Volume 2009, Article ID 491952,20pages doi:10.1155/2009/491952
Research Article
Existence of Solutions for Fourth-Order Four-Point
Boundary Value Problem on Time Scales
Dandan Yang,
1Gang Li,
1and Chuanzhi Bai
21School of Mathematical Science, Yangzhou University, Yangzhou 225002, China
2Department of Mathematics, Huaiyin Normal University, Huaian, Jiangsu 223300, China
Correspondence should be addressed to Chuanzhi Bai,czbai8@sohu.com
Received 11 April 2009; Revised 8 July 2009; Accepted 28 July 2009
Recommended by Irena Rach ˚unkov´a
We present an existence result for fourth-order four-point boundary value problem on time scales. Our analysis is based on a fixed point theorem due to Krasnoselskii and Zabreiko.
Copyrightq2009 Dandan Yang 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.
1. Introduction
Very recently, Karaca1investigated the following fourth-order four-point boundary value problem on time scales:
yΔ4t−qtyΔ2σt ft, yσt, yΔ2t,
yσ4b0, αya−βyΔa 0,
γyΔ2ξ1−δyΔ
3
ξ1 0, ζyΔ
2
ξ2 ηyΔ
3
ξ2 0,
1.1
fort∈a, b⊂T, a≤ξ1≤ξ2≤σb,andf ∈Ca, b×R×R×R.And the author made the following assumptions:
A1α, β, γ, δ, ζ, η≥0,anda≤ξ1≤ξ2≤σb,
A2qt≥0.Ifqt≡0,thenγζ >0.
Lemma 1.1 see 1, Lemma 2.5. Assume that conditions (A1) and (A2) are satisfied. If h ∈
Ca, b,then the boundary value problem
yΔ4t−qtyΔ2σt ht, t∈a, b,
yσ4b0, αya−βyΔa 0,
γyΔ2ξ1−δyΔ
3
ξ1 0, ζyΔ
2
ξ2 ηyΔ
3 ξ2 0
1.2
has a unique solution
yt
σ4b
a
G1t, ξ ξ2
ξ1
G2t, shsΔsΔξ, 1.3
where
G1t, s 1
d
⎧ ⎨ ⎩
σ4b−σs αt−a β , t≤s,
σ4b−t ασs−a β , t≥σs, 1.4
G2t, s 1
D
⎧ ⎨ ⎩
ψσsϕt, t≤s,
ψtϕσs, t≥σs.
1.5
HereD ζφξ1−ηψΔξ1 δϕξ2 γϕξ2, d βασ4b−a, andϕt, ψtare given as follows:
ϕt ηζt−ξ1 t
ξ1
τ
ξ1
qsϕσsΔsΔτ,
ψt δγξ2−t ξ2
t
ξ2
τ
qsψσsΔsΔτ.
1.6
α1, β0, ft, y, yΔ2≡ft, y,then1.1reduces to the following boundary value problem:
yt ft, yt , 0< t <1, y0 y1 0,
γyξ1−δyξ1 0, ζyξ2 ηyξ2 0.
1.7
The counterexample is given by [2], from which one can see clearly that [1, Lemma 2.5] is wrong. If one takesqt q, hereq >0is a constant, then1.1reduces to the following fourth-order four-point boundary value problem on time scales:
yΔ4t−qyΔ2σt ft, yσt, yΔ2t, t∈a, b⊂T,
yσ4b0, αya−βyΔa 0,
γyΔ2ξ1−δyΔ
3
ξ1 0, ζyΔ
2
ξ2 ηyΔ
3
ξ2 0.
1.8
The purpose of this paper is to establish some existence criteria of solution for BVP
1.8which is a special case of1.1. The paper is organized as follows. InSection 2, some basic time-scale definitions are presented and several preliminary results are given. InSection 3, by employing a fixed point theorem due to Krasnoselskii and Zabreiko, we establish existence of solutions criteria for BVP1.8.Section 4is devoted to an example illustrating our main result.
2. Preliminaries
The study of dynamic equations on time scales goes back to its founder Hilger 3 and it is a new area of still fairly theoretical exploration in mathematics. In the recent years boundary value problem on time scales has received considerable attention4–6. And an increasing interest in studying the existence of solutions to dynamic equations on time scales is observed, for example, see7–16.
