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Volume 2011, Article ID 894135, 20 pagesdoi:10.1155/2011/894135 Research Article Solutions to a Three-Point Boundary Value Problem Jin Liang1 and Zhi-Wei Lv2, 3 1 Department of Mathemati

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Volume 2011, Article ID 894135, 20 pages

doi:10.1155/2011/894135

Research Article

Solutions to a Three-Point Boundary Value Problem

Jin Liang1 and Zhi-Wei Lv2, 3

1 Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China

2 Department of Mathematics and Physics, Anyang Institute of Technology, Anyang, Henan 455000, China

3 Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China

Correspondence should be addressed to Jin Liang,jinliang@sjtu.edu.cn

Received 25 November 2010; Accepted 19 January 2011

Academic Editor: Toka Diagana

Copyrightq 2011 J Liang and Z.-W Lv This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

By using the fixed-point index theory and Leggett-Williams fixed-point theorem, we study the

exis-tence of multiple solutions to the three-point boundary value problem ut atft, ut, ut 

0, 0 < t < 1; u0 u0 0; u1 − αuη λ, where η ∈ 0, 1/2, α ∈ 1/2η, 1/η are constants,

λ ∈ 0, ∞ is a parameter, and a, f are given functions New existence theorems are obtained, which

extend and complement some existing results Examples are also given to illustrate our results

1 Introduction

It is known that when diﬀerential equations are required to satisfy boundary conditions at more than one value of the independent variable, the resulting problem is called a multipoint boundary value problem, and a typical distinction between initial value problems and multipoint boundary value problems is that in the former case one is able to obtain the solutions depend only on the initial values, while in the latter case, the boundary conditions

at the starting point do not determine a unique solution to start with, and some random choices among the solutions that satisfy these starting boundary conditions are normally not to satisfy the boundary conditions at the other specified points As it is noticed

point boundary value problem has deep physical and engineering background as well as realistic mathematical model For the development of the research of multi point boundary value problems for diﬀerential equations in last decade, we refer the readers to, for example,

1,4 9 and references therein

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In this paper, we study the existence of multiple solutions to the following three-point boundary value problem for a class of third-order diﬀerential equations with inhomogeneous three-point boundary values,

ut atft, u t, ut 0, 0 < t < 1,

u 0 u0 0, u1 − αu

η

where η ∈ 0, 1/2, α ∈ 1/2η, 1/η, λ ∈ 0, ∞, and a, f are given functions To the authors’

knowledge, few results on third-order diﬀerential equations with inhomogeneous three-point boundary values can be found in the literature Our purpose is to establish new existence

Let X be an Banach space, and let Y be a cone in X A mapping β is said to be a nonnegative continuous concave functional on Y if β : Y → 0, ∞ is continuous and

β

tx 1 − ty≥ tβx 1 − tβy

, x, y ∈ Y, t ∈ 0, 1. 1.2

Assume that

H

a ∈ C 0, 1, 0, ∞, 0 <

1 0

1 − ssasds < ∞,

f ∈ C 0, 1 × 0, ∞ × 0, ∞, 0, ∞.

1.3

Define

v → 0max

t∈0,1 sup

u∈0,∞

f t, u, v

v ,

v → 0min

t∈0,1 inf

u∈0,∞

f t, u, v

v ,

v → ∞max

t∈0,1 sup

u∈0,∞

f t, u, v

v ,

v → ∞min

t∈0,1 inf

u∈0,∞

f t, u, v

v .

1.4

Trang 3

2 Lemmas

where

u max

t∈0,1 |ut|, u max

t∈0,1

Lemma 2.1 Let u ∈ C10, 1 be the unique solution of 1.1 Then

u t 

1 0

G t, sasfs, u s, usds

2

1− αη

1 0

G1

η, s

a sfs, u s, usds λt

2

1− αη ,

2.3

where

G t, s  1

2

⎧

⎨

⎩

2t − t2− ss, s ≤ t,

G1t, s ∂G t, s

∂t

⎧

⎨

⎩

1 − ts, s ≤ t,

1 − st, t ≤ s.

