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Boundary Value Problems
Volume 2011, Article ID 569191, 13 pages
doi:10.1155/2011/569191
Research Article
A Fourth-Order Boundary Value Problem with
One-Sided Nagumo Condition
Wenjing Song
1, 2
and Wenjie Gao
1
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
Copyright q 2011 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
u
4
boundary conditions considered depends on how the beam is supported at the two endpoints
1, 2. We consider the separated boundary conditions
u
0
u
1
0,
au
0
− bu
0
A,
cu
1
du
,u
t
, 0 <t<1,
u
0
u
1
u
0
u
1
,
1.3
the authors in 3 obtained the existence of solutions with the assumption that f satisfies 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
t
, ∀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:
i α
4
t ≥ ft, αt,α
t,α
t,α
t, β
4
t ≤ ft, βt,β
t,β
t,β
t,
ii α0 ≤ 0,α1 ≤ 0,aα
0 − bα
t
≤ β
t
, ∀t ∈
0, 1
, 2.2
that is, lower and upper functions, and their first derivatives are also well ordered.
To have an a priori estimate on u
, we need a one-sided Nagumo-type growth
condition, which is defined as follows.
Boundary Value Problems 3
Definition 2.3. Given a set E ⊂ 0, 1 ×
4
, a continuous f : E → is said to satisfy the
one-sided Nagumo-type condition in E if there exists a real continuous function h
E
:
0
→
k, ∞,forsomek>0, such that
∞
0
s
h
E
s
ds ∞.
2.4
Lemma 2.4. Let Γ
i
t,γ
i
t ∈ C0, 1, satisfy
Γ
i
t
≥ γ
i
t
, ∀t ∈
0, 1
t
,i 0, 1, 2
. 2.6
Let f : 0, 1 ×
4
→ be a continuous function satisfying one-sided Nagumo-type condition in E.
Then, for every ρ>0,thereexistsanR>0 such that for every solution ut of problems 1.1
and 1.2 with
u
0
≤ ρ, u
1
≥−ρ, 2.7
γ
i
t
≤ u
i
t
− γ
2
1
. 2.9
Assume that ρ ≥ η, and suppose, for contradiction, that |u
t| >ρfor every t ∈ 0, 1.
If u
t >ρfor every t ∈ 0, 1, then we obtain the following contradiction:
Γ
2
1
− γ
2
0
≥ u
1
− u
t < −ρ for every t ∈ 0, 1, a similar contradiction can be derived. So there is a
t ∈ 0, 1
such that |u
t|≤ρ.By2.4 we can take R
1
>ρsuch that
R
1
ρ
s
h
E
s
ds > max
t∈0,1
Γ
2
t
− min
t∈0,1
<t
0
≤ 1suchthat
u
t
0
−ρ, u
t
< −ρ, ∀t ∈
t
1
,t
0
. 2.12
Applying a convenient change of variable, we have, by 2.3 and 2.11,
−u
t
1
−u
−u
4
t
dt
t
0
t
1
−u
t
h
E
−u
t
f
t, u
dt u
t
1
− u
t
0
≤ max
t∈0,1
Γ
2
t
− min
t∈0,1
γ
2
t
<
R
1
t <R
1
, for every t ∈ 0, 1 if u
t >ρ.
Therefore,
u
t
<R
1
, ∀t ∈
0, 1
. 2.14
Consider now the case η>ρ,andtakeR
2
>ηsuch that
R
2
η
<R
2
, ∀t ∈
0, 1
. 2.16
Taking R ma x{R
1
,R
2
},wehaveu
∞
<R.
Remark 2.5. Observe that the estimation R depends only on the functions h
E
, γ
2
, Γ
2
,andρ and
it does not depend on the boundary conditions.
3. Existence and Location Result
In the p resence o f a n ordered pair of lower and upper functions, the existence and location
results for problems 1.1 and 1.2 can be obtained.
