Fourth-Order Four-Point Boundary Value Problem: A Solutions Funnel Approach

Abstract

We investigate the existence of positive or a negative solution of several classes of four-point boundary-value problems for fourth-order ordinary differential equations. Although these problems do not always admit a (positive) Green's function, the obtained solution is still of definite sign. Furthermore, we prove the existence of an entire continuum of solutions. Our technique relies on the continuum property (connectedness and compactness) of the solutions funnel (Kneser's Theorem), combined with the corresponding vector field.

1. Introduction

In recent years, boundary-value problems for second and higher order differential equations have been extensively studied. They are used to describe a large number of physical, biological, and chemical phenomena. The work of Timoshenko [1] on elasticity, the monograph by Soedel [2], the paper by Palamides [3] on deformation of elastic membrane, and the work of Dulàcska [4] on the effects of soil settlement are rich sources of such applications.

Pietramala [5] presented some results on the existence of multiple positive solutions of a fourth-order differential equation subject to nonlocal and nonlinear boundary conditions that models a particular stationary state of an elastic beam with nonlinear controllers.

In [6], Erbe and Wang by using a Green's function and Krasnoselskii's fixed point theorem in a cone proved the existence of a positive solution of the following boundary value problem: 𝑥^′′(𝑡)((𝑡(0)=𝑓𝑡,𝑥)),0≤𝑡≤1,𝑎𝑥−𝑏𝑥^′(0)=0,𝑐𝑥(1)+𝑑𝑥^′(1)=0, (1.1) under positivity and sub or superlinearity of the nonlinearity and moreover 𝛿=𝑎𝑑+𝑏𝑐+𝑎𝑐>0.

The literature for previous BVP is voluminous. Suggestively, we refer to [7–11] and the references therein. The monograph of Agarwal et al. [12] contains excellent surveys of known results.

Recently, an increasing interest in studying the existence of solutions and positive solutions to boundary-value problems for higher order differential equations is observed, see for example [13–17]. Especially, Graef and Yang [15] and Hao et al. [18] proved existence results on nonlinear boundary-value problem for fourth-order equations.

Also, Ge and Bai [19] by using a fixed point theorem due to Krasnoselskii and Zabreiko in [20] investigated the fourth-order nonlinear boundary value problem as 𝑢⁽⁴⁾(𝑡)=−𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)𝑢(0)(1),0<𝑡<1,=𝑢=0,𝑎𝑢^′′𝜉₁−𝑏𝑢^′′′𝜉₁=0,𝑐𝑢^′′𝜉₂+𝑑𝑢^′′′𝜉₂=0. (1.2) Precisely, they proved the next result.

Theorem 1.1. Assume that(𝐻₁)𝑎,𝑏,𝑐,𝑎𝑛𝑑𝑑 are nonnegative constants satisfying 𝜌=𝑎𝑑+𝑏𝑐+𝑎𝑐(𝜉₂−𝜉₁)≠0, 𝑏−𝑎𝜉₁≥0 and 0≤𝜉₁<𝜉₂≤1;(𝐻₂) the nonlinearity can be separated as 𝑓(𝑡,𝑢,𝑣)=𝑝(𝑡)𝑔(𝑢)+𝑞(𝑡)ℎ(𝑣), where 𝑔,ℎ∶ℝ→ℝ are continuous; lim_𝑢→∞𝑔(𝑢)𝑢=𝜆,lim_𝑣→∞ℎ(𝑣)𝑣=𝜇, (1.3) and 𝑝,𝑞∈𝐶[0,1]. Moreover, there exists some 𝑡₀∈[0,1] such that 𝑝(𝑡₀)𝑔(0)+𝑞(𝑡₀)ℎ(0)≠0, and there exists a continuous nonnegative function 𝑤∶[0,1]→ℝ⁺ such that |𝑝(𝑠)|+|𝑞(𝑠)|≤𝑤(𝑠) for each 𝑠∈[0,1];(𝐻₃)max{|𝜆|,|𝜇|}<min{1/𝐿₁,1/𝐿₂}, where 𝐿_𝑖(𝑖=1,2) are constants depending on 𝑝(𝑡) and 𝑞(𝑡). Then the BVP (1.2) admits at least one nontrivial solution 𝑢∈𝐶²[0,1].

Similarly Cui and Zou [21], on the base of a fixed point theorem, studied the BVP 𝑥⁽⁴⁾(𝑡)(𝑡)𝑓(𝑥(𝑡(),𝑥(0)(1)=𝑝)),𝑡∈0,1=𝑥=𝑥^′(0)=𝑥^′(1)=0, (1.4) to the case where 𝑓(𝑥) is monotone and either superlinear or sublinear, mainly under the assumption ¹₀𝑝(𝑠)𝑓𝑠2(1−𝑠)²𝑑𝑠<1. (1.5) Finally, Infante and Pietramala [22] proved some results on the existence of positive solutions for some cantilever fourth-order differential equation subject to nonlocal and nonlinear boundary conditions.

Restricting our consideration on the linear case, notice as far as the author is awake, that only the conditions in (1.2) have been studied, where the constants 𝑎,𝑏,𝑐,and𝑑 are nonnegative.

