# Solutions to a Three Point Boundary Value Problem

20

## Full text

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doi:10.1155/2011/894135

## Solutions to a Three-Point Boundary Value Problem

1

### and Zhi-Wei Lv

2, 3

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

2Department of Mathematics and Physics, Anyang Institute of Technology, Anyang, Henan 455000, China 3Department 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

Copyrightq2011 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 problemut atft, ut, ut 0, 0< t <1;u0 u0 0;u1−αuη λ, whereη∈0,1/2,α∈1/2η,1/ηare constants, λ∈0,∞is a parameter, anda,fare 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 elsewhere see, e.g., Agarwal 1, Bisplinghoﬀ and Ashley 2, and Henderson3, multi 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,

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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, ut, ut0, 0< t <1,

u0 u0 0, u1−αuηλ,

1.1

whereη ∈0,1/2,α∈1/2η,1,λ∈0,∞, anda,fare 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 theorems for1.1which extend and complement some existing results.

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 onY ifβ:Y → 0,∞is continuous and

βtx 1−tytβx 1−tβy, x, yY, t∈0,1. 1.2

Assume that

H

aC0,1,0,, 0<

1

0

1−ssasds <,

fC0,1×0,∞×0,,0,.

1.3

Define

maxf0 lim

v→0tmax∈0,1usup0,

ft, u, v v ,

minf0 lim

v→0tmin∈0,1u∈inf0,

ft, u, v v ,

maxfvlimmax t∈0,1usup0,

ft, u, v v ,

minfvlimmin t∈0,1u∈inf0,

ft, u, v v .

1.4

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### 2. Lemmas

LetEC10,1be a Banach Space with norm

u1max

u,u, 2.1

where

u max

t∈0,1|ut|, u

max

t∈0,1u

t.

2.2

It is not hard to see Lemmas2.1and2.2.

Lemma 2.1. LetuC10,1be the unique solution of 1.1. Then

ut

1

0

Gt, sasfs, us, usds

αt2

21−αη

1

0

G1

η, sasfs, us, usds λt

2

21−αη,

2.3

where

Gt, s 1

2 ⎧ ⎨ ⎩

2tt2ss, st,

1−st2, ts,

G1t, s ∂Gt, s

∂t

⎧ ⎨ ⎩

1−ts, st,

1−st, ts.

2.4

Lemma 2.2. One has the following.

i0≤G1t, s≤1−ss, 1/2t21−ssGt, sG1, s 1/21−ss.

iiG1t, s≥1/4G1s, s 1/41−ss, fort∈1/4,3/4, s∈0,1.

iii G11/2, s≥1/21−ss, fors∈0,1.

Lemma 2.3. LetuC10,1be the unique solution of 1.1. Thenutis nonnegative and satisfies u1u.

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iFortη,

ut 1

21−αη

t

0

2tss21−αηt2−1asfs, us, usds

η

t

t21−αηt2−1asfs, us, usds

1

η

t21−sasfs, us, usdsλt2

,

ut 1

21−αη

t

0

2s1−αη2tsα−1asfs, us, usds

η

t

2t1−αη2tsα−1asfs, us, usds

1

η

2t1−sasfs, us, usds2λt

,

2.5

that is,utut.

iiFortη,

ut 1

21−αη

η

0

2tss21−αηt2−1asfs, us, usds

t

η

2tss21−αηt2αηsasfs, us, usds

1

t

t21sasfs, us, usdsλt2

,

ut 1

21−αη

η

0

2s1−αη2tsα−1asfs, us, usds

t

η

2s1−αη2tαηsasfs, us, usds

1

t

2t1−sasfs, us, usds2λt

.

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On the other hand, forηst, we have

2s1−αη2tαηs−2tss21−αηt2αηs

αηts2st s1−t2−t sst

αηts

2sts αη

s1−t2−t.

2.7

Sinceα∈1/2η,1,

2s1−αη2tαηs≥2tss21αηt2αηs. 2.8

So,utut. Therefore,utut, which means

uu, u1u. 2.9

The proof is completed.

