RESEARCH ON ORDINARY DIFFERENTIAL EQUATION AND FRACTIONAL DIFFERENTIAL … · 2014-07-05 · iv...

RESEARCH ON ORDINARY DIFFERENTIAL

EQUATION AND FRACTIONAL

DIFFERENTIAL EQUATION

QU HAIDONG and LIU XUAN

Department of Mathematics and StatisticsHanshan Normal University

HIKARI LT D

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QU HAIDONG and LIU XUANRESEARCH ON ORDINARY DIFFERENTIAL EQUATION AND FRAC-TIONAL DIFFERENTIAL EQUATION

Copyright 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 978-954-91999-7-0

Typeset using LATEX.

Mathematics Subject Classification: 34A08, 34B05, 34B15, 34B18

Keywords: Ordinary differential equation, Fractional differential equa-tion, HAM, Positive solution

Published by Hikari Ltd

iii

Hanshan Normal University

iv Preface

PREFACE

This book contains 4 papers of the authors, and all of which are on the research of differential equations with boundary value problem. The results obtained from these papers are new.

I wish first of all to thank the teachers of the Hanshan Normal University, in particular Prof. Ke Hansong, Prof. Xu Shaoyuan, Prof. Lin Wenxian and Prof. Xiao Gang for their invaluable aid during the writing of this work, the detailed explanations, the patience and the precision in the suggestions, the supplied solutions, the competence and the kindness. Thanks also to my students Dong Lige, Luo Xiaodan and all the people who have discussed with me on the problem of the differential equations, prodigal of precious observations and good advices.

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 University

01-04-2014

Contents

1 Ordinary differential equations 11.1 The symmetric positive solutions of four-point boundary value

problems for nonlinear second-order differential equations . . . . 11.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Preliminary Notes . . . . . . . . . . . . . . . . . . . . . . 21.1.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . 61.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 The symmetric positive solutions of three-point boundary valueproblems for nonlinear second-order differential equations . . . . 101.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . 121.2.3 Existence of positive solutions . . . . . . . . . . . . . . . 161.2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Fractional Differential Equations 202.1 Positive Solution for Boundary Value Problem of Fractional D-

ifferential Equation . . . . . . . . . . . . . . . . . . . . . . . . . 202.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 202.1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . 212.1.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . 262.1.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.2 HAM for A Class of Time Fractional Partial Differential Equations 282.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 282.2.2 Basic definitions and Lemmas . . . . . . . . . . . . . . . 292.2.3 HAM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.2.4 Applying HAM . . . . . . . . . . . . . . . . . . . . . . . 32

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

v

Chapter 1

Ordinary differential equations

This chapter contains three papers which are on the integer-order ordinarydifferential equations for boundary value problem.

1.1 The symmetric positive solutions of four-

point boundary value problems for nonlin-

ear second-order differential equations

Abstract: 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: 34B10,34B15,34B18

Keywords: Boundary value problem; Symmetric positive solution; Cones

1.1.1 Introduction

Recently, there are many results about the existence and multiplicity of pos-itive 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 dif-ferential equations. Yao (see[6]) considered the existence and iteration of nsymmetric positive solutions for a singular two-point boundary value prob-lem(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 symmetric

1

2 Ordinary differential equations

positive solutions of second-order four-point differential equations as follows, −u′′(t) = f(t, v),

−v′′(t) = g(t, u), 0 ≤ t ≤ 1,(1.1)

subject to the boundary conditionsu(t) = u(1− t), u′(0)− u′(1) = u(ξ1) + u(ξ2),

v(t) = v(1− t), v′(0)− v′(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).

1.1.2 Preliminary Notes

In 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 ‖u ‖= 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

u′′(t) + y(t) = 0, 0 < t < 1,

u(t) = u(1− t), u′(0)− u′(1) = u(ξ1) + u(ξ2),(1.3)

has a unique symmetric solution u(t) =∫ 1

0G(t, s)y(s)ds, where G(t, s) =

G1(t, s) +G2(s), here

Ordinary differential equations 3

G1(t, s) =

t(1− s), 0 ≤ t ≤ s ≤ 1,

s(1− t), 0 ≤ s ≤ t ≤ 1,

G2(s) =

12[(ξ1 − s) + (ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1], 0 ≤ s ≤ ξ1,

12[(ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1], ξ1 ≤ s ≤ ξ2,

12[−ξ1(1− s)− ξ2(1− s) + 1], ξ2 ≤ s ≤ 1.

Proof .From (3), we have u′′(t) = −y(t). For t ∈ [0, 1], integrating from 0 to twe get

u′(t) = −∫ t

0

y(s)ds+ A1, (1.4)

since u′(t) = −u′(1− t), we obtain that −∫ t

0y(s)ds + A1 =

∫ 1−t0

y(s)ds− A1,which leads to

A1 =1

2

∫ t

0

y(s)ds+1

2

∫ 1−t

0

y(s)ds

=1

2

∫ t

0

y(s)ds− 1

2

∫ 1−t

0

y(1− s)d(1− s)

=

∫ 1

0

(1− s)y(s)ds.

Integrating again we obtain

u(t) = −∫ t

0

(t− s)y(s)ds+ t

∫ 1

0

(1− s)y(s)ds+ A2.

From (3) and (4) we have∫ 1

0

y(s)ds = −∫ ξ1

0

(ξ1 − s)y(s)ds+ ξ1

∫ 1

0

(1− s)y(s)ds+ A2

−∫ ξ2

0

(ξ2 − s)y(s)ds+ ξ2

∫ 1

0

(1− s)y(s)ds+ A2.

Thus

A2 =1

2

∫ ξ1

0

[(ξ1 − s) + (ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1]y(s)ds

+1

2

∫ ξ2

ξ1

[(ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1]y(s)ds

+1

2

∫ 1

ξ2

[−ξ1(1− s)− ξ2(1− s) + 1]y(s)ds.


From above we can obtain the BVP(3) has a unique symmetric solution

u(t) = −∫ t

0

(t− s)y(s)ds+ t

∫ 1

0

(1− s)y(s)ds

+1

2

∫ ξ1

0

[(ξ1 − s) + (ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1]y(s)ds

+1

2

∫ ξ2

ξ1

[(ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1]y(s)ds

+1

2

∫ 1

ξ2

[−ξ1(1− s)− ξ2(1− s) + 1]y(s)ds

=

∫ 1

0

G1(t, s)y(s)ds+

∫ 1

0

G2(s)y(s)ds =

∫ 1

0

G(t, s)y(s)ds.

This complete the proof.Lemma 2.2. Let mG2 = minG2(ξ1), G2(ξ2),L =

4mG2

4mG2+1

,then the function

G(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 have

G(t, s) = G1(t, s) +G2(s) ≥ G2(s) =1

4mG2 + 1G2(s) +

4mG2

4mG2 + 1G2(s)

≥ 1

4· 4mG2

4mG2 + 1+

4mG2

4mG2 + 1G2(s) ≥ s(1− s) 4mG2

4mG2 + 1+

4mG2

4mG2 + 1G2(s)

≥ LG1(s, s) + LG2(s) = LG(s, 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 u′′(t) = −y(t) ≤ 0, t ∈ [0, 1], weknow that the graph of u(t) is concave on [0, 1]. From (3). We have that

u(0) = u(1) =1

2

∫ ξ1

0

[(ξ1 − s) + (ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1]y(s)ds

+1

2

∫ ξ2

ξ1

[(ξ2 − s)− ξ1(1− s)− ξ2(1− s) + 1]y(s)ds

+1

2

∫ 1

ξ2

[−ξ1(1− s)− ξ2(1− s) + 1]y(s)ds ≥ 0.

Note 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


BVP (3) satisfies

mint∈[0,1]

u(t) ≥ L ‖ u ‖ . (1.5)

Proof . For any t ∈ [0, 1], on one hand, from Lemma 2.2 we have that u(t) =∫ 1

0G(t, s)y(s)ds ≤

∫ 1

0G(s, s)y(s)ds. Therefore,

‖ u ‖≤∫ 1

0

G(s, s)y(s)ds. (1.6)

On the other hand, for any t ∈ [0, 1], from Lemma 2.2 we can obtain that

u(t) =

∫ 1

0

G(t, s)y(s)ds ≥ L

∫ 1

0

G(s, s)y(s)ds ≥ L ‖ u ‖ . (1.7)

From (6) and (7) we know that (5) holds. Obviously, (u, v) ∈ C2[0, 1]×C2[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

u(t) =∫ 1

0G(t, s)f(s, v(s))ds,

v(t) =∫ 1

0G(t, s)g(s, u(s))ds.

