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Hindawi Publishing Corporation
Boundary Value Problems
Volume 2009, Article ID 393259, 9 pages
doi:10.1155/2009/393259
Research Article
Positive Solutions for Some Beam Equation
Boundary Value Problems
Jinhui Liu
1, 2
and Weiya Xu
3
1
Department of Civil Engineering, Hohai University, Nanjing 210098, China
2
Zaozhuang Coal Mining Group Co., Ltd, Jining 277605, China
3
Graduate School, Hohai University, Nanjing 210098, China
Correspondence should be addressed to Jinhui Liu,
Received 2 September 2009; Accepted 1 November 2009
Recommended by Wenming Zou
A new fixed point theorem in a cone is applied to obtain the existence of positive solutions of
some fourth-order beam equation boundary value problems with dependence on the first-order
derivative u
iυ
tft, ut,u
t, 0 <t<1,u0u1u
0u
10, where f : 0, 1 ×
0, ∞ × R → 0, ∞ is continuous.
Copyright q 2009 J. Liu and W. Xu. 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.
1. Introduction
It is well known that beam is one of the basic structures in architecture. It is greatly used
in the designing of bridge and construction. Recently, scientists bring forward the theory of
combined beams. That is to say, we can bind up some stratified structure copings into one
global combined beam with rock bolts. The deformations of an elastic beam in equilibrium
state, whose two ends are simply supported, can be described by following equation of
deflection curve:
d
2
dx
2
EI
z
d
2
v
dx
2
q
x
, 1.1
where E is Yang’s modulus constant, I
z
is moment of inertia with respect to z axes,
determined completely by the beam’s shape cross-section. Specially, I
z
bh
3
/12 if the cross-
section is a rectangle with a height of h and a width of b. Also, qx is loading at x.Ifthe
2 Boundary Value Problems
loading of beam considered is in relation to deflection and rate of change of deflection, we
need to research the more general equation
u
4
x
f
x, u
x
,u
x
.
1.2
According to the forms of supporting, various boundary conditions should be considered.
Solving corresponding boundary value problems, one can obtain the expression of deflection
curve. It is the key in design of constants of beams and rock bolts.
Owing to its importance in physics and engineering, the existence of solutions to this
problem has been studied by many authors, see 1–10. However, in practice, only its positive
solution is significant. In 1, 9, 11, 12, Aftabizadeh, Del Pino and Man
´
asevich, Gupta, and Pao
showed the existence of positive solution for
u
iv
t
f
t, u
t
,u
t
1.3
under some growth conditions of f and a nonresonance condition involving a two-parameter
linear eigenvalue problem. All of these results are based on the Leray-Schauder continuation
method and topological degree.
The lower and upper solution method has been studied for the fourth-order problem
by several authors 2, 3, 7, 8, 13, 14. However, all of these authors consider only an equation
of the form
u
iv
t
f
t, u
t
,
1.4
with diverse kind of boundary conditions. In 10, Ehme et al. gave some sufficient conditions
for the existence of a solution of
u
iv
t
f
t, u
t
,u
t
,u
t
,u
t
1.5
with some quite general nonlinear boundary conditions by using the lower and upper
solution method. The conditions assume the existence of a strong upper and lower solution
pair.
Recently, Krasnosel’skii’s fixed point theorem in a cone has much application in
studying the existence and multiplicity of positive solutions for differential equation
boundary value problems, see 3, 6. With this fixed point theorem, Bai and Wang 6
discussed the existence, uniqueness, multiplicity, and infinitely many positive solutions for
the equation of the form
u
iv
t
λf
t, u
t
,
1.6
where λ>0 is a constant.
Boundary Value Problems 3
In this paper, via a new fixed point theorem in a cone and concavity of function, we
show the existence of positive solutions for the following problem:
u
iv
t
f
t, u
t
,u
t
, 0 <t<1,
u
0
u
1
u
0
u
1
0,
1.7
where f : 0, 1 × 0, ∞ × R → 0, ∞ is continuous.
We point out that positive solutions of 1.7 are concave and this concavity provides
lower bounds on positive concave functions of their maximum, which can be used in defining
a cone on which a positive operator is defined, to which a new fixed point theorem in a cone
due to Bai and Ge 5 can be applied to obtain positive solutions.
2. Fixed Point Theorem in a Cone
Let X be a Banach space and P ⊂ X a cone. Suppose α, β : X → R
are two continuous
nonnegative functionals satisfying
α
λx
≤
|
λ
|
α
x
,β
λx
≤
|
λ
|
β
x
, for x ∈ X, λ ∈
0, 1
,
M
1
max
α
x
,β
x
≤
x
≤ M
2
max
α
x
,β
x
, for x ∈ X,
2.1
where M
1
,M
2
are two positive constants.
