Introduction Multipoint boundary value problems BVPs for second-order differential equations in a finite interval have been studied extensively and many results for the existence of solut
Trang 1Volume 2009, Article ID 609143, 11 pages
doi:10.1155/2009/609143
Research Article
Problems on a Half-Line
Changlong Yu, Yanping Guo, and Yude Ji
College of Sciences, Hebei University of Science and Technology, Shijiazhuang 050018, Hebei, China
Correspondence should be addressed to Changlong Yu,changlongyu@126.com
Received 5 April 2009; Accepted 6 July 2009
Recommended by Paul Eloe
By using the Leray-Schauder continuation theorem, we establish the existence of solutions for
m-point boundary value problems on a half-line xt ft, xt, xt 0, 0 < t < ∞, x0
m−2
i1 α i xη i , lim t → ∞ xt 0, where α i ∈ R,m−2
i1 α i / 1 and 0 < η1 < η2< · · · < η m−2 < ∞ are
given
Copyrightq 2009 Changlong Yu 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
Multipoint boundary value problems BVPs for second-order differential equations in a finite interval have been studied extensively and many results for the existence of solutions, positive solutions, multiple solutions are obtained by use of the Leray-Schauder continuation theorem, Guo-Krasnosel’skii fixed point theorem, and so on; for details see 1 4 and the references therein
In the last several years, boundary value problems in an infinite interval have been arisen in many applications and received much attention; see 5, 6 Due to the fact that
an infinite interval is noncompact, the discussion about BVPs on the half-line is more complicated, see 5 14 and the references therein Recently, in 15, Lian and Ge studied the following three-point boundary value problem:
xt ft, x t, xt 0, 0 < t < ∞,
x 0 αxη
t → ∞ xt 0,
1.1
Trang 2where α ∈ R, α / 1, and η ∈ 0, ∞ are given In this paper, we will study the following
m-point boundary value problems:
xt ft, x t, xt 0, 0 < t < ∞,
x0 m−2
i1
α i x
η i
where α i ∈ R, m−2
i1 α i / 1, α i have the same signal, and 0 < η1 < η2 < · · · < η m−2 < ∞ are
given We first present the Green function for second-order multipoint BVPs on the half-line and then give the existence results for1.2 using the properties of this Green function and the Leray-Schauder continuation theorem
We use the space C1
∞0, ∞ {x ∈ C10, ∞, lim t → ∞ xt exists, lim t → ∞ xt exists}
with the normx max{x∞, x∞}, where · ∞is supremum norm on the half-line, and
L10, ∞ {x : 0, ∞ → R is absolutely integrable on 0, ∞} with the norm x L1
∞
0|xt|dt.
We set
P
∞
0
p sds, P1
∞
0
sp sds, Q
∞
0
q sdt, 1.3
and we suppose α i , i 1, 2, , m − 2 are the same signal in this paper and we always assume
α m−2
i1 α i
2 Preliminary Results
In this section, we present some definitions and lemmas, which will be needed in the proof
of the main results
Definition 2.1see 15 It holds that f : 0, ∞ × R2 −→ R is called an S-Carath´eodory
function if and only if
i for each u, v ∈ R2, t → ft, u, v is measurable on 0, ∞,
ii for almost every t ∈ 0, ∞, u, v → ft, u, v is continuous on R2,
iii for each r > 0, there exists ϕ r t ∈ L10, ∞ with tϕ r t ∈ L10, ∞, ϕ r t > 0 on
0, ∞ such that max{|u|, |v|} ≤ r implies |ft, u, v| ≤ ϕ r t, for a.e t ∈ 0, ∞.
i1 α i / 1, if for any vt ∈ L10, ∞ with tvt ∈ L10, ∞, then the
BVP,
xt vt 0, 0 < t < ∞,
x0 m−2
i1
α i x
η i
Trang 3
has a unique solution Moreover, this unique solution can be expressed in the form
x t
∞
0
G t, svsds, 2.2
where Gt, s is defined by
G t, s Λ1
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
m−2
i1
m−2
i1
i
k1
α k η k m−2
ki1
α k s Λs, 0 < η i ≤ s ≤ η i1 , s ≤ t, i 1, 2, , m − 3,
i
k1
α k η k m−2
ki1
α k s Λt, 0 < η i ≤ s ≤ η i1 , t ≤ s, i 1, 2, , m − 3,
m−2
i1
α i η i Λs, s ≥ η m−2 , s ≤ t,
m−2
i1
α i η i Λt, s ≥ η m−2 , t ≤ s,
2.3
here noteΛ 1 −m−2
i1 α i Proof Integrate the di fferential equation from t to ∞, noticing that vt, tvt ∈ L10, ∞, then from 0 to t and one has
x t x0
t
0
∞
s
v τdτ ds. 2.4
Since x0 m−2
i1 α i xη i, from 2.4, it holds that
x t 1
1−m−2
i1 α i
m−2
i1
α i η i
∞
η i
v sds m−2
i1
α i
η i
0
sv sds t
∞
t
v sds
t
0
sv sds.