Boundary Value Problems 5
0
≤ β
t
,α
t
≤ x
1
≤ β
t
,α
t
≤ x
2
≤ β
t
t, β
t
,β
t
,x
2
,x
3
3.2
for t, x
2
,x
3
∈ 0 , 1 ×
2
and
α
t
,α
,y
1
means x
0
≤ y
0
and x
1
≤ y
1
,thenproblems1.1 and 1.2 has at least one
solution ut ∈ C
4
0, 1 satisfying
α
t
≤ u
t
≤ β
t
,α
t
i
t, x
i
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
α
i
t
,x
i
<α
i
t
,
u
4
t
λf
t, δ
0
t, u
t
,δ
1
t, u
t
,δ
2
t, u
t
1
0,
u
0
λ
b
au
0
− A
,
u
1
λ
d
t
,α
t
,α
t
, 0
− r
1
− α
t
< 0, 3.9
f
t, β
t
,β
<r
1
.
3.11
Step 1. Every solution ut of problems 3.6 and 3.7 satisfies
u
i
t
<r
1
, ∀t ∈
0, 1
3.12
for i 0, 1, 2, for some r
1
independent of λ ∈ 0, 1.
Assume, for contradiction, that the above estimate does not hold for i 2. So there
exist λ ∈ 0, 1, t ∈ 0, 1,andasolutionu of 3.6 and 3.7 such that |u
t|≥r
0andu
4
t
0
≤ 0. Then, by 3.2 and 3.10,forλ ∈ 0, 1,
the following co ntradiction is obtained:
0 ≥ u
4
t
0
λf
t
0
,δ
0
t
0
,u
t
0
,δ
1
t
t
0
− λδ
2
t
0
,u
t
0
λf
t
0
,δ
0
t
0
,u
t
0
,δ
≥ λf
t
0
,β
t
0
,β
t
0
,β
t
0
, 0
u
t
0
r
1
− β
t
0
u
t
0
− λr
1
> 0.
3.14
For λ 0,
0 ≥ u
4
t
0
u
u
0 ≤ 0. If λ 0, then u
00andsou
4
0 ≤ 0. Therefore, the above
computations with t
0
replaced by 0 yield a contradiction. For λ ∈ 0, 1,by3.11,wegetthe
Boundary Value Problems 7
following contradiction:
0 ≥ u
0
λ
b
au
0
− A
≥
λ
t
ξ
u
s
ds
<r
1
|
t − ξ
|
≤ r
1
.
3.18
Step 2. There is an R>0 such that for every solution ut of problems 3.6 and 3.7
u
t
<R, ∀t ∈
0, 1
, 3.19
with R independent of λ ∈ 0, 1.
Consider the set
E
r
1
t, x
0
,x
1
,x
,x
1
,x
2
,x
3
λf
t, δ
0
t, x
0
,δ
1
t, x
1
,δ
2
t, x
2
,x
3
λf
t, δ
0
t, x
0
,δ
1
t, x
1
,δ
2
t, x
2
,x
3
x
2
− λδ
2
t, x
1
th
E
∗
|x
3
|2r
1
in
0
,weseethatF
λ
verifies 2.3 with E and h
E
replaced
by E
r
1
and h
E
r
1
, respectively. The condition 2.4 is also verified since
∞
0
s
h
E
8 Boundary Value Problems
Therefore, F
λ
satisfies the one-sided Nagumo-type condition in E
r
1
with h
E
replaced by h
E
r
1
,
with r
1
independent of λ ∈ 0, 1.
Moreover, for
ρ : max
ar
1
|
A
|
b
,
|
B
|
A
|
≤ ρ,
u
1
λ
d
B − cu
1
≥−
λ
d
|
B
|
cr
1
≥−ρ.
3.25
Define
E
r
1
do not
depend on λ,weseethatR is maybe independent of λ.
Step 3. For λ 1, the problems 3.6 and 3.7 has at least one solution u
1
t.
Define the operators
L : C
4
0, 1
⊂ C
3
0, 1
−→ C
0, 1
×
4
3.27
by
Lu
u
N
λ
u
λf
t, δ
0
t, u
t
,δ
1
t, u
t
,δ
2
t, u
t
0
− A
,
B
λ
:
λ
d
B − cu
1
.
