Hindawi Publishing Corporation
Boundary Value Problems
Volume 2010, Article ID 281908, 13 pages
doi:10.1155/2010/281908
Research Article
Multiple Positive Solutions of the Singular
Boundary Value Problem for Second-Order
Impulsive Differential Equations on the Half-Line
Jing Xiao,1 Juan J. Nieto,2 and Zhiguo Luo1
1
2
Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China
Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela,
15782 Santiago de Compostela, Spain
Correspondence should be addressed to Zhiguo Luo, luozg@hunnu.edu.cn
Received 17 November 2009; Revised 22 January 2010; Accepted 21 February 2010
Academic Editor: Claudianor Alves
Copyright q 2010 Jing Xiao 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.
This paper uses a fixed point theorem in cones to investigate the multiple positive solutions of a
boundary value problem for second-order impulsive singular differential equations on the halfline. The conditions for the existence of multiple positive solutions are established.
1. Introduction
Consider the following nonlinear singular Sturm-Liouville boundary value problems for
second-order impulsive differential equation on the half-line:
ptu t ft, u 0,
Δu tk Ik utk ,
∀t ∈ J ,
k 1, 2, . . . , n,
αu0 − β lim ptu t 0,
1.1
t→0
γu∞ δ lim ptu t 0,
t → ∞
where J 0, ∞, 0 < t1 < · · · < tn , J 0, ∞, J J \ {t1 , . . . , tn }, f ∈ CJ × J , J ,
∞
p ∈ CJ, J ∩ C1 J , J with p > 0 on J , and 0 1/psds < ∞; α, β, γ, δ ≥ 0 with
s
ρ βγ αδ αγB0, ∞ > 0, in which Bt, s t 1/pσdσ. Δu tk u tk − u t−k ,
2
Boundary Value Problems
where u t−k and u tk are, respectively, the left and right limits of u t at tk , k 1, . . . , n,
1 ≤ n < ∞.
The theory of singular impulsive differential equations has been emerging as an
important area of investigation in recent years. For the theory and classical results, we refer
the monographs to 1, 2 and the papers 3–19 to readers. We point out that in a second-order
differential equation u ft, u, u , one usually considers impulses in the position u and the
velocity u . However, in the motion of spacecraft one has to consider instantaneous impulses
depending on the position that result in jump discontinuities in velocity, but with no change
in position 20. The impulses only on the velocity occur also in impulsive mechanics 21.
In recent paper 3, by using the Krasnoselskii’s fixed point theorem, Kaufmann
has discussed the existence of solutions for some second-order boundary value problem
with impulsive effects on an unbounded domain. In 22 Sun et al. and 23 Liu et al.,
respectively, discussed the existence and multiple positive solutions for singular SturmLiouville boundary value problems for second-order differential equation on the half-line.
But the Multiple positive solutions of this case with both singularity and impulses are not to
be studied. The aim of this paper is to fill up this gap.
The rest of the paper is organized as follows. In Section 2, we give several important
lemmas. The main theorems are formulated and proved in Section 3. And in Section 4, we
give an example to demonstrate the application of our results.
2. Several Lemmas
Lemma 2.1 see 23. If conditions
value problem
∞
0
1/psds < ∞ and ρ > 0 are satisfied, then the boundary
ptu t νt 0,
∀t ∈ J ,
αu0 − β lim ptu t 0,
2.1
t→0
γu∞ δ lim ptu t 0
t → ∞
has a unique solution for any ν ∈ LJ , R. Moreover, this unique solution can be expressed in the
form
ut ∞
2.2
Gt, sνsds,
0
where Gt, s is defined by
⎧
1 ⎨ β αB0, s δ γBt, ∞ ,
Gt, s ρ ⎩β αB0, tδ γBs, ∞,
0 ≤ s ≤ t < ∞,
0 ≤ t ≤ s < ∞.
