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Hindawi Publishing Corporation
Boundary Value Problems
Volume 2007, Article ID 73176, 16 pages
doi:10.1155/2007/73176
Research Article
Extremal Solutions of Periodic Boundary Value
Problems for First-Order Impulsive Integrodifferential
Equations of Mixed-Type on Time Scales
Yongkun Li and Hongtao Zhang
Received 12 October 2006; Accepted 21 May 2007
Recommended by Ivan Kiguradze
We consider the existence of minimal and maximal solutions of periodic boundary value
problems for first-order impulsive integrodifferential equations of mixed-type on time
scales by establishing a comparison result and using the monotone iterative technique.
Copyright © 2007 Y. Li and H. Zhang. 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
The theory of calculus on time scales (see [1, 2] a nd references cited therein) was ini-
tiated by Stefan Hilger in his Ph.D. thesis in 1990 [3] in order to unify continuous and
discrete analyses, and it has a tremendous potential for applications and has recently re-
ceived much attention since his foundational work. In this paper, we will study the peri-
odic boundary value problem for the first-order impulsive integrodifferential equations
of mixed-type (PBVP):
u
Δ
(t) = f

t,u(t),[Tu](t),[Su](t)

, t = t

k
, t ∈ J
T
,
u

t
+
k

−
u

t
−
k

=
I
k

u

t
−
k

, k = 1,2, , p,
u(0)
= u(T),

(1.1)
where
T is a time scale which has the subspace topology inherited from the standard
topology on
R.ForeachintervalJ of R, we denote by J
T
= J ∩ T, f ∈ C[J
T
× R × R ×
R
,R], J = [0,T], I
k
∈ C[R,R], where u(t
+
k
)andu(t
−
k
) represent right and left limits of
u(t)att
= t
k
(k = 1,2, , p) in the sense of time scales, and in addition, if t
k
is right
scattered, then y(t
+
k
) = y(t
k

), whereas if t
k
is left scattered, then y(t
−
k
) = y(t
k
),
2 Boundary Value Problems
0 <t
1
<t
2
< ··· <t
k
< ··· <t
p
<T,
[Tu](t)
=

t
0
k(t,s)u(s)Δs,[Su](t) =

T
0
h(t,s)u(s)Δs, (1.2)
k(t,s)
∈ C[D,R

+
], D ={(t,s) ∈ J
T
× J
T
: t ≥ s}, h(t,s) ∈ C[J
T
× J
T
,R
+
], R
+
= [0,+∞),
k
0
= max{k(t,s):(t,s) ∈ D}, h
0
= max{h(t,s):(t,s) ∈ J
T
× J
T
}.
The study of impulsive dynamic equations on time scales has been initiated by Hender-
son [4], Benchohra et al. [5], and Atici and Biles [6]. Extremal solutions of PBVP for im-
pulsive differential equations and difference equations has been studied by some authors
(see [7, 8]). In this paper, we will obtain an inequality on time scales. And then, using
this inequalit y, a comparison result is obtained. At last, we obtain an existence theorem
of minimal and maximal solutions of PBVP (1.1) by using monotone iterative technique
(see [7–9]).

2. Preliminaries and comparison principle
In this section, we will first recall some basic definitions and lemmas, which are used in
what follows.
Let
T be a nonempty closed subset (time scale) of R.Theforwardandbackwardjump
operators σ,ρ :
T → T, and the graininess μ : T → R
+
are defined, respectively, by
σ(t)
= inf{s ∈ T : s>t}, ρ(t) = sup{s ∈ T : s<t}, μ(t) = σ(t) − t. (2.1)
Apointt
∈ T is called left dense if t>inf T and ρ(t) = t,leftscatteredifρ(t) <t, right
dense if t<sup
T and σ(t) = t, and right scattered if σ(t) >t.IfT has a left-scattered
maximum m,then
T
k
=
T \{
m}; otherwise T
k
=
T
.IfT has a right-scattered minimum
m,then
T
k
=
T \{

m}; otherwise T
k
=
T
.
A function f :
T → R is rig ht-dense continuous provided it is continuous at right-
dense point in
T and its left-side limits exist at left-dense points in T.If f is continuous at
each right-dense point and each left-dense point, then f is said to be continuous function
on
T.
For y :
T → R and t ∈ T
k
, we define the delta derivative of y(t), y
Δ
(t)tobethenumber
(if it exists) with the property that for a given ε>0, there exists a neighborhood U of t
such that



y

σ(t)

−
y(s)


−
y
Δ
(t)

σ(t) − s



<ε


σ(t) − s


(2.2)
for all s
∈ U.
If y is continuous, then y is right-dense continuous, and if y is delta differentiable at
t,theny is continuous at t.
Lemma 2.1 (see [1]). Assume that f ,g :
T → R are delta differentiable at t ∈ T
k
.Then,
( fg)
Δ
(t) = f
Δ
(t)g(t)+ f


σ(t)

g
Δ
(t) = f (t)g
Δ
(t)+ f
Δ
(t)g

σ(t)