In a recent paper, Kelevedjev et al. [23] proved the existence of a positive and/or a negative solution for the boundary value problem (1.2), mainly under superlinearity conditions on the nonlinearity. Moreover, they exhibited existence results to the following differential equation: 𝑢⁽⁴⁾(𝑡)=−𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡),0<𝑡<1, (1.6) subject to the boundary conditions as 𝑢(0)=𝑢(1)=0,𝑎𝑢^′′𝜉₁+𝑏𝑢^′′′𝜉₁=0,𝑐𝑢^′′𝜉₂−𝑑𝑢^′′′𝜉₂=0. (1.7) Moreover, in an interesting paper [24], Anderson and Avery, by applying a generalization of the Leggett-Williams fixed point theorem, proved the existence of at least three positive solutions to the BVP as 𝑥⁽⁴⁾(𝑡)(𝑥(𝑡𝑥(0)=−𝑓)),0≤𝑡≤1,=𝑥^′𝑞=𝑥^′′(𝑟)=𝑥^′′′(1)=0,0<𝑞<𝑟<1. (1.8)

In this paper, we relax the assumptions in [23] and extend the above results to the case where the constants 𝑐 and 𝑑 are not necessarily positive and the nonlinearity is asymptotically linear and not necessarily separated, that is we study the BVP 𝑢⁽⁴⁾(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)𝑢(0)(1),0<𝑡<1,=𝑢=0,𝑎𝑢^′′𝜉₁+𝑏𝑢^′′′𝜉₁=0,𝑐𝑢^′′𝜉₂+𝑑𝑢^′′′𝜉₂=0, (1.9) with 𝑎,𝑏,𝑐,and𝑑≥0. Instead of the assumptions 𝜌=𝑎𝑑+𝑏𝑐+𝑎𝑐(𝜉₂−𝜉₁)≠0 and/or 𝑏−𝑎𝜉₁≥0 in [20], we need only the condition𝜌^∗∶=𝑎𝑐𝜉₂−𝜉₁+𝑎𝑑−𝑏𝑐>0. (1.10) Also the constants 𝜇 and 𝜇^∗ (see the assumptions 𝐵₁–𝐵^∗₃ below) satisfy different (see (2.7)) inequalities than the following one: 𝜇>48𝜉₂−𝜉₁², (1.11) in [23]. Moreover, the obtained solutions are of definite sign.

Similarly, we give existence results for several boundary value problems of type 𝑢⁽⁴⁾(𝑡)=±𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)𝑢(0)(1),0<𝑡<1,=𝑢=0,𝑎𝑢^′′𝜉₁+𝑏𝑢^′′′𝜉₁=0,𝑐𝑢^′′𝜉₂+𝑑𝑢^′′′𝜉₂=0, (1.12) where the constants 𝑎,𝑏,𝑐,and𝑑≥0 are chosen suitable positive. At these cases, we do not use the related Green's function. Thus, our approach is applicable in cases where we cannot construct a Green's function or at least it is not of definite sign. For example, instead of the above linear boundary conditions, we could choose the following nonlinear one: 𝑢(0)=𝑢(1)=0,𝑎𝑢^′′𝜉₁+𝑏𝑢^′′′𝜉₁^5/8=0,𝑐𝑢^′′𝜉₂+𝑑𝑢^′′′𝜉₂=0. (1.13) At such cases, the Krasnoselskii's fixed point theorem in a cone seems not to be applicable. Furthermore, we prove the existence of an entire continuum of solutions  (Remark 2.3) and, to the best of our knowledge, this is the first result of this type of multiplicity.

Remark 1.2. Assume the nonlinearity is nonnegative. The differential Equation (1.9) defines a vector field, the properties of which will be crucial for our study. More specifically, assuming that 𝑎𝑑−𝑏𝑐>0, we set 𝑣(𝑡)=𝑢^′′(𝑡). Let us now look (see the Figure 1, case “Th 2.2”) at the (𝑣,𝑣^′) phase quadrant {(𝑣,𝑣^′)∶𝑣≤0and𝑣^′≥0}. By the sign condition on 𝑓, we obtain that 𝑣^′′>0. Thus any trajectory (𝑣(𝑡),𝑣^′(𝑡)),𝑡≥0, crossing the semiline as 𝐸₀∶=𝑣,𝑣^′∶𝑎𝑣+𝑏𝑣^′=0,𝑣<0, (1.14) at time 𝑡=𝜉₁, “evolutes” naturally, toward the negative semiline 𝐸₁∶=𝑣,𝑣^′∶𝑐𝑣+𝑑𝑣^′=0,𝑣<0. (1.15) Setting a certain growth rate on 𝑓 (say superlinearity), we can control the vector field, so that some trajectory reaches on 𝐸₁ at the time 𝑡=𝜉₂. Whenever the nonlinearity is nonpositive or/and the sign of constants different, trajectories have an analogous behavior.

Figure 1 

Vector fields and Trajectories of BVP’s solutions.

The technique presented here is different to those in the above mentioned papers. Actually, we rely on the above “properties of the vector field” and the Kneser's property (continuum) of the cross-sections of the solutions funnel. For completeness, we restate the well-known Kneser's theorem.

Theorem 1.3 (see [25]). Consider a system (∗)𝑥^′=𝑓(𝑡,𝑥),(𝑡,𝑥)∈Ω∶=[𝑎,𝑏]×ℝ^𝑛, with 𝑓 continuous. Let 𝐸₀ be a continuum (compact and connected) in Ω₀∶={(𝑡,𝑥)∈Ω∶𝑡=𝑎} and let 𝒳(𝐸₀) be the family of all solutions of (∗) emanating from 𝐸₀. If any solution 𝑥∈𝒳(𝐸₀) is defined on the interval [𝑎,𝜏], then the set (cross-section) 𝒳𝜏;𝐸₀∶=𝑥(𝜏)∶𝑥∈𝒳𝐸₀ (1.16) is a continuum in ℝ^𝑛.

