≥0. Hence (2.16) implies that 𝑦(𝑡)≥0,∀𝑡∈𝑅. (2.21)
(ii) We prove that 𝑦(𝑡) is concave with respect to 𝜏 on 𝑅. It is easy to see that 𝜏∈𝐶(𝑅,(0,1)) and 𝑑𝜏𝑑𝑡=Φ⁻¹1𝜌(𝑡)1∫^+∞_−∞Φ⁻¹1/𝜌(𝑠)𝑑𝑠>0. (2.22) Thus 𝑑𝑦=𝑑𝑡𝑑𝑦𝑑𝜏𝑑𝜏=𝑑𝑡𝑑𝑦Φ𝑑𝜏⁻¹1𝜌(𝑡)1∫^+∞_−∞Φ⁻¹(𝑠).1/𝜌𝑑𝑠 (2.23) It follows that 𝜌(𝑡)Φ𝑑𝑦𝑑𝑡=Φ𝑑𝑦Φ1𝑑𝜏∫^+∞_−∞Φ⁻¹(𝑠).1/𝜌𝑑𝑠 (2.24) Hence 𝜌(𝑡)Φ𝑑𝑦𝑑𝑡^′=Φ^′𝑑𝑦𝑑𝑑𝜏²𝑦𝑑𝜏²𝑑𝜏Φ1𝑑𝑡∫^+∞_−∞Φ⁻¹(𝑠).1/𝜌𝑑𝑠 (2.25) So 𝑑²𝑦𝑑𝜏²=Φ^+∞_−∞Φ⁻¹1𝜌(𝑠)𝜌(𝑡)Φ𝑑𝑠𝑑𝑦/𝑑𝑡^′).Φ′𝑑𝑦/𝑑𝜏(𝑑𝜏/𝑑𝑡 (2.26) Since [𝜌(𝑡)Φ(𝑦′(𝑡))]^′≤0,Φ^′(𝑦)>0(𝑦≠0) and (𝑑𝜏/𝑑𝑡)>0, we get that (𝑑²𝑦/𝑑𝜏²)≤0. Hence 𝑦(𝑡) is concave with respect to 𝜏 on 𝑅.
(iii) Now, we prove that min_{[]𝑡∈−𝑘,𝑘}𝑦(𝑡)≥𝜇sup_𝑡∈𝑅𝑦(𝑡). (2.27) Since 𝑑𝜏/𝑑𝑡>0 for all 𝑡∈𝑅, there exists the inverse function of 𝜏=𝜏(𝑡). Denote the inverse function of 𝜏=𝜏(𝑡) by 𝑡=𝑡(𝜏).
It follows from (2.13) that sup_𝑡∈𝑅𝑦(𝑡)=𝑦(𝜏₀). One sees min_{[]𝑡∈−𝑘,𝑘}𝑦(𝑡)=min𝑦(−𝑘),𝑦(𝑘). (2.28) If min{𝑦(−𝑘),𝑦(𝑘)}=𝑦(𝑘)=𝑦(𝑡(𝜏(𝑘))), note 𝜏(𝑘)<1, then for 𝑡∈[−𝑘,𝑘], one has 𝑦(𝑡)(𝑡(𝜏(𝑘𝑡(𝑘)𝜏≥𝑦)))=𝑦1−𝜏+𝜏₀1+𝜏𝜏₀𝜏(𝑘)1−𝜏(𝑘)+𝜏𝜏₀+𝜏(𝑘)1+𝜏𝜏₀𝜏𝜏₀. (2.29) Noting that 1>𝜏(𝑘) and 𝑦(𝑡) is concave with respect to 𝜏, then, for 𝑡∈[−𝑘,𝑘], 𝑦(𝑡)≥1−𝜏(𝑘)+𝜏𝜏₀1+𝜏𝜏₀𝑦𝑡𝜏(𝑘)1−𝜏(𝑘)+𝜏𝜏₀+𝜏(𝑘)1+𝜏𝜏₀𝑦𝑡𝜏𝜏₀≥^𝑘_−∞Φ⁻¹1𝜌(𝑠)1𝑑𝑠2∫^+∞_−∞Φ⁻¹(𝑠)𝑦𝜏1/𝜌𝑑𝑠⁰=𝜇sup_𝑡∈𝑅𝑦(𝑡). (2.30) Similarly, if min{𝑦(−𝑘),𝑦(𝑘)}=𝑦(−𝑘)=(𝑦(𝑡(𝜏(−𝑘))), note 𝜏(−𝑘)<1, for 𝑡∈[−𝑘,𝑘], one has 𝑦(𝑡)(𝑡(𝜏(𝑡𝜏≥𝑦−𝑘)))=𝑦1+𝜏₀−𝜏(−𝑘)1+𝜏𝜏₀𝜏(−𝑘)1+𝜏𝜏₀()+𝜏()−𝜏−𝑘−𝑘1+𝜏𝜏₀𝜏𝜏₀≥1+𝜏𝜏₀−𝜏(−𝑘)1+𝜏𝜏₀𝑦𝑡𝜏(−𝑘)1+𝜏𝜏₀()+𝜏()−𝜏−𝑘−𝑘1+𝜏𝜏₀𝑦𝑡𝜏𝜏₀≥^−𝑘_−∞Φ⁻¹1𝜌(𝑠)1𝑑𝑠2∫^+∞_−∞Φ⁻¹(𝑠)𝑦𝜏1/𝜌𝑑𝑠⁰>𝜇sup_𝑡∈𝑅𝑦(𝑡). (2.31) Hence (2.27) holds.
(iv) Finally, we prove the uniqueness of 𝐴_𝑥. Define 𝐻_𝑥(𝑐)=^+∞_−∞𝑔(𝑡)1−∫^+∞_−∞ℎ(𝑠)𝑑𝑠−ℎ(𝑡)1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠×^𝑡_−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝑐+_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)+𝑑𝑢𝑑𝑠𝑑𝑡_+∞−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝑐+_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)𝑑𝑢𝑑𝑠. (2.32) Then (H1) implies that 𝐻_𝑥∈𝐶(𝑅,𝑅) is increasing on 𝑅 and 𝐻_𝑥(0)≥0. Let 𝑐=−^+∞_−∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)𝑑𝑢, (2.33) then 𝐻_𝑥(𝑐)≤0. By mean value theorem, there exists an unique 𝐴_𝑥∈[𝑐,0] satisfying 𝐻_𝑥(𝐴_𝑥)=0. Then (iv) holds. This completes the proof of the lemma.

