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 Zhang1
1
2
School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, Jilin, China
School of Mathematics and Computer, Harbin University, Harbin 150086, Heilongjiang, China
Correspondence should be addressed to Chengjun Yuan, ycj7102@163.com
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 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 higherorder 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 discrete boundary value problems.
In this paper, we state a unique fixed point theorem for a class of mixed monotone operators, 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 tλqtgyhy 0,
y i 0 y n−2 1 0,
0 < t < 1, λ > 0,
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
Δn yiλqin−1gyin−1 hyin−1 0,
Δk y0 Δn−2 yT 1 0,
i ∈ N {0, 1, 2, . . . , T − 1}, λ > 0,
0 ≤ k ≤ n − 2,
1.2
where n ≥ 2, N {0, 1, 2, . . . , T n}, 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 cone of a Banach space E, and e ∈ P with e ≤ 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 definition see 18.
Definition 2.1 see 18. Assume A : Qe × Qe → Qe . A is said to be mixed monotone if Ax, y
is nondecreasing in x and nonincreasing in y, that is, if x1 ≤ x2 x1 , x2 ∈ Qe implies Ax1 , y ≤
Ax2 , y for any y ∈ Qe , and y1 ≤ y2 y1 , y2 ∈ Qe implies Ax, y1 ≥ Ax, y2 for any x ∈ Qe .
x∗ ∈ Qe is said to be a fixed point of A if Ax∗ , 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
1
A tx, y ≥ tα Ax, y, for x, y ∈ Qe , 0 < t < 1.
2.2
t
Then A has a unique fixed point x∗ ∈ Qe . Moreover, for any x0 , y0 ∈ Qe × Qe ,
xn A xn−1 , yn−1 , yn A yn−1 , xn−1 , n 1, 2, . . . ,
2.3
satisfy
xn −→ x∗ ,
where
xn − x∗ o 1 − r αn ,
yn −→ x∗ ,
2.4
yn − x∗ o 1 − r αn ,
2.5
0 < r < 1, r is a constant from x0 , y0 .
Theorem 2.3 see 6, 14, 18. Suppose that A: Qe × Qe → Qe is a mixed monotone operator and ∃ a
constant α ∈ 0, 1 such that 2.2 holds. If xλ∗ is a unique solution of equation
Ax, x λx,
in Qe , then
xλ∗
− xλ∗ 0 λ > 0
→ 0, λ → λ0 . If 0 < α < 1/2, then 0 < λ1 < λ2 implies
lim xλ∗ ∞.
lim xλ∗ 0,
λ→∞
λ→0
2.6
xλ∗ 1
≥
xλ∗ 2 ,
xλ∗ 1
/ xλ∗ 2 ,
and
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, y0 y1 0, we note that
⎧
⎨t1 − s, 0 ≤ t ≤ s ≤ 1,
Gt, s 3.1
⎩s1 − t, 0 ≤ s ≤ t ≤ 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
xt y n−2 t,
3.3
from y i 0 y n−2 1 0, 0 ≤ i ≤ n−2, and Taylor Formula, we define operator T : C2 0, 1 →
Cn 0, 1, by
t
t − sn−3
yt T xt xsds, for 3 ≤ n,
3.4
0 n − 3!
yt T xt xt,
for n 2..
Then we have
x2 t λft, T xt 0,
0 < t < 1, λ > 0,
x0 x1 0.
3.5
Then from 3.4, we have the next lemma.
Lemma 3.1. If xt is a solution of 3.5, then yt is a solution of 1.1.
Further, if yt is a solution of 1.1, imply that xt 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 ≥ tα gx,
h t−1 x ≥ tα hx,
for any t ∈ 0, 1 and x > 0, and q ∈ C0, 1, 0, ∞ satisfies
1
−α
n−1
s n − 2s qsds < ∞.
3.6
3.7
3.8
0
Then 1.1 has a unique positive solution yλ∗ t. And moreover, 0 < λ1 < λ2 implies yλ∗1 ≤ yλ∗2 , yλ∗1 /
yλ∗2 .
If α ∈ 0, 1/2, then
lim yλ∗ 0,
lim yλ∗ ∞.
3.9
λ→0
λ→∞
4
Boundary Value Problems
Proof. Since 3.7 holds, let t−1 x y, one has
hy ≥ tα hty.
3.10
Then
hty ≤
1
hy,
tα
for t ∈ 0, 1, y > 0.
3.11
Let y 1. The above inequality is
ht ≤
1
h1,
tα
for t ∈ 0, 1.
