This paper is concerned with the following nonlinear third-order three-point boundary value problem

\[\left\{
\begin{array}{l}
u^{\prime \prime \prime }(t)+f\left( t,u\left( t\right) ,u^{\prime}\left(t\right) \right) =0,\, t\in \left[ 0,1\right], \\
u\left( 0\right) =u^{\prime }\left( 0\right) =0,\, u^{\prime}\left( 1\right) =\alpha u^{\prime }\left( \eta \right),\label{1.1}
\end{array}
\right.\]

where $0<\eta <1$ and $0\leq \alpha <1.$ A new maximum principle is established and some existence criteria are obtained for the above problem by using the upper and lower solution method

Ren, Qiu-Yan

Sun, Jian-Ping

Zhao, Ya-Hong

Repository of the Academy's Library

Electronic Journal of Qualitative Theory of Differential Equations2010, No. 26, 1-8; http://www.math.u-szeged.hu/ejqtde/The upper and lower solution method for nonlinearthird-order three-point boundary value problem∗Jian-Ping Sun, Qiu-Yan Ren and Ya-Hong ZhaoDepartment of Applied Mathematics, Lanzhou University of Technology,Lanzhou, Gansu, 730050, People’s Republic of ChinaE-mail: jpsun@lut.cn (Sun)AbstractThis paper is concerned with the following nonlinear third-order three-point boundary valueproblem {u′′′(t) + f (t, u (t) , u′ (t)) = 0, t ∈ [0, 1] ,u (0) = u′ (0) = 0, u′ (1) = αu′ (η) ,where 0 < η < 1 and 0 ≤ α < 1. A new maximum principle is established and some existencecriteria are obtained for the above problem by using the upper and lower solution method.Keywords: Third-order three-point boundary value problem; Upper and lower solution method;Maximum principle; Existence2000 AMS Subject Classification: 34B10, 34B151 IntroductionThird-order differential equations arise in a variety of different areas of applied mathematics andphysics, e.g., in the deflection of a curved beam having a constant or varying cross section, a three-layerbeam, electromagnetic waves or gravity driven flows and so on [6].Recently, third-order boundary value problems (BVPs for short) have received much attention.For instance, [3, 4, 5, 8, 9, 13] discussed some third-order two-point BVPs, while [2, 7, 10, 11, 12]studied some third-order three-point BVPs. In particular, Feng and Liu [5] employed the upper andlower solution method to prove the existence of solution for the third-order two-point BVP{u′′′(t) + f (t, u (t) , u′ (t)) = 0, t ∈ [0, 1] ,u (0) = u′ (0) = u′ (1) = 0.In 2008, Guo, Sun and Zhao [7] established some existence results for at least one positive solution tothe third-order three-point BVP{u′′′(t) + a (t) f (u (t)) = 0, t ∈ (0, 1) ,u (0) = u′ (0) = 0, u′ (1) = αu′ (η) .