Solvability of a Third-Order Singular Generalized Left Focal Problem in Banach Spaces

Abstract

We consider the existence of positive solution for a third-order singular generalized left focal boundary value problem with full derivatives in Banach spaces. Green’s function and its properties, explicit a priori, estimates will be presented. By means of the theories of the fixed point in cones, we establish some new and general results on the existence of single and multiple positive solutions to the third-order singular generalized left focal boundary value problem. Our results are generalizations and extensions of the results of the focal boundary value problem. An example is included to illustrate the results obtained.

1. Introduction

Third-order differential equation describes many phenomena in applied mathematics, physical science, aeronautics, and applied mechanics such as the study of steady flows produced by free jets, wall jets, liquid jets, the flow past a stretching plate, and Blasius flow [1–6]. For two-dimensional flow of a fluid with small viscosity adhering to the flat plate, the simplified version, which has been derived by Blasius, only uses two unknowns 𝑢 and 𝑣. Laminar flat plate flow across a flat plate can be expressed using a boundary-layer equation defined by Blasius. Simplified momentum equation is given by 𝑢𝜕𝑢𝜕𝑥+𝑣𝜕𝑣𝜕𝜕𝑦=𝑣²𝑢𝜕𝑦²,𝑢(𝑥,0)=𝑣(𝑥,0)=0,𝑢(𝑥,∞)=𝑈_∞, (1) where 𝑢=𝑢(𝑥,𝑦) is the velocity measured, 𝑣=𝑣(𝑥,𝑦) is the transverse velocity component, 𝑥 is length of the plate, 𝑦 is the distance away from the plate, and 𝑈_∞ is free-stream velocity. Assume that the leading edge of the plate is 𝑥=0 and the plate is infinity long; this equation can be simplified as −𝑈²_∞2𝑥𝜍𝑓^′𝑓^′′+𝑈²_∞2𝑥𝜍𝑓^′𝑈−𝑓=𝑣^2∞𝑓𝑥𝑣^′′′, (2) where 𝑓=𝑓(𝜍) is the dimensionless stream function. For further simplification, we have the Blasius differential equation 𝑓^′𝑓^′′+2𝑓^′′′=0, (3) where the boundary conditions are 𝜍=0,𝑓=0, and 𝑓^′=0 or 𝜍=∞,𝑓^′=0. For other related application results of the problem we refer to [7–10]. Recently, increasing attention is paid to the question of the solution for the third-order focal boundary value problem, especially for right focal problem. The main means are the Leray-Schauder continuation theorem, iteration of monotone mapping, upper and lower solutions method, fixed point theory, and so on. Many applications of the above tools of various nonlocal boundary conditions include recent works [11–13] and the reference therein. It is well known that a powerful tool for proving the existence of the solution to the focal boundary value problem is the fixed point theorem. In many cases, it is possible to find single or multiple solutions for the given problem. We would like to mention some results of [14, 15]. Anderson [14] has investigated the existence of a solution to a third-order generalized right focal problem 𝑥^′′′(𝑡)=𝑓(𝑡,𝑥(𝑡)),𝑡₁<𝑡<𝑡₃,𝑥𝑡₁=𝑥^′𝑡₂=0,𝛾𝑥𝑡₃+𝛿𝑥^′′𝑡₃𝑡=0,₁<𝑡₂<𝑡₃, (4) for certain constraints placed on nonnegative 𝑓; the main approach is the Leggett and Williams fixed point theorem. Minhós [15] has proved an Ambrosetti-Prodi type result for the third-order fully nonlinear right focal problem 𝑢^′′′(𝑡)+𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡)(𝑡),𝑢(0)=𝑠𝑝0<𝑡<1,=𝐴,𝑎𝑢^′(0)−𝑏𝑢^′′(0)=𝐵,𝑐𝑢^′(1)+𝑑𝑢^′′(1)=𝐶, (5) where 𝑓∶[0,1]×ℝ→ℝ and 𝑝∶[0,1]→ℝ⁺ are continuous functions and 𝑠∈ℝ. The author has studied the existence, nonexistence, and multiplicity of solutions as variable 𝑠. Further, we may refer to recent contributions of Agarwal [16] and Agarwal et al. [17], Cabada and Heikkilä [18], and Wong and Agarwal [19] for the right focal problem. Accordingly, a third-order left focal problem for differential equation plays a role in many related fields too, even less attention has been focused, however, on third-order left focal problem. This paper will consider the existence of a positive solution for the following third-order singular generalized left focal boundary value problem with full derivatives in Banach spaces 𝑢^′′′(𝑡)+ℎ(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡)=0,0<𝑡₁<𝑡<𝑡₂,𝑎𝑢𝑡₁−𝑏𝑢^′′𝑡₁𝑢=0,_′𝜂=0,𝑡₁<𝜂<𝑡₂,𝑐𝑢𝑡₂+𝑑𝑢^′𝑡₂=0, 𝒫_𝑠 where 𝑎,𝑏,𝑐,𝑑,𝑡₁,𝜂,𝑡₂,ℎ, and 𝑓 satisfy (𝐻₁)𝑎,𝑏,𝑐,𝑑∈]0,∞[ and 𝑘=2𝑏𝑐+2𝑎𝑑(𝑡₂−𝜂)+𝑎𝑐(𝑡₂−𝑡₁)(𝑡₁+𝑡₂−2𝜂)>0,(𝐻₂)𝑡₁<𝜂<𝑡₂ with 𝑡₂−𝜂>𝜂−𝑡₁,(𝐻₃)ℎ∈𝐶(]𝑡₁,𝑡₂[,[0,∞[) may be singular at 𝑡=𝑡₁ and/or 𝑡=𝑡₂ and ∫^𝑡2_𝑡1ℎ(𝑠)𝑑𝑠<∞, (𝐻₄)𝑓∈𝐶([𝑡₁,𝑡₂]×[0,∞[×]−∞,∞[²,[0,∞[).

The novelty of our problem is to show the existence of solution of the left focal boundary value problem; we will concentrate on the solvability of the third-order singular generalized left focal problem for differential equation. To the best of our knowledge, it is the first time that the third-order left focal problem has been discussed successfully. As one might expect, the technique of finding Green’s function for the corresponding homogeneous left focal problem, as the kernel of an integral equation of Hammerstein type, has proven to be an effective method in most of third-order left focal problems. For this, we will employ known sign properties of Green’s function and its partial derivative cleverly. Note that the solution 𝑢 of the problem 𝒫_𝑠 lacks concavity, which is slightly different from right focal problem. In order to overcome this difficulty, we will establish the appropriate a priori estimates of the solution. On the other hand, the condition on nonlinearity required in this paper is different from those of existing papers, which motivated us to consider the above left focal problem, since 𝑓 may be superlinear (sublinear). As usual, the function 𝑓 in the problem 𝒫_𝑠 is called superlinear if 𝑓₀=0 and 𝑓_∞=∞; it is called sublinear when 𝑓_∞=0 and 𝑓₀=∞, where 𝑓₀=liminf_𝜇→0+min_{𝑡∈[𝑡1,𝑡2]}𝑓𝑡,𝜇,𝜈,𝜛𝜇,𝑓_∞=limsup_𝜇→∞max_{𝑡∈[𝑡1,𝑡2]}𝑓𝑡,𝜇,𝜈,𝜛𝜇,𝑓_∞=liminf_𝜇→∞min_{𝑡∈[𝑡1,𝑡2]}𝑓𝑡,𝜇,𝜈,𝜛𝜇,𝑓₀=limsup_𝜇→0+max_{𝑡∈[𝑡1,𝑡2]}𝑓𝑡,𝜇,𝜈,𝜛𝜇, (6) for 𝜈,𝜛∈]−∞,∞[. Based on the known results of the singular multipoint boundary value problem and high-order right focal problem for differential equation, this study is devoted to proving the existence of solution for the third-order singular generalized left focal problem 𝒫_𝑠, and the full derivatives are involved in the nonlinear term explicitly. Under the suitable limit and growth conditions on the nonlinearity 𝑓 in cones, we will establish some new and general results on the existence of single and multiple positive solutions to the problem 𝒫_𝑠; our results develop some results of the third-order focal problem and improve the results of third-order singular problem. The author believes that those results are useful in many applications for temporary new fields. Now we state the fixed point theorems.

Lemma 1 (see [20, 21]). Let 𝐾 be a cone in a Banach space 𝐸. Assume that 𝒦(𝑉₁),𝒦(𝑉₂) are open subsets of 𝐸 with 0∈𝒦(𝑉₁),𝒦(𝑉₁)⊂𝒦(𝑉₂). If 𝑇∶𝐾∩(𝒦(𝑉₂)⧵𝒦(𝑉₁))→𝐾 is a completely continuous operator such that either (𝐴₁)‖𝑇𝑥‖≤‖𝑥‖ for all 𝑥∈𝐾∩𝜕𝒦(𝑉₁) and ‖𝑇𝑥‖≥‖𝑥‖ for all 𝑥∈𝐾∩𝜕𝒦(𝑉₂) or(𝐴₂)‖𝑇𝑥‖≥‖𝑥‖ for all 𝑥∈𝐾∩𝜕𝒦(𝑉₁) and ‖𝑇𝑥‖≤‖𝑥‖ for all 𝑥∈𝐾∩𝜕𝒦(𝑉₂).
Then, 𝑇 has a fixed point in 𝐾∩(𝒦(𝑉₂)⧵𝒦(𝑉₁)).

Let 𝛾 and 𝜃 be nonnegative continuous convex functionals on a cone 𝐾,  let 𝛼 be a nonnegative continuous concave functional on 𝐾,  let 𝜓 be a nonnegative continuous functional on 𝐾, and let 𝑚₁,𝑚₂,𝑚₃, and 𝑚₄ be positive numbers. Define the convex sets as follows: 𝑃𝛾,𝑚₄=𝑢∈𝐾∶𝛾(𝑢)<𝑚₄,𝑃𝛾,𝛼,𝑚₂,𝑚₄=𝑢∈𝐾∶𝑚₂≤𝛼(𝑢),𝛾(𝑢)≤𝑚₄,𝑃𝛾,𝜃,𝛼,𝑚₂,𝑚₃,𝑚₄=𝑢∈𝐾∶𝑚₂≤𝛼(𝑢),𝜃(𝑢)≤𝑚₃,𝛾(𝑢)≤𝑚₄, (7) and a closed set 𝑄𝛾,𝜓,𝑚₁,𝑚₄=𝑢∈𝐾∶𝑚₁≤𝜓(𝑢),𝛾(𝑢)≤𝑚₄. (8)

Lemma 2 (see [22]). Let 𝐾 be a cone in a real Banach space 𝐸. Let 𝛾 and 𝜃 be nonnegative continuous convex functionals on 𝐾,  let 𝛼 be a nonnegative continuous concave functional on 𝐾, and let 𝜓 be a nonnegative continuous functional on 𝐾 satisfying 𝜓(𝜆𝑢)≤𝜆𝜓(𝑢) for 0≤𝜆≤1 such that for some positive numbers 𝜖 and 𝑚₄𝛼(𝑢)≤𝜓(𝑢),‖𝑢‖≤𝜖𝛾(𝑢), (9) for all 𝑢∈𝑃(𝛾,𝑚₄). Suppose that 𝑇∶𝑃(𝛾,𝑚₄)→𝑃(𝛾,𝑚₄) is completely continuous and there exist positive numbers 𝑚₁, 𝑚₂, and 𝑚₃ with 𝑚₁<𝑚₂ such that (𝐵₁){𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₃,𝑚₄)∶𝛼(𝑢)>𝑚₂}≠0/ and 𝛼(𝑇𝑢)>𝑚₂ for 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₃,𝑚₄),(𝐵₂)𝛼(𝑇𝑢)>𝑚₂ for 𝑢∈𝑃(𝛾,𝛼,𝑚₂,𝑚₄) with 𝜃(𝑇𝑢)>𝑚₃,(𝐵₃)0∉𝑄(𝛾,𝜓,𝑚₁,𝑚₄) and 𝜓(𝑇𝑢)<𝑚₁ for 𝑢∈𝑄(𝛾,𝜓,𝑚₁,𝑚₄) with 𝜓(𝑢)=𝑚₁.
Then, 𝑇 has at least three fixed points 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝛾,𝑚₄) such that 𝛾𝑢_𝑖≤𝑚₄𝑚𝑓𝑜𝑟𝑖=1,2,3,₂<𝛼𝑢₁,𝑚₁<𝜓𝑢₂𝑤𝑖𝑡ℎ𝛼𝑢₂<𝑚₂,𝜓𝑢₃<𝑚₁. (10)

2. Preliminaries

We will employ several lemmas; these lemmas are based on the corresponding linear third-order generalized left focal boundary value problem.