For convenience, we first recall some definitions and calculus on time scales, so that the paper is self-contained. For the further details concerning the time scales, please see17–19
which are excellent works for the calculus of time scales.
A time scaleTis an arbitrary nonempty closed subset of real numbersR. The operators
σandρfromTtoT
σt inf{τ∈T:τ > t}, ρt sup{τ∈T:τ < t} 2.1
For allt ∈T,we assume throughout thatThas the topology that it inherits from the standard topology onR.The notationsa, b,a, b,and so on, will denote time-scale intervals
a, b {t∈T:a≤t≤b}, 2.2
wherea, b∈Twitha < ρb.
Definition 2.1. Fixt∈T.Lety :T → R.Then we defineyΔtto be the numberif it exists with the property that givenε >0 there is a neighborhoodUoftwith
yσt−ys−yΔtσt−s< ε|σt−s| ∀s∈U. 2.3
ThenyΔis called derivative ofyt.
Definition 2.2. IfFΔt ftthen we define the integral by t
a
fτΔτFt−Fa. 2.4
We say that a functionp:T → Rnis regressive provided
1μtpt/0, t∈T, 2.5
whereμt σt−t,which is called graininess function. Ifpis a regressive function, then the generalized exponential functionepis defined by
ept, s exp t
s
ξμτpτ Δτ
, 2.6
fors, t∈T, ξhzis the cylinder transformation, which is defined by
ξhz
⎧ ⎪ ⎨ ⎪ ⎩
log1hz
h , h /0,
z, h0.
2.7
Letp, qbe two regressive functions, then define
p⊕qpqμpq, q− q
1μq, pqp⊕
q p−q
1μq. 2.8
Lemma 2.3see18. Assume thatp, qare two regressive functions, then
ie0t, s≡1andept, t≡1;
iiepσt, s 1μtptept, s;
iiiept, seps, r ept, r;
iv1/ept, s ept, s;
vept, s 1/eps, t eps, t;
viept, seqt, s ep⊕qt, s;
viiept, s/eqt, s epqt, s.
The following well-known fixed point theorem will play a very important role in proving our main result.
Theorem 2.4see20. LetX be a Banach space, and letF :X → Xbe completely continuous. Assume thatA:X → Xis a bounded linear operator such that1is not an eigenvalue ofAand
lim
x → ∞
Fx−Ax
x 0. 2.9
ThenFhas a fixed point inX.
Throughout this paper, letEC2a, bbe endowed with the norm by
y0maxy,yΔ2, 2.10
whereymaxt∈a,b|yt|.And we make the following assumptions:
H1α, β, γ, δ, ζ, η≥0,anda≤ξ1≤ξ2≤σb,
H2q >0,andr1√q, r2 −√q,
H3dβασ4b−a>0.
Set
For convenience, we denote t
a
ep1s, aΔslt, a,
A
γ−δp2ξ1 ep2ξ1, a
ξ1
a
ep1s, aΔs−δp2σξ1ep2σξ1, aep1ξ1, a
×ζηp2ξ2 ep2ξ2, a−
γ−δp2ξ1 ep2ξ1, a
×
ζηp2ξ2 ep2ξ2, a
ξ2
a
ep1s, aΔsηp2σξ2ep2σξ2, aep1ξ2, a
,
A11δep2σξ1, aep1ξ1, a
ζηp2ξ2 ep2ξ2, a
γ−δp2ξ1 ep2ξ1, aηep2σξ2, aep1ξ2, a,
A12
γ−δp2ξ1 ep2ξ1, a
ζηp2ξ2 ep2ξ2, a,
A13
γ−δp2ξ1 ep2ξ1, aηep2σξ2, aep1ξ2, a,
B11
ζηp2ξ2 ep2ξ2, a ξ2
a
ep1s, aΔsηp2σξ2ep2σξ2, aep1ξ2, a
×δp2ξ1−γ ep2ξ1, a
γ−δp2ξ1 ep2ξ1, a ξ1
a
ep1s, aΔs−δp2σξ1ep2σξ1, aep1ξ1, a
×ηp2ξ2 ζ ep2ξ2, a,
B12
ζηp2ξ2 ep2ξ2, a ξ2
a
ep1s, aΔsηp2σξ2ep2σξ2, aep1ξ2, a
×δep2σξ1, aep1ξ1, a
γ−δp2ξ1 ep2ξ1, a ξ1
a
ep1s, aΔs−δp2σξ1ep2σξ1, aep1ξ1, a
×ηep2σξ2, aep1ξ2, a,
B13
γ−δp2ξ1 ep2ξ1, a
ξ1
a
ep1s, aΔs−δp2σξ1ep2σξ1, aep1ξ1, a
×ηp2ξ2 ζ ep2ξ2, a,
B14
γ−δp2ξ1 ep2ξ1, a
ξ1
a
ep1s, aΔs−δp2σξ1ep2σξ1, aep1ξ1, a
×ηep2σξ2, aep1ξ2, a.