2.3
Boundary Value Problems
3
Remark 2.2. It is easy to prove that Gt, s has the following properties:
1 Gt, s is continuous on J × J ,
2 Gt, s is continuous differentiable on J × J , except t s,
3 ∂t Gt, s|ts − ∂t Gt, s|ts− ps−1 ,
4 Gt, s ≤ Gs, s ≤ ρ−1 β αB0, sδ γBs, ∞ < ∞,
5 Gs limt → ∞ Gt, s < ∞,
6 for all t ∈ a, b ⊂ 0, ∞, s ∈ 0, ∞, Gt, s ≥ ωGs, s, where
β αBb, ∞ δ γBb, ∞
,
.
w min
β αB0, ∞ δ γB0, ∞
2.4
Obviously, 0 < ω < 1.
For the interval a, b, 0 < a < t1 , tn < b < ∞, and the corresponding ω in Remark 2.2,
we define P C1 J, R {u ∈ CJ, R : u ∈ CJ , R, u t−k and u tk exist, and u tk u t−k }.
BP C1 J, R {u ∈ P C1 J, R : limt → ∞ ut exists}. K {u ∈ BP C1 J, R : ut > 0, t ∈ J
and mint∈a,b ut ≥ ωu}. It is easy to see that BP C1 J, R is a Banach space with the norm
u supt∈J |ut|, and K is a positive cone in BP C1 J, R. For details of the cone theory, see
1. u ∈ P C1 J, R ∩ C2 J R is called a positive solution of BVP 1.1 if ut > 0 for all t ∈ J
and ut satisfies 1.1.
As we know that the Ascoli-Arzela Theorem does not hold in infinite interval J, we
need the following compactness criterion:
Lemma 2.3 see 22. Let M ⊂ BP C1 J, R. Then M is relatively compact in BP C1 J, R if the
following conditions hold.
i M is uniformly bounded in BP C1 J, R.
ii The functions from M are equicontinuous on any compact interval of 0, ∞.
iii The functions from M are equiconvergent, that is, for any given ε > 0, there exists a
T T ε > 0 such that |ft − f∞| < ε, for any t > T , f ∈ M.
The main tool of this work is a fixed point theorem in cones.
Lemma 2.4 see 4. Let X be a Banach space and K is a positive cone in X. Assume that Ω1 , Ω2
are open subsets of X with 0 ∈ Ω1 , Ω1 ⊂ Ω2 . Let T : K ∩ Ω2 \ Ω1 → K be a completely continuous
operator such that
i T u ≤ u for all u ∈ K ∩ ∂Ω1 .
ii there exists a Φ ∈ K such that u / T u λΦ, for all u ∈ K ∩ ∂Ω2 and λ > 0.
Then T has a fixed point in K ∩ Ω2 \ Ω1 .
Remark 2.5. If i is satisfied for u ∈ K ∩ ∂Ω2 and ii is satisfied for u ∈ K ∩ ∂Ω1 , then
Lemma 2.4 is still true.
4
Boundary Value Problems
Lemma 2.6 see 3. The function u ∈ K ∩ C2 J , R is a solution of the BVP 1.1 if and only if
u ∈ K satisfies the equation
ut ∞
Gt, sfs, usds 0
n
Gt, tk ptk Ik utk ,
t ∈ J.
2.5
k1
The proof of this result is based on the properties of the Green function, so we omit it
as elementary.
Define
T ut ∞
Gt, sfs, usds 0
n
Gt, tk ptk Ik utk ,
t ∈ J.
2.6
k1
Obviously, the BVP 1.1 has a solution u if and only if u ∈ K is a fixed point of the operator
T defined by 2.6.
Let us list some conditions as follows.
A1 There exist two nonnegative functions: a ∈ CJ , J, g ∈ CJ, J such that ft, u ≤
atgu. ft, u, at may be singular at t 0. Ik : J → J, k 1, . . . , n, are
continuous.
∞
A2 0 < 0 Gs, sasds < ∞, 0 < Gtk , tk ptk < ∞, k 1, . . . , n.
Lemma 2.7. If A1 and A2 are satisfied, then for any bounded open set Ω ⊂ BP C1 J, R, T :
Ω ∩ K → K is a completely continuous operator.