2. Main Results

Consider the boundary value problem following: 𝑢⁽⁴⁾(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡),0<𝑡<1,𝑢(0)=𝑢(1)=0,𝑎𝑢^′′𝜉₁+𝑏𝑢^′′′𝜉₁=0,𝑐𝑢^′′𝜉₂+𝑑𝑢^′′′𝜉₂=0. E₊C₊₊

Setting 𝑣(𝑡)=𝑢^′′(𝑡), the boundary value problem E₊-C₊₊ reduces to 𝑣^′′(𝑡)((𝑡)(𝑡[],𝜉=𝑓𝑡,𝑢,𝑣)),𝑡∈0,1𝑎𝑣₁+𝑏𝑣^′𝜉₁=0,𝑐𝑣𝜉₂+𝑑𝑣^′𝜉₂=0, (2.1) where 𝑢(𝑡)=^𝑡₀𝑠(𝑡−1)𝑣(𝑠)𝑑𝑠+¹_𝑡𝑡(𝑠−1)𝑣(𝑠)𝑑𝑠,0≤𝑡≤1. (2.2)

Remark 2.1. We note that for all 𝑡∈[0,1], we have (𝑡)(𝑡)≤𝑀−𝑀≤𝑣≤0⟹0≤𝑢8(𝑡)(𝑡)≥𝑀,𝑣≤−𝑀⟹𝑢8;(𝑡)𝑀0≤𝑣≤𝑀⟹−8(𝑡)(𝑡)(𝑡)𝑀≤𝑢≤0,𝑣≥𝑀⟹𝑢≤−8. (2.3) Moreover, whenever we are interested in nonnegative solutions, without loss of generality, we may extend the nonlinearity as 𝑓(𝑡,𝑢,𝑣)=𝑓(𝑡,0,0),𝑢≤0,𝑣≥0, (2.4) and if we are asking for nonpositive solutions, we may set 𝑓(𝑡,𝑢,𝑣)=𝑓(𝑡,0,0),𝑢≥0,𝑣≤0. (2.5)

We will use the following assumptions.

The nonlinearity is a continuous and positive function, that is, 𝑓(𝑡,𝑢,𝑣)∈𝐶([0,1]×(−∞,0]×[0,+∞),[0,+∞)); 𝐵₁ it is asymptotically linear at the origin and at infinity, that is, 𝑓₀=lim_{𝑢→0−,𝑣→0+}min_0≤𝑡≤1𝑓(𝑡,𝑢,𝑣)𝑣=𝜇;𝑓_∞=lim_{𝑢→−∞,𝑣→+∞}max_0≤𝑡≤1𝑓(𝑡,𝑢,𝑣)−𝑣=𝜇^∗. 𝐵₂𝐵₃ Similarly, 𝑓 is a continuous and positive function, that is, 𝑓(𝑡,𝑢,𝑣)∈𝐶([0,1]×[0,+∞)×(−∞,0],[0,+∞)); 𝐵^∗₁ it is asymptotically linear at the origin and at infinity, that is, 𝑓₀=lim_{𝑢→0+,𝑣→0−}max_0≤𝑡≤1𝑓(𝑡,𝑢,𝑣)−𝑣=𝜇;𝑓_∞=lim_{𝑢→+∞,𝑣→−∞}max_0≤𝑡≤1𝑓(𝑡,𝑢,𝑣)−𝑣=𝜇^∗. 𝐵^∗₂𝐵^∗₃

Consider now the boundary value problem E₊-C₊₊.

Theorem 2.2. Assume 𝐵^∗₁–𝐵^∗₃ hold and furthermore that 𝑎𝑐𝜉₂−𝜉₁+𝑎𝑑−𝑏𝑐<0. (2.6) Then the boundary value problem E₊-C₊₊ admits a positive and concave solution 𝑢(𝑡),0≤𝑡≤1, provided that 2()𝜉𝜇<𝑏𝑐−𝑎𝑑−2𝑎𝑐₂−𝜉₁2𝑏𝑑+𝑏𝑐𝜉₂−𝜉₁𝜉₂−𝜉₁∶=𝜇₀,𝜇^∗>max𝑏𝑐−𝑎𝑑𝑏𝑑𝜉₂−𝜉₁,8𝑐(𝑏𝑐−𝑎𝑑)𝑏𝑑²∶=𝜇^∗₀. (2.7)(2.8) Furthermore, 𝑢^′′(𝑡)≤0,𝑢^′′′(𝑡)≥0,0≤𝑡≤1. (2.9)