Lemma 2.4. LetuC10,1be the unique solution of 1.1. Then

min

t∈1/4,3/4u

t 1

4u1. 2.10

Proof. From2.3, it follows that

ut

1

0

G1t, sasf

s, us, usds

αt

1−αη

1

0

G1

η, sasfs, us, usds λt

1−αη

≤ 1

0

1−ssasfs, us, usds

α

1−αη

1

0

G1

η, sasfs, us, usds λ

1−αη.

2.11

Hence,

u1u≤ 1

0

1−ssasfs, us, usds

α

1−αη

1

0

G1

η, sasfs, us, usds λ 1−αη.

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

min

t∈1/4,3/4u

t min

t∈1/4,3/4 1

0

G1t, sasf

s, us, usds

αt

1−αη

1

0

G1

η, sasfs, us, usds λt

1−αη

≥ 1 4

1

0

1−ssasfs, us, usds

α

1−αη

1

0

G1

η, sasfs, us, usds λ

1−αη

.

2.13

Thus,

min

t∈1/4,3/4u

t 1

4u1. 2.14

Define a cone by

K

uE:u≥0, min

t∈1/4,3/4u

t 1

4u1

. 2.15

Set

Kr {uK:u1< r}, ∂Kr{uK:u1r}, r >0,

Kr {uK:u1≤r}, K

β, r, suK:rβu, u1s, s > r >0.

2.16

Define an operatorTby

Tut

1

0

Gt, sasfs, us, usds

αt2

21−αη

1

0

G1

η, sasfs, us, usds λt

2

21−αη.

2.17

Lemma2.1implies that1.1has a solutionuutif and only ifuis a fixed point ofT.

From Lemmas2.1and2.2and the Ascoli-Arzela theorem, the following follow.

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Theorem 2.6see10. LetEbe a real Banach Space, letKEbe a cone, andΩr {uK:u

r}. Let operatorT :K∩ΩrKbe completely continuous and satisfyTx /x, for allxΩr. Then iifTxx, for all xΩr, theniT,Ωr, K 1,

iiifTxx, for all xΩr, theniT,Ωr, K 0.

Theorem 2.7see8. LetT :PcPcbe a completely continuous operator andβa nonnegative continuous concave functional onP such thatβxx for allxPc. Suppose that there exist

0< d0< a0 < b0≤csuch that

a{xPβ, a0, b0:βx> a0}/andβTx> a0forxPβ, a0, b0,

bTx< d0forxd0,

cβTx> a0forxPβ, a0, cwithTx> b0.

Then,T has at least three fixed pointsx1, x2, and x3inPcsatisfying

x1< d0, a0< βx2, x3> d0, βx3< 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

1

0

1−ssasds α

1−αη

1

0

G1

η, sasds

1

,

Λ2

3/4

1/4

1−ssasds α

1−αη

3/4

1/4

G1

η, sasds

1

.

3.1

Theorem 3.1. Assume that

H1minf0minf∞ ∞;

H2there exists a constantρ1 >0such thatft, u, v≤1/2Λ1ρ1, fort∈0,1,u∈0, ρ1 andv∈0, ρ1.

Then, the problem1.1has at least two positive solutionsu1andu2such that

0<u11< ρ1<u21, 3.2

forλsmall enough.

Proof. Since

minf0 lim

v→0tmin∈0,1u∈inf0,

ft, u, v

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there isρ0∈0, ρ1such that

By Theorem2.6, we have

iT,Ωρ0, K

0. 3.9

On the other hand, since

minfvlimmin t∈0,1u∈inf0,

ft, u, v

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there existρ0∗,ρ0> ρ1such that