Proof. For any bounded open set Ω ⊂ BP C1 J, R, there exists a constant M > 0 such that
u ≤ M for any u ∈ Ω.
First, we show that T : Ω ∩ K → K is well defined. Let u ∈ Ω ∩ K. From A1 , we
have SM max{S1 , S2 }, where S1 sup{gu : 0 ≤ u ≤ M}, S2 sup{Ik u : 0 ≤ u ≤ M,
k 1, . . . , n}, and
∞
Gt, sfs, usds 0
≤ SM
∞
0
n
Gt, tk ptk Ik utk k1
n
Gs, sasds Gtk , tk ptk < ∞.
2.7
k1
Hence, T is well defined. For any t1 , t2 ∈ J, we have
∞
0
∞
Gs, sasds < ∞.
|Gt1 , s − Gt2 , s|asds ≤ 2
0
2.8
Boundary Value Problems
5
Thus, by the Lebesgue dominated convergence theorem and the fact that Gs, t is continuous
on t, we have, for any t1 , t2 ∈ J, u ∈ Ω ∩ K,
|T ut1 − T ut2 |
≤
∞
|Gt1 , s − Gt2 , s|fs, usds
0
n
|Gt1 , tk − Gt2 , tk |ptk Ik utk 2.9
k1
≤ SM
∞
|Gt1 , s − Gt2 , s|asds 0
−→ 0,
n
|Gt1 , tk − Gt2 , tk |ptk k1
t1 −→ t2 .
Therefore, T u ∈ CJ, R. By the property 3 of Gs, t, it is easy to get T u ∈ P C1 J, R.
On the other hand, by 2.6 we have, for any u ∈ Ω ∩ K and t ∈ J ,
∞
T ut −
Gsfs, usds
0
≤
∞ n Gt, s − Gsfs, usds Gt, tk − Gtk ptk Ik utk 0
≤ SM
k1
∞
2.10
n Gt, s − Gsasds Gt, tk − Gtk ptk .
0
k1
Then by A2 , the property 5 of Remark 2.2 and the Lebesgue dominated convergence
theorem, we have
lim T ut t → ∞
Thus T u ∈ BP C1 J, R.
∞
0
Gsfs, usds n
Gtk ptk Ik utk < ∞.
k1
2.11
6
Boundary Value Problems
For any u ∈ Ω ∩ K, we get
T ut ≤
∞
Gt, sfs, usds n
Gt, tk ptk Ik utk 0
k1
∞
n
Gs, sfs, usds 0
2.12
Gtk , tk ptk Ik utk .
k1
So
T u ≤
∞
Gs, sfs, usds 0
n
Gtk , tk ptk Ik utk .
2.13
k1
On the other hand, for t ∈ a, b we obtain
T ut ≥ ω
∞
n
Gs, sfs, usds Gtk , tk ptk Ik utk 0
≥ ωT u.
2.14
k1
Thus T : Ω ∩ K → K.
Next, we prove that T is continuous. Let un → u0 in Ω ∩ K, then un ≤ M n 1, 2, . . .. We prove that T un → T u0 . For any ε > 0, by A2 , there exists a constant A0 > 0 such
that
∞
SM
Gs, sasds ≤
A0
ε
.
6
2.15
On the other hand, by the continuities of ft, u on 0, A0 × 0, M and the continuities of Ik
on J, for the above ε > 0, there exists a δ > 0 such that, for any u, v ∈ 0, M, |u − v| < δ,
ft, u − ft, v < ε
3
A
−1
0
Gs, sds
,
t ∈ 0, A0 ,
0
2.16
Gtk , tk ptk |Ik utk − Ik vtk | <
ε
.
3n
From un − u0 → 0, for the above δ, there exists a sufficiently large number N such that,
when n > N, we have
|un t − u0 t| ≤ un − u0 < δ,
t ∈ 0, A0 ,
2.17
|un tk − u0 tk | ≤ un − u0 < δ.
