• 
    

    
    

      99热精品在线国产_美女午夜性视频免费_国产精品国产高清国产av_av欧美777_自拍偷自拍亚洲精品老妇_亚洲熟女精品中文字幕_www日本黄色视频网_国产精品野战在线观看 ?

      Triple Positive Solutions for a Third-order Three-point Boundary Value Problem

      2019-10-30 10:17:52WUHongping

      WU Hong-ping

      (College of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070, P.R.China)

      Abstract: In this paper, we study the existence of triple positive solutions for the nonlinear third-order three-point boundary value problemwhere η ∈[0, )is a constant, by using a fixed-point theorem due to Avery and Peterson, we establish results of triple positive solutions to the boundary value problem, and an example is given to illustrate the importance of result obtained.

      Key words: Third-order three-point boundary value problem; Positive solutions; Fixedpoint theorem

      §1. Introduction

      Third-order differential equations arise in a variety of different areas of applied mathematics and physics. In recent years, the existence and positive solutions for nonlinear third-order ordinary differential equations with a three-point boundary value problem have been studied by several authors. For example, Agarwal[1], Anderson[2], Ma[3], Guo[4], Palamides[5], Sun[6],for more existence results see[7-10].

      In [5], Palamides et.al considered the following boundary value problem (BVP)and found a Green’s function, although this Green’s function is not positive, the obtained solutions is still positive and increasing.

      In this paper, we investigate the existence of triple positive solutions for the nonlinear third-order three-point boundary value problem

      In the rest of paper, we make the assumptions:

      §2. Preliminaries and Lemmas

      Definition 2.1Let E be a Banach space over R, a nonempty convex closed set K ?E is said to be a cone provided that

      i) au ∈K for all u ∈K and a>0;

      ii) u,?u ∈K implies u=0.

      Definition 2.2Suppose K is a cone in a Banach space E, the map ψ said to be a nonnegative continuous concave functional on K provided that ψ : K →[0,∞) is continuous and

      Similarly we say the map ? is nonnegative continuous convex functional on K provided that ?:K →[0,∞)is continuous and

      Definition 2.3Suppose K is a cone, let α a nonnegative continuous concave functional on K, γ and β be nonnegative continuous convex functional on K , and ? a nonnegative continuous functional on K. For positive real number a,b,c,d, we defined the following convex sets:

      and a closed set

      Consider the Bananch space C[0,1] equipped with the standard normand let

      Lemma 2.1[5]The BVP

      admits the unique solution u(t)=, where G(t,s) is Green’s function.

      Further, for s ≥η

      for s ≤η

      Thus we obtain

      also

      Consequently

      Lemma 2.2[5]Let y ∈K0, then the BVP

      admits the unique solution u(t)=which is monotonic.

      Remark 2.1From the proof of Lemma 2.2[5],we know the solution of BVP(2.4)satisfiesit is show that u(t) strictly increasing and convex for t ∈[0,η] , and concave for t ∈[η,1].

      Lemma 2.3[5]For any y ∈K0, the unique solution u(t) of (2.4) also belongs to the cone K0and furthermore it satisfiesis arbitrary.

      Theorem 2.1[11]Let E be a Banach space, K ?E be a cone in E. Let γ and β are nonnegative continuous convex functionals on K, α is a nonnegative continuous concave functionals on K,and ? is a nonnegative continuous functionals on K satisfying ?(λx)≤λ?(x)for λ ∈[0,1], such that for some positive numbers M and d

      (C1) {x ∈P(γ,β,α,b,c,d)|α(x)>and α(Tx)>b for x ∈P(γ,β,α,b,c,d);

      (C2) α(Tx)>b for x ∈P(γ,α,b,d) with β(Tx)>b;

      Then T has at least three fixed points x1,x2,x3∈such that γ(xi)≤d for i=1,2,3; b<α(x1); a

      §3. Main Result

      Define

      Note that for u ∈K, we have

      Therefore

      Define functionals

      Then α,β,γ,?:K →[0,∞) are nonnegative continuous functionals , α is concave and β,γ are convex. Consequently, we have

      Define the integral operator T :K →C[0,1] by

      Lemma 3.1[5]T :K →K is completely continuous.

      For convenience, we denote the constants by

      Theorem 3.1 Suppose (H1)-(H2) hold, assume there exist 0 < a < b < θ?d such that the following condition hold

      then the BVP(1.1) has at least three positive solutions u1,u2and u3, satisfying

      Proof We now show that all the conditions of Theorem 2.1 are satisfied.

      i.e. α(Tu) > b for all u ∈P(γ,β,α,b,d). This show that condition (C1) of Theorem 2.1 is satisfied.

      Secondly, from (3.2) we have

      α(Tu) ≥θ?β(Tu) ≥= b for all u ∈P(γ,α,b,d) with β(Tu) >Thus condition(C2) of Theorem 2.1 is satisfied.

      Finally, we check out condition (C3) of Theorem 2.1 hold too.

      Clearly, as ?(0)=0

      So condition(C3)of Theorem 2.1 is also satisfied. Therefore,an application of Theorem 2.1 implies there exist three positive solutions of BVP(1.1), such that

      In addition, β(u1)

      §4. Example

      We present an example to check out our results. Consider the BVP

      Then all assumptions of Theorem 3.1 hold. Thus,BVP(4.1)has at least three positive solutions u1,u2,u3, and

      偏关县| 金华市| 开阳县| 邯郸市| 英德市| 博乐市| 玉门市| 平谷区| 行唐县| 海淀区| 章丘市| 信阳市| 房山区| 察雅县| 弥渡县| 文昌市| 武定县| 石阡县| 嘉鱼县| 清新县| 江城| 南江县| 雷山县| 乐业县| 望谟县| 安阳市| 增城市| 布拖县| 松江区| 论坛| 忻城县| 宜春市| 留坝县| 崇明县| 宁强县| 保亭| 宜黄县| 渭南市| 镇江市| 华容县| 齐齐哈尔市|