Hindawi Publishing Corporation
Boundary Value Problems
Volume 2007, Article ID 68758, 10 pages
doi:10.1155/2007/68758
Research Article
Existence of Positive Solutions for Fourth-Order
Three-Point Boundary Value Problems
Chuanzhi Bai
Received 11 July 2007; Accepted 7 November 2007
Recommended by Jean Mawhin
We are concerned with the nonlinear fourth-order three-point boundary value problem u(4) (t) = a(t) f (u(t)), 0 < t < 1, u(0) = u(1) = 0, αu (η) − βu (η) = 0, γu (1) +
δu (1) = 0. By using Krasnoselskii’s fixed point theorem in a cone, we get some existence results of positive solutions.
Copyright © 2007 Chuanzhi Bai. 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
As is pointed out in [1, 2], boundary value problems for second- and higher-order differential equations play a very important role in both theory and applications. Recently, an
increasing interest in studying the existence of solutions and positive solutions to boundary value problems for fourth-order differential equations is observed; see, for example,
[3–8].
In this paper, we are concerned with the existence of positive solutions for the following fourth-order three-point boundary value problem (BVP):
u(4) (t) = a(t) f u(t) ,
0 < t < 1,
u(0) = u(1) = 0,
αu (η) − βu (η) = 0,
(1.1)
γu (1) + δu (1) = 0,
where α, β, γ, and δ are nonnegative constants satisfying αδ + βγ + αγ > 0; 0 < η < 1,
a ∈ C[0,1], and f ∈ C([0, ∞),[0, ∞)). We use Krasnoselskii’s fixed point theorem in cones
to establish some simple criteria for the existence of at least one positive solution to BVP
(1.1). To the best of our knowledge, no paper in the literature has investigated the existence of positive solutions for BVP (1.1).
2
Boundary Value Problems
The paper is formulated as follows. In Section 2, some definitions and lemmas are
given. In Section 3, we prove some existence theorems of the positive solutions for BVP
(1.1).
2. Preliminaries and lemmas
In this section, we introduce some necessary definitions and preliminary results that will
be used to prove our main results.
First, we list the following notations and assumptions:
f0 = lim
u→0
f (u)
,
u
f∞ = lim
u→∞
f (u)
.
u
(2.1)
(H1 ) f : [0, ∞)→[0, ∞) is continuous;
(H2 ) a ∈ C[0,1], a(t) ≤ 0, for all t ∈ [0,η], a(t) ≥ 0, for all t ∈ [η,1], and a(t) ≡ 0, for
all t ∈ (p,η) ∪ (η, q) (0 < p < η < q < 1).
By routine calculation, we easily obtain the following lemma.
Lemma 2.1. Suppose that α, β, γ, δ are nonnegative constants satisfying αδ + βγ + αγ > 0.
If h ∈ C[0,1], then the boundary value problem
v (t) = h(t),
t ∈ [0,1],
(2.2)
γv(1) + δv (1) = 0
αv(η) − βv (η) = 0,
has a unique solution
v(t) =
t
η
(t − s)h(s)ds +
1
σ
1
η
α(η − t) − β γ(1 − s) + δ h(s)ds,
(2.3)
where σ = αδ + βγ + αγ(1 − η) > 0.
Let G(t,s) be the Green’s function of the differential equation
−u (t) = 0,
t ∈ (0,1),
(2.4)
subject to the boundary condition
u(0) = u(1) = 0.
(2.5)
In particular,
⎧
⎨s(1 − t),
G(t,s) = ⎩
t(1 − s),
0 ≤ s ≤ t ≤ 1,
0 ≤ t < s ≤ 1.
(2.6)
Chuanzhi Bai 3
It is rather straightforward that
0 ≤ G(t,s) ≤ G(s,s),
G(t,s) ≥ mG(s,s),
0 ≤ t, s ≤ 1,
(2.7)
t ∈ [p, q], s ∈ [0,1],
(2.8)
where 0 < p < q < 1, and 0 < m = min { p,1 − q} < 1.
