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Hindawi Publishing Corporation
Boundary Value Problems
Volume 2008, Article ID 123823, 11 pages
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,
1
and You Zhang
1
1
School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, Jilin, China
2
School of Mathematics and Computer, Harbin University, Harbin 150086, Heilongjiang, China
Correspondence should be addressed to Chengjun Yuan,
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.
Copyright q 2008 Chengjun Yuan et al. This is an open access article distributed under the Creative
Commons Attribution License, which p ermits 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,
see 6, 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
problems 1.1 and 1.2 by using mixed monotone method. We first discuss the existence and
uniqueness of solutions for the following singular higher-order continuous boundary value
problem
y
n
tλqtgyhy0, 0<t<1,λ>0,
y
i
0y
n−2
10, 0 ≤ i ≤ n − 2,
1.1
2 Boundary Value Problems
where n ≥ 2, qt ∈ C0, 1, 0, ∞, g : 0, ∞ → 0, ∞ is continuous and nondecreasing;
h : 0, ∞ → 0, ∞ is continuous and nonincreasing, and h may be singular at y 0.
Next, we consider the existence and uniqueness of solutions for the following singular
higher-order discrete boundary value problem
Δ
n
yiλqin−1gyin−1 hyin−1 0,i∈ N {0, 1, 2, ,T − 1},λ>0,
Δ
k
y0Δ
n−2
yT 10, 0 ≤ k ≤ n − 2,
1.2
where n ≥ 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 at y 0. Throughout this paper, the topology on N
will be the discrete topology.
2. Preliminaries
Let P be a normal c one of a Banach space E,ande ∈ P with e≤1,e
/
θ. Define
Q
e
{x ∈ P | x
/
θ, there exist constants m, M > 0 such that me ≤ x ≤ Me}. 2.1
Now we give a definition see 18.
Definition 2.1 see 18. Assume A : Q
e
× Q
e
→ Q
e
. A is said to be mixed monotone if Ax, y
is nondecreasing in x and nonincreasing in y,thatis,ifx
1
≤ x
2
x
1
,x
2
∈ Q
e
implies Ax
1
,y ≤
Ax
2
,y for any y ∈ Q
e
,andy
1
≤ y
2
y
1
,y
2
∈ Q
e
implies Ax, y
1
≥ Ax, y
2
for any x ∈ Q
e
.
x
∗
∈ Q
e
is said to be a fixed point of A if Ax
∗
,x
∗
x
∗
.
Theorem 2.2 see 6, 14. Suppose that A: Q
e
× Q
e
→ Q
e
is a mixed monotone operator and ∃
a constant α, 0 ≤ α<1, such that
A
tx,
1
t
y
≥ t
α
Ax, y,forx,y∈ Q
e
, 0 <t<1. 2.2
Then A has a unique fixed point x
∗
∈ Q
e
. Moreover, for any x
0
,y
0
∈ Q
e
× Q
e
,
x
n
A
x
n−1
,y
n−1
,y
n
A
y
n−1
,x
n−1
,n 1, 2, , 2.3
satisfy
x
n
−→ x
∗
,y
n
−→ x
∗
, 2.4
where
x
n
− x
∗
o
1 − r
α
n
,
y
n
− x
∗
o
1 − r
α
n
, 2.5
0 <r<1, r is a constant from x
0
,y
0
.
Theorem 2.3 see 6, 14, 18. Suppose that A: Q
e
× Q
e
→ Q
e
is a mixed monotone operator and ∃ a
constant α ∈ 0, 1 such that 2.2 holds. If x
∗
λ
is a unique solution of equation
Ax, xλx, λ>02.6
in Q
e
,thenx
∗
λ
− x
∗
λ
0
→0,λ→ λ
0
. If 0 <α<1/2, then 0 <λ
1
<λ
2
implies x
∗
λ
1
≥ x
∗
λ
2
, x
∗
λ
1
/
x
∗
λ
2
,and
lim
λ→∞
x
∗
λ
0, lim
λ→0
x
∗
λ
∞. 2.7
Chengjun Yuan et al. 3
3. Uniqueness positive solution of differential equations 1.1
This section discusses singular higher-order boundary value problem 1.1. Throughout this
section, we let Gt, s be the Green’s function to −y
0,y0y10, 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, for Gt, s ≤ Gs, s, t, s ∈ 0, 1 × 0, 1.
