By using the Guo-Krasnoselskii fixed point theorem, we investigate the following third-order three-point boundary value problem
\[
\left\{
\begin{array}{l}
u'''(t)=f(t,u(t)),\ t\in [0,1], \\
u'(0)=u(1)=0,\ u''(\eta)+\alpha u(0)=0,
\end{array}
\right.
\]
where $\alpha \in [0,2)$ and $\eta\in[\frac{\sqrt{121+24\alpha}-5}{3(4+\alpha)},1)$. The emphasis is mainly that although the corresponding Green's function is sign-changing, the solution obtained is still positive

Kong, Fang-Di

Li, Xing-Long

Sun, Jian-Ping

Electronic journal of qualitative theory of differential equations

Xing-Long Li

Jian-Ping Sun

Fang-Di Kong

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Existence of positive solution for a third-order three-point BVP with sign-changing Green's function

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Electronic Journal of Qualitative Theory of Differential Equations2013, No. 30, 1-11; http://www.math.u-szeged.hu/ejqtde/Existence of positive solution for a third-orderthree-point BVP with sign-changing Green’sfunction∗Xing-Long Li, Jian-Ping Sun†, Fang-Di KongDepartment of Applied Mathematics, Lanzhou University of Technology,Lanzhou, Gansu 730050, People’s Republic of ChinaAbstractBy using the Guo-Krasnoselskii fixed point theorem, we investigate the following third-order three-point boundary value problemu′′′(t) = f(t, u(t)), t ∈ [0, 1],u′(0) = u(1) = 0, u′′(η) + αu(0) = 0,where α ∈ [0, 2) and η ∈ [√121+24α−53(4+α) , 1). The emphasis is mainly that although the cor-responding Green’s function is sign-changing, the solution obtained is still positive.Keywords: Third-order three-point boundary value problem; Sign-changing Green’sfunction; Positive solution; Existence; Fixed point2010 AMS Subject Classification: 34B10, 34B18∗Supported by the Natural Science Foundation of Gansu Province of China (1208RJZA240).†Corresponding author. E-mail: jpsun@lut.cnEJQTDE, 2013 No. 30, p. 11 IntroductionThird-order differential equations arise from a variety of different areas of applied mathe-matics and physics, e.g., in the deflection of a curved beam having a constant or varying crosssection, a three-layer beam, electromagnetic waves or gravity driven flows and so on [3].Recently, the existence of single or multiple positive solutions to some third-order three-pointboundary value problems (BVPs for short) has received much attention from many authors,see [1, 2, 5, 12, 15, 16] and the references therein.However, all the above-mentioned papers are achieved when the corresponding Green’sfunctions are positive, which is a very important condition. A natural question is that whetherwe can obtain the existence of positive solutions to some third-order three-point BVPs whenthe corresponding Green’s functions are sign-changing.In 2008, Palamides and