Volume 2008, Article ID 123823,11pages doi:10.1155/2008/123823
Research Article
Existence and Uniqueness of Solutions for
Singular Higher Order Continuous and Discrete
Boundary Value Problems
Chengjun Yuan,1, 2 Daqing Jiang,1and You Zhang1
1School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, Jilin, China 2School of Mathematics and Computer, Harbin University, Harbin 150086, Heilongjiang, China
Correspondence should be addressed to Chengjun Yuan,[email protected]
Received 4 July 2007; Accepted 31 December 2007
Recommended by Raul Manasevich
By mixed monotone method, the existence and uniqueness are established for singular higher-order continuous and discrete boundary value problems. The theorems obtained are very general and complement previous known results.
Copyrightq2008 Chengjun Yuan 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
In recent years, the study of higher-order continuous and discrete boundary value problems has been studied extensively in the literature see 1–17 and their references. Most of the results told us that the equations had at least single and multiple positive solutions.
Recently, some authors have dealt with the uniqueness of solutions for singular higher-order continuous boundary value problems by using mixed monotone method, for example, see6,14,15. However, there are few works on the uniqueness of solutions for singular dis-crete boundary value problems.
In this paper, we state a unique fixed point theorem for a class of mixed monotone op-erators, see 6, 14, 18. In virtue of the theorem, we consider the existence and uniqueness of solutions for the following singular higher-order continuous and discrete boundary value problems1.1and1.2by using mixed monotone method. We first discuss the existence and uniqueness of solutions for the following singular higher-order continuous boundary value problem
yntλqtgyhy0, 0< t <1, λ >0,
wheren≥2,qt∈C0,1,0,∞,g :0,∞→ 0,∞is continuous and nondecreasing;
h:0,∞→0,∞is continuous and nonincreasing, andhmay be singular aty0.
Next, we consider the existence and uniqueness of solutions for the following singular higher-order discrete boundary value problem
Δnyiλqin−1gyin−1 hyin−1 0, i∈N {0,1,2, . . . , T−1}, λ >0,
Δky0 Δn−2yT 1 0, 0≤k≤n−2,
1.2 wheren≥2,N {0,1,2, . . . , Tn},qi∈CN,0,∞,g:0,∞→0,∞is continuous and nondecreasing; h : 0,∞ → 0,∞is continuous and nonincreasing, and h may be singular aty0. Throughout this paper, the topology onNwill be the discrete topology.
2. Preliminaries
LetPbe a normal cone of a Banach spaceE, ande∈Pwithe ≤1, e /θ.Define
Qe{x∈P |x /θ, there exist constants m, M >0 such that m e≤x≤Me}. 2.1
Now we give a definitionsee18.
Definition 2.1see18. AssumeA:Qe×Qe→Qe.Ais said to be mixed monotone ifAx, y
is nondecreasing inxand nonincreasing iny, that is, ifx1≤x2x1, x2∈QeimpliesAx1, y≤
Ax2, yfor anyy ∈Qe, andy1 ≤ y2y1, y2 ∈QeimpliesAx, y1≥ Ax, y2for anyx∈ Qe.
x∗∈Qeis said to be a fixed point ofAifAx∗, x∗ x∗.
Theorem 2.2 see 6, 14. Suppose that A: Qe×Qe → Qe is a mixed monotone operator and ∃
a constantα,0≤α <1, such that
A
tx,1 ty
≥tαAx, y, for x, y∈Qe, 0< t <1. 2.2
ThenAhas a unique fixed pointx∗∈Qe. Moreover, for anyx0, y0∈Qe×Qe,
xnA
xn−1, yn−1
, ynA
yn−1, xn−1
, n1,2, . . . , 2.3
satisfy
xn −→x∗, yn−→x∗, 2.4
where
xn−x∗o1−rαn
, yn−x∗o
1−rαn, 2.5
0< r <1,ris a constant fromx0, y0.
Theorem 2.3see6,14,18. Suppose that A:Qe×Qe→ Qeis a mixed monotone operator and∃a
constantα∈0,1such that2.2holds. Ifx∗λis a unique solution of equation
Ax, x λx, λ >0 2.6
inQe, thenxλ∗−x∗λ0 →0, λ→λ0.If0< α <1/2,then0< λ1< λ2impliesxλ∗1 ≥xλ∗2,x∗λ1/x∗λ2, and
lim
λ→∞
x∗
λ0, λlim→0x
∗
3. Uniqueness positive solution of differential equations1.1
This section discusses singular higher-order boundary value problem 1.1. Throughout this section, we letGt, sbe the Green’s function to−y0, y0 y1 0,we note that
Gt, s ⎧ ⎨ ⎩
t1−s, 0≤t≤s≤1,
s1−t, 0≤s≤t≤1, 3.1
and one can show that
Gt, tGs, s≤Gt, s≤Gt, t, forGt, s≤Gs, s, t, s∈0,1×0,1. 3.2
Suppose thatyis a positive solution of1.1. Let
xt yn−2t, 3.3
fromyi0 yn−21 0, 0≤i≤n−2, and Taylor Formula, we define operatorT :C20,1→ Cn0,1, by
yt Txt t
0
t−sn−3
n−3! xsds, for 3≤n,
yt Txt xt, for n2..