. (2.3)
Y. Li and H. Z hang 3
Let y be right-dense continuous. If Y
Δ
(t) = y(t), then we define the delta integral by

t
a
y(s)Δs = Y (t) − Y(a). (2.4)
A function r :
T → R is called regressive if
1+μ(t)r(t)
= 0 (2.5)
for all t
∈ T
k
.
If r is regressive function, then the generalized exponential function e

r
is defined by
e
r
(t,s) = exp


t
s
ξ
μ(τ)

r(τ)

Δτ

for s,t ∈ T (2.6)
with the cylinder transformation
ξ
h
(z) =
⎧
⎪
⎨
⎪
⎩
Log(1 + hz)
h
if h
= 0,

z if h
= 0.
(2.7)
Let p,q :
T → R be two regressive functions, we define
p
⊕ q := p + q + μpq, p :=−
p
1+μp
, p
 q := p ⊕ (q). (2.8)
Then, the generalized exponential function has the following properties.
Lemma 2.2 (see [1]). Assume that p,q :
T → R are two regressive funct ions, then
(i) e
0
(t,s) ≡ 1 and e
p
(t,t) ≡ 1;
(ii) e
p
(σ(t),s) = (1 + μ(t)p(t))e
p
(t,s);
(iii) e
p
(t,σ(s)) = e
p
(t,s)/(1 + μ(s)p(s));
(iv) 1/e

p
(t,s) = e
p
(t,s);
(v) e
p
(t,s) = 1/e
p
(s,t) = e
p
(s,t);
(vi) e
p
(t,s)e
p
(s,r) = e
p
(t,r);
(vii) e
p
(t,s)e
q
(t,s) = e
p⊕q
(t,s);
(viii) e
p
(t,s)/e
q
(t,s) = e

pq
(t,s).
Lemma 2.3 [1]. Let r :
T → R be right-dense continuous and regressive, a ∈ T,andy
a
∈ R.
The unique solution of the initial value problem
y
Δ
(t) = r(t)y(t)+h(t), y(a) = y
a
, (2.9)
is given by
y(t)
= e
r
(t,a)y
a
+

t
a
e
r

t,σ(s)

h(s)Δs. (2.10)
Throughout this paper, we assume that, for each k
= 1, , p, the points of impulse t

k
are right dense. For convenience, we introduce the notation PC[J
T
,R] ={u : J
T
→ R,u(t)
4 Boundary Value Problems
is continuous everywhere except some t
k
at which u(t
−
k
)andu(t
+
k
) exist and u(t
−
k
) =
u(t
k
)}. Evidently, PC[J
T
,R] is a Banach space with norm u
PC
= sup
t∈J
T
|u(t)|.LetJ


T
=
J
T
\{t
1
,t
2
, ,t
p
}, C
1
[J

T
,R] ={u
Δ
(t)iscontinuousonJ

T
}, Ω = PC[J
T
,R] ∩ C
1
[J

T
,R],
T
+

=
T ∩ R
+
, PC
1
[T
+
,R] = PC[T
+
,R] ∩ C
1
[T
+
,R]. A function u ∈ Ω is called a solution
of PBVP (1.1) if it satisfies (1.1).
Next, we combine [10, 11] to obtain an inequality as fol lows.
Lemma 2.4. Assume that
(A
0
) the sequence {t
k
} satisfies 0 ≤ t
0
<t
1
<t
2
< ···<t
k
< ··· with lim

k→+∞
t
k
= +∞,
(A
1
) m ∈ PC
1
[T
+
,R] is right-dense continuous at t
k
for k = 1,2, ,
(A
2
)inf
t∈J
T
{μ(t)p(t)} > −1.Fork = 1,2, ,t ≥ t
0
,
m
Δ
(t) ≥ p(t)m(t)+q(t), t = t
k
, m

t
+
k


≥
d
k
m

t
k

+ b
k
, (2.11)
where p,q
∈ C(T
+
,R), d
k
≥ 0,andb
k
are real constants. Then,
m(t)
≥ m

t
0


t
0
<t

k
<t
d
k
e
p

t,t
0

+

t
t
0

s<t
k
<t
d
k
e
p

t,σ(s)

q(s)Δs
+

t

0
<t
k
<t

t
k
<t
j
<t
d
j
e
p

t,t
k

b
k
.
(2.12)
Proof. By condition (A
2
), we know that e
p
(σ(t),t
0
) ≥ 0fort ∈ [t
0

,+∞)
T
. For the follow-
ing inequality:
m
Δ
(t) ≥ p(t)m(t)+q(t), (2.13)
on multiplying e
p
(σ(t),t
0
) and arranging the terms, we obtain
e
p

σ(t),t
0

m
Δ
(t) − p(t)m(t)e
p

σ(t),t
0

≥
e
p


σ(t),t
0

q(t), (2.14)
which is the same as

e
p

t,t
0

m(t)

