Electronic Journal of Differential Equations, Vol. 2010(2010), No. 166, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
IMPULSIVE BOUNDARY-VALUE PROBLEMS FOR
FIRST-ORDER INTEGRO-DIFFERENTIAL EQUATIONS
XIAOJING WANG, CHUANZHI BAI
Abstract. This article concerns boundary-value problems of first-order nonlinear impulsive integro-differential equations:
y 0 (t) + a(t)y(t) = f (t, y(t), (T y)(t), (Sy)(t)),
t ∈ J0 ,
∆y(tk ) = Ik (y(tk )), k = 1, 2, . . . , p,
Z c
y(0) + λ
y(s)ds = −y(c), λ ≤ 0,
0
where J0 = [0, c] \ {t1 , t2 , . . . , tp }, f ∈ C(J × R × R × R, R), Ik ∈ C(R, R),
a ∈ C(R, R) and a(t) ≤ 0 for t ∈ [0, c]. Sufficient conditions for the existence
of coupled extreme quasi-solutions are established by using the method of lower
and upper solutions and monotone iterative technique. Wang and Zhang [18]
studied the existence of extremal solutions for a particular case of this problem,
but their solution is incorrect.
1. Introduction
In recent years, many authors have paid attention to the research of differential
equations with impulsive boundary conditions, because of their potential applications; see for example [4, 6, 9, 12, 13, 15, 17]. First-order and second-order impulsive
differential equations with anti-periodic boundary conditions have also drawn much
attention; see [1, 2, 3, 5, 7, 8, 14, 16, 19].
Recently, Wang and Zhang [18] studied the existence of extremal solutions of
the following nonlinear anti-periodic boundary value problem of first-order integrodifferential equation with impulse at fixed points
y 0 (t) = f (t, y(t), (T y)(t), (Sy)(t)),
∆y(tk ) = Ik (y(tk )),
t ∈ J0 ,
k = 1, 2, . . . , p,
(1.1)
y(0) = −y(T ),
where J = [0, T ], J0 = J \ {t1 , t2 , . . . , tp }, 0 < t1 < t2 < · · · < tp < T , f ∈
−
C(J × R × R × R, R), Ik ∈ C(R, R), ∆y(tk ) = y(t+
k ) − y(tk ) denotes the jump of
2000 Mathematics Subject Classification. 34A37, 34B15.
Key words and phrases. Impulsive integro-differential equation;
coupled lower-upper quasi-solutions; monotone iterative technique.
c
2010
Texas State University - San Marcos.
Submitted November 1, 2010. Published November 17, 2010.
Supported by grant 10771212 from the National Natural Science Foundation of China.
1
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X. WANG, C. BAI
EJDE-2010/166
−
y(t) at t = tk ; y(t+
k ) and y(tk ) represent the right and left limits of y(t) at t = tk ,
respectively.
Z t
Z T
(T y)(t) =
k(t, s)y(s)ds, (Sy)(t) =
h(t, s)y(s)ds,
0
0
+
k ∈ C(D, R ), D = {(t, s) ∈ J × J : t ≥ s}, h ∈ C(J × J, R+ ). Unfortunately, their
extremal solutions y∗ (t), y ∗ (t) are wrong. In fact, by [18, Theorem 3.1] we obtain
y∗0 (t) = f (t, y∗ (t), (T y∗ )(t), (Sy∗ )(t)),
∆y∗ (tk ) = Ik (y∗ (tk )),
t ∈ J0 ,
k = 1, 2, . . . , p,
∗
y∗ (0) = −y (T ),
and
y ∗0 (t) = f (t, y ∗ (t), (T y ∗ )(t), (Sy ∗ )(t)),
∗
∗
∆y (tk ) = Ik (y (tk )),
t ∈ J0 ,
k = 1, 2, . . . , p,
∗
y (0) = −y∗ (T ),
which implies that y∗ (t), y ∗ (t) are not solutions of (1.1). So the conclusions of [18]
are reconsidered here, for a more general equation.
In this paper, we investigate the following integral boundary value problem for
first-order integro-differential equation with impulses at fixed points
y 0 (t) + a(t)y(t) = f (t, y(t), (T y)(t), (Sy)(t)),
t ∈ J0 ,
∆y(tk ) = Ik (y(tk )), k = 1, 2, . . . , p,
Z c
y(0) + λ
y(s)ds = −y(c), λ ≤ 0,
(1.2)
0
where J = [0, c], J0 = J \ {t1 , t2 , . . . , tp }, 0 < t1 < t2 < · · · < tp < c, f ∈
C(J × R × R × R, R), Ik ∈ C(R, R), a ∈ C(R, R) and a(t) ≤ 0 for t ∈ J.