Lemma 3. Let (𝐻₁), (𝐻₂) hold. Green’s function 𝐺(𝑡,𝑠) for the boundary value problem 𝑢^′′′(𝑡)=0,𝑡₁<𝑡<𝑡₂,𝑎𝑢𝑡₁−𝑏𝑢^′′𝑡₁𝑢=0,_′𝜂=0,𝑡₁<𝜂<𝑡₂,𝑐𝑢𝑡₂+𝑑𝑢^′𝑡₂=0 (11) is given as 𝐺()=𝑡𝑡,𝑠𝑠∈₁𝑥,𝜂¹()𝑦𝑡,𝑠,𝑠≤𝑡,₁(𝑡,𝑠),𝑡≤𝑠,𝑠∈𝜂,𝑡₂𝑥₂()𝑦𝑡,𝑠,𝑠≤𝑡,₂(𝑡,𝑠),𝑡≤𝑠, (12) where 𝑥₁()=1𝑡,𝑠𝑘𝑎𝑏+2𝑡−𝑡₁²−𝑎𝜂−𝑡₁𝑡−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠−2𝑐𝑡₂−𝑡₁++𝑑𝜂−𝑠𝑡−𝑡₁−1𝜂−𝑠2(𝑡−𝑠)²,𝑦₁()=1𝑡,𝑠𝑘𝑎𝑏+2𝑡−𝑡₁²−𝑎𝜂−𝑡₁𝑡−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠−2𝑐𝑡₂−𝑡₁++𝑑𝜂−𝑠𝑡−𝑡₁,𝑥𝜂−𝑠₂()=1𝑡,𝑠𝑘𝑎𝑏+2𝑡−𝑡₁²−𝑎𝜂−𝑡₁𝑡−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−1−𝑠2(𝑡−𝑠)²,𝑦₂()=1𝑡,𝑠𝑘𝑎𝑏+2𝑡−𝑡₁²−𝑎𝜂−𝑡₁𝑡−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠. (13)

Proof. First, we check that 𝐺(𝑡,𝑠) is defined for each (𝑡,𝑠)∈[𝑡₁,𝑡₂]×[𝑡₁,𝑡₂]. Obviously, 𝑥₁(𝑠,𝑠)=𝑦₁(𝑠,𝑠), 𝑥₂(𝑠,𝑠)=𝑦₂(𝑠,𝑠). At 𝑠=𝜂, 𝑥₁=1𝑡,𝜂𝑘𝑎𝑏+2𝑡−𝑡₁²−𝑎𝜂−𝑡₁𝑡−𝑡₁×𝑐𝑡₂−𝜂²+2𝑑𝑡₂−1−𝜂2𝑡−𝜂²=𝑥₂,𝑦𝑡,𝜂₁=1𝑡,𝜂𝑘𝑎𝑏+2𝑡−𝑡₁²−𝑎𝜂−𝑡₁𝑡−𝑡₁×𝑐𝑡₂−𝜂²+2𝑑𝑡₂−𝜂=𝑦₂𝑡,𝜂. (14) Next, we check that 𝐺(𝑡,𝑠) satisfies the boundary conditions of problem (11). For convenience, we note that 𝜕𝐺()=𝑡𝜕𝑡𝑡,𝑠𝑠∈₁𝑎,𝜂𝑘𝑡−𝜂𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑎+𝑑𝜂−𝑠𝜂−𝑡,𝑠≤𝑡,𝑘𝑡−𝜂𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁++𝑑𝜂−𝑠𝜂−𝑠,𝑡≤𝑠,𝑠∈𝜂,𝑡₂𝑎𝑘×𝑐𝑡𝑡−𝜂₂−𝑠²+2𝑑𝑡₂−(),𝑎−𝑠𝑡−𝑠𝑠≤𝑡,𝑘𝑡−𝜂𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠,𝑡≤𝑠,𝜕²𝜕𝑡²𝐺()=𝑡𝑡,𝑠𝑠∈₁𝑎,𝜂𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁𝑎+𝑑𝜂−𝑠−1,𝑠≤𝑡,𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠,𝑡≤𝑠,𝑠∈𝜂,𝑡₂𝑎𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑎−𝑠−1,𝑠≤𝑡,𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠,𝑡≤𝑠. (15)(16)
For 𝑠∈[𝑡₁,𝜂], by (12) and (16), we have 𝑎𝐺𝑡₁𝜕,𝑠−𝑏²𝜕𝑡²𝐺𝑡₁,𝑠=𝑎𝑦₁𝑡₁𝜕,𝑠−𝑏²𝜕𝑡²𝑦₁𝑡₁=,𝑠𝑎𝑏𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠−2𝑐𝑡₂−𝑡₁−+𝑑𝜂−𝑠𝑎𝑏𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠−2𝑐𝑡₂−𝑡₁+𝑑𝜂−𝑠=0; (17) for 𝑠∈[𝜂,𝑡₂], we get 𝑎𝐺𝑡₁𝜕,𝑠−𝑏²𝜕𝑡²𝐺𝑡₁,𝑠=𝑎𝑦₂𝑡₁𝜕,𝑠−𝑏²𝜕𝑡²𝑦₂𝑡₁=,𝑠𝑎𝑏𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂−−𝑠𝑎𝑏𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠=0. (18)
For 𝑠∈[𝑡₁,𝜂], by (12) and (15), we have (𝜕/𝜕𝑡)𝐺(𝜂,𝑠)=(𝜕/𝜕𝑡)𝑥₁(𝜂,𝑠)=0; for 𝑠∈[𝜂,𝑡₂], we get (𝜕/𝜕𝑡)𝐺(𝜂,𝑠)=(𝜕/𝜕𝑡)𝑦₂(𝜂,𝑠)=0.
For 𝑠∈[𝑡₁,𝜂], by (12) and (15), we have 𝑐𝐺𝑡₂𝜕,𝑠+𝑑𝐺𝑡𝜕𝑡²,𝑠=𝑐𝑥₁𝑡₂𝜕,𝑠+𝑑𝑥𝜕𝑡¹𝑡₂=𝑐,𝑠𝑘𝑎𝑏+2𝑡₂−𝑡₁²−𝑎𝜂−𝑡₁𝑡₂−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠−2𝑐𝑡₂−𝑡₁𝑡+𝑑𝜂−𝑠+𝑐₂−𝑡₁−𝑐𝜂−𝑠2𝑡₂−𝑠²+𝑎𝑑𝑘𝑡₂×𝑐𝑡−𝜂₂−𝑠²+2𝑑𝑡₂−𝑠−2𝑐𝑡₂−𝑡₁+𝑑𝜂−𝑠+𝑑𝜂−𝑡₂=𝑐2𝑡₂−𝑠²+𝑑𝑡₂−𝑐𝑡−𝑠₂−𝑡₁+𝑑𝜂−𝑠+𝑐𝑡₂−𝑡₁−𝑐𝜂−𝑠2𝑡₂−𝑠²+𝑑𝜂−𝑡₂=0; (19) for 𝑠∈[𝜂,𝑡₂], we get 𝑐𝐺𝑡₂𝜕,𝑠+𝑑𝐺𝑡𝜕𝑡²,𝑠=𝑐𝑥₂𝑡₂𝜕,𝑠+𝑑𝑥𝜕𝑡²𝑡₂=𝑐,𝑠𝑘𝑎𝑏+2𝑡₂−𝑡₁²−𝑎𝜂−𝑡₁𝑡₂−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑐−𝑠2𝑡₂−𝑠²+𝑎𝑑𝑘𝑡₂×𝑐𝑡−𝜂₂−𝑠²+2𝑑𝑡₂−𝑠+𝑑𝑠−𝑡₂=𝑐2𝑡₂−𝑠²+𝑑𝑡₂−𝑐−𝑠2𝑡₂−𝑠²+𝑑𝑠−𝑡₂=0. (20) Hence, 𝐺(𝑡,𝑠) satisfies the boundary conditions of problem (11).

Lemma 4. Let (𝐻₁), (𝐻₂) hold. If 𝑏>(𝑎/2)(𝑡₂−𝑡₁)², then Green’s function as given in (12) satisfies 𝐺()𝐺()𝑡,𝑠≥0,𝑡,𝑠≥𝜏₁𝐺,𝜕𝜂,𝑠𝐺(𝜕𝑡𝑡,𝑠)≥𝜏₂𝜕𝐺,()∈𝑡𝜕𝑡𝜂,𝑠𝑡,𝑠₁,𝑡₂², (21) where 𝜏₁=min𝜏₁₁,𝜏₁₂∈][,𝜏0,1₁₁=minmin_{𝑡𝑠∈1,𝜂}𝑥₁𝑡₁,𝑡₁𝑥₁𝜂,𝑠,min_{𝑠∈𝜂,𝑡2}𝑥₂𝑡₂,𝑠𝑥₂,𝜏𝜂,𝑠₁₂=minmin_{𝑡𝑠∈1,𝜂}𝑦₁𝜂,𝜂𝑦₁𝑡₁,𝑠,min_{𝑠∈𝜂,𝑡2}𝑦₂𝜂,𝑠𝑦₂𝑡₂,𝜏,𝑠₂=minmin_{𝑡𝑠∈1,𝜂}(𝜕/𝜕𝑡)𝐺(𝑠,𝑠)(𝜕/𝜕𝑡)𝐺𝑡₁,,𝑠min_{𝑠∈𝜂,𝑡2}(𝜕/𝜕𝑡)𝐺(𝑠,𝑠)(𝜕/𝜕𝑡)𝐺𝑡₂∈][.,𝑠0,1 (22)