Smyrlis [11] studied the existence of at least one positive solutionto the singular third-order three-point BVP with an indefinitely signed Green’s functionu′′′(t) = a(t)f(t, u(t)), t ∈ (0, 1),u(0) = u(1) = u′′(η) = 0,where η ∈(1724, 1). Their technique was a combination of the Guo-Krasnoselskii fixed pointtheorem and properties of the corresponding vector field.In 2012, by using the Guo-Krasnoselskii and Leggett-Williams fixed point theorems, Sun andZhao [13, 14] discussed the third-order three-point BVP with sign-changing Green’s functionu′′′(t) = f(t, u(t)), t ∈ [0, 1],u′(0) = u(1) = u′′(η) = 0,(1.1)where η ∈ (12, 1). They obtained the existence of single or multiple positive solutions to theBVP (1.1) and proved that the obtained solutions were concave on [0, η] and convex on [η, 1].It is worth mentioning that there are other type of works on sign-changing Green’s functionswhich prove the existence of sign-changing solutions, positive in some cases, see Infante andWebb’s papers [6–8].EJQTDE, 2013 No. 30, p. 2In this paper we study the following third-order three-point BVPu′′′(t) = f(t, u(t)), t ∈ [0, 1],u′(0) = u(1) = 0, u′′(η) + αu(0) = 0.(1.2)Throughout this paper, we always assume that α ∈ [0, 2) and η ∈ [√121+24α−53(4+α), 1). Obviously,the BVP (1.1) is a special case of the BVP (1.2). However, it is necessary to point out thatthis paper is not a simple extension of [13]. In fact, if we let α = 0, then η ∈ [12, 1), which isdifferent from the restriction in [13]. On the other hand, compared with [13], we can only provethat the obtained solution is concave on [0, η].Our main tool is the following well-known Guo-Krasnoselskii fixed point theorem [4, 9]:Theorem 1.1 Let E be a Banach space and K be a cone in E. Assume that Ω1 and Ω2 arebounded open subsets of E such that 0 ∈ Ω1, Ω1 ⊂ Ω2, and let T : K ∩ (Ω2\Ω1) → K be acompletely continuous operator such that either(1) ‖Tu‖ ≤ ‖u‖ for u ∈ K ∩ ∂Ω1 and ‖Tu‖ ≥ ‖u‖ for u ∈ K ∩ ∂Ω2, or(2) ‖Tu‖ ≥ ‖u‖ for u ∈ K ∩ ∂Ω1 and ‖Tu‖ ≤ ‖u‖ for u ∈ K ∩ ∂Ω2.Then T has a fixed point in K ∩ (Ω2 \ Ω1).2 PreliminariesFor the BVP u′′′(t) = 0, t ∈ [0, 1],u′(0) = u(1) = 0, u′′(η) + αu(0) = 0,(2.1)we have the following lemma.Lemma 2.1 The BVP (2.1) has only trivial solution.Proof. It is simple to check. In the remainder of this paper, we always assume that Banach space C [0, 1] is equippedwith the norm ‖u‖ = maxt∈[0,1]|u(t)|.EJQTDE, 2013 No. 30, p. 3Now, for any y ∈ C [0, 1], we consider the BVPu′′′(t) = y(t), t ∈ [0, 1],u′(0) = u(1) = 0, u′′(η) + αu(0) = 0.