Z t
Z c
(T y)(t) =
k(t, s)y(s)ds, (Sy)(t) =
h(t, s)y(s)ds,
0
0
k ∈ C(D, R+ ), D = {(t, s) ∈ J × J : t ≥ s}, h ∈ C(J × J, R+ ).
Remark 1.1. If a(t) ≡ 0 and λ ≡ 0, then (1.2) reduces to (1.1).
We will give the concept of coupled quasi-solutions of BVP (1.2) in next section. It is well known that the monotone iterative technique offers an approach for
obtaining approximate solutions of nonlinear differential equations, for details, see
[10, 11] and the references therein. The aim of this paper is to investigate the existence of coupled quasi-solutions of (1.2) by using the method of upper and lower
solutions combined with a monotone iterative technique. Our result correct and
generalize the main result of [18].
2. Preliminaries
In this section, we present some definitions needed for introducing the concept
of quasi-solutions for (1.2). Let
P C(J) = {y : J → R : y is continuous at t ∈ J0 ;
−
−
y(0+ ), y(T − ), y(t+
k ), y(tk ) exist and y(tk ) = y(tk ), k = 1, . . . , p},
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IMPULSIVE BOUNDARY-VALUE PROBLEMS
3
P C 1 (J) = {y ∈ P C(J) : y is continuously differentiable for t ∈ J0 ;
0 −
y 0 (0+ ), y 0 (T − ), y 0 (t+
k ), y (tk ) exist, k = 1, . . . , p},
The sets P C(J) and P C 1 (J) are Banach spaces with the norms
kykP C(J) = sup{|y(t)| : t ∈ J},
kykP C 1 (J) = kykP C(J) + ky 0 kP C(J) .
Definition 2.1. Functions α0 , β0 ∈ P C 1 (J) are said to be coupled lower-upper
quasi-solutions to the problem (1.2) if
α00 (t) + a(t)α0 (t) ≤ f (t, α0 (t), (T α0 )(t), (Sα0 )(t)),
t ∈ J0 ,
∆α0 (tk ) ≤ Ik (α0 (tk )), k = 1, 2, . . . , p,
Z c
α0 (0) + λ
α0 (s)ds ≤ −β0 (c), λ ≤ 0,
0
β00 (t) + a(t)β0 (t) ≥ f (t, β0 (t), (T β0 )(t), (Sβ0 )(t)),
t ∈ J0 ,
(2.1)
∆β0 (tk ) ≥ Ik (β0 (tk )), k = 1, 2, . . . , p,
Z c
β0 (0) + λ
β0 (s)ds ≥ −α0 (c), λ ≤ 0.
0
Note that if α0 (c) = β0 (c), then the above definition reduces to the notion of
lower and upper solutions of (1.2).
Definition 2.2. Functions v, w ∈ P C 1 (J) are said to be coupled quasi-solutions
to (1.2) if
v 0 (t) + a(t)v(t) = f (t, v(t), (T v)(t), (Sv)(t)),
t ∈ J0 ,
∆v(tk ) = Ik (v(tk )), k = 1, 2, . . . , p,
Z c
v(0) + λ
v(s)ds = −w(c), λ ≤ 0,
0
w0 (t) + a(t)w(t) = f (t, w(t), (T w)(t), (Sw)(t)),
t ∈ J0 ,
(2.2)
∆w(tk ) = Ik (w(tk )), k = 1, 2, . . . , p,
Z c
w(0) + λ
w(s)ds = −v(c), λ ≤ 0.
0
1
Let α0 , β0 ∈ P C (J) and α0 (t) ≤ β0 (t) for t ∈ J0 . In what follows we define the
segment
[α0 , β0 ] = {u ∈ P C 1 (J) : α0 (t) ≤ u(t) ≤ β0 (t), t ∈ J}.
Definition 2.3. Let u, v be coupled quasi-solutions of (1.2) such as u(t) ≤ v(t)
for t ∈ J0 . Assume that α0 , β0 ∈ P C 1 (J) and α0 (t) ≤ β0 (t) for t ∈ J0 . Coupled
quasi-solutions u, v of (1.2) are called coupled minimal-maximal quasi-solutions in
segment [α0 , β0 ] if α0 (t) ≤ u(t), v(t) ≤ β0 (t) for t ∈ J0 and for any U, V coupled
quasi-solutions of (1.2), such as α0 (t) ≤ U (t), V (t) ≤ β0 (t) for t ∈ J0 we have
u(t) ≤ U (t) and V (t) ≤ v(t), t ∈ J0 .
