Hindawi Publishing Corporation
Fixed Point Theory and Applications
Volume 2011, Article ID 604046, 11 pages
doi:10.1155/2011/604046
Research Article
Existence of Positive Solutions for
Nonlocal Fourth-Order Boundary Value Problem
with Variable Parameter
Xiaoling Han, Hongliang Gao, and Jia Xu
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to Xiaoling Han, [email protected]
Received 26 November 2010; Accepted 14 January 2011
Academic Editor: M. Furi
Copyright q 2011 Xiaoling Han et al. 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.
By using the Krasnoselskii’s fixed point theorem and operator spectral theorem, the existence of
positive solutions for the nonlocal fourth-order boundary value problem with variable parameter
u
4
tBtu
tλft, ut,u
t,0<t<1, u0u1
1
0
psusds, u
0u
t, u
t
,u
t
, 0 <t<1,
u
0
u
1
1
0
p
s
u
A1 λ>0and0<β<π
2
,
A2 f ∈ C0, 1×0, ∞×−∞, 0, 0, ∞, p, q ∈ L
1
0, 1, ps ≥ 0, qs ≥ 0,
1
0
psds < 1,
1
0
qs sin
βsds
1
0
qs sin
β1 −sds < sin
β.
In this paper, we study the above generalizing form with variable parameters BVP
u
4
t
1
0
p
s
u
s
ds,
u
0
u
1
1
0
q
s
0
qs sin
βsds
1
0
qs sin
β1 −sds < sin
β.
2. T he Preliminary Lemmas
Set λ
1
0, −π
2
<λ
2
−β<0and
δ
1
1 −
1
0
p
s
By H1, H2,wegetδ
i
/
0, i 1, 2. Denote by K
1
t, s the Green’s function of the problem
−u
t
λ
1
u
t
0, 0 <t<1,
u
0
u
1
1
0
u
1
1
0
q
s
u
s
ds.
2.3
Then, carefully calculation yield
K
1
t, s
G
1
t, s
ρ
1
0
G
2
s, x
q
x
dx,
G
1
t, s
⎧
⎨
⎩
t
1 −s
, 0 ≤ t ≤ s ≤ 1,
s
1 −t
β sin
β
, 0 ≤ t ≤ s ≤ 1,
sin
βs sin
β
1 −t
β sin
β
, 0 ≤ s ≤ t ≤ 1,
ρ
1
1
δ
1
,ρ
2
t
0
u
1
1
0
p
s
u
s
ds,
u
0
u
1
max
0≤t≤1
|ut|,foru ∈ Y .
X {u ∈ C
2
0, 1 : u0u1
1
0
psusds, u
0u
1
1
0
qsu
sds}, u
1
u
0
,
u
2
u
0
G
i
t, tG
i
s, s, G
i
t, s ≤ C
i
G
i
s, s for t, s ∈ 0, 1, i 1, 2,
where C
1
1, b
1
1; C
2
1/ sin
β, b
2
β sin
β.
Denote
d
i
min
t∈0,1
1
0
K
i
t, s
ds
i 1, 2
.
2.6
Computations yield the following results.
Lemma 2.4 see 3. D
1
i
max
t∈0,1
1
0
G
i
t, sds > 0 i 1, 2
i when λ
i
> 0, D
i max
t∈0,1
ρ
2
tρ
2
1/2,
ii 0 <d
i
< 1, 0 <ξ<1.
By Lemmas 2.4 and 2.5, D
2
max
t∈0,1
1
0
K
2
1/2,sds.
Take θ min{d
1
,d
2
ξ/C
2
},byLemma 2.5,0<θ<1.
Define
Th
Ah
t
Th
t
−
1
0
K
2
t, τ
h
τ
dτ, t ∈
0, 1
.
ii Ax≥x,x ∈ P ∩ ∂Ω
1
and Ax≤x, x ∈ P ∩ ∂Ω
2
holds. Then, A has a fixed point in P ∩ Ω
2
\ Ω
1
.
3. The Main Results
Suppose that K
1
, K
2
, G
2
, ρ
2
, C
2
, θ,andD
2
,aredefinedasinSection 2,weintroducesome
notations as follows:
A
1
0
1
G
2
s, x
q
x
dx
ds,
K sup
t∈0,1
B
t
− β
,L D
2
K, η
0
1 −L
A C
2
B
v
|
,f
0
lim inf
|u||v|→0
min
t∈1/4,3/4
f
t, u, v
|
u
|
|
v
|
,
f
∞
lim sup
|u||v|→∞
max
t∈0,1
f
t, u, v
i
f
0
< 1/λη
0
, f
∞
> 1/λη
1
,
ii f
0
> 1/λη
1
, f
∞
< 1/λη
0
.
Proof. For any h ∈ Y , consider the following BVP:
u
4
t
B
t
u
u
0
u
1
1
0
q
s
u
s
ds.
