Theoretical Mathematics & Applications, vol.4, no.2, 2014, 79-91
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2014
Existence of positive solutions to second-order
periodic boundary value problems
with impulse actions
Ying He1
Abstract
In this paper we consider the existence of positive solutions for
second-order periodic boundary value problems with impulse actions.
By constructing a cone K1 × K2 , which is the Cartesian product of two
cones in the space C[0, 2π] and computing the fixed point index in the
K1 × K2 , we establish the existence of positive solutions for the system.
Mathematics Subject Classification: 34B15
Keywords: Periodic boundary value problem; Second-order impulsive differential equations; Fixed point index in cones.
1
School of Mathematics and Statistics, Northeast Petroleum University, Daqing 163318,
P.R.China.
Project supported by Heilongjiang province education department natural science
research item, China. (12541076).
E-mail: heying65338406@163.com
Article Info: Received : June 1, 2014. Revised : June 30, 2014.
Published online : June 30, 2014.
80
1
Existence of positive solutions to second-order...
Introduction
This paper is devoted to study the existence of positive solutions for the
following periodic boundary value problem with impulse effects:

−u00 + M u = g1 (x, u, v),
x ∈ I 0,





−v 00 + M v = g2 (x, v, u),
x ∈ I 0,



 −∆u0 |
x=xk = I1,k (u(xk )), ∆u|x=xk = I 1,k (u(xk )) k = 1, 2, · · · , l,
0

−∆v |x=xk = I2,k (v(xk )), ∆v|x=xk = I 2,k (v(xk )) k = 1, 2, · · · , l,





u(0) = u(2π), u0 (0) = u0 (2π),



v(0) = v(2π), v 0 (0) = v 0 (2π).
(1.1)
here I = [0, 2π],0 = x0 < x1 < x2 < · · · < xl < xl+1 = 2π, M > 0, I 0 =
I \ {x1 , x2 , · · · , xl } are given, R+ = [0, +∞), gi ∈ C(I × R+ , R+ ), Ii,k , I i,k ∈
C(R+ , R+ ) with − m1 Ii,k (u) < I i,k (u) < m1 Ii,k (u)(i = 1, 2), x ∈ R+ , m =
√
+
−
0 −
M , ∆u0 |x=xk = u0 (x+
∆v 0 |x=xk =
k ) − u (xk ), ∆u|x=xk = u(xk ) − u(xk )
+
−
+
+
0 −
0 +
0 +
v 0 (x+
k ) − v (xk ), ∆v|x=xk = v(xk ) − v(xk ), u (xk ), u(xk ), v (xk ) and v(xk ),
−
−
0 −
0
(u0 (x−
k ), u(xk ), v (xk ), v(xk )) denote the right limit ( left limit) of u (x), u(x),
v 0 (x) and v(x) at x = xk respectively.
It is well known that there are abundant results about the existence of positive solutions of boundary value problems for second order ordinary differential
equations. Some works can be found in [1 − 3] and references therein.They
