Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2011, Article ID 831730, 8 pages
doi:10.1155/2011/831730
Research Article
Positive Solutions for a Class of Fourth-Order
Boundary Value Problems in Banach Spaces
Jingjing Cai1 and Guilong Liu2
1
2
Department of Mathematics, Tongji University, Shanghai 200092, China
College of Electrical Engineering, Xinjiang University, Urumqi 830008, China
Correspondence should be addressed to Jingjing Cai, cjjing1983@163.com
Received 3 October 2010; Accepted 9 January 2011
Academic Editor: Yong Zhou
Copyright q 2011 J. Cai and G. Liu. 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.
Using a specially constructed cone and the fixed point index theory, this work shows existence
and nonexistence results of positive solutions for fourth-order boundary value problem with two
different parameters in Banach spaces.
1. Introduction
In this paper, we study the existence of positive solutions for the following fourth-order
boundary value problem BVP with singularity in Banach space E, · :
x4 t βx t λft, x μgt, x,
x0 x1 θ,
t ∈ J,
x 0 x 1 θ,
1.1
where J 0, 1, β < π 2 . g, f ∈ CJ, 0, ∞, R 0, ∞, and θ is the zero element of E.
Fourth-order boundary value problems are studied not only in mathematics but also
in physics and many other fields. For example, some models for bridge, underground water
flow, and plasma physics can be reduced to fourth-order boundary value problems. In the
recent years, some authors cf. 1–7 studied fourth-order boundary value problems, but
these papers mainly dealt with the problems in real space or the problems without singularity.
As far as we know, the fourth-order ordinary differential equation with singularity has
been seldom studied in Banach spaces. Also, not so much is known about the case that the
nonlinear term has two different parameters or has two parts with different properties.
2
Abstract and Applied Analysis
In 7, Yao studied the following two-point boundary value problem:
u4 t − λhtfut 0,
1.2
u0 u1 u 0 u 1 0,
where λ > 0 and f ∈ C0, ∞. The author obtained positive solutions of BVP 1.2 in real
space not abstract space.
In 4, the authors studied the following boundary value problem in real space:
x4 t βx t λft, x,
1.3
x0 x1 x 0 x 1 0,
where f ∈ C0, 1 × 0, ∞, β < π 2 , and λ > 0. With the method of monotone iterative, the
existence and uniqueness of positive solution of BVP 1.3 are obtained.
Comparing with the results in 5–7, this paper has the following features. Firstly, we
discuss positive solutions of BVP 1.1 in abstract space, not E R as in 5, 7. Secondly, BVP
1.1 has two parameters which may have different domains. Fourthly, f and g may have
different properties, and the main tool we used is the fixed point theorem on cone. Thirdly,
we talk about both the existence and nonexistence of positive solutions, but 5, 9 only study
the existence of positive solutions.
Basic facts about ordered Banach space E can be found in 8. Here we recall some
of them. The cone P in E is said to be normal if there exists a positive constant N such that
θ ≤ x ≤ y implies x ≤ Ny. In this paper, we always suppose that P is normal in E,
and without loss of generality, suppose that the normal constant N 1. E∗ is dual space of
E. Denote P ∗ {ψ ∈ E∗ | ψx ≥ 0, for all x ∈ P }, then P ∗ is a dual cone of cone P . The
noncompactness of Kuratowski is denoted by α·.
We study BVP 1.1 in CJ, E. Evidently, CJ, E, · c is a Banach space under the
norm xc maxt∈J xt. x ∈ CJ, E is a positive solution of BVP1.1 if it satisfies BVP
1.1 and xt > θ, for all t ∈ J.
2. Preliminaries and Lemmas
In this paper, we make the following conditions.
H1 f, g ∈ CJ × P, P , ft, Pr {ft, u, u ∈ Pr } is relatively compact and gt, Pr is
also relatively compact, where Pr {u ∈ P : u ≤ r}. For any u ∈ P , there exist
ai t, bi t ∈ LJ, R , w1 x, w2 x ∈ CJ, R such that
ft, u ≤ a1 t b1 tw1 u,
gt, u ≤ a2 t b2 tw2 u,
a.e. t ∈ J.
2.1
H2 There exist a function m ∈ LJ, 0, ∞, such that ψft, u ≥ mtu or
ψgt, u ≥ mtu, for t ∈ J, u ∈ P , where ψ ∈ P ∗ , ψ 1.
Abstract and Applied Analysis
3
Lemma 2.1 see 8. Let H {x | x : J → E, x is strong measurable} be countable, and there
exists M ∈ LJ, R such that xt ≤ Mt, a.e. t ∈ J, x ∈ H, then αHt ∈ LJ, R and
α
xtdt : x ∈ H
≤ 2 αHtdt.
J
2.2
J
Lemma 2.2 see 8. Let P be a cone in Banach space E. For r > 0, define Pr {x ∈ P : x < r}.