3.4
Then we have
x2t λft, Txt 0, 0< t <1, λ >0,
x0 x1 0. 3.5
Then from3.4, we have the next lemma.
Lemma 3.1. Ifxtis a solution of 3.5, thenytis a solution of 1.1.
Further, ifytis a solution of1.1, imply thatxtis a solution of3.5.
LetP {x∈C0,1|xt≥0, for allt∈0,1}. Obviously,P is a normal cone of Banach spaceC0,1.
Theorem 3.2. Suppose that there existsα∈0,1such that
gtx≥tαgx, 3.6
ht−1x≥tαhx, 3.7
for anyt∈0,1andx >0, andq∈C0,1,0,∞satisfies 1
0
sn−1n−2s−αqsds <∞. 3.8
Then1.1has a unique positive solutionyλ∗t.And moreover,0< λ1< λ2impliesy∗λ1 ≤y∗λ2, yλ∗1/yλ∗2. Ifα∈0,1/2, then
lim
λ→0 y∗
Proof. Since3.7holds, lett−1xy,one has
hy≥tαhty. 3.10
Then
hty≤ 1
tαhy, for t∈0,1, y >0. 3.11
Lety1.The above inequality is
ht≤ 1
tαh1, for t∈0,1. 3.12
From3.7,3.11, and3.12, one has
ht−1x≥tαhx, h
1
t
≥tαh1, htx≤ 1
tαhx, ht≤
1
tαh1,for t∈0,1, x >0.
3.13
Similarly, from3.6, one has
gtx≥tαgx, gt≥tαg1, fort∈0,1, x >0. 3.14
Lett 1/x, x >1,one has
gx≤xαg1, forx≥1. 3.15
Letet Gt, t t1−t, and we define
Qe
x∈C0,1| 1
MGt, t≤xt≤MGt, t, t∈0,1
, 3.16
whereM >1 is chosen such that
M >max
λg1 1
0
qsdsλh1 1
0
sn−1n−2s n!
−α
qsds
1/1−α
,
λg1
1
0 Gs, s
sn−1n−2s n!
α
qsdsλh1 1
0
Gs, sqsds
−1/1−α
.
3.17
First, from3.4and3.16, for anyx∈Qe, we have the following.
When 3≤n,
1
M
tn−1n−2t
n! ≤ t
0
1
MGs, s
t−sn−3
n−3! ds≤Txt
≤ t
0
MGs, stn−−sn−3
3! ds≤M
tn−1n−2t
n! ≤M, for t∈0,1,
whenn2,
1
M
tn−1n−2t
n! ≤Txt xt≤M
tn−2t
n! ≤M, for t∈0,1, 3.19
then
1
M
tn−1n−2t
n! ≤Txt≤M
tn−1n−2t
n! ≤M, fort∈0,1. 3.20
For anyx, y∈Qe,we define
Aλx, yt λ
1
0
Gt, sqsgTxs hTysds, fort∈0,1. 3.21
First, we show thatAλ :Qe×Qe→Qe.
Letx, y ∈Qe,from3.14,3.15, and3.20, we have
gTxt≤gM≤Mαg1, fort∈0,1, 3.22
and from3.13, we have
hTyt≤h
1
M
tn−1n−2t n!
≤
tn−1n−2t n!
−α
h
1
M
≤Mα
tn−1n−2t n!
−α
h1, fort∈0,1.
3.23
Then, from3.2,3.21,3.22and3.23, we have
On the other hand, for anyx, y∈Qe, from3.13and3.14, we have
gTxt≥g
1
M
tn−1n−2t n!
≥
tn−1n−2t n!
α
g
1
M
≥
tn−1n−2t n!
α
1
Mαg1,
hTyt≥hM h
1 1/M
≥ 1
Mαh1, for t∈0,1.
3.24
Thus, from3.2,3.21and3.24, we have
Aλx, yt
≥λGt, t 1
0
Gs, sqsM−α
sn−1n−2s
n!
α
g1ds 1
0
Gs, sqsM−αh1ds
≥ 1
MGt, t, for t∈0,1.
3.25
Next, for anyl∈0,1, one has
Aλ
lx, l−1yt λ 1
0
Gt, sqsglTxs hl−1Tysds
≥λ 1
0
Gt, sqslαgTxs lαhTysds
lαAλx, yt, for t∈0,1.