(1.8)

Integral equations (8) can be transferred to the nonlinear integral equation

u(t) =

∫ 1

0

G(t, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds (1.9)

Let P = u ∈ C+[0, 1] : u(t) is symmetric, concave on [0, 1] and min0≤t≤1u(t) ≥L ‖ u ‖. It is obvious that P is a positive cone in C[0, 1]. Define an integraloperator A : P → C by

Au(t) =

∫ 1

0

G(t, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds. (1.10)

It 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)′′(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,

Au(t) =

∫ 1

0

G(t, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds

≤∫ 1

0

G(s, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds,


then

‖ Au ‖≤∫ 1

0

G(s, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds.

For another hand,

Au ≥ L

∫ 1

0

G(s, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds ≥ L ‖ Au ‖ .

Thus, 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 that 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 addition

suppose either(I) ‖ Au ‖≤‖ u ‖,∀u ∈ P

⋂∂Ω1 and ‖ Au ‖≥‖ u ‖,∀u ∈ P

⋂∂Ω2 or

(II) ‖ Au ‖≤‖ u ‖,∀u ∈ P⋂∂Ω2 and ‖ Au ‖≥‖ u ‖,∀u ∈ P

⋂∂Ω1

holds. 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. As-sume that Ω1, Ω2 and Ω3 are open subsets of E with 0 ∈ Ω1 , Ω1 ⊂ Ω2,Ω2 ⊂ Ω3

and let A : P⋂

(Ω3\Ω1)→ P be a completely continuous operator. In additionsuppose either(I) ‖ Au ‖≥‖ u ‖,∀u ∈ P

⋂∂Ω1;

(II) ‖ Au ‖≤‖ u ‖,Au 6= u, ∀u ∈ P⋂∂Ω2;

(III) ‖ Au ‖≥‖ u ‖,∀u ∈ P⋂∂Ω3

holds.Then A has at least two fixed-points x1,x2 in P⋂

(Ω3\Ω1), and further-more x1 ∈ P

⋂(Ω2\Ω1),x2 ∈ P

⋂(Ω3\Ω2).

1.1.3 Main Results

In this section, we study the existence of positive solutions for BVP (1) and(2). First we give the following assumptions:

(H1) limu→0+ sup0≤t≤1f(t, u)u = 0,limu→0+ sup0≤t≤1

g(t, u)u = 0;

(H2) limu→∞ inf0≤t≤1f(t, u)u =∞,limu→∞ inf0≤t≤1

g(t, u)u =∞;

(H3) limu→0+ inf0≤t≤1f(t, u)u =∞,limu→0+ inf0≤t≤1

g(t, u)u =∞;

(H4) limu→∞ sup0≤t≤1f(t, u)u = 0,limu→∞ sup0≤t≤1

g(t, u)u = 0;

(H5)There exists a constant R1 > 0, such that f(s, u) ≤ R1∫ 1

0

G(s, s)ds

for

every (s, u) ∈ [0, 1]× [LR1, R1].


Theorem 3.1. If (H1) and (H2) are satisfied, then BVP (1) and (2) have atleast one symmetric positive solution (u, v) ∈ C2([0, 1], R+) × C2([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, 1] × (0, N1), one has f(s, u) ≤ η1u, g(s, u) ≤ η1u, where η1 > 0 satisfies

η1

∫ 1

0G(s, s)ds ≤ 1, for every u ∈ P and ‖ u ‖= N1

2, note that∫ 1

0G(s, ξ)g(ξ, u(ξ))dξ ≤

∫ 1

0G(ξ, ξ)g(ξ, u(ξ))dξ ≤

∫ 1

0η1G(ξ, ξ)u(ξ)dξ ≤‖ u ‖=

N1

2< N1,then

Au(x) ≤∫ 1

0

G(t, s)f(s,

∫ 1

0


≤ η1

∫ 1

0

G(s, s)

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξds

≤ η21

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)u(ξ)dξds ≤‖ u ‖ .

Let

Ω1 = u ∈ C+[0, 1], ‖ u ‖< N1

2,

then

‖ Au ‖≤‖ u ‖, u ∈ P⋂

∂Ω1. (1.11)

From (H2) there is a number N2 >√LN1 for each (s, u) ∈ [0, 1]× (N2,+∞),

one has f(s, u) ≥ η2u, g(s, u) ≥ η2u where η2 > 0 satisfies η2L32

∫ 1

0G(s, s)ds ≥

1, then, for every u ∈ P and ‖ u ‖= 2N2√L

, from Lemma 2.2 and Lemma 2.4, wehave ∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ ≥ L2

∫ 1

0

η2G(ξ, ξ) ‖ u ‖ dξ

≥ 2√L ‖ u ‖= 2N2 > N2,

then

‖ Au ‖ =

∫ 1

0

G(t, s)f(s,

∫ 1

0


≥ L2η22

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)u(ξ)dξds

≥ L3η22

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ) ‖ u ‖ dξds ≥‖ u ‖ .


Let

Ω2 = u ∈ C+[0, 1], ‖ u ‖< 2N2√L,

then

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω2. (1.12)

Thus from(11),(12) and Lemma 2.6,we know that the operator A has a fixedpoint 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) ∈ C2([0, 1], R+) × C2([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] × (0, N3), one has f(s, u) ≥ η3u, g(s, u) ≥ η3u where η3 > 0 satisfies

L32η3

∫ 1

0G(s, s)ds ≥ 1. From g(x, 0) ≡ 0 and the continuity of g(s, u), we

know that there exists number N3 ∈ (0, N3) such that g(s, u) ≤ N3∫ 10 G(s,s)ds

for

each (s, u) ∈ [0, 1]× (0, N3]. Then for every u ∈ P and ‖ u ‖= N3, note that

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ ≤∫ 1

0

G(ξ, ξ)N3∫ 1

0G(s, s)ds

dξ = N3.

Thus

Au(x) =

∫ 1

0

G(t, s)f(s,

∫ 1

0


≥ Lη3

∫ 1

0

G(s, s)

∫ 1

0


≥ L3η23

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ) ‖ u ‖ dξds ≥‖ u ‖ .

Let

Ω3 = u ∈ C+[0, 1], ‖ u ‖< N3,

then

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω3. (1.13)

From (H4), there exist C1 > 0 and C2 > 0 such that f(s, u) ≤ η4u +C1, g(s, u) ≤ η4u + C2 for ∀(s, u) ∈ [0, 1] × (0,∞), where η4 > 0, and


η4

∫ 1

0G(ξ, ξ)dξ ≤ 1. Then, for u ∈ C+[0, 1] we have

Au =

∫ 1

0

G(s, t)f(s,

∫ 1

0


≤∫ 1

0

G(s, s)(η4

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ + C1)ds

≤ η4

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)g(ξ, u(ξ))dξds+ C3

≤ η4

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)(η4u+ C2)dξds+ C3

≤ (η4)2

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ) ‖ u ‖ dξds+ C4 ≤‖ u ‖ +C4

Thus ‖ Au ‖≤‖ u ‖ with ‖ u ‖→ ∞.Let Ω4 = u ∈ E, ‖ u ‖< N4. For each u ∈ P and ‖ u ‖= N4 > N3 largeenough, we have

‖ Au ‖≤‖ u ‖, u ∈ P⋂

∂Ω4. (1.14)

Thus from(13),(14) and Lemma 2.6,we know that the operator A has a fixedpoint in P


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) ∈ C2([0, 1], R+) ×C2([0, 1], R+) satisfying u1(t) > 0, v1(t) > 0, u2(t) > 0, v2(t) > 0.Proof .Let

Ω5 = u ∈ C+[0, 1], ‖ u ‖< R1,

then ∀u ∈ P⋂∂Ω5, we have u(s) ∈ [LR1, R1]. From lemma 2.2, Lemma 2.4

and (6) we can obtain∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ ≥ L

∫ 1

0

G(ξ, ξ)g(ξ, u(ξ))dξ ≥ L ‖ u ‖

and ∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ ≤∫ 1

0

G(ξ, ξ)g(ξ, u(ξ))dξ

≤∫ 1

0

G(ξ, ξ)dξR1

G(s, s)ds= R1.