2.12
Lemma 2.5. Assume thatH1andH2are satisfied. Ifh∈Ca, b,thenu∈C2a, bis a solution of the following boundary value problem (BVP):
yΔ2t−qyσt ht, t∈a, b, γyξ1−δyΔξ1 0, ζyξ2 ηyΔξ2 0,
2.13
if and only if
yt
σb
a
Gt, shsΔs, t∈a, b, 2.14
where the Green’s function of 2.13is as follows:
Gt, s ep2t, a
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
−B11
A
ξ1
s
ep1ξ, ae−r1s, aΔξ
A12lt, a−B13
A
ξ2
ξ1
ep1ξ, ae−r1s, aΔξ
A11lt, a−B12
A
e−r1s, a
t
s
ep1ξ, ae−r1s, aΔξ, a≤σs≤min{t, ξ1},
−B11
A
ξ1
s
ep1ξ, ae−r1s, aΔξ
A12lt, a−B13
A
ξ2
ξ1
ep1ξ, ae−r1s, aΔξ
A11lt, a−B12
A
e−r1s, a, a≤t≤s≤ξ1,
A12lt, a−B13
A
ξ2
s
ep1ξ, ae−r1s, aΔξ
−B14
A e−r1s, a A13
A lt, ae−r2s, a
t
s
ep1ξ, ae−r1s, aΔξ, ξ1≤σs≤min{t, ξ2},
A12lt, a−B13
A
ξ2
s
ep1ξ, ae−r1s, aΔξ
−B14
A e−r1s, a A13
A lt, ae−r2s, a, max{ξ1, t} ≤s≤ξ2,
t
s
ep1ξ, ae−r1s, aΔξ, ξ2≤σs≤t≤b,
0, max{t, ξ2} ≤s≤b,
2.15
Proof. Ify∈C2a, bis a solution of2.13, setting
us yΔs−r2yσs, t∈a, b, 2.16
then it follows from the first equation of2.13that
uΔs−r1uσs hs, t∈a, b. 2.17
Multiplying2.17bye−r1s, aand integrating fromatot,we get
ut e−r1t, a
ua
t
a
e−r1s, ahsΔs
, t∈a, b. 2.18
Similarly, by2.18, we have
yt e−r2t, a
ya
t
a
e−r2s, ausΔs
, t∈a, b. 2.19
Then substituting2.18into2.19, we get for eacht∈a, bthat
yt ep2t, aya ep2t, aua
t
a
ep1s, aΔs
ep2t, a
t
a
ep1s, a
s
a
e−r1ξ, ahξΔξΔs.
2.20
Substituting this expression for yt into the boundary conditions of 2.13. By some calculations, we get
ua 1
A
A11 ξ1
a
e−r1s, ahsΔsA12
ξ2
ξ1
ep1s, a
s
a
e−r1ξ, ahξΔξΔs
A13 ξ2
ξ1
e−r2s, ahsΔs
,
ya −1 A
B11 ξ1
a
ep1s, a
s
a
e−r1ξ, ahξΔξΔsB12 ξ1
a
e−r1s, ahsΔs
B13 ξ2
ξ1
ep1s, a
s
a
e−r1ξ, ahξΔξΔsB14 ξ2
ξ1
e−r1s, ahsΔs
.