Choose 1≥𝑘>0 and ∫𝜇=^{−𝑘−∞}Φ⁻¹1/𝜌(𝑠)𝑑𝑠2∫^∞_−∞Φ⁻¹(𝑠).1/𝜌𝑑𝑠 (2.34) Define the cone 𝑃⊆𝑋 by 𝑥(𝑡)𝑃=𝑥∈𝑋∶≥0,𝑡∈𝑅,min_{[]𝑡∈−𝑘,𝑘}𝑥(𝑡)≥𝜇max_𝑡∈𝑅𝑥(𝑡). (2.35) It is easy to see that 𝑃 is a cone in 𝑋.

Define the operator 𝑇∶𝑃→𝑋 by (𝑡)=∫𝑇𝑥)(^+∞−∞𝑔(𝑢)∫^𝑢_−∞Φ⁻¹1/𝜌(𝑠)Φ⁻¹𝐴_𝑥+∫_𝑠^+∞𝑓𝑟,𝑥(𝑟),𝑥^′(𝑟)𝑑𝑟𝑑𝑠𝑑𝑢1−∫^+∞_−∞𝑔(𝑠)+𝑑𝑠_𝑡−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑟,𝑥(𝑟),𝑥^′(𝑟)𝑑𝑟𝑑𝑠, (2.36) where 𝐴_𝑥 satisfies ^+∞_−∞𝑔(𝑢)1−∫^+∞_−∞ℎ(𝑠)𝑑𝑠−ℎ(𝑢)1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠×^𝑢_−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑟,𝑥(𝑟),𝑥^′(𝑟)+𝑑𝑟𝑑𝑠𝑑𝑢_+∞−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑟,𝑥(𝑟),𝑥^′(𝑟)𝑑𝑟𝑑𝑠=0. (2.37) It follows from Lemma 2.6(iv) that 𝑇∶𝑃→𝑋 is well defined and 𝐴_𝑥∈[−∫^+∞_−∞𝑓(𝑠,𝑥(𝑠),𝑥′(𝑠))𝑑𝑠,0]. It is easy to show that 𝜌(𝑡)Φ(𝑇𝑥)^′(𝑡)^′+𝑓𝑡,𝑥(𝑡)𝑥^′(𝑡)()==0,𝑡∈𝑅,𝑇𝑥)(−∞^+∞_−∞𝑔(𝑠𝑠)()=)(𝑇𝑥)(𝑑𝑠,𝑇𝑥)(+∞^+∞_−∞ℎ(𝑠)(𝑇𝑥)(𝑠)𝑑𝑠. (2.38) It follows from Lemma 2.6(i) and (iii) that 𝑇𝑥∈𝑃 for all 𝑥∈𝑃. Then 𝑇∶𝑃→𝑃 is well defined.

Lemma 2.7. Suppose (H1)–(H3) hold. Then 𝑇 is completely continuous.

