A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition

Volume 2011, Article ID 569191,13pages doi:10.1155/2011/569191

Research Article

A Fourth-Order Boundary Value Problem with

One-Sided Nagumo Condition

Wenjing Song

_{and Wenjie Gao}

1_{Institute of Mathematics, Jilin University, Changchun 130012, China}

2_{Institute of Applied Mathematics, Jilin University of Finance and Economics, Changchun 130017, China}

Correspondence should be addressed to Wenjing Song,[email protected]

Received 10 January 2011; Accepted 9 March 2011

Academic Editor: I. T. Kiguradze

Copyrightq2011 W. Song and W. Gao. 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.

The aim of this paper is to study a fourth-order separated boundary value problem with the right-hand side function satisfying one-sided Nagumo-type condition. By making a series of a priori estimates and applying lower and upper functions techniques and Leray-Schauder degree theory, the authors obtain the existence and location result of solutions to the problem.

1. Introduction

In this paper we apply the lower and upper functions method to study the fourth-order nonlinear equation

u4t ft, ut, ut, ut, ut, 0< t <1, 1.1

withf:0,1×Ê

4 _→

Êbeing a continuous function.

This equation can be used to model the deformations of an elastic beam, and the type of boundary conditions considered depends on how the beam is supported at the two endpoints 1,2. We consider the separated boundary conditions

u0 u1 0, au0−bu0 A,

cu1 du1 B

1.2

witha, b, c, d∈Ê

_{0,∞,A, B∈}

For the fourth-order differential equation

u4t ft, ut, ut, ut, ut, 0< t <1, u0 u1 u0 u1,

1.3

the authors in3obtained the existence of solutions with the assumption thatfsatisfies the two-sided Nagumo-type conditions. For more related works, interested readers may refer to 1–14. The one-sided Nagumo-type condition brings some difficulties in studying this kind of problem, as it can be seen in15–18.

Motivated by the above works, we consider the existence of solutions when f

satisfies one-sided Nagumo-type conditions. This is a generalization of the above cases. We apply lower and upper functions technique and topological degree method to prove the existence of solutions by making a priori estimates for the third derivative of all solutions of problems1.1and1.2. The estimates are essential for proving the existence of solutions.

The outline of this paper is as follows. InSection 2, we give the definition of lower and upper functions to problems1.1and1.2and obtain some a priori estimates.Section 3 will be devoted to the study of the existence of solutions. InSection 4, we give an example to illustrate the conclusions.

2. Definitions and A Priori Estimates

Upper and lower functions will be an important tool to obtain a priori bounds onu,u, and

u. For this problem we define them as follows.

Definition 2.1. The functionsα, β∈C40,1∩C30,1verifying

αt≤βt, ∀t∈0,1, 2.1

define a pair of lower and upper functions of problems 1.1 and 1.2 if the following conditions are satisfied:

iα4_{t≥ft, αt, αt, αt, αt,β}4_{t≤ft, βt, βt, βt, βt,}

iiα0 ≤ 0, α1 ≤ 0, aα0−bα0 ≤ A, cα1 dα1 ≤ B,β0 ≥ 0, β1 ≥ 0, aβ0−bβ0≥A, cβ1 dβ1≥B,

iiiα0−β0≤min{β0−β1, α1−α0,0}.

Remark 2.2. By integration, fromiiiand2.1, we obtain

αt≤βt, αt≤βt, ∀t∈0,1, 2.2

that is, lower and upper functions, and their first derivatives are also well ordered.

Definition 2.3. Given a setE ⊂ 0,1×Ê

4_{, a continuous f : E →}

Ê is said to satisfy the

one-sided Nagumo-type condition inEif there exists a real continuous functionhE :Ê

0 →

k,∞, for somek >0, such that

ft, x0, x1, x2, x3≤hE|x3|, ∀t, x0, x1, x2, x3∈E, 2.3

with

_∞

0

s

hEsds ∞. 2.4

Lemma 2.4. LetΓit, γit∈C0,1,Êsatisfy

Γit≥γit, ∀t∈0,1, i0,1,2, 2.5

and consider the set

Et, x0, x1, x2, x3∈0,1×Ê

4 _:γit≤x

i≤Γit, i0,1,2

. 2.6

Letf:0,1×Ê

4 _→

Êbe a continuous function satisfying one-sided Nagumo-type condition inE.

Then, for everyρ >0, there exists anR >0such that for every solutionutof problems1.1

and1.2with

u0≤ρ, u1≥ −ρ, 2.7

γit≤uit≤Γit, 2.8

fori0,1,2and everyt∈0,1, one hasu∞< R.