For convenience, we assume the following conditions are satisfied
(H1) Functions α0 (t), β0 (t) are coupled lower-upper quasi-solutions of (1.2) such
that α0 (t) ≤ β0 (t) for t ∈ J0 .
(H2) There exist M > 0, N, N1 ≥ 0 such that
f (t, x1 , y1 , z1 ) − f (t, x2 , y2 , z2 ) ≥ −M (x1 − x2 ) − N (y1 − y2 ) − N1 (z1 − z2 ),
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X. WANG, C. BAI
EJDE-2010/166
for α0 ≤ x2 ≤ x1 ≤ β0 , T α0 ≤ y2 ≤ y1 ≤ T β0 , Sα0 ≤ z2 ≤ z1 ≤ Sβ0 ,
t ∈ J.
(H3) There exist 0 ≤ Lk < 1, k = 1, 2, . . . , p, satisfy
Ik (x) − Ik (y) ≥ −Lk (x − y),
for α0 ≤ y ≤ x ≤ β0 , t ∈ J.
Now we consider the problem
y 0 (t) + M y(t) + N (T y)(t) + N1 (Sy)(t) = σ(t),
∆y(tk ) = −Lk y(tk ) + bk ,
t ∈ J0 ,
k = 1, 2, . . . , p,
(2.3)
y(0) = b,
where M > 0, N, N1 ≥ 0, Lk < 1, k = 1, 2, . . . , p.
Lemma 2.4. If y ∈ P C 1 (J), M > 0, N, N1 ≥ 0, Lk < 1, k = 1, 2, . . . , p, and
k̄ + h̄ +
p
X
Li < 1,
(2.4)
i=1
where

−1
−M c

),
if M > 1,
k0 cM (1 − e
−1
k̄ = k0 cM (1 − M e−M c ), if 0 < M ≤ 1,

1
2
if M = 0.
2 k0 c ,
(
h0 cM −1 (1 − e−M c ), if M > 0,
h̄ =
h0 c2 ,
if M = 0,
where k0 = max0≤s≤t≤c k(t, s) and h0 = max0≤t,s≤c h(t, s). Then (2.3) has a
unique solution.
Proof. If y ∈ P C 1 (J) is a solution of (2.3), then, by integrating, we obtain
Z t
−M t
y(t) = be
+
e−M (t−s) [σ(s) − N (T y)(s) − N1 (Sy)(s)]ds
0
X
+
e−M (t−ti ) (−Li y(ti ) + bi ).
(2.5)
0<ti <t
Conversely, if y(t) ∈ P C(J) is solution of the above-mentioned integral equation
(2.5), then it is easy to check that y 0 (t) = −M y(t) − N (T y)(t) − N1 (Sy)(t) + σ(t),
t 6= tk , ∆y(tk ) = −Lk y(tk ) + bk , k = 1, 2, . . . , p, and y(0) = b. So (2.3) is equivalent
to the integral equation (2.5). Now, we define operator B : P C(J) → P C(J) as
Z t
(By)(t) = be−M t +
e−M (t−s) [σ(s) − N (T y)(s) − N1 (Sy)(s)]ds
0
(2.6)
X
+
e−M (t−ti ) (−Li y(ti ) + bi ).
0<ti <t
For each u, v ∈ P C(J), we have
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IMPULSIVE BOUNDARY-VALUE PROBLEMS
5
Z t
|(Bu)(t) − (Bv)(t)| ≤N e−M (t−s) (T u − T v)(s)ds
0
Z t
+ N1 e−M (t−s) (Su − Sv)(s)ds
0
X
+
Li |e−M (t−ti ) (u(ti ) − v(ti ))|.
(2.7)
0<ti <t
We easily check that
Z
t −M (t−s)
e
(T u − T v)(s)ds
0

−1
−M t

)ku − vkP C ,
if M > 1,
k0 tM (1 − e
−1
≤ k0 tM (1 − M e−M t )ku − vkP C , if 0 < M ≤ 1,

 1 2
if M = 0,
k0 2 t ku − vkP C ,
(2.8)
and
(
h0 cM −1 (1 − e−M t )ku − vkP C , if M > 0,
e
h0 ctku − vkP C ,
if M = 0.
