doi:10.1155/2010/971540
Research Article
Existence and Uniqueness of Positive Solutions for
Discrete Fourth-Order Lidstone Problem with
a Parameter
Yanbin Sang,
1, 2Zhongli Wei,
2, 3and Wei Dong
41Department of Mathematics, North University of China, Taiyuan, Shanxi 030051, China 2School of Mathematics, Shandong University, Jinan, Shandong 250100, China
3Department of Mathematics, Shandong Jianzhu University, Jinan, Shandong 250101, China 4Department of Mathematics, Hebei University of Engineering, Handan, Hebei 056021, China
Correspondence should be addressed to Yanbin Sang,[email protected] Received 9 January 2010; Revised 23 March 2010; Accepted 26 March 2010 Academic Editor: A. Pankov
Copyrightq2010 Yanbin Sang 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.
This work presents sufficient conditions for the existence and uniqueness of positive solutions for a discrete fourth-order beam equation under Lidstone boundary conditions with a parameter; the iterative sequences yielding approximate solutions are also given. The main tool used is monotone iterative technique.
1. Introduction
In this paper, we are interested in the existence, uniqueness, and iteration of positive solutions for the following nonlinear discrete fourth-order beam equation under Lidstone boundary conditions with explicit parameterβgiven by
Δ4yt−2−βΔ2yt−1 htf 1
ytf2
yt, t∈ a1, b−1Z, 1.1
ya 0 Δ2ya−1, yb 0 Δ2yb−1, 1.2
whereΔis the usual forward difference operator given byΔyt yt1−yt,Δnyt
Δn−1Δyt, c, d
Z:{c, c1, . . . , d−1, d}, andβ >0 is a real parameter.
1–7and references therein. Especially, there was much attention focused on the existence and multiplicity of positive solutions of fourth-order problem, for example, 8–10, and in particular the discrete problem with Lidstone boundary conditions 11–17. However, very little work has been done on the uniqueness and iteration of positive solutions of discrete fourth-order equation under Lidstone boundary conditions. We would like to mention some results of Anderson and Minh ´os 11and He and Su 12, which motivated us to consider the BVP1.1and1.2.
In 11, Anderson and Minh ´os studied the following nonlinear discrete fourth-order equation with explicit parametersβandλgiven by
Δ4yt−2−βΔ2yt−1 λft, yt, t∈ a1, b−1
Z, 1.3
with Lidstone boundary conditions1.2, whereβ > 0 andλ > 0 are real parameters. The authors obtained the following result.
Theorem 1.1see 11. Assume that the following condition is satisfied
A1ft, y gtwy, where g : a1, b−1Z → 0,∞with bz−a11gz > 0,w :
0,∞ → 0,∞is continuous and nondecreasing, and there existsθ ∈ 0,1such that
wκy≥κθwyforκ∈0,1andy∈ 0,∞,
then, for anyλ∈0,∞, the BVP1.3and1.2has a unique positive solutionyλ. Furthermore,
such a solutionyλsatisfies the following properties:
ilimλ→0yλ0andlimλ→ ∞yλ∞; iiyλis nondecreasing inλ;
iiiyλis continuous inλ, that is, ifλ → λ0, thenyλ−yλ0 → 0.
Very recently, in 12, He and Su investigated the existence, multiplicity, and nonexistence of nontrivial solutions to the following discrete nonlinear fourth-order boundary value problem
Δ4ut−2 ηΔ2ut−1−ξut λft, ut, t∈Z a1, b1,
ua 0 Δ2ua−1, ub2 0 Δ2ub1,
1.4
whereΔdenotes the forward difference operator defined byΔut ut1−ut,Δnut
ΔΔn−1ut,Z a1, b1is the discrete interval given by{a1, a2, . . . , b1}withaand
ba < bintegers,η, ξ, λare real parameters and satisfy
η <8 sin2 π
2b−a2, η
24ξ≥0, ξ4ηsin2 π
2b−a2 <16 sin
4 π
2b−a2, λ >0.
1.5
For the functionf, the authors imposed the following assumption:
B1ft, x gthx, where g : Z a1, b1 → 0,∞ with b1
ta1gt > 0, h : R → 0,∞is continuous and nondecreasing, and there existsθ ∈0,1such that
Their main result is the following theorem.
Theorem 1.2see 12. Assume thatB1holds. Then for anyλ ∈ 0,∞, the BVP1.4has a
unique positive solutionuλ. Furthermore, such a solutionuλsatisfies the properties (i)–(iii) stated in
Theorem 1.1.
