Some existence criteria are established for a class of fourth-order $m$-point boundary value problem by using the upper and lower solution method and the Leray-Schauder continuation principle

Ren, Qiu-Yan

Sun, Jian-Ping

Electronic journal of qualitative theory of differential equations

Jian-Ping Sun

Qiu-Yan Ren

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Existence of solutions for a class of fourth-order m-point boundary value problems

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Electronic Journal of Qualitative Theory of Differential Equations2010, No. 14, 1-8; http://www.math.u-szeged.hu/ejqtde/Existence of solutions for a class of fourth-orderm-point boundary value problems∗Jian-Ping Sun, Qiu-Yan RenDepartment of Applied Mathematics, Lanzhou University of Technology,Lanzhou, Gansu, 730050, People’s Republic of ChinaE-mail: jpsun@lut.cn (Sun)AbstractSome existence criteria are established for a class of fourth-order m-point boundary value prob-lem by using the upper and lower solution method and the Leray-Schauder continuation principle.Keywords: Fourth-order m-point boundary value problem; Upper and lower solution method;Leray-Schauder continuation principle; Nagumo condition2000 AMS Subject Classification: 34B10, 34B151 IntroductionBoundary value problems (BVPs for short) of fourth-order differential equations have been used todescribe a large number of physical, biological and chemical phenomena. For example, the deforma-tions of an elastic beam in the equilibrium state can be described as some fourth-order BVP. Recently,fourth-order BVPs have received much attention. For instance, [3, 5, 6, 7] discussed some fourth-ordertwo-point BVPs, while [1, 2, 4, 9] studied some fourth-order three-point or four-point BVPs. It isworth mentioning that Ma, Zhang and Fu [7] employed the upper and lower solution method to provethe existence of solutions for the BVP{u(4)(t) = f (t, u (t) , u′′ (t)) , t ∈ (0, 1),u (0) = u′ (1) = u′′ (0) = u′′′ (1) = 0,and Bai [3] considered the existence of a solution for the BVP{u(4)(t) = f (t, u (t) , u′ (t) , u′′ (t) , u′′′ (t)) , t ∈ (0, 1),u (0) = u′ (1) = u′′ (0) = u′′′ (1) = 0by using the upper and lower solution method and Schauder′s fixed point theorem.Although there are many works on fourth-order two-point, three-point or four-point BVPs, alittle work has been done for more general fourth-order m-point BVPs [8]. Motivated greatly by the∗Supported by the National Natural Science Foundation of China (10801068).EJQTDE, 2010 No. 14, p. 1above-mentioned excellent works, in this paper, we will investigate the following fourth-order m-pointBVP u(4)(t) + f(t, u (t) , u′ (t) , u′′ (t) , u′′′ (t)) = 0, t ∈ [0, 1] ,u (0) =m−2∑i=1aiu (ηi) , u′ (1) = 0,u′′ (0) =m−2∑i=1biu′′ (ηi) , u′′′ (1) = 0.