Δ
≥ e
p

σ(t),t
0

q(t). (2.15)
Integrating (2.15)fromt
0
to t
1
,then
e
p


t
1
,t
0

m

t
1

≥
m

t
0

+

t
1
t
0
e
p

σ(s),t
0

q(s)Δs. (2.16)
Again integrating (2.15)fromt

1
to t,wheret ∈ (t
1
,t
2
], then
e
p

t,t
0

m(t) ≥ e
p

t
1
,t
0

m

t
+
1

+

t
t

1
e
p

σ(s),t
0

q(s)Δs
≥ e
p

t
1
,t
0

d
1
m

t
1

+ b
1

+

t
t

1
e
p

σ(s),t
0

q(s)Δs
≥ d
1

m

t
0

+

t
1
t
0
e
p

σ(s),t
0

q(s)Δs


+ b
1
e
p

t
1
,t
0

+

t
t
1
e
p

σ(s),t
0

q(s)Δs,
(2.17)
Y. Li and H. Z hang 5
that is,
m(t)
≥ m

t
0


d
1
e
p

t,t
0

+

t
t
0

s<t
k
<t
d
k
e
p

t,σ(s)

q(s)Δs + b
1
e
p


t,t
1

. (2.18)
Repeating the above procession for t
∈ [t
0
,+∞)
T
,wehave
m(t)
≥ m

t
0


t
0
<t
k
<t
d
k
e
p

t,t
0


+

t
t
0

s<t
k
<t
d
k
e
p

t,σ(s)

q(s)Δs
+

t
0
<t
k
<t

t
k
<t
j
<t

d
j
e
p

t,t
k

b
k
.
(2.19)
Thus the proof of Lemma 2.4 is complete.

The following comparison result plays an important role in this paper.
Lemma 2.5. Let t
0
= 0, t
p+1
= T. Assume that u ∈ Ω satisfies
u
Δ
(t) ≥−a(t)u(t) − b(t)[Tu](t) − c(t)[Su](t), t = t
k
, t ∈ J
T
,
u

t

+
k

−
u

t
k

≥−
L
k
u

t
k

, k = 1,2, , p,
u(0) ≥ u(T),
(2.20)
where a,b, c
∈ C[J
T
,R
+
], a is not identically vanishing, and sup
t∈J
T
{μ(t)a(t)} < 1, 0 ≤ L
k

<
1(k
= 1,2, , p).If

Bk
0
+ Ch
0

e
(−a)
(T,0) ≤


0<t
k
<T

1 − L
k


2

T
0

s<t
k
<T


1 − L
k

Δs
(2.21)
with B
= sup
t∈J
T
{b(t)

t
0
e
(−a)
(σ(t),s)Δs} and C = sup
t∈J
T
{c(t)

T
0
e
(−a)
(σ(t),s)Δs},
then u(t)
≥ 0 for t ∈ J
T
.

Proof. Let p(t)
= u(t)e
(−a)
(t,0) for t ∈ J
T
.Thenp ∈ Ω satisfies
p
Δ
(t) ≥−b(t)

t
0
e
(−a)

σ(t),s

k(t,s)p(s)Δs
− c(t)

T
0
e
(−a)

σ(t),s

h(t,s)p(s)Δs, t = t
k
, t ∈ J

T
,
p

t
+
k

−
p

t
k

≥−
L
k
p

t
k

, k = 1,2, , p,
p(0) ≥ e
(−a)
(T,0)p(T).
(2.22)
We now prove
p(t)
≥ 0fort ∈ J

T
. (2.23)
6 Boundary Value Problems
Assume that (2.23) is not true. Then, there are two cases:
(a) there exists t
∗
1
∈ J
T
such that p(t
∗
1
) < 0andp(t) ≤ 0fort ∈ J
T
;
(b) there exists t
∗
1
,t
∗
2
∈ J
T
such that p(t
∗
1
) < 0andp(t
∗
2
) > 0.

In case (a), (2.22) implies that
p
Δ
(t) ≥ 0, t = t
k
, t ∈ J
T
,
p

t
+
k

−
p

t
k

≥
0, k = 1,2, , p.
(2.24)
This means that p(t) is nondecreasing in J
T
; therefore,
p(0)
≤ p

t

∗
1

< 0,
p(0) ≤ p(T) ≤ 0,
(2.25)
which contradicts p(T)
≤ e
(−a)
(T,0)p(0) < 0.
In case (b) let sup
t∈J
T
p(t) = λ.Then,λ>0 and there exists t
i
<t
∗
0
≤ t
i+1
for some i such
that p(t
∗
0
) = λ or p(t
+
i
) = λ. We may assume that p(t
∗
0

) = λ (since, in case of p(t
+
i
) = λ,
the proof is similar). From (2.22), we have
p
Δ
(t) ≥−λk
0
b(t)

t
0
e
(−a)

σ(t),s

Δs − λh
0
c(t)

T
0
e
(−a)

σ(t),s

Δs

≥−λ

Bk
0
+ Ch
0

, t = t
k
, t ∈ J
T
.
(2.26)
For t
∈ [t
∗
0
,T]
T
, k = i +1,i +2, , p,
p
Δ
(t) ≥−λ

Bk
0
+ Ch
0

, t = t

k
, p

t
+
k

≥

1 − L
k

p

t
k

. (2.27)
By Lemma 2.4,wehave
p(t)
≥ p

t
∗
0


t
∗
0

<t
k
<t

1 − L
k

+

t
t
∗
0

s<t
k
<t

1 − L
k

−
λ

Bk
0
+ Ch
0

Δs. (2.28)