2.5 For 0≤ t ≤ η1, the unique solution of2.1 can be stated by
x t
t
0
m−2
i1 α i s
1−m−2
i1 α i
s
v sds
η1
t
m−2
i1 α i s
1−m−2 i1 α i
t
v sds
m−3
i1
η i1
η i
i
k1 α k η km−2
ki1 α k s Λt
1−m−2 i1 α i
v sds
∞
η m−2
m−2
i1 α i η i
1−m−2 i1 α i
t
v sds.
2.6
Trang 4If η i ≤ t ≤ η i1 , 1 ≤ i ≤ m − 3, the unique solution of 2.1 can be stated by
x t
η1
0
m−2
i1 α i s
1−m−2 i1 α i
s
v sds
i−1
j1
η j1
η j
⎛
⎝
j k1 α k η km−2
kj1 α k s Λs
1−m−2 i1 α i
⎞
⎠vsds
t
η i
i
k1 α k η km−2
ki1 α k s Λs
1−m−2 i1 α i
v sds
η i1
t
i k1 α k η km−2
ki1 α k s Λt
1−m−2 i1 α i
v sds
m−3
ji1
η j1
η j
⎛
⎝
j k1 α k η km−2
kj1 α k s Λt
1−m−2 i1 α i
⎞
⎠vsds
∞
η m−2
m−2
i1 α i η i
1−m−2 i1 α i
t
v sds.
2.7
If η m−2 ≤ t < ∞, the unique solution of 2.1 can be stated by
x t
η1
0
m−2
i1 α i s
1−m−2
i1 α i
s
v sds m−3
i1
η i1
η i
i k1 α k η km−2
ki1 α k s Λs
1−m−2 i1 α i
v sds
t
η m−2
m−2
i1 α i η i
1−m−2 i1 α i
s
v sds
∞
t
m−2
i1 α i η i
1−m−2 i1 α i
t
v sds.
2.8
We noteΛ 1 −m−2
i1 α i , then
G t, s Λ1
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
m−2
i1
m−2
i1
i
k1
α k η k m−2
ki1
α k s Λs, 0 < η i ≤ s ≤ η i1 , s ≤ t, i 1, 2, , m − 3,
i
k1
α k η k m−2
ki1
α k s Λt, 0 < η i ≤ s ≤ η i1 , t ≤ s, i 1, 2, , m − 3,
m−2
i1
α i η i Λs, s ≥ η m−2 , s ≤ t,
m−2
i1
α i η i Λt, s ≥ η m−2 , t ≤ s.
2.9
Trang 5Therefore, the unique solution of 2.1 is xt ∞0 Gt, svsds, which completes the
proof
Remark of Lemma 2.2 Obviously Gt, s satisfies the properties of a Green function, so we call Gt, s the Green function of the corresponding homogeneous multipoint BVP of 2.1 on the half-line
Lemma 2.3 For all t, s ∈ 0, ∞, it holds that
|Gt, s| ≤
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
s,
m−2
i1
α i < 0, s
m−2
i1
α i < 1,
max
⎧
⎨
⎩
m−2
i1 α i s
−Λ ,
m−2
i1 α i η m−2
−Λ
⎫
⎬
⎭,
m−2
i1
α i > 1.
2.10
Proof For each s ∈ 0, ∞, Gt, s is nondecreasing in t Immediately, we have
min
m−2
i1 α i s
Λ ,
i
k1 α k η km−2
ki1 α k s
m−2
i1 α i η i
Λ
≤ Gt, s ≤ Gs, s
Λ1
⎧
⎪
⎪
⎪
⎪
⎪
⎪
i
k1
α k η k
m−2
ki1
α k Λ
s, η i ≤ s ≤ η i1 < ∞, i 1, 2, , m − 3,
m−2
i1
α i η i Λs, s ≥ η m−2
2.11
Further, we have
m−2
i1 α i s
Λ ≤ Gt, s ≤ s, m−2
i1
α i < 0,
0 < min
m−2
i1 α i s
Λ ,
m−2
i1 α i η1
Λ
≤ Gt, s ≤ Λs , 0≤m−2
i1
α i < 1,
min
m−2
i1 α i s
Λ ,
m−2
i1 α i η m−2
Λ
≤ Gt, s ≤ s, m−2
i1
α i > 1.