Boundary Value Problems
7
Therefore, by 2.15–2.17, we have, for n > N,
T un − T u0 ≤
∞
Gs, sfs, un s − fs, u0 sds
0
n
Gtk , tk ptk |Ik un tk − Ik u0 tk |
k1
≤ 2SM
∞
Gs, sasds A0
A0
Gs, sfs, un s − fs, u0 sds
2.18
0
n
Gtk , tk ptk |Ik utk − Ik u0 tk |
k1
≤
ε ε ε
ε.
3 3 3
This implies that the operator T is continuous.
Finally we show that T : Ω ∩ K → K is a compact operator. In fact for any bounded
set D ⊂ Ω, there exists a constant M1 > 0 such that u ≤ M1 for any u ∈ D ∩ K. Hence, we
obtain
T u ≤ SM1
∞
n
Gs, sasds Gtk , tk ptk 0
< ∞.
2.19
k1
Therefore, T D ∩ K is uniformly bounded in BP C1 J, R.
Given r > 0, for any u ∈ D∩K, as the proof of 2.9, we can get that {T u : u ∈ D∩K} are
equicontinuous on 0, r. Since r > 0 is arbitrary, {T u : u ∈ D ∩ K} are locally equicontinuous
on J . By 2.6, A1 , A2 , and the Lebesgue dominated convergence theorem, we have
|T ut − T u∞|
≤ SM1
∞
n Gt, s − Gsasds Gt, tk − Gtk ptk 0
−→ 0,
k1
2.20
t −→ ∞.
Hence, the functions from {T u : u ∈ D ∩ K} are equiconvergent. By Lemma 2.3, we have that
{T u : u ∈ D ∩K} is relatively compact in BP C1 J, R. Therefore, T : Ω ∩ K → K is completely
continuous. This completed the proof of Lemma 2.7.
8
Boundary Value Problems
3. Main Results
For convenience and simplicity in the following discussion, we use the following notations:
ft, u
,
t∈a,b
u
g0 lim inf
ft, u
,
u
g∞ lim inf
f0 lim inf min
u→0
f∞ lim inf min
u → ∞ t∈a,b
u→0
u→∞
ptk Ik u
,
I k lim sup
u
u→q
q
g q lim sup
u→q
gu
,
u
g
∞
I0 k lim inf
gu
,
u
I∞ k lim inf
gu
,
lim sup
u
u→∞
g 0 lim sup
u→0
ptk Ik u
,
u
gu
,
u
gu
,
u
u→0
u→∞
ptk Ik u
,
u
ptk Ik u
I k lim sup
,
u
u→∞
∞
I 0 k lim sup
u→0
3.1
ptk Ik u
,
u
Theorem 3.1. Let A1 and A2 hold. Then the BVP 1.1 has at least two positive solutions
satisfying 0 < u1 < q < u2 if the following conditions hold:
b
b
H1 ωf0 a Gs, sds nk1 Gtk , tk I0 k > 1, ωf∞ a Gs, sds nk1 Gtk , tk ·I∞ k >
1,
H2 there exists a q > 0 such that g q
ωq ≤ u ≤ q, a.e. t ∈ 0, ∞.
∞
0
Gs, sasds n
k1
Gtk , tk I q k < 1, for all
Proof. By the definition of f0 and I0 , for any ε > 0, there exist r ∈ 0, q such that
ft, u ≥ 1 − εf0 u,
∀u ≤ r, t ∈ a, b,
ptk Ik u ≥ 1 − εI0 ku,
b
n
1 − εω f0 Gs, sds Gtk , tk I0 k ≥ 1,
a
3.2
∀u ≤ r.
k1
Define the open sets
Ωr u ∈ BP C1 J, R : u < r .
3.3
Let Φ ≡ 1, then Φ ∈ K. Now we prove that
u/
T u λΦ,
∀u ∈ K ∩ ∂Ωr , λ > 0.