mainly investigated the case without impulse actions.Recently, Dirichlet boundary problems of second order impulsive differential equations have been studied in [4 − 6].Motivated by the work above, this paper attempts to study the
existence of positive solutions for periodic boundary value problems. By constructing a cone K × K, which is the Cartesian product of two cones in the
space C[0, 2π],and computing the fixed-point index in the K × K ,we establish
the existence of positive solutions for the impulsive differential system (1.1).
To conclude the introduction,we introduce the following notation:
Ii,k (u)
gi (x, u, v)
, Ii,0 (k) = lim inf
,
x∈[0,2π]
u→0+
u
u
gi,0 (v) = lim inf
min
+
u→0
Ii,k (u)
gi (x, u, v)
, Ii,∞ (k) = lim inf
,
u→+∞
x∈[0,2π]
u
u
gi,∞ (v) = lim inf min
u→+∞
Ii,k (u)
gi (x, u, v) ∞
, Ii (k) = lim sup
,
x∈[0,2π]
u
u
u→+∞
gi∞ (v) = lim sup max
u→+∞
81
Ying He
gi (x, u, v) 0
Ii,k (u)
, Ii (k) = lim sup
,
x∈[0,2π]
u
u
u→0+
gi0 (v) = lim sup max
u→0+
where v ∈ R+ and i = 1, 2.
Moreover,for the simplicity in the following discussion,we introduce the
following hypotheses.
(H1 ) :
[ inf+ g1,0 (z)2π +
z∈R
l
X
I1,0 (k)]σ > 2πM,
k=1
sup
z∈R+
g1∞ (z)2π
+
l
X
I1∞ (k) < 2πσM.
k=1
(H2 ) :
sup
z∈R+
g20 (z)2π
+
l
X
I20 (k)
< 2πσM,
k=1
G(0)
1
, e2mπ
}, G(0) =
where σ = min{ G(π)
2
[ inf+ g2,∞ (z)2π +
z∈R
m
X
I2,∞ (k)]σ > 2πM.
k=1
e2mπ +1
, G(π)
2m(e2mπ −1)
=
2emπ
,m
2m(e2mπ −1)
=
√
M.
Preliminary
In this paper, we shall consider the following space
+
P C(I, R) = {u ∈ C(I, R); u|(xk ,xk+1 ) ∈ C(xk , xk+1 ), u(x−
k ) = u(xk ), ∃ u(xk ), k =
1, 2, · · · , l} P C 0 (I, R) = {u ∈ C(I, R); u|(xk ,xk+1 ) , u0 |(xk ,xk+1 ) ∈ C(xk , xk+1 ), u(x−
k) =
+
0 −
0
0 +
u(xk ), u (xk ) = u (xk ), ∃ u(xk ), u (xk ), k = 1, 2, · · · , l} with the norm
kukP C = sup |u(x)|, kukP C 0 = max{kukP C , ku0 kP C }, Then P C(I, R),P C 0 (I, R)
x∈[0,2π]
are Banach spaces.
Definition 2.1. A couple function (u, v) ∈ P C 0 (I, R)∩C 2 (I 0 , R)×P C 0 (I, R)∩
C 2 (I 0 , R) is called a solution of system (1.1) if it satisfies system (1.1)
Lemma 2.2. The vector (u, v) ∈ P C 0 (I, R) ∩ C 2 (I 0 , R) × P C 0 (I, R) ∩
C 2 (I 0 , R) is a solution of differential system (1.1)if and only if (u, v) ∈ P C 0 (I, R)×
P C 0 (I, R) is a solution of the following integral system

l
l
R 2π
P
P

∂G(x,y)

|y=xk I 1,k (u(xk )),
G(x, xk )I1,k (u(xk )) +
 u(x) = 0 G(x, y)g1 (y, u(y), v(y))dy +
∂y
k=1
k=1
l
l
R 2π
P
P