3.2
Suppose that y is a positive solution of 1.1.Let
xty
n−2
t, 3.3
from y
i
0y
n−2
10, 0 ≤ i ≤ n−2, and Taylor Formula, we define operator T : C
2
0, 1 →
C
n
0, 1,by
ytTxt
t
0
t − s
n−3
n − 3!
xsds, for 3 ≤ n,
ytTxtxt, for n 2
3.4
Then we have
x
2
tλft, Txt 0, 0 <t<1,λ>0,
x0x10.
3.5
Then from 3.4, we have the next lemma.
Lemma 3.1. If xt is a solution of 3.5,thenyt is a solution of 1.1.
Further, if yt is a solution of 1.1,implythatxt is a solution of 3.5.
Let P {x ∈ C0, 1 | xt ≥ 0, for all t ∈ 0, 1
}. Obviously, P is a normal cone of Banach
space C0, 1.
Theorem 3.2. Suppose that there exists α ∈ 0, 1 such that
gtx ≥ t
α
gx, 3.6
h
t
−1
x
≥ t
α
hx, 3.7
for any t ∈ 0, 1 and x>0,andq ∈ C0, 1, 0, ∞ satisfies
1
0
s
n−1
n − 2s
−α
qsds < ∞. 3.8
Then 1.1 has a unique positive solution y
∗
λ
t. And moreover, 0 <λ
1
<λ
2
implies y
∗
λ
1
≤ y
∗
λ
2
,y
∗
λ
1
/
y
∗
λ
2
.
If α ∈ 0, 1/2,then
lim
λ→0
y
∗
λ
0, lim
λ→∞
y
∗
λ
∞. 3.9
4 Boundary Value Problems
Proof. Since 3.7 holds, let t
−1
x y, one has
hy ≥ t
α
hty. 3.10
Then
hty ≤
1
t
α
hy, for t ∈ 0, 1,y>0. 3.11
Let y 1. The above inequality is
ht ≤
1
t
α
h1, for t ∈ 0, 1. 3.12
From 3.7, 3.11,and3.12, one has
h
t
−1
x
≥ t
α
hx,h
1
t
≥ t
α
h1,htx ≤
1
t
α
hx,ht ≤
1
t
α
h1, for t ∈ 0, 1,x>0.
3.13
Similarly, from 3.6, one has
gtx ≥ t
α
gx,gt ≥ t
α
g1, for t ∈ 0, 1,x>0. 3.14
Let t 1/x, x > 1, one has
gx ≤ x
α
g1, for x ≥ 1. 3.15
Let etGt, tt1 − t, and we define
Q
e
x ∈ C0, 1 |
1
M
Gt, t ≤ xt ≤ MGt, t,t∈ 0, 1
, 3.16
where M>1 is chosen such that
M>max
λg1
1
0
qsds λh1
1
0
s
n−1
n − 2s
n!
−α
qsds
1/1−α
,
λg1
1
0
Gs, s
s
n−1
n − 2s
n!
α
qsds λh1
1
0
Gs, sqsds
−1/1−α
.
3.17
First, from 3.4 and 3.16, for any x ∈ Q
e
,wehavethefollowing.
When 3 ≤ n,
1
M
t
n−1
n − 2t
n!
≤
t
0
1
M
Gs, s
t − s
n−3
n − 3!
ds ≤ Txt
≤
t
0
MGs, s
t − s
n−3
n − 3!
ds ≤ M
t
n−1
n − 2t
n!
≤ M, for t ∈ 0, 1,
3.18
Chengjun Yuan et al. 5
when n 2,
1
M
t
n−1
n − 2t
n!
≤ Txtxt ≤ M
tn − 2t
n!
≤ M, for t ∈ 0, 1, 3.19
then
1
M
t
n−1
n − 2t
n!
≤ Txt ≤ M
t
n−1
n − 2t
n!
≤ M, for t ∈ 0, 1. 3.20
For any x, y ∈ Q
e
, we define
A
λ
x, ytλ
1
0
Gt, sqsgTxs hTysds, for t ∈ 0, 1. 3.21
First, we show that A
λ
: Q
e
× Q
e
→ Q
e
.
Let x, y ∈ Q
e
, from 3.14, 3.15,and3.20,wehave
gTxt ≤ gM ≤ M
α
g1, for t ∈ 0, 1, 3.22
and from 3.13,wehave
hTyt ≤ h
1
M
t
n−1
n − 2t
n!
≤
t
n−1
n − 2t
n!