Thus Au =∫ 1

0G(s, t)f(s,

∫ 1

0G(s, ξ)g(ξ, u(ξ))dξ)ds ≤

∫ 1

0G(s, s) R1∫ 1

0 G(ξ,ξ)dξds =

R1 =‖ u ‖. Then

‖ Au ‖≤‖ u ‖, u ∈ P⋂

∂Ω5. (1.15)

For another hand, from (H2) and (H3), we can choose two right numbers

N2 ∈ (R1,∞), N3 ∈ (0, R1) satisfy

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω2, (1.16)

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω3, (1.17)

where Ω2 = u ∈ C+[0, 1], ‖ u ‖< N2, Ω3 = u ∈ C+[0, 1], ‖ u ‖< N3.Then from Lemma 2.7, (15), (16) and (17), A has at least two fixed points in

P⋂

(Ω2\Ω5) and P⋂

(Ω5\Ω3) , respectively. The proof is complete.

1.1.4 Examples

In this section, we give three examples to illustrate our results.

Examples 4.1. Let f(t, v) = v2 + [1+t(1−t)]v21+v2

,g(t, u) = 2u2 + 2[1+t(1−t)]u21+u2

, ξ1 =14, ξ2 = 1

2, we can choose L = 5

16, then conditions of Theorem 3.1 are satisfied.

From Theorem 3.1, BVP (1) and (2) have at least one symmetric positivesolution.Examples 4.2. Let f(t, v) = v

12 + [1+t(1−t)]v2

1+v2,g(t, u) = 2u

12 + 2[1+t(1−t)]u2

1+u2,

ξ1 = 14, ξ2 = 1


16,, then conditions of Theorem 3.2 are

satisfied. From Theorem 3.2, BVP (1) and (2) have at least one symmetricpositive solution.Examples 4.3. Let f(t, v) = 45[t(1−t)+1]

32(v

12 + v2),g(t, u) = 43[t(1−t)+1]

32(u

12 +

u2), ξ1 = 14, ξ2 = 1


16and R1 = 1, then conditions of

Theorem 3.3 are satisfied. From Theorem 3.3, BVP (1) and (2) have at leasttwo symmetric positive solutions.

1.2 The symmetric positive solutions of three-

point boundary value problems for nonlin-

ear second-order differential equations

Abstract: The paper investigates the problem of existence of positive solu-tions of nonlinear third-order differential equations. Under the suitable con-ditions, the existence and multiplicity of positive solutions are established byusing Krasnoselskii’s fixed-point theorem of cone.


Mathematics Subject Classification: 34B18, 34B27,35K35

Keywords: Boundary value problem; Positive solution; Third-order

1.2.1 Introduction

Most 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 au-thor know, there are few results on the symmetric positive solutions. It shouldbe mentioned that Sun[8] discussed the following boundary value problem:

u′′(t) + a(t)f(t, u(t)) = 0, 0 < t < 1,

u(0) = u(1− t), u′(0)− u′(1) = u(1

2),

(1.18)

by using Krasnoselskii’s fixed-point theorems, the existence of symmetric posi-tive solutions is shown under certain conditions on f . Yang and Sun consideredthe boundary value problem of differential equations

−u′′(x) = f(x, v),

−v′′(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 estab-lished. Motivated by the work of Sun and Yang, we concern with the existenceof symmetric positive solutions of the boundary value problems.

−u′′(t) = f(t, v),

−v′′(t) = g(t, u),

u(t) = u(1− t), αu′(0)− βu′(1) = γu(1

2),

v(t) = v(1− t), αv′(0)− βv′(1) = γv(1

2),

(1.20)

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 |≤| γ

2|, 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 in defining certain cones. A fixed point theorem due to Krasnoselskii is applied to 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 (1.20). Finally, wegive some examples to illustrate our results in Section 4.


1.2.2 Preliminaries

In 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 Banach space. A nonempty closed set P ⊂ E is called a

cone of E if it satisfies the following conditions:(1) x ∈ P, λ > 0 implies λx ∈ P ;(2) x ∈ P,−x ∈ P implies x = 0.

Definition 2.2.The function u is said to be concave on [0, 1] if u(rt1 +(1−r)t2) ≥ ru(t1)+

(1− r)u(t2), r, t1, t2 ∈ [0, 1].Definition 2.1. The function u is said to be symmetric on [0, 1] if u(t) =

u(1− t), t ∈ [0, 1].Definition 2.1. The function (u, v) is called a symmetric positive solution

of the BVP (1.20) if u and v are symmetric and positive on [0, 1], and satisfythe BVP (1.20).

We shall consider the real Banach space C[0, 1], equipped with norm ‖u ‖= 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 three-pointBVP

u′′(t) + y(t) = 0, 0 < t < 1,

u(t) = u(1− t), αu′(0)− βu′(1) = γu(1

2),

(1.21)

has a unique symmetric solution u(t) =∫ 1

0G(t, s)y(s)ds, where G(t, s) =

G1(t, s) +G2(s), here

G1(t, s) =

t(1− s), 0 ≤ t ≤ s ≤ 1,

s(1− t), 0 ≤ s ≤ t ≤ 1,

G2(s) =

(1

2− s)− 1

2(1− s) +

(α− β)(1− s)γ

+β

γ, 0 ≤ s ≤ 1

2,

−1

2(1− s) +

(α− β)(1− s)γ

+β

γ,1

2≤ s ≤ 1.

Proof.From (1.21), we have u′′(t) = −y(t). For t ∈ [0, 1], integrating from0 to t we get

u′(t) = −∫ t

0

y(s)ds+ A1, (1.22)

since u′(t) = −u′(1−t), we can find that −∫ t

0y(s)ds+A1 = −

∫ 1−t0

y(s)ds−A1,


which leads to A1 = 12

∫ t0y(s)ds − 1

2

∫ 1−t0

y(s)ds = 12

∫ t0y(s)ds + 1

2

∫ 1−t0

y(1 −s)d(1− s) = 1

2

∫ t0y(s)ds+ 1

2

∫ 1

ty(s)ds = 1

2

∫ 1

0y(s)ds =

∫ 1

0(1− s)y(s)ds.

Integrating again we obtain

u(t) = −∫ t

0

(t− s)y(s)ds+ t

∫ 1

0

(1− s)y(s)ds+ A2.

From (1.21) and (1.22) we have

(α− β)A1 + β

∫ 1

0

y(s)ds = γ(−∫ 1

2

0

(1

2− s)ds+

1

2

∫ 1

0

(1− s)y(s)ds+ A2).

Thus

A2 =

∫ 12

0

[(1

2− s) +

α− βγ

(1− s) +β

γ− 1

2(1− s)]y(s)ds+∫ 1

12

[α− βγ

(1− s) +β

γ− 1

2(1− s)]y(s)ds.

From above we can obtain the BVP (1.21) has a unique symmetric solution

u(t) = −∫ t

0

(t− s)y(s)ds+ t

∫ 1

0

(1− s)y(s)ds

+

∫ 12

0

[(1

2− s) +

α− βγ

(1− s) +β

γ− 1

2(1− s)]y(s)ds

+

∫ 1

12

[α− βγ

(1− s) +β

γ− 1

2(1− s)]y(s)ds

=

∫ 1

0

G1(t, s)y(s)ds+

∫ 1

0

G2(s)y(s)ds =

∫ 1

0

[G1(t, s) +G2(s)]y(s)ds.

This completes the proof.Lemma 2.2. The function G(t, s) satisfies 3

4G(s, s) ≤ G(t, s) ≤ G(s, s) for

t, s ∈ [0, 1] if α, β, γ are defined in (1.20).Proof. For any t ∈ [0, 1] and s ∈ [0, 1

2], we have

G(t, s) = G1(t, s) +G2(s) ≥ G2(s) =1

4G2(s) +

3

4G2(s)

=1

4[(

1

2− s) +

α− βγ

(1− s) +β

γ− 1

2(1− s)]+

3

4[(

1

2− s) +

α− βγ

(1− s) +β

γ− 1

2(1− s)]

≥ s(1− s)G2(s) +3

4G2(s).


Note that | β − α |≤| γ2|,α + β ≥ 2γ,γ 6= 0, we obtain G2(s) ≥ 3

4.