0
(2.9)
Substituting (2.8) and (2.9) into (2.7), we obtain
Z
t
−M (t−s)
(Su − Sv)(s)ds ≤
kBu − BvkP C ≤ (k̄ + h̄ +
p
X
Li )ku − vkP C .
i=1
This indicates that B is a contraction mapping (by (2.4)). Then there is one unique
y ∈ P C(J) such that By = y, that is, (2.3) has a unique solution.
Lemma 2.5 ([18]). Assume that y ∈ P C 1 (J) satisfies
y 0 (t) + M y(t) + N (T y)(t) + N1 (Sy)(t) ≤ 0,
∆y(tk ) ≤ −Lk y(tk ),
t ∈ J0 ,
k = 1, 2, . . . , p,
(2.10)
y(0) ≤ 0,
where M > 0, N, N1 ≥ 0, Lk < 1, k = 1, 2, . . . , p, and
Z c
p
Y
q(s)ds ≤
(1 − L̄j )
0
with L̄k = max{Lk , 0}, k = 1, 2, . . . , p,
Z t
Z
Y
q(t) = N
k(t, s)eM (t−s)
(1−Lk )ds+N1
0
(2.11)
j=1
s<tk <c
0
c
h(t, s)eM (t−s)
Y
(1−Lk )ds,
s<tk <c
then y ≤ 0.
3. Main result
Theorem 3.1. If (H1),(H2),(H3) are satisfied, and, in addition, if there exist M >
0, N, N1 ≥ 0, Lk < 1, k = 1, 2, . . . , p, such that (2.4) and (2.11) hold, then (1.2)
has, in segment [α0 , β0 ] the coupled minimal-maximal quasi-solutions.
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X. WANG, C. BAI
EJDE-2010/166
Proof. For convenience, let (Kφ)(t) = N (T φ)(t) + N1 (Sφ)(t). We now construct
two sequences {αn (t)} and {βn (t)} that satisfy the following problems
αi0 (t) + a(t)αi−1 (t) + M αi (t) + (Kαi )(t)
= f (t, αi−1 (t), (T αi−1 )(t), (Sαi−1 )(t)) + M αi−1 (t) + (Kαi−1 )(t),
t ∈ J0 ,
∆αi (tk ) = Ik (αi−1 (tk )) − Lk (αi (tk ) − αi−1 (tk )), k = 1, 2, . . . , p,
Z c
αi (0) + λ
αi−1 (s)ds = −βi−1 (c),
(3.1)
0
and
βi0 (t) + a(t)βi−1 (t) + M βi (t) + (Kβi )(t)
t ∈ J0 ,
= f (t, βi−1 (t), (T βi−1 )(t), (Sβi−1 )(t)) + M βi−1 (t) + (Kβi−1 )(t),
∆βi (tk ) = Ik (βi−1 (tk )) − Lk (βi (tk ) − βi−1 (tk )), k = 1, 2, . . . , p,
Z c
βi (0) + λ
βi−1 (s)ds = −αi−1 (c).
(3.2)
0
For each φ, ψ ∈ [α0 , β0 ], we consider the equation
y 0 (t) + M y(t) + (Ky)(t)
= f (t, φ(t), (T φ)(t), (Sφ)(t)) − a(t)φ(t) + M φ(t) + (Kφ)(t),
t ∈ J0 ,
∆y(tk ) = Ik (φ(tk )) − Lk (y(tk ) − φ(tk )), k = 1, 2, . . . , p,
Z c
y(0) + λ
φ(s)ds = −ψ(c).
(3.3)
0
By condition (2.4) and Lemma 2.4, we know that (3.3) has a unique solution
y(t) ∈ P C 1 (J). Define the operator A : P C 1 (J) × P C 1 (J) → P C 1 (J) as A(φ, ψ) =
y. Let αn (t) = A(αn−1 , βn−1 )(t) and βn (t) = A(βn−1 , αn−1 )(t), n = 1, 2, . . . , we
will prove that {αn }, {βn } have the following properties.
(i) αi−1 ≤ αi , βi ≤ βi−1 ;
(ii) αi ≤ βi , i = 1, 2, ....
Firstly, we prove that α0 ≤ α1 . Set p(t) = α0 (t) − α1 (t), it follows that
p0 (t) + M p(t) + N (T p)(t) + N1 (Sp)(t) = p0 (t) + M p(t) + (Kp)(t) ≤ 0,
∆p(tk ) ≤ −Lk p(tk ),
k = 1, 2, . . . , p,
(3.4)
p(0) ≤ 0.