2.12
Therefore, we get the result
Trang 6Lemma 2.4 For the Green function Gt, s, it holds that
lim
t → ∞ G t, s Gs
Λ1
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
i
k1
α k η k
m−2
ki1
α k Λ
s, η i ≤ s ≤ η i1 < ∞, i 1, 2, , m − 3
m−2
i1
α i η i Λs, s ≥ η m−2
2.13
x0 m−2
i1
α i x
η i
2.14
and α i i 1, 2, , m − 2 have the same signal, 0 < η1 < η2 < · · · < η m−2 < ∞, then there exists
η ∈ η1, η m−2 satisfying
x 0 αxη
where α m−2
i1 α i
Proof Let α i i 1, 2, , m − 2 are positive, and note M∗ max{xt | t ∈ η1, η m−2 }, m∗ min{xt | t ∈ η1, η m−2 }, then for every i i 1, 2, , m − 2, we have m∗ ≤ xη i ≤ M∗,
so m∗m−2
i1 α i ≤ m−2
i1 α i xη i ≤ M∗m−2
i1 α i , that is, m∗ ≤ m−2
i1 α i xη i /m−2
i1 α i x ≤ M∗.
Because xt is continuous on the interval η1, η m−2 , there exists η ∈ η1, η m−2 satisfying
x0 αxη, where α m−2
i1 α i
relatively compact in X if the following conditions hold:
a M is uniformly bounded in C∞0, ∞;
b the functions from M are equicontinuous on any compact interval of 0, ∞;
c the functions from M are equiconvergent, that is, for any given > 0, there exists a T
T > 0 such that |ft − f∞| < , for any t > T, f ∈ M.
3 Main Results
Consider the space X {x ∈ C1
∞0, ∞, x0 m−2
i1 α i xη i , lim t → ∞ xt 0} and define the operator T : X × 0, 1 → X by
T x, λt λ
∞
0
G t, sfs, x s, xsds, 0≤ t < ∞. 3.1 The main result of this paper is following
Trang 7Theorem 3.1 Let f : 0, ∞ × R2 → R be an S-Carath´eodory function Suppose further that there
exists functions pt, qtrt ∈ L10, ∞ with tpt, tqttrt ∈ L10, ∞ such that
f t, u, v ≤ pt|u| qt|v| rt 3.2
for almost every t ∈ 0, ∞ and all u, v ∈ R2 Then1.2 has at least one solution provided:
η m−2 P P1 Q < 1, α < 0,
αη m−2
1− α P P1 Q < 1, 0 ≤ α < 1,
max
αη m−2
α − 1 P P1 Q, αP1
α − 1αη m−2 P
α − 1
< 1, α > 1.
3.3
0, 1, Tx, λ is completely continuous in X.
Proof First we show T is well defined Let x ∈ X; then there exists r > 0 such that x ≤ r For
each λ ∈ 0, 1, it holds that
T x, λt λ
∞
0
G t, sfs, x s, xsds
≤
∞
0
|Gt, s|ϕ r sds < ∞, ∀t ∈ 0, ∞.
3.4
Further, Gt, s is continuous in t so the Lebesgue dominated convergence theorem implies
that
|Tx, λt1 − Tx, λt2| ≤ λ
∞
0
|Gt1, s − Gt2, s|f
s, x s, xsds
≤ λ
∞
0
|Gt1, s − Gt2, s |ϕ r sds
−→ 0, as t1 −→ t2,
3.5
T x, λt1 − Tx, λt2 ≤ λt2
t1
f
s, x s, xsds
≤
t2
t1
ϕ r sds −→ 0 as t1−→ t2,
3.6
where 0≤ t1, t2< ∞ Thus, Tx ∈ C10, ∞.
Trang 8Obviously, Tx, λ0 m−2
i1 α i Tx, λη i Notice that
lim
t → ∞ T x, λt lim
t → ∞
∞
t
f
s, x s, xsds 0, 3.7
so we can get Tx, λt ∈ X.
We claim that Tx, λ is completely continuous in X, that is, for each λ ∈ 0, 1, Tx, λ
is continuous in X and maps a bounded subset of X into a relatively compact set.
Let x n → x as n → ∞ in X Next we prove that for each λ ∈ 0, 1, Tx n , λ → Tx, λ
as n → ∞ in X Because f is a S-Carath´eodory function and
∞
0
G sf
s, x n s, x
n s− fs, x s, xsds
≤ 2∞
0
Gsϕ
r0sds < ∞, 3.8
where r0> 0 is a real number such that max{max n∈N\{0} x n , x} ≤ r0, we have
|Tx n , λ ∞ − Tx, λ∞| ≤ λ
∞
0
Gsf
s, x n s, x
n s− fs, x s, xsds
−→ 0, as n −→ ∞.