 v(x) = 0 G(x, y)g2 (y, v(y), u(y))dy +
G(x, xk )I2,k (v(xk )) +
k=1
k=1
∂G(x,y)
|y=xk I 2,k (v(xk )).
∂y
(2.1)
82
Existence of positive solutions to second-order...
where G(x, y) is the Green’s function to the priodic boundary value problem
−u00 + M u = 0, u(0) = u(2π), u0 (0) = u0 (2π), and
1
G(x, y) :=
Γ
(
em(x−y) + em(2π−x+y) ,
em(y−x) + em(2π−y+x) ,
0 ≤ y ≤ x ≤ 2π,
0 ≤ x ≤ y ≤ 2π.
here Γ = 2m(e2mπ − 1).
One can find that
2emπ
e2mπ + 1
=
G(π)
≤
G(x,
y)
≤
G(0)
=
.
2m(e2mπ − 1)
2m(e2mπ − 1)
(2.2)
For every positive solution of problem (1.1),one has
kukP C = sup |u(x)|
x∈[0,2π]
Without loss of generality, we assume lim |u(x)| = kukP C , ξ ∈ [xk , xk+1 ], k ∈
x→ξ
83
Ying He
{0, 1 . . . , l}, then by(2.2)
Z
kukP C
≤
2π
G(0)
g1 (y, u(y), v(y))dy + lim
t→ξ
0
+
l
X
∂G(x, y)
i=1
∂y
Z
¾
|y=xi I 1,i (u(xi ))
½X
l
G(x, xi )I1,i (u(xi ))
i=1
2π
=
G(0)
g1 (y, u(y), v(y))dy
0
( k
)
l
X
1 X m(ξ−xi )
m(2π−ξ+xi )
m(xi −ξ)
m(2π−xi +ξ)
+
[e
+e
]I1,i (u(xi )) +
[e
+e
]I1,i (u(xi ))
Γ i=1
i=k+1
)
( k
1 X
+
[−mem(ξ−xi ) + mem(2π−ξ+xi ) ]I 1,i (u(xi ))
Γ i=1
( l
)
1 X
+
[mem(xi −ξ) − mem(2π−xi +ξ) ]I 1,i (u(xi ))
Γ i=k+1
Z 2π
g1 (y, u(y), v(y))dy
= G(0)
0
+
1
Γ
½X
k
[em(ξ−xi ) (I1,i (u(xi )) − mI 1,i (u(xi ))) + em(2π−ξ+xi ) (I1,i (u(xi ))
i=1
¾
+ mI 1,i (u(xi )))]
1
+
Γ
½ X
l
[em(xi −ξ) (I1,i (u(xi )) + mI 1,i (u(xi ))) + em(2π−xi +ξ) (I1,i (u(xi ))
i=k+1
¾
− mI 1,i (u(xi )))]
It follows from − m1 I1,i (u) < I 1,i (u) <
mI 1,i (u) > 0. So
Z
2π
kukP C ≤ G(0)
0
1
I (u),
m 1,i
thatI1,i (u)−mI 1,i (u) > 0, I1,i (u)+
l
2e2mπ X
g1 (y, u(y), v(y))dy +
I1,i (u(xi )).
Γ i=1
(2.3)
84
Existence of positive solutions to second-order...
For any x ∈ [0, 2π], without loss of generality, we assume that x ∈ [xk , xk+1 ),,then
Z
2π
u(x) ≥ G(π)
g1 (y, u(y), v(y))dy +
Z
0
l
X
G(x, xi )I1,i (u(xi )) +
i=1
l
X
∂G(x, y)
i=1
∂y
|y=xi I 1,i (u(xi ))
2π
= G(π)
g1 (y, u(y), v(y))dy
0
k
1 X m(x−xi )
[e
(I1,i (u(xi )) − mI 1,i (u(xi ))) + em(2π−x+xi ) (I1,i (u(xi )) + mI 1,i (u(xi )))]
