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Hindawi Publishing Corporation
Boundary Value Problems
Volume 2011, Article ID 212980, 22 pages
doi:10.1155/2011/212980
Research Article
Multiple Positive Solutions of a Singular
Emden-Fowler Type Problem for Second-Order
Impulsive Differential Systems
Eun Kyoung Lee and Yong-Hoon Lee
Department of Mathematics, Pusan National University, Busan 609-735, Republic of Korea
Correspondence should be addressed to Yong-Hoon Lee,
Received 14 May 2010; Accepted 26 July 2010
Academic Editor: Feliz Manuel Minh
´
os
Copyright q 2011 E. K. Lee and Y H. Lee. 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 studies the existence, and multiplicity of positive solutions of a singular boundary
value problem for second-order differential systems with impulse effects. By using the upper and
lower solutions method and fixed point index arguments, criteria of the multiplicity, existence and
nonexistence of positive solutions with respect to parameters given in the system are established.
1. Introduction
In this paper, we consider systems of impulsive differential equations of the form
u
t
λh
1
t
f
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
v
t
μh
2
t
g
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
Δu
|
tt
1
I
u
u
t
1
, Δv
|
tt
1
I
v
v
t
1
,
Δu
tt
1
N
u
u
t
1
, Δv
tt
1
N
v
v
t
1
,
u
0
a ≥ 0,v
0
b ≥ 0,u
1
c ≥ 0,v
1
d ≥ 0,
P
where λ, μ are positive real parameters, Δu|
tt
1
ut
1
− ut
1
,andΔu
|
tt
1
u
t
1
− u
t
−
1
.
Throughout this paper, we assume f, g ∈ CR
2
, R
with f0, 00 g0, 0 and fu, v >
0,gu, v > 0 for all u, v
/
0, 0,I
u
,I
v
∈ CR
, R satisfying I
u
00 I
v
0,N
u
,N
v
∈
CR
, −∞, 0, and h
i
∈ C0, 1, 0, ∞,i 1, 2. Here we denote R
0, ∞. We note that
2 Boundary Value Problems
h
i
may be singular at t 0and/or1. Let J 0, 1,J
0, 1 \{0, 1,t
1
},PC0, 1{u |
u : 0, 1 → R be continuous at t
/
t
1
, left continuous at t t
1
, and its right-hand limit at
t t
1
exists } and X PC0, 1 × PC0, 1. Then PC0, 1 and X are Banach spaces with norm
u sup
t∈0,1
|ut| and u, v u v, respectively. The solution of problem P means
u, v ∈ X ∩ C
2
J
× C
2
J
which satisfies P .
Recently, several works have been devoted to the study of second-order impulsive
differential systems. See, for example 1–6, and references therein. In Particular, E.K. Lee
and Y.H. Lee 3 studied problem P when f and g satisfy f0, 0 > 0andg0, 0 > 0. More
precisely, let us consider the following assumptions.
D
1
1
0
s1 − sh
i
sds < ∞, for i 1, 2.
D
2
t
1
N
u
u ≤ I
u
u ≤−1 − t
1
N
u
u and t
1
N
v
v ≤ I
v
v ≤−1 − t
1
N
v
v.
D
3
u I
u
u and v I
v
v are nondecreasing.
D
4
N
u,∞
lim
u →∞
|N
u
u|/u < 1andN
v,∞
lim
v →∞
|N
v
v|/v < 1.
D
5
f
∞
lim
uv →∞
fu, v/u v ∞ and g
∞
lim
uv →∞
gu, v/u v ∞.
D
6
f and g are nondecreasing on R
2
, that is, fu
1
,v
1
≤ fu
2
,v
2
and gu
1
,v
1
≤
gu
2
,v
2
whenever u
1
,v
1
≤ u
2
,v
2
, where inequality on R
2
can be understood
componentwise.
Under the above assumptions, they proved that there exists a continuous curve Γ splitting
R
2
\{0, 0} into two disjoint subsets O
1
and O
2
such that problem 3.20 has at least two
positive solutions for λ, μ ∈O
1
, at least one positive solution for λ, μ ∈ Γ, and no solution
for λ, μ ∈O
2
.
The aim of this paper is to study generalized Emden-Fowler-type problem for P,
that is, f and g satisfy f0, 00andg0, 00, respectively. In this case, we obtain two
interesting results. First, for Dirichlet boundary condition, that is, a b c d 0, assuming
D
1
, D
2
and
D
4
N
u,0
lim
u → 0
|N
u
u|/u < 1/2andN
v,0
lim
v → 0
|N
v
v|/v < 1/2,
D
5
f
∞
∞,g
∞
∞ and f
0
lim
uv → 0
fu, v/uv 0,g
0
lim
uv → 0
gu, v/uv 0,
we prove that problem P has at least one positive solution for all λ, μ ∈ R
2
\{0, 0}. On the
other hand, for two-point boundary condition, that is, c>aand d>b,assuming D
1
∼ D
6
,
we prove that there exists a continuous curve Γ
0
splitting R
2
\{0, 0} into two disjoint subsets
O
0,1
and O
0,2
and there exists a subset O⊂O
0,1
such that problem P has at least two positive
solutions for λ, μ ∈O, at least one positive solution for λ, μ ∈ O
0,1
\O ∪ Γ
0
, and no
solution for λ, μ ∈O
0,2
.
Our technique of proofs is mainly employed by the upper and lower solutions method
and several fixed point index theorems.
The paper is organized as follows: in Section 2, we introduce and prove two types of
upper and lower solutions and related theorems, one for singular systems with no impulse
effect and the other for singular impulsive systems and then introduce several fixed point
index theorems for later use. In Section 3, we prove an existence result for Dirichlet boundary
value problems and existence and nonexistence part of the result for two-point boundary
value problems. In Section 4, we prove the existence of t he second positive solution for
two point boundary value problems. Finally, in Section 5, we apply main results to prove
some theorems of existence, nonexistence, and multiplicity of positive radial solutions for
impulsive semilinear elliptic problems.
Boundary Value Problems 3
2. Preliminary
In this section, we introduce two types of fundamental theorems of upper and lower solutions
method for a singular system with no impulse effect and an impulsive system and then
introduce several well-known fixed point index theorems. We first give definition s of
somewhat general type of upper and lower solutions for the following singular system:
u
t
F
t, u
t
,v
t
0,t∈
0, 1
,
v
t
G
t, u
t
,v
t
0,t∈
0, 1
,
u
0
A, u
1
C, v
0
B, v
1
D,
H
where F, G : 0, 1 × R × R → R are continuous.
Definition 2.1. We say that α
u
,α
v
∈ C0, 1 × C0, 1 is a G-lower solution of H if α
u
,α
v
∈
C
2
0, 1 × C
2
0, 1 except at finite points τ
1
, ,τ
n
with 0 <τ
1
< < τ
n
< 1 such that
L
1
at each τ
i
, there exist α
u
τ
i
−
,α
v
τ
i
−
, α
u
τ
i
,α
v
τ
i
such that α
u
τ
i
−
≤
α
u
τ
i
,α
v
τ
i
−
≤ α
v
τ
i
, and
L
2
α
u
t
F
t, α
u
t
,α
v
t
≥ 0,
α
v
t
G
t, α
u
t
,α
v
t
≥ 0,t∈
0, 1
{
τ
1
, ,τ
n
}
,
α
u
0
≤ A, α
u
1
≤ C,
α
v
0
≤ B, α
v
1
≤ D.