3.12
From 3.7, 3.11, and 3.12, one has
1
≥ tα h1,
h
t
h t−1 x ≥ tα hx,
htx ≤
1
hx,
tα
ht ≤
1
h1, for t ∈ 0, 1, x > 0.
tα
3.13
Similarly, from 3.6, one has
gtx ≥ tα gx,
gt ≥ tα g1,
for t ∈ 0, 1, x > 0.
3.14
Let t 1/x, x > 1, one has
gx ≤ xα g1,
for x ≥ 1.
3.15
Let et Gt, t t1 − t, and we define
Qe 1
x ∈ C0, 1 |
Gt, t ≤ xt ≤ MGt, t, t ∈ 0, 1 ,
M
3.16
where M > 1 is chosen such that
M > max
1
λg1
qsds λh1
0
1
0
sn−1 n − 2s
n!
−α
1/1−α
qsds
,
α
−1/1−α 1
1
sn−1 n − 2s
λg1 Gs, s
qsds λh1 Gs, sqsds
.
n!
0
0
3.17
First, from 3.4 and 3.16, for any x ∈ Qe , we have the following.
When 3 ≤ n,
t
1
t − sn−3
Gs, s
ds ≤ T xt
n − 3!
0M
t
tn−1 n − 2t
t − sn−3
ds ≤ M
≤ M,
≤ MGs, s
n − 3!
n!
0
1 tn−1 n − 2t
≤
M
n!
3.18
for t ∈ 0, 1,
Chengjun Yuan et al.
5
when n 2,
tn − 2t
1 tn−1 n − 2t
≤ T xt xt ≤ M
≤ M,
M
n!
n!
for t ∈ 0, 1,
3.19
then
1 tn−1 n − 2t
tn−1 n − 2t
≤ T xt ≤ M
≤ M,
M
n!
n!
for t ∈ 0, 1.
For any x, y ∈ Qe , we define
1
Aλ x, yt λ Gt, sqsgT xs hT ysds,
for t ∈ 0, 1.
3.20
3.21
0
First, we show that Aλ : Qe × Qe → Qe .
Let x, y ∈ Qe , from 3.14, 3.15, and 3.20, we have
gT xt ≤ gM ≤ Mα g1,
and from 3.13, we have
for t ∈ 0, 1,
−α 1 tn−1 n − 2t
1
tn−1 n − 2t
hT yt ≤ h
h
≤
M
n!
n!
M
−α
n−1
n − 2t
α t
≤M
h1, for t ∈ 0, 1.
n!
Then, from 3.2, 3.21, 3.22 and 3.23, we have
−α
1
1
sn−1 n − 2s
α
α
Aλ x, yt ≤ λGt, t
M g1qsds M
h1qsds
n!
0
0
≤ MGt, t,
3.22
3.23
3.24
for t ∈ 0, 1.
On the other hand, for any x, y ∈ Qe , from 3.13 and 3.14, we have
α α
tn−1 n − 2t
1 tn−1 n − 2t
1
1
tn−1 n − 2t
≥
gT xt ≥ g
g
g1,
≥
M
n!
n!
M
n!
Mα
1
1
≥
h1, for t ∈ 0, 1.
hT yt ≥ hM h
1/M
Mα
3.25
Thus, from 3.2, 3.21 and 3.25, we have
Aλ x, yt
1
≥ λGt, t
Gs, sqsM
−α
0
≥
1
Gt, t,
M
sn−1 n − 2s
n!
α
g1ds for t ∈ 0, 1.
So, Aλ is well defined and Aλ Qe × Qe ⊂ Qe .
1
0
Gs, sqsM−α h1ds
3.26
6
Boundary Value Problems
Next, for any l ∈ 0, 1, one has
1
−1
Aλ lx, l y t λ Gt, sqs glT xs h l−1 T ys ds
0
1
≥ λ Gt, sqs lα gT xs lα hT ys ds
3.27
0
l Aλ x, yt,
α
for t ∈ 0, 1.
So the conditions of Theorems 2.2 and 2.3 hold. Therefore, there exists a unique xλ∗ ∈ Qe
such that Aλ x∗ , x∗ xλ∗ . It is easy to check that xλ∗ is a unique positive solution of 3.5 for
given λ > 0. Moreover, Theorem 2.3 means that if 0 < λ1 < λ2 , then xλ∗ 1 t ≤ xλ∗ 2 t, xλ∗ 1 t /
xλ∗ 2 t
and if α ∈ 0, 1/2, then
lim xλ∗ 0,
λ→0
lim xλ∗ ∞.
λ→∞
3.28
Next, from Lemma 3.1 and 3.4, we get that yλ∗ T xλ∗ is a unique positive solution
of 1.1 for given λ > 0. Moreover, if 0 < λ1 < λ2 , then yλ∗1 t ≤ yλ∗2 t, yλ∗1 t /
yλ∗2 t and if
α ∈ 0, 1/2, then
lim yλ∗ 0,
λ→0
lim yλ∗ ∞.