∗Supported by the National Natural Science Foundation of China (10801068).EJQTDE, 2010 No. 26, p. 1Their main tool was the well-known Guo-Krasnoselskii fixed point theorem.Motivated greatly by [5, 7], in this paper, we will investigate the following nonlinear third-orderthree-point BVP {u′′′(t) + f (t, u (t) , u′ (t)) = 0, t ∈ [0, 1] ,u (0) = u′ (0) = 0, u′ (1) = αu′ (η) ,(1.1)where 0 < η < 1 and 0 ≤ α < 1. A new maximum principle is established and some existence criteriaare obtained for the BVP (1.1) by using the upper and lower solution method. In order to obtain ourmain results, we need the following fixed point theorem [1].Theorem 1.1 Let (E,K) be an ordered Banach space and [a, b] be a nonempty interval in E. IfT : [a, b]→ E is an increasing compact mapping and a ≤ Ta, T b ≤ b, then T has a fixed point in [a, b].2 PreliminariesIn this section, we will present some fundamental definitions and several important lemmas.Definition 2.1 If x ∈ C3 [0, 1] satisfies{x′′′(t) + f (t, x (t) , x′ (t)) ≥ 0, t ∈ [0, 1] ,x (0) = 0, x′ (0) ≤ 0, x′ (1) ≤ αx′ (η) ,then x is called a lower solution of the BVP (1.1) .Definition 2.2 If y ∈ C3 [0, 1] satisfies{y′′′(t) + f (t, y (t) , y′ (t)) ≤ 0, t ∈ [0, 1] ,y (0) = 0, y′ (0) ≥ 0, y′ (1) ≥ αy′ (η) ,then y is called an upper solution of the BVP (1.1).Let G (t, s) be the Green′s function of the second-order three-point BVP{−u′′ (t) = 0, t ∈ [0, 1] ,u (0) = 0, u (1) = αu (η) .ThenG(t, s) =11− αηs (1− αη) + st (α− 1) , s ≤ min {η, t} ,t (1− αη) + st (α− 1) , t ≤ s ≤ η,s (1− αη) + t (αη − s) , η ≤ s ≤ t,t (1− s) , max {η, t} ≤ s.For G (t, s), we have the following two lemmas.Lemma 2.1 G(t, s) ≥ 0 for (t, s) ∈ [0, 1] × [0, 1] .Lemma 2.2 Let M =: maxt∈[0,1]∫ 10 G (t, s) ds. Then M <12.EJQTDE, 2010 No. 26, p. 2Proof. Since a simple computation shows that∫ 10G (t, s) ds = −12t2 +1− αη22 (1− αη)t,it is easy to obtain thatM =18(1− αη21− αη)2, αη (2− η) ≤ 1,αη (1− η)2 (1− αη), αη (2− η) ≥ 1,which implies that M <12. Lemma 2.3 Assume that λ1 and λ2 are two nonnegative constants with λ1 + λ2 ≤ 2. If m ∈ C2 [0, 1]satisfiesm′′ (t) ≥ λ1∫ t0m (s) ds+ λ2m (t) for t ∈ [0, 1]andm (0) ≤ 0, m (1) ≤ αm (η) ,then m (t) ≤ 0 for t ∈ [0, 1].Proof. We consider two cases: λ1 = 0 and λ1 6= 0.Case 1. λ1 = 0. If λ2 = 0, then m′′ (t) ≥ 0 for t ∈ [0, 1], which implies that the graph of m (t) isconcave up. Sincem (0) ≤ 0, we only need to prove m (1) ≤ 0. Suppose on the contrary thatm (1) > 0.Then m (0) ≤ 0 < m (1) ≤ αm (η) < m (η) . So,0 <m (η)−m (0)η≤m (1)−m (η)1− η< 0,which is a contradiction. Therefore, m (1) ≤ 0.If λ2 6= 0, then m′′ (t) ≥ λ2m (t) for t ∈ [0, 1]. Suppose on the contrary that there exists t0 ∈ [0, 1]such that m (t0) = maxt∈[0,1]m (t) > 0. Obviously, t0 6= 0. If t0 = 1, then 0 < m (1) ≤ αm (η) <m (η) ≤ m (1) , which is a contradiction. Consequently, t0 ∈ (0, 1), m′ (t0) = 0 and m′′ (t0) ≤ 0, whichcontradicts m′′ (t0) ≥ λ2m (t0) > 0.Case 2. λ1 6= 0. Suppose on the contrary that there exists t0 ∈ [0, 1] such that m0 = m (t0) =maxt∈[0,1]m (t) > 0. Similarly, we can obtain that t0 ∈ (0, 1), m′ (t0) = 0 and m′′ (t0) ≤ 