(1.1)Throughout this paper, we always assume that 0 < η1 < η2 < · · · < ηm−2 < 1, ai and bi (i =1, 2, · · ·,m − 2) are nonnegative constants and f : [0, 1] × R4 → R is continuous. Some existencecriteria are established for the BVP (1.1) by using the upper and lower solution method and theLeray-Schauder continuation principle.2 PreliminariesLet E = C [0, 1] be equipped with the norm ‖v‖∞= maxt∈[0,1]|v (t)| andK = {v ∈ E| v (t) ≥ 0 for t ∈ [0, 1]} .Then K is a cone in E and (E,K) is an ordered Banach space. For Banach space X = C1 [0, 1] , weuse the norm ‖v‖ = max {‖v‖∞, ‖v′‖∞} .Lemma 2.1 Letm−2∑i=1ai 6= 1. Then for any h ∈ E, the second-order m-point BVP−u′′(t) = h (t) , t ∈ [0, 1] ,u (0) =m−2∑i=1aiu (ηi) , u′ (1) = 0(2.1)has a unique solutionu (t) =∫ 10G1 (t, s)h (s) ds,whereG1 (t, s) = K (t, s) +11−m−2∑i=1aim−2∑i=1aiK (ηi, s) ,hereK (t, s) ={s, 0 ≤ s ≤ t ≤ 1,t, 0 ≤ t ≤ s ≤ 1is Green’s function of the second-order two-point BVP{−u′′(t) = 0, t ∈ [0, 1] ,u (0) = u′ (1) = 0.EJQTDE, 2010 No. 14, p. 2Proof. If u is a solution of the BVP (2.1), then we may suppose thatu (t) =∫ 10K(t, s)h(s)ds +At+B.By the boundary conditions in (2.1), we know thatA = 0 and B =11−m−2∑i=1aim−2∑i=1ai∫ 10K(ηi, s)h (s) ds.Therefore, the unique solution of the BVP (2.1)u (t) =∫ 10K(t, s)h(s)ds +11−m−2∑i=1aim−2∑i=1ai∫ 10K(ηi, s)h (s) ds=∫ 10G1 (t, s)h (s) ds.In the remainder of this paper, we always assume thatm−2∑i=1ai < 1 andm−2∑i=1bi < 1, which implythat G1 (t, s) and G2 (t, s) are nonnegative on [0, 1] × [0, 1], whereG2 (t, s) = K (t, s) +11−m−2∑i=1bim−2∑i=1biK (ηi, s) .Now, we define operators A and B : E → E as follows:(Av) (t) = −∫ 10G1 (t, s) v (s) ds, t ∈ [0, 1] (2.2)and(Bv) (t) = −∫ 1tv (s) ds, t ∈ [0, 1] . (2.3)Remark 2.1 A and B are decreasing operators on E.Lemma 2.2 If the following BVPv′′(t) + f(t, (Av) (t) , (Bv) (t) , v (t) , v′ (t)) = 0, t ∈ [0, 1] ,v (0) =m−2∑i=1biv (ηi) , v′ (1) = 0(2.4)has a solution, then does the BVP (1.1).Proof. Suppose that v is a solution of the BVP (2.4). Then it is easy to prove that u = Av is asolution of the BVP (1.1). EJQTDE, 2010 No. 14, p. 3Definition 2.1 If α ∈ C2 [0, 1] satisfiesα′′(t) + f(t, (Aα) (t) , (Bα) (t) , α (t) , α′ (t)) ≥ 0, t ∈ [0, 1] ,α (0) ≤m−2∑i=1biα (ηi) , α′ (1) ≤ 0,(2.5)then α is called a lower solution of the BVP (2.4).Definition 2.2 If β ∈ C2 [0, 1] satisfiesβ′′(t) + f(t, (Aβ) (t) , (Bβ) (t) , β (t) , β′ (t)) ≤ 0, t ∈ [0, 1] ,β (0) ≥m−2∑i=1biβ (ηi) , β′ (1) ≥ 