Thus G(t, s) ≥ 34[G1(s, s) + G2(s)] = 3

4G(s, s). As in the same way we can

conclude G(t, s) ≥ 34G(s, s) for any t ∈ [0, 1] and s ∈ [1

2, 1]. 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

the BVP (1.20) is nonnegative on [0, 1].Proof. Let y ∈ C+[0, 1]. From the fact that u′′(t) = −y(t) ≤ 0, t ∈ [0, 1], weknow that the graph of u(t) is concave on [0, 1]. From (1.21). We have that

u(0) = u(1) =

∫ 12

0

[(1

2− s) +

α− βγ

(1− s) +β

γ− 1

2(1− s)]y(s)ds+∫ 1

12

[α− βγ

(1− s) +β

γ− 1

2(1− s)]y(s)ds ≥ 0.

Note that u(t) is concave, thus u(t) ≥ 0 for t ∈ [0, 1]. This completes theproof.

Lemma 2.4. Let y ∈ C+[0, 1], then the unique symmetric solution u(t) ofBVP (1.20) satisfies

mint∈[0,1]

u(t) ≥ 3

4‖ u ‖ . (1.23)

Proof. For any t ∈ [0, 1], on the one hand, from lemma 2.2 we have that

u(t) =∫ 1

0G(t, s)y(s)ds ≤

∫ 1

0G(s, s)y(s)ds. Therefore,

‖ u ‖≤∫ 1

0

G(s, s)y(s)ds. (1.24)

On the other hand, for any t ∈ [0, 1], from lemma 2.2 we obtain that

u(t) =

∫ 1

0

G(t, s)y(s)ds ≥ 3

4

∫ 1

0

G(s, s)y(s)ds ≥ 3

4‖ u ‖ . (1.25)

From (1.25) and (1.24) we find that (1.23) holds.Obviously, (u, v) ∈ C2[0, 1]×C2[0, 1] is the solution of (1.20) if and only if

(u, v) ∈ C[0, 1]× C[0, 1] is the solution of integral equationsu(t) =

∫ 1

0

G(t, s)f(s, v(s))ds,

v(t) =

∫ 1

0

G(t, s)g(s, u(s))ds.

(1.26)

Integral equations (1.26) can be transferred to the nonlinear integral equation

u(t) =

∫ 1

0

G(t, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds (1.27)


Define an integral operator A : C → C by

Au(t) =

∫ 1

0

G(t, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds. (1.28)

It is easy to see that the BVP (1.20) has a solution u = u(t) if and only if u isa fixed point of the operator A defined by (1.28).Let P = u ∈ C+[0, 1] : u(t) is symmetric, concave on [0, 1] and min0≤t≤1u(t) ≥34‖ u ‖. It is obvious that P is a positive cone in C[0, 1].

Lemma 2.5. If the operator A is defined as in (1.28), then A : P → P iscompletely continuous.Proof. It is obvious that Au is symmetric on [0, 1]. Note that (Au)′′(t) −f(t, v(t)) ≤ 0, so we have that Au is concave. Thus from lemma 2.2 andnon-negativity of f and g,

Au(t) =

∫ 1

0

G(t, s)f(s,

∫ 1

0


≤∫ 1

0

G(s, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds,

then

‖ Au ‖≤∫ 1

0

G(s, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds.

On the other hand,

Au ≥ 3

4

∫ 1

0

G(s, s)f(s,

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ)ds ≥ 3

4‖ Au ‖ .

Thus, A(P ) ⊂ P . Since G(t, s), f(t, u) and g(t, u) are continuous, it is easy tosee 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 that 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 addition

suppose either(1) ‖ Au ‖≤‖ u ‖, ∀u ∈ P

⋂∂Ω1 and ‖ Au ‖≥‖ u ‖, ∀u ∈ P

⋂∂Ω2 or

(2) ‖ Au ‖≤‖ u ‖, ∀u ∈ P⋂∂Ω2 and ‖ Au ‖≥‖ u ‖, ∀u ∈ P

⋂∂Ω1

holds. 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, Ω2 and Ω3 are open subsets of E with 0 ∈ Ω1 , Ω1 ⊂ Ω2,Ω2 ⊂ Ω3 and let A : P

⋂(Ω3\Ω1) → P be a completely continuous operator.

In addition suppose either(1) ‖ Au ‖≥‖ u ‖, ∀u ∈ P

⋂∂Ω1;

(2) ‖ Au ‖≤‖ u ‖, Au 6= u, ∀u ∈ P⋂∂Ω2;

(3) ‖ Au ‖≥‖ u ‖, ∀u ∈ P⋂∂Ω3

holds.Then A has at least two fixed-points x1,x2 in P⋂

(Ω3\Ω1), and further-more x1 ∈ P

⋂(Ω2\Ω1),x2 ∈ P

⋂(Ω3\Ω2).


1.2.3 Existence of positive solutions

In this section, we study the existence of positive solutions for BVP (1.20).First we give the following assumptions:

(H1) limu→0+ sup0≤x≤1f(t, u)u = 0,limu→0+ sup0≤x≤1

g(t, u)u = 0;

(H2) limu→∞ inf0≤x≤1f(t, u)u =∞,limu→∞ inf0≤x≤1

g(t, u)u =∞;

(H3) limu→0+ inf0≤x≤1f(t, u)u =∞,limu→0+ inf0≤x≤1

g(t, u)u =∞;

(H4) limu→∞ sup0≤x≤1f(t, u)u = 0,limu→∞ sup0≤x≤1

g(t, u)u = 0;

(H5)There exists a constant R1 > 0, such that f(s, u) ≤ R1∫ 1

0

G(s, s)ds

for

every (s, u) ∈ [0, 1]× [34R1, R1]

Theorem 3.1. If (H1) and (H2) are satisfied, then (1.20) has at least onesymmetric positive solution (u, v) ∈ C2([0, 1], R+) × C2([0, 1], R+) satisfyingu(t) > 0, v(t) > 0.Proof. From (H1) there is a number N1 ∈ (0, 1) such that for each (s, u) ∈[0, 1] × (0, N1), one has f(s, u) ≤ η1u, g(s, u) ≤ η1u, where η1 > 0 satisfies

η1

∫ 1

0G(s, s)ds ≤ 1, for every u ∈ P and ‖ u ‖= N1

2, note that∫ 1


∫ 1

0G(ξ, ξ)g(ξ, u(ξ))dξ ≤

∫ 1

0η1G(ξ, ξ)u(ξ)dξ ≤‖ u ‖=

N1

2< N1,then

Au(x) ≤∫ 1

0

G(t, s)f(s,

∫ 1

0


≤ η1

∫ 1

0

G(s, s)

∫ 1

0


≤ η21

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)u(ξ)dξds ≤‖ u ‖ .

Let

Ω1 = u ∈ C+[0, 1], ‖ u ‖< N1

2,

then

‖ Au ‖≤‖ u ‖, u ∈ P⋂

∂Ω1. (1.29)

From (H2) there is a number N2 > 2N1 for each (s, u) ∈ [0, 1] × (N2,+∞),

one has f(s, u) ≥ η2u,g(s, u) ≥ η2u where η2 > 0 satisfies η2

√278

∫ 1

0G(s, s)ds ≥

1,then, for every u ∈ P and ‖ u ‖= 2N2, from lemma 2.2 and lemma 2.4, we


have ∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ ≥ 9

16

∫ 1

0

η2G(ξ, ξ) ‖ u ‖ dξ

≥ 9

6√

3‖ u ‖=

√3N2 > N2,

then

‖ Au ‖=∫ 1

0

G(t, s)f(s,

∫ 1

0


≥ 3

4η2

∫ 1

0

G(s, s)

∫ 1

0


≥ (3

4)2η2

2

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)u(ξ)dξds

≥ (3

4)3η2

2

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ) ‖ u ‖ dξds ≥‖ u ‖ .

Let

Ω2 = u ∈ C+[0, 1], ‖ u ‖< 2N2,

then

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω2. (1.30)

Thus from(1.29),(1.30) and Lemma 2.6,we see that the operator A has a fixedpoint in P


Theorem 3.2. If (H3) and (H4) are satisfied, then (1.20) has at least onesymmetric positive solution (u, v) ∈ C2([0, 1], R+) × C2([0, 1], R+) satisfyingu(t) > 0, v(t) > 0.Proof. From (H3), there is a number N3 ∈ (0, 1) such that for each (x, u) ∈[0, 1] × (0, N3), one has f(s, u) ≥ η3u, g(s, u) ≥ η3u where η3 > 0 satisfies√

278η3

∫ 1

0G(s, s)ds ≥ 1. From g(x, 0) ≡ 0 and the continuity of g(s, u), we

know that there exists number N3 ∈ (0, N3) such that g(s, u) ≤ N3∫ 10 G(s,s)ds

for

each (s, u) ∈ [0, 1]× (0, N3]. Then for every u ∈ P and ‖ u ‖= N3, note that∫ 1


∫ 1

0G(ξ, ξ) N3∫ 1

0 G(s,s)dsdξ = N3. Thus


Au(x) =

∫ 1

0

G(t, s)f(s,

∫ 1

0


≥ 3

4η3

∫ 1

0

G(s, s)

∫ 1

0


≥ (3

4)2η2

3

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)u(ξ)dξds

≥ (3

4)3η2

3

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ) ‖ u ‖ dξds ≥‖ u ‖ .