ft, u, v≥8Λ2v, fort∈0,1, u∈0,, v≥ 1 4ρ

0. 3.11

LetΩρ0{uK:u1 < ρ0}. Then, by a argument similar to that above, we obtain

Tu1≥ u1,uΩρ0. 3.12

By Theorem2.6,

iT,Ωρ0, K

0. 3.13

Finally, letΩρ1 {uK :u1 < ρ1}, and letλsatisfy 0< λ≤1/21−αηρ1for any

uΩρ1. Then,H2implies

Tut

1

0

G1t, sasf

s, us, usds

αt

1−αη

1

0

G1

η, sasfs, us, usds λt

1−αη

≤ 1

0

1−ssas1

2Λ1ρ1ds

α

1−αη

1

0

G1

η, sas1

2Λ1ρ1ds

λ

1−αη

≤ 1 2Λ1

1

0

1−ssasds α

1−αη

1

0

G1

η, sasds

ρ1

1/21−αηρ1 1−αη

1

2ρ1 1 2ρ1

u1,

3.14

which means thatTuu1. Thus,Tu1 ≤ u1, for alluΩρ1. Using Theorem2.6, we get

iT,Ωρ1, K

1. 3.15

From3.9–3.15andρ0< ρ1 < ρ0, it follows that

iT,Ωρ0ρ1, K

−1, iT,Ωρ1\Ωρ0, K

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Therefore,Thas fixed pointu1 ∈Ωρ1\Ωρ0and fixed pointu2 ∈Ωρ∗0\Ωρ1. Clearly,u1,u2are both positive solutions of the problem1.1and

0<u11< ρ1<u21. 3.17

The proof of Theorem3.1is completed.

Theorem 3.2. Assume that

H3maxf0maxf∞0;

H4there exists a constantρ2 >0such thatft, u, v≥2Λ2ρ2, fort∈0,1, u∈0, ρ2and

v∈1/4ρ2, ρ2.

Then, the problem1.1has at least two positive solutionsu1andu2such that

0<u11< ρ2<u21 3.18

forλsmall enough.

Proof. By

maxf0 lim

v→0tmax∈0,1usup0,

ft, u, v

v 0, 3.19

we see that there existsρ∗∈0, ρ2such that

ft, u, v≤ 1

2Λ1v, fort∈0,1, u∈0,, v

0, ρ. 3.20

Put

Ωρ∗

uK:u1 < ρ, 3.21

and letλsatisfy

0< λ≤ 1

2

1−αηρ. 3.22

Then Lemmas2.2and2.3and3.20implies that for anyuΩρ∗,

Tut

1

0

G1t, sasf

s, us, usds

αt

1−αη

1

0

G1

η, sasfs, us, usds λt

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1

0

1−ssas1

2Λ1u

sds α

1−αη

1

0

G1

η, sas1

2Λ1u

sds λ

1−αη

≤ 1 2Λ1

1

0

1−ssasds α

1−αη

1

0

G1

η, sasds

u1

1−αηρ

21−αη

≤ 1 2u1

1 2ρ

u1.

3.23

SoTuu1. Hence,Tu1≤ u1, for alluΩρ∗.

Applying Theorem2.6, we have

iT,Ωρ∗, K

1. 3.24

Next, by

maxfvlimmax t∈0,1usup0,

ft, u, v

v 0, 3.25

we know that there existsr0 > ρ2such that

ft, u, v≤ 1

2Λ1v, fort∈0,1, u∈0,, vr0. 3.26

Case 1. maxt∈0,1ft, u, vis unbounded.

Define a functionf∗:0,∞ → 0,∞by

fρmaxft, u, v:t∈0,1, u, v∈0, ρ. 3.27

Clearly,f∗is nondecreasing and limρ→∞fρ/ρ0, and

fρ≤ 1

2Λ1ρ, forρ > r0. 3.28

Takingρ∗≥max{2r0,2λ/1−αη,2ρ2}, it follows from3.26–3.28that

ft, u, vfρ∗≤ 1

2Λ1ρ

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

In this case, there exists anM >0 such that

max Therefore, in both cases, taking

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we get

Tu1≤ u1,uΩρ. 3.34

By Theorem2.6, we have

iT,Ωρ, K

1. 3.35

Finally, putΩρ2{uK:u1< ρ2}. ThenH4implies that

Tu

1 2

1

0

G1

1 2, s

asfs, us, usds

α

21−αη

1

0

G1

η, sasfs, us, usds λ

21−αη

3/4

1/4 1

21−ssas2Λ2ρ2ds

α

21−αη

3/4

1/4

G1

η, sas2Λ2ρ2ds

≥Λ2ρ2 3/4

1/4

1−ssasds α

1−αη

3/4

1/4

G1

η, sasds

ρ2,

3.36

that is,Tuρ2, and thenTu1≥ u1, for alluΩρ2. By virtue of Theorem2.6, we have

iT,Ωρ2, K

0. 3.37

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

iT,Ωρ∗\Ωρ2, K

1, iT,Ωρ2\Ωρ∗, K

−1. 3.38

Hence,T has fixed pointu1 ∈Ωρ2\Ωρ∗and fixed pointu2∈Ωρ∗\Ωρ2. Obviously,u1,u2are both positive solutions of the problem1.1and

0<u11< ρ2<u21. 3.39

The proof of Theorem3.2is completed.

