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RESEARCH ON ORDINARY DIFFERENTIALEQUATION AND FRACTIONALDIFFERENTIAL EQUATIONQU HAIDONG and LIU XUANDepartment of Mathematics and StatisticsHanshan Normal UniversityHIKARI LT D

HIKARI LTDHikari Ltd is a publisher of international scientific journals and books.www.m-hikari.comQU HAIDONG and LIU XUANRESEARCH ON ORDINARY DIFFERENTIAL EQUATION AND FRACTIONAL DIFFERENTIAL EQUATIONCopyright c 2014 Qu Haidong and Liu Xuan. This is an open access bookdistributed under the Creative Commons Attribution License, which permitsunrestricted use, distribution, and reproduction in any medium, provided theoriginal work is properly cited.ISBN 978-954-91999-7-0Typeset using LATEX.Mathematics Subject Classification: 34A08, 34B05, 34B15, 34B18Keywords: Ordinary differential equation, Fractional differential equation, HAM, Positive solutionPublished by Hikari Ltd

iiiHanshan Normal University

ivPrefacePREFACEThis book contains 4 papers of the authors, and all of which are on theresearch of differential equations with boundary value problem. The resultsobtained from these papers are new.I wish first of all to thank the teachers of the Hanshan Normal University, inparticular Prof. Ke Hansong, Prof. Xu Shaoyuan, Prof. Lin Wenxian and Prof.Xiao Gang for their invaluable aid during the writing of this work, the detailedexplanations, the patience and the precision in the suggestions, the suppliedsolutions, the competence and the kindness. Thanks also to my students DongLige, Luo Xiaodan and all the people who have discussed with me on theproblem of the differential equations, prodigal of precious observations and goodadvices.Finally, thanks to My father-in-law (Liu Yingjun) and My mother-in-law(Li Yingjie), who gave me a lot of support for this work. In particular thanksto my virtuous wife and my cute daughter.Qu HaidongHanshan Normal University01-04-2014

Contents1 Ordinary differential equations1.1 The symmetric positive solutions of four-point boundary valueproblems for nonlinear second-order differential equations . . .1.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . .1.1.2 Preliminary Notes . . . . . . . . . . . . . . . . . . . . .1.1.3 Main Results . . . . . . . . . . . . . . . . . . . . . . .1.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . .1.2 The symmetric positive solutions of three-point boundary valueproblems for nonlinear second-order differential equations . . .1.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . .1.2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . .1.2.3 Existence of positive solutions . . . . . . . . . . . . . .1.2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . .1. 1. 1. 2. 6. 10.10111216192 Fractional Differential Equations2.1 Positive Solution for Boundary Value Problem of Fractional Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . .2.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . .2.1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . .2.1.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . .2.1.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2 HAM for A Class of Time Fractional Partial Differential Equations2.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . .2.2.2 Basic definitions and Lemmas . . . . . . . . . . . . . . .2.2.3 HAM . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2.4 Applying HAM . . . . . . . . . . . . . . . . . . . . . . .Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20v2020212627282829303235

Chapter 1Ordinary differential equationsThis chapter contains three papers which are on the integer-order ordinarydifferential equations for boundary value problem.1.1The symmetric positive solutions of fourpoint boundary value problems for nonlinear second-order differential equationsAbstract: In this paper, we are concerned with the existence of symmetricpositive solutions for second-order differential equations. Under the suitableconditions, the existence and symmetric positive solutions are established byusing Krasnoselskii’s fixed-point theorems.Mathematics Subject Classification: 34B10,34B15,34B18Keywords: Boundary value problem; Symmetric positive solution; Cones1.1.1IntroductionRecently, there are many results about the existence and multiplicity of positive solutions for nonlinear second-order differential equations [7], [11], [12].Henderson and Thompson(see[4]), Li and Zhang (see[2]) studied the multiplesymmetric positive and nonnegative solutions of second-order ordinary differential equations. Yao (see[6]) considered the existence and iteration of nsymmetric positive solutions for a singular two-point boundary value problem(BVP). Sun(see[8]) considered the existence and multiplicity of symmetricpositive solutions for three-point boundary value problem. Inspired by theworks mentioned above, in this paper, we study the existence of symmetric1