Then substituting2.21into2.20, we get
yt −ep2t, a A
B11 ξ1
a
ep1s, a
s
a
e−r1ξ, ahξΔξΔsB12
ξ1
a
e−r1s, ahsΔs
B13 ξ2
ξ1
ep1s, a
s
a
e−r1ξ, ahξΔξΔsB14
ξ2
ξ1
e−r1s, ahsΔs
ep2t, a A
t
a
ep1s, aΔs
A11 ξ1
a
e−r1s, ahsΔsA12
ξ2
ξ1
ep1s, a
s
a
e−r1ξ, ahξΔξΔs
A13 ξ2
ξ1
e−r2s, ahsΔs
ep2t, a
t
a
ep1s, a
s
a
e−r1ξ, ahξΔξΔs.
2.22
By interchanging the order of integration and some rearrangement of2.22, we obtain
yt ep2t, a
×
ξ1
a
−B11
A
ξ1
s
ep1ξ, ae−r1s, aΔξ
A12lt, a−B13
A
ξ2
ξ1
ep1ξ, ae−r1s, aΔξ
e−r1s, a
A11lt, a−B12
A
hsΔs
ξ2
ξ1
A
12lt, a−B13
A
ξ2
s
ep1ξ, ae−r1s, aΔξ
−B14
A e−r1s, a
A13
A lt, ae−r2s, a
hsΔs
t
a
t
s
ep1ξ, ae−r1s, aΔξ
hsΔs
.
2.23
Thus, we obtain2.14consequently.
On the other hand, ifysatisfies2.14, then direct differentiation of2.14yields
yΔ2t−qyΔσt ht, t∈a, b. 2.24
Corollary 2.6. IfTR,then BVP2.13reduces to the following problem:
yt−qyt ht, t∈a, b,
γyξ1−δyξ1 0, ζyξ2 ηyξ2 0.
2.25
FromLemma 2.5, BVP2.25has a unique solution
yt
b
a
Gt, shsds, 2.26
where the Green’s function of 2.25is as follows:
Gt, s 1
r1−r2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 Δ1
er1t−aM
1s−er2t−aM2s er1t−s−er2t−s, a≤s≤min{t, ξ1}, 1
Δ1
er1t−aM
1s−er2t−aM2s , a≤t≤s≤ξ1,
1
Δ1
er1t−aM3s−er2t−aM4s er1t−s−er2t−s, ξ1≤s≤min{t, ξ2},
1
Δ1
er1t−aM
3s−er2t−aM4s , max{ξ1, t} ≤s≤ξ2,
er1t−s−er2t−s, ξ
2≤s≤t≤b,
0, max{t, ξ2} ≤s≤b,
2.27
where
Δ1
γ−δr2 ζηr1 er1ξ2−ar2ξ1−a−
γ−δr1 ζηr2 er1ξ1−ar2ξ2−a, 2.28
M1s
δr2−γ ηr1ζ er1ξ2−sr2ξ1−a−
δr1−γ ηr2ζ er2ξ2−ar1ξ1−s,
M2s
γ−δr1 ηr2ζ er2ξ2−sr1ξ1−a
δr2−γ ηr1ζ er1ξ2−ar2ξ1−s,
M3s
γ−δr2 ηr2ζ er2ξ2−sr2ξ1−a−
γ−δr2 ηr1ζ er2ξ1−ar1ξ2−s,
M4s
γ−δr1 ηr2ζ er2ξ2−sr1ξ1−a−
γ−δr1 ηr1ζ er1ξ1−ar1ξ2−s.
Proof. Ify∈C2a, bis a solution of2.25, takeTR,thenp
1 r1−r2, p2r2.Hence, from
2.20we have
yt ep2t, aya ep2t, aua
t
a
ep1s, aΔs
ep2t, a
t
a
ep1s, a
s
a
e−r1ξ, ahξΔξΔs
er2t−aya er2t−aua
t
a
er1−r2s−ads
er2t−a
t
a
s
a
er1−r2s−ae−r1ξ−ahξdξ ds
er2t−aya ua r1−r2
er1t−a−er2t−a
1
r1−r2 t
a
er1t−s−er2t−shsds.