3.26
So the conditions of Theorems2.2and2.3hold. Therefore, there exists a uniquex∗λ ∈Qe
such thatAλx∗, x∗ xλ∗. It is easy to check thatx∗λ is a unique positive solution of 3.5for
givenλ >0. Moreover,Theorem 2.3means that if 0< λ1< λ2,thenx∗λ
1t≤x
∗
λ2t,x
∗
λ1t/x
∗
λ2t and ifα∈0,1/2, then
lim
λ→0x
∗
λ0, λlim→∞xλ∗ ∞. 3.27
Next, from Lemma 3.1 and 3.4, we get that y∗λ Tx∗λ is a unique positive solution of 1.1for given λ > 0. Moreover, if 0 < λ1 < λ2,then yλ∗1t ≤ y∗λ2t, yλ∗1t/yλ∗2t and if
α∈0, 1/2, then
lim
λ→0 y∗
λ0, λlim→∞y∗λ ∞. 3.28
This completes the proof.
Example 3.3. Consider the following singular boundary value problem:
ynt λμyat y−bt0, t∈0,1,
yi0 yn−21 0, 0≤i≤n−2,
3.29
whereλ, a, b >0,μ≥0, max{a, b}< 1/n−1.
ApplyingTheorem 3.2, letα max{a, b} < 1/n−1,qt 1,gy μya,hy y−b, then
gty≥tαgy, ht−1≥tαhy, 1
0
sn−1n−2s−αds <∞. 3.30
Thus all conditions in Theorem 3.2are satisfied. We can find3.29has a unique positive so-lution yλ∗t. In addition, 0 < λ1 < λ2 impliesyλ∗
1 ≤ y
∗
λ2, y
∗
λ1/y
∗
λ2. If α max{a, b} ∈ 0,1/2, then
lim
λ→0 y∗
4. Uniqueness positive solution of difference equations1.2
This section discusses singular higher-order boundary value problem 1.2. Throughout this section, we letKi, jbe Green’s function to−Δ2yi ui1 0,i∈N,y0 yT 1 0,
we note that
Ki, j ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
jT 1−i
T1 , 0≤j≤i−1,
iT1−j
T1 , i≤j≤T1,
4.1
and one can show that
Ki, i≥Ki, j, Kj, j≥Ki, j, Ki, j≥ Ki, i
T1 , for 0≤i≤T1, 1≤j≤T. 4.2
Suppose thatyis a positive solution of1.2. Let
xi Δn−2yi, for 0≤i≤T1. 4.3
FromΔiy0 Δn−2yT1 0, 0≤i≤n−2, andΔmyi−1 Δm−1yi−Δm−1yi−1, so we
define operatorT, by
Txi yin−1
i1
l1
Cni−−l2n−1xl, for 0≤i≤T. 4.4
Then
Δ2xi λFin−1, Txi 0, 0≤i≤T−1, λ >0,
x0 xT 1 0. 4.5
Lemma 4.1. Ifxiis a solution of 4.5, thenyiis a solutionn of 1.2.
Proof. Since we remark thatxiis a solution of4.5, if and only if
xi T
j1
Ki, jλFjn−1, Txj, for 0≤i≤T1. 4.6
Let
Txi yin−1, for 0≤i≤T. 4.7
From4.4we findΔiy0 Δn−2yT1 0, 0≤i≤n−2, andxi Δn−2yi, so thatyiis
a solution of1.2.
Further, ifyiis a solution of1.2, imply thatxiis a solution of4.5.
Theorem 4.2. Suppose that there existsα∈0,1such that
gtx≥tαgx,
ht−1x≥tαgx, 4.8
for anyt∈0,1andx >0, andq∈CN,0,∞.
Then1.2has a unique positive solutiony∗λi.And moreover,0 < λ1 < λ2impliesyλ∗
1 ≤ y
∗
λ2,
y∗λ
1/y
∗
λ2. Ifα∈0,1/2, then
lim
λ→0y
∗
λ0, λlim→∞y∗λ ∞. 4.9
Proof. The proof is the same as that ofTheorem 3.2, from4.12and4.13, one has
ht−1x≥tαhx, h
1
t
≥tαh1, htx≤ 1
tαhx, ht≤
1
tαh1, fort∈0,1, x >0;
gtx≥tαgx, gt≥tαg1, fort∈0,1, x >0.
4.10
gtx≥tαgx, gt≥tαg1, for t∈0,1, x >0. 4.11
Lett1/x, x >1,one has
gx≤xαg1, forx≥1. 4.12
Letei Ki, i/T1,and we define
Qe
x∈P | 1
Mei≤xi≤Mei, for 0≤i≤T1
, 4.13
whereM >1 is chosen such that
M >max
λT1g1
T
j1
qjn−1 j1
l1
Cjn−−l2n−1 α
λT 11αh1
T
j1
K−αj, jqjn−1
1/1−α
;
λg1
T
j1
qjn−1
Kj, j
T 1 α
λh1
T
j1
qjn−1 j1
l1
Cnj−−l2n−1
−α−1/1−α
.