2.1
We also say that β
u
,β
v
∈ C0, 1 × C0, 1 is a G-upper solution of the problem H if
β
u
,β
v
∈ C
2
0, 1 × C
2
0, 1 except at finite points σ
1
, ,σ
m
with 0 <σ
1
< ··· <σ
m
< 1
such that
U
1
at each σ
i
, there exist β
u
σ
j
−
,β
v
σ
j
−
, β
u
σ
j
,β
v
σ
j
such that β
u
σ
j
−
≥
β
u
σ
j
,β
v
σ
j
−
≥ β
v
σ
j
, and
U
2
β
u
t
F
t, β
u
t
,β
v
t
≤ 0,
β
v
t
G
t, β
u
t
,β
v
t
≤ 0,t∈
0, 1
{
σ
1
, ,σ
n
}
,
β
u
0
≥ A, β
u
1
≥ C,
β
v
0
≥ B, β
v
1
≥ D.
2.2
For the proof of the fundamental theorem on G-upper and G-lower solutions for
problem H, we need the following lemma. One may refer to 7 for the proof.
4 Boundary Value Problems
Lemma 2.2. Let F, G : D → R be continuous functions and D ⊂ 0, 1 × R × R. Assume that there
exist h
F
,h
G
∈ C0, 1, R
such that
|
F
t, u, v
|
≤ h
F
t
,
|
G
t, u, v
|
≤ h
G
t
, 2.3
for all t, u, v ∈ 0, 1 × R × R, and
1
0
s
1 − s
h
F
s
ds
1
0
s
1 − s
h
G
s
ds < ∞. 2.4
Then problem H has a solution.
Let D
β
α
{t, u, v | α
u
t,α
v
t ≤ u, v ≤ β
u
t,β
v
t,t ∈ 0, 1}. Then the
fundamental theorem of G-upper and G-lower solutions for singular problem H is given
as follows.
Theorem 2.3. Let α
u
,α
v
and β
u
,β
v
be a G-lower solution and a G-upper solution of problem
H, respectively, such that
a
1
α
u
t,α
v
t ≤ β
u
t,β
v
t for all t ∈ 0, 1.
Assume also that there exist h
F
,h
G
∈ C0, 1, R
such that
a
2
|Ft, u, v|≤h
F
t and |Gt, u, v|≤h
G
t for all t, u, v ∈ D
β
α
;
a
3
1
0
s1 − sh
F
sds
1
0
s1 − sh
G
sds < ∞;
a
4
Ft, u, v
1
≤ Ft, u, v
2
, whenever v
1
≤ v
2
and Gt, u
1
,v ≤ Gt, u
2
,v, whenever u
1
≤
u
2
.
Then problem H has at least one solution u, v such that
α
u
t
,α
v
t
≤
u
t
,v
t
≤
β
u
t
,β
v
t
, ∀ t ∈
0, 1
. 2.5
Proof. Define a modified function of F as follows:
F
∗
t, u, v
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
F
t, β
u
t
,v
−
u − β
u
t
1
u
2
if u>β
u
t
,
F
t, u, v
if α
u
t
≤ u ≤ β
u
t
,
F
t, α
u
t
,v
−
u − α
u
t
1
u
2
if u<α
u
t
,
Boundary Value Problems 5
F
∗
t, u, v
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
F
∗
t, u, β
v
t
if v>β
v
t
,
F
∗
t, u, v
if α
v
t
≤ v ≤ β
v
t
,
F
∗
t, u, α
v
t
if v<α
v
t
,
G
∗
t, u, v
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
G
t, β
u
t
,v
if u>β
u
t
,
G
t, u, v
if α
u
t
≤ u ≤ β
u
t
,
G
t, α
u
t
,v
if u<α
u
t
,
2.6
G
∗
t, u, v
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
G
∗
t, u, β
v
t
,
−
v − β
v
t
1
v
2
if v>β
v
t
,
G
∗
t, u, v
if α
v
t
≤ v ≤ β
v
t
,
G
∗
t, u, α
v
t
−
v − α
v
t
1
v
2
if v<α
v
t
.
2.7
Then F
∗
,G
∗
: 0, 1 × R × R → R are continuous and
|
F
∗
t, u, v
|
≤ m
α
u
,β
u
h
F
t
,
|
G
∗
t, u, v
|
≤ m
α
v
,β
v
h
G
t
,
2.8
for all t, u, v ∈ 0, 1 × R × R, where mα, βα β 1. For the problem
u
t
F
∗
t, u
t
,v
t
0,
v
t
G
∗
t, u
t
,v
t
0,t∈
0, 1
u
0
A, u
1
C, v
0
B, v
1
D,
M
Lemma 2.2 guarantees the existence of solutions of problem M and thus it is enough to
prove that any solution u, v of problem M satisfies
α
u
t
,α
v
t
≤
u
t
,v
t
≤
β
u
t
,β
v
t
, ∀t ∈
0, 1
. 2.9
Suppose, on the contrary, α
u
,α
v
/
≤ u, v, so we consider the case α
u
/
≤ u. Let α
u
− ut
0
max
t∈0,1
α
u
− ut > 0. If t
0
∈ 0, 1 \{τ
1
, ,τ
n
}, then α
u
− u
t
0
≤ 0. We consider two
6 Boundary Value Problems
cases.First,ifα
v
t
0
≤ vt
0
, then by α
u
t
0
>ut
0
and condition a
4
,
0 ≥
α
u
− u
t
0
α
u
t
0
F
∗
t
0
,u
t
0
,v
t
0
α
u
t
0
F
t
0
,α
u
t
0
,v
t
0
−
u
t
0
− α
u
t
0
1 u
2
t
0
≥ α
u
t
0
F
t
0
,α
u
t
0
,α
v
t
0
−
u
t
0
− α
u
t
0
1 u
2
t
0
≥
α
u
t
0
− u
t
0
1 u
2
t
0
> 0,
2.10
which is a contradiction. Next, if α
v
t
0
>vt
0
, then by the definition of F
∗
,
0 ≥
α
u
− u
t
0
α
u
t
0
F
∗
t
0
,u
t
0
,v
t
0
α
u
t
0
F
t
0
,α
u
t
0
,α
v
t
0
−
u
t
0
− α
u
t
0
1 u
2
t
0
> 0,
2.11
which is also a contradiction. If t
0
τ
i
for some i 1, ,n,then since α
u
−u attains its positive
maximum at τ
i
,
α
u
− u
τ
i
−
≥ 0,
α
u
− u
τ
i
≤ 0. 2.12
If α
u
− u
τ
i
−
> 0, then
0 <
α
u
− u
τ
i
−
−
α
u
− u
τ
i
α
u
τ
i
−
− α
u
τ
i
. 2.13
This leads a contradiction to the definition of G-lower solution. If α
u
− u
τ
i
−
0, then there
exists δ>0 such that for all t ∈ τ
i
− δ, τ
i
,
α
u
− u
t
> 0,
α
u
− u
t
≥ 0,
α
u
− u
t
≤ 0. 2.14
For t ∈ τ
i
− δ, τ
i
, if α
v
t ≤ vt, then by α
u
t >ut and condition a
4
,
0 ≥
α
u
− u
t
α
u
t
F
∗
t, u
t
,v
t
α
u
t
F
t, α
u
t
,v
t
−
u
t
− α
u
t
1 u
2
t
≥ α
u
t
F
t, α
u
t
,α
v
t
−
u
t
− α
u
t
1 u
2
t
≥
α
u
t
− u
t
1 u
2
t
> 0,
2.15
Boundary Value Problems 7
which is a contradiction. If α
v
t >vt, then by definition of F
∗
,
0 ≥
α
u
− u
t
α
u
t
F
∗
t, u
t
,v
t
α
u
t
F
t, α
u
t
,α
v
t
−
u
t
− α
u
t
1 u
2
t
> 0,
2.16
which is a contradiction. If t
0
0or1, then
0 <
α
u
− u
0
α
u
0
− A ≤ 0,
0 <
α
u
− u
1
α
u
1
− C ≤ 0,
2.17
which is a contradiction. Similarly, we get contradictions for the case α
v
/
≤ v. The proof for
u, v ≤ β
u
,β
v
can be done by similar fashion.