2Ordinary differential equationspositive solutions of second-order four-point differential equations as follows, u00 (t) f (t, v),(1.1) v 00 (t) g(t, u), 0 t 1,subject to the boundary conditions u(t) u(1 t), u0 (0) u0 (1) u(ξ1 ) u(ξ2 ),v(t) v(1 t), v 0 (0) v 0 (1) v(ξ1 ) v(ξ2 ), 0 ξ1 ξ2 1,(1.2)where f, g : [0, 1] R R are continuous, both f (·, u) and g(·, u) aresymmetric on [0, 1],f (x, 0) g(x, 0) 0. To the best of author’s knowledge,there is no such result involving this problem. In this paper, we intend to fillin such gaps in the literature. The arguments for establishing the symmetricpositive solutions of (1) and (2) involve the properties of the functions inLemma 2.1 that play a key role in defining some cones. A fixed point theoremdue to Krasnoselskii is applied to yield the existence of symmetric positivesolutions of (1) and (2).1.1.2Preliminary NotesIn this section, we present some necessary definitions and preliminary lemmasthat will be used in the proof of the results.Definition 2.1. Let E be a real Bananch space. A nonempty closed set P Eis called a cone of E if it satisfies the following conditions:(I) x P, λ 0 implies λx P ;(II) x P, x P implies x 0.Definition 2.2. The function u is called to be concave on [0, 1] if u(rt1 (1 r)t2 ) ru(t1 ) (1 r)u(t2 ), r, t1 , t2 [0, 1].Definition 2.3. The function u is symmetric on [0, 1] if u(t) u(1 t), t [0, 1].Definition 2.4. The function (u, v) is called a symmetric positive solution ofthe equation (1) if u and v are symmetric and positive on [0, 1], and satisfythe equation (2).We shall consider the real Banach space C[0, 1], equipped with norm ku k max0 t 1 u(t) . Denote C [0, 1] {u C[0, 1] : u(t) 0, t [0, 1]}.Lemma 2.1. Let y C[0, 1] be symmetric on [0, 1], then the four-point BVP u00 (t) y(t) 0, 0 t 1,u(t) u(1 t), u0 (0) u0 (1) u(ξ1 ) u(ξ2 ),has a unique symmetric solution u(t) G1 (t, s) G2 (s), hereR10(1.3)G(t, s)y(s)ds, where G(t, s)

3Ordinary differential equations G1 (t, s) t(1 s), 0 t s 1,s(1 t), 0 s t 1, 1 2 [(ξ1 s) (ξ2 s) ξ1 (1 s) ξ2 (1 s) 1], 0 s ξ1 ,1[(ξ2 s) ξ1 (1 s) ξ2 (1 s) 1], ξ1 s ξ2 ,G2 (s) 2 1 [ ξ (1 s) ξ (1 s) 1], ξ s 1.1222Proof .From (3), we have u00 (t) y(t). For t [0, 1], integrating from 0 to twe getZ t0y(s)ds A1 ,(1.4)u (t) 0RtR 1 tsince u0 (t) u0 (1 t), we obtain that 0 y(s)ds A1 0 y(s)ds A1 ,which leads toZZ1 1 t1 ty(s)ds y(s)dsA1 2 02 0ZZ1 t1 1 t y(s)ds y(1 s)d(1 s)2 02 0Z 1 (1 s)y(s)ds.0Integrating again we obtainZ tZ 1u(t) (t s)y(s)ds t(1 s)y(s)ds A2 .00From (3) and (4) we haveZ 1Z ξ1Z 1y(s)ds (ξ1 s)y(s)ds ξ1(1 s)y(s)ds A2000Z ξ2Z 1 (ξ2 s)y(s)ds ξ2(1 s)y(s)ds A2 .00Thus1A2 2Z0ξ1[(ξ1 s) (ξ2 s) ξ1 (1 s) ξ2 (1 s) 1]y(s)dsZ1 ξ2 [(ξ2 s) ξ1 (1 s) ξ2 (1 s) 1]y(s)ds2 ξ1Z1 1 [ ξ1 (1 s) ξ2 (1 s) 1]y(s)ds.2 ξ2