Let

Ω3 = u ∈ C+[0, 1], ‖ u ‖< N3,

then

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω3. (1.31)

From (H4), there exist C1 > 0 and C2 > 0 such that f(s, u) ≤ η4u +C1, g(s, u) ≤ η4u + C2 for ∀(s, u) ∈ [0, 1] × (0,∞), where η4 > 0, and

η4

∫ 1

0G(ξ, ξ)dξ ≤ 1. Then, for u ∈ C+[0, 1] we have

Au =

∫ 1

0

G(s, t)f(s,

∫ 1

0


≤∫ 1

0

G(s, s)(η4

∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ + C1)ds

≤ η4

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)g(ξ, u(ξ))dξds+ C3

≤ η4

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ)(η4u+ C2)dξds+ C3

≤ (η4)2

∫ 1

0

G(s, s)

∫ 1

0

G(ξ, ξ) ‖ u ‖ dξds+ C4 ≤‖ u ‖ +C4

Thus ‖ Au ‖≤‖ u ‖ as ‖ u ‖→ ∞.

Let Ω4 = u ∈ E, ‖ u ‖< L4. For each u ∈ P and ‖ u ‖= L4 > L3 largeenough, we have

‖ Au ‖≤‖ u ‖, u ∈ P⋂

∂Ω4. (1.32)

Thus from(1.31),(1.32) and Lemma 2.6,we know that the operator A has afixed point in P


Theorem 3.3. If (H2), (H3)and (H5) are satisfied, then (1.20) has at leasttwo symmetric positive solutions (u1, v1), (u2, v2) ∈ C2([0, 1], R+)×C2([0, 1], R+)


satisfying u1(t) > 0, v1(t) > 0, u2(t) > 0, v2(t) > 0.Proof.Let

Ω5 = u ∈ C+[0, 1], ‖ u ‖< R1,

then ∀u ∈ P⋂∂Ω5, we have u(s) ∈ [3

4R1, R1]. From lemma 2.2,lemma 2.4 and

(1.24) we can obtain∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ ≥ 3

4

∫ 1

0

G(ξ, ξ)g(ξ, u(ξ))dξ ≥ 3

4‖ u ‖∫ 1

0

G(s, ξ)g(ξ, u(ξ))dξ ≤∫ 1

0

G(ξ, ξ)g(ξ, u(ξ))dξ

≤∫ 1

0

G(ξ, ξ)dξR1

G(s, s)ds= R1.

(1.33)

Thus Au =∫ 1

0G(s, t)f(s,

∫ 1

0G(s, ξ)g(ξ, u(ξ))dξ)ds ≤

∫ 1

0G(s, s) R1∫ 1

0 G(ξ,ξ)dξds =

R1 =‖ u ‖. Then

‖ Au ‖≤‖ u ‖, u ∈ P⋂

∂Ω5. (1.34)

For another hand, from (H2) and (H3), we can choose two right numbers

N2 ∈ (R1,∞), N3 ∈ (0, R1) satisfy

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω2, (1.35)

‖ Au ‖≥‖ u ‖, u ∈ P⋂

∂Ω3, (1.36)

where Ω2 = u ∈ C+[0, 1], ‖ u ‖< N2, Ω3 = u ∈ C+[0, 1], ‖ u ‖< N3.Then from lemma 2.7, (1.34), (1.35) and (1.36), A has at least two fixed points

in P⋂

(Ω2\Ω5) and P⋂

(Ω5\Ω3) , respectively. The proof is complete.

1.2.4 Examples

In this section, we give three examples to illustrate our results.

Example 4.1. Let f(t, v) = v2 + [1+t(1−t)]v21+v2

,g(t, u) = 2u2 + 2[1+t(1−t)]u21+u2

,α = β = γ = 1, then the conditions of Theorem 3.1 are satisfied. FromTheorem 3.1, BVP(1.20) has at least one symmetric positive solution.

Example 4.2. Let f(t, v) = v12 + [1+t(1−t)]v2

1+v2,g(t, u) = 2u

12 + 2[1+t(1−t)]u2

1+u2,

α = β = γ = 1, then conditions of Theorem 3.2 are satisfied. From Theorem3.2, BVP(1.20) has at least one symmetric positive solution.

Example 4.3. Let f(t, v) = t(1−t)+14

(v12 + v2),g(t, u) = t(1−t)+1

5(u

12 + u2),

α = β = γ = 1, then conditions of Theorem 3.3 are satisfied. From Theorem3.3, BVP(1.20) has at least two symmetric positive solutions.

Chapter 2

Fractional Differential Equations

2.1 Positive Solution for Boundary Value Prob-

lem of Fractional Differential Equation

Abstract: In this paper, we prove the existence of the solution for boundaryvalue problem(BVP) of fractional differential equations of order q ∈ (2, 3].The Krasnoselskii’s fixed point theorem is applied to establish the results. Inaddition, we give an detailed example to demonstrate the main result.

Mathematics Subject Classification: 30E25, 34A08

Keywords:Fractional differential equation; Krasnoselskii’s fixed point the-orem; Boundary value problem

2.1.1 Introduction

Fractional differential equations are the generalization of ordinary differentialequation to arbitrary non-integer order, and have received more and moreinterest due to their wide applications in various sciences, such as physics,chemistry, biophysics, capacitor theory, blood flow phenomena, electrical cir-cuits, control theory, etc, also recent investigations have demonstrated that thedynamics of many systems are described more accurately by using fractionaldifferential equations. So fractional differential equations have attracted manyauthors.In [25], Nickolai was concerned with the nonlinear differential equation of frac-tional order

Dq0+u(t) = f(t, u(t), u′(t)) a.e. t ∈ (0, 1),

where Dq0+ is Riemann-Liouville(R-L) fractional order derivative, subject to

the boundary conditions u(0) = u(1) = 0. The author obtained the existenceof at least one solution by using the Leray-Schauder Continuation Principle.

20

Fractional Differential Equations 21

In [16], Zhang has given the existence of positive solution to the equationcDqu(t) + f(t, u(t)) = 0, 0 < t < 1,u(0) + u′(0) = u(1) + u′(1) = 0,

by the use of classical fixed point theorems, wherecDq denotes Caputo frac-tional derivative with 1 < q ≤ 2. Very recently, Chen (see[17]) considered theexistence of three positive solutions to three-point boundary value problem ofthe following fractional differential equation

Dq0+u(t) + f(t, u(t)) = 0, 0 < t < 1,

u(0) = 0, Dp0+u(t) |t=1= αDp

0+u(t) |t=ξ,

where 1 < q ≤ 2, 0 < p < 1, 1 + p ≤ q, and Dq0+ is the R-L fractional order

derivative. The multiplicity results of positive solutions to the equations areobtained by using the well-known Leggett-Williams fixed-point theorem on aconvex cone. The other excellent studies of fractional differential equationscan be founded in [47, 19, 20, 21, 22].Motivated by the paper mentioned above, we study the existence of positivesolution to two-point BVP of nonlinear fractional equation

Dq0+u(t) + λf(t, u(t)) = 0, 0 < t < 1,

u(0) = Dp0+u(t) |t=0= Dp

0+u(t) |t=1= 0,(2.1)

where q, p ∈ R, 2 < q ≤ 3, 1 < p ≤ 2, 1+p ≤ q, Dq0+ is the R-L fractional order

derivative, and f ∈ C([0, 1] × [0,∞), [0,∞)), λ > 0. By using Krasnoselskii’sfixed point theorem, the positive solution to the equations (1.1) is obtained.

2.1.2 Preliminaries

In this section, we present some definitions and preliminary results.Definition 2.1. (see equation (2.1.1) in [47]) The R-L fractional integralsIp0+f of order p ∈ R (p > 0) is defined by

Ip0+f(x) :=1

Γ(p)

∫ x

0

f(t)dt

(x− t)1−p , (x > 0).