Let X be the Banach space C[0,1] endowed with the norm
u = max u(t).
(2.9)
0≤t ≤1
We define the operator T : X →X by
Tu(t) =
1
0
s
G(t,s)
η
(τ − s)a(τ) f u(τ) dτ
1
1
+
σ
(2.10)
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ ds,
η
where G(t,s) as in (2.6). From Lemma 2.1, we easily know that u(t) is a solution of the
fourth-order three-point boundary value problem (1.1) if and only if u(t) is a fixed point
of the operator T.
Define the cone K in X by
K = u ∈ X | u ≥ 0, min u(t) ≥ mu ,
(2.11)
t ∈[p,q]
where 0 < p < η < q < 1, and
0 < m = max { p,1 − q} < 1.
(2.12)
Lemma 2.2. Assume that (H1 ) and (H2 ) hold. If β ≥ αη, then T : K →K is completely continuous.
Proof. For any u ∈ K, we know from (2.10), (H1 ), (H2 ), and β ≥ αη that
(Tu)(t) =
η
0
η
G(t,s)
s
+
1
+
η
1
1
σ
s
G(t,s)
(s − τ)a(τ) f u(τ) dτ
η
η
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
(τ − s)a(τ) f u(τ) dτ
+
1
σ
+
1
σ
s
η
1
s
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
4
Boundary Value Problems
=
η
η
G(t,s)
0
1
+
σ
1
+
η
s
1
1
σ
η
1
σ
+
≥ 0,
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
η
s
G(t,s)
(s − τ)a(τ) f u(τ) dτ
αδ(τ − η) + βγ(1 − s) + αγ(1 − s)(τ − η) + βδ a(τ) f u(τ) dτ
1
β + α(s − η) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
s
t ∈ [0,1].
(2.13)
Hence, in view of (2.13) and (2.7), we have
Tu = max (Tu)(t) ≤
η
t ∈[0,1]
η
G(s,s)
0
s
+
1
+
η
s
G(s,s)
1
σ
+
η
1
σ
1
σ
(s − τ)a(τ) f u(τ) dτ
1
η
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
αδ(τ − η) + βγ(1 − s) + αγ(1 − s)(τ − η) + βδ a(τ) f u(τ) dτ
1
s
β + α(s − η) γ(1 − τ) + δ a(τ) f u(τ) dτ ds.
(2.14)
Thus from (2.8), (2.13), and (2.14), we get
min (Tu)(t)
t ∈[p,q]
≥m
η
0
1
+
σ
+m
η
G(s,s)
1
η
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
1
η
s
(s − τ)a(τ) f u(τ) dτ
s
G(s,s)
1
σ
η
1
+
σ
αδ(τ − η) + βγ(1 − s) + αγ(1 − s)(τ − η) + βδ a(τ) f u(τ) dτ
1
s
β + α(s − η) γ(1 − τ) + δ a(τ) f u(τ) dτ ds = mTu,
(2.15)
where m as in (2.12). So T : K →K. Moreover, it is easy to check by the Arzela-Ascoli
theorem that the operator T is completely continuous.
Remark 2.3. By σ = αδ + βγ + αγ(1 − η) > 0 and β ≥ αη, we have β > 0.
Recently, Krasnoselskii’s theorem of cone expansion/compression type has been used
to study the existence of positive solutions of boundary value problems in many papers;
Chuanzhi Bai 5
see, for example, Liu [7], Ma [9], Torres [10], and the references contained therein. The
following lemma (Krasnoselskii’s fixed point theorem) will play an important role in the
proof of our theorem.
Lemma 2.4 [11]. Let X be a Banach space, and let K ⊂ X be a cone in X. Assume that Ω1 , Ω2
are open subsets of X with 0 ∈ Ω1 , Ω1 ⊂ Ω2 and let A : K ∩ (Ω2 \ Ω1 )→K be a completely
operator such that either
(i) Au ≤ u, u ∈ K ∩ ∂Ω1 and Au ≥ u, u ∈ K ∩ ∂Ω2 ; or
(ii) Au ≥ u, u ∈ K ∩ ∂Ω1 and Au ≤ u, u ∈ K ∩ ∂Ω2 .