Proof. First, we prove that 𝑇 is continuous.
We claim that the function 𝐴_𝑥∶𝑃→[0,+∞) is continuous in 𝑥.
Let {𝑥_𝑛}∈𝑃 with 𝑥_𝑛→𝑥₀∈𝑃 as 𝑛→∞ in 𝑃. Let {𝐴_𝑥𝑛}(𝑛=1,2,…) be the constants determined by ^+∞_−∞𝑔(𝑡)1−∫^+∞_−∞ℎ(𝑠)𝑑𝑠−ℎ(𝑡)1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠×^𝑡_−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥𝑛+_𝑠^+∞𝑓𝑟,𝑥_𝑛(𝑟),𝑥^′_𝑛(𝑟)+𝑑𝑟𝑑𝑠𝑑𝑡_+∞−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥𝑛+_𝑠^+∞𝑓𝑟,𝑥_𝑛(𝑟),𝑥^′_𝑛(𝑟)𝑑𝑟𝑑𝑠=0 (2.39) Since 𝑥_𝑛→𝑥₀ in 𝑃 as 𝑛→∞, there exists an 𝑟>0 such that ‖𝑥_𝑛‖≤𝑟. The fact 𝑓 is a Carathédory function means that there exists a nonnegative function 𝜙_𝑟∈𝐿¹(−∞,+∞) such that 0≤𝑓𝑡,𝑥_𝑛(𝑡),𝑥^′_𝑛(𝑡)=𝑓𝑡,𝑥_𝑛(𝑡),1Φ⁻¹𝜌(𝑡)Φ⁻¹𝜌(𝑡)𝑥^′_𝑛(𝑡)≤𝜙_𝑟(𝑡),𝑡∈𝑅. (2.40) Then ^+∞_−∞𝑓𝑠,𝑥_𝑛(𝑠),𝑥^′_𝑛(𝑠)𝑑𝑠isconvergent. (2.41) So 𝐴_𝑥𝑛∈−^+∞_−∞𝑓𝑠,𝑥_𝑛(𝑠),𝑥^′_𝑛(𝑠)𝑑𝑠,0⊆−^+∞_−∞𝜙_𝑟(𝑠)𝑑𝑠,0, (2.42) which means that {𝐴_𝑥𝑛} is uniformly bounded.
Suppose that {𝐴_𝑥𝑛} does not converge to 𝐴_𝑥0. By the bounded property, we know that there exist two subsequences {𝐴_𝑥⁽¹⁾_𝑛𝑘} and {𝐴_𝑥⁽²⁾_𝑛𝑘} of {𝐴_𝑥𝑛𝑘} with 𝐴_𝑥⁽¹⁾_𝑛𝑘→𝑐₁ and 𝐴_𝑥⁽²⁾_𝑛𝑘→𝑐₂ and 𝑐₁≠𝑐₂.
By the construction of 𝐴_𝑥𝑛, (𝑛=1,2,…), we have ^+∞_−∞𝑔(𝑡)1−∫^+∞_−∞ℎ(𝑠)𝑑𝑠−ℎ(𝑡)1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠×^𝑡_−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥⁽¹⁾_𝑛𝑘+_𝑠^+∞𝑓𝑢,𝑥_𝑛𝑘(𝑢),𝑥^′_𝑛𝑘(𝑢)+𝑑𝑢𝑑𝑠𝑑𝑡_+∞−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥⁽¹⁾_𝑛𝑘+_𝑠^+∞𝑓𝑢,𝑥_𝑛𝑘(𝑢),𝑥^′_𝑛𝑘(𝑢)𝑑𝑢𝑑𝑠=0. (2.43) Let 𝑘→∞, using Lebesgue's dominated convergence theorem, the above equality implies ^+∞_−∞𝑔(𝑡)1−∫^+∞_−∞ℎ(𝑠)𝑑𝑠−ℎ(𝑡)1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠×^𝑡_−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝑐₁+_𝑠^+∞𝑓𝑢,𝑥₀(𝑢),𝑥^′_𝑜(𝑢)+𝑑𝑢𝑑𝑠𝑑𝑡_+∞−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝑐₁+_𝑠^+∞𝑓𝑢,𝑥₀(𝑢),𝑥^′_𝑜(𝑢)𝑑𝑢𝑑𝑠=0. (2.44) By Lemma 2.6 (iv), we get 𝑐₁=𝐴_𝑥0. Similarly, 𝑐₂=𝐴_𝑥0. Thus 𝑐₁=𝑐₂, a contradiction. So, for any 𝑥_𝑛→𝑥₀, one has 𝐴_𝑥𝑛→𝐴_𝑥0, which means 𝐴_𝑥∶𝑃→𝑅 is continuous.
Since 𝐴_𝑥 is continuous, together with the continuity of (𝑥,𝑦)→𝑓(𝑡,𝑥,(1/Φ⁻¹(𝜌(𝑡)))𝑦), we get that 𝑇 is continuous.
Second, we show that 𝑇 is maps bounded subsets into bounded sets.
Let 𝐷⊆𝑃 be bounded. Then, there exists 𝑟>0 such that 𝐷⊆{𝑥∈𝑃∶‖𝑥‖≤𝑟}. Hence 𝑟=maxsup_𝑡∈𝑅|𝑥(𝑡)|,sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)<+∞. (2.45) Then there exists a nonnegative function 𝜙_𝑟∈𝐿¹(−∞,+∞) such that 0≤𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)(𝑡),1=𝑓𝑡,𝑥Φ⁻¹𝜌(𝑡)Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)≤𝜙_𝑟(𝑡),𝑡∈𝑅. (2.46) Then ^+∞_−∞𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡≤^+∞_−∞𝜙_𝑟(𝑡)𝑑𝑡∶=𝐿. (2.47) Thus |𝐴_𝑥|≤𝐿 for all 𝑥∈𝐷. Therefore, (𝑡)=∫𝑇𝑥)(^+∞−∞𝑔(𝑡)∫^𝑡_−∞Φ⁻¹1/𝜌(𝑠)Φ⁻¹𝐴_𝑥+∫_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)𝑑𝑢𝑑𝑠𝑑𝑡1−∫^+∞_−∞𝑔(𝑠)+𝑑𝑠_𝑡−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢≤∫)𝑑𝑢𝑑𝑠_+∞−∞𝑔(𝑡)∫^𝑡_−∞Φ⁻¹1/𝜌(𝑠)𝑑𝑠𝑑𝑡1−∫^+∞_−∞𝑔(𝑠)+𝑑𝑠^𝑡−∞Φ⁻¹1𝜌(𝑠)𝑑𝑠Φ⁻¹()≤∫2𝐿_+∞−∞𝑔(𝑡)∫^𝑡_−∞Φ⁻¹1/𝜌(𝑠)𝑑𝑠𝑑𝑡1−∫^+∞_−∞𝑔(𝑠)+𝑑𝑠^+∞−∞Φ⁻¹1𝜌(𝑠)𝑑𝑠Φ⁻¹(2𝐿)=∶𝑀₁. (2.48) On the other hand, we have Φ⁻¹𝜌(𝑡)(𝑇𝑥)^′(𝑡)=Φ⁻¹𝐴_𝑥+^∞_𝑡𝑓𝑢,𝑥(𝑢),𝑥′(𝑢)𝑑𝑢≤Φ⁻¹(2𝐿)=∶𝑀₂. (2.49) Then ‖(𝑇𝑥)‖=maxsup_𝑡∈𝑅(𝑇𝑥)(𝑡),sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)(𝑇𝑥)^′(𝑡)<∞. (2.50) So, {𝑇𝑥∶𝑥∈𝐷} is bounded.
Third, given a bounded set 𝐷⊆𝑃, we prove that both {𝑇𝑥∶𝑥∈𝐷} and {Φ⁻¹(𝜌(𝑡))(𝑇𝑥)′∶𝑥∈𝐷} are equicontinuous on each finite subinterval on 𝑅.
Then, there exists 𝑟>0 such that 𝐷⊆{𝑥∈𝑃∶‖𝑥‖≤𝑟}. Hence 𝑟=maxsup_𝑡∈𝑅|𝑥(𝑡)|,sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)<+∞. (2.51) Then there exists a nonnegative function 𝜙_𝑟∈𝐿¹(−∞,+∞) such that 0≤𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)(𝑡),1=𝑓𝑡,𝑥Φ⁻¹𝜌(𝑡)Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)≤𝜙_𝑟(𝑡),𝑡∈𝑅. (2.52) Then 𝐴_𝑥+_𝑡^+∞𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡≤^+∞_−∞𝜙_𝑟(𝑡)𝑑𝑡∶=𝐿. (2.53)