Here Γ(p) is the Gamma function.Definition 2.2.(see equation (2.1.5) in [47]) The R-L fractional derivative

Dp0+f of order p ∈ R (p > 0) is defined by

Dp0+f(x) =(

d

dx)nIn−p0+ f(x)

=1

Γ(n− p)(d

dx)n∫ x

0

f(t)dt

(x− t)p−n+1, (n = [p] + 1, x > 0),

22 Fractional Differential Equations

where [p] means the integral part of p.Lemma 2.1.(see Lemma 2.4 and property 2.2 in [47]) If q1 > q2 > 0, then,

for f(x) ∈ Lp(0, 1), (1 ≤ p ≤ ∞), the relations

Dq20+I

q10+f(x) =Iq1−q20+ f(x),

Iq10+Iq20+f(x) = Iq1+q2

0+ f(x) and Dq10+I

q10+f(x) = f(x)

hold a.e. on [0,1].Lemma 2.2.(see Lemma 2.5 in [47]) Let q > 0, n = [q]+1, f(x) ∈ L1(0, 1),

then the equality

Iq0+Dq0+f(x) = f(x) +

n∑i=1

Citq−n.

Lemma 2.3. Let y ∈ C[0, 1], 2 < q ≤ 3, 1 < p ≤ 2, 1 + p ≤ q, then theproblem

Dq0+u(t) + y(t) = 0, 0 < t < 1, (2.2)

subject to the boundary conditions

u(0) = Dp0+u(t) |t=0= Dp

0+u(t) |t=1= 0, (2.3)

has the unique solution u(t) =∫ 1

0G(t, s)ds, where

G(t, s) =1

Γ(q)

tq−1(1− s)q−p−1 − (t− s)q−1, 0 ≤ s ≤ t ≤ 1,tq−1(1− s)q−p−1, 0 ≤ t ≤ s ≤ 1.

And that G(t, s) has the following propertiesI) G(t, s) ∈ C([0, 1]× [0, 1]), and G(t, s) > 0 for t, s ∈ (0, 1), and max

0≤t≤1G (t, s)

=G(s, s), s ∈ (0, 1).II)There exists a positive function ϕ ∈ C((0, 1)× (τ,+∞)) such that

min14≤t≤ 3

4

G (t, s) = ϕ(s)G(s, s) ≥ inf0<s<1

ϕ (s) max0≤t≤1

G (t, s) = τG(s, s),

where

G(s, s) =sq−p(1− s)q−p−1

Γ(q), s, τ ∈ (0, 1), τ = inf

0<s<1ϕ (s) .

Proof. Applying the operator Iq0+ to both sides of the equation (2.1), and usingLemma 2, we have

u(t) = −Iq0+y(t) + C1tq−1 + C2t

q−2 + C3tq−3. (2.4)

In view of the boundary condition u(0) = 0, we find that C3 = 0, hence

u(t) = −Iq0+y(t) + C1tq−1 + C2t

q−2,


then, noting the relation Dq20+I

q10+f(x) = Iq1−q20+ f(x) in Lemma 1, we obtain

Dp0+u(t) = −Iq−p0+ y(t) + C1

Γ(q)

Γ(q − p)tq−p−1 + C2

Γ(q − 1)

Γ(q − p− 1)tq−p−2,

in accordance with the equations (2.2) ,we can calculate out that

C1 =1

Γ(q)

∫ 1

0

(1− s)q−p−1y(s)ds, C2 = 0.

Substituting the values of C1,C2 and C3 in (2.3), we have

u(t) =− 1

Γ(q)

∫ t

0

(t− s)q−1y(s)ds+tq−1

Γ(q)

∫ 1

0

(1− s)q−p−1y(s)ds

=1

Γ(q)∫ t

0

[tq−1(1− s)q−p−1 − (t− s)q−1]y(s)ds

+

∫ 1

t

[tq−1(1− s)q−p−1]y(s)ds

=

∫ 1

0

G(t, s)y(s)ds.

Next we prove the properties of G(t, s).For a given s ∈ (0, 1), G(t, s) is decreasing with respect to t for s ≤ t whileincreasing for t ≤ s. Thus, we have

max0≤t≤1

G (t, s) = G(s, s) =sq−1(1− s)q−p−1

Γ(q)≤ sq−p(1− s)q−p−1

Γ(q)= G(s, s),

for s ∈ (0, 1). Then we set

g1(t, s) =tq−1(1− s)q−p−1 − (t− s)q−1

Γ(q), g2(t, s) =

tq−1(1− s)q−p−1

Γ(q),

from the two equation above we have

min14≤t≤ 3

4

G (t, s) =1

Γ(q)

0.75q−1(1− s)q−p−1 − (0.75− s)q−1, 0 < s ≤ r,0.25q−1(1− s)q−p−1, r ≤ s < 1,

where 14< r < 3

4is the unique solution of the equation

0.75q−1(1− s)q−p−1 − (0.75− s)q−1 = 0.25q−1(1− s)q−p−1.

Finally, we consider a function ϕ(s) defined by

ϕ(s) =

min14≤t≤ 3

4

G (t, s)

G(s, s)=

0.75q−1(1−s)q−p−1−(0.75−s)q−1

sq−p(1−s)q−p−1 , 0 < s ≤ r,0.25q−1

sq−p , r ≤ s < 1.


When q > p− 1 we find from the continuity of ϕ(s) and lims→0+

ϕ (s) = +∞ that

there exists r small enough such that ϕ′(s) < 0 for s ∈ (0, r], hence, we set

0 < τ = inf0<s<1

ϕ (s) = minϕ(r),m,1

4q−1 < 1,

here, m = minr≤s≤r

ϕ (s).

When q = p− 1, we have lims→0+

ϕ (s) = 43(q − 1), then we set

0 < τ = inf0<s<1

ϕ (s) = min inf0<s≤r

ϕ (s) ,4

3(q − 1),

1

4q−1 < 1.

Thus,

min14≤t≤ 3

4

G (t, s) ≥ ϕ(s)G(s, s) ≥ inf0<s<1

ϕ (s) max0≤t≤1

G (t, s) = τG(s, s).

This completes the proof. Therefore, the solution u ∈ C[0,1] of the problem(1.1) can be written by

u(t) = λ

∫ 1

0

G(t, s)f(s, u(s))ds.

Lemma 2.4.(see[23]) Let E be a Banach space and P ⊂ E is a cone inE. Assume that 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 either(1) ‖Au‖ ≤ ‖u‖, ∀u ∈ P ∩ ∂Ω1 and ‖Au‖ ≥ ‖u‖, ∀u ∈ P ∩ ∂Ω2 or(2) ‖Au‖ ≤ ‖u‖, ∀u ∈ P ∩ ∂Ω2 and ‖Au‖ ≥ ‖u‖, ∀u ∈ P ∩ ∂Ω1

holds. Then A has a fixed point in P ∩ (Ω2\Ω1).Define P to be a cone in C[0,1](with norm ‖u‖ = max

0≤t≤1|u (t) |) by

P = u ∈ C[0,1] | u(t) ≥ 0, t ∈ [0, 1] and min14≤t≤ 3

4

u (t) ≥ τ‖u‖,

and the operator A : P → C[0,1] by

Au(t) = λ

∫ 1

0

G(t, s)f(s, u(s))ds. (2.5)

Lemma 2.5. If A is defined by (2.4), then A : P → P is completelycontinuous. Proof. First, assume that f ∈ C([0, 1] × [0,∞), [0,∞)), u ∈ P ,


and from Lemma 3, we have

min14≤t≤ 3

4

Au (t) = min14≤t≤ 3

4

λ

∫ 1

0

G(t, s)f(s, u(s))ds

≥ max0≤t≤1

λ

∫ 1

0

inf0<s<1

ϕ (s)G(t, s)f(s, u(s))ds

= τ max0≤t≤1

λ

∫ 1

0

G(t, s)f(s, u(s))ds

= τ‖Au(t)‖,

thus A : P → P .Second, ∀N > 0, Let Ω = Ω ⊂ P : ‖u‖ ≤ N, u ∈ Ω, M = max

(t,u)∈[0,1]×[0,N ]f(t, u(t)),

and noting the property (II) of G(t, s), we can easily obtain A(Ω) is bounded.Third, for each u ∈ Ω, let t1, t2 ∈ [0, 1] such that t1 < t2, then we have

|Au(t2)− Au(t1)| =λ|∫ 1

0

G(t2, s)f(s, u(s))ds−∫ 1

0

G(t1, s)f(s, u(s))ds|

=λ

Γ(q)|∫ t1

0

[tq−12 (1− s)q−p−1 − (t2 − s)q−1]

− [tq−11 (1− s)q−p−1 − (t1 − s)q−1]f(s, u(s))ds

+

∫ t2

t1

[tq−12 (1− s)q−p−1 − (t2 − s)q−1]

− tq−11 (1− s)q−p−1f(s, u(s))ds

+

∫ 1

t2

[tq−12 (1− s)q−p−1 − tq−1

1 (1− s)q−p−1]f(s, u(s))ds|

<λ

Γ(q)

∫ 1

0

(tq−12 − tq−1

1 )(1− s)q−p−1f(s, u(s))ds

<λM

Γ(q)(tq−1

2 − tq−11 )

∫ 1

0

(1− s)q−p−1ds

=λM

Γ(q)(q − p)(tq−1

2 − tq−11 )

=λM(q − 1)

Γ(q)(q − p)[t1 + θ(t2 − t1)]q−2(t2 − t1), (0 < θ < 1)

<λM(q − 1)

Γ(q)(q − p)2q−2(t2 − t1).