The aim of this work is to relax the assumptionsA1andB1on the nonlinear term,
without demanding the existence of upper and lower solutions, we present conditions for the BVP1.1and1.2to have a unique solution and then study the convergence of the iterative sequence. The ideas come from Zhai et al. 18,19and Liang 20.
Let Bdenote the Banach space of real-valued functions on a−1, b1Z, with the supremum norm
y sup
t∈a−1,b1Z
yt. 1.6
Throughout this paper, we need the following hypotheses:
H1fi: 0,∞ → 0,∞are continuous andfiy>0 fory >0i1,2;
H2h: a1, b−1Z → 0,∞withbz−a11hz>0;
H3f1 : 0,∞ → 0,∞is nondecreasing,f2 : 0,∞ → 0,∞is nonincreasing,
and there existϕτ, ψτon interval a1, b−1Zwithϕ: a1, b−1Z → 0,1, for alle0∈0,1, there existsτ0∈ a1, b−1Zsuch thatϕτ0 e0, andψτ> ϕτ,for
allτ ∈ a1, b−1Zwhich satisfy
f1
ϕτy≥ψτf1
y, f2
1
ϕτy
≥ψτf2
y, ∀τ∈ a1, b−1Z, y≥0. 1.7
2. Two Lemmas
To prove the main results in this paper, we will employ two lemmas. These lemmas are based on the linear discrete fourth-order equation
Δ4yt−2−βΔ2yt−1 ut, t∈ a1, b−1
Z, 2.1
with Lidstone boundary conditions1.2.
Lemma 2.1see 11. Letu: a1, b−1Z → Rbe a function. Then the nonhomogeneous discrete fourth-order Lidstone boundary value problem2.1,1.2has solution
yt
b
sa b−1
za1
whereG2t, sgiven by
G2t, s
1
1,0b, a
⎧ ⎨ ⎩
t, ab, s: t≤s,
s, ab, t: s≤t,
t, s∈ a−1, b1Z× a, bZ 2.3
witht, s μt−s−μs−tforμ β2ββ4/2, is the Green’s function for the second-order
discrete boundary value problem
−Δ2yt−1−βyt0, t∈ a, b
Z,
ya 0yb,
2.4
andG1s, zgiven by
G1s, z 1
b−a
⎧ ⎨ ⎩
s−ab−z: s≤z,
z−ab−s: z≤s, s, z∈ a, bZ× a1, b−1Z 2.5
is the Green’s function for the second-order discrete boundary value problem
−Δ2xs−1 0, s∈ a1, b−1
Z,
xa 0xb.
2.6
Lemma 2.2see 11. Let
m: 1,0b, a1
b−a2b, a , M:
b−a2b/2, a/2
41,0b, a . 2.7
Then, fort, s, z∈ a1, b−13Z, one has
m≤G2t, sG1s, z≤M. 2.8
3. Main Results
Theorem 3.1. Assume thatH1–H3hold. Then, the BVP1.1and1.2has a unique solution
y∗tinD, where
Moreover, for anyx0, y0 ∈D, constructing successively the sequences fixed point of the operator equation
conditionH2impliesL >0. Sincefiy>0 fory >0i1,2, byLemma 2.2, we have
AL, L
b−1
sa1 b−1
za1
G2t, sG1s, zhz
f1L f2L
≥mf1L f2L b−1
sa1 b−1
za1
hz
mf1L f2L
L
3.6
formin2.1andLin3.5. Moreover, we obtain
AL, L≤Mf1L f2L
L 3.7
forMin2.1. Thus
mf1L f2L
L≤AL, L≤Mf1L f2L
L. 3.8
Therefore, we can choose a sufficiently small numbere1∈0,1such that
e1L≤AL, L≤
L e1
, 3.9
which together withH3implies that there existsτ1∈ a1, b−1Zsuch thatϕτ1 e1, so
ϕτ1L≤AL, L≤ L
ϕτ1. 3.10
Sinceψτ1/ϕτ1>1, we can take a sufficiently large positive integerksuch that
ψτ 1
ϕτ1 k
≥ 1
ϕτ1
. 3.11
It is clear that
ϕτ 1
ψτ1 k
Thus, we have
Av0, u0 A
L
ϕτ1 k,
ϕτ1 k
L
A
L ϕτ1
ϕτ1
k−1, ϕτ1
ϕτ1 k−1
L
≤ 1
ψτ1
A
L
ϕτ1 k−1,
ϕτ1 k−1
L
≤ · · ·
≤ 1
ψτ1
kAL, L≤
1
ψτ1 k
L ϕτ1
.