Proof. By the asymptotic linearity of 𝑓(𝑡,𝑢,𝑣) at 𝑣=0 (see the assumption 𝐵^∗₂ and (2.3)–(2.7), for any 𝜆∈(𝜇,𝜇₀) there is an 𝜂>0 such that 𝜂−𝜂≤𝑣<0,0<𝑢≤8,implymax_0≤𝑡≤1𝑓(𝑡,𝑢,𝑣)<−𝜆𝑣. (2.10) Consider any positive number 𝜀, such that 𝑎𝑑+𝜇𝑏𝑑𝜉₂−𝜉₁𝜉𝑏𝑐<𝜀<2𝑏−𝜇𝑏₂−𝜉₁+2𝑎𝜉₂−𝜉₁2𝑏<1, (2.11) and choose a positive 𝜆, where 𝜆<min𝜀𝑏𝑐−𝑎𝑑𝑏𝑑𝜉₂−𝜉₁,2𝑏𝜉₂−𝜉₁²𝑏(1−𝜀)−𝑎𝜉₂−𝜉₁. (2.12) We assert that for𝑃^∗₀=(𝑣₀,−(𝑎/𝑏)𝑣₀) and any solution 𝑣∈𝒳(𝑃^∗₀), where 𝑣₀=−𝜂, it follows that −𝜂≤𝑣(𝑡)≤−𝜀𝜂,𝑡∈𝜉₁,𝜉₂. (2.13) Indeed in view of Remark 1.2, let us assume that there exists 𝑡^∗∈(𝜉₁,𝜉₂] such that −𝜂≤𝑣(𝑡)≤−𝜀𝜂,𝑣^′(𝑡)>0,𝜉₁≤𝑡<𝑡^∗,𝑣𝑡^∗=−𝜀𝜂. (2.14) Then by the Taylor's formula, (2.10) and (2.14), we get 𝑡∈(0,𝑡^∗), such that −𝜀𝜂=𝑣𝑡^∗=𝑣₀𝑎1−𝑏𝑡^∗−𝜉₁+𝑡^∗−𝜉₁²𝑓2!𝑡,𝑢𝑡,𝑣𝑡𝑎𝑡≤−𝜂1−^∗−𝜉₁𝑏−𝑡^∗−𝜉₁²2𝜆𝑣𝑡𝑎≤−𝜂1−𝑏𝜉₂−𝜉₁+𝜉₂−𝜉₁²2𝜆𝜂. (2.15) Consequently, 2𝜆≥𝑏𝜉₂−𝜉₁²𝑏(1−𝜀)−𝑎𝜉₂−𝜉₁, (2.16) contrary to the choice of 𝜆 in (2.12). Hence, the assertion 𝑣(𝑡)≤−𝜀𝜂,𝑡∈[𝜉₁,𝜉₂] in (2.13) is proved. Moreover, if there is 𝑡^∗∈(𝜉₁,𝜉₂) such that 𝑎𝜂𝑏≤𝑣^′(𝑡)≤𝑐𝜀𝜂𝑑,𝜉₁≤𝑡<𝑡^∗,𝑣^′𝑡^∗=𝑐𝜂𝜀𝑑, (2.17) then 𝑐𝜂𝜀𝑑=𝑣^′𝑡^∗=−𝑣₀𝑎𝑏+𝑡^∗−𝜉₁𝑓𝑡,𝑢𝑡,𝑣𝑡≤𝑎𝜂𝑏−𝑡^∗−𝜉₁𝜆𝑣𝑡≤𝑎𝜂𝑏−𝑡^∗−𝜉₁𝜆𝑣𝜉₁=𝑎𝜂𝑏+𝑡^∗−𝜉₁𝜆𝜂≤𝑎𝜂𝑏+𝜉₂−𝜉₁𝜆𝜂, (2.18) a contradiction to (2.12). Thus 𝑎𝜂𝑏≤𝑣^′(𝑡)≤𝑐𝜀𝜂𝑑,𝜉₁≤𝑡<𝜉₂. (2.19) Consequently, 𝐺^∗𝑃^∗₀=𝑐𝑣𝜉₂+𝑑𝑣^′𝜉₂≤𝑐−𝜂𝜀+𝑑𝑐𝜀𝜂𝑑=0. (2.20)
On the other hand, in order to prove the opposite inequality, we assume on the contrary that for every solution 𝑣∈𝒳(𝑃^∗₁),𝑃^∗₁=(−𝐻,(𝑎/𝑏)𝐻), we have 𝐺^∗𝑡;𝑃^∗₀=𝑐𝑣(𝑡)+𝑑𝑣^′(𝑡)<0,𝜉₁≤𝑡<𝜉₂. (2.21) Then, 𝑣^′(𝑡)𝑐<−𝑑𝑣(𝑡),𝜉₁≤𝑡<𝜉₂. (2.22) In view of the assumption 𝐵^∗₃, we chose 𝐾>0 and 𝜃∈(0,1) such that max𝑏𝑐−𝑎𝑑𝑏𝑑𝜉₂−𝜉₁𝜃,2𝑐(𝑏𝑐−𝑎𝑑)𝑏𝑑²(1−𝜃)𝜃<𝐾<𝜇^∗. (2.23) Then, there exists 𝐻>0, such that 𝑓(𝑡,𝑢,𝑣)>−𝐾𝑣,𝑣≤−𝜃𝐻. (2.24) Assume that there is a solution 𝑣∈𝒳(𝑃^∗₁) and 𝑡^∗∈(𝜉₁,𝜉₂) such that −𝐻≤𝑣(𝑡)≤−𝜃𝐻,𝜉₁≤𝑡<𝑡^∗,𝑣𝑡^∗=−𝜃𝐻. (2.25) We assert first that 𝑡^∗≥𝜉₁+𝑑𝑐(1−𝜃). (2.26) Indeed, by (2.22) and (2.25), we obtain −𝜃𝐻=𝑣𝑡^∗𝑐≤−𝐻+𝑑𝐻𝑡^∗−𝜉₁, (2.27) and then (2.26) follows.