Now we introduce definition and fundamental theorem of upper and lower solutions
for impulsive differential systems of the form
u
t
F
t, u
t
,v
t
0,t
/
t
1
,t∈
0, 1
,
u
t
G
t, u
t
,v
t
0,t
/
t
1
,t∈
0, 1
,
Δu
|
tt
1
I
u
u
t
1
, Δv
|
tt
1
I
v
v
t
1
,
Δu
tt
1
N
u
u
t
1
, Δv
tt
1
N
v
v
t
1
,
u
0
a, v
0
b, u
1
c, v
1
d,
S
where F, G ∈ C0, 1 × R × R, R, I
u
,I
v
∈ CR
, R satisfying I
u
00 I
v
0 and N
u
,N
v
∈
CR
, −∞, 0.
Definition 2.4. α
u
,α
v
∈ X ∩ C
2
J
× C
2
J
is called a lower solution of problem S if
α
u
t
F
t, α
u
t
,α
v
t
≥ 0,t
/
t
1
,
α
v
t
G
t, α
u
t
,α
v
t
≥ 0,t
/
t
1
,
Δα
u
|
tt
1
I
u
α
u
t
1
, Δα
v
|
tt
1
I
v
α
v
t
1
,
Δα
u
tt
1
≥ N
u
α
u
t
1
, Δα
v
tt
1
≥ N
v
α
v
t
1
,
α
u
0
≤ a, α
v
0
≤ b, α
u
1
≤ c, α
v
1
≤ d.
2.18
We also define an upper solution β
u
,β
v
∈ X∩C
2
J
×C
2
J
if β
u
,β
v
satisfies the reverses
of the above inequalities.
The following existence theorem for upper and lower solutions method is proved in
3.
8 Boundary Value Problems
Theorem 2.5. Let α
u
,α
v
and β
u
,β
v
be lower and upper solutions of problem S, respectively,
satisfying a
1
. Moreover, we assume a
2
∼ a
4
and D
3
. Then problem S has at least one solution
u, v such that
α
u
t
,α
v
t
≤
u
t
,v
t
≤
β
u
t
,β
v
t
, ∀ t ∈
0, 1
. 2.19
The following theorems are well known cone theoretic fixed point theorems. See
Lakshmikantham 8 for proofs and details.
Theorem 2.6. Let X be a Banach space and K a cone in X. Assume that Ω
1
and Ω
2
are bounded open
subsets in X with 0 ∈ Ω
1
and Ω
1
⊂ Ω
2
.LetT : K∩Ω
2
\ Ω
1
→Kbe a completely continuous
such that either
i Tu≤u for u ∈K∩∂Ω
1
and Tu≥u for u ∈K∩∂Ω
2
or
ii Tu≥u for u ∈K∩∂Ω
1
and Tu≤u for u ∈K∩∂Ω
2
.
Then T has a fixed point in K∩
Ω
2
\ Ω
1
.
Theorem 2.7. Let X be a Banach space, K a cone in X and Ω bounded open in X. Let 0 ∈ Ω and
T : K∩
Ω →Kbe condensing. Suppose that Tx
/
νx, for all x ∈K∩∂Ω and all ν ≥ 1. Then
i
T, K∩Ω, K
1. 2.20
3. Existence
In this section, we prove an existence theorem of positive solutions for problem P with
Dirichlet boundary condition and the existence and nonexistence part of the result for
problem P with two-point boundary condition. Let us consider the following second-order
impulsive differential systems.
u
t
λh
1
t
f
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
v
t
μh
2
t
g
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
Δu
|
tt
1
I
u
u
t
1
, Δv
|
tt
1
I
v
v
t
1
Δu
tt
1
N
u
u
t
1
, Δv
tt
1
N
v
v
t
1
,
u
0
a ≥ 0,v
0
b ≥ 0,u
1
c ≥ 0,v
1
d ≥ 0,
P
where λ, μ are positive real p arameters, f, g ∈ CR
2
, 0, ∞ with f0, 00,g0, 00,
and fu, v > 0,gu, v > 0 for all u, v
/
0, 0,I
u
,I
v
∈ CR
, R satisfying I
u
00
I
v
0,N
u
,N
v
∈ CR
, −∞, 0, and h
1
,h
2
∈ C0, 1, 0, ∞ may be singular at t 0and/or
1.
Boundary Value Problems 9
We first set up an equivalent operator equatio for problem P.LetusdefineA
λ
: X →
PC0, 1 and B
μ
: X → PC0, 1 by taking
A
λ
u, v
t
a
c − a
t λ
1
0
K
t, s
h
1
s
f
u
s
,v
s
ds W
u
t, u
,
B
μ
u, v
t
b
d − b
t μ
1
0
K
t, s
h
2
s
g
u
s
,v
s
ds W
v
t, v
,
3.1
where
K
t, s
⎧
⎨
⎩
t
−I
v
v
t
1
−
1 − t
1
N
v
v
t
1
, 0 ≤ t ≤ t
1
,
1 − t
I
v
v
t
1
− t
1
N
v
v
t
1
,t
1
<t≤ 1.
W
u
t, u
t
⎧
⎨
⎩
t
−I
u
u
t
1
−
1 − t
1
N
u
u
t
1
, 0 ≤ t ≤ t
1
,
1 − t
I
u
u
t
1
− t
1
N
u
u
t
1
,t
1
<t≤ 1,
W
v
t, u
t
⎧
⎨
⎩
t
−I
v
v
t
1
−
1 − t
1
N
v
v
t
1
, 0 ≤ t ≤ t
1
,
1 − t
I
v
v
t
1
− t
1
N
v
v
t
1
,t
1
<t≤ 1.