2.4

Lemma 2.2 One has the following.

i 0 ≤ G1t, s ≤ 1 − ss, 1/2t21 − ss ≤ Gt, s ≤ G1, s 1/21 − ss.

ii G1t, s ≥ 1/4G1s, s 1/41 − ss, for t ∈ 1/4, 3/4, s ∈ 0, 1.

iii G11/2, s ≥ 1/21 − ss, for s ∈ 0, 1.

Lemma 2.3 Let u ∈ C10, 1 be the unique solution of 1.1 Then ut is nonnegative and satisfies

u1 u.

Proof Let u ∈ C10, 1 be the unique solution of 1.1 Then it is obvious that ut is

Trang 4

i For t ≤ η,

2

1− αη

t 0

2ts − s2

1− αη t2s α − 1a sfs, u s, usds

η

t

t2

1

η

t21 − sasfs, u s, usds λt2

,

2

1− αη

t 0

2s

1− αη 2tsα − 1a sfs, u s, usds

η

t

2t

1

η

2t1 − sasfs, u s, usds 2λt

,

2.5

that is, ut ≤ ut.

ii For t ≥ η,

2

1− αη

η

0

2ts − s2

t

η

2ts − s2

1− αη t2

αη − s

a sfs, u s, usds

1

t

t21 − sasfs, u s, usds λt2

,

2

1− αη

η 0

2s

t

η

2s

1− αη 2tαη − s

a sfs, u s, usds

1

t

2t1 − sasfs, u s, usds 2λt

.

2.6

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On the other hand, for η ≤ s ≤ t, we have

2s

1− αη 2tαη − s

1− αη t2

αη − s

 αηt − s2 s − t s1 − t2 − t ss − t

 αηt − s

2 s − t − s

αη

 s1 − t2 − t.

2.7

Since α ∈ 1/2η, 1/η,

2s

1− αη 2tαη − s

≥2ts − s2

1− αη t2

αη − s

So, ut ≤ ut Therefore, ut ≤ ut, which means

The proof is completed

Lemma 2.4 Let u ∈ C10, 1 be the unique solution of 1.1 Then

min

t∈1/4,3/4 ut ≥ 1

Proof From2.3, it follows that

ut

1 0

G1t, sasfs, u s, usds

1 0

G1

η, s

≤

1 0

1 − ssasfs, u s, usds

1 0

G1

η, s

a sfs, u s, usds λ

1− αη .

2.11

Hence,

u1u ≤1

0

1− αη

1 0

G1

η, s

1− αη .

2.12

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By Lemmas2.2and2.3, we get, for any t ∈ 1/4, 3/4,

min

t∈1/4,3/4 ut min

t∈1/4,3/4

0

1 0

G1

η, s

4

0

1 0

G1

η, s

.

2.13

Thus,

min

t∈1/4,3/4 ut ≥ 1

Define a cone by

K 

u ∈ E : u ≥ 0, min

t∈1/4,3/4 ut ≥ 1

4u1

Set

K r {u ∈ K : u1< r }, ∂K r {u ∈ K : u1 r}, r > 0,

K r {u ∈ K : u1≤ r}, Kβ, r, s

u ∈ K : r ≤ β u, u1≤ s, s > r > 0.

2.16

Define an operator T by

Tu t 

1 0

2

1− αη

1 0

G1

η, s

2

1− αη .

2.17

Lemma 2.5 The operator defined in 2.17 is completely continuous and satisfies TK ⊂ K.

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Theorem 2.6 see 10 Let E be a real Banach Space, let K ⊂ E be a cone, and Ω r {u ∈ K : u ≤

r} Let operator T : K ∩Ω r → K be completely continuous and satisfy Tx / x, for all x ∈ ∂Ω r Then

i if Tx ≤ x, for all x ∈ ∂Ω r , then iT, Ω r , K 1,

ii if Tx ≥ x, for all x ∈ ∂Ω r , then iT, Ω r , K 0.