0. Consequently,0 ≥ m′′ (t0) ≥ λ1∫ t00 m (s) ds+λ2m (t0) , which implies that∫ t00 m (s) ds ≤ 0. So, there exists t1 ∈ [0, t0)such that m1 = m (t1) = mint∈[0,t0]m (t) < 0. It follows from Taylor′s formula that there exists ξ ∈ (t1, t0)such thatm1 = m (t1) = m (t0) +m′ (t0) (t1 − t0) +m′′ (ξ)2(t1 − t0)2 .Noting that m1 < 0, we obtainm′′ (ξ) =2 (m1 −m0)(t1 − t0)2 <2m1(t1 − t0)2 < 2m1.EJQTDE, 2010 No. 26, p. 3And so,2m1 > m′′ (ξ) ≥ λ1∫ ξ0m (s) ds+ λ2m (ξ) ≥ λ1ξm1 + λ2m1,which implies that λ1 + λ2 > 2. This contradicts the fact that λ1 + λ2 ≤ 2. 3 Main resultsIn the remainder of this paper, we always assume that the following condition is satisfied:(H) f : [0, 1] × R2 → R is continuous and there exist two nonnegative constants λ1 and λ2 withλ1 + λ2 ≤ 2 such thatf (t, u1, v1)− f (t, u2, v2) ≥ −λ1 (u1 − u2)− λ2 (v1 − v2)for t ∈ [0, 1] , u1 ≥ u2 and v1 ≥ v2.Theorem 3.1 If the BVP (1.1) has a lower solution x and an upper solution y with x′ (t) ≤ y′ (t) fort ∈ [0, 1] , then the BVP (1.1) has a solution u ∈ C3 [0, 1], which satisfiesx′ (t) ≤ u′ (t) ≤ y′ (t) for t ∈ [0, 1] .Proof. Let v (t) = u′ (t) . Then the BVP (1.1) is equivalent to the following BVP{v′′ (t) + f(t,∫ t0 v (s) ds, v (t))= 0, t ∈ [0, 1] ,v (0) = 0, v (1) = αv (η) .(3.1)Let 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. Now, if we define operators L : D ⊂E → E and N : E → E as follows:Lv = −v′′ (t) + λ1∫ t0v (s) ds+ λ2v (t)andNv = f(t,∫ t0v (s) ds, v (t))+ λ1∫ t0v (s) ds+ λ2v (t) ,where D = {v ∈ E : v′′ ∈ E, v (0) = 0 and v (1) = αv (η)} , then it is easy to see that the BVP (3.1)is equivalent to the operator equationLv = Nv. (3.2)Now, we shall show that the operator equation (3.2) is solvable. The proof will be given in severalsteps.Step 1. L : D ⊂ E → E is invertible.EJQTDE, 2010 No. 26, p. 4Suppose h ∈ E. We will find unique v ∈ D such that Lv = h. Since Lv = h is equivalent to theintegral equationv (t) =∫ 10G (t, s)[h (s)−(λ1∫ s0v (r) dr + λ2v (s))]ds, t ∈ [0, 1],we define a mapping A : E → E by(Av) (t) =∫ 10G (t, s)[h (s)−(λ1∫ s0v (r) dr + λ2v (s))]ds, t ∈ [0, 1].Noting that M <12and 0 ≤ λ1+λ2 ≤ 2, it is easy to verify that A : E → E is a contraction mapping.And so, there exists unique v ∈ D such that Av = v, which implies that Lv = h. This shows that Lis invertible.Step 2. L−1 : E → E is continuous.Assume that {hn}∞n=1 ⊂ E, h ∈ E and limn→∞ hn = h. Denote L−1hn = vn and L−1h = v. Thenvn (t) =∫ 10G (t, s)[hn (s)−(λ1∫ s0vn (r) dr + λ2vn (s))]ds, t ∈ [0, 1]andv (t) =∫ 10G (t, s)[h (s)−(λ1∫ s0v (r) dr + λ2v (s))]ds, t ∈ [0, 1].So,‖vn − v‖ = maxt∈[0,1]∣∣∣∣∫ 10G (t, s)[(hn (s)− h (s))− λ1∫ s0(vn (r)− v (r)) dr − λ2 (vn (s)− v (s))]ds∣∣∣∣≤ maxt∈[0,1]∫ 10G (t, s) [‖hn − h‖+ (λ1 + λ2) ‖vn − v‖] ds≤ M ‖hn − h‖+ 2M ‖vn − v‖ .This shows that ‖vn − v‖ ≤M1− 2M‖hn − h‖ , which together with limn→∞ hn = h implies thatlimn→∞ vn = v. This indicates that L−1 : E → E is continuous.Step 3. L−1N : E → E is completely continuous.Since f and L−1 are continuous, we only need to prove that L−1 : E → E is compact. Let X be abounded subset in E. Then there exists a constant C > 0 such that ‖h‖ ≤ C for any h ∈ X. For anyv ∈ L−1 (X), there exists an h ∈ X such that v = L−1h. So,v (t) =∫ 10G (t, s)[h (s)−(λ1∫ s0v (r) dr + λ2v (s))]ds, t ∈ [0, 1].On the one hand, for any v ∈ L−1 (X), we have‖v‖ = maxt∈[0,1]∣∣∣∣∫ 10G (t, s)[h (s)−(λ1∫ s0v (r) dr + λ2v (s))]ds∣∣∣∣≤ maxt∈[0,1]∫ 10G (t, s) [‖h‖+ (λ1 + λ2) ‖v‖] ds≤ M ‖h‖+ 2M ‖v‖ ,EJQTDE, 2010 No. 26, p. 5which implies that ‖v‖ ≤M1− 2M‖h‖ ≤MC1− 2M. This shows that L−1 (X) is uniformly bounded.On the other hand, in view of the uniform continuity of G (t, s) , we know that for any ǫ > 0, thereexists a δ > 0 such that for any t1, t2 ∈ [0, 1] and |t1 − t2| < δ, |G (t1, s)−G (t2, s)| <1− 2MCǫ forany s ∈ [0, 1]. Then for any v ∈ L−1 (X) , t1, t2 ∈ [0, 1] and |t1 − t2| < δ, we have|v (t1)− v (t2)| =∣∣∣∣∫ 10(G (t1, s)−G (t2, s))[h (s)−(λ1∫ s0v (r) dr + λ2v (s))]ds∣∣∣∣≤∫ 10|G (t1, s)−G (t2, s)| [‖h‖+ (λ1 + λ2) ‖v‖] ds≤C1− 2M∫ 10|G (t1, s)−G (t2, s)| ds< ǫ,which shows that L−1 (X) is equicontinuous.By the Arzela-Ascoli theorem, we know that L−1 (X) is relatively compact, which implies thatL−1 : E → E is a compact mapping.Step 4. L−1N : E → E is increasing.Suppose h1, h2 ∈ E and h1 ≤ h2. Then the condition (H) implies that Nh1 ≤ Nh2. Denotev1 = L−1Nh1 and v2 = L−1Nh2. Then Lv1 = Nh1 ≤ Nh2 = Lv2. It follows from Lemma 2.3 thatv1 ≤ v2. Therefore, L−1N : E → E is increasing.Step 5. Let β0 = x′ and γ0 = y′. Then β0 ≤ L−1Nβ0 and L−1Nγ0 ≤ γ0.Since x is a lower solution of the BVP (1.1), we have−β′′0 (t) + λ1∫ t0β0 (s) ds+ λ2β0 (t) ≤ (Nβ0) (t) , t ∈ [0, 1], β0 (0) ≤ 0, β0 (1) ≤ αβ0 (η) . (3.3)Let β∗ = L−1Nβ0. Then Lβ∗ = Nβ0, that is,−(β∗)′′ (t) + λ1∫ t0β∗ (s) ds + λ2β∗ (t) = (Nβ0) (t) , t ∈ [0, 1], β∗ (0) = 0, β∗ (1) = αβ∗ (η) . (3.4)Denote q (t) = β0 (t)− β∗ (t) . In view of (3.3) and (3.4), we know that−q′′ (t) + λ1∫ t0q (s) ds+ λ2q (t) ≤ 0, t ∈ [0, 1], q (0) ≤ 0, q (1) ≤ αq (η) .By Lemma 2.3, we get q (t) ≤ 0 for t ∈ [0, 1], i.e., β0 ≤ β∗ = L−1Nβ0. Similarly, we can obtain thatL−1Nγ0 ≤ γ0.It follows from Theorem 1.1 that L−1N : E → E has a fixed point v ∈ [β0, γ0], which solves the BVP(3.1). Therefore, u (t) =∫ t0 v (s) ds, t ∈ [0, 1] is a solution of the BVP (1.1) and x′ (t) ≤ u′ (t) ≤ y′ (t)for t ∈ [0, 1] . Corollary 3.2 (1) If mint∈[0,1]f (t, 0, 0) ≥ 0 and there exists c > 0 such thatmax{f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, c] ×[0,3c2]}≤ 3c,EJQTDE, 2010 No. 26, p. 6then the BVP (1.1) has a nonnegative solution u with ‖u‖ ≤ c. Moreover, if there exists tn ∈ (0, 1](n = 1, 2, · · · ) satisfying limn→∞ tn = 0 such that f (tn, 0, 0) > 0 (n = 1, 2, · · · ), then u (t) > 0 fort ∈ (0, 1] .