3.2
Also define
T
λ,μ
u, v
A
λ
u, v
,B
μ
u, v
. 3.3
Then T
λ,μ
: X → X is well defined on X and problem P is equivalent to the fixed-point
equation
T
λ,μ
u, v
u, v
in X. 3.4
Mainly due to D
1
,T
λ,μ
is completely continuous see 3 for the proof.Letu
0
sup
t∈0,t
1
|ut|, u
1
sup
t∈t
1
,1
|ut|,S
0
t
1
/4, 3t
1
/4,S
1
3t
1
1/4,t
1
3/4, P {u, v ∈
X | u,v ≥ 0}, and K {u, v ∈P | min
t∈S
0
utvt ≥ t
1
/4u
0
v
0
, min
t∈S
1
ut
vt ≥ 1 − t
1
/4u
1
v
1
}. Then u max{u
0
, u
1
} and P, K are cones in X. By using
concavity of T
λ,μ
u with u ∈P, we can easily show that T
λ,μ
P ⊂K.
10 Boundary Value Problems
We now prove the existence theorem of positive solutions for Dirichlet boundary value
problem
u
t
λh
1
t
f
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
v
t
μh
2
t
g
u
t
,v
t
0,t∈
0, 1
t
/
t
1
,
Δu
|
tt
1
I
u
u
t
1
, Δv
|
tt
1
I
v
v
t
1
,
Δu
tt
1
N
u
u
t
1
, Δv
tt
1
N
v
v
t
1
,
u
0
0,v
0
0,u
1
0,v
1
0.
P
D
Theorem 3.1. Assume D
1
, D
2
, D
4
, and D
5
. Then problem P
D
has at least one positive
solution for all λ, μ ∈ R
2
\{0, 0}.
Proof. First, we consider case λ>0andμ>0. By the fact N
u,0
< 1/2andN
v,0
< 1/2, we
may choose c
1
,m
1
> 0 such that max{N
u,0
,N
v,0
} <c
1
< 1/2, |N
u
u |≤ c
1
u for u ≤ m
1
and
|N
v
v|≤c
1
v for v ≤ m
1
. Also choose η
λ
and η
μ
satisfying 0 <η
λ
< 1 − 2c
1
/2λ
1
0
s1 −
sh
1
sds and 0 <η
μ
< 1 − 2c
1
/2μ
1
0
s1 − sh
2
sds. Since f
0
0andg
0
0, there exist
m
2
,m
3
> 0 such that fu, v ≤ η
λ
u v for u v ≤ m
2
and gu, v ≤ η
μ
u v for u v ≤ m
3
.
Let Ω
1
B
M
1
{u, v ∈ X |u, v <M
1
} with M
1
min{m
1
,m
2
,m
3
}. Then for u, v ∈
K∩∂Ω
1
, we obtain by using D
2
A
λ
u, v
t
λ
1
0
K
t, s
h
1
s
f
u
s
,v
s
ds W
u
t, u
≤ λη
λ
1
0
s
1 − s
h
1
s
u
s
v
s
ds
|
N
u
u
t
1
|
≤
λη
λ
1
0
s
1 − s
h
1
s
ds c
1
u, v
≤
1
2
u, v
,
3.5
for all t ∈ 0, 1. Similarly, we obtain
B
μ
u, v
t
≤
1
2
u, v
3.6
for all t ∈ 0, 1. Thus
T
λ, μ
u, v
A
λ
u, v
B
μ
u, v
≤
u, v
. 3.7
Boundary Value Problems 11
On the other hand, let us choose η
1
and η
2
such that
1
η
2
<μmin
⎧
⎪
⎨
⎪
⎩
t
1
8
min
t∈S
0
S
0
K
t, s
h
2
s
ds,
1 − t
1
8
min
t∈S
1
S
1
K
t, s
h
2
s
ds
⎫
⎪
⎬
⎪
⎭
.
1
η
2
<μmin
⎧
⎪
⎨
⎪
⎩
t
1
8
min
t∈S
0
S
0
K
t, s
h
2
s
ds,
1 − t
1
8
min
t∈S
1
S
1
K
t, s
h
2
s
ds
⎫
⎪
⎬
⎪
⎭
.
3.8
Also by D
5
, we may choose R
f
and R
g
such that fu, v ≥ η
1
u v for u v ≥ R
f
and
gu, v ≥ η
2
u v for u v ≥ R
g
. Let Ω
2
{u, v ∈ X |u, v <M
2
}, where M
2
max{8R
f
/t
1
, 8R
f
/1 − t
1
, 8R
g
/t
1
, 8R
g
/1 − t
1
,M
1
1}. Then Ω
1
⊂ Ω
2
. Let u, v ∈K∩∂Ω
2
,
then we have the following four cases: u≥v and u u
0
, u≥v and u
u
1
, u≤v and v v
0
, and u≤v and v v
1
. We consider the first case,
the rest of them can be considered in a similar way. So let u≥v and u u
0
; then for
t ∈ S
0
, we have
u
t
v
t
≥ u
t
≥
t
1
8
2
u
0
≥
t
1
8
u
v
t
1
8
u, v
≥ R
f
. 3.9
Thus fut,vt ≥ η
1
utvt for t ∈ S
0
. Since W
u
t, u ≥ 0, we get for t ∈ S
0
,
A
λ
u, v
t
λ
1
0
K
t, s
h
1
s
f
u
s
,v
s
ds W
u
t, u
≥ λ
S
0
K
t, s
h
1
s
f
u
s
,v
s
ds
≥ λη
1
S
0
K
t, s
h
1
s
u
s
v
s
ds
≥ λη
1
t
1
8
S
0
K
t, s
h
1
s
ds
2
u
0
2
v
0
≥ λη
1
t
1
8
min
t∈S
0
S
0
K
t, s
h
1
s
ds
u, v
>
u, v
.
3.10
Therefore,
T
λ,μ
u, v
≥
A
λ
u, v
>
u, v
, 3.11
and by Theorem 2.6, T
λ,μ
has a fixed point in K∩Ω
2
\ Ω
1
.
Second, consider case λ>0andμ 0. Taking c
1
,η
λ
,m
1
, and m
2
as above and using
the same computation, we may show
A
λ
u, v
≤
1
2
u, v
, 3.12
12 Boundary Value Problems
for all u, v ∈K∩∂Ω
1
, where Ω
1
B
M
1
with M
1
min{m
1
,m
2
}. Since μ 0,
B
μ
u, v
t
W
v
t, v
≤
|
N
v
v
t
1
|
≤ c
1
u, v
≤
1
2
u, v
, 3.13
for all t ∈ 0, 1. Thus
T
λ,μ
u, v
≤
A
λ
u, v
B
μ
u, v
≤
u, v
, 3.14
for u, v ∈K∩∂Ω
1
. Now, let us choose η
1
and R
f
as above and let Ω
2
{u, v ∈ X |
u, v <M
2
}, where M
2
max{8R
f
/t
1
, 8R
f
/1 − t
1
,M
1
1}. Then Ω
1
⊂ Ω
2
and we can
show by the same computation as above,
T
λ,μ
u, v
≥
A
λ
u, v
>
u, v
, 3.15
for u, v ∈K∩∂Ω
2
and thus T
λ,μ
has a fixed point in K∩Ω
2
\ Ω
1
.