(2.2)After a direct computation, one may obtain the expression of Green’s function G(t, s) ofthe BVP (2.2) as follows:G(t, s) = g1(t, s) + g2(t, s) + g3(η, t, s),whereg1(t, s) = −(2− αt2)(1− s)22(2− α), (t, s) ∈ [0, 1]× [0, 1],g2(t, s) =0, 0 ≤ t ≤ s ≤ 1,(t−s)22, 0 ≤ s ≤ t ≤ 1andg3(η, t, s) =0, s ≥ η,1−t22−α , s < η.It is not difficult to verify that the G(t, s) has the following properties:G(t, s) ≥ 0 for 0 ≤ s ≤ η and G(t, s) ≤ 0 for η ≤ s ≤ 1.Moreover, for s ≥ η,max{G(t, s) : t ∈ [0, 1]} = G(1, s) = 0,min{G(t, s) : t ∈ [0, 1]} = G(0, s) = −(1− s)22− αand for s < η,max{G(t, s) : t ∈ [0, 1]} = G(0, s) =2s− s22− α,min{G(t, s) : t ∈ [0, 1]} = G(1, s) = 0.LetK0 = {y ∈ C [0, 1] : y(t) is nonnegative and decreasing on [0, 1]} .Then K0 is a cone in C [0, 1].EJQTDE, 2013 No. 30, p. 4Lemma 2.2 Let y ∈ K0 and u(t) =∫ 10G(t, s)y(s)ds, t ∈ [0, 1]. Then u is the unique solutionof the BVP (2.2) and u ∈ K0. Moreover, u(t) is concave on [0, η].Proof. For 0 ≤ t ≤ η, we haveu(t) =∫ t0[g1(t, s) +(t− s)22+1− t22− α]y(s)ds+∫ ηt[g1(t, s) +1− t22− α]y(s)ds+∫ 1ηg1(t, s)y(s)ds.Since η ≥√121+24α−53(4+α)implies that η ≥ 2α3α+6, we getu′(t) = −αt2− α∫ η0(2s− s2)y(s)ds−∫ t0sy(s)ds− t∫ ηty(s)ds+αt2− α∫ 1η(1− s)2y(s)ds≤ y(η)[−αt2− α∫ η0(2s− s2)ds−∫ t0sds− t∫ ηtds+αt2− α∫ 1η(1− s)2ds]= ty(η)[α(1− 3η)3(2− α)− η +t2]≤ ty(η)[α(1− 3η)3(2− α)−η2]≤ 0.At the same time, η ≥√121+24α−53(4+α)> 13shows thatu′′(t) = −α2− α∫ η0(2s− s2)y(s)ds−∫ ηty(s)ds+α2− α∫ 1η(1− s)2y(s)ds≤ −αy(η)2− α∫ η0(2s− s2)ds− y(η)∫ ηtds+αy(η)2− α∫ 1η(1− s)2ds≤αy(η)(1− 3η)3(2− α)≤ 0.For η < t ≤ 1, we haveu(t) =∫ η0[g1(t, s) +(t− s)22+1− t22− α]y(s)ds+∫ tη[g1(t, s) +(t− s)22]y(s)ds+∫ 1tg1(t, s)y(s)ds.EJQTDE, 2013 No. 30, p. 5Since η ≥√121+24α−53(4+α)implies that η ≥ 6−α12, we getu′(t) = −αt2− α∫ η0(2s− s2)y(s)ds+∫ tη(t− s)y(s)ds−∫ η0sy(s)ds+αt2− α∫ 1η(1− s)2y(s)ds≤ −αty(η)2− α∫ η0(2s− s2)ds+y(η)(η − t)22− y(η)∫ η0sds+αty(η)(1− η)33(2− α)= ty(η)[α(1− 3η)3(2− α)+t− 2η2]≤ 0.Obviously, u′′′(t) = y(t) for t ∈ [0, 1], u′(0) = u(1) = 0 and u′′(η) + αu(0) = 0. This showsthat u is a solution of the BVP (2.2). The uniqueness follows immediately from Lemma 2.1.Since u′(t) ≤ 0 for t ∈ [0, 1] and u(1) = 0, we have u(t) ≥ 0 for t ∈ [0, 1]. So, u ∈ K0. In viewof u′′(t) ≤ 0 for t ∈ [0, η], we know that u(t) is concave on [0, η]. Lemma 2.3 Let y ∈ K0. Then the unique solution u of the BVP (2.2) satisfiesmint∈[0,θ]u(t) ≥ θ∗ ‖u‖ ,where θ ∈ (0, 13] and θ∗ = η−θη.Proof. By Lemma 2.2, we know that u(t) is concave on [0, η], thus for t ∈ [0, η],u(t) ≥ (1−tη)u(0) +tηu(η). (2.3)In view of u ∈ K0, we know that ‖u‖ = u(0), which together with (2.3) implies thatu(t) ≥η − tη‖u‖ , 0 ≤ t ≤ η.Consequently,mint∈[0,θ]u(t) = u(θ) ≥η − θη‖u‖ = θ∗ ‖u‖ .EJQTDE, 2013 No. 30, p. 63 Main resultsFor convenience, we denoteA =∫ η0G(0, s)ds and B =∫ θ0G(η, s)ds.Then it is obvious that 0 < B < A.Theorem 3.1 Assume that f : [0, 1] × [0,+∞) → [0,+∞) is continuous and satisfies thefollowing conditions:(H1) For each u ∈ [0,+∞), the mapping t 7→ f(t, u) is decreasing;(H2) For each t ∈ [0, 1], the mapping u 7→ f(t, u) is increasing;(H3) There