Theorem 3.3. Let there existd0,a0,b0, andcwith

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such that

ft, u, v≤ 1

4Λ1d0, t∈0,1, u∈0, d0, v∈0, d0, 3.41

ft, u, v≥35Λ2a0, t∈0,1, ua0, b0, va0, b0, 3.42

ft, u, v≤1

2Λ1c, t∈0,1, u∈0, c, v∈0, c. 3.43

Then problem1.1has at least three positive solutionsu1,u2,u3satisfying

u11< d0, a0< βu2, u31> d0, βu3< a0, 3.44

forλ≤1/21−αηd0.

Proof. Let

βu min

t∈1/4,3/4|ut|, uK. 3.45

Then,βis a nonnegative continuous concave functional onKandβuu1for eachuK. Letube inKc. Equation3.43implies that

Tut

1

0

G1t, sasf

s, us, usds

αt

1−αη

1

0

G1

η, sasfs, us, usds λt

1−αη

≤ 1 2Λ1

1

0

1−ssasds α

1−αη

1

0

G1

η, sasds

c1/2

1−αηc

1−αη

c.

3.46

Hence,Tu1≤c. This means thatT :KcKc.

Take

u0t 1

2a0b0, t∈0,1. 3.47

Then,

u0∈

uKβ, a0, b0

:βu> a0

/

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By3.42, we have, for anyuKβ, a0, b0,

Therefore,ain Theorem2.7holds.

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≥ min

So,cin Theorem2.7holds. Thus, by Theorem2.7, we know that the operatorT has at least three positive fixed pointsu1, u2, u3∈Kcsatisfying

u11< d0, a0< βu2, u31> d0, βu3< a0. 3.53

### 4. Examples

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

Example 4.1. Consider the problem

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whereη1/2,α1. Set

So, the conditionH1is satisfied. Observe

Λ1

Thus, conditionH2is satisfied.

Therefore, by Theorem3.1, the problem4.1has at least two positive solutionsu1and

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Example 4.2. Consider the problem

ut 2×581t2sinutut25−ut0, 0< t <1,

u0 u0 0, u1−u

1 2

λ,

4.9

whereη1/2,α1. Set

at 2, ft, u, v 581t2sinutut25−ut. 4.10

Then,

maxf0maxf∞0, 4.11

that is, the conditionH3is satisfied. Moreover,

Λ2

3/4

1/4

1−ssasds α

1−αη

3/4

1/4

G1

η, sasds

−1

3/4

1/4

1−ss2ds 1

1−1/2 1/2

1/4

1−1 2

s2ds

3/4

1/2 1

21−s2ds −1

48

29.

4.12

Taking

ρ28, fort∈0,1, u

0, ρ2

, v

1 4ρ2, ρ2

, 4.13

we get

ft, u, v≥58825−8828ρ2>2Λ2ρ2 96

29ρ2. 4.14

Thus, conditionH4is satisfied.

Consequently, by Theorem3.2, we see that for

0< λ≤ 1

2

1−αηρ∗≤ 14ρ2, 4.15

the problem4.9has at least two positive solutionsu1andu2such that

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That is, the conditions of Theorem3.3are satisfied. Consequently, the problem1.1has at least three positive solutionsu1, u2, u3∈Kcfor

λ≤ 1

2

1−αηd0 1

4 4.21

satisfying

u11<1, 2< βu2, u31>1, βu3<2. 4.22

### Acknowledgments

This paper was supported partially by the NSF of China 10771202and the Specialized Research Fund for the Doctoral Program of Higher Education of China2007035805.

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