Hindawi Publishing Corporation
Boundary Value Problems
Volume 2008, Article ID 194234, 8 pages
doi:10.1155/2008/194234
Research Article
Existence and Uniqueness of Solutions
for Boundary Value Problems to the Singular
One-Dimension p-Laplacian
Xiaoning Lin,1 Weizhi Sun,2 and Daqing Jiang3
1
School of Business, Northeast Normal University, Changchun 130024, China
Department of Mathematics, Changchun University of Science and Technology,
Changchun 130022, China
3
School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China
2
Correspondence should be addressed to Xiaoning Lin, linxiaoning@126.com
Received 26 June 2007; Accepted 29 January 2008
Recommended by Kanishka Perera
In this paper, We study the existence and uniqueness of solutions for boundary value problems to
the singular one-dimension p-Laplacian by using mixed monotone method. Our results improve
several recent results established in the literature.
Copyright q 2008 Xiaoning Lin 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 this paper, we discuss the existence and uniqueness of solution to the boundary value problem
φ x
λqtfx 0, t ∈ 0, 1, λ > 0,
1.1
x0 x1 0,
where φs |s|p−2 s, p > 1, and f may be singular at x 0.
By a solution x to 1.1 we mean a function x ∈ C1 0, 1, φx ∈ AC0, 1 such that
x satisfies 1.1 and the boundary condition; here AC0, 1 denotes the space of absolutely
continuous functions defined on 0, 1.
It is of interest to note here that the existence of positive solutions to problem 1.1 has
been studied in great detail in the literature, see 1–10. However, there are few works on the
uniqueness of solutions for boundary problems to the singular one-dimension p-Laplacian.
In this paper, we present a new existence and uniqueness theory by using mixed monotone
method which has been used in 11, 12.
2
Boundary Value Problems
2. Preliminaries
Let E C0, 1, with the norm u maxt∈0,1 |ut|, so E is a Banach space. Also, we define
P {u ∈ E : u is concave on 0, 1 and u0 u1 0}.
2.1
One may readily verify that P is a cone in E. 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}.
2.2
Now we give a definition see 13.
Definition 2.1. 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 11. Suppose that A : Qe × Qe → Qe is a mixed monotone operator and there
exists a constant β, 0 ≤ β < 1 such that
1
A tx, y ≥ tβ Ax, y,
t
∀x, y ∈ Qe , 0 < t < 1.
2.3
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, . . . ,
2.4
satisfy
xn −→ x∗ ,
yn −→ x∗ ,
2.5
yn − x ∗ o 1 − r β n ,
2.6
where
xn − x ∗ o 1 − r β n ,
0 < r < 1, r is a constant from x0 , y0 .
Theorem 2.3 see 11, 13. Suppose that A : Qe × Qe → Qe is a mixed monotone operator and there
exists a constant β ∈ 0, 1 such that 2.3 holds. If xλ∗ is a unique solution of equation
Ax, x λx
λ > 0
2.7
in Qe , then xλ∗ − xλ∗ 0 → 0, λ → λ0 . If 0 < β < 1/2, then 0 < λ1 < λ2 implies xλ∗ 1 ≥ xλ∗ 2 , xλ∗ 1 /
xλ∗ 2 ,
and
lim xλ∗ 0,
λ→∞
lim xλ∗ ∞.
λ→0
2.8
Xiaoning Lin et al.
3
3. Existence and uniqueness
In this section, we discuss the singular one-dimension p-Laplacian
qtfx 0,
φ x
t ∈ 0, 1,
3.1
x0 x1 0.
Throughout this section we assume that
fx gx hx,
3.2
where
g : 0, ∞ −→ 0, ∞ is continuous and nondecreasing ;
h : 0, ∞ −→ 0, ∞ is continuous and nonincreasing .
3.3
Theorem 3.1. Suppose that there exists α ∈ 0, p − 1 such that
3.4
gtx ≥ tα gx,
h t−1 x ≥ tα hx,
3.5
for any t ∈ 0, 1 and x > 0, and q ∈ C0, 1, 0, ∞ satisfies
1
t−α 1 − t−α qtdt < ∞,
0 < α < p − 1.
3.6
0
Then 3.1 has a unique positive solution xλ∗ t.