4Ordinary differential equationsFrom above we can obtain the BVP(3) has a unique symmetric solutiontZZ(t s)y(s)ds tu(t) 01 2Z1(1 s)y(s)ds0ξ1[(ξ1 s) (ξ2 s) ξ1 (1 s) ξ2 (1 s) 1]y(s)dsZ1 ξ2[(ξ2 s) ξ1 (1 s) ξ2 (1 s) 1]y(s)ds 2 ξ1Z1 1 [ ξ1 (1 s) ξ2 (1 s) 1]y(s)ds2 ξ2Z 1Z 1Z 1 G1 (t, s)y(s)ds G2 (s)y(s)ds G(t, s)y(s)ds.0000This complete the proof.4m 2,then the functionLemma 2.2. Let mG2 min{G2 (ξ1 ), G2 (ξ2 )},L 4mGG 12G(t, s) satisfies LG(s, s) G(t, s) G(s, s) for t, s [0, 1].Proof . For any t [0, 1] and s [0, 1], we have4mG2G2 (s)4mG2 14mG2 114mG24mG24mG24mG2 · G2 (s) s(1 s) G2 (s)4 4mG2 1 4mG2 14mG2 1 4mG2 1 LG1 (s, s) LG2 (s) LG(s, s).G(t, s) G1 (t, s) G2 (s) G2 (s) 1G2 (s) It is obvious that G(s, s) G(t, s) for t, s [0, 1]. The proof is complete.Lemma 2.3. Let y C [0, 1], then the unique symmetric solution u(t) of theBVP (3) is nonnegative on [0, 1].Proof . Let y C [0, 1]. From the fact that u00 (t) y(t) 0, t [0, 1], weknow that the graph of u(t) is concave on [0, 1]. From (3). We have that1u(0) u(1) 2Z0ξ1[(ξ1 s) (ξ2 s) ξ1 (1 s) ξ2 (1 s) 1]y(s)dsZ1 ξ2 [(ξ2 s) ξ1 (1 s) ξ2 (1 s) 1]y(s)ds2 ξ1Z1 1 [ ξ1 (1 s) ξ2 (1 s) 1]y(s)ds 0.2 ξ2Note that u(t) is concave, thus u(t) 0 for t [0, 1]. This complete the proof.Lemma 2.4. Let y C [0, 1], then the unique symmetric solution u(t) of

5Ordinary differential equationsBVP (3) satisfiesmin u(t) L k u k .(1.5)t [0,1]Proof. For any t R [0, 1], on one hand, from Lemma 2.2 we have that u(t) R11G(t, s)y(s)ds 0 G(s, s)y(s)ds. Therefore,0Z 1G(s, s)y(s)ds.(1.6)k u k 0On the other hand, for any t [0, 1], from Lemma 2.2 we can obtain thatZ 1Z 1G(s, s)y(s)ds L k u k .(1.7)G(t, s)y(s)ds Lu(t) 00From (6) and (7) we know that (5) holds. Obviously, (u, v) C 2 [0, 1] C 2 [0, 1]is the solution of (1) and (2) if and only if (u, v) C[0, 1] C[0, 1] is the solutionof integral equations(R1u(t) 0 G(t, s)f (s, v(s))ds,R1(1.8)v(t) 0 G(t, s)g(s, u(s))ds.Integral equations (8) can be transferred to the nonlinear integral equationZ 1Z 1u(t) G(t, s)f (s,G(s, ξ)g(ξ, u(ξ))dξ)ds(1.9)00Let P {u C [0, 1] : u(t) is symmetric, concave on [0, 1] and min0 t 1 u(t) L k u k}. It is obvious that P is a positive cone in C[0, 1]. Define an integraloperator A : P C byZ 1Z 1G(t, s)f (s,G(s, ξ)g(ξ, u(ξ))dξ)ds.(1.10)Au(t) 00It is easy to see that the BVP (1) and (2) has a solution u u(t) if and onlyif u is a fixed point of the operator A defined by (10).Lemma 2.5. If the operator A is defined as (10), then A : P P is completelycontinuous.Proof . It is obvious that Au is symmetric on [0, 1]. Note that (Au)00 (t) f (t, v(t)) 0, we have that Au is concave, and from Lemma 2.3, it is easilyknown that Au C [0, 1]. Thus from lemma 2.2 and non-negativity of f andg,Z 1Z 1Au(t) G(t, s)f (s,G(s, ξ)g(ξ, u(ξ))dξ)ds00Z 1Z 1 G(s, s)f (s,G(s, ξ)g(ξ, u(ξ))dξ)ds,00