For any 𝜖>0, since Φ⁻¹ is uniformly continuous on [−𝐿,𝐿], there exists 𝛿₁>0 such that Φ⁻¹(𝑢)−Φ⁻¹(𝑣)<𝜖,𝑢,𝑣∈[−𝐿,𝐿],|𝑢−𝑣|<𝛿₁. (2.54) For any 𝑎,𝑏∈𝑅,𝑡₁,𝑡₂∈[𝑎,𝑏] and 𝑥∈𝐷 with 𝑡₁≤𝑡₂, we have 𝐴_𝑥+_𝑡^+∞₁𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡−𝐴_𝑥−_𝑡^+∞₂𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡≤^𝑡2_𝑡1𝜙_𝑟(𝑠)𝑑𝑠. (2.55) Then there exists 𝛿>0 such that 𝐴_𝑥+_𝑡^+∞₁𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡−𝐴_𝑥−_𝑡^+∞₂𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡<𝛿₁,𝑡₁,𝑡₂∈[𝑎,𝑏],𝑡₁−𝑡₂<𝛿. (2.56) Hence 𝑡₁,𝑡₂∈[𝑎,𝑏],|𝑡₁−𝑡₂|<𝛿 imply that Φ⁻¹𝜌𝑡₁(𝑇𝑥)𝑡₁−Φ⁻¹𝜌𝑡₂(𝑇𝑥)𝑡₂=Φ⁻¹𝐴_𝑥+_𝑡^+∞₁𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡−Φ⁻¹𝐴_𝑥+_𝑡^+∞₂𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡<𝜖. (2.57) It follows that {𝑇𝑥∶𝑥∈𝐷} is equicontinuous on each finite subinterval on 𝑅.
On the other hand, we have (𝑇𝑥)𝑡₁−(𝑇𝑥)𝑡₂=^𝑡2_𝑡1Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)≤𝑑𝑢𝑑𝑠_𝑡2𝑡1Φ⁻¹1𝜌(𝑠)𝑑𝑠Φ⁻¹(2𝐿)→0uniformlyas𝑡₁→𝑡₂. (2.58) Then {Φ⁻¹(𝜌(𝑡))(𝑇𝑥)′∶𝑥∈𝐷} is equicontinuous on each finite subinterval on 𝑅.
At last given a bounded set 𝐷⊆𝑃, we show that both {𝑇𝑥∶𝑥∈𝐷} and {Φ⁻¹(𝜌(𝑡))(𝑇𝑥)′∶𝑥∈𝐷} are equiconvergent at ±∞, respectively.
Then, there exists 𝑟>0 such that 𝐷⊆{𝑥∈𝑃∶‖𝑥‖≤𝑟}. Hence 𝑟=maxsup_𝑡∈𝑅|𝑥(𝑡)|,sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑥′(𝑡)<+∞. (2.59) Then there exists a nonnegative function 𝜙_𝑟∈𝐿¹(−∞,+∞) such that 0≤𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)(𝑡),1=𝑓𝑡,𝑥Φ⁻¹𝜌(𝑡)Φ⁻¹𝜌(𝑡)𝑥^′(𝑡)≤𝜙_𝑟(𝑡),𝑡∈𝑅. (2.60) Then 𝐴_𝑥+_𝑡^+∞𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑑𝑡≤^+∞_−∞𝜙_𝑟(𝑡)𝑑𝑡∶=𝐿. (2.61)
For any 𝜖>0, since Φ⁻¹ is uniformly continuous on [−𝐿,𝐿], there exists 𝛿₁>0 such that Φ⁻¹(𝑢)−Φ⁻¹(𝑣)<𝜖,𝑢,𝑣∈[−𝐿,𝐿],|𝑢−𝑣|<𝛿₁. (2.62) Since 𝜌(𝑡)Φ(𝑇𝑥)′(𝑡)−𝐴_𝑥=^∞_𝑡𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)𝑑𝑢≤^∞_𝑡𝜙_𝑟(𝑢)𝑑𝑢→0 (2.63) uniformly as 𝑡→∞, we get that there exists 𝑇>0 such that 𝜌(𝑡)Φ(𝑇𝑥)^′(𝑡)−𝐴_𝑥<𝛿₁,𝑡≥𝑇. (2.64) Hence 𝑡>𝑇 implies that Φ⁻¹𝜌(𝑡)(𝑇𝑥)^′(𝑡)−Φ⁻¹𝐴_𝑥=Φ⁻¹𝜌(𝑡)Φ(𝑇𝑥)^′(𝑡)−Φ⁻¹𝐴_𝑥<𝜖. (2.65) Furthermore, we get (𝑡)−∫𝑇𝑥)(^+∞−∞𝑔(𝑡)∫^𝑡_−∞Φ⁻¹1/𝜌(𝑠)Φ⁻¹𝐴_𝑥+∫_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)𝑑𝑢𝑑𝑠𝑑𝑡1−∫^+∞_−∞𝑔(𝑠)+𝑑𝑠_+∞−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+_𝑠^+∞𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)=𝑑𝑢𝑑𝑠_𝑡+∞Φ⁻¹1𝜌(𝑠)Φ⁻¹𝐴_𝑥+^∞_𝑠𝑓𝑢,𝑥(𝑢),𝑥^′(𝑢)≤𝑑𝑢𝑑𝑠_𝑡+∞Φ⁻¹1𝜌(𝑠)𝑑𝑠Φ⁻¹(2𝐿)→0uniformlyas𝑡→+∞. (2.66) Hence {𝑇𝑥∶𝑥∈𝐷} and {Φ⁻¹(𝜌(𝑡))(𝑇𝑥)′∶𝑥∈𝐷} are equiconvergent at +∞.
Similarly we can how that {𝑇𝑥∶𝑥∈𝐷} and {Φ⁻¹(𝜌(𝑡))(𝑇𝑥)′∶𝑥∈𝐷} are equiconvergent at −∞. We omit the details.
Therefore, 𝑇∶𝑃→𝑃 is equiconvergent at ±∞. So the operator 𝑇∶𝑃→𝑃 is completely continuous.
Define the functionals on 𝑃 by 𝛾𝑦=sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑦^′(𝑡𝛽𝑦),𝑦∈𝑃,=sup_𝑡∈𝑅𝑦(𝑡𝜃𝑦),𝑦∈𝑃,=sup_𝑡∈𝑅𝑦(𝑡𝛼𝑦),𝑦∈𝑃,=min_{[]𝑡∈−𝑘,𝑘}𝑦(𝑡𝜓𝑦),𝑦∈𝑃,=min_{[]𝑡∈−𝑘,𝑘}𝑦(𝑡),𝑦∈𝑃. (2.67) It is easy to see that 𝛼,𝜓 are two nonnegative continuous concave functionals on the cone 𝑃, 𝛾,𝛽,𝜃 are three nonnegative continuous convex functionals on the cone 𝑃 and 𝛼(𝑦)≤𝛽(𝑦) for all 𝑦∈𝑃.
For 𝑒₁,𝑒₂,𝑐>0 and 1≥𝑘>0, define ∫𝑀=max^+∞−∞1/Φ⁻¹𝜌(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠),𝐿𝑑𝑠,1₁=⁰_−𝑘Φ⁻¹1𝜌(𝑠)Φ⁻¹2|𝑠|𝐿𝑑𝑠,₂=^𝑘₀Φ⁻¹1𝜌(𝑠)Φ⁻¹2√𝑠Φ(𝑐)𝑑𝑠,𝑄=min,Φ(𝑐)4+𝜋Φ4+𝜋1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠∫^+∞_−∞1/Φ⁻¹𝜌(𝑠),𝑒𝑑𝑠𝑊=Φ²𝜇Φ1max𝐿₁1,Φ𝐿₂,Φ𝑒𝐸=¹,𝛿(𝑡)=14+𝜋𝑡²,|𝑡|1≥1,|𝑡|,|𝑡|∫≤1,𝜇=^{−𝑘−∞}Φ⁻¹1/𝜌(𝑠)𝑑𝑠2∫^+∞_−∞Φ⁻¹(𝑠).1/𝜌𝑑𝑠 (2.68)