Then A has a fixed point in K ∩ (Ω2 \ Ω1 ).
3. Main result
We are now in a position to present and prove our main result.
Theorem 3.1. Let β ≥ αη. Assume that (H1 )-(H2 ) hold. If f0 = ∞ and f∞ = 0, then (1.1)
has at least a positive solution.
Proof. Since f0 = ∞, we can choose r > 0 sufficiently small so that
f (u) ≥ εu
for 0 ≤ u ≤ r,
(3.1)
where ε satisfies
⎧
6
⎪
η
⎪
,
⎪
⎪
⎨ m(1 − η) 0 −a(τ)τ 3 dτ
ε≥ ⎪
⎪
⎪
⎪
⎩
if a t0 < 0, for some t0 ∈ (p,η),
σ
, if a t1 > 0, for some t1 ∈ (η, q).
βmη η (τ − η)(1 − τ) γ(1 − τ) + δ a(τ)dτ
(3.2)
1
Set Ωr = {u ∈ K | u < r }. From condition (H2 ), we consider two cases as follows.
Case 1. If a(t0 ) < 0 for some t0 ∈ (p,η), then, for u ∈ ∂Ωr , we have from (2.13), (3.1),
and (3.2) that
(Tu)(η) ≥
η
0
η
G(η,s)
s
+
≥
η
η
0
1
σ
G(η,s)
≥ mεu
= mεu
η
0
s
1
η
G(η,s)
η
s
0
η
0
η
G(η,s)
(s − τ)a(τ)dτ ds = mεu
τ
a(τ)dτ
β − α(η − s) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
(s − τ)a(τ) f u(τ) dτ ds ≥ ε
η
0
(s − τ)a(τ) f u(τ) dτ
η
(1 − η)s(s − τ)ds = mεu
0
s
(s − τ)a(τ)u(τ)dτ ds
τ
a(τ)dτ
1−η
6
η
0
0
G(η,s)(s − τ)ds
−a(τ)τ 3 dτ ≥ u,
(3.3)
6
Boundary Value Problems
which implies
Tu ≥ u,
∀u ∈ ∂Ωr .
(3.4)
Case 2. If a(t1 ) > 0 for some t1 ∈ (η, q), then, for u ∈ ∂Ωr , we have from (2.13), (3.1),
and (3.2) that
(Tu)(η) ≥
s
1
η
1
σ
G(η,s)
≥
1
σ
1
G(η,s)
η
1
s
1
≥
β
σ
≥
εβm
u σ
G(η,s)
η
s
1
η
εβm
u σ
=
εβm
u σ
≥
εβηm
u σ
1
σ
1
s
β + α(s − η) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
γ(1 − τ) + δ a(τ) f u(τ) dτ ds
1
s
γ(1 − τ) + δ a(τ)dτ
γ(1 − τ) + δ a(τ)dτ
η
β + α(s − η) γ(1 − τ) + δ a(τ) f u(τ) dτ ds
G(η,s)ds
1
=
αδ(τ − η) + βγ(1 − s) + αγ(1 − s)(τ − η) + βδ a(τ) f u(τ) dτ
η
+
1
τ
1
η
η(1 − s)ds
1
η(τ − η) 1 − (τ + η) γ(1 − τ) + δ a(τ)dτ
2
η
1
η
(τ − η)(1 − τ) γ(1 − τ) + δ a(τ)dτ ≥ u,
(3.5)
that is,
Tu ≥ u,
∀u ∈ ∂Ωr .
(3.6)
Next, define a function f ∗ (v) : [0, ∞)→[0, ∞) by
f ∗ (v) = max f (u).
(3.7)
0≤u≤v
It is easy to see that f ∗ (v) is nondecreasing. Since f∞ = 0, we have lim v→∞ f ∗ (v)/v = 0.