exist two positive constants r and R with r 6= R such thatf(0, r) ≤rAand f(θ, θ∗R) ≥RB.Then the BVP (1.2) has a positive and decreasing solution u satisfying min{r, R} ≤ ‖u‖ ≤max{r, R}. Moreover, the obtained solution u(t) is concave on [0, η].Proof. LetK ={u ∈ K0 : mint∈[0,θ]u(t) ≥ θ∗ ‖u‖}.Then it is easy to see that K is a cone in C [0, 1]. Now, we define an operator T on K by(Tu)(t) =∫ 10G(t, s)f(s, u(s))ds, t ∈ [0, 1].Obviously, if u is a fixed point of T in K, then u is a nonnegative and decreasing solution ofthe BVP (1.2). In what follows, we will seek a fixed point of T in K by using Theorem 1.1.First, by Lemma 2.2 and Lemma 2.3, we know that T : K → K. Furthermore, althoughG(t, s) is not continuous, it follows from known textbook results, for example see [10], thatT : K → K is completely continuous.Next, for any u ∈ K, we claim that∫ ηθG(η, s)f(s, u(s))ds+∫ 1ηG(η, s)f(s, u(s))ds ≥ 0. (3.1)EJQTDE, 2013 No. 30, p. 7In fact, if u ∈ K, recall that G(t, s) ≥ 0 for 0 ≤ s ≤ η and G(t, s) ≤ 0 for η ≤ s ≤ 1, thenit follows from η ≥√121+24α−53(4+α)that∫ ηθG(η, s)f(s, u(s))ds+∫ 1ηG(η, s)f(s, u(s))ds≥ f(η, u(η))[∫ ηθG(η, s)ds+∫ 1ηG(η, s)ds]= f(η, u(η))[∫ ηθ(g1(η, s) +(η − s)22+1− η22− α)ds+∫ 1ηg1(η, s)ds]=(1− η)f(η, u(η))6(2− α)[(4 + α)η2 + (4 + αθ3 − 3αθ2)η − 6θ2 + αθ3 − 2]≥(1− η)f(η, u(η))6(2− α)[(4 + α)η2 +103η −83]≥ 0.Now, without loss of generality, we assume that r < R. LetΩ1 = {u ∈ C [0, 1] : ‖u‖ < r} and Ω2 = {u ∈ C [0, 1] : ‖u‖ < R} .For any u ∈ K ∩ ∂Ω1, we get 0 ≤ u(s) ≤ r for s ∈ [0, 1], which together with (H3) impliesthat0 ≤ (Tu)(t) ≤∫ η0maxt∈[0,1]G(t, s)f(s, u(s))ds+∫ 1ηmaxt∈[0,1]G(t, s)f(s, u(s))ds=∫ η0G(0, s)f(s, u(s))ds≤∫ η0G(0, s)f(0, r)ds≤ r = ‖u‖ , t ∈ [0, 1].This shows that‖Tu‖ ≤ ‖u‖ for u ∈ K ∩ ∂Ω1. (3.2)For any u ∈ K ∩ ∂Ω2, we get θ∗R ≤ u(s) ≤ R for s ∈ [0, θ], which together with (3.1) andEJQTDE, 2013 No. 30, p. 8(H3) implies thatTu(η) =∫ 10G(η, s)f(s, u(s))ds=∫ θ0G(η, s)f(s, u(s))ds+∫ ηθG(η, s)f(s, u(s))ds+∫ 1ηG(η, s)f(s, u(s))ds≥∫ θ0G(η, s)f(s, u(s))ds≥∫ θ0G(η, s)f(θ, θ∗R)ds≥ R = ‖u‖ ,This indicates that‖Tu‖ ≥ ‖u‖ for u ∈ K ∩ ∂Ω2. (3.3)Therefore, it follows from Theorem 1.1, (3.2) and (3.3) that the operator T has a fixed pointu ∈ K ∩ (Ω2 \ Ω1), which is a desired positive and decreasing solution of the BVP (1.2) withr ≤ ‖u‖ ≤ R. Moreover, similar to the proof of Lemma 2.2, we can prove that the obtainedsolution u(t) is concave on [0, η]. Example 3.2 We consider the BVPu′′′(t) = u2(t)4+ 9(1−t2)2, t ∈ [0, 1],u′(0) = u(1) = 0, u′′(12) + u(0) = 0.