Theorem 2.7 see 8 Let T : P c → P c be a completely continuous operator and β a nonnegative continuous concave functional on P such that βx ≤ x for all x ∈ P c Suppose that there exist

0 < d0< a0 < b0≤ c such that

a {x ∈ Pβ, a0, b0 : βx > a0} / ∅ and βTx > a0for x ∈ P β, a0, b0,

b Tx < d0for x ≤ d0,

c βTx > a0for x ∈ P β, a0, c with Tx > b0.

Then, T has at least three fixed points x1, x2, and x3in P c satisfying

x1 < d0, a0< β x2, x3 > d0, β x3 < a0. 2.18

3 Main Results

In this section, we give new existence theorem about two positive solutions or three positive solutions for1.1

Write

Λ1

0

1 0

G1

η, s

a sds

−1

,

Λ2

1/4

3/4

1/4

G1

η, s

−1

.

3.1

Theorem 3.1 Assume that

H2 there exists a constant ρ1 > 0 such that ft, u, v ≤ 1/2Λ1ρ1, for t ∈ 0, 1, u ∈ 0, ρ1

and v ∈ 0, ρ1.

Then, the problem1.1 has at least two positive solutions u1and u2such that

for λ small enough.

Proof Since

v → 0min

t∈0,1 inf

u∈0,∞

f t, u, v

Trang 8

there is ρ0∈ 0, ρ1 such that

f t, u, v ≥ 8Λ2v, for t ∈ 0, 1, u ∈ 0, ∞, v ∈0, ρ0

, Λ2 > 0. 3.4 Let

Ωρ0 u ∈ K : u1< ρ0

Tu

1

2

1

0

G1

1

2, s

a sfs, u s, usds

2

1− αη

1 0

G1

η, s

2

1− αη

≥

1

0

G1

1

2, s

a sfs, u s, usds

2

1− αη

1 0

G1

η, s

a sfs, u s, usds

2

3/4

1/4

2

1− αη

3/4

1/4

G1

η, s

a s8Λ2usds

≥ 4Λ2

1/4

2

1− αη

3/4

1/4

G1

η, s

1

4u11

 u1.

3.6

Hence,

So

i

T, Ω ρ0 , K

On the other hand, since

v → ∞min

t∈0,1 inf

u∈0,∞

f t, u, v

Trang 9

there exist ρ0∗, ρ∗0> ρ1such that

f t, u, v ≥ 8Λ2v, for t ∈ 0, 1, u ∈ 0, ∞, v ≥ 1

LetΩρ∗0 {u ∈ K : u1 < ρ∗0} Then, by a argument similar to that above, we obtain

i

T, Ω ρ∗0, K

Finally, letΩρ1 {u ∈ K : u1 < ρ1}, and let λ satisfy 0 < λ ≤ 1/21 − αηρ1for any

u ∈ ∂Ω ρ1 Then,H2 implies

Tut

1

0

1 0

G1

η, s

≤

1

0

1 − ssas1

1 0

G1

η, s

a s1

2Λ1

0

1 0

G1

η, s

a sds

1− αηρ1

2ρ1 1

2ρ1

 u1,

3.14

which means thatTu ≤ u1 Thus,Tu1 ≤ u1, for all u ∈ ∂Ω ρ1

i

T, Ω ρ1 , K

From3.9–3.15 and ρ0< ρ1 < ρ∗0, it follows that

i

T, Ω ρ∗\ Ωρ1 , K

T, Ω ρ1\ Ωρ0 , K

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Therefore, T has fixed point u1 ∈ Ωρ1\ Ωρ0 and fixed point u2 ∈ Ωρ∗0\ Ωρ1 Clearly, u1, u2are

Theorem 3.2 Assume that

H4 there exists a constant ρ2 > 0 such that ft, u, v ≥ 2Λ2ρ2, for t ∈ 0, 1, u ∈ 0, ρ2 and

v ∈ 1/4ρ2, ρ2.