Finally, consider case λ 0andμ>0. Taking c
1,
η
μ
,m
1
, and m
3
as the first case, we
may show by similar argument,
B
μ
u, v
≤
1
2
u, v
,
A
λ
u, v
≤
1
2
u, v
, 3.16
for all u, v ∈K∩∂Ω
1
, where Ω
1
B
M
1
with M
1
min{m
1
,m
3
} . Thus
T
λ,μ
u, v
≤
u, v
, 3.17
for u, v ∈K∩∂Ω
1
. Now, let us choose η
2
and R
g
as the first case and let Ω
2
{u, v ∈ X |
u, v <M
2
}, where M
2
max{8R
g
/t
1
, 8R
g
/1 − t
1
,M
1
1}. Then Ω
1
⊂ Ω
2
and we also
show similarly, as before,
T
λ,μ
u, v
≥
B
μ
u, v
>
u, v
, 3.18
for u, v ∈K∩∂Ω
2
. Therefore, T
λ,μ
has a fixed point in K∩Ω
2
\ Ω
1
and this completes the
proof.
Now let us consider two point boundary value problems given as follows:
u
t
λh
1
t
f
u
t
,v
t
0,t∈
0, 1
, t
/
t
1
,
v
t
μh
2
t
g
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
Δu
|
tt
1
I
u
u
t
1
, Δv
|
tt
1
I
v
v
t
1
,
Δu
tt
1
N
u
u
t
1
, Δv
Δv
tt
1
N
v
v
t
1
,
u
0
a ≥ 0,v
0
b ≥ 0,u
1
c>a, v
1
d>b.
P
T
Boundary Value Problems 13
Lemma 3.2. Assume D
5
. Let R be a compact subset of R
2
\{0, 0}. Then there exists a constant
b
R
> 0 such that for all λ, μ ∈Rfor possible positive solutions u, v of problem 3.20 at λ, μ,
one has u, v <b
R
.
Proof. Suppose on the contrary that there is a sequence u
n
,v
n
of positive solutions of 3.20
at λ
n
,μ
n
such that λ
n
,μ
n
∈Rfor all n and u
n
,v
n
→∞. Since 0, 0
/
∈R, there is a
subsequence, say again {λ
n
,μ
n
}, such that α min{λ
n
} > 0orβ min{μ
n
} > 0. First, we
assume α>0. From u
n
,v
n
→∞,weknowu
n
0
v
n
0
→∞or u
n
1
v
n
1
→∞.
Suppose u
n
0
v
n
0
→∞. Then by the concavity of u
n
and v
n
, we have
u
n
t
v
n
t
≥
t
1
4
u
n
0
v
n
0
, 3.19
for t ∈ S
0
. Let us choose η
1
> 2π
2
/t
2
1
αh
1
, where h
1
min
t∈S
0
h
1
t. Then by D
5
, there exists
R
f
> 0 such that
f
u, v
≥ η
1
u v
∀u v ≥ R
f
. 3.20
Since u
n
0
v
n
0
> 4/t
1
R
f
for sufficiently large n, 3.19 implies u
n
tv
n
t >R
f
for
t ∈ S
0
.Thusfort ∈ S
0
,
f
u
n
t
,v
n
t
>η
1
u
n
t
v
n
t
≥ η
1
u
n
t
. 3.21
Hence we have for t ∈ S
0
,
0 u
n
t
λ
n
h
1
t
f
u
n
t
,v
n
t
>u
n
t
α
h
1
η
1
u
n
t
. 3.22
If we multiply by φtsin2π/t
1
t−t
1
/4 both sides in the above inequality and integrate
on S
0
, then by the facts φ
t
1
/4 > 0,φ
3t
1
/4 < 0 and integration by part, we obtain
0 >
3t
1
/4
t
1
/4
u
n
t
φ
t
dt α
h
1
η
1
3t
1
/4
t
1
/4
u
n
t
φ
t
dt
≥−
2π
t
1
2
3t
1
/4
t
1
/4
u
n
t
φ
t
dt α
h
1
η
1
3t
1
/4
t
1
/4
u
n
t
φ
t
dt.
3.23
Thus 2π/t
1
2
/αh
1
≥ η
1
which is a contradiction to the choice of η
1
. Suppose u
n
1
v
n
1
→
∞, then we also get a contradiction by a similar calculation with η
2
> 2π
2
/1 − t
1
2
α
h
1
,
where
h
1
min
t∈S
1
h
1
t. Finally, the case β>0 can also be proved by similar way using the
condition g
∞
∞.
Lemma 3.3. Assume D
1
, D
3
, and
Q fu, v
1
≤ fu, v
2
, whenever v
1
≤ v
2
, gu
1
,v ≤ gu
2
,v, whenever u
1
≤ u
2
.
14 Boundary Value Problems
If problem 3.20 has a positive solution at
λ, μ. Then the problem also has a positive solution at
λ, μ for all λ, μ ≤
λ, μ.
Proof. Let
u, v be a positive solution of problem 3.20 at λ, μ and let λ, μ ∈ R
2
\{0, 0}
with λ, μ ≤
λ, μ. Then u, v is an upper solution of 3.20 at λ, μ. Define α
u
,α
v
by
α
u
t
⎧
⎪
⎨
⎪
⎩
0,t∈
0,t
1
,
c
1 − t
1
t − t
1
,t∈
t
1
, 1
,
α
v
t
⎧
⎪
⎨
⎪
⎩
0,t∈
0,t
1
,
d
1 − t
1
t − t
1
,t∈
t
1
, 1
.
3.24
Then α
u
,α
v
is a lower solution of problem 3.20 at λ, μ. By the concavity of u, v, u, v ≥
α
u
,α
v
. Therefore, Theorem 2.5 implies that problem 3.20 has a positive solution at λ, μ.
Lemma 3.4. Assume D
1
∼ D
4
and Q. Then there exists λ
∗
,μ
∗
> 0, 0 such that problem
3.20 has a positive solution for all λ, μ ≤ λ
∗
,μ
∗
.
Proof. It is not hard to see that the following problem:
u
t
h
1
t
0,t∈
0, 1
,t
/
t
1
,
v
t
h
2
t
0,t∈
0, 1
,t
/
t
1
,
Δu
|
tt
1
I
u
u
t
1
, Δv
|
tt
1
I
v
v
t
1
,
Δu
tt
1
N
u
u
t
1
, Δv
tt
1
N
v
v
t
1
,
u
0
a ≥ 0,v
0
b ≥ 0,u
1
c>a, v
1
d>b
3.25
has a positive solution so let β
u
,β
v
be a positive solution. Let M
f
sup
t∈0,1
fβ
u
t,β
v
t
and M
g
sup
t∈0,1
gβ
u
t,β
v
t. Then M
f
,M
g
> 0andforλ
∗
,μ
∗
1/M
f
, 1/M
g
, we
get
β
u
λ
∗
h
1
t
f
β
u
t
,β
v
t
h
1
t
λ
∗
f
β
u
t
,β
v
t
− 1
≤ 0,
β
v
μ
∗
h
2
t
g
β
u
t
,β
v
t
h
2
t
μ
∗
g
β
u
t
,β
v
t
− 1
≤ 0.
3.26
This shows that β
u
,β
v
is an upper solution of 3.20 at λ
∗
,μ
∗
. On the other hand,
α
u
,α
v
given in Lemma 3.3 is obviously a lower solution and α
u
,α
v
≤ β
u
,β
v
. Thus by
Theorem 2.5, 3.20 has a positive solution at λ
∗
,μ
∗
and the proof is done by Lemma 3.3.