2.30
Substituting this expression for yt into the boundary conditions of 2.25. By some calculations, we obtain
ua 1
Δ1 ξ1
a
δr2−γ ηr1ζ er1ξ2−sr2ξ1−a−
δr1−γ ηr2ζ er2ξ2−ar1ξ1−s
hsds
ξ2
ξ1
γ−δr2 ηr2ζ er2ξ2−sr2ξ1−a−
γ−δr2 ηr1ζ er2ξ1−ar1ξ2−s
hsds
,
ya − 1
r1−r2Δ1
×
ξ1
a
γ−δr1 ηr2ζ er1ξ1−ar2ξ2−s
ζηr1 δr2−γ er1ξ2−ar2ξ1−s
γ−δr2 ηr1ζ er1ξ2−sr2ξ1−a
δr1−γ ηr2ζ er2ξ2−ar1ξ1−s
hsds
ξ2
ξ1
γ−δr1 ηr2ζ er2ξ2−sr1ξ1−a−
γ−δr1 ηr1ζ er1ξ1−ar1ξ2−s
−γ−r2δ ηr2ζ er2ξ1−ar2ξ2−s
γ−r2δ ηr1ζ er2ξ1−ar1ξ2−s
hsds
,
whereΔ1is given as2.28. Then substituting2.31into2.30, we get
yt 1
r1−r2 ξ1
a
er1t−a Δ1
M1s−
er2t−a Δ1
M2s
hsds
ξ2
ξ1
er1t−a
Δ1 M3s−
er2t−a Δ1 M4s
hsds
t
a
er1t−s−er2t−s
hsds
,
2.32
whereMis i {1,2,3,4}are as in2.29, respectively. By some rearrangement of2.32, we obtain2.26consequently.
From the proof ofCorollary 2.6, ifTR,takeγζ1, δη0, aξ10, bξ21, we get the following result.
Corollary 2.7. The following boundary value problem:
−yt qyt ht, t∈0,1,
y0 y1 0
2.33
has a unique solution
yt
1
0
Gt, shsds, 2.34
where the Green’s function of 2.33is as follows:
Gt, s −1 r1−r2
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
1
Δ1
er1tM
3s−er2tM4s er1t−s−er2t−s, 0≤t≤s≤1, 1
Δ1
er1tM
3s−er2tM4s , 0≤s≤t≤1,
2.35
where
Δ1er1−er2, M3s M4s er21−s−er11−s. 2.36
After some rearrangement of 2.35, one obtains
Gt, s
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
sinhr1tsinhr11−s
r1sinhr1
, 0≤t≤s≤1,
sinhr1ssinhr11−t
r1sinhr1
, 0≤s≤t≤1.
Remark 2.8. Green function2.37associated with BVP2.33which is a special case of2.13
is coincident with part of21, Lemma 1.
Lemma 2.9. Assume that conditions (H1)–(H3) are satisfied. Ifh ∈ Ca, b,then boundary value problem
yΔ4t−qyΔ2σt ht, t∈a, b,
yσ4b0, αya−βyΔa 0,
γyΔ2ξ1−δyΔ
3
ξ1 0, ζyΔ
2
ξ2 ηyΔ
3 ξ2 0
2.38
has a unique solution
yt
σ4b
a
G1t, ξ σb
a
Gξ, shsΔsΔξ, 2.39
where
G1t, s 1
d
⎧ ⎨ ⎩
σ4b−σs αt−a β , t≤s,
σ4b−t ασs−a β , t≥σs, 2.40
andGt, sis given inLemma 2.5.
Proof. Consider the following boundary value problem:
yΔ2t−qyΔσt
σb
a
Gt, shsΔs, t∈a, σ2b,
yσ4b0, αya−βyΔa 0.
2.41
The Green’s function associated with the BVP2.41isG1t, s. This completes the proof. Remark 2.10. In1, Lemma 2.5, the solution of1.2is defined as
yt
σ4b
a
G1t, ξ ξ2
ξ1
whereG1t, sandG2t, sare given as1.4and1.5, respectively. In fact,ytis incorrect. Thus, we give the right form ofytas the special caseqt qin ourLemma 2.9.
3. Main Results
Theorem 3.1. Assume (H1)–(H3) are satisfied. Moreover, suppose that the following condition is satisfied:
H4ft, yσt, yΔ
2
t mtgu nthv,whereg, h : R → Rare continuous, mt, nt∈Ca, b,with
lim
u→ ∞ gu
u λ, ulim→ ∞ hv
v μ, 3.1
and there exists a continuous nonnegative functionw : a, b → R such that|ms||ns| ≤ ws, s∈a, b.If
max|λ|,μ<min
1
L1
, 1 L2
, 3.2
where
L1max
a≤t≤b
σ4b
a
G1t, ξ σb
a
|Gξ, s|wsΔsΔξ
,
L2max
a≤t≤b
σb
a
|Gt, s|wsΔs,
3.3
then BVP1.8has a solutiony∈C2a, b.