Proof. Letube a solution of problems1.1and1.2such that2.7and2.8hold. Define

η:maxΓ21−γ20,Γ20−γ21

. 2.9

Assume thatρ≥η, and suppose, for contradiction, that|ut|> ρfor everyt∈0,1. Ifut> ρfor everyt∈0,1, then we obtain the following contradiction:

Γ21−γ20≥u1−u0

₁

0

utdt >

₁

0

ρ dt≥Γ21−γ20. 2.10

Ifut<−ρfor everyt∈0,1, a similar contradiction can be derived. So there is at∈0,1 such that|ut| ≤ρ. By2.4we can takeR1 > ρsuch that

R1

ρ

s

If|ut| ≤ρfor everyt∈0,1, then we have trivially|ut|< R1. If not, then we can

taket1 ∈0,1such thatut1<−ρort1 ∈0,1such thatut1> ρ. Suppose that the first

case holds. By2.7we can considert1< t0≤1 such that

ut0 −ρ, ut<−ρ, ∀t∈t1, t0. 2.12

Applying a convenient change of variable, we have, by2.3and2.11,

_−ut 1

−ut0

s hEsds

_t1

t0

−ut hE−ut

−u4t

dt

_t0

t1

−ut hE−utf

t, ut, ut, ut, utdt

≤

_t0

t1

−utdtut1−ut0

≤ max

t∈0,1Γ2t−tmin∈0,1γ2t<

R1

ρ

s hEsds.

2.13

Hence,ut1>−R1. Sincet1can be taken arbitrarily as long asut1<−ρ, we conclude that

ut>−R1for everyt∈0,1provided thatut<−ρ.

In a similar way, it can be proved thatut < R1, for every t ∈ 0,1ifut > ρ.

Therefore,

_{ut< R}

1, ∀t∈0,1. 2.14

Consider now the caseη > ρ, and takeR2> ηsuch that

_R2

η

s

hEsds >tmax∈0,1Γ2t−tmin∈0,1γ2t. 2.15

In a similar way, we may show that

_{ut< R}

2, ∀t∈0,1. 2.16

TakingRmax{R1, R2}, we haveu∞< R.

Remark 2.5. Observe that the estimationRdepends only on the functionshE,γ2,Γ2, andρand

it does not depend on the boundary conditions.

3. Existence and Location Result

Theorem 3.1. Suppose that there exist lower and upper functionsαtand βtof problems1.1

and1.2, respectively. Letf :0,1×Ê

4 _→

Ê be a continuous function satisfying the one-sided

Nagumo-type conditions2.3and2.4in

E∗

t, x0, x1, x2, x3∈0,1×Ê

4 _:αt≤x

0≤βt, αt≤x1≤βt, αt≤x2≤βt

3.1

Iffverifies

ft, αt, αt, x2, x3

≥ft, x0, x1, x2, x3≥f

t, βt, βt, x2, x3

3.2

fort, x2, x3∈0,1×Ê

2 _and

αt, αt≤x0, x1≤

βt, βt, 3.3

wherex0, x1≤y0, y1meansx0 ≤y0 andx1 ≤y1, then problems1.1and1.2has at least one

solutionut∈C4_{0,1satisfying}

αt≤ut≤βt, αt≤ut≤βt, αt≤ut≤βt 3.4

fort∈0,1.

Proof. Define the auxiliary functions

δit, xi

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

αi_{t, x}

i< αit,

xi, αit≤xi≤βit,

βi_{t, x}

i> βit.

i0,1,2, 3.5

Forλ∈0,1, consider the homotopic equation

u4t λft, δ0t, ut, δ1

t, ut, δ2

t, ut, utut−λδ2

t, ut, 3.6

with the boundary conditions

u0 u1 0,

u0 λ

b

au0−A,

u1 λ

d

B−cu1.

Taker1>0 large enough such that, for everyt∈0,1,

−r1 < αt≤βt< r1, 3.8

ft, αt, αt, αt,0−r1−αt<0, 3.9

ft, βt, βt, βt,0r1−βt>0, 3.10

|A| a < r1,

|B|

c < r1. 3.11

Step 1. Every solutionutof problems3.6and3.7satisfies

uit< r1, ∀t∈0,1 3.12

fori0,1,2, for somer1independent ofλ∈0,1.