Proof. Condition 𝑏>(𝑎/2)(𝑡₂−𝑡₁)² implies that 𝑏>(𝑎/2)(𝜂−𝑡₁)² and 𝑏+(𝑎/2)(𝑡₂−𝑡₁)²−𝑎(𝜂−𝑡₁)(𝑡₂−𝑡₁)>0 for 𝑡₁<𝜂<𝑡₂. To prove the nonnegative property of Green’s function, combining with (15) and (16), we divide it into two cases.
Case 1. For 𝑠≤𝑡, 𝜕𝐺()=𝜕𝜕𝑡𝑡,𝑠𝑥𝜕𝑡¹()=𝑎𝑡,𝑠𝑘×𝑐𝑡𝑡−𝜂₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+=𝑐+𝑑𝜂−𝑠𝜂−𝑡𝑡−𝜂𝑘𝑎𝑠−𝑡₁²𝜕−2𝑏≥0,₂𝜕𝑡²𝐺()=𝜕𝑡,𝑠²𝜕𝑡²𝑥₁()=𝑎𝑡,𝑠𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠−1≤0 (23) imply that 𝑥₁(𝑡,𝑠) is nondecreasing and concave in 𝑡 on [𝑡₁,𝜂]; it follows that 𝐺(𝜂,𝑠)≥𝐺(𝑡,𝑠) for each (𝑡,𝑠)∈[𝑡₁,𝜂]×[𝑡₁,𝜂]: 𝐺𝑡₁,𝑠=𝑥₁𝑡₁=𝑏,𝑠𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁−1+𝑑𝜂−𝑠2𝑡₁−𝑠²=𝑏𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁𝜕+𝑑𝜂−𝑠≥0,𝐺()𝜕𝜕𝑡𝑡,𝑠>0,𝐺()𝜕𝜕𝑡𝑠,𝑠>0,𝐺𝑡𝜕𝑡¹,𝑠>0. (24) Likewise, 𝜕𝐺()=𝜕𝜕𝑡𝑡,𝑠𝑥𝜕𝑡²()=𝑎𝑡,𝑠𝑘𝑡−𝜂𝑐𝑡₂−𝑠²+2𝑑𝑡₂−()=𝑎−𝑠𝑡−𝑠𝑡−𝜂𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂+𝜕−𝑠−1𝑠−𝜂≤0,₂𝜕𝑡²𝐺()=𝜕𝑡,𝑠²𝜕𝑡²𝑥₂()=𝑎𝑡,𝑠𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂=1−𝑠−1𝑘𝑎𝑐𝑡₂−𝑠²+2𝑎𝑑𝑡₂−𝑠−𝑘≤0 (25) guarantee that 𝑥₂(𝑡,𝑠) is nonincreasing and concave in 𝑡 on [𝜂,𝑡₂]; we have 𝐺(𝜂,𝑠)≥𝐺(𝑡,𝑠) for each (𝑡,𝑠)∈[𝜂,𝑡₂]×[𝜂,𝑡₂]: 𝐺𝑡₂,𝑠=𝑥₂𝑡₂=1,𝑠𝑘𝑎𝑏+2𝑡₂−𝑡₁²−𝑎𝜂−𝑡₁𝑡₂−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−1−𝑠2𝑡₂−𝑠²=1𝑡2𝑘2𝑏𝑐+𝑎𝑐₂−𝑡₁𝑡₁+𝑡₂−2𝜂−𝑘𝑡₂−𝑠²+𝑑𝑘2𝑏+𝑎𝑡₂−𝑡₁𝑡₁+𝑡₂−2𝜂𝑡₂≥𝑡−𝑠𝑎𝑑₂−𝑠𝑘𝑠−𝜂𝑡₂𝜕−𝜂≥0,𝐺()𝜕𝜕𝑡𝑡,𝑠<0,𝐺()𝜕𝜕𝑡𝑠,𝑠>0,𝐺𝑡𝜕𝑡²,𝑠<0. (26) From the above argument, we obtain that 𝐺(𝑡,𝑠)≥0,𝐺(𝑡,𝑠)≥minmin_{𝑡𝑠∈1,𝜂}𝑥₁𝑡₁,𝑡₁𝑥₁𝜂,𝑠,min_{𝑠∈𝜂,𝑡2}𝑥₂𝑡₂,𝑠𝑥₂𝐺;𝜂,𝑠𝜂,𝑠𝜕𝐺(𝜕𝑡𝑡,𝑠)≥min_{𝑠∈𝜂,𝑡2}(𝜕/𝜕𝑡)𝐺(𝑠,𝑠)(𝜕/𝜕𝑡)𝐺𝑡₂𝜕,𝑠𝐺,𝜕𝑡𝜂,𝑠 (27)(28)(29) we have (29), since (𝜕/𝜕𝑡)𝐺(𝑡,𝑠)≥min_{𝑠∈[𝑡1,𝜂]}((𝜕/𝜕𝑡)𝐺(𝑠,𝑠)/(𝜕/𝜕𝑡)𝐺(𝑡₁,𝑠))(𝜕/𝜕𝑡)𝐺(𝜂,𝑠)=0.
Case 2. For 𝑡≤𝑠, 𝜕𝐺()=𝜕𝜕𝑡𝑡,𝑠𝑦𝜕𝑡¹()=𝑎𝑡,𝑠𝑘𝑡−𝜂𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+=+𝑑𝜂−𝑠𝜂−𝑠𝑡−𝜂𝑘×𝑎𝑐𝑡₂−𝑠²+2𝑎𝑑𝑡₂𝑐𝑡−𝑠−2𝑎₂−𝑡₁+𝜕+𝑑𝜂−𝑠𝜂−𝑠≤𝑡−𝑠≤0,₂𝜕𝑡²𝐺()=𝜕𝑡,𝑠²𝜕𝑡²𝑦₁()=𝑎𝑡,𝑠𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠≥0 (30) imply that 𝑦₁(𝑡,𝑠) is nonincreasing and convex in 𝑡 on [𝑡₁,𝜂]; it follows that 𝐺(𝑡,𝑠)≥𝐺(𝜂,𝑠) for each (𝑡,𝑠)∈[𝑡₁,𝜂]×[𝑡₁,𝜂]: 𝐺𝜂,𝑠=𝑦₁=1𝜂,𝑠𝑘𝑎𝑏−2𝜂−𝑡₁²×𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠+𝜂−𝑡₁𝜕𝜂−𝑠≥0,𝐺()𝜕𝜕𝑡𝑡,𝑠<0,𝐺()𝜕𝜕𝑡𝑠,𝑠>0,𝐺𝑡𝜕𝑡¹,𝑠<0. (31) Likewise, 𝜕𝐺()=𝜕𝜕𝑡𝑡,𝑠𝑦𝜕𝑡²()=𝑎𝑡,𝑠𝑘𝑡−𝜂𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝜕−𝑠≥0,₂𝜕𝑡²𝐺()=𝜕𝑡,𝑠²𝜕𝑡²𝑦₂()=𝑎𝑡,𝑠𝑘𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠≥0 (32) guarantee that 𝑦₂(𝑡,𝑠) is nondecreasing and convex in 𝑡 on [𝜂,𝑡₂]; it follows that 𝐺(𝑡,𝑠)≥𝐺(𝜂,𝑠) for each (𝑡,𝑠)∈[𝜂,𝑡₂]×[𝜂,𝑡₂]: 𝐺𝜂,𝑠=𝑦₂=1𝜂,𝑠𝑘𝑎𝑏−2𝜂−𝑡₁²×𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝜕−𝑠≥0,𝐺()𝜕𝜕𝑡𝑡,𝑠>0,𝐺()𝜕𝜕𝑡𝑠,𝑠>0,𝐺𝑡𝜕𝑡²,𝑠>0. (33) So we get 𝐺(𝑡,𝑠)≥0;𝐺(𝑡,𝑠)≥minmin_{𝑡𝑠∈1,𝜂}𝑦₁𝜂,𝜂𝑦₁𝑡₁,𝑠,min_{𝑠∈𝜂,𝑡2}𝑦₂𝜂,𝑠𝑦₂𝑡₂𝐺,,𝑠𝜂,𝑠𝜕𝐺(𝜕𝑡𝑡,𝑠)≥min_{𝑡𝑠∈1,𝜂}(𝜕/𝜕𝑡)𝐺(𝑠,𝑠)(𝜕/𝜕𝑡)𝐺𝑡₁𝜕,𝑠𝐺,𝜕𝑡𝜂,𝑠 (34)(35)(36) we have (35), since (𝜕/𝜕𝑡)𝐺(𝑡,𝑠)≥min_{𝑠∈[𝜂,𝑡2]}(((𝜕/𝜕𝑡)𝐺(𝑠,𝑠))/((𝜕/𝜕𝑡)𝐺(𝑡₂,𝑠)))(𝜕/𝜕𝑡)𝐺(𝜂,𝑠)=0.
It is easy to show that 𝜏₁,𝜏₂∈]0,1[. Therefore, we obtain that 𝐺(𝑡,𝑠)≥0 for each (𝑡,𝑠)∈[𝑡₁,𝑡₂]×[𝑡₁,𝑡₂]; (27)–(36) imply the desired conclusion.

3. A Priori Estimates

Lemma 5. Let (𝐻₁), (𝐻₂) hold. If 𝑔∈𝐶[𝑡₁,𝑡₂] with 𝑔≥0, then a solution of the problem 𝑢^′′′(𝑡)=𝑔(𝑡),0<𝑡₁<𝑡<𝑡₂,𝑎𝑢𝑡₁−𝑏𝑢^′′𝑡₁𝑢=0,_′𝜂=0,𝑡₁<𝜂<𝑡₂,𝑐𝑢𝑡₂+𝑑𝑢^′𝑡₂=0 (37) satisfies 𝑢(𝑡)≥0,𝑡∈𝑡₁,𝑡₂,min_{𝑡∈[𝜉,𝜔]}𝑢(𝑡)≥𝜏₁max_{𝑡𝑡∈1,𝑡2}|𝑢(𝑡)|, (38) where [𝜉,𝜔]⊆[𝑡₁,𝑡₂], (𝜉<𝜔).

Proof. A solution of problem (37) is 𝑢(𝑡)=∫^𝑡2_𝑡1𝐺(𝑡,𝑠)𝑔(𝑠)𝑑𝑠, where 𝐺(𝑡,𝑠) is defined by (12); it is not difficult to show that the results hold by Lemma 4 and we omit the proof process.

Lemma 6. Let (𝐻₁), (𝐻₂) hold. If 𝑔∈𝐶[𝑡₁,𝑡₂] with 𝑔≥0, then a solution of problem (37) satisfies max_{𝑡∈[𝑡1,𝑡2]}𝑢(𝑡)≤𝜏₃max_{𝑡𝑡∈1,𝑡2}𝑢^′(𝑡),max_{𝑡∈[𝑡1,𝑡2]}𝑢^′(𝑡)≤𝜏₄max_{𝑡𝑡∈1,𝑡2}𝑢^′′(𝑡), (39) where 𝜏₃𝑎=max𝑏−2𝜂−𝑡₁²𝑐𝑡₂−𝑡₁²+2𝑑𝑡₂−𝑡₁+𝑘𝜂−𝑡₁²×𝜂−𝑡₁𝑘−𝑎𝑐𝑡₂−𝑡₁²+2𝑑𝑡₂−𝑡₁⁻¹,𝑏−(𝑎/2)𝜂−𝑡₁²𝑐𝑡₂−𝜂²+2𝑑𝑡₂−𝜂𝑡₂−𝜂𝑘−𝑎𝑐𝑡₂−𝜂²+2𝑑𝑡₂,𝑏𝑐𝑡−𝜂₂−𝑡₁²+2𝑑𝑡₂−𝑡₁×inf_{𝑠∈[𝑡1,𝜂]}𝑎𝜂−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠+𝑘𝑠−𝜂⁻¹,𝑏+(𝑎/2)𝑡₂−𝑡₁²𝑎𝑡₂,𝜏−𝜂₄=𝑡₂−𝜂. (40)