Thus, ∀ε > 0, ∃δ = ε Γ(q)(q−p)2q−2λM(q−1)

, we have |Au(t2)−Au(t1)| < ε for t2− t1 < δ.

Therefore, A(Ω) is equivalent-continuous, so the Arzela-Ascoli theorem impliesthat the operator A : P → P is completely continuous. This completes theproof.


2.1.3 Main Results

In this section, we study the existence of the positive solution to BVP ofequations (1.1). Suppose

(H1) limu→0+

sup0≤t≤1

f(t, u)

u= 0, (H2) lim

u→+∞inf

0≤t≤1

f(t, u)

u= +∞,

(H3) limu→0+

inf0≤t≤1

f(t, u)

u= +∞, (H4) lim

u→+∞sup

0≤t≤1

f(t, u)

u= 0.

Theorem 3.1. If (H1) and (H2) hold, then for all λ > 0, the equations(1.1) have a positive solution.

Theorem 3.2. If (H3) and (H4) hold, then for all λ > 0, the equations(1.1) have a positive solution.

The Proof of Theorem 1. From (H1), there exists L1 ∈ (0, 1) such that

f(t, u) ≤ η1u for (t, u) ∈ [0, 1]×(0, L1], where η1 > 0 satisfying λη1

∫ 1

0G(s, s)ds ≤

1. Then let Ω1 = u ∈ P : ‖u‖ < L1, ∂Ω1 = u ∈ P : ‖u‖ = L1, for u ∈ ∂Ω1,we have

Au(t) = λ

∫ 1

0

G(t, s)f(s, u(s))ds

≤ λ max0≤t≤1

∫ 1

0

G(t, s)f(s, u(s))ds

≤ λη1

∫ 1

0

G(s, s)u(s)ds

≤ λη1

∫ 1

0

G(s, s)ds‖u‖ ≤ ‖u‖,

which implies that

‖Au‖ ≤ ‖u‖, for u ∈ ∂Ω1. (2.6)

On the other hand, from (H2), there exists L2 > L1 such that f(t, u) ≥ η2u for

(t, u) ∈ [0, 1]× [L2,∞), where η2 > 0 satisfying λη2τ2∫ 3

414

G(s, s)ds ≥ 1. Then

let Ω2 = u ∈ P : ‖u‖ < L2, ∂Ω2 = u ∈ P : ‖u‖ = L2, for u ∈ ∂Ω2, we


have

min14≤t≤ 3

4

Au (t) = min14≤t≤ 3

4

λ

∫ 1

0

G(t, s)f(s, u(s))ds

≥λτ∫ 1

0

G(s, s)f(s, u(s))ds

≥λτη2

∫ 1

0

G(s, s)u(s)ds

≥λτη2

∫ 34

14

G(s, s)u(s)ds

≥λτ 2η2

∫ 34

14

G(s, s)ds‖u‖ ≥ ‖u‖,

which implies that‖Au‖ ≥ ‖u‖, for u ∈ ∂Ω2. (2.7)

Then from (3.1), (3.2) and Lemma 4, the operator A has a fixed point inP ∩ (Ω2\Ω1).The Proof of Theorem 2. By the similar method of the proof of Theorem 1,we can easily obtain Ω3 = u ∈ P : ‖u‖ < L3, ∂Ω3 = u ∈ P : ‖u‖ = L3,and Ω4 = u ∈ P : ‖u‖ < L4, ∂Ω4 = u ∈ P : ‖u‖ = L4, and satisfying

‖Au‖ ≥ ‖u‖, for u ∈ ∂Ω3, (2.8)

and‖Au‖ ≤ ‖u‖, for u ∈ ∂Ω4, (2.9)

respectively. Then from (3.3), (3.4) and Lemma 4, we obtain a fixed point ofoperator A in P ∩ (Ω4\Ω3).

2.1.4 Example

We consider the following problemD

520+u(t) + (t+ 1)u2 = 0, 0 < t < 1,

u(0) = D320+u(t) |t=0= D

320+u(t) |t=1= 0,

(2.10)

Then f(t, u) = (t+1)u2, λ = 1, and limu→0+

sup0≤t≤1

(t+ 1)u2

u= 0, lim

u→+∞inf

0≤t≤1

(t+ 1)u2

u=

∞, so the condition (H1) and (H2) hold. On the other hand, substituting theequations q = 5

2and p = 3

2in G(t, s) and ϕ(s), we have

G(t, s) =1

Γ(52)

t32 − (t− s) 3

2 , 0 ≤ s ≤ t ≤ 1,

t32 , 0 ≤ t ≤ s ≤ 1,


and

ϕ(s) =

(0.75)

32−(0.75−s)

32

s, 0 < s ≤ r,

(0.25)32

s, r ≤ s < 1,

where r is the unique solution of the equation

(0.75)32 − (0.75− s)

32 = (0.25)

32 .

By calculating the minimum of ϕ(s), we obtain τ = 18. Thus, we set L1 = 1

2,

η1 = 2 ≤ 1∫ 10 G(s,s)ds

= 52Γ(5

2), then f(t, u) = (t + 1)u2 ≤ 2u2 ≤ η1u, for

(t, u) ∈ [0, 1]× [0, L1]. Therefore, we derive

Ω1 = u ∈ P : ‖u‖ < 1

2. (2.11)

Next we set L2 =5120Γ( 5

2)

352−1

, η2 ≥ 1

τ2∫ 0.750.25 G(s,s)ds

=5120Γ( 5

2)

352−1

, then f(t, u) = (t +

1)u2 ≥ u2 ≥ η2u, for (t, u) ∈ [0, 1]× [L2,+∞). Therefore, we derive

Ω2 = u ∈ P : ‖u‖ <5120Γ(5

2)

352 − 1

. (2.12)

According to (4.2) and (4.3), from Theorem 1, we obtain a positive solution u

of (4.1) such that 12≤ ‖u‖ ≤ 5120Γ( 5

2)

352−1

.

2.2 HAM for A Class of Time Fractional Par-

tial Differential Equations

Abstract: In this paper, we investigate the two-dimensional time fractionalpartial differential equation(FPDE) by using homotopy analysis method(HAM).Series solution is obtained for the FPDE with initial and boundary conditions.Numerical results and computer graphics show that the HAM is efficient insolving FPDE.

Mathematics Subject Classification: 26A33; 34A08

Keywords:Fractional partial differential equation; Homotopy analysis method;Series solution

2.2.1 Introduction

Recently, fractional differential equations interested many researchers due totheir widely application in physics[36], [37], [38], chemistry[39], engineering,


signal processing[40] and systems identification etc. These applications in in-terdisciplinary sciences motivate us to try to find numerical or analytic solu-tions for fractional differential equations.

The HAM was first proposed by Liao in 1992[41], [42]. This method hasbeen successfully applied to solve various linear or nonlinear problems [43],[44], [45], [46]. In this paper, we use this method to solve the two-dimensionaltime fractional partial differential equation which is given by

cDα0,tu(x, y, t) =

1

a2 + b2(uxx(x, y, t) + uyy(x, y, t)), (2.13)

where 1 < α ≤ 2, It is worth to point out that HAM gives rapidly convergentsuccessive approximates of exact solution.