+
Γ i=1
l
1 X m(xi −x)
+
[e
(I1,i (u(xi )) + mI 1,i (u(xi ))) + em(2π−xi +x) (I1,i (u(xi )) − mI 1,i (u(xi )))]
Γ i=k+1
It follows from − m1 I1,i (u) < I 1,i (u) <
0, I1,i (u) + mI 1,i (u) > 0. So
Z
that I1,i (u) − mI 1,i (u) >
l
2π
u(x) ≥ G(π)
g1 (y, u(y), v(y))dy +
0
G(π)
≥
• G(0)
G(0)
≥ min{
1
I (u),
m 1,i
Z
2π
0
2X
I1,i (u(xi ))
Γ i=1
l
1 2e2mπ X
g1 (y, u(y), v(y))dy + 2mπ
I1,i (u(xi ))
e
Γ i=1
G(π) 1
,
}kukP C := σkukP C .
G(0) e2mπ
(2.4)
Similarly, v(x) ≥ σkvkP C .
In applications below, we take E = C(I, R) and define
K = {u ∈ C(I, R) : u(x) ≥ σkukP C , x ∈ [0, 2π]}.
It is easy to see that K is a closed convex cone in E. For r > 0, let Kr = {u ∈
K : ||u|| < r} and ∂Kr = {u ∈ K : ||u|| = r}. For any (u, v) ∈ K × K,define
mappings Φv : K → C(I, R+ ), Ψu : K → C(I, R+ ),and T : K × K →
C(I, R+ ) × C(I, R+ ) as follows
Φv (u)(x) =
Ψu (v)(x) =
R 2π
0
R 2π
0
G(x, y)g1 (y, u(y), v(y))dy +
G(x, y)g2 (y, v(y), u(y))dy +
T (u, v)(x) = (Φv (u)(x), Ψu (v)(x)),
l
P
k=1
l
P
G(x, xk )I1,k (u(xk )) +
G(x, xk )I2,k (v(xk )) +
k=1
l
P
k=1
l
P
k=1
∂G(x,y)
|y=xk I 1,k (u(xk )),
∂y
∂G(x,y)
|y=xk I 2,k (v(xk )),
∂y
x ∈ [0, 2π].
(2.5)
85
Ying He
Lemma 2.3.
T : K × K → K × K is completely continuous.Moreover,
T (K × K) ⊂ K × K.
Proof It is easy to see that T : K × K → K × K is completely continuous.
Thus we only need to show T (K × K) ⊂ K × K,
For any (u, v) ∈ K × K, we prove T (u, v) ∈ K × K,i.e. Φv (u) ∈ K and
Ψu (v) ∈ K. By using inequalities (2.3) and (2.4), we have that
Z
2π
kΦv (u)k ≤ G(0)
0
Φv (u)(x) ≥ min{
l
2e2mπ X
I1,i (u(xi ))
g1 (y, u(y), v(y))dy +
Γ i=1
G(π) 1
,
}kΦv (u)kP C := σkΦv (u)kP C , x ∈ [0, 2π]
G(0) e2mπ
Similarly, Ψu (v)(x) ≥ σkΨu (v)kP C . Thus, Φv (u)(x) ∈ K and Ψu (v)(x) ∈ K.
Consequently, T (K × K) ⊂ K × K
Lemma 2.4. Let Φ : K → K be a completely continuous mapping with
µΦu 6= u for every u ∈ ∂Kr and 0 < µ ≤ 1. Then i(Φ, Kr , K) = 1.