3.4
Boundary Value Problems
9
If not, then there exist u0 ∈ K ∩ ∂Ωr and λ0 > 0 such that u0 T u0 λ0 Φ. Let μ mint∈a,b u0 t,
then for any t ∈ a, b, we have
u0 t T u0 t λ0
∞
n
Gt, sfs, u0 sds Gt, tk ptk Ik u0 tk λ0
0
≥ω
k1
∞
Gs, sfs, u0 sds ω
0
n
Gtk , tk ptk Ik u0 tk λ0
3.5
k1
b
n
> 1 − εμω f0 Gs, sds Gtk , tk I0 k λ0
a
k1
≥ μ λ0 .
This implies μ > μ λ0 , a contradiction. Therefore, 3.4 holds.
That by the definition of f∞ and I∞ , for any ε > 0 there exist R > q such that
ft, u ≥ 1 − εf∞ u,
∀u ≥ R, t ∈ a, b,
ptk Ik u ≥ 1 − εI∞ ku,
b
n
1 − εω f∞ Gs, sds Gtk , tk I∞ k ≥ 1,
3.6
a
∀u ≥ R.
k1
Define the open sets:
ΩR u ∈ BP C1 J, R : u < R .
3.7
As the proof of 3.4, we can get that
u
/ T u λΦ,
3.8
∀x ∈ K ∩ ∂ΩR , λ > 0.
On the other hand, for any ε > 0, choose q in H2 such that
1 ε g
q
∞
0
n
Gs, sasds Gtk , tk I q k
≤ 1,
ωq ≤ u ≤ q.
3.9
k1
By the definition of g q , I q , for the above ε > 0, there exists δ > 0, when u ∈ q − δ, q δ; thus,
we have
gu ≤ 1 εg q u,
ptk Ik u ≤ 1 εI q ku.
3.10
10
Boundary Value Problems
Define
Ωq u ∈ BP C1 J, R : u < q .
3.11
Then, for any u ∈ K ∩ ∂Ωq and t ∈ 0, ∞, we can obtain
T ut ∞
Gt, sfs, usds 0
≤
∞
n
Gt, tk ptk Ik utk k1
Gs, sasgusds 0
≤ 1 ε g q
n
Gtk , tk ptk Ik utk k1
∞
0
Gs, sasds 3.12
n
Gtk , tk I q k u
k1
≤ u.
Therefore, T u ≤ u.
Thus, we can obtain the existence of two positive solutions u1 and u2 satisfying 0 <
u1 < q < u2 by using Lemma 2.4 and Remark 2.5, respectively.
Using a similar proof of Theorem 3.1, we can get the following conclusions.
Theorem 3.2. Let A1 and A2 hold. Then the BVP 1.1 has at least two positive solutions
satisfying 0 < u1 < q < u2 if the following conditions hold:
∞
∞
H3 g 0 0 Gs, sasds nk1 Gtk , tk I 0 k <1, g ∞ 0 Gs, sasds nk1 Gtk I ∞ k <
1,
b
H4 there exists q > 0 such that ωfq a Gs, sds nk1 Gtk , tk Iq k > 1, for all ωq ≤ u ≤
q, a.e. t ∈ 0, ∞.
Corollary 3.3. In Theorems 3.1 and 3.2, if conditions H1 and H3 are replaced by H1∗ and H3∗ ,
respectively, then the conclusions also hold.
H1∗ f0 ∞, or nk1 I0 k ∞; f∞ ∞ or nk1 I∞ k ∞,
H3∗ g ∞ 0, nk1 I ∞ k 0, g 0 0, nk1 I 0 k 0.
Remark 3.4. Notice that, in the above conclusions, we suppose that the singularity only exist
in ft, u, that is, ft, u → ∞ as t → 0. If we permit ft, u → ∞ as t → 0 or
u → 0 and Ik uk → ∞ as uk → 0 , then the discussion will be much more complex.
Now we state the corresponding results.
Let us define the following.