4.14
From4.4and4.13, for anyx∈Qe, we have
1
Mej≤Txj
j1
l1
Cnj−−l2n−1xl≤Mej
j1
l1
For anyx, y∈Qe,we define
Aλx, yi λ T
j1
Ki, jqjn−1gTxj hTyj, for 0≤i≤T 1. 4.16
First we show thatAλ:Qe×Qe→Qe.
Letx, y ∈Qe,from4.11and4.12, we have
gTxj≤g
Mejj1
l1 Cn−2
j−ln−1
≤g
M
j1
l1 Cn−2
j−ln−1
≤Mα
j1
l1 Cn−2
j−ln−1 α
g1, for 1≤j≤T ,
4.17 and from4.10, we have
hTyj≤h
1
Mej
≤e−αjh
1
M
≤Mαe−αjh1, for 1≤j≤T. 4.18
Then, from4.2and the above, we have
Aλx, yi≤λKi, i T
j1
qjn−1gTxj T1hTyj
≤eiMαλT 1
g1
T
j1
qjn−1 j1
l1
Cnj−−l2n−1 α
h1
T
j1
e−αjqjn−1
≤eiMαλT 1
g1
T
j1
qjn−1 j1
l1
Cnj−−l2n−1 α
h1
T
j1
Kj, j
T 1 −α
qjn−1
≤Mei, for 0≤i≤T 1.
4.19
On the other hand, for anyx, y∈Qe, from4.10and4.12, we have
gxj≥g
1
Mej
≥eαj 1
Mαg1, for 1≤j ≤T, 4.20
hyj≥h
Mej
j1
l1
Cnj−−l2n−1
≥M−α j1
l1
Cnj−−l2n−1 −α
h1, for 1≤j ≤T. 4.21
Thus, from4.2and4.16, we have
Aλx, yi≥λei
T
j0
qjn−1gTxj
T
j0
qjn−1hTyj
≥λeiM−α
g1
T
j0 qj
Ki, i
T1 α
h1
T
j0
qjn−1 j1
l1
Cjn−−l2n−1 −α
≥ 1
Mei, for 0≤i≤T1.
So,Aλis well defined andAλQe×Qe⊂Qe.
Next, for anyl∈0,1, one has
Aλ
lx, l−1yi λ
T
j1
Ki, jqjn−1gTlxj hTl−1yj
λ
T
j1
Ki, jqjn−1glTxj hl−1Tyj
≥λ
T
j1
Ki, jqjn−1lαgTxj lαhTyjds
lαAλx, yi, for 0≤i≤T1.
4.23
So the conditions of Theorems2.2and2.3hold. Therefore, there exists a uniquexλ∗ ∈Qesuch
thatAλx∗, x∗ x∗λ.It is easy to check thatx∗λ is a unique positive solution of4.5for given
λ > 0. Moreover,Theorem 2.3means that if 0 < λ1 < λ2,thenxλ∗
1t≤x
∗
λ2t,x
∗
λ1t/x
∗
λ2tand ifα∈0,1/2, then
lim
λ→0x
∗
λ0, λlim→∞xλ∗ ∞. 4.24
Next, on usingLemma 3.1, from4.5, we get thatyλ∗Tx∗λis a unique positive solution of 1.2for given λ > 0. Moreover, if 0 < λ1 < λ2,then yλ∗
1t ≤ y
∗
λ2t, y
∗
λ1t/y
∗
λ2t and if
α∈0,1/2, then
lim
λ→0y
∗
λ0, λlim→∞y∗λ ∞. 4.25
This completes the proof.
Example 4.3. Consider the following singular boundary value problem:
Δnyi−1 λμyai y−bi0, i∈N,
Δiy0 Δn−2y1 0, 0≤i≤n−2, 4.26
whereλ, a, b >0,μ≥0, max{a, b}<1.
Letqi 1,gy μya,hy y−b,αmax{a, b}<1, then
gty≥tαgy, ht−1y≥tαhy, 4.27
thus all conditions inTheorem 4.2are satisfied. We can find 4.26has a unique positive so-lution yλ∗t. In addition, 0 < λ1 < λ2 impliesy∗λ1 ≤ yλ∗2, y∗λ1/yλ∗2. If α max{a, b} ∈ 0,1/2, then
lim
λ→0 y∗
Acknowledgments
The work was supported by the National Natural Science Foundation of China Grants no. 10571021 and 10701020. The work was supported by Subject Foundation of Harbin University
Grant no. HXK200714.
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