Proof. According to the nonnegative solution 𝑢 of problem (37), we establish a priori estimates as follows. For 𝑡₁≤𝑠≤𝑡≤𝜂, we can check that max_{𝑡∈[𝑡1,𝜂]}𝑢(𝑡)=𝑥₁(𝜂,𝑠),max_{𝑡∈[𝑡1,𝜂]}|𝑢^′(𝑡)|=(𝜕/𝜕𝑡)𝑥₁(𝑡₁,𝑠),max_{𝑡∈[𝑡1,𝜂]}|𝑢^′′(𝑡)|=−(𝜕²/𝜕𝑡²)𝑥₁(𝜂,𝑠); note that (𝑎/𝑘)(𝑐(𝑡₂−𝑠)²+2𝑑(𝑡₂−𝑠)−2(𝑐(𝑡₂−𝑡₁)+𝑑)(𝜂−𝑠))−1≤0, so max_{𝑡∈[𝑡1,𝜂]}𝑢(𝑡)max_{𝑡∈[𝑡1,𝜂]}𝑢^′(𝑡≤𝑎)𝑏−2𝜂−𝑡₁²×𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠−2𝑐𝑡₂−𝑡₁+𝑑𝜂−𝑠+𝑘𝜂−𝑡₁²×𝜂−𝑡₁𝑘−𝑎𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠⁻¹≤𝑎𝑏−2𝜂−𝑡₁²×𝑐𝑡₂−𝑡₁²+2𝑑𝑡₂−𝑡₁+𝑘𝜂−𝑡₁²×𝜂−𝑡₁𝑘−𝑎𝑐𝑡₂−𝑡₁²+2𝑑𝑡₂−𝑡₁⁻¹,max_{𝑡∈[𝑡1,𝜂]}𝑢^′(𝑡)max_{𝑡∈[𝑡1,𝜂]}𝑢^′′(𝑡≤𝑐𝑡)𝜂−𝑡𝑘−𝑎₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁×𝑐𝑡+𝑑𝜂−𝑠𝑘−𝑎₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠⁻¹≤𝜂−𝑡₁. (41)
For 𝜂≤𝑠≤𝑡≤𝑡₂, we can check that max_{𝑡∈[𝜂,𝑡2]}𝑢(𝑡)=𝑥₂(𝜂,𝑠),max_{𝑡∈[𝜂,𝑡2]}|𝑢^′(𝑡)|=−(𝜕/𝜕𝑡)𝑥₂(𝑡₂,𝑠),max_{𝑡∈[𝜂,𝑡2]}|𝑢^′′(𝑡)|=−(𝜕²/𝜕𝑡²)𝑥₂(𝑡₂,𝑠); notice that (𝑎/𝑘)(𝑐(𝑡₂−𝑠)²+2𝑑(𝑡₂−𝑠))−1≤0 and thus max_{𝑡∈[𝜂,𝑡2]}𝑢(𝑡)max_{𝑡∈[𝜂,𝑡2]}𝑢^′(𝑡≤())𝑏−𝑎/2𝜂−𝑡₁²𝑐𝑡₂−𝜂²+2𝑑𝑡₂−𝜂𝑡₂−𝜂𝑘−𝑎𝑐𝑡₂−𝑠²+2𝑑𝑡₂≤(−𝑠𝑏−𝑎/2)𝜂−𝑡₁²𝑐𝑡₂−𝜂²+2𝑑𝑡₂−𝜂𝑡₂−𝜂𝑘−𝑎𝑐𝑡₂−𝜂²+2𝑑𝑡₂,−𝜂max_{𝑡∈[𝜂,𝑡2]}𝑢^′(𝑡)max_{𝑡∈[𝜂,𝑡2]}𝑢^′′(𝑡≤𝑘𝑐𝑡)𝑡−𝜂−𝑎𝑡−𝜂₂−𝑠²+2𝑑𝑡₂−𝑠𝑘−𝑎𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠≤𝑡₂−𝜂. (42)
For 𝑡₁≤𝑡≤𝑠≤𝜂, we can check that max_{𝑡∈[𝑡1,𝜂]}𝑢(𝑡)=𝑦₁(𝑡₁,𝑠),max_{𝑡∈[𝑡1,𝜂]}|𝑢^′(𝑡)|=−(𝜕/𝜕𝑡)𝑦₁(𝑡₁,𝑠),max_{𝑡∈[𝑡1,𝜂]}|𝑢^′′(𝑡)|=(𝜕²/𝜕𝑡²)𝑦₁(𝑡₁,𝑠); note that (𝑎/𝑘)(𝑐(𝑡₂−𝑠)²+2𝑑(𝑡₂−𝑠)−2(𝑐(𝑡₂−𝑡₁)+𝑑)(𝜂−𝑠))−1≤0 and so max_{𝑡∈[𝑡1,𝜂]}𝑢(𝑡)max_{𝑡∈[𝑡1,𝜂]}𝑢^′(𝑡)≤𝑏𝑐𝑡₂−𝑡₁²+2𝑑𝑡₂−𝑡₁×𝑎𝜂−𝑡₁𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠+𝑘𝑠−𝜂⁻¹≤𝑏𝑐𝑡₂−𝑡₁²+2𝑑𝑡₂−𝑡₁×inf_{𝑠∈]𝑡1,𝜂]}𝑎𝜂−𝑡₁×𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠+𝑘𝑠−𝜂⁻¹,max_{𝑡∈[𝑡1,𝜂]}𝑢^′(𝑡)max_{𝑡∈[𝑡1,𝜂]}𝑢^′′(𝑡≤𝑎)𝜂−𝑡₁𝑐𝑡₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁×𝑎𝑐𝑡+𝑑𝜂−𝑠+𝑘𝑠−𝜂₂−𝑠²+2𝑑𝑡₂𝑐𝑡−𝑠−2₂−𝑡₁+𝑑𝜂−𝑠⁻¹≤𝜂−𝑡₁. (43)
For 𝜂≤𝑡≤𝑠≤𝑡₂, we can check that max_{𝑡∈[𝜂,𝑡2]}𝑢(𝑡)=𝑦₂(𝑡₂,𝑠),max_{𝑡∈[𝜂,𝑡2]}|𝑢^′(𝑡)|=(𝜕/𝜕𝑡)𝑦₂(𝑡₂,𝑠),max_{𝑡∈[𝜂,𝑡2]}|𝑢^′′(𝑡)|=(𝜕²/𝜕𝑡²)𝑦₂(𝜂,𝑠); then max_{𝑡∈[𝜂,𝑡2]}𝑢(𝑡)max_{𝑡∈[𝜂,𝑡2]}𝑢^′(𝑡=()𝑡)𝑏+𝑎/2₂−𝑡₁²−𝑎𝜂−𝑡₁𝑡₂−𝑡₁𝑎𝑡₂≤()𝑡−𝜂𝑏+𝑎/2₂−𝑡₁²𝑎𝑡₂,−𝜂max_{𝑡∈[𝜂,𝑡2]}𝑢^′(𝑡)max_{𝑡∈[𝜂,𝑡2]}𝑢^′′(𝑡≤𝑎𝑡₂)−𝜂𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠𝑎𝑐𝑡₂−𝑠²+2𝑑𝑡₂−𝑠=𝑡₂−𝜂. (44) From the above argument, we can obtain the desired result (39).

Let 𝐸 be the Banach space 𝐶²[𝑡₁,𝑡₂] with the norm ‖𝑢‖=max{‖𝑢‖₀,‖𝑢^′‖₀,‖𝑢^′′‖₀}, where ‖⋅‖₀=max_{𝑡∈[𝑡1,𝑡2]}|⋅(𝑡)|. Set a cone in 𝐸𝑡𝐾=𝑢∈𝐸∶𝑢isnonnegativeon₁,𝑡₂,andsatisfies𝑎𝑢𝑡₁−𝑏𝑢^′′𝑡₁𝑢=0,_′𝜂=0,𝑐𝑢𝑡₂+𝑑𝑢^′𝑡₂=0,min_{𝑡∈[𝜉,𝜔]}𝑢(𝑡)≥𝜏₁max_{𝑡𝑡∈1,𝑡2}|𝑢(𝑡.)| (45)In view of the assumptions (𝐻₁) and (𝐻₂), take 𝑢₀(𝑡)=−𝑚𝑡³+𝜆₂𝜆₁𝑚𝑡²+𝜆₃𝜆₁+𝑡𝑚𝑡₃₁−6𝑏𝑎𝑡₁+2𝑏𝑎−𝑡²₁𝜆₂𝜆₁−𝑡₁𝜆₃𝜆₁𝑚, (46) where 𝜆₁=2𝜂(𝑎𝑐𝑡₁−𝑎𝑐𝑡₂−𝑎𝑑)−𝑎𝑐𝑡²₁+2𝑏𝑐+𝑎𝑐𝑡²₂+2𝑎𝑑𝑡₂, 𝜆₂=3𝜂²(𝑎𝑐𝑡₁−𝑎𝑐𝑡₂−𝑎𝑑)−𝑎𝑐𝑡³₁+6𝑏𝑐𝑡₁+𝑎𝑐𝑡³₂+3𝑎𝑑𝑡²₂,  𝜆₃=3𝜂²(−𝑎𝑐𝑡²₁+𝑎𝑐𝑡²₂+2𝑎𝑑𝑡₂)+2𝜂(𝑎𝑐𝑡³₁−6𝑏𝑐𝑡₁−𝑎𝑐𝑡³₂−3𝑎𝑑𝑡²₂),  and 𝑚 is an arbitrary positive constant. By some simplifying, we obtain that 𝜆₁,𝜆₂≥0. Set a constant 𝜗∶=max−𝑡³₁+𝜆₂𝜆₁𝑡²₁+𝜆₃𝜆₁𝑡₁+𝜆₄,−𝜂³+𝜆₂𝜆₁𝜂²+𝜆₃𝜆₁𝜂+𝜆₄,−𝑡³+𝜆₂𝜆₁𝑡²+𝜆₃𝜆₁𝑡+𝜆₄,−𝑡³₂+𝜆₂𝜆₁𝑡²₂+𝜆₃𝜆₁𝑡₂+𝜆₄, (47) where 𝑡=(𝜆₂+3𝜆₁𝜆₃+𝜆²₂)/3𝜆₁,𝜆₄=𝑡³₁−(6𝑏/𝑎)𝑡₁+((2𝑏/𝑎)−𝑡²₁)(𝜆₂/𝜆₁)−𝑡₁(𝜆₃/𝜆₁). It is not difficult to check that 𝑢₀(𝑡) is nonnegative on [𝑡₁,𝑡₂] and satisfies 𝑎𝑢₀(𝑡₁)−𝑏𝑢₀^′′(𝑡₁)=0, 𝑢^′₀(𝜂)=0, 𝑐𝑢₀(𝑡₂)+𝑑𝑢^′₀(𝑡₂)=0. Under the assumption min{−𝜉³+(𝜆₂/𝜆₁)𝜉²+(𝜆₃/𝜆₁)𝜉+𝜆₄,−𝜔³+(𝜆₂/𝜆₁)𝜔²+(𝜆₃/𝜆₁)𝜔+𝜆₄,−𝜂³+(𝜆₂/𝜆₁)𝜂²+(𝜆₃/𝜆₁)𝜂+𝜆₄}≥𝜏₁𝜗, we have min_{𝑡∈[𝜉,𝜔]}𝑢₀(𝑡)≥𝜏₁max_{𝑡∈[𝑡1,𝑡2]}|𝑢₀(𝑡)|; that is, 𝑢₀∈𝐾, so 𝐾⧵{0}≠0/. By assumption (𝐻₃), both 0<∫^𝑡2_𝑡1𝐺(𝑡₁,𝑠)ℎ(𝑠)𝑑𝑠<∞,0<∫^𝑡2_𝑡1𝐺(𝜂,𝑠)ℎ(𝑠)𝑑𝑠<∞ and 0<∫^𝑡2_𝑡1𝐺(𝑡₂,𝑠)ℎ(𝑠)𝑑𝑠<∞ hold. Thus, we can define an operator 𝑇 by (𝑇𝑢)(𝑡)∶=^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),𝑢^′′(𝑠)𝑡𝑑𝑠,𝑡∈₁,𝑡₂. (48)

Lemma 7. Let (𝐻₁)--(𝐻₄) hold. Then, 𝑇∶𝐾→𝐾 is completely continuous.