3.9
Also, we can get
|Tx n , λ t − Tx n , λ ∞| ≤ λ
∞
0
Gt, s − Gsf
s, x n s, x
n sds
≤
∞
0
Gt, s − Gsϕ
r0sds
−→ 0, as t −→ ∞,
3.10
T x n , λt − Tx n , λ∞ ≤∞
t
f
s, x n s, x
n sds
≤
∞
t
ϕ r0sds −→ 0, as t −→ ∞.
3.11
Similarly, we have
|Tx, λt − Tx, λ∞| −→ 0, as t −→ ∞,
T x, λt − Tx, λ∞ −→ 0, as t −→ ∞ 3.12
Trang 9For any positive number T0< ∞, when t ∈ 0, T0, we have
|Tx n , λ t − Tx, λt| ≤
∞
0
|Gt, s|f
s, x n s, x
n s− fs, x s, xsds
−→ 0, as n −→ ∞,
T x n , λt − Tx, λt ≤∞
t
f
s, x n s, x
n s− fs, x s, xsds
−→ 0, as n −→ ∞.
3.13
Combining3.9–3.13, we can see that T·, λ is continuous Let B ⊂ X be a bounded subset; it is easy to prove that TB is uniformly bounded In the same way, we can prove
3.5,3.6, and 3.12, we can also show that TB is equicontinuous and equiconvergent Thus,
Proof of Theorem 3.1 In view ofLemma 2.2, it is clear that x ∈ X is a solution of the BVP 1.2 if
and only if x is a fixed point of T·, 1 Clearly, Tx, 0 0 for each x ∈ X If for each λ ∈ 0, 1 the fixed points T·, λ in X belong to a closed ball of X independent of λ, then the Leray-Schauder continuation theorem completes the proof We have known T·, λ is completely
continuous byLemma 3.2 Next we show that the fixed point of T·, λ has a priori bound M independently of λ Assume x Tx, λ and set
Q1
∞
0
sq sds, R
∞
0
r sds, R1
∞
0
sr sdt. 3.14
According toLemma 2.5, we know that for any x ∈ X, there exists η ∈ η1, η m−2 satisfying
x0 αxη Hence, there are three cases as follow.
Case 1 α < 0 For any x ∈ X, x0xη ≤ 0 holds and, therefore, there exists a t0∈ 0, η such that xt0 0 Then, we have
|xt|
t
t0
xsds
≤
t η x
∞≤t η m−2 x
∞, t ∈ 0, ∞, 3.15
and so it holds that
x
∞≤ λft, x, x
L1≤ ft, x, x
L1
≤ pt|xt| qt|xt| rt
L1
≤η m−2 P P1 Q x
∞ R,
3.16
therefore,
x ∞≤ R
Trang 10At the same time, we have
|xt| ≤ λ
∞
0
G t, sfs, x s, xsds
≤
∞
0
sf
s, x s, xsds
≤ P1x∞ Q1M1 R1, t ∈ 0, ∞,
3.18
and so
x∞≤ Q1M1 R1
1− P1 M1. 3.19
Set M max{M1, M1}, which is independent of λ.
Case 2 0 ≤ α < 1 For any x ∈ X, we have
|xt|
αx
η
t
0
xsds
≤ αx
η t x
∞, t ∈ 0, ∞, 3.20
which implies that|xt| ≤ αη/1 − α tx∞≤ αη m−2 /1 − α tx∞for all t ∈ 0, ∞.
In the same way as for Case1, we can get
x ∞≤ 1 − αR
1 − α1 − P1− Q − αη m−2 P M
2,
x∞≤ Q1M2 R1
1− α − P1 M2.
3.21
Set M max{M2, M2}, which is independent of λ and is what we need.
Case 3 α > 1 For x ∈ X, we have
|xt|
x
η
t
η
xsds
≤
1
α |x0| t − η x
∞, t ∈ 0, ∞, 3.22
and so|xt| ≤ αη/α − 1 tx∞≤ αη m−2 /α − 1 tx∞for all t ∈ 0, ∞.
Similarly, we obtain
x
∞≤ α − 1R
α − 11 − P1− Q − αη m−2 P M
3,
x∞≤ α
Q1M3 R1
αη m−2
QM3 R
3.23
Set M max{M3, M3} and which is we need So 1.2 has at least one solution
Trang 11The Natural Science Foundation of Hebei ProvinceA2009000664 and the Foundation of Hebei University of Science and TechnologyXL200759 are acknowledged
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... M3} and which is we need So 1.2 has at least one solution Trang 11The Natural Science...
Trang 9For any positive number T0< ∞, when t ∈ 0, T0, we have
|Tx...
Trang 4If η i ≤ t ≤ η i1 , ≤ i ≤ m − 3, the unique solution of 2.1 can