0,(2.6)then β is called an upper solution of the BVP (2.4).Remark 2.2 If the inequality in Definition (2.1)α′′(t) + f(t, (Aα) (t) , (Bα) (t) , α (t) , α′ (t)) ≥ 0, t ∈ [0, 1]is replaced byα′′(t) + f(t, (Aα) (t) , (Bα) (t) , α (t) , α′ (t)) > 0, t ∈ [0, 1] ,then α is called a strict lower solution of the BVP (2.4). Similarly, we can also give the definition ofa strict upper solution for the BVP (2.4).Definition 2.3 Assume that f ∈ C([0, 1]×R4, R), α, β ∈ E and α(t) ≤ β(t) for t ∈ [0, 1]. Wesay that f satisfies Nagumo condition with respect to α and β provided that there exists a functionh ∈ C ([0,+∞) , (0,+∞)) such that|f(t, x1, x2, x3, x4)| ≤ h (|x4|) ,for all (t, x1, x2, x3, x4) ∈ [0, 1] × [(Aβ) (t) , (Aα) (t)]× [(Bβ) (t) , (Bα) (t)]× [α(t), β(t)] ×R, and∫ +∞λsh (s)ds > maxt∈[0,1]β (t)− mint∈[0,1]α (t) , (2.7)where λ = max {|β (1)− α (0)| , |β (0)− α (1)|} .Lemma 2.3 Assume that α and β are, respectively, the lower and the upper solution of the BVP(2.4) with α (t) ≤ β (t) for t ∈ [0, 1] , and f satisfies the Nagumo condition with respect to α and β.Then there exists N > 0 (depending only on α and β) such that any solution ω of the BVP (2.4) lyingin [α, β] satisfies ∣∣ω′ (t)∣∣ ≤ N, t ∈ [0, 1].Proof. It follows from the definition of λ and the mean-value theorem that there exists t0 ∈ (0, 1)such that ∣∣ω′ (t0)∣∣ = |ω (1)− ω (0)| ≤ λ. (2.8)By (2.7), we know that there exists N > λ such that∫ Nλsh (s)ds > maxt∈[0,1]β (t)− mint∈[0,1]α (t) . (2.9)EJQTDE, 2010 No. 14, p. 4Now, we will prove that |ω′ (t)| ≤ N for any t ∈ [0, 1] . Suppose on the contrary that there existst1 ∈ [0, 1] such that ∣∣ω′ (t1)∣∣ > N. (2.10)In view of (2.8) and (2.10), we know that there exist t2, t3 ∈ (0, 1) with t2 < t3 such that one of thefollowing cases holds:Case 1. λ < ω′ (t) < N for t ∈ (t2, t3), ω′ (t2) = λ and ω′ (t3) = N ;Case 2. λ < ω′ (t) < N for t ∈ (t2, t3), ω′ (t2) = N and ω′ (t3) = λ;Case 3. −N < ω′ (t) < −λ for t ∈ (t2, t3), ω′ (t2) = −N and ω′ (t3) = −λ;Case 4. −N < ω′ (t) < −λ for t ∈ (t2, t3), ω′ (t2) = −λ and ω′ (t3) = −N.Since the others is similar, we only consider Case 1. By the Nagumo condition, we have∣∣ω′′ (t)∣∣ · ω′ (t) = ∣∣f(t, (Aω) (t) , (Bω) (t) , ω (t) , ω′ (t))∣∣ · ω′ (t)≤ h(∣∣ω′ (t)∣∣) · ω′ (t) , t ∈ [t2, t3] .So,|ω′′ (t)| · ω′ (t)h (ω′ (t))≤ ω′ (t) , t ∈ [t2, t3] ,and so, ∣∣∣∣∫ t3t2ω′′ (t) · ω′ (t)h (ω′ (t))dt∣∣∣∣ ≤∫ t3t2∣∣∣∣ω′′ (t) · ω′ (t)h (ω′ (t))∣∣∣∣ dt ≤∫ t3t2ω′ (t) dt,which implies that ∫ Nλsh (s)ds ≤ ω (t3)− ω (t2) ≤ maxt∈[0,1]β (t)− mint∈[0,1]α (t) ,which contradicts with (2.9) and the proof is complete. 