Thus, there exists R > r such that
f ∗ (R) ≤ θR,
where θ satisfies
θ
1
12
η
0
−a(τ)τ 3 dτ +
(3.8)
1
(1 − η)σ + βδ 1 − η2
6σ
1
β + α(1 − η)
+
6σ
1
η
1
η
a(τ)dτ
γ(1 − τ) + δ a(τ)dτ ≤ 1.
(3.9)
Chuanzhi Bai 7
Hence, we obtain
f (u) ≤ f ∗ (R) ≤ θR,
0 ≤ u ≤ R.
(3.10)
Thus from (2.14) and (3.10), for all u ∈ ∂ΩR , we have
Tu ≤ θR
η
0
η
G(s,s)
s
1
+
s
G(s,s)
η
(s − τ)a(τ)dτ +
1
σ
+
≤ θR
η
0
+
+
= θR
1
12
+
τ
a(τ)dτ
1
σ
1
σ
1
η
1
η
0
η
0
η
1
σ
1
1
σ
η
s(1 − s)ds
1
s
1
η
1
η
−a(τ)τ 3 dτ +
β
σ
1
≤ θR
1
12
+
0
−a(τ)τ 3 dτ +
γ(1 − τ) + δ a(τ)dτ
η
0
s(1 − s)ds
αδ(1 − η) + βγ(1 − η) + αγ(1 − η)2 + βδ a(τ)dτ
β + α(1 − η) γ(1 − τ) + δ a(τ)dτ
β 2
3η − 2η3
6σ
1
η
γ(1 − τ) + δ a(τ)dτ
1
a(τ)dτ
η
1
η
1
η
γ(1 − τ) + δ a(τ)dτ
1
(1 − η)σ + βδ 1 − η2
6σ
1
β + α(1 − η)
6σ
η
1
(1 − η)σ + βδ 1 − 3η2 + 2η3
6σ
η
β + α(s − η) γ(1 − τ) + δ a(τ)dτ ds
1
β + α(1 − η) 1 − 3η2 + 2η3
+
6σ
αδ(τ − η) + βγ(1 − s) + αγ(1 − s)(τ − η) + βδ a(τ)dτ
s(1 − s)(s − τ)ds +
s(1 − s)ds
(β − αη + αs) γ(1 − τ) + δ a(τ)dτ ds
1
η
a(τ)dτ
γ(1 − τ) + δ a(τ)dτ ≤ R = u,
(3.11)
that is,
Tu ≤ u,
for u ∈ ∂ΩR .
(3.12)
Hence, from (3.6), (3.12), and Lemma 2.4, T has a fixed point u ∈ ΩR \ Ωr , which means
that u is a positive solution of BVP (1.1).
Theorem 3.2. Let β ≥ αη. Assume that (H1 )-(H2 ) hold. If f0 = 0 and f∞ = ∞, then (1.1)
has at least a positive solution.
8
Boundary Value Problems
Proof. Since f∞ = ∞, we can choose R1 > 0 sufficiently large so that
f (u) ≥ Au,
u ≥ R1 ,
(3.13)
where A satisfies
⎧
6
⎪
⎪
η
,
⎪
⎪
⎪
m(1
−
η)
−
a(τ)(τ
−
p) τ 2 + τ p − τ p2 dτ
⎨
p
if a t0 < 0, for some t0 ∈ (p,η),
σ
⎪
⎪
⎪
⎪
⎩ βmη q (τ − η)(1 − τ)γ(1 − τ) + δ a(τ)dτ ,
if a t1 > 0, for some t1 ∈ (η, q).
A≥ ⎪
η
(3.14)
Choose
R≥
R1
,
m
(3.15)
where m > 0 as in (2.12). Let u ∈ ∂ΩR . Since u(t) ≥ mu = mR ≥ R1 for t ∈ [p, q], from
(3.13), we see that
f u(t) ≥ Au(t) ≥ AmR,
∀t ∈ [p, q], u ∈ ∂ΩR .