Then by condition (2.11) and Lemma 2.5, we get p(t) ≤ 0, which implies that
α0 (t) ≤ α1 (t), for all t ∈ J0 . In a similar way, it can be proved that β1 (t) ≤ β0 (t),
for all t ∈ J0 . Now we prove that α1 (t) ≤ β1 (t), for all t ∈ J0 . In fact, setting
p(t) = α1 (t) − β1 (t) and using assumption, we obtain
p0 (t) + M p(t) + N (T p)(t) + N1 (Sp)(t)
= α10 (t) − β10 (t) + M (α1 (t) − β1 (t)) + N (T α1 (t) − T β1 (t)) + N1 (Sα1 (t) − Sβ1 (t))
= f (t, α0 (t), (T α0 )(t), (Sα0 )(t)) − a(t)α0 (t) + M α0 (t) + N (T α0 )(t) + N1 (Sα0 )(t)
− f (t, β0 (t), (T β0 )(t), (Sβ0 )(t)) + a(t)β0 (t) − M β0 (t) − N (T β0 )(t) − N1 (Sβ0 )(t)
≤ a(t)(β0 (t) − α0 (t)) ≤ 0,
t ∈ J0 ,
and
∆p(tk ) = −Lk p(tk ) + Ik (α0 (tk )) − Ik (β0 (tk )) + Lk α0 (tk ) − Lk β0 (tk ) ≤ −Lk p(tk ),
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IMPULSIVE BOUNDARY-VALUE PROBLEMS
7
c
Z
p(0) = α1 (0) − β1 (0) = λ
(β0 (s) − α0 (s))ds + α0 (c) − β0 (c) ≤ 0.
0
Again by Lemma 2.5, we obtain p(t) ≤ 0, that is, α1 (t) ≤ β1 (t) for all t ∈ J0 . Thus
we have α0 (t) ≤ α1 (t) ≤ β1 (t) ≤ β0 (t) for all t ∈ J0 . Continuing this process, by
induction, one can obtain monotone sequence {αn (t)} and {βn (t)} such that
α0 (t) ≤ α1 (t) ≤ · · · ≤ αn (t) ≤ · · · ≤ βn (t) ≤ . . . β1 (t) ≤ β0 (t),
t ∈ J0 ,
where each αi (t), βi (t) ∈ P C 1 (J) satisfies (3.1) and (3.2). As the sequences {αn },
{βn } are uniformly bounded and equi-continuous, by employing the standard arguments Ascoli-Arzela criterion [12], we conclude that the sequences {αn } and {βn }
converge uniformly on J0 with
lim αn (t) = y∗ (t),
n→∞
lim βn (t) = y ∗ (t).
n→∞
Obviously, y∗ (t), y ∗ (t) are coupled lower-upper quasi-solutions of (1.2). Now we
have to prove that (y∗ , y ∗ ) are coupled minimal-maximal quasi-solutions of problem
(1.2) in segment [α0 , β0 ]. Let x, z be coupled quasi-solutions of (1.2) such that
αn (t) ≤ x(t),
z(t) ≤ βn (t),
t ∈ J0
for some n ∈ N. Put q(t) = αn+1 (t) − x(t), for t ∈ J0 . Form definition of αn+1 and
properties of quasi-solution x(t), we obtain
q 0 (t) + M q(t) + N (T q)(t) + N1 (Sq)(t)
= f t, αn (t), (T αn )(t), (Sαn )(t) − a(t)αn (t) + M αn (t) + N (T αn )(t)
+ N1 (Sαn )(t) − f t, x(t), (T x)(t), (Sx)(t) + a(t)x(t) − M x(t)
− N (T x)(t) − N1 (Sx)(t)
≤ a(t)(x(t) − αn (t)) ≤ 0,
t ∈ J0 ,
and
∆q(tk ) = −Lk q(tk ) + Ik (αn (tk )) − Ik (x(tk )) + Lk αn (tk ) − Lk x(tk ) ≤ −Lk q(tk ),
Z c
q(0) = αn+1 (0) − x(0) = λ
(x(s) − αn (s))ds + z(c) − βn (c) ≤ 0.
0
By Lemma 2.5, we have q(t) ≤ 0 for all t ∈ J0 , that is αn+1 (t) ≤ x(t). Similarly,
we can prove that z(t) ≤ βn+1 (t) for all t ∈ J0 .
By induction, we obtain
αm (t) ≤ x(t),
z(t) ≤ βm (t),
t ∈ J0 ,
for m ∈ N.
If m → ∞, it yields
y∗ (t) ≤ x(t),
z(t) ≤ y ∗ (t),
t ∈ J0 .