3 Main resultTheorem 3.1 Assume that α and β are, respectively, the strict lower and the strict upper solution ofthe BVP (2.4) with α (t) ≤ β (t) for t ∈ [0, 1] , and f satisfies the Nagumo condition with respect to αand β. Then the BVP (2.4) has a solution v0 andα (t) ≤ v0 (t) ≤ β (t) for t ∈ [0, 1] .Proof. It follows from Lemma 2.3 that there exists N > 0 such that any solution ω of the BVP (2.4)lying in [α, β] satisfies ∣∣ω′ (t)∣∣ ≤ N for t ∈ [0, 1] .We denote C = max{N, maxt∈[0,1]|α′ (t)| , maxt∈[0,1]|β′ (t)|}and define the auxiliary functions f1, f2, f3and F : [0, 1] ×R4 → R as follows:f1(t, x1, x2, x3, x4) =f(t, x1, x2, x3, C), x4 > C,f(t, x1, x2, x3, x4), − C ≤ x4 ≤ C,f(t, x1, x2, x3,−C), x4 < −C;EJQTDE, 2010 No. 14, p. 5f2(t, x1, x2, x3, x4) =f1(t, (Aα) (t) , x2, x3, x4), x1 > (Aα) (t) ,f1(t, x1, x2, x3, x4), (Aβ) (t) ≤ x1 ≤ (Aα) (t) ,f1(t, (Aβ) (t) , x2, x3, x4), x1 < (Aβ) (t) ;f3(t, x1, x2, x3, x4) =f2(t, x1, (Bα) (t) , x3, x4), x2 > (Bα) (t) ,f2(t, x1, x2, x3, x4), (Bβ) (t) ≤ x2 ≤ (Bα) (t) ,f2(t, x1, (Bβ) (t) , x3, x4), x2 < (Bβ) (t)andF (t, x1, x2, x3, x4) =f3(t, x1, x2, β (t) , x4), x3 > β (t) ,f3(t, x1, x2, x3, x4), α (t) ≤ x3 ≤ β (t) ,f3(t, x1, x2, α (t) , x4), x3 < α (t) .Consider the following auxiliary BVPv′′(t) + F (t, (Av) (t) , (Bv) (t) , v (t) , v′ (t)) = 0, t ∈ [0, 1] ,v (0) =m−2∑i=1biv (ηi) , v′ (1) = 0.(3.1)If we define an operator T : X → X by(Tv) (t) =∫ 10G2 (t, s)F (s, (Av) (s) , (Bv) (s) , v (s) , v′ (s))ds, t ∈ [0, 1] ,then it is obvious that fixed points of T are solutions of the BVP (3.1). Now, we will apply theLeray-Schauder continuation principle to prove that the operator T has a fixed point. Since it is easyto verify that T : X → X is completely continuous by using the Arzela-Ascoli theorem, we only needto prove that the set of all possible solutions of the homotopy group problem v = λTv is a prioribounded in X by a constant independent of λ ∈ (0, 1) . Denoteαm = mint∈[0,1]α (t) , βM = maxt∈[0,1]β (t) ,(Aβ)m = mint∈[0,1](Aβ) (t) , (Aα)M = maxt∈[0,1](Aα) (t) ,(Bβ)m = mint∈[0,1](Bβ) (t) , (Bα)M = maxt∈[0,1](Bα) (t) ,L = sup {|f(t, x1, x2, x3, x4)| : (t, x1, x2, x3, x4) ∈ [0, 1] × [(Aβ)m , (Aα)M ]× [(Bβ)m , (Bα)M ]× [αm, βM ]× [−C,C]} .Let v = λTv. Then we have|v (t)| = |λ (Tv) (t)|= λ∣∣∣∣∫ 10G2 (t, s)F (s, (Av) (s) , (Bv) (s) , v (s) , v′ (s))ds∣∣∣∣≤∫ 10K (t, s) + 11−m−2∑i=1bim−2∑i=1biK (ηi, s)∣∣F (s, (Av) (s) , (Bv) (s) , v (s) , v′ (s))∣∣ ds≤L1−m−2∑i=1bi=: R, t ∈ [0, 1]EJQTDE, 2010 No. 14, p. 6and∣∣v′ (t)∣∣ = ∣∣λ (Tv)′ (t)∣∣= λ∣∣∣∣∫ 10∂G2 (t, s)∂tF (s, (Av) (s) , (Bv) (s) , v (s) , v′ (s))ds∣∣∣∣≤∫ 1t∣∣F (s, (Av) (s) , (Bv) (s) , v (s) , v′ (s))∣∣ ds≤ L ≤ R, t ∈ [0, 1] ,which imply