Proof. Define an operatorF :C2a, b → C2a, bby
Fyt
σ4b
a
G1t, ξ σb
a
Gξ, smsgys nshyΔ2sΔsΔξ, 3.4
Step 1F : C2a, b → C2a, bis continuous. Let{yn}∞
n1 be a sequence such thatyn →
yn → ∞,then we have
Fyn t−
Fy t
σ4b
a
G1t, ξ σb
a
Gξ, smsgyns nshyΔn2sΔsΔξ
−
σ4b
a
G1t, ξ σb
a
Gξ, smsgys nshyΔ2sΔsΔξ
σ4b
a
G1t, ξ σb
a
Gξ, smsgyns −gys nshynΔ2s−hyΔ2sΔsΔξ
≤
σ4b
a
|G1t, ξ| σb
a
|Gξ, sms|gyns −g
ys Δξ
σ4b
a
|G1t, ξ| σb
a
|Gξ, sms|hynΔ2s−hyΔ2sΔξ,
Fyn Δ
2
t−Fy Δ2t
σb
a
Gξ, smsgyns nshyΔn2sΔs
−
σb
a
Gξ, smsgys nshyΔ2sΔs
σb
a
Gξ, smsgyns −gys nshyΔn2s−hyΔ2sΔs
≤
σb
a
|Gξ, sms|gyns −g
ys Δs
σb
a
|Gξ, sms|hynΔ2s−hyΔ2sΔs. 3.5
Sinceg, hare continuous, we have |Fynt−Fyt| → 0,which yieldsFyn−Fy →
Step 2Fmaps bounded sets into bounded sets inC2a, b. LetΩ⊂ C2a, bbe a bounded set. Then, fort∈a, band anyy∈Ω,we have
Fyt
σ4b
a
G1t, ξ σb
a
Gξ, smsgys nshyΔ2sΔsΔξ
≤
σ4b
a
|G1t, ξ| σb
a
|Gξ, s|msgys nshyΔ2sΔsΔξ.
3.6
By virtue of the continuity ofg and h, we conclude that Fuis bounded uniformly, and so
FΩis a bounded set.
Step 3F maps bounded sets into equicontinuous sets ofC2a, b. Lett
1, t2 ∈a, b, y ∈Ω, then
Fy t1−
Fy t2
σ4b
a
G1t1, ξ−G1t2, ξ σb
a
Gξ, smsgys nshyΔ2sΔsΔξ
≤
σ4b
a
|G1t1, ξ−G1t2, ξ| σb
a
Gξ, smsgys nshyΔ2sΔsΔξ.
3.7
The right hand side tends to uniformly zero ast1−t2 → 0.Consequently, Steps1–3together with the Arzela-Ascoli theorem show thatFis completely continuous.
Now we consider the following boundary value problem:
yΔ4t−qyΔ2σt λmtyt μntyΔ2t, t∈a, b,
yσ4b0, αya−βyΔa 0,
γyΔ2ξ1−δyΔ
3
ξ1 0, ζyΔ
2
ξ2 ηyΔ
3
ξ2 0.
3.8
Define
Ayt
σ4b
a
G1t, ξ σb
a
Gξ, sλmsys μnsyΔ2sΔsΔξ. 3.9
Obviously,Ais a completely continuous bounded linear operator. Moreover, the fixed point ofAis a solution of the BVP3.8and conversely.
We are now in the position to claim that 1 is not an eigenvalue ofA.
Ifλ /0 orμ /0,suppose that the BVP3.8has a nontrivial solutionyandy0 >0, then we have
Ayt≤σ4b a
G1t, ξ σb
a
Gξ, sλmsys μnsyΔ2sΔsΔξ
≤
σ4b
a
G1t, ξ σb
a
|Gξ, s||λ||ms|ysμ|ns|yΔ2sΔsΔξ
≤max
a≤t≤b
σ4b
a
G1t, ξ σb
a
|Gξ, s||λ||ms|μ|ns|y0ΔsΔξ
≤max|λ|,μmax
a≤t≤b
σ4b
a
G1t, ξ σb
a
|Gξ, s|wsΔsΔξ
y0,
3.10
which yields
Ayt≤max
|λ|,μL1y0y0. 3.11
On the other hand, we have
Ay Δ2t≤
σb
a
Gt, sλmsys μnsyΔ2sΔs
≤max
a≤t≤b
σb
a
|Gt, s||λ||ms|μ|ns|y0Δs
≤max|λ|,μmax
a≤t≤b
σb
a
|Gt, s|wsΔsy0
≤max|λ|,μL2y0< 1
L2
L2y0y0.