Assume that T : P r → P is a completely continuous map such that x / T x for x ∈ ∂Pr .
i If x ≤ T x for x ∈ ∂Pr , then iT, Pr , P 0.
ii If x ≥ T x for x ∈ ∂Pr , then iT, Pr , P 1.
Lemma 2.3 see 4. For any ϕ ∈ CJ, P , the following fourth-order boundary value problem:
x4 t βx t ϕt,
2.3
x0 x1 x 0 x 1 θ
has the solution
xt 1 1
0
2.4
G1 t, τG2 τ, sϕsds dτ,
0
where G2 t, s is defined in Proposition 2.4 and
G1 t, s ⎧
⎨t1 − s, 0 ≤ t ≤ s ≤ 1,
2.5
⎩s1 − t, 0 ≤ s ≤ t ≤ 1.
Proposition 2.4. For all t, s ∈ 0, 1 × 0, 1 one has the following properties:
t1 − tG1 s, s ≤ G1 t, s ≤ G1 s, s s1 − s,
G1 t, s ≤ t1 − t;
G1 t, s ≤
1
.
4
2.6
Let ω β. G2 t, s is explicitly given by the following
If β < 0, then
G2 t, s ⎧
sinh ωt sinh ω1 − s
⎪
⎪
,
⎪
⎪
⎨
ω sinh ω
0 ≤ t ≤ s ≤ 1,
⎪
⎪
⎪
sinh ωs sinh ω1 − t
⎪
⎩
,
ω sinh ω
0 ≤ s ≤ t ≤ 1.
If β 0, then G2 t, s G1 t, s.
2.7
4
Abstract and Applied Analysis
If 0 < β < π 2 , then
⎧
sin ωt sin ω1 − s
⎪
⎪
, 0 ≤ t ≤ s ≤ 1,
⎪
⎨
ω sin ω
G2 t, s ⎪
⎪ sin ωs sin ω1 − t
⎪
⎩
0 ≤ s ≤ t ≤ 1.
ω sin ω
2.8
Proposition 2.5. For any t, s ∈ 0, 1, one has G1 t, s > 0, G2 t, s > 0, and G1 t, s ≥ δG1 u, s,
for t ∈ Jδ , s, u ∈ 0, 1, where δ ∈ 0, 1/2.
Proposition 2.6. It is easy to see that
⎧1
⎪
⎪
,
⎪
⎪
4
⎪
⎪
⎪
⎨ 2ω
e
G2 t, s ≤
,
⎪
⎪ 4
⎪
⎪
⎪
⎪
⎪
⎩ sin ω ,
ω
β 0,
2.9
β < 0,
β > 0.
Let d max{1/4, e2ω /4, sin ω/ω}, Q {x ∈ CJ, E : xt ≥ θ, t ∈ J}, and K {x ∈ Q : xt ≥ δxs, t ∈ Jδ , s ∈ 0, 1}. It is easy to know that K is a cone in CJ, E. Let
Kr {x ∈ K : x ≤ r} ⊂ K, K ⊂ Q. Define T as
T xt 1 1
0
G1 t, τG2 τ, s λfs, xs μgs, xs ds dτ,
2.10
0
then x is the solution of BVP1.1 if and only if x is the fixed point of T .
Lemma 2.7. Suppose that condition (H1 holds, then T : K → K is completely continuous.
Proof. By the continuity of G1 t, s and H1 , we have T x ∈ Q. And for any t ∈ Jδ , x ∈ K, by
Proposition 2.5, we get
T xt δT xu.
2.11
Next we prove that T is compact. Let V {xn }∞
n1 ⊂ K be any bounded set, and we
suppose that xn ≤ r, for some r > 0. Let Mr max0≤v≤r {wi v, i 1, 2}, then by condition
H1 ,
ft, xn t ≤ a1 t b1 tMr ,
gt, xn t ≤ a2 t b2 tMr ,
a.e. t ∈ J, xn ∈ V.
2.12
Abstract and Applied Analysis
5
We obtain from Lemma 2.1
α{T xn t : xn ∈ V }
1
G1 t, τG2 τ, s λfs, xn s μgs, xn s ds dτ : xn ∈ V
α
0
≤ 2d
1
2.13
α λfs, xn s μgs, xn s : xn ∈ V ds
0
0.
Hence, T V is relatively compact, so there exists a subsequence {xni } of {xn } such that {T xni }
converges to some v ∈ CJ, E. So T : K → K is completely continuous.
In this paper, for u ∈ P we denote
ψf
ν
ψ ft, u
lim inf min
,
u → ν t∈J
u
ψg
ν
ψ gt, u
lim inf min
,
u → ν t∈J
u
2.14
where ν denotes 0 or ∞, ψ ∈ P ∗ , ψ 1.
3. Main Results
Theorem 3.1. Assume that (H1 ) holds and the following conditions are satisfied:
ψf
∞
∞,
ψg
0
∞.
3.1
Then BVP1.1 has at least two positive solutions when λ and μ are sufficiently small.