And moreover, 0 < λ1 < λ2 implies that xλ∗ 1 ≤ xλ∗ 2 , xλ∗ 1 /
xλ∗ 2 . If α/p − 1 ∈ 0, 1/2, then
lim xλ∗ 0,
lim xλ∗ ∞.
λ→0
Lemma 3.2. Let u, v be solutions to
λqtt−α 1 − t−α 0,
φ u
3.7
λ→∞
t ∈ 0, 1, λ > 0, qtt−α 1 − t−α ∈ L1 0, 1,
u0 u1 0,
φ v
λqttα 1 − tα 0, t ∈ 0, 1, λ > 0, qttα 1 − tα ∈ L1 0, 1,
3.8
v0 v1 0,
then there exist positive constants Cu , Cv such that
t1 − tu ≤ ut ≤ Cu t1 − t,
t1 − tv ≤ vt ≤ Cv t1 − t.
3.9
Proof. Because qtt−α 1 − t−α qttα 1 − tα ≥ 0, then u, v is concave and positive on 0, 1. As
u, v ∈ C1 0, 1, thus
lim
t→0
ut
u 0,
t
lim
t→1
ut
−u 1,
1−t
lim
t→0
vt
v 0,
t
lim
t→1
vt
−v 1.
1−t
3.10
4
Boundary Value Problems
Let
ut
,
t∈0,1 t1 − t
vt
,
t∈0,1 t1 − t
Cu sup
Cv sup
3.11
ut ≤ Cu t1 − t,
vt ≤ Cv t1 − t.
3.12
ut ≥ t1 − tu,
vt ≥ t1 − tv.
3.13
t1 − tv ≤ vt ≤ Cv t1 − t.
3.14
then 0 < Cu , Cv < ∞, and
Since u, v is concave, then
So we have
t1 − tu ≤ ut ≤ Cu t1 − t,
Lemma 3.3 see 7. If u, v ∈ C1 0, 1 satisfies
− φ u
≥ − φ v ,
u0 ≥ v0,
a.e. t ∈ 0, 1,
u1 ≥ v1,
3.15
then ut ≥ vt for all t ∈ 0, 1.
Proof of Theorem 3.1. Since 3.5 holds, let t−1 x y, one has
hy ≥ tα hty.
3.16
Then
hty ≤
1
hy,
tα
∀t ∈ 0, 1, y > 0.
3.17
Let y 1. The above inequality is
ht ≤
1
h1,
tα
∀t ∈ 0, 1.
3.18
From 3.5, 3.17, and 3.18, one has
h t−1 x ≥ tα hx,
1
≥ tα h1,
h
t
htx ≤
1
hx,
tα
ht ≤
1
h1,
tα
t ∈ 0, 1, x > 0.
3.19
Similarly, from 3.4, one has
gtx ≥ tα gx,
gt ≥ tα g1,
t ∈ 0, 1, x > 0.
3.20
Xiaoning Lin et al.
5
Let t 1/x, x > 1, one has
gx ≤ xα g1,
x ≥ 1.
3.21
Let et t1 − t, and we define
1
Qe x ∈ C0, 1 |
t1 − t ≤ xt ≤ Mt1 − t, t ∈ 0, 1 ,
M
3.22
where M > 1 is chosen such that
p−1/p−1−α 1/p−1−α
−1/p−1−α , v−p−1/p−1−α g1 h1
M > max Cu
g1 h1
. 3.23
For any fixed x, y ∈ Qe ,consider the following boundary value problem:
φ w t λqt g xt h yt 0,
t ∈ 0, 1, λ > 0;
w0 w1 0.
3.24
By 3.18–3.21, for x, y ∈ Qe , we can obtain
g xt ≤ g Mt1 − t ≤ gM ≤ Mα g1, t ∈ 0, 1,
1
1
h yt ≤ h
t1 − t ≤ t−α 1 − t−α h
M
M
≤ Mα t−α 1 − t−α h1,
3.25
t ∈ 0, 1.