6Ordinary differential equationsthenZk Au k 1ZG(s, s)f (s,0G(s, ξ)g(ξ, u(ξ))dξ)ds.0For another hand,ZZ 1G(s, s)f (s,Au L011G(s, ξ)g(ξ, u(ξ))dξ)ds L k Au k .0Thus, A(P ) P . Since G(t, s), f (t, u) and g(t, u) are continuous, it is easy toknow that A : P P is completely continuous. The proof is complete.Lemma 2.6.(see[1]) Let E be a Banach space and P E is a cone in E.Assume thatT Ω1 and Ω2 are open subsets of E with 0 Ω1 and Ω1 Ω2 .Let A : P (Ω2 \Ω1 ) P be a completely continuous operator. In additionsuppose eitherTT(I) k Au k k u k, u P T Ω1 and k Au k k u k, u P T Ω2 or(II) k Au k k u k, u P Ω2 and kTAu k k u k, u P Ω1holds. Then A has a fixed point in P (Ω2 \Ω1 ).Lemma 2.7.(see[1]) Let E be a Banach space and P E is a cone in E. Assume that Ω1 ,TΩ2 and Ω3 are open subsets of E with 0 Ω1 , Ω1 Ω2 ,Ω2 Ω3and let A : P (Ω3 \Ω1 ) P be a completely continuous operator. In additionsuppose eitherT(I) k Au k k u k, u P Ω1 ; T(II) k Au k k u k,Au 6 u, T u P Ω2 ;(III) k Au k k u k, u P Ω3Tholds.Then AThas at least two fixed-pointsx,xinP(Ω3 \Ω1 ), and further12Tmore x1 P (Ω2 \Ω1 ),x2 P (Ω3 \Ω2 ).1.1.3Main ResultsIn this section, we study the existence of positive solutions for BVP (1) and(2). First we give the following assumptions:f (t, u)g(t, u)(H1 ) limu 0 sup0 t 1 u 0,limu 0 sup0 t 1 u 0;f (t, u)g(t, u)(H2 ) limu inf 0 t 1 u ,limu inf 0 t 1 u ;f (t, u)g(t, u)(H3 ) limu 0 inf 0 t 1 u ,limu 0 inf 0 t 1 u ;g(t, u)f (t, u)(H4 ) limu sup0 t 1 u 0,limu sup0 t 1 u 0;(H5 )There exists a constant R1 0, such that f (s, u) Z 1 R1forG(s, s)dsevery (s, u) [0, 1] [LR1 , R1 ].0

7Ordinary differential equationsTheorem 3.1. If (H1 ) and (H2 ) are satisfied, then BVP (1) and (2) have atleast one symmetric positive solution (u, v) C 2 ([0, 1], R ) C 2 ([0, 1], R )satisfying u(t) 0, v(t) 0.Proof . From (H1 ) there is a number N1 (0, 1) such that for each (s, u) [0,R1] (0, N1 ), one has f (s, u) η1 u, g(s, u) η1 u, where η1 0 satisfies1η1 0 G(s, s)ds 1, for every u P and k u k N21 , note thatR1R1R1G(s, ξ)g(ξ, u(ξ))dξ 0 G(ξ, ξ)g(ξ, u(ξ))dξ 0 η1 G(ξ, ξ)u(ξ)dξ k u k 0N1 N1 ,then2Z1Z1Au(x) G(t, s)f (s,G(s, ξ)g(ξ, u(ξ))dξ)ds00Z 1Z 1 η1G(s, s)G(s, ξ)g(ξ, u(ξ))dξds00Z 1Z 12 η1G(s, s)G(ξ, ξ)u(ξ)dξds k u k .00LetΩ1 {u C [0, 1], k u k N1},2thenk Au k k u k, u P\ Ω1 .(1.11) From (H2 ) there is a number N2 LN1 for each (s, u) [0, 1]R (N2 , ),31one has f (s, u) η2 u, g(s, u) η2 u where η2 0 satisfies η2 L 2 0 G(s, s)ds 2 , from Lemma 2.2 and Lemma 2.4, we1, then, for every u P and k u k 2NLhaveZ 1Z 12G(s, ξ)g(ξ, u(ξ))dξ Lη2 G(ξ, ξ) k u k dξ00 2 L k u k 2N2 N2 ,thenZ1Z1k Au k G(t, s)f (s,G(s, ξ)g(ξ, u(ξ))dξ)ds0Z 1Z 12 2 L η2G(s, s)G(ξ, ξ)u(ξ)dξds00Z 1Z 13 2 L η2G(s, s)G(ξ, ξ) k u k dξds k u k .000