−α
h
1
M
≤ M
α
t
n−1
n − 2t
n!
−α
h1, for t ∈ 0, 1.
3.23
Then, from 3.2, 3.21, 3.22 and 3.23,wehave
On the other hand, for any x, y ∈ Q
e
,from3.13 and 3.14,wehave
gTxt ≥ g
1
M
t
n−1
n − 2t
n!
≥
t
n−1
n − 2t
n!
α
g
1
M
≥
t
n−1
n − 2t
n!
α
1
M
α
g1,
hTyt ≥ hMh
1
1/M
≥
1
M
α
h1, for t ∈ 0, 1.
3.24
Thus, from 3.2, 3.21 and 3.24,wehave
A
λ
x, yt
≥ λGt, t
1
0
Gs, sqsM
−α
s
n−1
n − 2s
n!
α
g1ds
1
0
Gs, sqsM
−α
h1ds
≥
1
M
Gt, t, for t ∈ 0, 1.
3.25
So, A
λ
is well defined and A
λ
Q
e
× Q
e
⊂ Q
e
.
6 Boundary Value Problems
Next, for any l ∈ 0, 1, one has
A
λ
lx, l
−1
y
tλ
1
0
Gt, sqs
glTxs h
l
−1
Tys
ds
≥ λ
1
0
Gt, sqs
l
α
gTxs l
α
hTys
ds
l
α
A
λ
x, yt, for t ∈ 0, 1.
3.26
So the conditions of Theorems 2.2 and 2.3 hold. Therefore, there exists a unique x
∗
λ
∈ Q
e
such that A
λ
x
∗
,x
∗
x
∗
λ
. It is easy to check that x
∗
λ
is a unique positive solution of 3.5 for
given λ>0. Moreover, Theorem 2.3 means that if 0 <λ
1
<λ
2
, then x
∗
λ
1
t ≤ x
∗
λ
2
t, x
∗
λ
1
t
/
x
∗
λ
2
t
and if α ∈ 0, 1/2,then
lim
λ→0
x
∗
λ
0, lim
λ→∞
x
∗
λ
∞. 3.27
Next, from Lemma 3.1 and 3.4,wegetthaty
∗
λ
Tx
∗
λ
is a unique positive solution
of 1.1 for given λ>0. Moreover, if 0 <λ
1
<λ
2
, then y
∗
λ
1
t ≤ y
∗
λ
2
t, y
∗
λ
1
t
/
y
∗
λ
2
t and if
α ∈ 0, 1/2,then
lim
λ→0
y
∗
λ
0, lim
λ→∞
y
∗
λ
∞. 3.28
This completes the proof.
Example 3.3. Consider the following singular boundary value problem:
y
n
tλ
μy
a
ty
−b
t
0,t∈ 0, 1,
y
i
0y
n−2
10, 0 ≤ i ≤ n − 2,
3.29
where λ, a, b > 0, μ ≥ 0, max{a, b} < 1/n − 1.
Applying Theorem 3.2,letα max{a, b} < 1/n − 1, qt1, gyμy
a
, hyy
−b
,
then
gty ≥ t
α
gy,h
t
−1
≥ t
α
hy,
1
0
s
n−1
n − 2s
−α
ds < ∞.
3.30
Thus all conditions in Theorem 3.2 are satisfied. We can find 3.29 has a unique positive so-
lution y
∗
λ
t. In addition, 0 <λ
1
<λ
2
implies y
∗
λ
1
≤ y
∗
λ
2
,y
∗
λ
1
/
y
∗
λ
2
.Ifα max{a, b}∈0, 1/2,
then
lim
λ→0
y
∗
λ
0, lim
λ→∞
y
∗
λ
∞. 3.31
Chengjun Yuan et al. 7
4. Uniqueness positive solution of difference equations 1.2
This section discusses singular higher-order boundary value problem 1.2. Throughout this
section, we let Ki, j be Green’s function to −Δ
2
yiui 10, i ∈ N, y0yT 10,
we note that
Ki, j
⎧
⎪
⎪
⎨
⎪
⎪
⎩
jT 1 − i
T 1
, 0 ≤ j ≤ i − 1,
iT 1 − j
T 1
,i≤ j ≤ T 1,
4.1
and one can show that
Ki, i ≥ Ki, j,Kj, j ≥ Ki, j,Ki, j ≥
Ki, i
T 1
, for 0 ≤ i ≤ T 1, 1 ≤ j ≤ T. 4.2
Suppose that y is a positive solution of 1.2.Let
xiΔ
n−2
yi, for 0 ≤ i ≤ T 1. 4.3
From Δ
i
y0Δ
n−2
yT 10, 0 ≤ i ≤ n − 2, and Δ
m
yi − 1Δ
m−1
yi − Δ
m−1
yi − 1,sowe
define operator T,by
Txiyi n − 1
i1
l1
C
n−2
i−ln−1
xl, for 0 ≤ i ≤ T. 4.4
Then
Δ
2
xiλFi n − 1,Txi 0, 0 ≤ i ≤ T − 1,λ>0,
x0xT 10.