λ→∞
3.29
This completes the proof.
Example 3.3. Consider the following singular boundary value problem:
y n t λ μy a t y −b t 0,
y i 0 y n−2 1 0,
t ∈ 0, 1,
0 ≤ i ≤ n − 2,
3.30
where λ, a, b > 0, μ ≥ 0, max{a, b} < 1/n − 1.
Applying Theorem 3.2, let α max{a, b} < 1/n − 1, qt 1, gy μy a , hy y −b ,
then
gty ≥ tα gy,
h t−1 ≥ tα hy,
1
−α
n−1
s n − 2s ds < ∞.
3.31
0
Thus all conditions in Theorem 3.2 are satisfied. We can find 3.30 has a unique positive solution yλ∗ t. In addition, 0 < λ1 < λ2 implies yλ∗1 ≤ yλ∗2 , yλ∗1 / yλ∗2 . If α max{a, b} ∈ 0, 1/2,
then
lim yλ∗ 0,
λ→0
lim yλ∗ ∞.
λ→∞
3.32
Chengjun Yuan et al.
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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 −Δ2 yi ui 1 0, i ∈ N, y0 yT 1 0,
we note that
⎧
jT 1 − i
⎪
⎪
, 0 ≤ j ≤ i − 1,
⎨
T 1
Ki, j ⎪
⎪
⎩ iT 1 − j , i ≤ j ≤ T 1,
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
xi Δn−2 yi,
for 0 ≤ i ≤ T 1.
4.3
From Δi y0 Δn−2 yT 1 0, 0 ≤ i ≤ n − 2, and Δm yi − 1 Δm−1 yi − Δm−1 yi − 1, so we
define operator T , by
T xi yi n − 1 i1
n−2
Ci−ln−1
xl,
for 0 ≤ i ≤ T.
4.4
l1
Then
Δ2 xi λFi n − 1, T xi 0,
0 ≤ i ≤ T − 1, λ > 0,
x0 xT 1 0.
4.5
Lemma 4.1. If xi is a solution of 4.5, then yi is a solutionn of 1.2.
Proof. Since we remark that xi is a solution of 4.5, if and only if
xi T
Ki, jλFj n − 1, T xj,
for 0 ≤ i ≤ T 1.
4.6
j1
Let
T xi yi n − 1,
for 0 ≤ i ≤ T.
4.7
From 4.4 we find Δi y0 Δn−2 yT 1 0, 0 ≤ i ≤ n − 2, and xi Δn−2 yi, so that yi is
a solution of 1.2.
Further, if yi is a solution of 1.2, imply that xi 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 ≥ tα gx,
h t−1 x ≥ tα gx,
4.8
for any t ∈ 0, 1 and x > 0, and q ∈ CN , 0, ∞.
Then 1.2 has a unique positive solution yλ∗ i. And moreover, 0 < λ1 < λ2 implies yλ∗1 ≤ yλ∗2 ,
∗
yλ1 / yλ∗2 . If α ∈ 0, 1/2, then
lim yλ∗ 0,
lim yλ∗ ∞.
λ→0
4.9
λ→∞
Proof. The proof is the same as that of Theorem 3.2, from 4.12 and 4.13, one has
h t−1 x ≥ tα hx,
1
≥ tα h1,
h
t
htx ≤
1
hx,
tα
gtx ≥ tα gx, gt ≥ tα g1,
ht ≤
1
h1,
tα
for t ∈ 0, 1, x > 0;
for t ∈ 0, 1, x > 0.
4.10
gtx ≥ t gx, gt ≥ t g1,
α
for t ∈ 0, 1, x > 0.
α
4.11
Let t 1/x, x > 1, one has
gx ≤ xα g1,
for x ≥ 1.
4.12
Let ei Ki, i/T 1, and we define
Qe 1
ei ≤ xi ≤ Mei, for 0 ≤ i ≤ T 1 ,
x∈P |
M
4.13
where M > 1 is chosen such that
M > max
j1
α
T
n−2
λT 1g1 qj n − 1
Cj−ln−1
j1
λT 11α h1
l1
T
1/1−α
K −α j, jqj n − 1
;
j1
j1
−α −1/1−α T
T
Kj, j α
n−2
λg1 qj n − 1
λh1 qj n − 1
Cj−ln−1
.
T 1
j1
j1
l1
4.14
From 4.4 and 4.13, for any x ∈ Qe , we have
j1
j1
1
n−2
n−2
Cj−ln−1
xl ≤ Mej Cj−ln−1
,
ej ≤ T xj M
l1
l1
for 0 ≤ j ≤ T.