Let t
= T in (2.28), then
p(T)
≥ λ

t
∗
0
<t
k
<T

1 − L
k

−
λ

Bk
0
+ Ch
0


T
t
∗
0

s<t

k
<T
(1 − L
k
)Δs. (2.29)
If p(T) < 0, then (2.29)gives

Bk
0
+ Ch
0

>

t
∗
0
<t
k
<T

1 − L
k


T
t
∗
0


s<t
k
<T

1 − L
k

Δs
≥

0<t
k
<T

1 − L
k


T
0

s<t
k
<T

1 − L
k

Δs
, (2.30)

which contradicts (2.21), so, we have p(T)
≥ 0, and by (2.22), p(0) ≥ p(T)e
−a
(T,0) ≥ 0.
Hence, 0 <t
∗
1
<T.Lett
j
<t
∗
1
≤ t
j+1
for some j.Wefirstassumethatt
∗
0
<t
∗
1
,soi ≤ j.Let
t
= t
∗
1
in (2.28), we have
0 >p

t
∗

1

≥
λ

t
∗
0
<t
k
<t
∗
1

1 − L
k

+

t
∗
1
t
∗
0

s<t
k
<t
∗

1

1 − L
k

−
λ

Bk
0
+ Ch
0

Δs, (2.31)
Y. Li and H. Z hang 7
which gives

Bk
0
+ Ch
0

>

t
∗
0
<t
k
<t

∗
1

1 − L
k


t
∗
1
t
∗
0

s<t
k
<t
∗
1

1 − L
k

Δs
≥

0<t
k
<T


1 − L
k


T
0

s<t
k
<T

1 − L
k

Δs
, (2.32)
which contradicts (2.21).
Next we assume that t
∗
1
<t
∗
0
.Soj ≤ i.Fort ∈ J
T
, k = 1,2, , p,
p
Δ
(t) ≥−λ


Bk
0
+ Ch
0

, t = t
k
, p

t
+
k

≥

1 − L
k

p

t
k

. (2.33)
By Lemma 2.4,wehave
p(t)
≥ p(0)

0<t
k

<t

1 − L
k

+

t
0

s<t
k
<t

1 − L
k

−
λ

Bk
0
+ Ch
0

Δs. (2.34)
Let t
= t
∗
1

in (2.34), then
0 >p

t
∗
1

≥
p(0)

0<t
k
<t
∗
1

1 − L
k

−
λ

Bk
0
+ Ch
0


t
∗

1
0

s<t
k
<t
∗
1

1 − L
k

Δs, (2.35)
which implies
p(0)

0<t
k
<t
∗
1
(1 − L
k
) <λ

Bk
0
+ Ch
0



t
∗
1
0

s<t
k
<t
∗
1

1 − L
k

Δs. (2.36)
By (2.22), we obtain
λ

Bk
0
+ Ch
0


t
∗
1
0


s<t
k
<t
∗
1

1 − L
k

Δs>e
(−a)
(T,0)p(T)

0<t
k
<t
∗
1

1 − L
k

. (2.37)
From (2.29), (2.37), we have
λ

Bk
0
+ Ch
0



t
∗
1
0

s<t
k
<t
∗
1

1 − L
k

Δs
>e
(−a)
(T,0)

0<t
k
<t
∗
1

1 − L
k



λ

t
∗
0
<t
k
<T

1 − L
k

−
λ

Bk
0
+ Ch
0


T
t
∗
0

s<t
k
<T


1 − L
k

Δs

(2.38)
or

0<t
k
<t
∗
1

1 − L
k


t
∗
0
<t
k
<T

1 − L
k

<


Bk
0
+ Ch
0


0<t
k
<t
∗
1

1 − L
k


T
t
∗
0

s<t
k
<T

1 − L
k

Δs

+

Bk
0
+ Ch
0

e
(−a)
(T,0)

t
∗
1
0

s<t
k
<t
∗
1

1 − L
k

Δs.
(2.39)
8 Boundary Value Problems
Hence



0<t
k
<T

1 − L
k


2
≤

0<t
k
<t
∗
1

1 − L
k


t
∗
0
<t
k
<T

1 − L

k


0<t
k
<T

1 − L
k

<

Bk
0
+ Ch
0


0<t
k
<t
∗
1

1 − L
k


0<t
k

<T

1 − L
k


T
t
∗
0

s<t
k
<T

1 − L
k

Δs
+

Bk
0
+ Ch
0

e
(−a)
(T,0)


0<t
k
<T

1 − L
k


t
∗
1
0

s<t
k
<t
∗
1

1 − L
k

Δs
<

Bk
0
+ Ch
0


e
(−a)
(T,0)

T
0

s<t
k
<T

1 − L
k

Δs,
(2.40)
which contradicts (2.21).
Thus the proof of Lemma 2.5 is complete.