We introduce a known existence result for a singular boundary value problem with no
impulse effect.
Boundary Value Problems 15
Lemma 3.5 see9. Consider, D
1
, D
5
and Q. For problem
u
t
λh
1
t
f
u
t
,v
t
0,
v
t
μh
2
t
g
u
t
,v
t
0,t∈
0, 1
,
u
0
a ≥ 0,u
1
c>a,
v
0
b ≥ 0,v
1
d>b,
U
T
let A
T
{λ, μ ∈ R
2
\{0, 0}|3.27 has a positive solution at λ, μ}. Then A
T
, ≤ is bounded
above.
Define A {λ, μ ∈ R
2
\{0, 0}|3.20 has a positive solution at λ, μ}. Then A
/
∅ by
Lemma 3.4 and A, ≤ is a partially ordered set.
Lemma 3.6. Assume D
1
∼ D
6
. Then A, ≤ is bounded above.
Proof. Suppose on the contrary that there exists a sequence λ
n
,μ
n
∈Asuch that |λ
n
,μ
n
|→
∞. Let u
n
,v
n
be a positive solution of problem 3.20 at λ
n
,μ
n
. By condition D
2
, we may
choose sequences s
n
, t
n
in 0,t
1
∪ t
1
, 1 such that if I
u
u
n
t
1
> 0, then t
n
∈ t
1
, 1 and
I
u
u
n
t
1
t
n
− t
1
N
u
u
n
t
1
0,
I
u
u
n
t
1
t − t
1
N
u
u
n
t
1
> 0, on
t
1
,t
n
,
I
u
u
n
t
1
t − t
1
N
u
u
n
t
1
< 0, on
t
n
, 1
;
3.27
if I
u
u
n
t
1
< 0, then t
n
∈ 0,t
1
and
I
u
u
n
t
1
t
n
− t
1
N
u
u
n
t
1
0,
I
u
u
n
t
1
t − t
1
N
u
u
n
t
1
> 0, on
0,t
n
,
I
u
u
n
t
1
t − t
1
N
u
u
n
t
1
< 0, on
t
n
,t
1
;
3.28
if I
v
v
n
t
1
> 0, then s
n
∈ t
1
, 1 and
I
v
v
n
t
1
s
n
− t
1
N
v
v
n
t
1
0,
I
v
v
n
t
1
t − t
1
N
v
v
n
t
1
> 0, on
t
1
,s
n
,
I
v
v
n
t
1
t − t
1
N
v
v
n
t
1
< 0, on
s
n
, 1
;
3.29
if I
v
v
n
t
1
< 0, then s
n
∈ 0,t
1
and
I
v
v
n
t
1
s
n
− t
1
N
v
v
n
t
1
0,
I
v
v
n
t
1
t − t
1
N
v
v
n
t
1
> 0, on
0,s
n
,
I
v
v
n
t
1
t − t
1
N
v
v
n
t
1
< 0, on
s
n
,t
1
.
3.30
16 Boundary Value Problems
If I
u
u
n
t
1
> 0, define
u
n
t
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
u
n
t
, on
0,t
1
,
u
n
t
−
I
u
u
n
t
1
t − t
1
N
u
u
n
t
1
, on
t
1
,t
n
,
u
n
t
, on
t
n
, 1
,
3.31
and if I
u
u
n
t
1
< 0, define
u
n
t
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
u
n
t
, on
0,t
n
,
u
n
t
I
u
u
n
t
1
t − t
1
N
u
u
n
t
1
, on
t
n
,t
1
,
u
n
t
, on
t
1
, 1
.
3.32
Moreover, if I
v
v
n
t
1
> 0, define
v
n
t
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
v
n
t
, on
0,t
1
,
v
n
t
−
I
v
v
n
t
1
t − t
1
N
v
v
n
t
1
, on
t
1
,s
n
,
v
n
t
, on
s
n
, 1
,
3.33
and if I
v
v
n
t
1
< 0, define
v
n
t
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
v
n
t
, on
0,s
n
,
v
n
t
I
v
v
n
t
1
t − t
1
N
v
v
n
t
1
, on
s
n
,t
1
,
v
n
t
, on
t
1
, 1
.
3.34
Then we can easily see that u
n
, v
n
∈ C0, 1 × C0, 1 ∩ C
2
0, 1 × C
2
0, 1 except
t
1
,t
n
,s
n
. Furthermore, u
n
t
−
1
, v
n
t
−
1
u
n
t
1
, v
n
t
1
, u
n
t
−
n
, v
n
t
−
n
≥ u
n
t
n
, v
n
t
n
and
u
n
s
−
n
, v
n
s
−
n
≥ u
n
s
n
, v
n
s
n
. We also see u
n
t,v
n
t ≥ u
n
t, v
n
t on 0, 1. Thus by
D
6
, we get
u
n
t
λ
n
h
1
t
f
u
n
t
, v
n
t
u
n
t
λ
n
h
1
t
f
u
n
t
, v
n
t
λ
n
h
1
t
f
u
n
t
, v
n
t
− f
u
n
t
,v
n
t
≤ 0,
v
n
t
μ
n
h
2
t
g
u
n
t
, v
n
t
v
n
t
μ
n
h
2
t
g
u
n
t
, v
n
t
μ
n
h
2
t
g
u
n
t
, v
n
t
− g
u
n
t
,v
n
t
≤ 0.
3.35
We also get u
n
0u
n
0a, u
n
1u
n
1c, v
n
0v
n
0b, and v
n
1v
n
1d.
Thus u
n
, v
n
is a G-upper solution of problem U
T
at λ
n
,μ
n
. If I
u
u
n
t
1
0orI
v
v
n
t
1
0, then we consider u
n
u
n
or v
n
v
n
as a G-upper solution. Let α
u
t, α
v
t c −
at a, d − bt b, then α
u
, α
v
is the G-lower solution of 3.27 at λ
n
,μ
n
. Therefore,
Boundary Value Problems 17
by Theorem 2.3, problem 3.27 has a positive solution for all λ
n
,μ
n
. This contradicts to
Lemma 3.5 and the proof is done.
Lemma 3.7. Assume D
1
∼ D
6
. Then every nonempty chain in A has a unique supremum in A.
Proof. Let C be a chain in A. Without loss of generality, we may choose a distinct sequence
{λ
n
,μ
n
}⊂Csuch that λ
n
,μ
n
≤ λ
n1
,μ
n1
.ByLemma 3.6, there exists λ
C
,μ
C
such
that λ
n
,μ
n
→ λ
C
,μ
C
. If we show λ
C
,μ
C
∈A, then the proof is done. Since {λ
n
,μ
n
}
is bounded, Lemma 3.2 implies that there is a constant B such that u
n
,v
n
<B, where
u
n
,v
n
is a solution corresponding to λ
n
,μ
n
. By the compactness of T
λ,μ
, {u
n
,v
n
} has a
convergent subsequence converging to say, u
C
,v
C
. By Lebesgue Convergence theorem, we
see that u
C
,v
C
is a solution of 3.20 at λ
C
,μ
C
.Thusλ
C
,μ
C
∈A.
Theorem 3.8. Assume D
1
∼ D
6
. Then there exists a continuous curve Γ splitting R
2
\{0, 0}
into two disjoint subsets O
1
and O
2
such that problem P
T
has at least one positive solution for
λ, μ ∈O
1
∪ Γ and no solution for λ, μ ∈O
2
.