3.17
In accordance with3.12, we can see that
Av0, u0≤
L
ϕτ1
k v0. 3.18
Construct successively the sequences
unAun−1, vn−1, vnAvn−1, un−1, n1,2, . . . . 3.19
By the mixed monotonicity ofA, we haveu1 Au0, v0≤Av0, u0 v1. By induction, we
obtainun ≤vn, n 1,2, . . .. It follows from3.14,3.18, and the mixed monotonicity ofA
that
u0≤u1≤ · · · ≤un≤ · · · ≤vn≤ · · · ≤v1≤v0. 3.20
Note thatu0≥λv0, so we can getunt≥u0t≥λv0t≥λvnt, t∈ a, bZ, n1,2, . . .. Let
λnsup{λ >0|unt≥λvnt, t∈ a, bZ}, n1,2, . . . . 3.21
Thus, we have
unt≥λnvnt, t∈ a, bZ, n1,2, . . . , 3.22
and then
un1t≥unt≥λnvnt≥λnvn1t, t∈ a, bZ, n1,2, . . . . 3.23
Therefore,λn1 ≥λn,that is,{λn}is increasing with{λn} ⊂0,1. Setλlimn→ ∞λn. We can
show thatλ1. In fact, if 0<λ < 1, byH3, there existsτ2∈ a1, b−1Zsuch thatϕτ2 λ.
iThere exists an integerN such thatλN λ. In this case, we have λn λfor all
n≥Nholds. Hence, forn≥N, it follows from3.4and the mixed monotonicity ofAthat
un1Aun, vn≥A λv n,
1
λun
A ϕτ2vn,
1
ϕτ2
un
≥ψτ2Avn, un ψτ2vn1.
3.24
By the definition ofλn, we have
λn1λ≥ψτ2> ϕτ2 λ. 3.25
This is a contradiction.
iiFor all integern,λn < λ. In this case, we have 0 < λn/λ < 1. In accordance with
H3, there existsθn ∈ a1, b−1Zsuch thatϕθn λn/λ. Hence, combining3.4with the
mixed monotonicity ofA, we have
un1 Aun, vn≥A λnvn,
1
λnun
A
⎛ ⎜ ⎝λn
λ
λvn,
un
λn/λ
λ
⎞ ⎟
⎠A ϕθnϕτ2vn,
un
ϕθnϕτ2
≥ψθnA ϕτ2vn,
un
ϕτ2
≥ψθnψτ2Avn, un
ψθnψτ2vn1.
3.26
By the definition ofλn, we have
λn1≥ψθnψτ2> ϕθnψτ2
λn
λ ψτ2. 3.27
Letn → ∞, we haveλ≥λ/ λψτ2>λ/ λϕτ2 ϕτ2 λ, and this is also a contradiction.
Hence, limn→ ∞λn1.
Thus, combining3.20with3.22, we have
0≤unlt−unt≤vnt−unt≤vnt−λnvnt 1−λnvnt≤1−λnv0t 3.28
fort∈ a, bZ, wherelis a nonnegative integer. Thus,
unl−un ≤ vn−un ≤1−λnv0. 3.29
Therefore, there exists a functiony∗∈Dsuch that
lim
n→ ∞unt nlim→ ∞vnt y
∗t fort∈ a−1, b1
By the mixed monotonicity ofAand3.20, we have contrary, that there exist two nontrivial solutionsy1 andy2of the BVP1.1and1.2such
thatAy1t, y1t y1tandAy2t, y2t y2tfort∈ a−1, b1Z. According to3.9,
Thus, we can choose a sufficiently large positive integerksuch that
Obviously,u"0< x0, y0<v"0. Let
Similarly to the above proof, it follows that there existsy"∈Dsuch that
lim
This completes the proof of the theorem.
Remark 3.2. From the proof ofTheorem 3.1, we easily know that assumey Ay, x,x Ax, y, thus, lety0y,x0x, we have
yny, xnx, n1,2, . . . . 3.42
Thereforeyxy∗.
Theorem 3.3. Assume thatH2holds, and the following conditions are satisfied:
C3for fixedτ∈ a1, b−1Z, one has
iψτ, yis nonincreasing with respect toy, and there existsτ4∈ a1, b−1Zsuch
that
mfL≥ϕτ4,
ψτ4, L/ϕτ4
ϕτ4 ≥
MfL 3.44
or
iiψτ, yis nondecreasing with respect toy, and there existsτ5 ∈ a1, b−1Zsuch
that
mfL≥ ϕτ5 ψτ5, L
, 1 ϕτ5 ≥
MfL, 3.45
wherem, Mare defined in2.1,Lis defined in3.5. Then, the BVP
Δ4yt−2−βΔ2yt−1 htfyt, t∈ a1, b−1
Z, ya 0 Δ2ya−1, yb 0 Δ2yb−1
3.46
has a unique solutiony∗.