Lemma 2.5. Let Φ : K → K be a completely continuous mapping. Suppose that the following two conditions are satisfied:
(i) inf ||Φu|| > 0;
(ii) µΦu 6= u for every u ∈ ∂Kr and µ ≥ 1.
u∈∂Kr
Then, i(Φ, Kr , K) = 0.
Lemma 2.6. Let E be a Banach space and Ki ⊂ K (i = 1, 2) be a closed
set in E. For ri > 0 (i = 1, 2), denote Kri = {u ∈ Ki : ||u|| < ri } and
∂Kri = {u ∈ Ki : ||u|| = ri }. Suppose Φi : Ki → Ki is completely continuous.
If ui 6= Φi ui for any ui ∈ ∂Kri , then
i(Φ, Kr1 × Kr2 , K1 × K2 ) = i(Φ1 , Kr1 , K1 ) × i(Φ2 , Kr2 , K2 )
where Φ(u, v) =: (Φ1 u, Φ2 v) for any (u, v) ∈ K1 × K2 .
3
Main Results
Theorem 3.1. Assume that (H1 )−(H2 ) are satisfied. Then problem (1.1)
has at least one positive solution (u, v).
86
Existence of positive solutions to second-order...
To prove Theorem3.1,we first give the following lemmas.
Lemma 3.2.
If (H1 ) is satisfied, then i(Φv , KR1 \ K r1 , K) = 1.
Proof Since (H1 ) holds, then there exists 0 < ε < 1 such that
(1 − ε)[ inf+ g1,0 (z)2π +
z∈R
l
X
I1,0 (k)]σ > 2πM,
k=1
l
X
2πσM >
(I1∞ (k) + ε) + 2π( sup g1∞ (z) + ε).
(3.1)
z∈R+
k=1
By the definitions of g1,0 , I1,0 , one can find r0 > 0 such that for any x ∈
[0, 2π], 0 < u < r0 , v ∈ R+
g1 (x, u, v) ≥ g1,0 (v)(1 − ε)u, I1,k (u) ≥ I1,0 (k)(1 − ε)u.
Let r1 ∈ (0, r0 ), then for u ∈ ∂Kr1 , we have
u(x) ≥ σkuk = σr1 > 0. ∀x ∈ [0, 2π]
Thus
Z
Φv u(x)
2π
G(x, y)g1 (y, u(y), v(y))dy +
=
0
+
≥
∂y
Z
|y=xk I 1,k (u(xk ))
l
2π
G(π)
g1 (y, u(y), v(y))dy +
0
Z
≥
≥
G(x, xk )I1,k (u(xk ))
k=1
l
X
∂G(x, y)
k=1
l
X
2X
I1,k (u(xk ))
Γ k=1
l
X
2
I1,0 (k)u(xk )
G(π)(1 − ε)
g1,0 (v(y))u(y)dy + (1 − ε)
Γ
0
k=1
Ã
!
l
2X
(1 − ε)σr1 inf+ g1,0 (z)G(π)2π +
I1,0 (k)
z∈R
Γ k=1
2π
from which we see that inf ||Φv u||P C > 0, namely, hypothesis (i) of Lemma
u∈∂Kr1
2.5 holds. Next we show that µΦv u 6= u for any u ∈ ∂Kr1 , v ∈ K and µ ≥ 1.
87
Ying He
If this is not true, then there exist u0 ∈ ∂Kr1 and µ0 ≥ 1 such that
µ0 Φv u0 = u0 . Note that u0 (x) satisfies