Theorem 2.8. Suppose that (H1)–(H3) hold. Given positive constants 𝑒₁,𝑒₂,𝑐 and 𝑘∈(0,1), let 𝑄,𝑊, and 𝐸 be as above. If 𝑒𝑐≥²𝜇>𝑒₂>𝑒₁>0,𝑄≥𝑊 (2.69) and(A1)𝑓(𝑡,𝑢,(1/Φ⁻¹(𝜌(𝑡)))𝑣)≤𝛿(𝑡)𝑄  for all   𝑡∈𝑅,𝑢∈[0,𝑐],𝑣∈[−𝑐,𝑐];(A2)𝑓(𝑡,𝑢,(1/Φ⁻¹(𝜌(𝑡)))𝑣)≥𝛿(𝑡)𝑊  for all  𝑡∈[−𝑘,𝑘],𝑢∈[𝑒₂,𝑒₂/𝜇],𝑣∈[−𝑐,𝑐];(A3)𝑓(𝑡,𝑢,(1/Φ⁻¹(𝜌(𝑡)))𝑣)≤𝛿(𝑡)𝐸  for all  𝑡∈𝑅,𝑢∈[0,𝑒₁],𝑣∈[−𝑐,𝑐];
then BVP(1.2) has at least three positive solutions 𝑥₁,𝑥₂,𝑥₃ such that sup_𝑡∈𝑅𝑥₁(𝑡)<𝑒₁,min_{𝑡∈[−𝑘,𝑘]}𝑥₂(𝑡)>𝑒₂,sup_𝑡∈𝑅𝑥₃(𝑡)>𝑒₁,min_{𝑡∈[−𝑘,𝑘]}𝑥₃(𝑡)<𝑒₂. (2.70)