8Ordinary differential equationsLet2N2Ω2 {u C [0, 1], k u k },Lthenk Au k k u k, u P\ Ω2 .(1.12)Thus from(11),(12)and Lemma 2.6,we know that the operator A has a fixedTpoint in P (Ω2 \Ω1 ) . The proof is complete.Theorem 3.2. If (H3 ) and (H4 ) are satisfied, then BVP (1) and (2) have atleast one symmetric positive solution (u, v) C 2 ([0, 1], R ) C 2 ([0, 1], R )satisfying u(t) 0, v(t) 0.Proof . From (H3 ) there is a number N3 (0, 1) such that for each (x, u) [0, 1] R 1 (0, N3 ), one has f (s, u) η3 u, g(s, u) η3 u where η3 0 satisfies3L 2 η3 0 G(s, s)ds 1. From g(x, 0) 0 and the continuity of g(s, u), weN3know that there exists number N3 (0, N3 ) such that g(s, u) R 1 G(s,s)dsfor0each (s, u) [0, 1] (0, N3 ]. Then for every u P and k u k N3 , note thatZ1ZG(s, ξ)g(ξ, u(ξ))dξ 001G(ξ, ξ) R 10N3G(s, s)dsdξ N3 .ThusZ1Z1G(s, ξ)g(ξ, u(ξ))dξ)dsG(t, s)f (s,00Z 1Z 1 Lη3G(s, s)G(s, ξ)g(ξ, u(ξ))dξds00Z 1Z 13 2 L η3G(s, s)G(ξ, ξ) k u k dξds k u k .Au(x) 00LetΩ3 {u C [0, 1], k u k N3 },thenk Au k k u k, u P\ Ω3 .(1.13)From (H4 ), there exist C1 0 and C2 0 such that f (s, u) η4 u C1 , g(s, u) η4 u C2 for (s, u) [0, 1] (0, ), where η4 0, and

9Ordinary differential equationsη4R10G(ξ, ξ)dξ 1. Then, for u C [0, 1] we haveZ1Z1G(s, t)f (s,G(s, ξ)g(ξ, u(ξ))dξ)ds00Z 1Z 1G(s, s)(η4G(s, ξ)g(ξ, u(ξ))dξ C1 )ds00Z 1Z 1G(s, s)G(ξ, ξ)g(ξ, u(ξ))dξds C3η400Z 1Z 1η4G(s, s)G(ξ, ξ)(η4 u C2 )dξds C300Z 1Z 12G(ξ, ξ) k u k dξds C4 k u k C4G(s, s)(η4 )Au 00Thus k Au k k u k with k u k .Let Ω4 {u E, k u k N4 }. For each u P and k u k N4 N3 largeenough, we havek Au k k u k, u P\ Ω4 .(1.14)Thus from(13),(14)and Lemma 2.6,we know that the operator A has a fixedTpoint in P (Ω4 \Ω3 ) . The proof is complete.Theorem 3.3. If (H2 ), (H3 )and (H5 ) are satisfied, then BVP (1) and (2) haveat least two symmetric positive solutions (u1 , v1 ), (u2 , v2 ) C 2 ([0, 1], R ) C 2 ([0, 1], R ) satisfying u1 (t) 0, v1 (t) 0, u2 (t) 0, v2 (t) 0.Proof .LetΩ5 {u C [0, 1], k u k R1 },Tthen u P Ω5 , we have u(s) [LR1 , R1 ]. From lemma 2.2, Lemma 2.4and (6) we can obtainZ1ZG(s, ξ)g(ξ, u(ξ))dξ L01G(ξ, ξ)g(ξ, u(ξ))dξ L k u k0andZ01Z1G(s, ξ)g(ξ, u(ξ))dξ G(ξ, ξ)g(ξ, u(ξ))dξ0Z 1R1 R1 . G(ξ, ξ)dξG(s, s)ds0