For any δ(t) ∈ PC[J
T
,R]andη ∈ Ω, a,b,c ∈ C[J
T
,R
+
], a is not identically vanishing,
and 0
≤ L
k
< 1(k = 1,2, , p), I

k
∈ C[R,R](k = 1,2, , p), we consider the linear peri-
odic boundary value problem for a linear impulsive integrodifferential equation(PBVP):
u
Δ
(t)+a(t)u(t) =−b(t)[Tu](t) − c(t)[Su](t)+δ(t), t = t
k
, t ∈ J
T
,
u

t
+
k

−
u

t
k

=−
L
k
u

t
k


+ I
k

η

t
k

+ L
k
η

t
k

, k = 1,2 , p,
u(0)
= u(T).
(2.41)
Lemma 2.6. u
∈ Ω is a solution of PBVP (2.41)ifandonlyifu ∈ PC[J
T
,R] is a solution of
the following impulsive integral equation:
u(t)
=

T
0
G(t,s)


δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
+

0<t
k
<T
G

t,t
k

e
(−a)

σ

t
k

,t
k

−
L
k
u


t
k

+ I
k

η

t
k

+ L
k
η

t
k

, t ∈ J
T
,
(2.42)
where
G(t,s)
=
1
1 − e
(−a)
(T,0)
⎧

⎪
⎨
⎪
⎩
e
(−a)

t,σ(s)

,0≤ s<t≤ T,
e
(−a)
(T,0)e
(−a)

t,σ(s)

,0≤ t ≤ s ≤ T.
(2.43)
Proof. Assume that u
∈ Ω is a solution of (2.41). For the first equation of (2.41), using
Lemma 2.3 on t
∈ [0,t
1
]
T
,wehave
u(t)
= e
(−a)

(t,0)u(0) +

t
0
e
(−a)

t,σ(s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs. (2.44)
Y. Li and H. Z hang 9
Then
u

t
1

=
e
(−a)

t
1
,0

u(0) +

t

1
0
e
(−a)

t
1
,σ(s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs. (2.45)
Again using Lemma 2.3 on t
∈ (t
1
,t
2
]
T
,then
u(t)
= u

t
+
1

e
(−a)


t,t
1

+

t
t
1
e
(−a)

t,σ(s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
= u

t
1

e
(−a)

t,t
1

+

t

t
1
e
(−a)

t,σ(s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
+ e
(−a)

t,t
1

−
L
1
u

t
1

+ I
1

η

t

1

+ L
1
η

t
1

=
e
(−a)
(t,0)u(0) +

t
0
e
(−a)

t,σ(s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
+ e
(−a)

t,t
1


−
L
1
u

t
1

+ I
1

η

t
1

+ L
1
η

t
1

.
(2.46)
Repeating the above procession for t
∈ J
T
,wehave
u(t)

= u(0)e
(−a)
(t,0)+

t
0
e
(−a)

t,σ(s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
+

0<t
k
<t
e
(−a)

t,t
k

−
L
k
u


t
k

+ I
k

η

t
k

+ L
k
η

t
k

.
(2.47)
Setting t
= T in (2.47) and using the boundary condition u(0) = u(T), we obtain
u(0)
=
1
1 − e
(−a)
(T,0)



T
0
e
(−a)

T,σ(s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
+

0<t
k
<T
e
(−a)

T,t
k

−
L
k
u

t
k

+ I

k

η

t
k

+ L
k
η

t
k


.
(2.48)
Substituting (2.48)into(2.47), we see that u
∈ PC[J
T
,R] satisfies (2.42).
If u
∈ PC[J
T
,R]isasolutionof(2.42), then u ∈ C
1
(J

T
,R)and

u
Δ
(t)+a(t)u(t) =−b(t)[Tu](t) − c(t)[Su](t)+δ(t), t = t
k
, t ∈ J
T
,
u

t
+
k

−
u

t
k

=−
L
k
u

t
k

+ I
k


η

t
k

+ L
k
η

t
k

, k = 1,2 , p.
(2.49)
Setting t
= 0,T in (2.42), respectively, we have
u(T)
=
1
1 − e
(−a)
(T,0)


T
0
e
(−a)

T,σ(s)


δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
+

0<t
k
<T
e
(−a)

T,t
k

−
L
k
u

t
k

+ I
k

η

t
k


+ L
k
η

t
k


=
u(0).
(2.50)
Therefore, u
∈ Ω is a solution of (2.41). Thus Lemma 2.6 is proved. 
10 Boundary Value Problems
Lemma 2.7. Assume that a,b,c
∈ C[J
T
,R
+
] and 0 ≤ L
k
< 1(k = 1,2, , p), I
k
∈ C[R,R]
(k
= 1,2 , p), δ ∈ PC[J
T
,R], η ∈ Ω, and the following inequality holds:
1

1 − e
(−a)
(T,0)


T
0

k
0
sb(s)+Th
0
c(s)

Δs +
p

k=1
L
k

< 1. (2.51)
Then PBVP (2.41)possessesauniquesolutioninΩ.
Proof. For any u
∈ Ω, consider the operator F defined by the formula
(Fu)(t)
=

T
0

G(t,s)

δ(s) − b(s)[Tu](s) − c(s)[Su](s)

Δs
+

0<t
k
<T
G

t,t
k

e
(−a)

σ

t
k

,t
k

−
L
k
u


t
k

+ I
k

η

t
k

+ L
k
η

t
k

, t ∈ J
T
.
(2.52)
Then Fu
∈ Ω, that is, FΩ ⊂ Ω.
For every u,v
∈ Ω,t ∈ J
T
,wehave



(Fu)(t) − (Fv)(t)


≤

T
0
G(t,s)

b(s)


[Tu](s) − [Tv](s)


+ c(s)


[Su](s) − [Sv](s)



Δs
+

0<t
k
<T
G


t,t
k

e
(−a)