Proof. T is completely continuous by Lemma 2.7. For any l > 0, set
Fl d
max
4 x∈K, xc l
1
fs, xsds,
0
Gl d
max
4 x∈K, xc l
1
gs, xsds.
3.2
0
Since ψf∞ ∞, there exists r1 > 0 such that Fr1 > 0. If Gr1 0, let λ1 r1 /Fr1 ; then
for any λ ∈ 0, λ1 and x ∈ K, xc r1 , we have
d
T xt ≤
4
≤
d
4
1
λfs, xs μgs, xsds
0
1
λfs, xsds d
4
0
≤ λ1 Fr1 r1 xc .
1
0
μgs, xsds
3.3
6
Abstract and Applied Analysis
0, let μ1 r1 /2Gr1 and λ1 r1 /2Fr1 . For λ ∈ 0, λ1 , μ ∈ 0, μ1 , and
If Gr1 /
x ∈ K, xc r1 , we get
1
λfs, xsds d μgs, xsds
4 0
0
1
1
d
d
≤ λ1 λfs, xsds μ1 gs, xsds
4
4
0
0
T xt ≤
d
4
1
≤ r1 Fr1 μ1 Gr1 ≤
3.4
r1 r1
2
2
r1 xc ,
which means that T xc < r1 xc , for x ∈ K, xc r1 , and Lemma 2.2 implies that
iT, Kr1 , K 1.
Since ψf∞ ∞, there exists r 2 > 0 such that ψft, u ≥ ε1 u, u ∈ P, t ∈ J, u ≥ r 2 ,
1−δ
where ε1 > 0 satisfies ε1 λδ δ G1 1/2, τG2 τ, sds dτ > 1. Let r2 max{2r1 , r 2 /δ}, then for
any t ∈ Jδ , x ∈ K, xc r2 , by the definition of K we get xt ≥ δxc ≥ r 2 , and we have
T x 1 ≥ ψ T x 1
2 2
1 1
1
, τ G2 τ, sψ fs, xs ds dτ
2
0 0
1 1−δ 1
, τ G2 τ, sψ fs, xs ds dτ
G1
≥λ
2
0 δ
1 1−δ 1
, τ G2 τ, sxsds dτ
G1
≥ λε1
2
0 δ
1 1−δ 1
, τ G2 τ, sds dτ
G1
≥ λε1 δxc
2
0 δ
≥λ
G1
3.5
> xc ,
which implies that T xc > r2 , x ∈ Kr2 . By Lemma 2.2 we have iT, Kr2 , K 0.
On the other hand, since ψg0 ∞, there exists r 3 : 0 < r 3 < r1 such that ψgt, u ≥
ε2 u, for t ∈ J, u ∈ P, u ≤ r 3 , where ε2 > 0 and satisfies
1 1−δ
ε2 μδ
G1
0
δ
1
, τ G2 τ, sds dτ > 1.
2
3.6
Abstract and Applied Analysis
7
Accordingly, 0 < r3 < r 3 . Then for x ∈ K, xc r3 , similar to 3.5 we get T xc > r3 .
By Lemma 2.2 we have iT, Kr3 , K 0. Hence, from the additivity of fixed point index, we
obtain
i T,
Kr2
K r1
−1,
,K
i T,
Kr1
K r3
,K
1.
3.7
Hence, T has two fixed points x1 ∈ Kr2 /K r1 and x2 ∈ Kr1 /K r3 . Therefore, BVP1.1 has at
least two fixed points x1 and x2 .
Corollary 3.2. If (H1 ), ψg∞ ∞, and ψf0 ∞ hold, then BVP1.1 has at least two positive
solutions when λ and μ are sufficiently small.
Theorem 3.3. Suppose that (H1 ) and (H2 ) hold, and if λ is sufficiently large, then BVP1.1 has no
positive solutions.
Proof. First by H2 we can suppose that ψft, u ≥ mtu, for t ∈ J. suppose
that BVP1.1 has a positive solution x, and choose λ sufficiently large satisfying
1−δ
λδ δ G1 1/2, τG2 τ, smsds dτ > 1. Then we have
1 ≥ψ x 1
x
xc ≥ 2 2
≥λ
1 1
0
≥λ
G1
0
1 1−δ
G1
0
≥ λδ
1
, τ G2 τ, sψ fs, xs ds dτ
2
δ
1 1−δ
G1
0
δ
1
, τ G2 τ, smsxsds dτ
2
3.8
1
, τ G2 τ, smsds dτxc
2
> xc ,
which is a contradiction; so BVP1.1 has no positive solutions. Similarly, if ψgt, u ≥
mtu, for t ∈ J, and μ is sufficiently large, then BVP1.1 has no positive
solutions.
Acknowledgments
The authors thank Professor Lishan Liu and Professor Bendong Lou for many useful
discussions and helpful suggestions. This work was partially supported by NSFC 10971155
and Innovation Program of Shanghai Municipal Education Commission 09ZZ33.
8
Abstract and Applied Analysis
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