A∗1 There exist four nonnegative functions a, g ∈ CJ , J, b, h ∈ CJ, J such that
bthu ≤ ft, u ≤ atgu, and hu is nondecreasing on J. Ik : J → J,
k 1, . . . , n, are continuous.
∞
∞
A∗2 0 < 0 Gs, sasds < ∞, 0 Gs, sbsds ≥ u∗ /ωh∗ , 0 < Gtk , tk ptk <
∞, k 1, . . . , n, where u∗ ∈ K, h∗ h0.
Boundary Value Problems
11
Theorem 3.5. Suppose A∗1 and A∗2 hold, then the BVP 1.1 has at least two positive solutions
satisfying u∗ < u1 < q < u2 if H1 and H2 hold.
Proof. Define Q {u ∈ K : ut ≥ u∗ , for all t ∈ J}. We only need to proove T : Ω ∩ Q → Q is
a completely continuous operator. Then the rest of the proof is the same as that Theorem 3.1.
Notice that
T ut ≥ ω
∞
Gs, sfs, usds ≥ ωh∗
0
∞
Gs, sbsds ≥ u∗ ,
3.13
0
and change S1 , S2 to S1 sup{gu : u∗ ≤ u ≤ M}, S2 sup{Ik u : u∗ ≤ u ≤ M, k 1 . . . , n},
then the same as the proof of Lemma 2.7, it is easy to compute that T : Ω ∩ Q → Q is a
completely continuous operator.
Corresponding to Theorem 3.2 and Corollary 3.3, there are Theorem 3.6 and
Corollary 3.7. We just list here without proof.
Theorem 3.6. Suppose A∗1 and A∗2 hold, then the BVP 1.1 has at least two positive solutions
satisfying u∗ < u1 < q < u2 , if H3 and H4 hold.
Corollary 3.7. In Theorems 3.5 and 3.6, if conditions H1 and H3 are replaced by H1∗ and H3∗ ,
respectively, then the conclusions also hold.
4. Example
To illustrate how our main results can be used in practice we present the following example.
Example 4.1. Consider the following boundary value problem:
et u t |ln t| 0, ∀t ∈ J , t /
1,
Δu t1 1 u2 1,
u0 0,
4.1
u∞ 0.
Conclusion 1. BVP 4.1 has at least two positive solutions u1 , u2 satisfying 0 < u1 < 1/2 <
u2 .
Proof. Let pt et , gu 1, ft, u at | ln t|, Iu u2 . Then by simple computation we
have
∞
⎧ s
⎪
−σ
⎪
e
dσ
e−σ dσ,
⎪
⎪
⎨ 0
t
Gt, s ∞
t
⎪
⎪
⎪
⎪
e−σ dσ,
⎩ e−σ dσ
0
s
0 ≤ s ≤ t < ∞,
4.2
0 ≤ t ≤ s < ∞,
12
Boundary Value Problems
where ρ 1. Furthermore,
∞
0
∞
0<
1/pσdσ Gs, sasds 0
∞
0
∞
e−σ dσ 1 < ∞ and
1 − e−s e−s |ln s|ds < ∞,
0
0 < Gt1 , t1 pt1 1 − e
−1
4.3
< ∞.
Let a, b 1, 2 ⊂ 0, ∞. Then ω e−2 . Thus A1 and A2 are satisfied. It is easy to get
that f0 ∞, I∞ 1 ∞. Let q 1/2. Then
g
q
∞
0
Gs, sasds n
Gtk , tk I q k < 1.
4.4
k1
Hence, H1∗ and H2 are satisfied. Therefore, by Corollary 3.3, problem 4.1 has at least two
positive solutions u1 , u2 satisfying 0 < u1 < 1/2 < u2 . The proof is completed.
Acknowledgment
This work is supported by the National Nature Science Foundation of P. R.China 10871063
and Scientific Research Fund of Hunan Provincial Education Department 07A038, partially
supported by Ministerio de Educacion y Ciencia and FEDER, Project MTM2007-61724, and
by Xunta de Galicia and FEDER, project no.PGIDIT06PXIB207023PR.
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Boundary Value Problems
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