σ

t
k

,t
k

L
k


u

t
k

−
v

t
k




<
1
1 − e
(−a)
(T,0)


T
0

k
0
sb(s)+Th
0
c(s)

Δs +
p

k=1
L
k


u − v
PC
.
(2.53)

Hence
Fu− Fv
PC
= sup
t∈J
T


(Fu)(t) − (Fv)(t)


≤
αu − v
PC
, (2.54)
where
α
=
1
1 − e
(−a)
(T,0)


T
0

k
0
sb(s)+Th

0
c(s)

Δs +
p

k=1
L
k

< 1. (2.55)
Thus the operator F is a contraction on Ω. That is, there is a unique element u
∈ Ω such
that u
= Fu. Therefore, u is the unique solution of PBVP (2.41). The proof of Lemma 2.7
is complete.

Lemma 2.8. u ∈ Ω is a solution of PBVP (1.1)ifandonlyifu
∈ PC[J
T
,R] is solution of the
following integral equation:
u(t)
=

T
0
G(t,s)

f


s,u(s),[Tu](s),[Su](s)

+ a(s)u(s)

Δs
+

0<t
k
<1
G

t,t
k

e
(−a)

σ

t
k

,t
k

I
k


u

t
k

,
(2.56)
Y. Li and H. Zhang 11
where
G(t,s)
=
1
1 − e
(−a)
(T,0)
⎧
⎨
⎩
e
(−a)

t,σ(s)

,0≤ s<t≤ T,
e
(−a)
(T,0)e
(−a)

t,σ(s)


,0≤ t ≤ s ≤ T.
(2.57)
The proof of Lemma 2.8 is similar to that of Lemma 2.6 andwewillomitithere.
3. Main results
In this section, we will use the monotone iterative technique to prove the existence of
minimal and maximal solutions of the PBVP (1.1).
Theorem 3.1. Assume that the following conditions hold.
(H
1
) There exist functions u
0
,v
0
∈ Ω, u
0
(t) ≤ v
0
(t) for all t ∈ J
T
such that
u
Δ
0
(t) ≤ f

t,u
0
(t),


Tu
0

(t),

Su
0

(t)

, t = t
k
, t ∈ J
T
,
u
0

t
+
k

−
u
0

t
k

≤

I
k

u
0

t
k

, k = 1,2, , p,
u
0
(0) ≤ u
0
(T),
v
Δ
0
(t) ≥ f

t,v
0
(t),

Tv
0

(t),

Sv

0

(t)

, t = t
k
, t ∈ J
T
,
v
0

t
+
k

−
v
0

t
k

≥
I
k

v
0


t
k

, k = 1,2, , p,
v
0
(0) ≥ v
0
(T).
(3.1)
(H
2
)Thefunction f ∈ C[J
T
× R × R × R,R] satisfies
f

t,u
2
,v
2
,w
2

−
f

t,u
1
,v

1
,w
1

≥−
a(t)

u
2
− u
1

−
b(t)

v
2
− v
1

−
c(t)

w
2
− w
1

, (3.2)
whenever u

0
(t) ≤ u
1
≤ u
2
≤ v
0
(t),[Tu
0
](t) ≤ v
1
≤ v
2
≤ [Tv
0
](t),[Su
0
](t) ≤ w
1
≤ w
2
≤ [Sv
0
](t), t ∈ J
T
, where for a,b, c ∈ C[J
T
,R
+
], sup

t∈J
T
{μ(t)a(t)} < 1, a is not identically
vanishing.
(H
3
)ThefunctionI
k
∈ C[R,R] satisfies
I
k
(x) − I
k
(y) ≥−L
k
(x − y), (3.3)
whenever u
0
(t
k
) ≤ y ≤ x ≤ v
0
(t
k
)(k = 1,2, , p),and0 ≤ L
k
< 1(k = 1,2, , p).
Further, assume that the inequalities (2.21)and(2.51) hold. Then PBVP (1.1)hasthe
minimal solution u
∗

and maximal v
∗
in [u
0
,v
0
]. Moreover, there exist monotone iteration
sequences
{u
n
(t)},{v
n
(t)}⊂[u
0
,v
0
] such that
u
n
(t) −→ u
∗
(t),v
n
(t) −→ v
∗
(t) as n −→ ∞ uniformly on t ∈ J
T
, (3.4)
12 Boundary Value Problems
where

{u
n
(t)},{v
n
(t)} satisfy
u
Δ
n
(t) = f

t,u
n−1
(t),

Tu
n−1

(t),

Su
n−1

(t)

−
a(t)

u
n
− u

n−1

(t)
− b(t)

T

u
n
− u
n−1

(t) − c(t)

S

u
n
− u
n−1

(t), t = t
k
, t ∈ J
T
,
u
n

t

+
k

−
u
n

t
k

=−
L
k
u
n

t
k

+ I
k

u
n−1

t
k

+ L
k

u
n−1

t
k

, k = 1,2, , p,
u
n
(0) = u
n
(T)(n = 1,2,3, ),
v
Δ
n
(t) = f

t,v
n−1
(t),

Tv
n−1

(t),

Sv
n−1

(t)


−
a(t)

v
n
− v
n−1

(t)
− b(t)