Lemma 2.1 see 5. Let r
2
>r
1
> 0,L
2
>L
1
> 0 are constants and
Ω
i
x ∈ X | α
x
<r
i
,β
x
<L
i
,i 1, 2 2.2
are two open subsets in X such that θ ∈ Ω
1
⊂ Ω
1
⊂ Ω
2
. In addition, let
C
i
x ∈ X | α
x
r
i
,β
x
≤ L
i
,i 1, 2;
D
i
x ∈ X | α
x
≤ r
i
,β
x
L
i
,i 1, 2.
2.3
Assume T : P → P is a completely continuous operator satisfying
S
1
αTx ≤ r
1
,x∈ C
1
∩ P ; βTx ≤ L
1
,x ∈ D
1
∩ P ; αTx ≥ r
2
,x ∈ C
2
∩ P ; βTx ≥
L
2
,x ∈ D
2
∩ P;
or
S
2
αTx ≥ r
1
,x ∈ C
1
∩ P; βTx ≥ L
1
,x ∈ D
1
∩ PαTx ≤ r
2
,x ∈ C
2
∩ P; βTx ≤
L
2
,x ∈ D
2
∩ P,
then T has at least one fixed point in
Ω
2
\ Ω
1
∩ P.
4 Boundary Value Problems
3. Existence of Positive Solutions
In this section, we are concerned with the existence of positive solutions for the fourth-order
two-point boundary value problem 1.7.
Let X C
1
0, 1 with u max{max
0≤t≤1
|ut|, max
0≤t≤1
|u
t|} be a Banach space,
P {u ∈ X | ut ≥ 0,u is concave on 0, 1}⊂X a cone. Define functionals
α
u
max
0≤t≤1
|
u
t
|
,β
u
max
0≤t≤1
u
t
, for u ∈ X,
3.1
then α, β : X → R
are two continuous nonnegative functionals such that
u
max
α
u
,β
u
3.2
and 2.1 hold.
Denote by Gt, s Green’s function for boundary value problem
−y
t
0, 0 <t<1,
y
0
y
1
0.
3.3
Then Gt, s ≥ 0, for 0 ≤ t, s ≤ 1, and
G
t, s
⎧
⎨
⎩
t
1 − s
, 0 ≤ t ≤ s ≤ 1,
s
1 − t
, 0 ≤ s ≤ t ≤ 1.
3.4
Let
M max
0≤t≤1
1
0
G
t, s
G
s, x
dx ds,
N max
0≤t≤1
1
0
3/4
1/4
G
t, s
G
s, x
dx ds,
A max
1
0
1 − s
G
s, x
dx ds,
1
0
sG
s, x
dx ds
,
B max
1
0
1−h
h
1 − s
G
s, x
dx ds,
1
0
1−h
h
sG
s, x
dx ds
.
3.5
However, 1.7 has a solution u ut if and only if u solves the operator equation
u
t
Tu
t
:
1
0
1
0
G
t, s
G
s, x
f
x, u
x
,u
x
dx
ds. 3.6
It is well know that T : P → P is completely continuous.
Boundary Value Problems 5
Theorem 3.1. Suppose there are four constants r
2
>r
1
> 0,L
2
>L
1
> 0 such that max{r
1
,L
1
}≤
min{r
2
,L
2
} and the following assumptions hold:
A
1
ft, x
1
,x
2
≥ max{r
1
/M, L
1
/A}, for t, x
1
,x
2
∈ 0, 1 × 0,r
1
× −L
1
,L
1
;
A
2
ft, x
1
,x
2
≤ min{r
2
/M, L
2
/A}, for t, x
1
,x
2
∈ 0, 1 × 0,r
2
× −L
2
,L
2
.
Then, 1.7 has at least one positive solution ut such that
r
1
≤ max
0≤t≤1
u
t
≤ r
2
or L
1
≤ max
0≤t≤1
u
t
≤ L
2
.
3.7
Proof. Let
Ω
i
u ∈ X | α
u
<r
i
,β
u
<L
i
,i 1, 2, 3.8
be two bounded open subsets in X. In addition, let
C
i
u ∈ X | α
u
r
i
,β
u
≤ L
i
,i 1, 2;
D
i
u ∈ X | α
u
≤ r
i
,β
u
L
i
,i 1, 2.
3.9
For u ∈ C
1
∩ P,byA
1
, there is
α
Tu
max
t∈
0,1
1
0
G
t, s
G
s, x
f
x, u
x
,u
x
dx ds
≥
r
1
M
· max
t∈
0,1
1
0
G
t, s
G
s, x
dx ds
r
1
.
3.10
For u ∈ P , because T : P → P,soTu ∈ P,thatistosayTu concave on 0, 1, it follows
that
max
t∈
0,1
Tu
t
max
Tu
0
,
Tu
1
.