Proof. λ
∗
,μ
∗
is given in Lemma 3.4.WeknowfromLemma 3.4 that 3.20 has a positive
solution at 0,s for all 0 <s≤ μ
∗
.Thus{0,s | s>0}∩Ais a nonempty chain in A and
by Lemma 3.7, it has unique supremum of the form 0,s
∗
in A. This implies that 3.20 has
a positive solution at 0,s for all 0 <s≤ s
∗
and no solution at 0,s for all s>s
∗
. Similarly,
there is r
∗
≥ λ
∗
such that 3.20 has a positive solution at r, 0 for all 0 <r≤ r
∗
and no
solution at r, 0 for all r>r
∗
. Define L : R → R
2
by taking Lt{r, s | s r t}. Then
for t ∈ −r
∗
,s
∗
, Lt ∩Ais a nonempty chain in A. Define Γt as the unique supremum
of Lt ∩A. Then Γ is well defined on −r
∗
,s
∗
and as a consequence of Lemma 3.3,wesee
that Γ is continuous on −r
∗
,s
∗
, Γ−r
∗
r
∗
, 0, and Γs
∗
0,s
∗
. Therefore, the curve
ΓΓ−r
∗
,s
∗
separates R
2
\{0, 0} into two disjoint subsets O
1
and O
2
, where O
1
is bounded
and O
2
is unbounded and we get the conclusion of this theorem for Γ, O
1
, and O
2
.
4. Multiplicity
In this section, we study existence of the second positive solution for two point boundary
value problem 3.20 with λ, μ in certain region of O
1
appeared in Theorem 3.8. For the
computation of fixed point index, we need to consider problems of the form
u
t
λh
1
t
f
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
v
t
μh
2
t
g
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
Δu
|
tt
1
I
u
u
t
1
, Δv
|
tt
1
I
v
v
t
1
,
Δu
tt
1
N
u
u
t
1
, Δv
tt
1
N
v
v
t
1
,
u
0
a ε, u
1
c ε,
v
0
b ε, v
1
d ε,
P
ε
T
where ε>0,c>a≥ 0andd>b≥ 0. Theorem 3.8 implies that there exists a continuous curve
Γ
ε
splitting R
2
\{0, 0} into two disjoint subsets O
ε,1
and O
ε,2
such that the problem 4.1 has
at least one positive solution for λ, μ ∈O
ε,1
∪Γ
ε
and no solution for λ, μ ∈O
ε,2
. Using upper
18 Boundary Value Problems
and lower solutions argument, we can easily show that if 0 <
ε<ε,then O
ε,1
∪ Γ
ε
⊂O
ε,1
∪ Γ
ε
.
Let O ∪
ε>0
O
ε,1
∪ Γ
ε
, then O⊂O
0,1
. We state the main theorem for two point boundary
value problem 3.20 as follows.
Theorem 4.1. Assume D
1
∼ D
6
. Then there exists a continuous curve Γ
0
splitting R
2
\{0, 0}
into two disjoint subsets O
0,1
and, O
0,2
and there exists a subset O⊂O
0,1
such that problem 3.20 has
at least two positive solutions for λ, μ ∈O, at least one positive solution for λ, μ ∈ O
0,1
\O ∪ Γ
0
,
and no solution for λ, μ ∈O
0,2
.
Proof. Let O ∪
ε>0
O
ε,1
∪ Γ
ε
and let λ, μ ∈O. It is enough to prove that problem 3.20 has
the second solution at λ, μ. By the definition of O, there exists ε>0 such that λ, μ ∈O
ε,1
∪Γ
ε
.
That is 4.1 has a positive solution at λ, μ. So let u
ε
,v
ε
be a positive solution of problem
4.1 at λ, μ and let Ω{u, v ∈ X |−ε<ut <u
ε
t, −ε<vt <v
ε
t for t ∈ 0, 1,ut
1
<
u
ε
t
1
,vt
1
<v
ε
t
1
}. Then Ω is bounded open in X, 0 ∈ Ω. Furthermore, T
λ,μ
: K∩Ω →
K is condensing, since it is completely continuous. We show that T
λ,μ
u, v
/
νu, v for all
u, v ∈K∩∂Ω and all ν ≥ 1. If it is not true, then there exist u, v ∈K∩∂Ω and ν
0
≥ 1 such
that T
λ,μ
u, vν
0
u, v. Thus u, v is a positive solution of the following equation
ν
0
u
t
λh
1
t
f
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
ν
0
v
t
μh
2
t
g
u
t
,v
t
0,t∈
0, 1
,t
/
t
1
,
ν
0
Δu
|
tt
1
I
u
u
t
1
,ν
0
Δv
|
tt
1
I
v
v
t
1
,
ν
0
Δu
tt
1
N
u
u
t
1
,ν
0
Δv
tt
1
N
v
v
t
1
,
ν
0
u
0
a, ν
0
v
0
b, ν
0
u
1
c, ν
0
v
1
d,
4.1
and we can consider the following two cases. The first case is ut
0
u
ε
t
0
or vt
0
v
ε
t
0
for some t
0
∈ 0, 1. The second case is ut
1
u
ε
t
1
or vt
1
v
ε
t
1
. First, let us consider
case ut
0
u
ε
t
0
for some t
0
∈ 0, 1. One may prove similarly for case vt
0
v
ε
t
0
.
If t
0
∈ J
, that is, t
0
/
t
1
, let mtu − u
ε
t, then m
t ≥ 0onJ
,m0 < 0,m1 < 0,
and mt
1
≤ 0. Thus on one of intervals 0,t
1
or t
1
, 1 containing t
0
, maximum principle
implies m ≡ 0 and this contradicts to the facts of m0 < 0andm1 < 0. If t
0
t
1
, then
ut
1
u
ε
t
1
,ut ≤ u
t and vt ≤ v
ε
t on 0, 1. Thus by D
6
and D
2
, we get the
following contradiction:
u
t
1
1
ν
0
A
λ
u, v
t
1
a
b − a
t
1
ν
0
λ
ν
0
1
0
K
t
1
,s
h
1
s
f
u
s
,v
s
ds
−t
1
I
u
u
t
1
1 − t
1
N
u
u
t
1
ν
0
<a ε
b − a
t
1
λ
1
0
K
t
1
,s
h
1
s
f
u
ε
s
,v
ε
s
ds
− t
1
I
u
u
ε
t
1
1 − t
1
N
u
u
ε
t
1
u
ε
t
1
u
t
1
.
4.2
Boundary Value Problems 19
Second, let us consider ut
1
u
ε
t
1
. Since ut ≤ u
t and vt ≤ v
ε
t,weget
u
t
1
1
ν
0
A
λ
u, v
t
1
a
b − a
t
1
ν
0
λ
ν
0
1
0
K
t
1
,s
h
1
s
f
u
s
,v
s
ds
1 − t
1
I
u
u
t
1
− t
1
N
u
u
t
1
ν
0
<a ε
b − a
t
1
λ
1
0
K
t
1
,s
h
1
s
f
u
ε
s
,v
ε
s
ds
1 − t
1
I
u
u
ε
t
1
− t
1
N
u
u
ε
t
1
u
ε
t
1
u
t
1
.