It shows that (y∗ , y ∗ ) are coupled minimal-maximal quasi-solutions of problem (1.2)
in segment [α0 , β0 ].
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X. WANG, C. BAI
EJDE-2010/166
Example 3.2. Consider the problem
Z
Z
t
1 t −(t−s)
5 1
y 0 (t) − (1 − e−t )y(t) = −y(t) −
te
y(s)ds −
y(s)ds,
4
8 0
6 0
t ∈ [0, t1 ) ∪ (t1 , 1],
1
1
∆y(t1 ) = − y(t1 ), t1 =
9
3
Z
1 1
y(0) −
y(s)ds = −y(1).
6 0
(3.5)
where a(t) = − 4t (1 − e−t ) ≤ 0, I1 (x) = − 19 x, L1 = 91 and λ = − 61 < 0. Let
f (t, x, y, z) = −M x − N y − N1 z, M = 1, N = 83 , N1 = 65 , J = [0, 1], c = 1,
k(t, s) = 3t e−(t−s) , h(t, s) = 1, then for t ∈ J, xi , yi , zi ∈ R, i = 1, 2, x1 ≥ x2 ,
y 1 ≥ y 2 , z1 ≥ z2 ,
3
5
f (t, x1 , y1 , z1 ) − f (t, x2 , y2 , z2 ) = −(x1 − x2 ) − (y1 − y2 ) − (z1 − z2 ).
8
6
Thus the condition (H2) holds. It is easy to see that k0 =
1
−1
) and
3 (1 − e
1
3,
h0 = 1, k̄ =
1
3 h̄
=
h̄ + k̄ + L1 = 0.9359 < 1.
Hence the condition (2.4) holds. Moreover, we have
Z
1
Z
5 1 (t−s)
t −(t−s) (t−s)
e
e
(1 − L1 )ds +
e
(1 − L1 )ds dt
8 0 3
6 0
0
Z 1 2
20
t
+ (1 − e−1 )et dt
=
18
27
0
1
20
=
+ (e + e−1 − 2) = 0.8231 < 0.8889 = 1 − L1 ,
54 27
Z
q(s)ds ≤
0
1
3 Z
t
which implies that the condition (2.11) holds. Let
5
α0 (t) = − ,
4
β0 (t) = 2 − t,
t ∈ [0, 1].
Then α0 (t) and β0 (t) are coupled lower-upper quasi-solutions of problem (??). In
fact,
5
5
t(1 − e−t ) ≤ 2 + t(1 − e−t )
16
32
Z t
Z
5
25 1
5
−(t−s)
< +
te
ds +
ds
4 32 0
24 0
= f (t, α0 (t), (T α0 )(t), (Sα0 )(t)),
5
= −L1 α0 (1/3)
∆α0 (1/3) = 0 <
36
Z 1
1
25
α0 (0) −
α0 (s)ds = −
< −1 = −β0 (1),
6 0
24
α00 (t) + a(t)α0 (t) =
EJDE-2010/166
IMPULSIVE BOUNDARY-VALUE PROBLEMS
9
and
1
β00 (t) + a(t)β0 (t) = −1 − t(1 − e−t )(2 − t)
4
1
≥ −1 − (1 − e−1 )
4
27 3 −1
>− + e
12 8
1
3
15
≥ t − 2 − t(3 − t) + te−t −
8
8
12
Z
Z
1 t −(t−s)
5 1
=t−2−
te
(2 − s)ds −
(2 − s)ds
8 0
6 0
= f (t, β0 (t), (T β0 )(t), (Sβ0 )(t)),
5
= −L1 β0 (1/3)
∆β0 (1/3) = 0 > −
27
Z
1 1
7
5
β0 (0) −
β0 (s)ds = > = −α0 (1).
6 0
4
4
Obviously, α0 (t) ≤ β0 (t). Thus, all the conditions of Theorem 3.1 are satisfied,
so problem (3.5) has the coupled minimal-maximal quasi-solutions in the segment
[α0 (t), β0 (t)].
Acknowledgments. The authors want to thank the anonymous referees for their
valuable comments and suggestions which improved the presentation of this article.
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Xiaojing Wang
Department of Mathematics, Huaiyin Normal University, Huaian, Jiangsi 223300, China
E-mail address: wangxj2010106@sohu.com
Chuanzhi Bai
Department of Mathematics, Huaiyin Normal University, Huaian, Jiangsi 223300, China
E-mail address: czbai8@sohu.com