Assume, for contradiction, that the above estimate does not hold fori 2. So there existλ∈0,1,t∈0,1, and a solutionuof3.6and3.7such that|ut| ≥r1. In the case

ut≥r1define

max t∈0,1u

_t:ut

0≥r1. 3.13

Ift0 ∈0,1, thenut0 0 andu4t0 ≤0. Then, by3.2and3.10, forλ ∈0,1,

the following contradiction is obtained:

0≥u4t0 λf

t0, δ0t0, ut0, δ1

t0, ut0

, δ2

t0, ut0

, ut0

ut0−λδ2

t0, ut0

λft0, δ0t0, ut0, δ1

t0, ut0

, βt0,0

ut0−λβt0

≥λft0, βt0, βt0, βt0,0

ut0−λβt0

λft0, βt0, βt0, βt0,0

r1−βt0

ut0−λr1>0.

3.14

Forλ0,

0≥u4t0 ut0≥r1 >0. 3.15

Ift00, then

max t∈0,1u

_t:u0≥r

1>0 3.16

andu0 u0 ≤ 0. Ifλ 0, thenu0 0 and so u4_{0 ≤ 0. Therefore, the above}

following contradiction:

0≥u0 λ

b

au0−A≥ λ

bar1−A>0. 3.17

The caset0 1 is analogous. Thus,ut< r1for everyt∈0,1. In a similar way, we

may prove thatut>−r1for everyt∈0,1.

By the boundary condition3.7there exists aξ ∈ 0,1, such thatuξ 0. Then by integration we obtain

_ut

t

ξ

usds

< r1|t−ξ| ≤r1,

|ut|

_t

0

usds< r1t≤r1.

3.18

Step 2. There is anR >0 such that for every solutionutof problems3.6and3.7

_{ut< R, ∀t∈0,1, 3.19}

withRindependent ofλ∈0,1. Consider the set

Er1

t, x0, x1, x2, x3∈0,1×Ê

4 _:−r

1≤xi≤r1, i0,1,2

3.20

and forλ∈0,1the functionFλ:Er1 → Êgiven by

Fλt, x0, x1, x2, x3 λft, δ0t, x0, δ1t, x1, δ2t, x2, x3 x2−λδ2t, x2. 3.21

In the following we will prove that the function Fλ satisfies the one-sided Nagumo-type conditions 2.3and 2.4in Er1 independently of λ ∈ 0,1. Indeed, asf verifies2.3 in

E∗, then

Fλt, x0, x1, x2, x3 λft, δ0t, x0, δ1t, x1, δ2t, x2, x3 x2−λδ2t, x2

≤hE∗|x3| r1−λαt≤hE∗|x3| 2r1.

3.22

So, defininghEr₁t hE∗|x3| 2r1inÊ

0, we see thatFλverifies2.3withEandhEreplaced

byEr1andhEr1, respectively. The condition2.4is also verified since _∞

0

s hEr1s

ds

_∞

0

s hE∗s2r1

ds≥ 1

12r1/k

_∞

0

s

Therefore,Fλsatisfies the one-sided Nagumo-type condition inEr1withhEreplaced byhEr1,

withr1independent ofλ∈0,1.

Moreover, for

ρ:max

ar1|A|

b ,

|B|cr1

d

, 3.24

every solutionuof3.6and3.7satisfies

u0 λ

b

au0−A≤λ

bar1|A|≤ρ,

u1 λ

d

B−cu1≥ −λ

d|B|cr1≥ −ρ.

3.25

Define

γit:−r1, Γit:r1, fori0,1,2. 3.26

The hypotheses ofLemma 2.4are satisfied withEreplaced byEr1. So there exists anR > 0,

depending onr1 and hEr1, such that|u

_{t| < Rfor every t ∈ 0,1. As r}

1 and hEr1 do not

depend onλ, we see thatRis maybe independent ofλ.

Step 3. Forλ1, the problems3.6and3.7has at least one solutionu1t.

Define the operators

L:C40,1⊂C30,1−→C0,1×Ê

4 _3.27

by

Luu4−u, u0, u1, u0, u1 3.28

and forλ∈0,1,Nλ:C3_{0,1 → C0,1×}

Ê

4 _by

Nλuλft, δ0t, ut, δ1

t, ut, δ2

t, ut, ut−λδ2

t, ut,0,0, Aλ, Bλ

, 3.29

with

Aλ: λ

b

au0−A,

Bλ: λ

d

B−cu1.

Observe thatLhas a compact inverse. Therefore, we can consider the completely continuous operator

Tλ:

C30,1,Ê

−→C30,1,Ê

3.31

given by

Tλu L−1Nλu. 3.32

ForRgiven byStep 2, take the set

Ω x∈C30,1:xi

∞< r1, i0,1,2, x

∞< R

. 3.33

By Steps1and2, degreedI−Tλ,Ω,0is well defined for everyλ∈0,1and by the invariance with respect to a homotopy

dI−T0,Ω,0 dI−T1,Ω,0. 3.34

The equationxT0xis equivalent to the problem

u4t ut,

u0 u1 u0 u1 0

3.35

and has only the trivial solution. Then, by the degree theory,

dI−T0,Ω,0 ±1. 3.36

So the equationT1x xhas at least one solution, and therefore the equivalent problem

u4t ft, δ0t, ut, δ1

t, ut, δ2

t, ut, utut−δ2

t, ut,

u0 u1 0, au0−bu0 A,

cu1 du1 B

3.37

has at least one solutionu1tinΩ.