Proof. Since conditions (𝐻₁)--(𝐻₄) hold, by Lemma 5, we observe that (𝑇𝑢)(𝑡)≥0 for 𝑡∈[𝑡₁,𝑡₂]; 𝑎(𝑇𝑢)(𝑡₁)−𝑏(𝑇𝑢)^′′(𝑡₁)=0, (𝑇𝑢)^′(𝜂)=0, 𝑐(𝑇𝑢)(𝑡₂)+𝑑(𝑇𝑢)^′(𝑡₂)=0,min_{𝑡∈[𝜉,𝜔]}(𝑇𝑢)(𝑡)≥𝜏₁max_{𝑡∈[𝑡1,𝑡2]}|(𝑇𝑢)(𝑡)| are satisfied, so 𝑇𝑢∈𝐾. Since ℎ(𝑡) may be singular at 𝑡=𝑡₁ and/or 𝑡=𝑡₂, we will take the arguments to show that the operator 𝑇 is completely continuous. Assume that 𝑢_𝑛,𝑢₀∈𝐾(𝑛=1,2,…) satisfy ‖𝑢_𝑛−𝑢₀‖→0(𝑛→∞); then there exists 𝑅>0 such that ‖𝑢_𝑛‖≤𝑅 for arbitrary positive integer 𝑛. One has that 𝑇𝑢_𝑛−𝑇𝑢₀₀=max_{𝑡∈[𝑡1,𝑡2]}𝑇𝑢_𝑛(𝑡)−𝑇𝑢₀(𝑡)≤max_{𝑡∈[𝑡1,𝜂]}^𝑡2_𝑡1𝐺()ℎ(𝑠)×𝑓𝑡,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+max_{𝑡∈𝜂,𝑡2}^𝑡2_𝑡1𝐺()ℎ(𝑠)×𝑓𝑡,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠≤max_{𝑡∈[𝑡1,𝜂]}^𝑡_𝑡1𝐺ℎ(𝑠)×𝑓𝜂,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+^𝜂_𝑡𝐺𝑡₁ℎ(𝑠)×𝑓,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+max_{𝑡∈[𝜂,𝑡2]}^𝑡_𝜂𝐺ℎ(𝑠)×𝑓𝜂,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′+𝑑𝑠_𝑡2𝑡𝐺𝑡₂ℎ(𝑠)×𝑓,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠≤^𝜂_𝑡1𝐺ℎ(𝑠)×𝑓𝜂,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′+𝑑𝑠_𝜂𝑡1𝐺𝑡₁ℎ(𝑠)×𝑓,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+^𝑡2_𝜂𝐺ℎ(𝑠)×𝑓𝜂,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′+𝑑𝑠_𝑡2𝜂𝐺𝑡₂ℎ(𝑠)×𝑓,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠,𝑇𝑢_𝑛^′−𝑇𝑢₀^′₀=max_{𝑡∈[𝑡1,𝑡2]}𝑇𝑢_𝑛^′(𝑡)−𝑇𝑢₀^′(𝑡)≤max_{𝑡∈[𝑡1,𝜂]}^𝑡2_𝑡1𝜕𝐺()ℎ(𝑠)×𝑓𝜕𝑡𝑡,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+max_{𝑡∈]𝜂,𝑡2]}^𝑡2_𝑡1𝜕𝐺()ℎ(𝑠)×𝑓𝜕𝑡𝑡,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠≤max_{𝑡𝑡∈1,𝜂}^𝑡2_𝑡1𝜕𝐺(ℎ(𝑠)×𝑓𝜕𝑡𝑡,𝑠)𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+max_{𝑡∈]𝜂,𝑡2]}^𝑡2_𝑡1𝜕𝐺(ℎ(𝑠)×𝑓𝜕𝑡𝑡,𝑠)𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠,𝑇𝑢_𝑛^′′−𝑇𝑢₀^′′₀=max_{𝑡∈[𝑡1,𝑡2]}𝑇𝑢_𝑛^′′(𝑡)−𝑇𝑢₀^′′(𝑡)≤max_{𝑡∈[𝑡1,𝜂]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺()ℎ(𝑠)×𝑓𝑡,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+max_{𝑡∈]𝜂,𝑡2]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺()ℎ(𝑠)×𝑓𝑡,𝑠𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠≤max_{𝑡𝑡∈1,𝜂}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺(ℎ(𝑠)×𝑓𝑡,𝑠)𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠+max_{𝑡∈]𝜂,𝑡2]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺(ℎ(𝑠)×𝑓𝑡,𝑠)𝑠,𝑢_𝑛,𝑢^′_𝑛,𝑢_𝑛^′′−𝑓𝑠,𝑢₀,𝑢^′₀,𝑢₀^′′𝑑𝑠. (49) In view of the continuity of 𝑓(𝑡,𝜇,𝜈,𝜛) in 𝜇,𝜈,𝜛, we can obtain that, for any 𝑡∈[𝑡₁,𝑡₂],‖𝑇𝑢_𝑛−𝑇𝑢₀‖=max{‖𝑇_𝑛𝑢−𝑇𝑢‖₀,‖(𝑇_𝑛𝑢)^′−(𝑇𝑢)^′‖₀,‖(𝑇_𝑛)^′′𝑢−(𝑇𝑢)^′′‖₀}→0(𝑛→∞). This means that the operator 𝑇∶𝐾→𝐾 is continuous. It follows from the properties of the function ℎ that we can choose two sequences {𝛼_𝑛}^∞_𝑛=1⊂[𝑡₁,𝜂[, {𝛽_𝑛}^∞_𝑛=1⊂]𝜂,𝑡₂]. Obviously 𝛼_𝑛<𝛽_𝑛 for any 𝑛≥1 and 𝛼_𝑛→𝑡₁,𝛽_𝑛→𝑡₂ as 𝑛→∞, respectively. Define ℎ_𝑛(𝑡)=infℎ(𝑡),ℎ𝛼_𝑛,𝑡∈𝑡₁,𝛼_𝑛,ℎ(𝑡),𝑡∈𝛼_𝑛,𝛽_𝑛,infℎ𝛽_𝑛,ℎ(𝑡),𝑡∈𝛽_𝑛,𝑡₂, (50) and an operator sequence 𝑇_𝑛 by 𝑇_𝑛𝑢(𝑡)=^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ_𝑛(𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),𝑢^′′(𝑠)𝑡𝑑𝑠,𝑡∈₁,𝑡₂. (51) Clearly, ℎ_𝑛(𝑡)∶[𝑡₁,𝑡₂]→[0,∞[ is a piecewise continuous function; the operator 𝑇_𝑛∶𝐾→𝐾 is well defined. Further, we have that 𝑇_𝑛∶𝐾→𝐾 is completely continuous.
Let 𝑟>0,𝔹_𝑟∶={𝑢∈𝐾∶‖𝑢^′′‖₀≤𝑟} and 𝑀_𝑟=max{𝑓(𝑡,𝜇,𝜈,𝜛)∶𝜇∈𝐾,𝑡∈[𝑡₁,𝑡₂],𝜇∈ [0,𝜏₃𝜏₄𝑟],𝜈∈[−𝜏₄𝑟,𝜏₄𝑟],𝜛∈[−𝑟,𝑟]}. We will show that 𝑇_𝑛 approaches 𝑇 uniformly on 𝔹_𝑟 as 𝑛→∞. From the absolute continuity of integral, we obtain lim_𝑛→∞max_{𝑡∈[𝑡1,𝑡2]𝑙(𝑛)}𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠=0, (52) where 𝑙(𝑛)=[𝑡₁,𝛼_𝑛]∪[𝛽_𝑛,𝑡₂]. Green’s functions 𝐺(𝑡,𝑠) and (52) imply that, for any 𝑢∈𝔹_𝑟,𝑡∈[𝑡₁,𝛼_𝑛]; we have 𝑇_𝑛𝑢−𝑇𝑢₀=max_{𝑡∈[𝑡1,𝛼𝑛]}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ_𝑛(𝑠)−ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠≤𝑀_𝑟max_{𝑡𝑡∈1,𝛼𝑛}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ_𝑛(𝑠)−ℎ(𝑠)𝑑𝑠≤𝑀_𝑟max_{𝑡𝑡∈1,𝛼𝑛𝑙(𝑛)}𝐺()ℎ(𝑠)().𝑡,𝑠𝑑𝑠⟶0𝑛⟶∞ (53) For any 𝑢∈𝔹_𝑟,𝑡∈[𝛽_𝑛,𝑡₂], we get that ‖𝑇_𝑛𝑢−𝑇𝑢‖₀→0(𝑛→∞) and, for any 𝑢∈𝔹_𝑟,𝑡∈[𝛼_𝑛,𝛽_𝑛], there is ‖𝑇_𝑛𝑢−𝑇𝑢‖₀=0(𝑛≥1).
Similarly, for any 𝑢∈𝔹_𝑟,𝑡∈[𝑡₁,𝛼_𝑛],𝑡∈[𝛼_𝑛,𝛽_𝑛], or 𝑡∈[𝛽_𝑛,𝑡₂], we can obtain that ‖(𝑇_𝑛𝑢)^′−(𝑇𝑢)^′‖₀→0(𝑛→∞) and ‖(𝑇_𝑛𝑢)^′′−(𝑇𝑢)^′′‖₀→0(𝑛→∞), respectively.
The above argument implies that ‖𝑇_𝑛𝑢−𝑇𝑢‖→0(𝑛→∞); that is, the sequence 𝑇_𝑛 uniformly approximates 𝑇 on any bounded subset of 𝐾. Hence, 𝑇∶𝐾→𝐾 is completely continuous.

4. Single Solution

Now, we are ready to present the sufficient conditions for the existence of at least one positive solution to the problem 𝒫_𝑠.

Theorem 8. Suppose (𝐻₁)--(𝐻₄) hold; 𝑓(𝑡,0,0,0)≢0 on any subinterval of [𝑡₁,𝑡₂] and(𝐶₁)there exist a constant 𝐿 and a function 𝐹(𝑡) which is integrable on [𝑡₁,𝑡₂], satisfying 𝑓²(𝑡),∈𝑡𝑡,𝜇,𝜈,𝜛≤𝐹𝑡,𝜇,𝜈,𝜛₁,𝑡₂×[𝐿,∞[×]−∞,∞[², (54)(𝐶₂)there exist constants 𝑚, 𝑀(𝑚<𝑀) such that [𝑚,𝑀]⊆[𝜉,𝜔] and satisfy liminf_𝜇→0+min_{𝑡∈[𝑚,𝑀]}ℎ𝑓𝑡,𝜇,𝜈,𝜛𝜇=∞. (55)
Then, the boundary value problem 𝒫_𝑠 has at least one positive solution.