4.5
Lemma 4.1. If xi is a solution of 4.5,thenyi is a solutionn of 1.2.
Proof. Since we remark that xi is a solution of 4.5,ifandonlyif
xi
T
j1
Ki, jλFj n − 1,Txj, for 0 ≤ i ≤ T 1. 4.6
Let
Txiyi n − 1, for 0 ≤ i ≤ T. 4.7
From 4.4 we find Δ
i
y0Δ
n−2
yT 10, 0 ≤ i ≤ n − 2, and xiΔ
n−2
yi,sothatyi is
asolutionof1.2.
Further, if yi is a solution o f 1.2,implythatxi is a solution of 4.5.
Let P {x ∈ CN
, 0, ∞ | xi ≥ 0, for all i ∈ N
}. Obviously, P is a normal cone of
Banach space CN
, 0, ∞.
8 Boundary Value Problems
Theorem 4.2. Suppose that there exists α ∈ 0, 1 such that
gtx ≥ t
α
gx,
h
t
−1
x
≥ t
α
gx,
4.8
for any t ∈ 0, 1 and x>0,andq ∈ CN
, 0, ∞.
Then 1.2 has a unique positive solution y
∗
λ
i. And moreover, 0 <λ
1
<λ
2
implies y
∗
λ
1
≤ y
∗
λ
2
,
y
∗
λ
1
/
y
∗
λ
2
.Ifα ∈ 0, 1/2,then
lim
λ→0
y
∗
λ
0, lim
λ→∞
y
∗
λ
∞. 4.9
Proof. The proof is the same as that of Theorem 3.2,from4.12 and 4.13, one has
h
t
−1
x
≥ t
α
hx,h
1
t
≥ t
α
h1,htx ≤
1
t
α
hx,ht ≤
1
t
α
h1, for t ∈ 0, 1,x>0;
gtx ≥ t
α
gx,gt ≥ t
α
g1, for t ∈ 0, 1,x>0.
4.10
gtx ≥ t
α
gx,gt ≥ t
α
g1, for t ∈ 0, 1,x>0. 4.11
Let t 1/x, x > 1, one has
gx ≤ x
α
g1, for x ≥ 1. 4.12
Let eiKi, i/T 1, and we define
Q
e
x ∈ P |
1
M
ei ≤ xi ≤ Mei, for 0 ≤ i ≤ T 1
, 4.13
where M>1 is chosen such that
M>max
λT 1g1
T
j1
qj n − 1
j1
l1
C
n−2
j−ln−1
α
λT 1
1α
h1
T
j1
K
−α
j, jqj n − 1
1/1−α
;
λg1
T
j1
qj n − 1
Kj, j
T 1
α
λh1
T
j1
qj n − 1
j1
l1
C
n−2
j−ln−1
−α
−1/1−α
.
4.14
From 4.4 and 4.13, for any x ∈ Q
e
,wehave
1
M
ej ≤ Txj
j1
l1
C
n−2
j−ln−1
xl ≤ Mej
j1
l1
C
n−2
j−ln−1
, for 0 ≤ j ≤ T. 4.15
Chengjun Yuan et al. 9
For any x, y ∈ Q
e
, we define
A
λ
x, yiλ
T
j1
Ki, jqj n − 1gTxj hTyj, for 0 ≤ i ≤ T 1. 4.16
First we show that A
λ
: Q
e
× Q
e
→ Q
e
.