(2) If maxt∈[0,1]f (t, 0, 0) ≤ 0 and there exists c > 0 such thatmin{f (t, u, v) : (t, u, v) ∈ [0, 1]× [−c, 0] ×[−3c2, 0]}≥ −3c,then the BVP (1.1) has a nonpositive solution u with ‖u‖ ≤ c. Moreover, if there exists tn ∈ (0, 1](n = 1, 2, · · · ) satisfying limn→∞ tn = 0 such that f (tn, 0, 0) < 0 (n = 1, 2, · · · ), then u (t) < 0 fort ∈ (0, 1] .(3) If there exists c > 0 such thatmax{|f (t, u, v)| : (t, u, v) ∈ [0, 1] × [−c, c] ×[−3c2,3c2]}≤ 3c,then the BVP (1.1) has a solution u with ‖u‖ ≤ c. Moreover, if f(t, 0, 0) is not identically zero on[0, 1], then u is nontrivial.Proof. Since the proof of (2) and (3) is similar, we only prove (1). Let x (t) ≡ 0 and y (t) =3c(t22−t36)for t ∈ [0, 1]. Then it is easy to verify that x and y are lower and upper solutions ofthe BVP (1.1), respectively. By Theorem 3.1, we know that the BVP (1.1) has a solution u satisfyingx′ (t) ≤ u′ (t) ≤ y′ (t) for t ∈ [0, 1] , which together with x(0) = u(0) = y(0) = 0 implies that u isnonnegative and ‖u‖ ≤ c.If there exists tn ∈ (0, 1] (n = 1, 2, · · · ) satisfying limn→∞ tn = 0 such that f (tn, 0, 0) > 0 (n =1, 2, · · · ), then it is not difficult to prove that for any ǫ ∈ (0, 1) , u(t) is not identically zero on [0, ǫ]. Inview of u(0) = 0 and u′ (t) ≥ 0 for t ∈ [0, 1] , we know that u (t) > 0 for t ∈ (0, 1] . 4 An exampleConsider the following BVP:{u′′′(t) + teu(t) + 8135 (u′(t))3 = 0, t ∈ [0, 1] ,u (0) = u′ (0) = 0, u′ (1) = 14u′(12).(4.1)Since f(t, u, v) = teu+ 8135v3, it is easy to verify that the condition (H) is fulfilled with λ1 = λ2 = 1.If we let c = 1, then all the conditions of Corollary 3.2 (1) are satisfied. It follows from Corollary 3.2(1) that the BVP (4.1) has a positive solution u and ‖u‖ ≤ 1.References[1] H. 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Kosmatov, Third-order boundary value problems with sign-changing solutions,Nonlinear Anal. 67 (2007) 126-137.[9] Z. Liu, L. Debnath, S. M. Kang, Existence of monotone positive solutions to a third-order two-point generalized right focal boundary value problem, Comput. Math. Appl. 55 (2008) 356-367.[10] R. Ma, Multiplicity results for a third order boundary value problem at resonance, NonlinearAnal. 32 (1998) 493-499.[11] Y. Sun, Positive solutions of singular third-order three-point boundary value problem, J. Math.Anal. Appl. 306 (2005) 589-603.[12] Q. Yao, Positive solutions of singular third-order three-point boundary value problems, J. Math.Anal. Appl. 354 (2009) 207-212.[13] Q. Yao, Y. Feng, The existence of solution for a third-order two-point boundary value problem,Appl. Math. Lett. 15 (2002) 227-232.(Received December 17, 2009)EJQTDE, 2010 No. 26, p. 8

The upper and lower solution method for nonlinear third-order three-point boundary value problem

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