Proof. By the definition of operator 𝑇 and Lemma 7, it suffices to show that the conditions of Lemma 1 hold with respect to 𝑇. Furthermore, from the fact that 𝐺(𝑡,𝑠) is Green’s function, it is not difficult to prove that a fixed point of 𝑇 is coincident with the solution of the boundary value problem 𝒫_𝑠, so we concentrate on the existence of the fixed point of the operator 𝑇.
First of all, we show that lim_𝑣→∞Φ(𝑣)𝑣=0, (56) where Φ(𝑣)∶=sup_{(𝑣)𝑢∈𝜕𝒦}maxmax_{𝑡𝑡∈1,𝑡2}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ(𝑠)×𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,max_{𝑡𝑡∈1,𝑡2}^𝑡2_𝑡1𝜕𝐺()ℎ(𝑠)𝜕𝑡𝑡,𝑠×𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺(𝑡,𝑠)ℎ(𝑠)×𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠, (57) and 𝒦(𝑣)={𝑢∈𝐾∶‖𝑢‖₀≤𝑣,𝑣∈]0,∞[}. For convenience, we introduce notation as follows: ℒ(𝑡)=𝑡∈𝑡₁,𝑡₂(𝑡),𝑀∶𝑢≤𝐿₁=max^𝑡2_𝑡1max𝐺𝑡₁,𝑠,𝐺𝜂,𝑠,𝐺𝑡₂,𝑠²×ℎ²(𝑠)𝑑𝑠^1/2,_𝑡2𝑡1𝜕max𝐺𝑡𝜕𝑡¹,𝜕,𝑠𝐺,𝜕𝜕𝑡𝜂,𝑠𝐺𝑡𝜕𝑡²,𝑠²×ℎ²(𝑠)𝑑𝑠^1/2,_𝑡2𝑡1𝜕max²𝜕𝑡²𝐺𝑡₁,𝜕,𝑠²𝜕𝑡²𝐺,𝜕𝜂,𝑠₂𝜕𝑡²𝐺𝑡₂,𝑠²×ℎ²(𝑠)𝑑𝑠^1/2,𝑀₂=max_{𝑡,𝑢,𝑢′∈[𝑡1,𝑡2]×[0,𝐿]×]−∞,∞[2}𝑓²𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡),𝑀₃=^𝑡2_𝑡1𝐹(𝑠)𝑀𝑑𝑠,₄=minmin_{[]𝑠∈𝑚,𝑀}𝐺𝜂,𝑠,min_{[]𝑠∈𝑚,𝑀}𝜕²𝜕𝑡²𝐺𝜂,𝑠. (58) By the well-known Hölder’s inequality, for any 𝑡∈[𝑡₁,𝑡₂] and 𝑢∈𝜕𝒦(𝑣), by using hypothesis (𝐶₁), we get max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′≤𝑑𝑠_𝑡2𝑡1max𝐺𝑡₁,𝑠,𝐺𝜂,𝑠,𝐺𝑡₂(𝑠)𝑓,𝑠×ℎ𝑠,𝑢,𝑢^′,𝑢^′′≤𝑑𝑠_𝑡2𝑡1max𝐺𝑡₁,𝑠,𝐺𝜂,𝑠,𝐺𝑡₂,𝑠²ℎ²(𝑠)𝑑𝑠^1/2×^𝑡2_𝑡1𝑓²𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠^1/2≤𝑀_{1[𝑡1,𝑡2]−ℒ(𝑡)}𝑓²𝑠,𝑢,𝑢^′,𝑢^′′+𝑑𝑠_ℒ(𝑡)𝑓²𝑠,𝑢,𝑢^′𝑑𝑠^1/2≤𝑀₁^𝑡2_𝑡1𝐹(𝑠)𝑑𝑠+_ℒ(𝑡)𝑓²𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠^1/2≤𝑀₁𝑀₂𝑡₂−𝑡₁+𝑀₃^1/2,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕𝐺()ℎ(𝑠)𝑓𝜕𝑡𝑡,𝑠𝑠,𝑢,𝑢^′,𝑢^′′≤𝑑𝑠_𝑡2𝑡1𝜕max𝐺𝑡𝜕𝑡¹,𝜕,𝑠𝐺,𝜕𝜕𝑡𝜂,𝑠𝐺𝑡𝜕𝑡²(𝑠)𝑓,𝑠×ℎ𝑠,𝑢,𝑢^′,𝑢^′′≤𝑑𝑠_𝑡2𝑡1𝜕max𝐺𝑡𝜕𝑡¹,𝜕,𝑠𝐺,𝜕𝜕𝑡𝜂,𝑠𝐺𝑡𝜕𝑡²,𝑠²×ℎ²(𝑠)𝑑𝑠^1/2𝑡2_𝑡1𝑓²𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠^1/2≤𝑀₁𝑀₂𝑡₂−𝑡₁+𝑀₃^1/2,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′≤𝑑𝑠_𝑡2𝑡1𝜕max²𝜕𝑡²𝐺𝑡₁,𝜕,𝑠²𝜕𝑡²𝐺,𝜕𝜂,𝑠₂𝜕𝑡²𝐺𝑡₂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′≤𝑑𝑠_𝑡2𝑡1𝜕max²𝜕𝑡²𝐺𝑡₁,𝜕,𝑠²𝜕𝑡²𝐺,𝜕𝜂,𝑠₂𝜕𝑡²𝐺𝑡₂,𝑠²×ℎ²(𝑠)𝑑𝑠^1/2𝑡2_𝑡1𝑓²𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠^1/2≤𝑀₁𝑀₂𝑡₂−𝑡₁+𝑀₃^1/2. (59) Consequently, lim_𝑣→∞(Φ(𝑣)/𝑣)≤lim_𝑣→∞(𝑀₁(𝑀₂(𝑡₂−𝑡₁)+𝑀₃)^1/2/𝑣)=0.
Next, we verify that lim_𝑣→0+Ψ(𝑣)𝑣=∞, (60) where Ψ(𝑣)∶=inf_{(𝑣)𝑢∈𝜕𝒦}maxmax_{𝑡𝑡∈1,𝑡2}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕𝐺()ℎ(𝑠)𝜕𝑡𝑡,𝑠×𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,max_{𝑡𝑡∈1,𝑡2}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺(𝑡,𝑠)ℎ(𝑠)×𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠. (61) In fact, hypothesis (𝐶₂) implies that, for any 𝑢∈𝐾, 𝑛∈ℕ, there exists 𝜖_𝑛>0 such that 𝜖_𝑛→0 and ℎ𝑓𝑡,𝜇,𝜈,𝜛𝜇∈[]×≥𝑛,𝑡,𝜇,𝜈,𝜛𝑚,𝑀0,𝜖_𝑛[×]−∞,∞². (62) Then by (62) and Lemma 4, we have max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠=max_{𝑡∈[𝜉,𝜔]}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠≥𝜏₁^𝑡2_𝑡1𝐺𝜂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠≥𝜏₁𝑀₄min_{𝑡∈[𝑚,𝑀]}𝑢(𝑡)^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝑑𝑠≥𝜏²₁𝑀₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝑑𝑠,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕𝐺()ℎ(𝑠)𝑓𝜕𝑡𝑡,𝑠𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠=max_{𝑡𝑡∈1,𝑡2}^𝑡2_𝑡1𝜕𝐺(ℎ(𝑠)𝑓𝜕𝑡𝑡,𝑠)𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠≥𝜏₂^𝑡2_𝑡1𝜕𝐺ℎ(𝑠)𝑓𝜕𝑡𝜂,𝑠𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠≥𝜏₂𝑀₄min_{𝑡∈[𝑚,𝑀]}𝑢(𝑡)^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝑑𝑠≥𝜏₁𝜏₂𝑀₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝑑𝑠,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠=max_{𝑡∈]𝑡1,𝑡2]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′=𝑑𝑠_𝑡2𝑡1𝜕²𝜕𝑡²𝐺𝜂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠≥𝑀₄min_{[]𝑡∈𝑚,𝑀}𝑢(𝑡)^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝑑𝑠≥𝑀₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢(𝑠)𝑑𝑠. (63) The argument guarantees that, for 𝑣∈]0,𝜖_𝑛[, Ψ(𝑣)𝑣=1𝑣inf_{(𝑣)𝑢∈𝜕𝒦}maxmax_{𝑡𝑡∈1,𝑡2}^𝑡2_𝑡1𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕𝐺()(𝑠)𝑓𝜕𝑡𝑡,𝑠×ℎ𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝑡1𝜕²𝜕𝑡²𝐺()(𝑠)𝑓𝑡,𝑠×ℎ𝑠,𝑢,𝑢^′,𝑢^′′≥1𝑑𝑠𝑣inf_{(𝑣)𝑢∈𝜕𝒦}max𝜏²₁𝑀₄‖𝑢‖₀×^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝜏𝑑𝑠,₁𝜏₂𝑀₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝑀𝑑𝑠,₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢1𝑑𝑠=max𝑣inf_{(𝑣)𝑢∈𝜕𝒦}𝜏²₁𝑀₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢1𝑑𝑠,𝑣inf_{(𝑣)𝑢∈𝜕𝒦}𝜏₁𝜏₂𝑀₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢1𝑑𝑠,𝑣inf_{(𝑣)𝑢∈𝜕𝒦}𝑀₄‖𝑢‖₀^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝜏𝑑𝑠≥max²₁𝑀₄^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝜏𝑑𝑠,₁𝜏₂𝑀₄^𝑀_𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝑀𝑑𝑠,_4𝑀𝑚ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑢𝜏𝑑𝑠≥max²₁𝑀₄(𝑀−𝑚)𝑛,𝜏₁𝜏₂𝑀₄()𝑀𝑀−𝑚𝑛,₄()𝑛𝜏𝑀−𝑚≥max₁,𝜏₂𝜏₁𝑀₄(𝑀−𝑚)𝑛. (64) Hence, (60) holds.
Finally, from the definitions of Φ and Ψ, it is easy to check that Φ(𝑣)=sup_{(𝑣)𝑢∈𝜕𝒦}‖𝑇𝑢‖,Ψ(𝑣)=inf_{(𝑣)𝑢∈𝜕𝒦}‖𝑇𝑢‖. (65) Note that (56) and (60) imply that there exist 𝑉₁, 𝑉₂ such that 𝑉₂≫𝑉₁ and Φ(𝑉₂)/𝑉₂≤1, Ψ(𝑉₁)/𝑉₁≥1, respectively. Thus, it follows that ‖𝑇𝑢‖≥‖𝑢‖,𝑢∈𝜕𝒦𝑉₁,‖𝑇𝑢‖≤‖𝑢‖,𝑢∈𝜕𝒦𝑉₂. (66)
By virtue of (66), the hypotheses of Lemma 1 are satisfied; we infer that there exists 𝑢^∗∈𝐾 such that 𝑇𝑢^∗=𝑢^∗, and 𝑉₁≤‖𝑢^∗‖≤𝑉₂. In addition, condition (𝐻₂) guarantees that the fixed point is positive, so the problem 𝒫_𝑠 has at least one positive solution. This completes the proof.

Remark 9. Since the definition of the operator 𝑇, a slight modification of 𝑀₄=min{min_{𝑠∈[𝑚,𝑀]} 𝐺(𝜂,𝑠)ℎ(𝑡),min_{𝑠∈[𝑚,𝑀]}|(𝜕²/𝜕𝑡²)𝐺(𝜂,𝑠)ℎ(𝑡)|} and(𝐶₂)there exist constants 𝑚, 𝑀(𝑚<𝑀) such that [𝑚,𝑀]⊆[𝜉,𝜔] and satisfy liminf_𝜇→0+min_{𝑡∈[𝑚,𝑀]}𝑓𝑡,𝜇,𝜈,𝜛𝜇=∞ (67)
which guarantee that the problem 𝒫_𝑠 has at least one positive solution.

Furthermore, 𝑓 may be superlinear or sublinear; we can give the following existence results for the problem 𝒫_𝑠 and the proof is omitted.