Then, the problem1.1 has at least two positive solutions u1and u2such that

for λ small enough.

Proof By

v → 0max

t∈0,1 sup

u∈0,∞

f t, u, v

we see that there exists ρ∗∈ 0, ρ2 such that

f t, u, v ≤ 1

2Λ1v, for t ∈ 0, 1, u ∈ 0, ∞, v ∈0, ρ∗

Put

Ωρ∗ u ∈ K : u1 < ρ∗

and let λ satisfy

0 < λ ≤ 1

2

Tut

1

0

1− αη

1 0

G1

η, s

1− αη

Trang 11

1

0

1 − ssas1

2Λ1usds α

1 0

G1

η, s

a s1

2Λ1usds λ

2Λ1

0

1 0

G1

η, s

u1

1− αηρ∗

2

1− αη

2u11

2ρ∗

 u1.

3.23

SoTu ≤ u1 Hence,Tu1≤ u1, for all u ∈ ∂Ω ρ∗

i

T, Ω ρ∗ , K

Next, by

v → ∞max

t∈0,1 sup

u∈0,∞

f t, u, v

we know that there exists r0 > ρ2such that

f t, u, v ≤ 1

Case 1 max t∈0,1 ft, u, v is unbounded.

f∗

ρ

maxf t, u, v : t ∈ 0, 1, u, v ∈0, ρ

f∗

ρ

Taking ρ∗≥ max{2r0, 2λ/1 − αη, 2ρ2}, it follows from 3.26–3.28 that

f t, u, v ≤ f∗

ρ∗

2Λ1ρ∗, for t ∈ 0, 1, u, v ∈0, ρ∗

Trang 12

By Lemmas2.2and2.3and3.28, we have

Tut

1

0

1 0

G1

η, s

≤

1

0

1 − ssas1

1 0

G1

η, s

a s1

2Λ1

0

1 0

G1

η, s

a sds

ρ∗ ρ∗ 2

 ρ∗.

3.30

SoTu ≤ ρ∗, and thenTu1≤ ρ∗

Case 2 max t∈0,1 ft, u, v is bounded.

In this case, there exists an M > 0 such that

max

t∈0,1 f t, u, v ≤ M, for t ∈ 0, 1, u, v ∈ 0, ∞. 3.31

Tut

1 0

G1

η, s

≤ M

0

1− αη

1 0

G1

η, s

a sds

ρ∗ 2

≤ ρ∗

2

 ρ∗,

3.32

which impliesTu ≤ ρ∗, and thenTu1≤ ρ∗

Therefore, in both cases, taking

Ωρ∗ u ∈ K : u1< ρ∗

Trang 13

we get

i

T, Ω ρ∗, K

Finally, putΩρ2 {u ∈ K : u1< ρ2} Then H4 implies that

Tu

1

2

1 0

G1

1

2, s

a sfs, u s, usds

2

1− αη

1 0

G1

η, s

2

1− αη

≥

3/4

1/4

1

2

1− αη

3/4

1/4

G1

η, s

a s2Λ2ρ2ds

≥ Λ2ρ2

1/4

3/4

1/4

G1

η, s

 ρ2,

3.36

that is,Tu ≥ ρ2, and thenTu1≥ u1, for all u ∈ ∂Ω ρ2 By virtue of Theorem2.6, we have

i

T, Ω ρ2 , K

From3.24, 3.35, 3.37, and ρ∗< ρ2< ρ∗, it follows that

i

T, Ω ρ∗\ Ωρ2 , K

T, Ω ρ2\ Ωρ∗ , K

Hence, T has fixed point u1 ∈ Ωρ2\ Ωρ∗ and fixed point u2∈ Ωρ∗\ Ωρ2 Obviously, u1, u2are

Theorem 3.3 Let there exist d0 , a0, b0, and c with

Trang 14

such that

f t, u, v ≤ 1

Then problem1.1 has at least three positive solutions u1, u2, u3satisfying

u11< d0, a0< β u2, u31> d0, β u3 < a0, 3.44

for λ ≤ 1/21 − αηd0.