By (2.24)-(2.25), we obtain −𝜃𝐻=𝑣𝑡^∗𝑎=−𝐻+𝑏𝐻𝑡^∗−𝜉₁+𝑡^∗−𝜉₁²𝑓2!𝑡,𝑢𝑡,𝑣𝑡𝑎≥−𝐻1−𝑏𝑡^∗−𝜉₁−𝑡^∗−𝜉₁²2!𝐾𝑣𝑡𝑎≥−𝐻1−𝑏𝑡^∗−𝜉₁+𝑡^∗−𝜉₁²2!𝐾𝜃𝐻. (2.28) Consequently, (2.26) yields 2)/𝑡𝐾≤(1−𝜃^∗−𝜉₁−𝑎/𝑏𝑡^∗−𝜉₁𝜃<2𝑐(𝑏𝑐−𝑎𝑑)𝑏𝑑²()𝜃,1−𝜃 (2.29) a contradiction to the choice of 𝐾. Thus, −𝐻≤𝑣(𝑡)≤−𝜃𝐻,𝜉₁≤𝑡<𝜉₂. (2.30) Consequently, by the choice of 𝐾 and (2.24), we get 𝑣^′𝜉₂=𝑎𝑏𝐻+𝜉₂−𝜉₁𝑓𝑡,𝑢𝑡,𝑣𝑡≥𝑎𝑏𝐻−𝜉₂−𝜉₁𝐾𝑣𝑡≥𝑎𝑏𝐻+𝜉₂−𝜉₁𝑎𝐾𝐻𝜃≥𝐻𝑏+𝜉₂−𝜉₁𝑏𝑐−𝑎𝑑𝑏𝑑𝜉₂−𝜉₁=𝑐𝑑𝐻. (2.31) Hence, we conclude that 𝐺^∗𝑃^∗₁=𝑐𝑣𝜉₂+𝑑𝑣^′𝜉₂()𝑐≥𝑐−𝐻+𝑑𝑑𝐻=0, (2.32) contrary to the assertion (2.21).
Finally, consider the segment (recall that 𝑣₀=−𝜂₀ and 𝑣₁=−𝐻) 𝑃^∗₀,𝑃^∗₁∶=𝑣,𝑣^′∈𝐸₀∶𝑣₀≤𝑣≤𝑣₁, (2.33) and furthermore the cross-section 𝒳1;𝑃^∗₀,𝑃^∗₁∶=𝑣(1),𝑣^′(1)∶𝑣∈𝒳(𝑃),𝑃∈𝑃^∗₀,𝑃^∗₁ (2.34) of the solutions funnel emanating from the segment [𝑃^∗₀,𝑃^∗₁]. By the definition of the function 𝐺^∗(𝑃₁)∶=𝑐𝑣(𝜉₂)+𝑑𝑣^′(𝜉₂), (2.20), and (2.32), it is clear (recall that 𝐸₁∶={(𝑣,𝑣^′)∶𝑐𝑣+𝑑𝑣^′=0,𝑣>0}) that 𝐸₁∩𝒳1;𝑃^∗₀,𝑃^∗₁≠∅. (2.35) This means that there is a point 𝑃∈[𝑃^∗₀,𝑃^∗₁] such that 𝐺^∗(𝑃)=0 and thus a solution 𝑣₀(𝑡)∈𝒳(𝑃) satisfying the second boundary condition in C₊₊.
Furthermore, by the above analysis, the obtained solution 𝑣₀(𝑡),𝜉₁≤𝑡≤𝜉₂ is negative. We extend 𝑣₀(𝑡) on the entire interval as follows: 𝑣(𝑡)=𝑣₀𝜉₁,0≤𝑡≤𝜉₁,𝑣₀(𝑡),𝜉₁≤𝑡≤𝜉₂,𝑣₀𝜉₂,𝜉₂≤𝑡≤1. (2.36) Then, the function 𝑣(𝑡),0≤𝑡≤1, is negative and continuous. In view of the transformation 𝑣(𝑡)=𝑢^′′(𝑡), we consider the boundary value problem as 𝑢^′′(𝑡),𝑢(0)(1).=𝑣=0=𝑢 (2.37) It is well known that the Green function of it is 𝐺()=𝑠()𝑡()𝑡,𝑠1−𝑡,0≤𝑠≤𝑡≤11−𝑠,0≤𝑡≤𝑠≤1. (2.38) Consequently (see (2.3)), the desired positive and concave solution of the boundary value problem E₊-C₊₊ is given by the formula as 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣(𝑠)𝑑𝑠=−^𝑡₀𝑠(𝑡−1)𝑣(𝑠)𝑑𝑠−¹_𝑡𝑡(𝑠−1)𝑣(𝑠)𝑑𝑠,0≤𝑡≤1. (2.39)