Theorem 10. Suppose (𝐻₁)--(𝐻₄) hold, 𝑓(𝑡,0,0,0)≢0 on any subinterval of [𝑡₁,𝑡₂] and (𝐷₁)𝑓₀=0 and 𝑓_∞=∞, or(𝐷₂)𝑓_∞=0 and 𝑓₀=∞.
Then, the problem 𝒫_𝑠 has at least one positive solution.

5. Multiple Solutions

Let the nonnegative continuous convex functionals 𝛾 and 𝜃, the nonnegative continuous concave functional 𝛼, and the nonnegative continuous functional 𝜓 be defined on the cone 𝐾 by 𝛾(𝑢)=max_{𝑡∈[𝑡1,𝑡2]}𝑢^′′(𝑡,𝜃(𝑢))=max_{𝑡∈[𝑡1,𝑡2]}|𝑢(𝑡,𝛼(𝑢))|=min_{𝑡∈[𝜉,𝜔]}𝑢(𝑡),𝜓(𝑢)=max_{𝑡∈[𝑡1,𝑡2]}𝑢^′(𝑡). (68)

To prove our main results, we recommend notation 𝑁₁=max^𝜂_𝑡1−𝜕²𝜕𝑡²𝑥₁𝜂,𝑠ℎ(𝑠)𝑑𝑠,_𝜂𝑡1𝜕²𝜕𝑡²𝑦₁𝑡₁,𝑠ℎ(𝑠)𝑑𝑠+max^𝑡2_𝜂−𝜕²𝜕𝑡²𝑥₂𝑡₂,𝑠ℎ(𝑠)𝑑𝑠,_𝑡2𝜂𝜕²𝜕𝑡²𝑦₂ℎ(𝑠),𝑁𝜂,𝑠𝑑𝑠₂=𝜏₁max^𝜂_𝑡1𝑥₁𝜂,𝑠ℎ(𝑠)𝑑𝑠,^𝜂_𝑡1𝑦₁𝑡₁,𝑠ℎ(𝑠)𝑑𝑠+max^𝑡2_𝜂𝑥₂𝜂,𝑠ℎ(𝑠)𝑑𝑠,_𝑡2𝜂𝑦₂𝑡₂ℎ(𝑠),𝑁,𝑠𝑑𝑠₃=max^𝜂_𝑡1𝜕𝑥𝜕𝑡¹𝑡₁,𝑠ℎ(𝑠)𝑑𝑠,^𝜂_𝑡1−𝜕𝑦𝜕𝑡¹𝑡₁,𝑠ℎ(𝑠)𝑑𝑠+max^𝑡2_𝜂−𝜕𝑥𝜕𝑡²𝑡₂,𝑠ℎ(𝑠)𝑑𝑠,_𝑡2𝜂𝜕𝑦𝜕𝑡²𝑡₂,𝑠ℎ(𝑠)𝑑𝑠. (69)

Now, we are ready to apply Avery and Peterson fixed point theorem to the operator 𝑇 to give the sufficient conditions for the existence of at least three positive solutions to the problem 𝒫_𝑠.

Theorem 11. Let (𝐻₁)--(𝐻₄) hold; 𝑓(𝑡,0,0,0)≢0 on any subinterval of [𝑡₁,𝑡₂]. Suppose that there exist numbers 𝑚₁, 𝑚₂, 𝑚₄ such that 0<𝑚₁<𝑚₂≤𝜏²₁𝜗/max{|6𝑡₁−(2𝜆₂/𝜆₁)|,|6𝑡₂−(2𝜆₂/𝜆₁)|}𝑚₄,min{−𝜉³+(𝜆₂/𝜆₁)𝜉²+(𝜆₃/𝜆₁)𝜉+𝜆₄,−𝜔³+(𝜆₂/𝜆₁)𝜔²+(𝜆₃/𝜆₁)𝜔+𝜆₄,−𝜂³+(𝜆₂/𝜆₁)𝜂²+(𝜆₃/𝜆₁)𝜂+𝜆₄}≥𝜏₁𝜗, and(𝐸₁)𝑓(𝑡,𝜇,𝜈,𝜛)≤𝑚₄/𝑁₁ for (𝑡,𝜇,𝜈,𝜛)∈[𝑡₁,𝑡₂]×[0,𝜏₃𝜏₄𝑚₄]×[−𝜏₄𝑚₄,𝜏₄𝑚₄]×[−𝑚₄,𝑚₄],(𝐸₂)𝑓(𝑡,𝜇,𝜈,𝜛)>𝑚₂/𝑁₂ for (𝑡,𝜇,𝜈,𝜛)∈[𝜉,𝜔]×[𝑚₂,𝑚₂𝜏₁⁻²]×[−𝜏₄𝑚₄,𝜏₄𝑚₄]×[−𝑚₄,𝑚₄],(𝐸₃)𝑓(𝑡,𝜇,𝜈,𝜛)≤𝑚₁/𝑁₃ for (𝑡,𝜇,𝜈,𝜛)∈[𝑡₁,𝑡₂]×[0,𝜏₃𝜏₄𝑚₄]×[−𝜏₄𝑚₄,𝑚₁]×[−𝑚₄,𝑚₄].Then, the 𝒫_𝑠 has at least three positive solutions 𝑢₁,𝑢₂,𝑢₃∈𝐾 such that max_{𝑡∈[𝑡1,𝑡2]}𝑢_𝑖^′′(𝑡)≤𝑚₄𝑚𝑓𝑜𝑟𝑖=1,2,3,₂<min_{𝑡∈[𝜉,𝜔]}𝑢₁(𝑡),𝑚₁<max_{𝑡∈[𝑡1,𝑡2]}𝑢^′₂(𝑡)𝑤𝑖𝑡ℎmin_{𝑡∈[𝜉,𝜔]}𝑢₂(𝑡)<𝑚₂,max_{𝑡∈[𝑡1,𝑡2]}𝑢^′₃(𝑡)<𝑚₁. (70)

Proof. By the definition of operator 𝑇 and its properties, it suffices to show that the conditions of Lemma 2 hold with respect to 𝑇. Furthermore, from the fact that 𝐺(𝑡,𝑠) is the Green function, it is not difficult to prove that a fixed point of 𝑇 is coincident with the solution of the boundary value problem 𝒫_𝑠, so we concentrate on the existence of the fixed point of the operator 𝑇.
The functionals are defined as (68); we also note that 𝛼, 𝛾, 𝜃, and 𝜓 are continuous nonnegative functionals on 𝐾 satisfying 𝜓(𝜆𝑢)≤𝜆𝜓(𝑢) for 0≤𝜆≤1 such that, for some positive numbers 𝜖 and 𝑚₄, 𝛼(𝑢)≤𝜃(𝑢)=𝜓(𝑢) and ‖𝑢‖₀≤𝜖𝛾(𝑢), for 𝑢∈𝑃(𝛾,𝑚₄), where 𝜖≤1. If 𝑢∈𝑃(𝛾,𝑚₄), then 𝛾(𝑢)=max_{𝑡∈[𝑡1,𝑡2]}|𝑢^′′(𝑡)|≤𝑚₄. We get max_{𝑡∈[𝑡1,𝑡2]}|𝑢(𝑡)|≤𝜏₃max_{𝑡∈[𝑡1,𝑡2]}|𝑢^′(𝑡)|≤𝜏₃𝜏₄𝑚₄ by (39). It follows from assumption (𝐸₁) that 𝑓(𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡))≤𝑚₄/𝑁₁. On the other hand, for 𝑢∈𝐾, we have proved that 𝑇𝑢∈𝐾 in Lemma 7 and thus 𝛾(𝑇𝑢)=max_{𝑡𝑡∈1,𝑡2}(𝑇𝑢)^′′(𝑡)≤max_{𝑡𝑡∈1,𝑡2}^𝜂_𝑡1𝜕²𝜕𝑡²𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠+max_{𝑡∈[𝑡1,𝑡2]}^𝑡2_𝜂𝜕²𝜕𝑡²𝐺(𝑡,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠=max^𝜂_𝑡1−𝜕²𝜕𝑡²𝑥₁𝜂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝜂𝑡1𝜕²𝜕𝑡²𝑦₁𝑡₁,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠+max^𝑡2_𝜂−𝜕²𝜕𝑡²𝑥₂𝑡₂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝑡2𝜂𝜕²𝜕𝑡²𝑦₂𝜂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′≤𝑚𝑑𝑠₄𝑁₁max^𝜂_𝑡1−𝜕²𝜕𝑡²𝑥₁𝜂,𝑠ℎ(𝑠)𝑑𝑠,_𝜂𝑡1𝜕²𝜕𝑡²𝑦₁𝑡₁,𝑠ℎ(𝑠)𝑑𝑠+max^𝑡2_𝜂−𝜕²𝜕𝑡²𝑥₂𝑡₂,𝑠ℎ(𝑠)𝑑𝑠,_𝑡2𝜂𝜕²𝜕𝑡²𝑦₂𝜂,𝑠ℎ(𝑠)𝑑𝑠=𝑚₄. (71) Therefore, 𝑇∶𝑃(𝛾,𝑚₄)→𝑃(𝛾,𝑚₄).
Firstly, to verify the condition (𝐵₁) of Lemma 2 holds, we choose 𝑢₀ which is defined by (46); (47) gives the expression of 𝜗. Set 𝑚=𝑚₂/(𝜏²₁𝜗); we have 𝛼𝑢₀=min_{𝑡∈[𝜉,𝜔]}𝑢₀(𝑡=)𝑚min−𝜉³+𝜆₂𝜆₁𝜉²+𝜆₃𝜆₁𝜉+𝜆₄,−𝜔³+𝜆₂𝜆₁𝜔²+𝜆₃𝜆₁𝜔+𝜆₄,−𝜂³+𝜆₂𝜆₁𝜂²+𝜆₃𝜆₁𝜂+𝜆₄≥𝜏₁𝑚𝑚𝜗=²𝜏₁>𝑚₂,𝜃𝑢₀=max_{𝑡∈[𝑡1,𝑡2]}𝑢₀(𝑡)=𝑚𝑚𝜗=²𝜏²₁>𝑚₂,𝛾𝑢₀=max_{𝑡∈[𝑡1,𝑡2]}𝑢₀^′′(𝑡=)𝑚max6𝑡₁−2𝜆₂𝜆₁,6𝑡₂−2𝜆₂𝜆₁=max6𝑡₁−2𝜆₂/𝜆₁,6𝑡₂−2𝜆₂/𝜆₁𝜗𝑚₂𝜏²₁≤𝑚₄. (72)(73) Equation (72) implies that 𝛼(𝑢₀)≥𝜏₁𝜃(𝑢₀); that is, min_{𝑡∈[𝜉,𝜔]}𝑢₀(𝑡)≥𝜏₁max_{𝑡∈[𝑡1,𝑡2]}|𝑢₀(𝑡)|. So 𝑢₀∈𝐾; it holds 𝑢₀∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄) and {𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄)∣𝛼(𝑢₀)>𝑚₂}≠0/. Hence, if 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄), then 𝑚₂≤𝑢(𝑡)≤𝑚₂𝜏₁⁻², |𝑢^′′(𝑡)|≤𝑚₄ for 𝑡∈[𝜉,𝜔]. In view of assumption (𝐸₂), we get 𝛼(𝑇𝑢)=min_{𝑡∈𝜉,𝜔}(𝑇𝑢)(𝑡)=min^𝜂_𝑡1𝑥₁(𝜉,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝜂𝑡1𝑦₁(𝜔,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝑡2𝜂𝑥₂(𝜔,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝑡2𝜂𝑦₂(𝜉,𝑠)ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝜂𝑡1𝑦₁𝜂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝑡2𝜂𝑦₂𝜂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠≥𝜏₁max^𝜂_𝑡1𝑥₁𝜂,𝑠ℎ(s)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝜂𝑡1𝑦₁𝑡₁,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠+max^𝑡2_𝜂𝑥₂𝜂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝑡2𝜂𝑦₂𝑡₂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′≥𝑚𝑑𝑠₂𝑁₂𝜏₁max^𝜂_𝑡1𝑥₁𝜂,𝑠ℎ(𝑠)𝑑𝑠,_𝜂𝑡1𝑦₁𝑡₁,𝑠ℎ(𝑠)𝑑𝑠+max^𝑡2_𝜂𝑥₂𝜂,𝑠ℎ(𝑠)𝑑𝑠,_𝑡2𝜂𝑦₂𝑡₂,𝑠ℎ(𝑠)𝑑𝑠=𝑚₂. (74) Thus, 𝛼(𝑇𝑢)>𝑚₂ for each 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄).
Next, we verify that the condition (𝐵₂) of Lemma 2 is satisfied. In fact, if 𝑢∈𝑃(𝛾,𝛼,𝑚₂,𝑚₄) with 𝜃(𝑇𝑢)>𝑚₂𝜏₁⁻², then 𝛼(𝑇𝑢)=min_{𝑡∈[𝜉,𝜔]}(𝑇𝑢)(𝑡)≥𝜏₁max_{𝑡∈[𝑡1,𝑡2]}(𝑇𝑢)(𝑡)=𝜏₁𝜃(𝑇𝑢)>𝑚₂. (75) So 𝛼(𝑇𝑢)>𝑚₂ for any 𝑢∈𝑃(𝛾,𝛼,𝑚₂,𝑚₄) with 𝜗(𝑇𝑢)>𝑚₂𝜏₁⁻².
Finally, we assert that the condition (𝐵₃) of Lemma 2 is fulfilled. It is clear that 0∉𝑄 (𝛾,𝜓,𝑚₁,𝑚₄). Suppose that 𝑢∈𝑄(𝛾,𝜓,𝑚₁,𝑚₄) with 𝜓(𝑢)=𝑚₁; by assumption (𝐸₃), we have 𝜓(𝑇𝑢)=max_{𝑡𝑡∈1,𝑡2}(𝑇𝑢)^′(𝑡)≤max_{𝑡𝑡∈1,𝑡2}^𝜂_𝑡1𝜕𝐺(ℎ(𝑠)𝑓𝜕𝑡𝑡,𝑠)𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠+max_{𝑡𝑡∈1,𝑡2}^𝑡2_𝜂𝜕𝐺(ℎ(𝑠)𝑓𝜕𝑡𝑡,𝑠)𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠=max^𝜂_𝑡1𝜕𝑥𝜕𝑡¹𝑡₁,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝜂𝑡1−𝜕𝑦𝜕𝑡¹𝑡₁,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠+max^𝑡2_𝜂−𝜕𝑥𝜕𝑡²𝑡₂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′𝑑𝑠,_𝑡2𝜂𝜕𝑦𝜕𝑡²𝑡₂,𝑠ℎ(𝑠)𝑓𝑠,𝑢,𝑢^′,𝑢^′′≤𝑚𝑑𝑠₁𝑁₃max^𝜂_𝑡1𝜕𝑥𝜕𝑡¹𝑡₁,𝑠ℎ(𝑠)𝑑𝑠,_𝜂𝑡1−𝜕𝑦𝜕𝑡¹𝑡₁,𝑠ℎ(𝑠)𝑑𝑠+max^𝑡2_𝜂−𝜕𝑥𝜕𝑡²𝑡₂,𝑠ℎ(𝑠)𝑑𝑠,_𝑡2𝜂𝜕𝑦𝜕𝑡²𝑡₂,𝑠ℎ(𝑠)𝑑𝑠=𝑚₁. (76) Hence, 𝜓(𝑇𝑢)≤𝑚₁ for 𝑢∈𝑄(𝛾,𝜓,𝑚₁,𝑚₄) with 𝜓(𝑢)=𝑚₁.
Hence, the hypotheses of Lemma 2 are satisfied. Therefore, we can conclude that the operator 𝑇 has three positive fixed points 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝛾,𝑚₄). In addition, condition (𝐴₂) guarantees that those fixed points are positive, so the problem 𝒫_𝑠 has at least three positive solutions 𝑢₁(𝑡), 𝑢₂(𝑡), 𝑢₃(𝑡) satisfying max_{𝑡∈[𝑡1,𝑡2]}|𝑢_𝑖^′′(𝑡)|≤𝑚₄, for 𝑖=1,2,3, 𝑚₂<min_{𝑡∈[𝜉,𝜔]}𝑢₁(𝑡),  and 𝑚₁<max_{𝑡∈[𝑡1,𝑡2]}𝑢^′₂(𝑡) with min_{𝑡∈[𝜉,𝜔]}𝑢₂(𝑡)<𝑚₂,max_{𝑡∈[𝑡1,𝑡2]}𝑢^′₃(𝑡)<𝑚₁.