T

v
n
− v
n−1

(t) − c(t)

S

v
n
− v
n−1

(t), t = t
k

, t ∈ J
T
,
v
n

t
+
k

−
v
n

t
k

=−
L
k
v
n

t
k

+ I
k

v

n−1

t
k

+ L
k
v
n−1

t
k

, k = 1,2, , p,
v
n
(0) = v
n
(T)(n = 1,2,3, ),
(3.5)
u
0
≤ u
1
≤···≤u
n
≤···≤u
∗
≤ v
∗

≤···≤v
n
≤···≤v
1
≤ v
0
.
(3.6)
Proof. For any u
n−1
,v
n−1
∈ Ω,byLemma 2.7,weknowthat(3.5) has unique solution u
n
and v
n
in Ω, respectively.
In the following, we will show by induction that
u
n−1
≤ u
n
≤ v
n
≤ v
n−1
, n = 1,2,3, (3.7)
By (3.5) and the conditions (H
1
), (H

2
), and (H
3
), we have

u
1
− u
0

Δ
(t) ≥−a(t)

u
1
− u
0

(t) − b(t)

T

u
1
− u
0

(t)
− c(t)


S

u
1
− u
0

(t), t = t
k
, t ∈ J
T
,

u
1
− u
0

t
+
k

−

u
1
− u
0

t

k

≥−
L
k

u
1
− u
0

t
k

, k = 1,2, , p,

u
1
− u
0

(0) ≥

u
1
− u
0

(T),


v
0
− v
1

Δ
(t) ≥−a(t)

v
0
− v
1

(t) − b(t)

T

v
0
− v
1

(t)
− c(t)

S

v
0
− v

1

(t), t = t
k
, t ∈ J
T
,

v
0
− v
1

t
+
k

−

v
0
− v
1

t
k

≥−
L
k


v
0
− v
1

t
k

, k = 1,2, , p,

v
0
− v
1

(0) ≥

v
0
− v
1

(T),

v
1
− u
1


Δ
(t) ≥−a(t)

v
1
− u
1

(t) − b(t)

T

v
1
− u
1

(t)
− c(t)

S

v
1
− u
1

(t), t = t
k
, t ∈ J

T
,

v
1
− u
1

t
+
k

−

v
1
− u
1

t
k

≥−
L
k

v
1
− u
1


t
k

, k = 1,2, , p,

v
1
− u
1

(0) =

v
1
− u
1

(T).
(3.8)
Thus, by Lemma 2.5,wehaveu
0
≤ u
1
≤ v
1
≤ v
0
.
Y. Li and H. Zhang 13

Now we assume that (3.7)istruefori>1, that is, u
i−1
≤ u
i
≤ v
i
≤ v
i−1
,andweprove
that (3.7)istruefori +1 too. Infact, by(3.5), and the conditions H
2
and H
3
,wehave
that

u
i+1
− u
i

Δ
(t) ≥−a(t)

u
i+1
− u
i

(t) − b(t)


T

u
i+1
− u
i

(t)
− c(t)

S

u
i+1
− u
i

(t), t = t
k
, t ∈ J
T
,

u
i+1
− u
i

t

+
k

−

u
i+1
− u
i

t
k

≥−
L
k

u
i+1
− u
i

t
k

, k = 1,2, , p,

u
i+1
− u

i

(0) =

u
i+1
− u
i

(T),

v
i+1
− v
i

Δ
(t) ≥−a(t)

v
i+1
− v
i

(t) − b(t)

T

v
i+1

− v
i

(t)
− c(t)

S

v
i+1
− v
i

(t), t = t
k
, t ∈ J
T
,

v
i+1
− v
i

t
+
k

−


v
i+1
− v
i

t
k

≥−
L
k

v
i+1
− v
i

t
k

, k = 1,2, , p,

v
i+1
− v
i

(0) =

v

i+1
− v
i

(T),

v
i+1
− u
i+1

Δ
(t) ≥−a(t)

v
i+1
− u
i+1

(t) − b(t)

T

v
i+1
− u
i+1

(t)
− c(t)


S

v
i+1
− u
i+1

(t), t = t
k
, t ∈ J
T
,

v
i+1
− u
i+1

t
+
k

−

v
i+1
− u
i+1


t
k

≥−
L
k

v
i+1
− u
i+1

t
k

, k = 1,2, , p,

v
i+1
− u
i+1

(0) =

v
i+1
− u
i+1

(T).

(3.9)
Thus, by Lemma 2.5,wehavethatu
i
≤ u
i+1
≤ v
i+1
≤ v
i
. So, by induction, (3.7)holdsfor
any positive integer n.
It is easy to know by (3.7)that
u
0
≤ u
1
≤···≤u
n
≤···≤v
n
≤···≤v
1
≤ v
0
. (3.10)
Furthermore, by (3.5), and Lemma 2.6,wehave
u
n
(t) =


T
0
G(t,s)

f

s,u
n−1
(s),

Tu
n−1

(s),

Su
n−1

(s)

+ a(s)u
n−1
(s)
− b(s)

T

u
n
− u

n−1

(s) − c(s)

S(u
n
− u
n−1

(s)

Δs
+

0<t
k
<T
G(t,t
k
)e
(−a)