3.11
6 Boundary Value Problems
Combined with A
1
and f ≥ 0, for u ∈ D
1
∩ P, there is
β
Tu
max
t∈
0,1
Tu
t
max
t∈
0,1
−
t
0
s
1
0
G
s, x
f
x, u
x
,u
x
dx ds
1
t
1 − s
1
0
G
s, x
f
x, u
x
,u
x
dx ds
max
1
0
1 − s
1
0
G
s, x
f
x, u
x
,u
x
dx ds,
1
0
s
1
0
G
s, x
f
x, u
x
,u
x
dx ds
≥
L
1
A
· max
1
0
1 − s
G
s, x
dx ds,
1
0
sG
s, x
dx ds
L
1
A
· A L
1
.
3.12
For u ∈ C
2
∩ P,byA
2
, there is
α
Tu
max
t∈
0,1
1
0
G
t, s
G
s, x
f
x, u
x
,u
x
dx ds
≤ max
t∈
0,1
1
0
G
t, s
G
s, x
·
r
2
M
dx ds
r
2
M
· max
t∈
0,1
1
0
G
t, s
G
s, x
dx ds r
2
.
3.13
For u ∈ D
2
∩ P,byA
2
, there is
β
Tu
max
1
0
1 − s
1
0
G
s, x
f
x, u
x
,u
x
dx ds,
1
0
s
1
0
G
s, x
f
x, u
x
,u
x
dx ds
≤
L
2
A
· max
1
0
1 − s
G
s, x
dx ds,
1
0
sG
s, x
dx ds
L
2
A
· A L
2
.
3.14
Boundary Value Problems 7
Now, Lemma 2.1 implies there exists u ∈
Ω
2
\ Ω
1
∩ P such that u Tu, namely, 1.7
has at least one positive solution ut such that
r
1
≤ α
u
≤ r
2
or L
1
≤ β
u
≤ L
2
, 3.15
that is,
r
1
≤ max
0≤t≤1
u
t
≤ r
2
or L
1
≤ max
0≤t≤1
u
t
≤ L
2
.
3.16
The proof is complete.
Theorem 3.2. Suppose there are five constants 0 <r
1
<r
2
, 0 <L
1
<L
2
, 0 ≤ h<1/2 such that
max{r
1
/N, L
1
/B}≤min{r
2
/M, L
2
/A}, and the following assumptions hold
A
3
ft, x
1
,x
2
≥ r
1
/N, for t, x
1
,x
2
∈ 1/4, 3/4 × r
1
/4,r
1
× −L
1
,L
1
;
A
4
ft, x
1
,x
2
≥ L
1
/B, for t, x
1
,x
2
∈ h, 1 − h × 0,r
1
× −L
1
,L
1
;
A
5
ft, x
1
,x
2
≤ min{r
2
/M, L
2
/A}, for t, x
1
,x
2
∈ 0, 1 × 0,r
2
× −L
2
,L
2
.
Then, 1.7 has at least one positive solution ut such that
r
1
≤ max
0≤t≤1
u
t
≤ r
2
or L
1
≤ max
0≤t≤1
u
t
≤ L
2
.
3.17
Proof. We just need notice the following difference to the proof of Theorem 3.1.
For u ∈ C
1
∩P, the concavity of u implies that ut ≥ 1/4αur
1
/4fort ∈ 1/4, 3/4.
By A
3
, there is
α
Tu
max
t∈
0,1
1
0
G
t, s
G
s, x
f
x, u
x
,u
x
dx ds
≥ max
t∈
0,1
1
0
3/4
1/4
G
t, s
G
s, x
f
x, u
x
,u
x
dx ds
≥ max
t∈
0,1
1
0
3/4
1/4
G
t, s
G
s, x
·
r
1
N
dx ds
r
1
N
· max
t∈
0,1
1
0
3/4
1/4
G
t, s
G
s, x
dx ds
r
1
.
3.18
8 Boundary Value Problems
For u ∈ D
1
∩ P,byA
4
, there is
β
Tu
max
1
0
1 − s
1
0
G
s, x
f
x, u
x
,u
x
dx ds,
1
0
s
1
0
G
s, x
f
x, u
x
,u
x
dx ds
≥ max
1
0
1 − s
1−h
h
G
s, x
f
x, u
x
,u
x
dx ds,
1
0
s
1−h
h
G
s, x
f
x, u
x
,u
x
dx ds
≥
L
1
B
· max
1
0
1−h
h
1 − s
G
s, x
dx ds,
1
0
1−h
h
sG
s, x
dx ds
L
1
B
· B L
1
3.19
The rest of the proof is similar to Theorem 3.1 and the proof is complete.
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Boundary Value Problems 9
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