Remark 2.3. Since we can extend the solution 𝑣₀(𝑡),𝜉₁≤𝑡≤𝜉₂ with infinite ways, we can immediately obtain an entire continuum of solutions for the boundary value problem E₊-C₊₊.

Consider the boundary conditions 𝑢(0)=𝑢(1)=0𝑎𝑢^′′𝜉₁−𝑏𝑢^′′′𝜉₁=0,𝑐𝑢^′′𝜉₂−𝑑𝑢^′′𝜉₂=0. C₋₋

Corollary 2.4. Assume 𝐵^∗₁–𝐵^∗₃ hold, where the interval [0,1] is replaced by [−1,0], and furthermore 𝑎𝑐𝜉₂−𝜉₁+𝑏𝑐−𝑎𝑑<0. (2.40) Then the boundary value problem E₊–C₋₋ has a positive and concave solution 𝑢(𝑡),0≤𝑡≤1, provided that 2()𝜉𝜇<𝑎𝑑−𝑏𝑐−2𝑎𝑐₂−𝜉₁2𝑏𝑑+𝑏𝑐𝜉₂−𝜉₁𝜉₂−𝜉₁,𝜇^∗()>max𝑎𝑑−𝑏𝑐𝑏𝑑𝜉₂−𝜉₁,8𝑎(𝑎𝑑−𝑏𝑐)𝑑𝑏². (2.41) Furthermore, 𝑢^′′(𝑡)≤0,𝑢^′′′(𝑡)≤0,0≤𝑡≤1. (2.42)

Proof. Consider the BVP 𝑦⁽⁴⁾(𝑡)=𝐹𝑡,𝑦(𝑡),𝑦^′′(𝑡)𝑢()(0),−1<𝑡<0,−1=𝑢=0,𝑐𝑢^′′−𝜉₂+𝑑𝑢^′′′−𝜉₂=0,𝑎𝑢^′′−𝜉₁+𝑏𝑢^′′−𝜉₁=0, (2.43) where 𝐹−𝑡,𝑢(𝑡),𝑢^′′(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡), (2.44) and 𝑢(𝑡),−1≤𝑡≤0 is any real map.
Since 𝑓 satisfies the conditions 𝐵^∗₁–𝐵^∗₃, it is obvious that the same conditions are fulfilled by 𝐹. Hence, in view of Theorem 2.2, the BVP (2.43) admits a positive and concave solution 𝑦(𝑡),−1≤𝑡≤0. Thus, setting 𝑢(𝑡)=𝑦(−𝑡),0≤𝑡≤1, (2.45) we get 𝑢⁽⁴⁾(𝑡)=𝑦(−𝑡)⁽⁴⁾=𝐹−𝑡,𝑦(−𝑡),𝑦^′′()(𝑡)−𝑡=𝐹−𝑡,𝑢,𝑢^′′(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡),0≤𝑡≤1. (2.46) Consequently, 𝑢(𝑡),0≤𝑡≤1, is a solution of the differential equation E₊. Furthermore the boundary conditions in (2.43) and the transformation 𝑢(𝑡)=𝑦(−𝑡),0≤𝑡≤1 guarantee that the boundary conditions C₋₋ also hold. Hence, 𝑢(𝑡) is the requite solution of E₊–C₋₋.

Remark 2.5. Obviously, we could give an analytical proof (similar to the one given at Theorem 2.2) to the above Corollary 2.4, as well as to the following ones.
Consider now the differential equation as 𝑢⁽⁴⁾(𝑡)=−𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡),0<𝑡<1, E₋ and then the boundary value problem E₋–C₋₋,

Corollary 2.6. Assume 𝐵₁–𝐵₃ hold and furthermore that 𝑎𝑐𝜉₂−𝜉₁+𝑏𝑐−𝑎𝑑>0. (2.47) Then the boundary value problem E₋–C₋₋ admits a negative and convex solution 𝑢(𝑡),0≤𝑡≤1, provided that (2.7)-(2.8) holds. Furthermore, 𝑢^′′(𝑡)≥0,𝑢^′′′(𝑡)≥0,0≤𝑡≤1. (2.48)

Proof. Consider the BVP 𝑦⁽⁴⁾(𝑡)=𝐹𝑡,𝑦(𝑡),𝑦^′′(𝑡)𝑦()(0),−1≤𝑡≤0,−1=𝑦=0,𝑐𝑦^′′−𝜉₂+𝑑𝑦^′′′−𝜉₂=0,𝑎𝑦^′′−𝜉₁+𝑏𝑦^′′′−𝜉₁=0, (2.49) where, for any real function 𝑢(𝑡),0≤𝑡≤1, 𝐹𝑡,𝑢(𝑡),𝑢^′′(𝑡)∶=𝑓−𝑡,−𝑢(−𝑡),−𝑢^′′(−𝑡),−1≤𝑡≤0. (2.50) Then, since the conditions 𝐵₁–𝐵₃ are performed by 𝑓, 𝐹 satisfies the assumptions 𝐵^∗₁–𝐵^∗₃ (applied on the interval [−1,0]). Hence, by Theorem 2.2, (2.49) admits a positive and concave solution 𝑦(𝑡),−1≤𝑡≤0. We set 𝑢(𝑡)=−𝑦(−𝑡),0≤𝑡≤1. (2.51) Then 𝑢⁽⁴⁾(𝑡)=−𝑦(−𝑡)⁽⁴⁾=−𝑦(−𝑡)⁽⁴⁾=−𝐹−𝑡,𝑦(−𝑡),𝑦^′′()(𝑡)−𝑡=−𝐹−𝑡,−𝑢,−𝑢^′′(𝑡)=−𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡),0≤𝑡≤1, (2.52) that is, the map 𝑢(𝑡),0≤𝑡≤1, is a solution of E₋. Furthermore, since 0=𝑐𝑦^′′−𝜉₂+𝑑𝑦^′′′−𝜉₂=−𝑐𝑢^′′𝜉₂+𝑑𝑢^′′′𝜉₂=0, (2.53) (and similarly for the other) the boundary conditions in C₋₋ are satisfied by the solution 𝑢(𝑡). Consequently, 𝑢(𝑡)=−𝑦(−𝑡),0≤𝑡≤1 is the desired solution of the boundary value problem E₋–C₋₋

Remark 2.7. By the following formula: 𝑦^∗(𝑡)=−¹₀𝐺(𝑡,𝑠)𝑣(𝑠)𝑑𝑠=^𝑡₀𝑠(𝑡−1)𝑣(𝑠)𝑑𝑠+¹_𝑡𝑡(𝑠−1)𝑣(𝑠)𝑑𝑠,0≤𝑡≤1, (2.54) it is clear that the map 𝑢^∗(𝑡),0≤𝑡≤1, is a negative solution of the BVP (2.49). Hence we get also the positive solution as 𝑢^∗(𝑡)=−𝑦^∗(−𝑡),0≤𝑡≤1, (2.55) of the boundary value problem E₋–C₋₋. The same reasoning folds for all the following results.