10Ordinary differential equationsThus Au R10G(s, t)f (s,R10G(s, ξ)g(ξ, u(ξ))dξ)ds R1 k u k. Thenk Au k k u k, u P\ Ω5 .R10R1G(s, s) R 1 G(ξ,ξ)dξds 0(1.15)For another hand, from (H2 ) and (H3 ), we can choose two right numbersf2 (R1 , ), Nf3 (0, R1 ) satisfyN\f2 ,k Au k k u k, u P Ω(1.16)k Au k k u k, u P\f3 , Ω(1.17)f2 {u C [0, 1], k u k Nf2 }, Ωf3 {u C [0, 1], k u k Nf3 }.where ΩThen from Lemma 2.7, (15), (16) and (17), A has at least two fixed points inT fTf3 ) , respectively. The proof is complete.P (Ω(Ω5 \Ω2 \Ω5 ) and P1.1.4ExamplesIn this section, we give three examples to illustrate our results.22,g(t, u) 2u2 2[1 t(1 t)]u, ξ1 Examples 4.1. Let f (t, v) v 2 [1 t(1 t)]v1 v 21 u2151, ξ 2 , we can choose L 16 , then conditions of Theorem 3.1 are satisfied.4 2From Theorem 3.1, BVP (1) and (2) have at least one symmetric positivesolution.2112[1 t(1 t)]u22 Examples 4.2. Let f (t, v) v 2 [1 t(1 t)]v,g(t,u) 2u,21 v1 u2151ξ1 4 , ξ2 2 , we can choose L 16 ,, then conditions of Theorem 3.2 aresatisfied. From Theorem 3.2, BVP (1) and (2) have at least one symmetricpositive solution.11Examples 4.3. Let f (t, v) 45[t(1 t) 1](v 2 v 2 ),g(t, u) 43[t(1 t) 1](u 2 32325u2 ), ξ1 41 , ξ2 12 , we can choose L 16and R1 1, then conditions ofTheorem 3.3 are satisfied. From Theorem 3.3, BVP (1) and (2) have at leasttwo symmetric positive solutions.1.2The symmetric positive solutions of threepoint boundary value problems for nonlinear second-order differential equationsAbstract: The paper investigates the problem of existence of positive solutions of nonlinear third-order differential equations. Under the suitable conditions, the existence and multiplicity of positive solutions are established byusing Krasnoselskii’s fixed-point theorem of cone.

11Ordinary differential equationsMathematics Subject Classification: 34B18, 34B27,35K35Keywords: Boundary value problem; Positive solution; Third-order1.2.1IntroductionMost of the recent results on the positive solutions are concerned with singleequation and simple boundary condition(see[?][7][11][12][13]). As far as the author know, there are few results on the symmetric positive solutions. It shouldbe mentioned that Sun[8] discussed the following boundary value problem:u00 (t) a(t)f (t, u(t)) 0, 0 t 1,(1.18)1u(0) u(1 t), u0 (0) u0 (1) u( ),2by using Krasnoselskii’s fixed-point theorems, the existence of symmetric positive solutions is shown under certain conditions on f . Yang and Sun consideredthe boundary value problem of differential equations u00 (x) f (x, v), v 00 (x) g(x, u),u(0) u(1) 0,v(0) v(1) 0.(1.19)using the degree theory, the existence of a positive solution of (1.19) is established. Motivated by the work of Sun and Yang, we concern with the existenceof symmetric positive solutions of the boundary value problems. u00 (t) f (t, v), v 00 (t) g(t, u),1(1.20)u(t) u(1 t), αu0 (0) βu0 (1) γu( ),21v(t) v(1 t), αv 0 (0) βv 0 (1) γv( ),2 where f, g : [0, 1] R R are continuous, both f (·, u) and g(·, u) aresymmetric on [0, 1], f (x, 0) 0, g(x, 0) 0, β α γ2 , β α 2γ, α, β 0, γ 6 0. The arguments for establishing the symmetric positive solutions of(1.20) involve properties of the functions in Lemma that play a key role indefining certain cones. A fixed point theorem due to Krasnoselskii is appliedto yield the existence of symmetric positive solutions of (1.20).This paper contains three sections besides the Introduction. In Section2, we present some necessary definitions and preliminary lemmas that will beused to prove our main results. In Section 3, we discuss the existence of at leastone and at least two symmetric positive solutions for BVP (1.

Ordinary di erential equations This chapter contains three papers which are on the integer-order ordinary di erential equations for boundary value problem. 1.1 The symmetric positive solutions of four-point boundary value problems for nonlin-ear second-order di erential equations Abstract: In this paper, we are concerned with the existence of .

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Introduction to Differential Equation Solving with DSolve The Mathematica function DSolve finds symbolic solutions to differential equations. (The Mathe- matica function NDSolve, on the other hand, is a general numerical differential equation solver.) DSolve can handle the following types of equations: † Ordinary Differential Equations

Note! The orderof a differential equation is the order of the highest derivative appearing in the equation. Example 1.3:Equation 1.1 is a first-order differential equation; 1.2, 1.4, and 1.5 are second-order differential equations. (Note in 1.4 that the or-der of the highest derivative appearing in the equation is two.)

DIFFERENTIAL – DIFFERENTIAL OIL DF–3 DF DIFFERENTIAL OIL ON-VEHICLE INSPECTION 1. CHECK DIFFERENTIAL OIL (a) Stop the vehicle on a level surface. (b) Using a 10 mm socket hexagon wrench, remove the rear differential filler plug and gasket. (c) Check that the oil level is between 0 to 5 mm (0 to 0.20 in.) from the bottom lip of the .