−u000 (x) + M u0 (x) = µ0 g1 (x, u0 (x), v(x)),
x ∈ I 0,




 −∆u00 |x=xk = µ0 I1,k (u0 (xk )), k = 1, 2, · · · , l,

(3.2)
∆u0 |x=xk = µ0 I 1,k (u0 (xk )), k = 1, 2, · · · , l,



u0 (0) = u0 (2π),



 u0 (0) = u0 (2π).
0
0
Integrate from 0 to2π, using integration by parts in the left side, notice that
Z
2π
0
[−u000 (x)
+ M u0 (x)]dx =
l
X
Z
∆u00 (xk )
= −µ0
u0 (x)dx
0
k=1
l
X
2π
+M
Z
2π
u0 (x)dx
I1,k (u0 (xk )) + M
0
k=1
So we obtain
Z
2π
M
x0 (t)dt = µ0
0
l
X
Z
2π
I1,k (u0 (xk )) + µ0
g1 (y, u0 (y), v(y))dy
0
k=1
l
X
[(I1,0 (k) + inf+ g1,0 (z)2π]σr1
≥ (1 − ε)
k=1
z∈R
l
X
2πM r1 ≥ (1 − ε)[ (I1,0 (k) + inf+ g1,0 (z)2π]σr1 ,
k=1
z∈R
which contradicts with (3.1) . Hence,from Lemma 2.5 we have
i(Φ, Kr1 , K) = 0.
∀v ∈ K
(3.3)
On the other hand, from (H1 ), there exists H > r1 such that for any x ∈
[0, 2π], u ≥ H, v ∈ R+
g1 (x, u, v) ≤ (g1∞ (v) + ε)u, I1,k (u) ≤ (I1∞ (k) + ε)u,
(3.4)
Choose R1 > R0 := max{ Hσ , r1 } and let u ∈ ∂KR1 ,v ∈ K. Since u(x) ≥
σ||u||P C = σR1 > H for x ∈ [0, 2π], v ∈ K. Now we show that µΦv u 6= u
for any u ∈ ∂KR1 , v ∈ K and 0 < µ ≤ 1. In fact, if there exist u0 ∈ ∂KR1
88
Existence of positive solutions to second-order...
and 0 < µ0 ≤ 1 such that µ0 Φv u0 = u0 , then u0 (x) satisfies equation (3.2).
Integrating from 0 to 2π,we obtain
Z
2π
M
Z
l
X
u0 (x)dx = µ0 [
I1,k (u0 (xk )) +
0
g1 (x, u0 (x), v(x))dx]
0
k=1
Z
l
X
∞
≤
(I1 (k) + ε)u0 (xk ) +
k=1
2π
2π
0
u0 (x)dx( sup g1∞ (z) + ε)
z∈R+
l
X
≤ R1 [ (I1∞ (k) + ε) + 2π( sup g1∞ (z) + ε)]
z∈R+
k=1
i.e.,
l
X
2πσM R1 ≤ R1 [ (I1∞ (k) + ε) + 2π( sup g1∞ (z) + ε)]
z∈R+
k=1
which is a contradiction with (3.1) .
Let R1 = max{r1 , Hσ }, then for any u ∈ ∂KR1 , v ∈ K and 0 < µ ≤ 1, we
have µΦv u 6= u. Thus
i(Φ, KR1 , K) = 1.
(3.5)
In view of (3.3) and (3.5), we obtain
i(Φ, KR1 \ K r1 , K) = 1.
Lemma 3.3. :
If (H2 ) is satisfied, then i(Ψu , KR2 \ K r2 , K) = −1.
Proof Since (H2 ) holds, there exists 0 < ε < 1 such that
2πσM >
l
X
(I20 (k) + ε) + 2π( sup g20 (z) + ε),
z∈R+
k=1
(1 − ε)[ inf+ g2,∞ (z)2π +
z∈R
l
X
I2,∞ (k)]σ > 2πM.
(3.6)
k=1
One can find r0 > 0 such that for any x ∈ [0, 2π], 0 ≤ v ≤ r0 , u ∈ R+
g2 (x, v, u) ≤ (g20 (u) + ε)v, I2,k (v) ≤ (I20 (k) + ε)v,
(3.7)
Let r2 ∈ (0, r0 ). Now we prove that µΨu v 6= v for any v ∈ ∂Kr2 , u ∈ K and
0 < µ ≤ 1. If this is not true, then there exist v0 ∈ ∂Kr2 and 0 < µ0 ≤ 1 such
89
Ying He
that µ0 Ψu v0 = v0 . Note that v0 (x) satisfies

00

x ∈ I 0,
 −v0 (x) + M v0 (x) = µ0 g2 (x, v0 (x), u(x)),


0


 −∆v0 |x=xk = µ0 I2,k (v0 (xk )), k = 1, 2, · · · , l,
∆v0 |x=xk = µ0 I 2,k (v0 (xk )), k = 1, 2, · · · , l,