σ

t
k

,t
k


−
L
k
u
n

t
k

+ I
k

u
n−1

t
k

+ L
k
u
n−1

t
k

, t ∈ J
T
,
v

n
(t) =

T
0
G(t,s)

f

s,v
n−1
(s),

Tv
n−1

(s),

Sv
n−1

(s)

+ a(s)v
n−1
(s)
− b(s)

T


v
n
− v
n−1

(s) − c(s)

S

v
n
− v
n−1

(s)

Δs
+

0<t
k
<T
G

t,t
k

e
(−a)


σ

t
k

,t
k

−
L
k
v
n

t
k

+ I
k

v
n−1

t
k

+ L
k
v
n−1


t
k

, t ∈ J
T
.
(3.11)
14 Boundary Value Problems
By (3.5) and the condition (H
2
), we have
f

t,u
0
(t),T

u
0

(t),S

u
0

(t)

−
a(t)


v
0
− u
0

(t)
− b(t)T

v
0
− u
0

(t) − c(t)S

v
0
− u
0

(t)
≤ u
Δ
n
(t) ≤ f

t,v
0
(t),T


v
0

(t),S

v
0

(t)

+ a(t)

v
0
− u
0

(t)+b(t)T

v
0
− u
0

(t)+c(t)S

v
0
− u

0

(t).
(3.12)
Thus,
{u
Δ
n
(t)} is uniformly bounded. Also, similarly to the above we can show that {v
Δ
n
(t)}
is uniformly bounded. Using Lemma 2.4 [12], we know that there exist u
∗
,v
∗
such that
lim
n→∞
u
n
(t) = u
∗
(t),lim
n→∞
v
n
(t) = v
∗
(t)uniformlyonJ

T
.
Taking limits as n
→∞,by(3.11), we have that
u
∗
(t) =

T
0
G(t,s)

f

s,u
∗
(s),

Tu
∗

(s),

Su
∗

(s)

+ a(s)u
∗

(s)

Δs
+

0<t
k
<1
G

t,t
k

e
(−a)

σ

t
k

,t
k

I
k

u
∗


t
k

,
v
∗
(t) =

T
0
G(t,s)

f

s,v
∗
(s),

Tv
∗

(s),

Sv
∗

(s)

+ a(s)v
∗

(s)

Δs
+

0<t
k
<1
G

t,t
k

e
(−a)

σ

t
k

,t
k

I
k

v
∗


t
k

.
(3.13)
From the above, by Lemma 2.8,weknowthatu
∗
and v
∗
are solutions of PBVP (1.1)in
[u
0
,v
0
].
Next we prove that u
∗
and v
∗
are the minimal and maximal solutions of PBVP (1.1)
in [u
0
,v
0
].
In fact, let w
∈ [u
0
,v
0

] be a solution of PBVP(1.1), that is,
w
Δ
(t) = f

t,w(t),[Tw](t),[Sw](t)

, t = t
k
, t ∈ J
T
,
w

t
+
k

−
w

t
k

=
I
k

w


t
k

, k = 1,2, , p,
w(0) = w(T).
(3.14)
Using induction, suppose that there exists a positive integer n such that u
n
(t) ≤ w(t) ≤
v
n
(t)onJ
T
.Then,

w − u
n+1

Δ
(t) = f

t,w(t),[Tw](t),[Sw](t)

−

f

t,u
n
(t),


Tu
n

(t),

Su
n

(t)

−
a(t)

u
n
− u
n+1

(t)
− b(t)

T

u
n
− u
n+1

(t) − c(t)


S

u
n
− u
n+1

(t)

≥−
a(t)

w(t) − u
n+1
(t)

−
b(t)

T

w − u
n+1

(t)
− c(t)

S


w − u
n+1

(t), t = t
k
, t ∈ J
T
,
Y. Li and H. Zhang 15

w − u
n+1

t
+
k

=

w − u
n+1

t
k

+ I
k

w


t
k

−

−
L
k
u
n+1

t
k

+ I
k

u
n

t
k

+ L
k
u
n

t
k


≥

1 − L
k

w − u
n+1

t
k

, k = 1,2, , p,

w − u
n+1

(0) =

w − u
n+1

(T).
(3.15)
By Lemma 2.5, it follows that w(t)
≥ u
n+1
(t)onJ
T
. Similarly, we obtain v

n+1
(t) ≥ w(t)on
J
T
.Sinceu
0
(t) ≤ w(t) ≤ v
0
(t)onJ
T
,byinductionweget
u
n+1
(t) ≤ w(t) ≤ v
n+1
(t), n = 1,2,3, (3.16)
Thus, letting n
→∞in (3.16), we have that
u
∗
≤ w ≤ v
∗
, (3.17)
that is, u
∗
and v
∗
are the minimal and maximal solutions of the PBVP (1.1) in the interval
[u
0

,v
0
].
The proof of Theorem 3.1 is complete.

Acknowledgments
The authors thank the referees for their many thoughtful suggestions that lead to an im-
proved exposition of manuscript. This work is supported by the National Natural Sci-
ences Foundation of China.
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Yongkun Li: Depar tment of Mathematics, Yunnan University, Kunming, Yunnan 650091, China
Email address:
Hongtao Zhang: Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, China
Email address: hatfl