3.2
6 Fixed Point Theory and Applications
It is easy to see that the above question is equivalent to the following question:
u
4
1
1
0
p
s
u
s
ds,
u
0
u
1
1
0
3.3 if and only if u ∈ X satisfies u TGu h,thatis,
u ∈ X,
I − TG
u Th. 3.4
Owing to G : X → Y and T : Y → X, the operator I −TG maps X into X.FromT≤D
2
by
Lemma 2.6 together with G≤K and condition L<1, applying operator spectral theorem,
we have that the I −TG
−1
exists and is bounded. Let H I −TG
−1
T,then3.4 is equivalent
to u Hh. By the Neumann expansion formula, H can be expressed by
H
I TG ···
TG
n
···
T T
TG
T ···
, 3.6
and so TGThtTGTht ≥ 0, t ∈ 0, 1.
Assume that for all h ∈ Y
, TG
k
Tht ≥ 0, t ∈ 0, 1,leth
1
GTh,by3.6 we have
h
1
∈ Y
,andsoTG
k1
ThtTG
k
TGThtTG
k
Th
1
t ≥ 0, t ∈ 0 , 1.Thusby
induction, it follows that TG
n
Tht ≥ 0, for all n ≥ 1, h ∈ Y
, t ∈ 0, 1.By3.5,forall
h ∈ Y
,wehave
Th
t
,t∈
0, 1
,
Hh
t
Ah
t
AG
Th
t
0, 1
,
3.7
and so H : Y
→ Y
∩ X.
Fixed Point Theory and Applications 7
On the other hand, for all h ∈ Y
,wehave
Hh
t
≤
Th
t
|
TG
|
Th
t
t ∈
0, 1
,
3.8
Hh
t
≤
|
Ah
t
|
|
AG
Th
|
Ah
t
|
··· L
n
|
Ah
t
|
···
1 L ··· L
n
···
|
Ah
t
|
1
1 − L
Th
Th
0
,
Hh
1
≥
Th
1
,
Hh
1
≤
1
1 −L
Th
1
.
3.10
For any u ∈ Y
u
t
≥
1 −L
d
1
u
0
, max
1/4≤t≤3/4
u
t
≤−
1 −L
d
2
ξ
C
2
Th
0
, max
1/4≤t≤3/4
Th
t
≤−
d
2
ξ
C
2
Th
0
.
3.13
8 Fixed Point Theory and Applications
By 3.7 and 3.10,
,
max
1/4≤t≤3/4
Qu
t
≤ max
1/4≤t≤3/4
TFu
t
≤−
d
2
ξ
C
2
TFu
, by the definition of f
0
,thereexistsr
1
> 0suchthat
max
0≤t≤1,
|
ut
|
|
u
t
|
≤r
1
f
t, u
t
,u
t
1
λ
η
0
,u∈ ∂Ω
r
1
,t∈
0, 1
.
3.16
So, by 3.10,weget
Qu
0
HFu
0
≤
1
1 −L
TFu
0
,u
τ
dτ ds
0
≤
r
1
η
0
1 −L
1
0
1
0
K
1
s, s
1
1 −L
1
0
G
2
τ, τ
ρ
2
1
2
1
0
G
2
τ, x
q
x
dx
2
HFu
2
≤
1
1 −L
TFu
2
≤
A BC
2
η
0
r
1
1 −L
r
1
u
|
u
t
|
|
u
t
|
≥
1
λ
η
1
.
3.19
Choose r
2
> 1/θr
2
,letΩ
r
2
{u ∈ P : u
2
2
λ
η
1
,u∈ ∂Ω
r
2
,t∈
1
4
,
3
4
.
TFu
1
2
K
2
1
2
,τ
f
τ, u
τ
,u
τ
dτ ≥ η
1
θr
2
3/4
1/4
K
2
1
1
2
≥ r
2
u
2
. 3.21
By the use of the Krasnoselskii’s fixed point theorem, we know there exists u
0
∈ Ω
2
\Ω
1
such
that Qu
0
u
0
,namely,u
0
is a solution of 1.2 and satisfied u
0
1
,fort ∈ 1/4, 3/4, |u| |v|≥θR
0
.
Proof. By the proof of Theorem 3.1, we know that 1 from the condition
f
0
< 1/λη
0
,there
exists Ω
r
1
{u ∈ P : u
2
<r
1
},suchthatQu
2
≤u
2
, u ∈ ∂Ω
r
1
, 2 from the condition
f
∞
< 1/λη
0
,thereexistsΩ
1
,suchthat
Qu
2
≥u
2
, u ∈ ∂Ω
r
3
. By the use of Krasnoselskii’s fixed point theorem, it is easy to know
that 1.2 has at least two positive solutions.
Corollary 3.3. Assume (H1), (H2) hold, and L<1.Thenproblem1.2 has at least two positive
solution, if f satisfy
i f
0
> 1/λη
1
, f
∞
> 1/λη
1
,
ii There exists R
0
> 0 such that ft, u, v ≤ θR
0
/λη
0
,fort ∈ 0, 1, |u| |v|≤R
0
− u
t
− 17.9sin
u
t
− u
t
, 0 <t<1,
u
0
u
1
1
2
√
2, β π
2
/4, K 1, D
2
4
√
2 −1/π
2
.Furthermore,weobtain
A 48 − 13π
2
/π
3
, B 2/π
2
,thenη
0
1 −Lπ
3
/48 − 11π, η
1
4π
2
/
√
2cosπ/8 − 1, so
f
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