2.2.2 Basic definitions and Lemmas

Definition II.1 [47] A real function f(x), x > 0, is said to be in the space Cµ,µ ∈ R, if there exists a real number p > µ, such that f(x) = xpf1(x), wheref1(x) ∈ C(0,∞), and it is said to be in the space Cn

µ , if and only if fn ∈ Cµ,n ∈ N .

Definition II.2 [47] The Riemann-Liouville fractional integral of orderα ∈ R, α > 0 of a function f(x) ∈ Cµ, µ ≥ −1 is defined as

(Iα0+f(t))(x) :=1

Γ(α)

∫ x

0

f(t)dt

(x− t)1−α , (x > 0).

Definition II.3 [47] The Riemann-Liouville fractional derivative of orderα ∈ R, α > 0, on the usual Lebesgue space L1[a, b] is given by

(Dα0+f(t))(x) :=

1

Γ(n− α)(d

dx)n∫ x

0

f(t)dt

(x− t)α−n+1,

where (n = [α] + 1, x > 0).Definition II.4 [47] The Caputo fractional derivative of f(x) ∈ C−1

µ , n ∈N , is defined as

cDα0,xf(x) :=

In−α0+ (

d

dx)nf(x), n− 1 < α < n,

(d

dx)nf(x), α = n.

Definition II.5 The classical Mittag-Leffler function[47] is defined by

Eα(Z) :=∞∑k=0

Zk

Γ(αk + 1), (Z ∈ C, α > 0).


Definition II.6 The functions Sinhα(Z) and Coshα(Z) (Z ∈ C, α > 0)are defined by

Sinhα(Z) =∞∑k=1

Z2k−1

Γ(α(2k − 1) + 1),

Coshα(Z) =∞∑k=0

Z2k

Γ(α(2k) + 1).

Obviously, hyperbolic sine’s or cosine’s form equation

Sinhα(Z) =1

2(Eα(Z)− Eα(−Z))

and

Coshα(Z) =1

2(Eα(Z) + Eα(−Z))

hold.Lemma II.1 [47] If α > 0, a ∈ R and λ ∈ C, then

(cDαa+Eα[λ(t− a)α])(x) = λEα[λ(x− a)α].

Lemma II.2 If Sinhα(Z) and Coshα(Z) are defined as in Definition II.6,then we have

(cDαa+Sinhα[λ(t− a)α])(x) = λCoshα[λ(x− a)α],

(cDαa+Coshα[λ(t− a)α])(x) = λSinhα[λ(x− a)α].

Proof The above two equations are proved directly by the term-by-term dif-ferentiation of the series in the Sinhα[λ(t− a)α])(x) and Coshα[λ(t− a)α])(x).

2.2.3 HAM

In this section, we consider a linear or nonlinear equation in a general form:

N [u(x, t)] = 0, (2.14)

where u(x, t) is an unknown function, x and t are independent variables. Letu0(x, t) denote an initial approximation of the solution of equation (2), h anonzero auxiliary parameter, H(x, t) a nonzero auxiliary function and L is anauxiliary linear operator. Then we construct the HAM deformation equationin the following form:

(1− q)L[Φ(x, t; q)− u0(x, t)] = qhH(x, t)N [Φ(x, t; q)] (2.15)


where q ∈ [0, 1] is an embedding parameter. Obviously, when q = 0 and q = 1,the above HAM deformation equation (3) has the solutions

Φ(x, t; 0) = u0(x, t), Φ(x, t; 1) = u(x, t),

respectively. Thus as q increases form 0 to 1, Φ(x, t; q) varies from the initialguesses Φ(x, t; 0) to the solution Φ(x, t; 1) of equation(2). Expanding Φ(x, t; q)in Taylor’s series with respect to q, we have

Φ(x, t; q) = u0(x, t) +∞∑m=1

um(x, t)qm,

where

um(x, t) =1

m!

∂mΦ(x, t; q)

∂qm|q=0.

For brevity, define a vector

−→um = u0, u1, . . . , um.

Differentiating the HAM deformation equation (3) m times with respect toq, then setting q = 0, and finally dividing it by m!, we obtain the mth-orderdeformation equation

L[um − χmum−1] = hH(x, t)Rm(−−−→um−1), (2.16)

where Rm(−−−→um−1) =1

(m− 1)!

∂m−1N [Φ(x, t; q)]

∂qm−1|q=0, and

χm =

0,m = 1,1,m > 1.

Operating the inverse operator of L on both sides of equation (4), we have

um(x, t) = χmum−1(x, t) + hH(x, t)L−1Rm(−−−−−−→um−1(x, t)).

In this way, it is easily to obtain u1(x, t), u2(x, t), . . . one after another, thenwe get an exact solution of the original equation (2)

u(x, t) =∞∑m=0

um(x, t).


2.2.4 Applying HAM

In this section, we consider the two-dimensional time fractional partial differ-ential equation (1) with initial and boundary conditions

u(x, y, 0) = sin(2π

ax)sin(

2π

by), ut(x, y, 0) = 0, (2.17)

andu(0, y, t) = u(a, y, t) = u(x, 0, t) = u(x, b, t) = 0. (2.18)

First, we choose the linear fractional order operator

L[ϕ(x, y, t; q)] = cDα0,tϕ(x, y, t; q).

Secondly, we define the linear operator as

N [ϕ(x, y, t; q)] = cDα0,tϕ(x, y, t; q)− ϕxx + ϕyy

a2 + b2.

Using above definitions, and with assumption H(x, t) = 1, we construct thezeroth-order deformation equation(ZDE).

(1− q)L[ϕ(x, y, t; q)− u0(x, y, t)] = qhN [ϕ(x, y, t; q)], (2.19)

obviously, when q = 0 and q = 1, it hold

ϕ(x, y, t; 0) = u0(x, y, t) = u(x, y, 0) = sin(2π

ax)sin(

2π

by),

andϕ(x, y, t; 1) = u(x, y, t).

Thirdly, differentiating the ZDE m times with respect to q, then setting q = 0,and dividing it by m!, we get the mth-order deformation equation.

L[um − χmum−1] = hN [Rm(−−−→um−1)], (2.20)

where

N [Rm(−−−→um−1)] = cDα0,tum−1 −

um−1,xx + um−1,yy

a2 + b2, (2.21)

um =1

m!

∂mϕ(x, y, t; q)

∂qm|q=0,

and

χm =

0,m = 1,1,m > 1.


Finally, operating the operator L−1 on both side of equation (8), we have

um = χmum−1 + hL−1N [Rm(−−−→um−1)].

Calculating one by one, we get

u0 = sin(2π

ax)sin(

2π

by),

u1 = h(2π

ab)2sin(

2π

ax)sin(

2π

by)

tα

Γ(1 + α),

u2 = (1 + h)u1 + h2(2π

ab)4sin(

2π

ax)sin(

2π

by)

t2α

Γ(1 + 2α),

u3 = (1 + h)u2 −h

a2 + b2Iα0 ((1 + h)u1,xx + (1 + h)u1,yy)

+h3(2π

ab)6sin(

2π

ax)sin(

2π

by)

t3α

Γ(1 + 3α),

u4 = (1 + h)u3 −h(Iα0 ((1 + h)u2,xx + (1 + h)u2,yy))

a2 + b2

+h2

(a2 + b2)2Iα0 ((Iα0 ((1 + h)u1,xx + (1 + h)u1,yy))xx

+ (Iα0 ((1 + h)u1,xx + (1 + h)u1,yy))yy)

+ h4(2π

ab)8sin(

2π

ax)sin(

2π

by)

t4α

Γ(1 + 4α),

......

By repeating this procedure for h = −1, we obtain the exact solution

u(x, y, t) =∞∑k=0

uk =∞∑k=0

u2k +∞∑k=1

u2k−1

= sin(2π

ax)sin(

2π

by)Coshα((

2π

ab)2tα)

−sin(2π

ax)sin(

2π

by)Sinhα((

2π

ab)2tα).

Fig.1, Fig.2, Fig.3 and Fig.4 show that the exact solution and the approximatesolution (23 terms) obtained by HAM for α = 2 and α = 1.9, respectively.Fig.5 and Fig.6 show the absolute error for the above two cases, respectively.If fixing y = 0.1, we see from Fig.3 or Fig.4 that the solution for the problemis a periodic degenerate function for 1 < α < 2, while, from Fig.1 or Fig.2, weknow that the solution is a periodic function for α = 2. Fig.7, Fig.8 show thepeak of the exact solution is on the decrease with respect to t changing from0 to 1.

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