3.12
From the above discussion 3.11 and 3.12, we have Ay0 < y0. This contradiction implies that boundary value problem 3.8 has no trivial solution. Hence, 1 is not an eigenvalue ofA.
At last, we show that
lim
x0→ ∞
Fx−Ax0
x0 0. 3.13
By limu→ ∞gu/u λ,limu→ ∞hv/v μ,then for anyε > 0,there exist aR > 0 such
that
gu−λu< ε|u|, hv−μv< ε|v|, |u|,|v|> R. 3.14
Denote
E1{t∈a, b:|ut| ≤R,|vt|> R}, E2{t∈a, b:|ut| ≤R,|vt|> R},
E3{t∈a, b:|ut|,|vt| ≤R}, E4{t∈a, b:|ut|,|vt|> R}.
3.15
Thus for anyy∈Eandy0> M,whent∈E1,it follows that
gut−λut≤gut|λ||ut| ≤R∗|λ|R < εM < εu0,
hvt−μvt< ε|vt| ≤εv0. 3.16
In a similar way, we also conclude that for anyt∈Ei, i2,3,4,
gut−λut< εu0, hvt−μvt< εv0. 3.17
Therefore, Fyt−Ayt
σ4b
a
G1t, ξ σb
a
Gξ, smsgys −λysnshyΔ2s−μyΔ2sΔsΔξ
≤max
a≤t≤b
σ4b
a
G1t, ξ σb
a
|Gξ, s|wsΔsΔξ
εy0εL1y0.
3.18
On the other hand, we get
Fy−Ay Δ2t≤
σb
a
Gt, smsgys −λysnshyΔ2s−μyΔ2sΔs
≤max
a≤t≤b
σb
a
|Gt, s|wsΔs
εy0
εL2y0.
3.19
Combining3.18with3.19, we have
lim
x0→ ∞
Fx−Ax0
x0 0. 3.20
Theorem 2.4 guarantees that boundary value problem 1.8 has a solution y∗ ∈ C2a, b.It is obvious thaty∗/0 whenmt
if mt0g0 nt0h0/0,then 0Δ4 −q0Δ
2
mt0g0 nt0h0/0 will lead to a contradiction, which completes the proof.
4. Application
We give an example to illustrate our result.
Example 4.1. Consider the fourth-order four-pint boundary value problem
y4t− 1
4y
t tsin 2πt t21 yt−
1 2te
costcosyt, 0< t <1,
y0 y1 0,
y
1 3
−y
1 3
0, y
2 3
y
2 3
0.
4.1
Notice thatTR.To show that4.1has at least one nontrivial solution we applyTheorem 3.1
withmt tsin 2πt/t21, nt 1/2tecost, gu u, hu cosu, α γ δ η ζ
1, β0, q1/4, ξ1 1/3,andξ22/3.Obviously,H1–H3are satisfied. And
mt0g0 nt0h0 1 2t0e
cost0/0, t
0∈0,1. 4.2
Since|ms||ns| ≤1/2e1s:ws,for eachs∈0,1,we have the following. By simple calculation we have
L1max 0≤t≤1
1
0
G1t, ξ 1
0
|Gξ, s|wsds dξ
≈0.05<1,
L2max 0≤t≤1
1
0
|Gt, s|wsds≈0.82<1.
4.3
On the other hand, we notice that
λ lim
u→ ∞ gu
u 1, μulim→ ∞ hu
u 0. 4.4
Hence,
maxλ, μ<1<min
1
L1
, 1 L2
. 4.5
Acknowledgments
The authors would like to thank the referees for their valuable suggestions and comments. This work is supported by the National Natural Science Foundation of China10771212and the Natural Science Foundation of Jiangsu Education Office06KJB110010.
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