So,
g xt h yt ≤ Mα g1 t−α 1 − t−α h1
≤ Mα t−α 1 − t−α g1 h1,
t ∈ 0, 1,
3.26
then λqtgxthyt ∈ L1 0, 1. It follows from 7 that, for each fixed x, y ∈ Qe , problem
3.22 has a solution w ∈ C1 0, 1, and 3.24 is equivalent to
wt λ
1/p−1
t
1
τ qr g xr h yr dr ds,
φ
−1
0
0 ≤ t ≤ 1,
3.27
s
where τ φw 1 is a solution of the equation
1
1
φ−1 τ qr g xr h yr dr ds 0.
0
3.28
s
For any x, y ∈ Qe , we define
Ax, yt wt λ1/p−1
t
0
1
φ−1 τ qr g xr h yr dr ds,
3.29
s
then Ax, yt is concave on 0, 1, for any x, y ∈ Qe × Qe , qtgxt hyt ∈ L1 0, 1.
6
Boundary Value Problems
First, we show for any x, y ∈ Qe , Ax, y ∈ Qe .
Let x, y ∈ Qe , from 3.19 and 3.20, we have
1
1
1
t1 − t ≥ tα 1 − tα g
≥ tα 1 − tα α g1,
g xt ≥ g
M
M
M
1
1
≥
h yt ≥ h Mt1 − t ≥ hM h
h1, t ∈ 0, 1.
1/M
Mα
3.30
Thus, we have
1 g xt h yt ≥
g1tα 1 − tα h1 .
α
M
3.31
So, we can obtain
∗ g xt h yt ≤ Mα g1 t−α 1 − t−α h1
≤ Mα t−α 1 − t−α g1 h1 ,
1 g1tα 1 − tα h1
∗∗ g xt h yt ≥
α
M
≥ M−α tα 1 − tα g1 h1 ,
t ∈ 0, 1,
3.32
t ∈ 0, 1.
So
− φ w ≤ qtMα t−α 1 − t−α g1 h1 ,
3.33
− φ w ≤ −Mα g1 h1 φ u .
3.34
− φ w ≥ qtM−α tα 1 − tα g1 h1 ,
3.35
− φ w ≥ −M−α g1 h1 φ v .
3.36
that is,
Similarly,
that is,
By Lemma 3.3,
1/p−1
wt ≤ Mα/p−1 g1 h1
ut
1/p−1 α/p−1
≤ Cu g1 h1
M
t1 − t ≤ Mt1 − t,
1/p−1
vt
wt ≥ M−α/p−1 g1 h1
1/p−1
1
t1 − t.
t1 − t ≥
≥ vM−α/p−1 g1 h1
M
So, the operator A is well defined.
3.37
Xiaoning Lin et al.
7
Next, for any l ∈ 0, 1, one has
qt g lxt h l−1 yt ≥ lα qt g xt h yt .
3.38
Then by Lemma 3.3 we have
A lx, l−1 y t ≥ lα/p−1 Ax, yt
0<β
α
<1 .
p−1
3.39
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 3.1
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/p − 1 ∈ 0, 1/2, then
lim xλ∗ 0,
λ→0
lim xλ∗ ∞.
λ→∞
3.40
This completes the proof.
Example 3.4. Consider the following singular p-Laplace boundary value problem:
λqt μxa x−b 0,
φ x
t ∈ 0, 1;
x0 x1 0,
3.41
where λ, a, b > 0, μ ≥ 0, q ∈ C0, 1, q > 0, t ∈ 0, 1, and
1
qtt−α 1 − t−α dt < ∞,
0 < α max{a, b} < p − 1.
3.42
0
Applying Theorem 3.1, we can find that 3.41 has a unique positive solution xλ∗ t. In
addition, 0 < λ1 < λ2 implies xλ∗ 1 ≤ xλ∗ 2 , xλ∗ 1 / xλ∗ 2 . If α/p − 1 ∈ 0, 1/2, then
lim xλ∗ 0,
λ→0
lim xλ∗ ∞.
λ→∞
3.43
To see that, we put
β
α
,
p−1
gx μxa ,
hx x−b .
3.44
Thus 0 < β < 1 and
gtx ta gx ≥ tα gx,
ht−1 x tb hx ≥ tα hx,
3.45
for any t ∈ 0, 1 and x > 0, thus all conditions in Theorem 3.1 are satisfied.
Acknowledgment
The work was supported by the NSFC of China no. 10571021and Key Laboratory for Applied
Statistics of MOEKLAS and the NSFC of China no. 10271023.
8
Boundary Value Problems
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