v0 (0) = v0 (2π),



 v 0 (0) = v 0 (2π).
0
0
(3.8)
Integrating from 0 to 2π,we obtain
Z
2π
M
v0 (x)dx =
0
≤
l
X
(I20 (k)
l
X
Z
k=1
g2 (x, v0 (x), u(x))dx
0
Z
2π
+ ε)v0 (xk ) +
0
k=1
2π
µ0 I2,k (v0 (xk )) + µ0
v0 (x)dx( sup g20 (z) + ε)
z∈R+
l
X
≤ r2 [ (I20 (k) + ε) + 2π( sup g20 (z) + ε)].
z∈R+
k=1
so
l
X
2πσM r2 ≤ r2 [ (I20 (k) + ε) + 2π( sup g20 (z) + ε)].
z∈R+
k=1
which is a contradiction with (3.6). By Lemma 2.4, we have
i(Ψu , Kr2 , K) = 1.
(3.9)
On the other hand, from (H2 ), there exists H > r2 such that for any x ∈
[0, 2π], v ≥ H, u ∈ R+
g2 (x, v, u) ≥ g2,∞ (u)(1 − ε)v, I2,k (v) ≥ I2,∞ (k)(1 − ε)v,
(3.10)
Choose R2 > R0 := max{ Hσ , r2 } and let v ∈ ∂KR2 ,u ∈ K. Since v(x) ≥
σ||v||P C = σR2 > H for x ∈ [0, 2π], u ∈ K, from (3.10) we see that
g2 (x, v(x), u(x)) ≥ g2,∞ (u(x))(1 − ε)v(x) ≥ σg2,∞ (u(x))(1 − ε)R2 ,
I2,k (v(xk ) ≥ σI2,∞ (k)(1 − ε)R2 ,
Essentially the same reasoning as above yields
inf ||Ψu v||P C > 0. Next we
v∈∂KR2
show that if R2 is large enough, then µΨu v 6= v for any v ∈ ∂KR2 , u ∈ K and
µ ≥ 1. In fact, if there exist v0 ∈ ∂KR2 and µ0 ≥ 1 such that µ0 Ψu v0 = v0 ,
90
Existence of positive solutions to second-order...
then v0 (x) satisfies equation (3.8) . Integrate from 0 to 2π, using integration
by parts in the left side to obtain
Z 2π
Z 2π
l
X
M
v0 (x)dx =
µ0 I2,k (v0 (xk )) + µ0
g2 (x, v0 (x), u(x))dx
0
0
k=1
l
X
≥ (1 − ε)[
I2,∞ (k) + inf+ g2,∞ (z)2π]σR2 .
k=1
z∈R
So we obtain
l
X
2πM R2 ≥ (1 − ε)[
I2,∞ (k) + inf+ g2,∞ (z)2π]σR2
z∈R
k=1
which contradicts with (3.6),too.
Hence hypothesis (ii) of Lemma 2.5 is satisfied and
i(Ψu , KR2 , K) = 0.
(3.11)
In view of (3.9) and (3.11), we obtain
i(Ψu , KR2 \ K r2 , K) = −1
Proof of Theorem 3.1. Since (H1 ) − (H2 ) are satisfied ,from Lemma2.3 we
get Φv : K → K, Ψu : K → K and T : K × K → K × K are completely
continuous. From Lemma3.2,3.3 and 2.6 we have
i(T, KR1 \K r1 ×KR2 \K r2 , K×K) = i(Φv , KR1 \K r1 , K)×i(Ψu , KR2 \K r2 , K) = −1
Thus, system (1.1)has at least one positive solution (u,v).
Corollary 3.4. The conclusion of Theorem 3.1 is valid if (H1 ) and (H2 )
are replaced by
(H1∗ )
inf g1,0 (z) = ∞ or
z∈R+
l
X
I1,0 (k) = ∞;
k=1
sup g1∞ (z) = 0 and I1∞ (k) = 0,
k = 1, 2, ...l.
z∈R+
(H2∗ ) sup g20 (z) = 0 and I20 (k) = 0, k = 1, 2, ...l;
z∈R+
inf g2,∞ (z) = ∞ or
z∈R+
l
X
k=1
I2,∞ (k) = ∞.
Ying He
91
References
[1]
X.Y. Cheng and C.K. Zhong, Existence of positive solutions for a secondorder ordinary differential system, J. Math. Anal. Appl., 312, (2005), 1423.
[2]
W. Ding and M. Han, Periodic boundary value problem for the second
order impulsive functional differential equations, Appl. Math. Comput.,
155, (2004), 709-726.
[3]
D. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones,
Academic Press, Boston, MA, 1988.
[4]
D. Guo, J. Sun and Z. Liu, Functional Methods in Nonlinear Ordinary
Differential Equations, Shandong Science and Technology Press,Jinan,
1995 (in Chinese).
[5] S.G. Hristova and D.D. Bainov, Monotone-iterative techniques of V. Lakshmikantham for a boundary value problem for systems of impulsive
differential-difference equations,J. Math. Anal. Appl., 197, (1996), 1-13.
[6] V. Lakshmikntham, D.D. Bainov and P.S. Simeonov, Theory of Impulsive
Differential Equations, World Scientific, Singapore, 1989.