Let x, y ∈ Q
e
, from 4.11 and 4.12,wehave
gTxj ≤ g
Mej
j1
l1
C
n−2
j−ln−1
≤g
M
j1
l1
C
n−2
j−ln−1
≤M
α
j1
l1
C
n−2
j−ln−1
α
g1, for 1≤j ≤ T
,
4.17
and from 4.10,wehave
hTyj ≤ h
1
M
ej
≤ e
−α
jh
1
M
≤ M
α
e
−α
jh1, for 1 ≤ j ≤ T. 4.18
Then, from 4.2 and the above, we have
A
λ
x, yi ≤ λKi, i
T
j1
qj n − 1gTxj T 1hTyj
≤ eiM
α
λT 1
g1
T
j1
qj n − 1
j1
l1
C
n−2
j−ln−1
α
h1
T
j1
e
−α
jqj n − 1
≤ eiM
α
λT 1
g1
T
j1
qj n − 1
j1
l1
C
n−2
j−ln−1
α
h1
T
j1
Kj, j
T 1
−α
qj n − 1
≤ Mei, for 0 ≤ i ≤ T 1.
4.19
On the other hand, for any x, y ∈ Q
e
,from4.10 and 4.12,wehave
gxj ≥ g
1
M
ej
≥ e
α
j
1
M
α
g1, for 1 ≤ j ≤ T, 4.20
hyj ≥ h
Mej
j1
l1
C
n−2
j−ln−1
≥ M
−α
j1
l1
C
n−2
j−ln−1
−α
h1, for 1 ≤ j ≤ T. 4.21
Thus, from 4.2 and 4.16,wehave
A
λ
x, yi ≥ λei
T
j0
qj n − 1gTxj
T
j0
qj n − 1hTyj
≥ λeiM
−α
g1
T
j0
qj
Ki, i
T 1
α
h1
T
j0
qj n − 1
j1
l1
C
n−2
j−ln−1
−α
≥
1
M
ei, for 0 ≤ i ≤ T 1.
4.22
10 Boundary Value Problems
So, A
λ
is well defined and A
λ
Q
e
× Q
e
⊂ Q
e
.
Next, for any l ∈ 0, 1, one has
A
λ
lx, l
−1
y
iλ
T
j1
Ki, jqj n − 1
gTlxj h
T
l
−1
yj
λ
T
j1
Ki, jqj n − 1
glTxj h
l
−1
Tyj
≥ λ
T
j1
Ki, jqj n − 1
l
α
gTxj l
α
hTyj
ds
l
α
A
λ
x, yi, for 0 ≤ i ≤ T 1.
4.23
So the conditions of Theorems 2.2 and 2.3 hold. Therefore, there exists a unique x
∗
λ
∈ Q
e
such
that A
λ
x
∗
,x
∗
x
∗
λ
. It is easy to check that x
∗
λ
is a unique positive solution of 4.5 for given
λ>0. Moreover, Theorem 2.3 means that if 0 <λ
1
<λ
2
, then x
∗
λ
1
t ≤ x
∗
λ
2
t, x
∗
λ
1
t
/
x
∗
λ
2
t and
if α ∈ 0, 1/2,then
lim
λ→0
x
∗
λ
0, lim
λ→∞
x
∗
λ
∞. 4.24
Next, on using Lemma 3.1,from4.5,wegetthaty
∗
λ
Tx
∗
λ
is a unique positive solution
of 1.2 for given λ>0. Moreover, if 0 <λ
1
<λ
2
, then y
∗
λ
1
t ≤ y
∗
λ
2
t, y
∗
λ
1
t
/
y
∗
λ
2
t and if
α ∈ 0, 1/2,then
lim
λ→0
y
∗
λ
0, lim
λ→∞
y
∗
λ
∞. 4.25
This completes the proof.
Example 4.3. Consider the following singular boundary value problem:
Δ
n
yi − 1λ
μy
a
iy
−b
i
0,i∈ N,
Δ
i
y0Δ
n−2
y10, 0 ≤ i ≤ n − 2,
4.26
where λ, a, b > 0, μ ≥ 0, max{a, b} < 1.
Let qi1, gyμy
a
, hyy
−b
, α max{a, b} < 1, then
gty ≥ t
α
gy,h
t
−1
y
≥ t
α
hy, 4.27
thus all conditions in Theorem 4.2 are satisfied. We can find 4.26 has a unique positive so-
lution y
∗
λ
t. In addition, 0 <λ
1
<λ
2
implies y
∗
λ
1
≤ y
∗
λ
2
,y
∗
λ
1
/
y
∗
λ
2
.Ifα max{a, b}∈0, 1/2,
then
lim
λ→0
y
∗
λ
0, lim
λ→∞
y
∗
λ
∞. 4.28
Chengjun Yuan et al. 11
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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