3.30
Boundary Value Problems 9
Observe that L has a compact inverse. Therefore, we can consider the completely continuous
operator
T
λ
:
C
3
For R given by Step 2,taketheset
Ω
x ∈ C
3
0, 1
:
x
i
∞
<r
1
,i 0, 1, 2,
x
∞
<R
t
,
u
0
u
1
u
0
u
1
0
3.35
and has only the trivial solution. Then, by the degree theory,
d
I − T
0
, Ω, 0
t, u
t
,u
t
u
t
− δ
2
t, u
t
,
u
0
1
t in Ω.
Step 4. The function u
1
t is a solution of the problems 1.1 and 1.2.
10 Boundary Value Problems
The proof will be finished if the above function u
1
t satisfies the inequalities
α
t
≤ u
1
t
≤ β
t
,α
t
≤ u
1
t,anddefine
max
t∈0,1
u
1
t
− β
t
: u
1
t
2
− β
t
2
f
t
2
,δ
0
t
2
,u
1
t
2
,δ
1
t
2
,u
1
t
2
,δ
t
2
,u
1
t
2
f
t
2
,δ
0
t
2
,u
1
t
2
,δ
1
t
2
t
2
≥ f
t
2
,β
t
2
,β
t
2
,β
t
2
,β
t
2
1
0
− β
0
> 0,
u
1
0
− β
0
u
1
0
− β
0
− A
≥ β
0
.
3.42
Then t
2
/
0 and, by similar arguments, we prove that t
2
/
1. Thus,
u
1
t
≤ β
t
On the other hand, by 1.2,
0 u
1
1
− u
1
0
1
0
u
1
t
dt
1
0
u
1
s
ds dt, 3.45
Boundary Value Problems 11
that is,
u
1
0
−
1
0
t
0
u
1
s
ds dt.
3.46
Applying the same technique, we have
−
1
β
0
,
3.47
and then by Definition 2.1 iii, 3.44 and 3.46,weobtain
α
0
≤ β
0
− β
1
β
0
−
β
0
≥ α
0
− α
1
α
0
−
1
0
t
0
α
s
0
≤ β
0
. 3.49
Since, by 3.44, β
t − u
1
t is nondecreasing, we have by 3.49
β
t
− u
1
t
≥ β
0
1
0
β
0
≥ 0, 3.51
and so βt ≥ u
1
t for every t ∈ 0, 1.
The inequalities u
1
t ≥ α
t and u
1
t ≥ αt for every t ∈ 0, 1 can be proved in the
same way. Then u
1
t is a solution of problems 1.1 and 1.2.
4. An Example
The following example shows the applicability of Theorem 3.1 when f satisfies only the one-
sided Nagumo-type condition.
12 Boundary Value Problems
Example 4.1. Consider now the problem
u
4
4
, 4.1
u
0
u
1
0,
u
0
− u
0
A,
u
1
u
− 2
2
−
x
3
4
4.3
is continuous in 0, 1 ×
4
.IfA, B ∈ −2, 2, then the functions α, β : 0, 1 → defined by
α
t
−t
2
− t, β
t
t
2
t 4.4
are, respectively, lower and upper functions of 4.1 and 4.2.Moreover,define
E
E
|x
3
|1,
in E.
Therefore, by Theorem 3.1, ther e is at least one solution ut of Problem 4.1 and 4.2
such that, for every t ∈ 0, 1,
−t
2
− t ≤ u
t
≤ t
2
t, −2t − 1 ≤ u
t
≤ 2t 1, −2 ≤ u
t
≤ 2.
4.6
Notice that the function
f
does not satisfy the two-sided Nagumo condition.
Acknowledgments
The authors would like to thank the referees for their valuable comments on and suggestions
regarding the original manuscript. This work was supported by NSFC 10771085,byKey
Lab of Symbolic Computation and Knowledge Engineering of Ministry of Education, and by
the 985 Program of Jilin University.
Boundary Value Problems 13
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