4.3
One may show the contradiction similarly for case vt
1
v
ε
t
1
. This contradiction shows
T
λ,μ
u, v
/
νu, v for all u, v ∈K∩∂Ω and ν ≥ 1. Therefore, by Theorem 2.7,weobtain
i
T
λ,μ
, K∩Ω, K
1. 4.4
On the other hand, by Lemma 3.6, we know that there is λ
1
,μ
1
such that 3.20 has no
positive solution at λ
1
,μ
1
.Thus for any open set U in X,weget
i
T
λ
1
,μ
1
, K∩U, K
0. 4.5
Let R be a compact rectangle containing λ, μ and λ
1
,μ
1
. By Lemma 3.2, for all λ, μ ∈R,
there exists b
R
> 0 such that all possible solutions u, v of P
T
at λ, μ satisfy u, v <b
R
and Ω ⊂ B
b
R
. Define h : 0, 1 × B
b
R
∩K →Kby
h
τ,
u, v
T
τλ
1
1−τλ,τμ
1
1−τμ
u, v
. 4.6
Then h0, u, v T
λ,μ
u, v,h1, u, v T
λ
1
,μ
1
u, v,h is completely continuous on 0, 1 ×
K, and hτ,u, v
/
u, v for all τ, u, v ∈ 0, 1 × ∂B
b
R
∩K. By the property of homotopy
invariance and 4.5, w e have
i
T
λ,μ
,B
b
R
∩K, K
i
T
λ
1
,μ
1
,B
b
R
∩K, K
0. 4.7
By the additive property and 4.4, 4.7, we have
i
T
λ,μ
,
B
b
R
Ω
∩K, K
−1. 4.8
Therefore, 3.20 has another positive solution in B
b
R
\Ω∩K at λ, μ ∈Oand this completes
the proof.
20 Boundary Value Problems
5. Applications
In this section, we apply the results in previous sections to study the existence and
multiplicity theorems of positive radial solutions for impulsive semilinear elliptic problems.
5.1. On an Annular Domain
Let us consider
Δu λk
1
|
x
|
f
u, v
0,
Δv μk
2
|
x
|
g
u, v
0, in Ω
l
1
,l
2
,
|
x
|
/
r
1
,
Δu
|
|x|r
1
I
u
u
|
|x|r
1
, Δv
|
|x|r
1
I
v
u
|
|x|r
1
,
Δ∂u
∂r
|x|r
1
−
r
1−n
1
N
u
u
|
|x|r
1
m
,
Δ∂v
∂r
|x|r
1
−
r
1−n
1
N
v
v
|
|x|r
1
m
,
u
x
a ≥ 0,v
x
b ≥ 0if
|
x
|
l
1
,
u
x
c>a, v
x
d>b if
|
x
|
l
2
,
P
A
where f0, 00,g0, 00, Δ is the Laplacian of u, 0 <l
1
<r
1
<l
2
, and Ωl
1
,l
2
{x ∈
R
n
| l
1
< |x| <l
2
} with n>2.∂u/∂r denotes the differentiation in the radial direction,
Δu|
|x|r
1
ur
1
− ur
1
, Δ∂u/∂r|
|x|r
1
∂u/∂rr
1
− ∂u/∂rr
−
1
and m −
l
2
l
1
t
1−n
dt.
Applying consecutive changes of variables, r |x|,s −
l
2
r
t
1−n
dt and t m − s/m, we may
transform problem 5.1 into problem 3.20, where t
1
r
2−n
1
− l
2−n
1
/l
2−n
2
− l
2−n
1
and h
i
can
be written as
h
i
t
m
2
r
m
1 − t
2n−1
k
i
r
m
1 − t
. 5.1
If k
i
: l
1
,l
2
→ 0, ∞ are continuous, then h
i
: 0, 1 → 0, ∞ are also continuous and
satisfies D
1
. We may apply Theorem 4.1 to obtain the following result.
Corollary 5.1. Assume D
2
∼ D
6
. Let k
i
∈ Cl
1
,l
2
, 0, ∞,i 1, 2. Then there exists a
continuous curve Γ
0
splitting R
2
\{0, 0} into two disjoint subsets O
0,1
and O
0,2
and there exists a
subset O⊂O
0,1
such that problem 5.1 has at least two positive solutions for λ, μ ∈O, at least one
positive solution for λ, μ ∈ O
0,1
\O ∪ Γ
0
, and no solution for λ, μ ∈O
0,2
.
If k
i
: l
1
,l
2
→ 0, ∞ are continuous and singular at r l
1
and/or l
2
, then h
i
are also
singular at t 0 and/or 1. In this case, we assume
l
2
l
1
r
2−n
− l
2−n
1
l
2−n
2
− r
2−n
k
i
r
dr < ∞, 5.2
then we can easily check that both h
i
satisfy D
1
and apply Theorem 4.1 to obtain the following
corollary.
Boundary Value Problems 21
Corollary 5.2. Assume D
2
∼ D
6
. If both k
i
∈ Cl
1
,l
2
, 0, ∞ satisfy
l
2
l
1
r
2−n
− l
2−n
1
l
2−n
2
− r
2−n
k
i
r
dr < ∞. 5.3
Then the conclusion of Corollary 5.1 is valid.
5.2. On an Exterior Domain
Let us consider
Δu λk
1
|
x
|
f
u, v
0,
Δv μk
2
|
x
|
g
u, v
0,
|
x
|
>r
0
,
|
x
|
/
r
1
,
Δu
|
|x|r
1
I
u
u
|
|x|r
1
, Δv
|
|x|r
1
I
v
v
|
|x|r
1
,
Δ∂u
∂r
|x|r
1
n − 2
r
0
r
1
r
0
1−n
N
u
u
|
|x|r
1
,
Δ∂v
∂r
|x|r
1
n − 2
r
0
r
1
r
0
1−n
N
v
v
|
|x|r
1
,
u
x
a ≥ 0,v
x
b ≥ 0, if
|
x
|
r
0
,
u
x
→ c>a, v
x
→ d>b, as
|
x
|
→∞,
P
E
where f0, 00,g0, 00, 0 <r
0
<r
1
, and n>2. Assume that both k
i
: r
0
, ∞ → 0, ∞
are continuous. Applying changes of variables, r |x| and t 1−r/r
0
2−n
, we may transform
problem 5.4 into problem 3.20, where t
1
1 − r
0
/r
1
n−2
and h
i
are written as
h
i
t
r
2n−1
0
n − 2
2
1 − t
−2n−1/n−2
k
i
r
0
1 − t
−1/n−2
. 5.4
We know that h
i
are singular at t 1 and can easily check that h
i
satisfy D
1
if k
i
satisfy
∞
r
0
rk
i
rdr < ∞ for i 1, 2. Thus by Theorem 4.1, we obtain the following result.
Corollary 5.3. Assume D
2
∼ D
6
. If both k
i
∈ Cr
0
, ∞, 0, ∞ satisfy
∞
r
0
rk
i
r
dr < ∞, 5.5
then the conclusion of Corollary 5.1 is valid for problem 5.4.
22 Boundary Value Problems
Acknowledgment
This work was supported for two years by Pusan National University Research Grant.
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