that‖v‖ = max{‖v‖∞,∥∥v′∥∥∞}≤ R.It is now immediate from the Leray-Schauder continuation principle that the operator T has a fixedpoint v0, which solves the BVP (3.1).Now, let us prove that v0 is a solution of the BVP (2.4). Therefor, we only need to verify thatα (t) ≤ v0 (t) ≤ β (t) and |v′0 (t)| ≤ C for t ∈ [0, 1] .First, we will verify that v0 (t) ≤ β (t) for t ∈ [0, 1] . Suppose on the contrary that there existst0 ∈ [0, 1] such thatv0 (t0)− β (t0) = maxt∈[0,1]{v0 (t)− β (t)} > 0.We consider the following three cases:Case 1: If t0 ∈ (0, 1) , then v0 (t0) > β (t0) , v′0 (t0) = β′ (t0) and v′′0 (t0) ≤ β′′ (t0) . Since β is astrict upper solution of the BVP (2.4), one hasv′′0 (t0) = −F (t0, (Av0) (t0) , (Bv0) (t0) , v0 (t0) , v′0 (t0))= −f3(t0, (Av0) (t0) , (Bv0) (t0) , β (t0) , β′ (t0))= −f2(t0, (Av0) (t0) , (Bβ) (t0) , β (t0) , β′ (t0))= −f1(t0, (Aβ) (t0) , (Bβ) (t0) , β (t0) , β′ (t0))= −f(t0, (Aβ) (t0) , (Bβ) (t0) , β (t0) , β′ (t0))> β′′ (t0) ,which is a contradiction.Case 2: If t0 = 0, then v0 (0) > β (0) . On the other hand, v0 (0) =m−2∑i=1biv0 (ηi) ≤m−2∑i=1biβ (ηi) ≤β (0) . This is a contradiction.Case 3: If t0 = 1, then v0 (1)− β (1) = maxt∈[0,1]{v0 (t)− β (t)} > 0, which shows that v′0 (1) ≥ β′ (1) .On the other hand, v′0 (1) = 0 ≤ β′ (1) . Consequently, v′0 (1) = β′ (1) , and so, v′′0 (1) ≤ β′′ (1) . Withthe similar arguments as in Case 1, we can obtain a contradiction also.Thus, v0 (t) ≤ β (t) for t ∈ [0, 1] . Similarly, we can prove that α (t) ≤ v0 (t) for t ∈ [0, 1] .Next, we will show that |v′0 (t)| ≤ C for t ∈ [0, 1] . In fact, since f satisfies the Nagumo conditionwith respect to α and β, with the similar arguments as in Lemma 2.3, we can obtain that∣∣v′0 (t)∣∣ ≤ N ≤ C for t ∈ [0, 1] .Therefore, v0 is a solution of the BVP (2.4) and α (t) ≤ v0 (t) ≤ β (t) for t ∈ [0, 1] . EJQTDE, 2010 No. 14, p. 7References[1] C. Z. Bai, Triple positive solutions of three-point boundary value problems for fourth-order dif-ferential equations, Computers and Mathematics with Applications 56 (2008) 1364-1371.[2] C. Z. Bai, D. D. Yang and H. B. Zhu, Existence of solutions for fourth order differential equationwith four-point boundary conditions, Appl. Math. Lett. 20 (2007) 1131-1136.[3] Z. B. 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Wang, Existence results for a fourth-order ordinary differentialequation with a four-point boundary condition, Appl. Math. Lett. 21 (2008) 465-470.(Received October 8, 2009)EJQTDE, 2010 No. 14, p. 8

Existence of solutions for a class of fourth-order m-point boundary value problems

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