(3.16)
For u ∈ ∂ΩR , we consider two cases as follows.
Case 1. If a(t0 ) < 0 for some t0 ∈ (p,η), then we have from (3.3), (3.14), and (3.16) that
(Tu)(η) ≥
≥
η
η
0
G(η,s)
η
η
p
s
G(η,s)
≥ AmR
s
η
G(η,s)
= AmR(1 − η)
=
(s − τ)a(τ) f u(τ) dτ ds
η
p
(s − τ)a(τ) f u(τ) dτ ds
s
(s − τ)a(τ)dτ ds
p
η
1
AmR(1 − η)
6
τ
a(τ)dτ
p
η
p
(3.17)
s(s − τ)ds
−a(τ)(τ − p) τ 2 + τ p − 2p2 dτ
≥ R = u ,
which implies
Tu ≥ u,
∀u ∈ ∂ΩR .
(3.18)
Chuanzhi Bai 9
Case 2. If a(t1 ) > 0 for some t1 ∈ (η, q), then we have from (3.5), (3.14), and (3.16) that
(Tu)(η) ≥
≥
β
σ
β
σ
1
η
1
G(η,s)
q
η
s
q
G(η,s)
≥ AmR
β
σ
= AmR
β
σ
q
η
q
η
γ(1 − τ) + δ a(τ) f u(τ) dτ ds
q
s
γ(1 − τ) + δ a(τ)dτ ds
η
γ(1 − τ) + δ a(τ) f u(τ) dτ ds
G(η,s)
q
βη
≥ AmR
σ
s
γ(1 − τ) + δ a(τ)dτ
τ
η
(3.19)
η(1 − s)ds
(τ − η)(1 − τ) γ(1 − τ) + δ a(τ)dτ ≥ R = u,
which implies
Tu ≥ u,
∀u ∈ ∂ΩR .
(3.20)
Since f0 = 0, we can choose 0 < r < R such that
f (u) ≤ θu,
0 ≤ u ≤ r,
(3.21)
where θ as in (3.9). For u ∈ ∂Ωr , we have from (3.11) and (3.21) that
Tu ≤ θ u
1
12
η
0
−a(τ)τ 3 dτ +
1
(1 − η)σ + βδ 1 − η2
6σ
1
β + α(1 − η)
+
6σ
1
η
1
η
a(τ)dτ
(3.22)
γ(1 − τ) + δ a(τ)dτ ≤ u.
So,
Tu ≤ u,
u ∈ ∂Ωr .
(3.23)
Therefore, from (3.20), (3.23), and Lemma 2.4, T has a fixed point u ∈ ΩR \ Ωr , which
means u is a positive solution of BVP (1.1).
Finally, we conclude this paper with the following example.
Example 3.3. Consider the following fourth-order three-point boundary value problem:
u(4) (t) = sinπ(1 + 2t)ur (t),
αu
0 < t < 1,
u(0) = u(1) = 0,
1
1
− βu
= 0,
2
2
(3.24)
γu (1) + δu (1) = 0,
where 0 < r < 1, α, β, γ, and δ are nonnegative constants satisfying αδ + βγ + αγ > 0 and
β ≥ (1/2)α. Then BVP (3.24) has at least one positive solution.
10
Boundary Value Problems
To see this, we will apply Theorem 3.1. Set
f (u) = ur ,
a(t) = sinπ(1 + 2t),
1
η= .
2
(3.25)
With the above functions f and a, we see that (H1 ) and (H2 ) hold. Moreover, it is easy to
see that
f 0 = ∞,
f∞ = 0,
β ≥ αη.
(3.26)
The result now follows from Theorem 3.1.
Acknowledgments
This work is supported by the National Natural Science Foundation of China (10771212)
and the Natural Science Foundation of Jiangsu Education Office (06KJB110010). The
author is grateful to the referees for their valuable suggestions and comments.
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Chuanzhi Bai: Department of Mathematics, Huaiyin Teachers College, Huaian 223300, China
Email address: czbai8@sohu.com