Proof. We prove that all conditions in Lemma 2.5 are satisfied.
(i) By the definitions, it is easy to show that 𝛼,𝜓 are two nonnegative continuous concave functionals on the cone 𝑃, 𝛾,𝛽,𝜃 are three nonnegative continuous convex functionals on the cone 𝑃 and 𝛼(𝑦)≤𝛽(𝑦) for all 𝑦∈𝑃. One sees 𝑥∈𝑃 is a positive solution of BVP (1.2) if and only if 𝑥 is a solution of the operator equation 𝑥=𝑇𝑥.
(ii) For 𝑦∈𝑃, we have 𝑦(𝑡)=^𝑡_−∞𝑦^′(𝑠)(≤𝑑𝑠+𝑦−∞)_𝑡−∞𝑦^′(𝑠𝑦(≤)𝑑𝑠+−∞)_𝑡−∞1Φ⁻¹𝜌(𝑠)Φ⁻¹𝜌(𝑠)𝑦^′(𝑠)𝑑𝑠+^+∞_−∞𝑔(𝑠𝑦(𝑠≤))𝑑𝑠_+∞−∞1Φ⁻¹𝜌(𝑠)𝑑𝑠sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑦^′(𝑡)^+∞_−∞𝑔(𝑠)𝑑𝑠sup_𝑡∈𝑅𝑦(𝑡). (2.71) It follows that sup_𝑡∈𝑅𝑦(𝑡)≤^+∞_−∞1Φ⁻¹𝜌(𝑠)𝑑𝑠sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑦^′(𝑡)^+∞_−∞𝑔(𝑠)𝑑𝑠sup_𝑡∈𝑅𝑦(𝑡). (2.72) Then sup_𝑡∈𝑅𝑦(𝑡≤∫)^+∞−∞1/Φ⁻¹𝜌(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑦^′(𝑡). (2.73) Hence 𝑦=maxsup_𝑡∈𝑅𝑦(𝑡),sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑦^′(𝑡∫)≤max^+∞−∞1/Φ⁻¹𝜌(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠,1sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑦^′(𝑡). (2.74) It follows that ||𝑦||≤𝑀𝛾(𝑦) for all 𝑦∈𝑃.
(iii) Corresponding to Lemma 2.5, 𝑐=𝑐,ℎ=𝜇𝑒₁,𝑑=𝑒₁,𝑎=𝑒₂𝑒,𝑏=²𝜇. (2.75) Now, we prove that all other conditions of Lemma 2.5 hold. One sees that 0<𝑑<𝑎. The remainder is divided into five steps.
Step 1. Prove that 𝑇∶𝑃_𝑐→𝑃_𝑐.
For 𝑦∈𝑃_𝑐, we have ||𝑦||≤𝑐. Then 0≤𝑦(𝑡)≤𝑐 and −𝑐≤Φ⁻¹(𝜌(𝑡))𝑦′(𝑡)≤𝑐 for all 𝑡∈𝑅. So (A1) implies that 𝑓𝑡,𝑦(𝑡),𝑦^′(𝑡)(𝑡),1=𝑓𝑡,𝑦Φ⁻¹𝜌(𝑡)Φ⁻¹𝜌(𝑡)𝑦^′(𝑡)≤𝛿(𝑡)𝑄,𝑡∈𝑅. (2.76) We have Φ⁻¹𝜌(𝑡)𝑇𝑦^′(𝑡)=Φ⁻¹𝐴_𝑦+_𝑡^+∞𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢≤Φ^−1+∞_−∞𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢≤Φ^−1+∞_−∞𝛿(𝑠)𝑄𝑑𝑟=Φ⁻¹(𝑄)Φ⁻¹(4+𝜋)≤𝑐. (2.77) Similarly to (ii), we can show that 0≤𝑇𝑦(𝑡)≤𝑀sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑇𝑦^′(𝑡)≤𝑀Φ^−1+∞_−∞𝛿(𝑠)𝑄𝑑𝑟=𝑀Φ⁻¹(𝑄)Φ⁻¹(4+𝜋)≤𝑐. (2.78) It follows that 𝑇𝑦=maxsup_𝑡∈𝑅𝑇𝑦(𝑡),sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑇𝑦^′(𝑡)≤𝑐. (2.79) Then 𝑇∶𝑃_𝑐→𝑃_𝑐.
Step 2. Prove that 𝑦∈𝑃𝛾,𝜃,𝛼;𝑎,𝑏,𝑐∣𝛼𝑦>𝑎=𝑦∈𝑃𝛾,𝜃,𝛼;𝑒₂,𝑒₂𝜇,𝑐∣𝛼𝑦>𝑒₂≠∅ (2.80) and 𝛼(𝑇𝑦)>𝑒₂forevery𝑦∈𝑃(𝛾,𝜃,𝛼;𝑒₂,𝑒₂/𝜇,𝑐).
Choose 𝑦(𝑡)=𝑒₂/2𝜇 for all 𝑡∈𝑅. Then 𝑦∈𝑃 and 𝛼𝑦=𝑒₂2𝜇>𝑒₂𝑦=𝑒,𝜃²≤𝑒2𝜇𝜇²,𝛾𝑦=0<𝑐. (2.81) It follows that {𝑦∈𝑃(𝛾,𝜃,𝛼;𝑎,𝑏,𝑐)∣𝛼(𝑦)>𝑎}≠∅.