6. Applications

The known Falkner-Skan equation 𝑢^′′′(𝑡)+12𝜚+1𝑢^′′(𝑡)𝑢(𝑡)+𝜌1−𝑢^′(𝑡)²=0, (77) whose solution is the similarity solution of the two-dimensional incompressible laminar boundary-layer equations [4, 5]. When 𝜚=1, which arises in the study of two-dimensional incompressible viscous flow past a thin semi-infinite flat plate 𝜌𝜋, 0≤𝜌≤1. The special case 𝜌=0 is Blasius’s equation, in which the wedge reduces to a flat plate. The special case 𝜌=1/2 is called Homann's equation, in which the wedge reduces to a flat plate. The special case 𝜌=1 is called Hiemenz’s equation. 𝜌>0 corresponds to flow toward the wedge, otherwise, to flow away from the wedge. When 𝜚≠1, taking 𝜚=𝜌, Blasius flow over a flat plate with a sharp edge as 𝜌=0, flow over a wedge with half angle 𝜃_1/2=𝜌𝜋/(𝜌+1), 0<𝜃_1/2<𝜋/2 as 0<𝜌<1, Hiemenz flow toward a plane stagnation point as 𝜌=1, flow into a corner with 𝜃_1/2>𝜋/2 as 1<𝜌<2, and no corresponding simple ideal flow as 𝜌>2 exist.

Now, we give a general equation of (77), subject to the boundary conditions. But it is even more difficult to solve, because it is singular; the solution 𝑢(𝑡) denotes the physically relevant solutions for 𝜚=1 and the dimensionless stream function for 𝜚≠1 of (77), respectively.

Consider 𝑢^′′′(𝑡)+ℎ(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡)𝑢(0)=0,0<𝑡<1,−𝑢^′′(0)=𝑢^′14=𝑢(1)+𝑢^′(1)=0, (78)whereℎ(𝑡)=1𝑡(),𝑓=√1−𝑡𝑡,𝜇,𝜈,𝜛1+𝑡²𝜇^2/3+2𝜋arctan1+𝜈²𝜛⁴√,0≤𝜇≤1,1+𝑡²21+𝜋arctan1+𝜈²𝜛⁴,𝜇>1. (79)

It is easy to check that hypotheses (𝐶₁), (𝐶₂) hold. Taking 𝐿=1, [𝑚,𝑀]=[1/3,2/5]⊆[𝜉,𝜔]=[1/4,1/2],𝐹(𝑡)=1, by some calculations, we have 𝑀₁=(33/16)√2𝜋, 𝑀₂=1, 𝑀₃=1, 𝑀₄=17/36, 𝜏₁=8/11, and 𝐺()=1𝑡,𝑠𝑠∈0,41162𝑡²−𝑡+4𝑠²1+2+𝑡4−1−𝑠2(𝑡−𝑠)²1,𝑠≤𝑡,162𝑡²−𝑡+4𝑠²1+2+𝑡41−𝑠,𝑡≤𝑠,𝑠∈41,1162𝑡²−𝑡+4𝑠²−1−4𝑠+32(𝑡−𝑠)²1,𝑠≤𝑡,162𝑡²−𝑡+4𝑠²−4𝑠+3,𝑡≤𝑠; (80) then 𝑓²(𝑡,𝜇,𝜈,𝜛)≤1 for (𝑡,𝜇,𝜈,𝜛)∈[0,1]×[1,∞[×]−∞,∞[² and liminf_𝜇→0⁺min_{𝑡∈[1/4,1/2]}(ℎ𝑓(𝑡,𝜇,𝜈,𝜛)/𝜇) =∞. By Theorem 8, problem (78) has at least one positive solution. These results accord with 𝜚=1; the solutions of (77) are known to exist [23].

On the other hand, let𝑓=𝑡,𝜇,𝜈,𝜛10⁻²3𝑡+4𝜇²+𝜈𝜛10¹⁰³,0<𝜇<3.78,10⁻²3𝑡+4×3.78²+𝜈𝜛10¹⁰³,𝜇≥3.78. (81)

It is easy to check that hypotheses (𝐸₁)--(𝐸₃) hold and 𝑓(𝑡,0,0,0)≢0 on [0,1]. By some calculations, we have 𝑘=4, 𝜏₁=8/11, 𝜏₃=8.992, 𝜏₄=3/4, 𝜆₁=4, 𝜆₂=29/8, 𝜆₃=−15/16, 𝜆₄=29/16, 𝜗=29/16, min{1.795,1.797,1.795}≥𝜏₁𝜗≈1.318, 𝑁₁≈2.238, 𝑁₂≈0.730, 𝑁₃≈0.505. If we choose 𝑚₁=1, 𝑚₂=2, 𝑚₄=10⁵, then 𝑓(𝑡,𝜇,𝜈,𝜛) satisfies 𝑓≤𝑚𝑡,𝜇,𝜈,𝜛⁴𝑁₁≈4.468×10⁴,∈[]×[]×𝑡,𝜇,𝜈,𝜛0,10,674400−7.5×10⁴,7.5×10⁴×−10⁵,10⁵,𝑓>𝑚𝑡,𝜇,𝜈,𝜛²𝑁₂∈1≈2.740,𝑡,𝜇,𝜈,𝜛4,12×[]×2,3.78−7.5×10⁴,7.5×10⁴×−10⁵,10⁵,𝑓≤𝑚𝑡,𝜇,𝜈,𝜛¹𝑁₃∈[]×[]×≈1.980,𝑡,𝜇,𝜈,𝜛0,10,1−7.5×10⁴,1×−10⁵,10⁵. (82) Then all hypotheses of Theorem 11 hold. Hence, problem (78) has at least three positive solutions 𝑢₁(𝑡), 𝑢₂(𝑡), and 𝑢₃(𝑡) such that max_0≤𝑡≤1|𝑢_𝑖^′′(𝑡)|≤10⁴(𝑖=1,2,3), 1<min_{1/4≤𝑡≤3/4}𝑢₁(𝑡), 1/2<max_0≤𝑡≤1𝑢^′₂(𝑡) with min_{1/4≤𝑡≤3/4}𝑢₂(𝑡)<1, max_0≤𝑡≤1𝑢^′₃(𝑡)<1/2 hold. These results coincide with 𝜚≠1; the solutions of (77) are the dimensionless stream functions.

Conflict of Interests

The author declares that there is no conflict of interests.

Authors’ Contribution

The author plotted this paper carefully, gave a rigorous derivation process, read, and approved the final paper.