Proof Let

Let u be in K c Equation3.43 implies that

Tut

1

0

1 0

G1

η, s

2Λ1

0

1− αη

1 0

G1

η, s

a sds

c 1/21− αηc

1− αη

 c.

3.46

Take

u0t 1

Then,

u0∈u ∈ K

β, a0, b0

: βu > a0

/

Trang 15

By3.42, we have, for any u ∈ Kβ, a0, b0,

t∈1/4,3/4

1 0

2

1− αη

1 0

G1

η, s

2

1− αη

≥ min

t∈1/4,3/4

1 0

1

2t2s 1 − sasfs, u s, usds

2

1− αη

1 0

G1

η, s

2

1− αη

≥

3/4

1/4

1 2

1 4

2

1− αη

1 4

23/4

1/4

G1

η, s

a s35Λ2a0ds

32a0

> a0.

3.49

By3.41, we see that for any u1≤ d0

Tut

1

0

1 0

G1

η, s

≤

1

0

1 − ssas1

1 0

G1

η, s

a s1

4Λ1d0ds 1/21− αηd0

1− αη

4d0.

3.50

So,Tu1 ≤ 3/4d0 < d0 This means thatb of Theorem2.7holds

Moreover, for any u ∈ Kβ, a0, c with Tu1> b0, we have

t∈1/4,3/4

1 0

2

1− αη

1 0

G1

η, s

2

1− αη

Trang 16

≥ min

t∈1/4,3/4

1 0

1

2t21 − ssasfs, u s, usds

2

1− αη

1 0

G1

η, s

2

1− αη

16

1 0

G1

η, s

32

1 0

1− αη

1 0

G1

η, s

32

1 0

1− αη

1 0

G1

η, s

32Tut,

3.51 which implies

32Tu1> 1

32b0 > a0. 3.52

three positive fixed points u1, u2, u3∈ K csatisfying

u11< d0, a0< β u2, u31> d0, β u3 < a0. 3.53

4 Examples

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

Example 4.1 Consider the problem

ut 1

102

t−1

 0, 0 < t < 1,

u 0 u0 0, u1 − u

1 2

 λ,

4.1

Trang 17

where η 1/2, α 1 Set

t, u t, ut 2t−1

Then,

Λ1

0

1 0

G1

η, s

−1

1

0

1− ηsa sds 

1

η

1

0

1 − ss 1

0

2

s 1

1

1/2

1

21 − s1

 24.

4.4

Taking

ρ1 4, for t ∈ 0, 1, u ∈0, ρ1

, v ∈

0, ρ1

we have

f t, u, v ≤ 1 12 16 36 9ρ1< 1

Thus, conditionH2 is satisfied

u2such that

for

0 < λ ≤ 1

2

1− αηρ1 1

Trang 18

Example 4.2 Consider the problem

ut 2 × 581 t2 sin utut2

5−ut 0, 0 < t < 1,

u 0 u0 0, u1 − u

1 2

where η 1/2, α 1 Set

a t 2, ft, u, v 581 t2 sin utut2

Then,

that is, the conditionH3 is satisfied Moreover,

Λ2

1/4

3/4

1/4

G1

η, s

a sds

−1

3/4

1/4

2

s2ds 

3/4

1/2

1

29.

4.12

Taking

ρ2 8, for t ∈ 0, 1, u ∈0, ρ2

, v ∈

1

4ρ2, ρ2

we get

f t, u, v ≥ 58825−8 82 8ρ2> 2Λ2ρ2 96

Thus, conditionH4 is satisfied

0 < λ ≤ 1

2

1− αηρ∗≤ 1

Trang 7

Theorem 2.6 see 10 Let E be a real Banach Space, let K ⊂ E be a cone, and Ω r... d0 , a< /i>0, b0, and c with

Trang 14

such that

f... > a< /i>0

/

Trang 15
By3.42, we have, for any u ∈ Kβ, a< /i>0,

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