For 𝑦∈𝑃(𝛾,𝜃,𝛼;𝑎,𝑏,𝑐), one has that 𝛼𝑦=min_{[]𝑡∈−𝑘,𝑘}𝑦(𝑡)≥𝑒₂,𝜃𝑦=sup_𝑡∈𝑅𝑦(𝑡)≤𝑒₂𝜇,𝛾𝑦=sup_𝑡∈𝑅𝑦^′(𝑡)≤𝑐. (2.82) Then 𝑒₂(𝑡)≤𝑒≤𝑦²𝜇,𝑡∈[−𝑘,𝑘],Φ⁻¹𝜌(𝑡)𝑦^′(𝑡)≤𝑐. (2.83) Thus (A2) implies that 𝑓𝑡,𝑦(𝑡),𝑦^′(𝑡)≥𝛿(𝑡)𝑊,𝑡∈[−𝑘,𝑘]. (2.84) Similarly to Lemma 2.6(i), we know that there exists 𝜏₀∈(0,1) such that (𝑇𝑦)′(𝜏₀)=0. Then (𝑡)=()𝑇𝑦𝑇𝑦+∞+∫_𝑡^+∞Φ⁻¹1𝜌(𝑠)Φ⁻¹∫^𝑠_𝜏0𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢𝑑𝑠,𝑡≥𝜏₀,𝑇𝑦(−∞)+∫^𝑡_−∞Φ⁻¹1𝜌(𝑠)Φ⁻¹∫^𝜏0_𝑠𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢𝑑𝑠,𝑡≤𝜏₀. (2.85) Since 𝛼𝑇𝑦=min_{[]𝑡∈−𝑘,𝑘}𝑇𝑦(𝑡)≥𝜇sup_t∈𝑅𝑇𝑦(𝑡)=𝜇𝑇𝑦𝜏₀, (2.86) if 𝜏₀≥0, we get 𝛼𝑇𝑦≥𝜇𝑇𝑦(−∞)+^𝜏0_−∞Φ⁻¹1𝜌(𝑠)Φ^−1𝜏0_𝑠𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢𝑑𝑠>𝜇⁰_−𝑘Φ⁻¹1𝜌(𝑠)Φ⁻¹⁰_𝑠𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢𝑑𝑠≥𝜇⁰_−𝑘Φ⁻¹1𝜌(𝑠)Φ⁻¹⁰_𝑠𝛿(𝑢)𝑊𝑑𝑢𝑑𝑠≥𝜇⁰_−𝑘Φ⁻¹1𝜌(𝑠)Φ⁻¹2|𝑠|𝑑𝑠Φ⁻¹(𝑊)≥𝑒₂. (2.87) If 𝜏₀<0, we get 𝛼𝑇𝑦≥𝜇𝑇𝑦(+∞)+_𝜏^+∞₀Φ⁻¹1𝜌(𝑠)Φ^−1𝑠_𝜏0𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢𝑑𝑠>𝜇^𝑘_𝜏0Φ⁻¹1𝜌(𝑠)Φ^−1𝑠_𝜏0𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢𝑑𝑠≥𝜇^𝑘₀Φ⁻¹1𝜌(𝑠)Φ^−1𝑠₀𝛿(𝑢)𝑊𝑑𝑢𝑑𝑠≥𝜇^𝑘₀Φ⁻¹1𝜌(𝑠)Φ⁻¹2√𝑡𝑑𝑠Φ⁻¹(𝑊)≥𝑒₂. (2.88) This completes Step 2.
Step 3. Prove that {𝑦∈𝑄(𝛾,𝜃,𝜓;ℎ,𝑑,𝑐)∣𝛽(𝑦)<𝑑}={𝑦∈𝑄(𝛾,𝜃,𝜓;𝜇𝑒₁,𝑒₁,𝑐)|𝛽(𝑦)<𝑒₁}≠∅ and 𝛽𝑇𝑦<𝑒₁forevery𝑦∈𝑄𝛾,𝜃,𝜓;ℎ,𝑑,𝑐=𝑄𝛾,𝜃,𝜓;𝜇𝑒₁,𝑒₁,𝑐. (2.89)
Choose 𝑦(𝑡)=𝜇𝑒₁. Then 𝑦∈𝑃, and 𝜓𝑦=𝜇𝑒₁≥ℎ,𝛽𝑦=𝜃𝑦=𝜇𝑒₁<𝑒₁=𝑑,𝛾𝑦=0≤𝑐. (2.90) It follows that {𝑦∈𝑄(𝛾,𝜃,𝜓;ℎ,𝑑,𝑐)∣𝛽(𝑦)<𝑑}≠∅.
For 𝑦∈𝑄(𝛾,𝜃,𝜓;ℎ,𝑑,𝑐), one has that 𝜓𝑦=min_{[]𝑡∈−𝑘,𝑘}𝑦(𝑡)≥ℎ=𝜇𝑒₁,𝜃𝑦=sup_𝑡∈𝑅𝑦(𝑡)≤𝑑=𝑒₁,𝛾𝑦=sup_𝑡∈𝑅𝑦^′(𝑡)≤𝑐. (2.91) Hence we get that 0≤𝑦(𝑡)≤𝑒₁,𝑡∈𝑅;−𝑐≤Φ⁻¹𝜌(𝑡)𝑦^′(𝑡)≤𝑐,𝑡∈𝑅. (2.92) Then (A3) implies that 𝑓𝑡,𝑦(𝑡),𝑦^′(𝑡)≤𝛿(𝑡)𝐸,𝑡∈𝑅. (2.93) So 𝛽𝑇𝑦=𝑀sup_𝑡∈𝑅Φ⁻¹𝜌(𝑡)𝑇𝑦^′(𝑡Φ)≤𝑀⁻¹𝐴_𝑦+_𝑡^+∞𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢≤𝑀Φ^−1+∞_−∞𝑓𝑢,𝑦(𝑢),𝑦^′(𝑢)𝑑𝑢≤𝑀Φ^−1+∞_−∞𝛿(𝑢)𝐸𝑑𝑢≤𝑀Φ⁻¹(4+𝜋)Φ⁻¹(𝐸)=𝑒₁=𝑑. (2.94) This completes Step 3.
Step 4. Prove that 𝛼(𝑇𝑦)>𝑎 for 𝑦∈𝑃(𝛾,𝛼;𝑎,𝑐) with 𝜃(𝑇𝑦)>𝑏;
For 𝑦∈𝑃(𝛾,𝛼;𝑎,𝑐)=𝑃(𝛾,𝛼;𝑒₂,𝑐) with 𝜃(𝑇𝑦)>𝑏=𝑒₂/𝜇, we have that 𝛼(𝑦)=min_{𝑡∈[−𝑘,𝑘]}𝑦(𝑡)≥𝑒₂ and 𝛾(𝑦)=sup_𝑡∈𝑅Φ⁻¹(𝜌(𝑡))|𝑦′(𝑡)|≤𝑐 and sup_𝑡∈𝑅(𝑇𝑦)(𝑡)>𝑒₂/𝜇. Then 𝛼𝑇𝑦=min_{[]𝑡∈−𝑘,𝑘}(𝑡)𝑒𝑇𝑦≥𝜇𝛽𝑇𝑦>𝜇²𝜇=𝑒₂=𝑎. (2.95) This completes Step 4.
Step 5. Prove that 𝛽(𝑇𝑦)<𝑑 for each 𝑦∈𝑄(𝛾,𝛽;𝑑,𝑐) with 𝜓(𝑇𝑦)<ℎ.
For 𝑦∈𝑄(𝛾,𝛽;𝑑,𝑐) with 𝜓(𝑇𝑦)<𝑑, we have 𝛾(𝑦)=sup_𝑡∈𝑅Φ⁻¹(𝜌(𝑡))|𝑦′(𝑡)|≤𝑐 and 𝛽(𝑦)=sup_𝑡∈𝑅𝑦(𝑡)≤𝑑=𝑒₁ and 𝜓(𝑇𝑦)=min_{𝑡∈[−𝑘,𝑘]}(𝑇𝑦)(𝑡)<ℎ=𝑒₁𝜇. Then 𝛽𝑇𝑦=sup_𝑡∈𝑅(𝑡)≤1𝑇𝑦𝜇min_{[]𝑡∈−𝑘,𝑘}(𝑡)<1𝑇𝑦𝜇𝑒₁𝜇=𝑒₁=𝑑. (2.96) This completes the Step 5.
Then Lemma 2.5 implies that 𝑇 has at least three fixed points 𝑥₁, 𝑥₂, and 𝑥₃ such that 𝛽𝑥₁<𝑒₁,𝛼𝑥₂>𝑒₂,𝛽𝑥₃>𝑒₁,𝛼𝑥₃<𝑒₂. (2.97) Hence BVP(1.2) has three decreasing positive solutions 𝑥₁,𝑥₂ and 𝑥₃ such that (2.70) holds. The proof is complete.