Consider the BVP E₊–C₊₊.

Corollary 2.8. Assume 𝐵₁–𝐵₃ hold, and furthermore that 𝑎𝑐𝜉₂−𝜉₁+𝑎𝑑−𝑏𝑐<0. (2.56) Then the boundary value problem E₊–C₊₊ admits a negative and convex solution 𝑢(𝑡),0≤𝑡≤1, provided that (2.7)-(2.8) holds. Furthermore, 𝑢^′′(𝑡)≥0,𝑢^′′′(𝑡)≤0,0≤𝑡≤1. (2.57)

Proof. Consider the BVP 𝑦⁽⁴⁾(𝑡)=𝐹𝑡,𝑦(𝑡),𝑦^′′(𝑡)𝑦(0)(1),0≤𝑡≤1,=𝑦=0,𝑎𝑦^′′𝜉₁+𝑏𝑦^′′𝜉₁=0,𝑐𝑦^′′𝜉₂+𝑑𝑦^′′′𝜉₂=0, (2.58) where 𝐹(𝑡,𝑢,𝑣)=−𝑓(𝑡,−𝑢,−𝑣),𝑢≥0,𝑣≤0. (2.59) Then, since the conditions 𝐵₁–𝐵₃ are performed by 𝑓, 𝐹 satisfies the assumptions 𝐵^∗₁–𝐵^∗₃ of Theorem 2.2. Hence, that BVP admits a positive and concave solution 𝑦(𝑡),0≤𝑡≤1. We set 𝑢(𝑡)=−𝑦(𝑡),0≤𝑡≤1. (2.60) Then (−𝑢(𝑡))⁽⁴⁾=𝑦⁽⁴⁾(𝑡)=𝐹𝑡,𝑦(𝑡),𝑦^′′(𝑡)=−𝑓𝑡,−𝑦(𝑡),−𝑦^′′(𝑡)=−𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡), (2.61) that is, the map 𝑢(𝑡),0≤𝑡≤1, is a solution of E₊. Furthermore, since 0=𝑎𝑦^′′𝜉₁+𝑏𝑦^′′𝜉₁=−𝑎𝑢^′′𝜉₁−𝑏𝑢^′′′𝜉₁=0, (2.62) the first (and similar the other) of the boundary conditions in C₊₊ is satisfied. Consequently, 𝑢(𝑡)=−𝑦(𝑡),0≤𝑡≤1, is the desired solution of the boundary value problem E₊–C₊₊.

Consider now the boundary value problem E₋–C₊₊.

Corollary 2.9. Assume 𝐵^∗₁–𝐵^∗₃ hold and furthermore that 𝑎𝑐(𝜉₂−𝜉₁)+𝑎𝑑−𝑏𝑐<0. Then the boundary value problem E₋–C₊₊ admits a positive and concave solution 𝑢(𝑡),0≤𝑡≤1, provided that (2.41) holds. Furthermore, 𝑢^′′(𝑡)≥0,𝑢^′′′(𝑡)≤0,0≤𝑡≤1. (2.63)

Proof. Consider any function 𝑢(𝑡),0≤𝑡≤1.We define a map 𝑓^∗ by 𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)=−𝑓^∗𝜉₁+𝜉₂−𝑡,𝑢𝜉₁+𝜉₂−𝑡,𝑢^′′𝜉₁+𝜉₂−𝑡,𝑡∈[0,1],𝑢≥0,𝑣≤0. (2.64) Since 𝑓 satisfies the conditions 𝐵^∗₁–𝐵^∗₃, we easily check that 𝑓^∗ suits the same conditions. In view of Theorem 2.2, consider a solution 𝑢^∗(𝑡),𝜉₁+𝜉₂−1≤𝑡≤𝜉₁+𝜉₂, of the following BVP: 𝑢^∗(4)(𝑡)=𝑓^∗𝑡,𝑢^∗(𝑡),𝑢^∗′′(𝑡),𝑢^∗(0)=𝑢^∗(1)=0,𝑐𝑢^∗′′𝜉₁+𝑑𝑢^∗′′′𝜉₁=0,𝑎𝑢^∗′′𝜉₂+𝑏𝑢^∗′′′𝜉₂=0. (2.65)(2.66) We notice that 𝜉₁+𝜉₂−1≤𝜉₁<𝜉₂≤𝜉₁+𝜉₂ and set 𝑢(𝑡)≡𝑢^∗𝜉₁+𝜉₂−𝑡,0≤𝑡≤1. (2.67) Then, we obtain (𝑢(𝑡))⁽⁴⁾=𝑢^∗𝜉₁+𝜉₂−𝑡⁽⁴⁾=𝑓^∗𝜉₁+𝜉₂−𝑡,𝑢^∗𝜉₁+𝜉₂−𝑡,𝑢^∗′′𝜉₁+𝜉₂−𝑡=−𝑓(𝑡,𝑢(𝑡),𝑣(𝑡)), (2.68) that is, the function 𝑢(𝑡),0≤𝑡≤1 is a solution of the differential equation E₋. Moreover, the boundary conditions C₊₊ are satisfied by the function 𝑢(𝑡), since the solution 𝑢^∗(𝑡) fulfils the boundary conditions (2.66). Consequently, 𝑢(𝑡) is the requite solution of E₋–C₊₊.