4.15
Chengjun Yuan et al.
9
For any x, y ∈ Qe , we define
Aλ x, yi λ
T
Ki, jqj n − 1gT xj hT yj,
for 0 ≤ i ≤ T 1.
4.16
j1
First we show that Aλ : Qe × Qe → Qe .
Let x, y ∈ Qe , from 4.11 and 4.12, we have
α
j1
j1
j1
n−2
n−2
n−2
Cj−ln−1 g1,
gT xj ≤ g Mej Cj−ln−1 ≤ g M Cj−ln−1 ≤ Mα
l1
l1
for 1 ≤ j ≤ T ,
l1
4.17
and from 4.10, we have
1
1
−α
ej ≤ e jh
≤ Mα e−α jh1, for 1 ≤ j ≤ T.
hT yj ≤ h
M
M
Then, from 4.2 and the above, we have
T
Aλ x, yi ≤ λKi, i qj n − 1gT xj T 1hT yj
j1
≤ eiM λT 1 g1
α
T
j1
qj n − 1
j1
h1
≤ eiMα λT 1 g1
T
e jqj n − 1
j1
qj n − 1
j1
h1
α
T 1
4.19
n−2
Cj−ln−1
l1
T Kj, j −α
j1
≤ Mei,
−α
j1
α
n−2
Cj−ln−1
l1
T
4.18
qj n − 1
for 0 ≤ i ≤ T 1.
On the other hand, for any x, y ∈ Qe , from 4.10 and 4.12, we have
1
1
ej ≥ eα j α g1, for 1 ≤ j ≤ T,
gxj ≥ g
M
M
j1
−α
j1
n−2
−α
n−2
Cj−ln−1
h1, for 1 ≤ j ≤ T.
hyj ≥ h Mej Cj−ln−1 ≥ M
l1
4.20
4.21
l1
Thus, from 4.2 and 4.16, we have
T
T
qj n − 1gT xj qj n − 1hT yj
Aλ x, yi ≥ λei
j0
≥ λeiM
≥
1
ei,
M
−α
j0
j1
−α T
T
Ki, i α
n−2
g1 qj
h1 qj n − 1
Cj−ln−1
T 1
j0
j0
l1
for 0 ≤ i ≤ T 1.
4.22
10
Boundary Value Problems
So, Aλ is well defined and Aλ Qe × Qe ⊂ Qe .
Next, for any l ∈ 0, 1, one has
T
Aλ lx, l−1 y i λ Ki, jqj n − 1 gT lxj h T l−1 yj
j1
T
λ Ki, jqj n − 1 glT xj h l−1 T yj
j1
4.23
T
≥ λ Ki, jqj n − 1 lα gT xj lα hT yj ds
j1
lα Aλ x, yi,
for 0 ≤ i ≤ T 1.
So the conditions of Theorems 2.2 and 2.3 hold. Therefore, there exists a unique xλ∗ ∈ Qe such
that Aλ x∗ , x∗ xλ∗ . It is easy to check that xλ∗ is a unique positive solution of 4.5 for given
λ > 0. Moreover, Theorem 2.3 means that if 0 < λ1 < λ2 , then xλ∗ 1 t ≤ xλ∗ 2 t, xλ∗ 1 t /
xλ∗ 2 t and
if α ∈ 0, 1/2, then
lim xλ∗ 0,
λ→0
lim xλ∗ ∞.
λ→∞
4.24
Next, on using Lemma 3.1, from 4.5, we get that yλ∗ T xλ∗ is a unique positive solution
of 1.2 for given λ > 0. Moreover, if 0 < λ1 < λ2 , then yλ∗1 t ≤ yλ∗2 t, yλ∗1 t /
yλ∗2 t and if
α ∈ 0, 1/2, then
lim yλ∗ 0,
λ→0
lim yλ∗ ∞.
λ→∞
4.25
This completes the proof.
Example 4.3. Consider the following singular boundary value problem:
Δn yi − 1 λ μy a i y −b i 0,
Δi y0 Δn−2 y1 0,
i ∈ N,
0 ≤ i ≤ n − 2,
4.26
where λ, a, b > 0, μ ≥ 0, max{a, b} < 1.
Let qi 1, gy μy a , hy y −b , α max{a, b} < 1, then
gty ≥ tα gy,
h t−1 y ≥ tα hy,
4.27
thus all conditions in Theorem 4.2 are satisfied. We can find 4.26 has a unique positive solution yλ∗ t. In addition, 0 < λ1 < λ2 implies yλ∗1 ≤ yλ∗2 , yλ∗1 /
yλ∗2 . If α max{a, b} ∈ 0, 1/2,
then
lim yλ∗ 0,
λ→0
lim yλ∗ ∞.
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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