3. Examples

Now, we present an example, whose three positive solutions cannot be obtained by theorems in known papers, to illustrate the main results.

Example 3.1. Consider the following BVP 𝑒^3𝑡2𝑥^′(𝑡)^3′+𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑥()𝑥()=0,𝑡∈𝑅,−∞=0,+∞=0. (3.1) Corresponding to BVP(1.2), one sees that  𝜙(𝑥)=𝑥³,𝜙⁻¹(𝑥)=𝑥^1/3, 𝑔(𝑡)=ℎ(𝑡)≡0, 𝜌(𝑡)=𝑒^3𝑡2, 𝑓∶𝑅×𝑅×𝑅→[0,∞) is nonnegative and continuous and is defined by 𝑓𝑡,𝑥,𝑦=𝛿(𝑡)𝑓₀(𝑥)+𝑔₀𝑒^𝑡2𝑦,𝛿(𝑡)=1𝑡²,|𝑡|1≥1,|𝑡|,|𝑡|≤1. (3.2)

Choose 𝑘=1, 𝑒₁=50,𝑒₂=250,𝑐=40000. By direct computation, we see that 𝑄,𝑊, and 𝐸 are given by ∫𝑀=max^+∞−∞1/Φ⁻¹𝜌(𝑠)𝑑𝑠1−∫^+∞_−∞𝑔(𝑠)=√𝑑𝑠,1𝜋2,𝐿₁=⁰_−𝑘Φ⁻¹1𝜌(𝑠)Φ⁻¹2|𝑠|𝑑𝑠=2^1/30₋₁𝑒^−𝑠2|𝑠|^1/6𝐿𝑑𝑠>0.06,₂=^𝑘₀Φ⁻¹1𝜌(𝑠)Φ⁻¹2√𝑠𝑑𝑠=2^1/31₀𝑒^−𝑠2𝑠^1/6∫𝑑𝑠>0.06,𝜇=^{−𝑘−∞}Φ⁻¹1/𝜌(𝑠)𝑑𝑠2∫^+∞_−∞Φ⁻¹(𝑠=∫1/𝜌)𝑑𝑠^−1−∞𝑒^−𝑠2𝑑𝑠√𝜋Φ(𝑐)>0.37,𝑄=min,Φ(𝑐)4+𝜋Φ4+𝜋1−∫^+∞_−∞𝑔(𝑠)𝑑𝑠∫^+∞_−∞1/Φ⁻¹𝜌(𝑠)=𝑑𝑠64×10¹²4+𝜋>8.53×10¹²,𝑒𝑊=Φ²𝜇Φ1max𝐿₁1,Φ𝐿₂<16.67³×675.7³<3.43×10¹¹,Φ𝑒𝐸=¹=4+𝜋25004+𝜋>333.33. (3.3) One can show that 𝑒𝑐≥²𝜇>𝑒₂>𝑒₁>0,𝑄≥𝑊. (3.4) Suppose that 𝑓₀(𝑥)=166.67,𝑥∈[0,50],166.67+44.36×10¹¹−166.67()[],250−50𝑥−50,𝑥∈50,25044.36×10¹¹,𝑥∈[250,40000],44.36×10¹¹𝑒^𝑥−40000𝑔,𝑥≥40000,₀𝑦≤10,∀𝑦∈𝑅. (3.5)

From 𝑓𝑡,𝑥,𝑒^−𝑡2𝑦=𝛿(𝑡)𝑓₀(𝑥)+𝑔₀𝑦, (3.6) it is easy to show that(A₁)𝑓(𝑡,𝑢,𝑒^−𝑡2𝑣)≤8.53×10¹²𝛿(𝑡) for all 𝑡∈𝑅,𝑢∈[0,400000],𝑣∈[−40000,40000];(A₂)𝑓(𝑡,𝑢,𝑒^−𝑡2𝑣)≥3.43×10¹¹𝛿(𝑡) for all 𝑡∈[−1,1],𝑢∈[250,1000],𝑣∈[−40000,40000];(A₃)𝑓(𝑡,𝑢,𝑒^−𝑡2𝑣)≤333.33𝛿(𝑡) for all 𝑡∈𝑅,𝑢∈[0,50],𝑣∈[−40000,40000];

then Theorem 2.8 implies that BVP(3.1) has at least three positive solutions 𝑥₁,𝑥₂,𝑥₃ such that sup_𝑡∈𝑅𝑥₁(𝑡)<50,min_{[]𝑡∈−1,1}𝑥₂(𝑡)>250,sup_𝑡∈𝑅𝑥₃(𝑡)>50,min_{𝑡∈[−1,1]}𝑥₃(𝑡)<250. (3.7)

Remark 3.2. Example 3.1 implies that there is a large number of functions that satisfy the conditions of Theorem 2.8. In addition, the conditions of Theorem 2.8 are also easy to check.

Acknowledgments

This work was supported by Natural Science Foundation of Guangdong provinces Grant (no S2011010001900) and the Foundation for High-level talents in Guangdong Higher Education Project.