Proof. For convenience, we still define the operator equationAby
Ayt
b
sa b−1
za1
G2t, sG1s, zhzf
yz, t∈ a−1, b1Z. 3.47
In the following, we consider the following two cases.
iFor fixedτ∈ a1, b−1Z,ψτ, yis nonincreasing with respect toy.
According to conditionC3andLemma 2.2, we can know that there existsτ4 ∈ a
1, b−1Zsuch that
ϕτ4L≤AL≤
ψτ4, L/ϕτ4
ϕτ4
L. 3.48
Sinceψτ4, L/ϕτ4>1, we can find a sufficiently large positive integerksuch that
ψτ 4, L
ϕτ4 k
≥ 1
ϕτ4
Fort∈ a1, b−1Z, we still define
u0t
ϕτ4 k
L, v0t L
ϕτ4 k,
unt Aun−1t, vnt Avn−1t, n1,2, . . . .
3.50
By the proof ofTheorem 3.1, it is sufficient to show that
u0 ≤u1≤v1 ≤v0. 3.51
Obviously,u0≤v0andu1≤v1.
In this case, it follows from conditionsC2,C3, and3.49that
u1 Au0Aϕτ4 kL
Aϕτ4
ϕτ4 k−1
L
≥ψτ4,
ϕτ4 k−1
LAϕτ4 k−1
L
ψτ4,
ϕτ4 k−1
LAϕτ4
ϕτ4 k−2
L
≥ψτ4,
ϕτ4 k−1
Lψτ4,
ϕτ4 k−2
LAϕτ4 k−2
L
≥ · · ·
≥ψτ4,
ϕτ4 k−1
Lψτ4,
ϕτ4 k−2
L· · ·ψτ4, LAL
≥ψτ4, L k
ϕτ4L
≥ϕτ4 k
Lu0.
3.52
In accordance with3.16, we have
A y ϕs
≤ 1
Fort∈ a1, b−1Z, we still define
u0t
ϕτ5 k
L, v0t
L
ϕτ5 k,
unt Aun−1t, vnt Avn−1t, n1,2, . . . .
3.57
We continue to prove that
u1≥u0, v1≤v0. 3.58
By3.52, combining3.55with the monotonicity ofψ, we have
u1 Au0Aϕτ5 k
L
≥ψτ5,
ϕτ5 k−1
Lψτ5,
ϕτ5 k−2
L· · ·ψτ5, LAL
≥ϕτ5 k−1ψτ
5, LAL ≥ϕτ5
k
Lu0.
3.59
In accordance with3.54, combining the monotonicity ofψand3.55, we get
v1Av0A
L
ϕτ5 k
≤ 1
ψτ5, L/
ϕτ5 k
1
ψτ5, L/
ϕτ5
k−1· · ·
1
ψτ5, L/ϕτ5 AL
≤ 1
ψτ5, L/ϕτ5 k
L ϕτ5.
3.60
An application of3.56yields
v1≤ 1
ϕτ5
kLv0. 3.61
Therefore, we obtain
u0 ≤u1≤v1 ≤v0. 3.62
Remark 3.4. InTheorem 3.1, the more general conditions are imposed on the nonlinear term thanTheorem 1.1. In particular, inTheorem 3.3,ψτ, ycontains the variabley; therefore, the more comprehensive functions can be incorporated.
4. An Example
Example 4.1. Consider the following discrete fourth-order Lidstone problem:
Δ4yt−2−Δ2yt−1 t
Moreover, for anyx0, y0∈D, constructing successively the sequences
ψτ ϕτ1/2. It is easy to see that
f1
ϕτy1ϕτy1/4≥ψτ1y1/4ψτf1
y, ∀τ∈ 3,6Z, y≥0,
f2
y ϕτ
2 1
y/ϕτ1/4 ≥ψτ
2 1
y1/4
, ∀τ∈ 3,6Z, y≥0.
4.5
The conclusion then follows fromTheorem 3.1.
Acknowledgments
The authors were supported financially by the National Natural Science Foundation of China
10971046, the Natural Science Foundation of Shandong ProvinceZR2009AM004, and the Youth Science Foundation of Shanxi Province2009021001-2.
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