Corollary 2.10. Assume 𝐵₁–𝐵₃ hold and furthermore that 𝑎𝑐(𝜉₂−𝜉₁)+𝑎𝑑−𝑏𝑐<0. Then the boundary value problem E₋–C₊₊ admits a positive and concave solution 𝑢(𝑡),0≤𝑡≤1, provided that (2.41) holds. Furthermore, 𝑢^′′(𝑡)≥0,𝑢^′′′(𝑡)≤0,0≤𝑡≤1. (2.69)

Proof. It follows by Corollary 2.9, via the following transformation: 𝑢(𝑡)=𝑦(−𝑡),0≤𝑡≤1. (2.70) Here 𝑦(𝑡),−1≤𝑡≤0, is a positive solution of the BVP as 𝑦⁽⁴⁾(𝑡)=𝐹𝑡,𝑦(𝑡),𝑦^′′(𝑡)𝑢(0)(1),−1≤𝑡≤0,=𝑢=0,𝑐𝑢^′′−𝜉₂+𝑑𝑢^′′′−𝜉₂=0,𝑎𝑢^′′−𝜉₁+𝑏𝑢^′′−𝜉₁=0, (2.71) where 𝐹𝑡,𝑢(−𝑡),𝑢^′′(−𝑡)=−𝑓−𝑡,𝑢(𝑡),𝑢^′′(𝑡),−1≤𝑡≤0. (2.72) Indeed, for 𝑡∈[0,1], 𝑢⁽⁴⁾(𝑡)=𝑦(−𝑡)⁽⁴⁾=𝐹−𝑡,𝑦(−𝑡),𝑦^′′(−𝑡)=𝐹−𝑡,𝑢(𝑡),𝑢^′′(𝑡)=−𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡), (2.73) that is, 𝑢(𝑡),𝑡∈[0,1], is a solution of differential equation E₋. Finally, 𝑢(𝑡)=𝑦(−𝑡),−1≤𝑡≤0, is a desired solution of BVP E₋-C₋₋.

Similarly, we may prove the next results.

Corollary 2.11. Assume 𝐵₁–𝐵₃ hold and furthermore, 𝑎𝑐𝜉₂−𝜉₁+𝑏𝑐−𝑎𝑑<0. (2.74) Then the boundary value problem E₊–C₋₋ has a negative and convex solution 𝑢(𝑡),0≤𝑡≤1, provided that (2.7)-(2.8) holds. Furthermore, 𝑢^′′(𝑡)≥0,𝑢^′′′(𝑡)≥0,0≤𝑡≤1. (2.75)

Corollary 2.12. Assume 𝐵₁–𝐵₃ hold and furthermore 𝑎𝑐𝜉₂−𝜉₁+𝑏𝑐−𝑎𝑑<0. (2.76) Then the boundary value problem E₋–C₊₊ has a negative and convex solution 𝑢(𝑡),0≤𝑡≤1, provided that (2.41) holds. Furthermore, 𝑢^′′(𝑡)≥0,𝑢^′′′(𝑡)≤0,0≤𝑡≤1. (2.77)

Corollary 2.13. Assume 𝐵₁–𝐵₃ hold and furthermore 𝑎𝑐𝜉₂−𝜉₁+𝑎𝑑−𝑏𝑐<0. (2.78) Then the boundary value problem E₋–C₋₋ has a negative and convex solution 𝑢(𝑡),0≤𝑡≤1, provided that (2.7)-(2.8) holds. Furthermore, 𝑢^′′(𝑡)≥0,𝑢^′′′(𝑡)≥0,0≤𝑡≤1. (2.79)

Corollary 2.14. Assume that 𝐵^∗₁ holds and the nonlinearity is a sublinear map, that is, 𝜇=0, 𝜇^∗=+∞. Then, the boundary value problem E₊-C₊₊ admits a positive and concave solution.

Proof. Obviously, 𝐵^∗₁–𝐵^∗₃ are fulfilled. Thus, Theorem 2.2, yields the result.

Remark 2.15. It is obvious that the results of all Corollaries, given above, still hold, under a sublinearity assumption of the nonlinearity (see [23]).

Remark 2.16. We could replace the following lines (initial and terminal points): 𝐸₀∶=𝑣,𝑣^′∶𝑎𝑣+𝑏𝑣^′=0,𝑣>0,𝐸₁∶=𝑣,𝑣^′∶𝑐𝑣+𝑑𝑣^′=0,𝑣>0, (2.80) with any suitable continuum 𝐾₀ and 𝐾₁, respectively. At such a case, the existence of the related Green's function is uncertain. For example, we may consider the following boundary conditions: 𝑎𝑢^′′𝜉₁³+𝑏𝑢^′′′𝜉₁=0,𝑐𝑢^′′𝜉₂^1/3+𝑏𝑢^′′′𝜉₂=0. (2.81) Then, the proof can be carried out word perfect.

Disclosure

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