(3.4)Since α = 1 and η = 12, if we choose θ = 13, then a simple calculation shows thatθ∗ =13, A =524and B =7108.Let f(t, u) = u24+ 9(1−t2)2, (t, u) ∈ [0, 1]× [0,+∞). Then (H1) and (H2) are satisfied. Moreover,it is easy to verify thatf(θ,θ∗4) ≥14B, f(0, 1) ≤1Aandf(0, 18) ≤18A, f(θ, 556θ∗) ≥556B.EJQTDE, 2013 No. 30, p. 9Therefore, it follows from Theorem 3.1 that the BVP (3.4) has positive and decreasing solutionsu1 and u2 satisfying14≤ ‖u1‖ ≤ 1 < 18 ≤ ‖u2‖ ≤ 556.References[1] D. Anderson, Green’s function for a third-order generalized right focal problem, J. Math.Anal. Appl. 288 (2003), no. 1, 1-14.[2] Z. Bai, X. Fei, Existence of triple positive solutions for a third order generalized right focalproblem, Math. Inequal. Appl. 9 (2006), no. 3, 437-444.[3] M. Gregus, Third Order Linear Differential Equations, Reidel, Dordrecht, 1987.[4] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press,New York, 1988.[5] L.-J. Guo, J.-P. Sun, Y.-H. Zhao, Existence of positive solution for nonlinear third-orderthree-point boundary value problem, Nonlinear Anal. 68 (2008), no. 10, 3151-3158.[6] G. Infante, J. R. L. Webb, Nonzero solutions of Hammerstein integral equations withdiscontinuous kernels, J. Math. Anal. Appl. 272 (2002), no. 1, 30-42.[7] G. Infante, J. R. L. Webb, Three-point boundary value problems with solutions that changesign, J. Integral Equations Appl. 15 (2003), no. 1, 37-57.[8] G. Infante, J. R. L. Webb, Loss of positivity in a nonlinear scalar heat equation. NoDEANonlinear Differential Equations Appl. 13 (2006), no. 2, 249-261.[9] M. Krasnoselskii, Positive Solutions of Operator Equations, Noordhoff, Groningen, 1964.[10] R. H. Martin, Nonlinear Operators & Differential Equations in Banach Spaces, Wiley, NewYork, 1976.EJQTDE, 2013 No. 30, p. 10[11] Alex P. Palamides, George Smyrlis, Positive solutions to a singular third-order three-pointboundary value problem with indefinitely signed Green’s function, Nonlinear Anal. 68(2008), no. 7, 2104-2118.[12] Alex P. Palamides, Nikolaos M. Stavrakakis, Existence and uniqueness of a positive solutionfor a third-order three-point boundary-value problem, Electron. J. Differential Equations2010 (2010), no. 155, 1-12.[13] J.-P. Sun, J. Zhao, Positive solution for a third-order three-point boundary value problemwith sign-changing Green’s function, Commun. Appl. Anal. 16 (2012), no. 2, 219-228.[14] J.-P. Sun, J. Zhao, Multiple positive solutions for a third-order three-point BVP withsign-changing Green’s function, Electron. J. Differential Equations 2012 (2012), no. 118,1-7.[15] Y. Sun, Positive solutions of singular third-order three-point boundary value problem, J.Math. Anal. Appl. 306 (2005), no. 2, 589-603.[16] Q. Yao, The existence and multiplicity of positive solutions for a third-order three-pointboundary value problem, Acta Math. Appl. Sin. Engl. Ser. 19 (2003), no. 1, 117-122.(Received February 3, 2013)EJQTDE, 2013 No. 30, p. 11

Existence of positive solution for a third-order three-point BVP with sign-changing Green's function

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