The proof will be finished if the above functionu1tsatisfies the inequalities

αt≤u1t≤βt, αt≤u₁t≤βt, αt≤u₁t≤βt. 3.38

Assume, for contradiction, that there is at∈0,1such thatu₁t> βt, and define

max t∈0,1

u1t−βt

:u1t2−βt2>0. _3.39

If t2 ∈ 0,1, then u₁t2 βt2 and u₁4t2 ≤ β4t2. Therefore, by 3.2 and

Definition 2.1, we obtain the contradiction

u₁4t2 f

t2, δ0t2, u1t2, δ1

t2, u₁t2

, δ2

t2, u₁t2

, u₁t2

u₁t2−δ2

t2, u₁t2

ft2, δ0t2, u1t2, δ1

t2, u1t2

, βt2, βt2

u₁t2−βt2

≥ft2, βt2, βt2, βt2, βt2

≥β4t2.

3.40

Ift20, then we have

max t∈0,1

u₁t−βt:u₁0−β0>0,

u₁0−β0 u₁0−β0≤0.

3.41

ByDefinition 2.1this yields a contradiction

u₁0 1

b

au₁0−A> 1 b

aβ0−A≥β0. 3.42

Thent2/0 and, by similar arguments, we prove thatt2/1. Thus,

u₁t≤βt, ∀t∈0,1. 3.43

Using an analogous technique, it can be deduced thatαt≤u₁tfor everyt∈0,1. So we have

αt≤u1t≤βt. 3.44

On the other hand, by1.2,

0u11−u10

₁

0

u₁tdt

₁

0

u₁0

_t

0

u₁sds

dtu₁0

₁

0

_t

0

that is,

u₁0 −

₁

0

_t

0

u₁sds dt. 3.46

Applying the same technique, we have

−

1

0

t

0

βsds dt−

1

0

βtdtβ0 β0−β1 β0, 3.47

and then byDefinition 2.1iii,3.44and3.46, we obtain

α0≤β0−β1 β0

−

1

0

t

0

βsds dt≤ −

1

0

t

0

u₁sds dtu₁0,

β0≥α0−α1 α0

−

₁

0

_t

0

αsds dt≥ −

₁

0

_t

0

u1sds dtu10,

3.48

that is,

α0≤u₁0≤β0. 3.49

Since, by3.44,βt−u₁tis nondecreasing, we have by3.49

βt−u1t≥β0−u10≥0, 3.50

and, therefore,βt≥u1tfor everyt∈0,1. By the monotonicity ofβt−u1t,

βt−u1t≥β0−u10 β0≥0, 3.51

and soβt≥u1tfor everyt∈0,1.

The inequalitiesu₁t≥αtandu1t≥αtfor everyt∈0,1can be proved in the

same way. Thenu1tis a solution of problems1.1and1.2.

4. An Example

Example 4.1. Consider now the problem

u4t −3uteutut−22−ut4, 4.1

u0 u1 0, u0−u0 A,

u1 u1 B

4.2

withA, B∈Ê. The nonlinear function

ft, x0, x1, x2, x3 −3x0ex1x2−22−x34 4.3

is continuous in0,1×Ê

4_{. IfA, B∈−2,2, then the functionsα, β:0,1 →}

Êdefined by αt −t2_{−t, βt t}2_{t 4.4}

are, respectively, lower and upper functions of4.1and4.2. Moreover, define

Et, x0, x1, x2, x3∈0,1×Ê

4 _:−t2_−t≤x

0 ≤t2t, −2t−1≤x1 ≤2t1, −2≤x2≤2

.

4.5

Thenfsatisfies condition3.2and the one-sided Nagumo-type condition withhE|x3| 1,

inE.

Therefore, byTheorem 3.1, there is at least one solutionutof Problem4.1and4.2 such that, for everyt∈0,1,

−t2−t≤ut≤t2t, −2t−1≤ut≤2t1, −2≤ut≤2. 4.6

Notice that the function

ft, x0, x1, x2, x3 −3x0ex1x2−22−x34 4.7

does not satisfy the two-sided Nagumo condition.

Acknowledgments

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