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Applied Mathematics E-Notes, 14(2014), 86-96 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
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Existence Of Positive Periodic Solutions For Two
Types Of Third-Order Nonlinear Neutral Di¤erential
Equations With Variable Delay
Abdelouaheb Ardjouniy, Ahcene Djoudiz, Ali Rezaiguiax
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Received 21 October 2013
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Abstract
In this article we study the existence of positive periodic solutions for two
types of third-order nonlinear neutral di¤erential equation with variable delay.
The main tool employed here is the Krasnoselskii’s …xed point theorem dealing
with a sum of two mappings, one is a contraction and the other is completely
continuous. The results obtained here generalize the work of Ren, Siegmund and
Chen [14].
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1
Introduction
In recent years, there have been a few papers written on the existence of periodic solutions, nontrivial periodic solutions and positive periodic solutions for several classes of
functional di¤erential equations with delays, which arise from a number of mathematical ecological models, economical and control models, physiological and population
models and other models, see [1–14], [16–18] and the references therein.
In this paper, we are interested in the analysis of qualitative theory of positive
periodic solutions of delay di¤erential equations. Motivated by the papers [2, 7, 8, 9,
10, 11, 12, 13, 14, 16, 17] and the references therein, we concentrate on the existence of
positive periodic solutions for the two types of third-order nonlinear neutral di¤erential
equation with variable delay
d3
(x (t)
dt3
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27
and
28
d3
(x (t)
dt3
g (t; x (t
g (t; x (t
(t)))) = a (t) x (t)
(t)))) =
f (t; x (t
a (t) x (t) + f (t; x (t
(t))) ;
(t))) ;
(1)
(2)
Mathematics Subject Classi…cations: 34K13, 34A34; Secondary 34K30, 34L30.
of Sciences and Technology, Department of Mathematics and Informatics, Univ Souk
Ahras, P.O. Box 1553, Souk Ahras, 41000, Algeria; Applied Mathematics Lab., Faculty of Sciences,
Department of Mathematics, Univ. Annaba, P.O. Box 12, Annaba 23000, Algeria
z Applied Mathematics Lab., Faculty of Sciences, Department of Mathematics, Univ. Annaba, P.O.
Box 12, Annaba 23000, Algeria
x Faculty of Sciences and Technology, Department of Mathematics and Informatics, Univ Souk
Ahras, P.O. Box 1553, Souk Ahras, 41000, Algeria; Applied Mathematics Lab., Faculty of Sciences,
Department of Mathematics, Univ. Annaba, P.O. Box 12, Annaba 23000, Algeria
y Faculty
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Ardjouni et al.
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44
where a; 2 C (R; (0; 1)) ; g 2 C (R [0; 1) ; R) ; f 2 C (R [0; 1) ; [0; 1)), and a;
; g (t; x) ; f (t; x) are T -periodic in t where T is a positive constant. To reach our
desired end we have to transform (1) and (2) into integral equations and then use
Krasnoselskii’s …xed point theorem to show the existence of positive periodic solutions.
The obtained equation splits into a sum of two mappings, one is a contraction and the
other is compact. In the special case g (t; x) = cx with jcj < 1, Ren et al. in [14] show
that (1) and (2) have a positive periodic solutions by using Krasnoselskii’s …xed point
theorem.
The organization of this paper is as follows. In Section 2, we introduce some notations and lemmas, and state some preliminary results needed in later sections, then we
give the Green’s function of (1) and (2), which plays an important role in this paper.
Also, we present the inversions of (1) and (2), and Krasnoselskii’s …xed point theorem.
For details on Krasnoselskii’s theorem we refer the reader to [15]. In Section 3 and
Section 4, we present our main results on existence of positive periodic solutions of (1)
and (2), respectively. The results presented in this paper generalize the main results
in [14].
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Preliminaries
For T > 0, let CT be the set of all continuous scalar functions x, periodic in t of period
T . Then (CT ; k k) is a Banach space with the supremum norm
kxk = sup jx (t)j = sup jx (t)j :
48
t2R
49
De…ne
CT+ = fx 2 CT : x > 0g ; CT = fx 2 CT : x < 0g :
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51
Denote
M = sup fa (t) : t 2 [0; T ]g ; m = inf fa (t) : t 2 [0; T ]g ;
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53
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55
56
57
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t2[0;T ]
=
and
F (t; x) = f (t; x (t
(t)))
a (t) g (t; x (t
LEMMA 2.1 ([14]). The equation
d3
y (t)
dt3
M y (t) = h (t) ; h 2 CT ;
has a unique T -periodic solution
y (t) =
Z
0
T
G1 (t; s) ( h (s)) ds;
(t))) :
p
3
M;
88
59
60
Existence of Positive Periodic Solutions
where if 0
G1 (t; s)
s
t
=
3
h
1 + exp(
and if 0
t
s
2 exp
T)
1
T
2
exp
61
62
2
T;
sin
G1 (t; s)
=
3
64
+
65
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68
73
T) +
p
6
3 T
2
!#
(s t T )
2
h
1 + exp( T ) 2 exp
"
!
p
3
sin
(t s + T ) +
2
6
2
p
3
(t
2
i sin
s) +
6
!
exp( (t s))
;
3 2 (exp( T ) 1)
+
T
2
cos
exp
p
3 T
2
1
T
2
i
sin
p
3
(t
2
s) +
6
!#
exp( (t + T s))
:
3 2 (exp( T ) 1)
p
R!
LEMMA 2.2 ([14]). 0 G1 (t; s)ds = 1=M and if 3 T < 4 =3 holds, then G1 (t; s) >
0 for all t 2 [0; T ] and s 2 [0; T ].
LEMMA 2.3 ([14]). The equation
d3
y (t)
dt3
a (t) y (t) = h (t) ; h 2 CT ;
has a unique positive T -periodic solution
(P1 h) (t) = (I
71
72
s
cos
"
T;
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T
2
2 exp
p
3
(t
2
2 exp
63
(s t)
2
T1 B1 )
1
T1 h (t) ;
where
(T1 h) (t) =
Z
T
G1 (t; s) ( h (s)) ds and (B1 y) (t) = [ M + a (t)] y (t) :
0
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LEMMA 2.4 ([14]). If
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3 T < 4 =3 holds, then P1 is completely continuous and
0 < (T1 h) (t)
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p
(P1 h) (t)
M
kT1 hk ; h 2 CT :
m
The following lemma is essential for our results on existence of positive periodic
solution of (1). The proof is similar to that of Section 6 of [14] and hence, we omit it.
LEMMA 2.5. If x 2 CT then x is a solution of equation (1) if and only if
x (t) = g (t; x (t
(t))) + P1 ( f (t; x (t
(t))) + a (t) g (t; x (t
(t)))) :
(3)
Ardjouni et al.
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LEMMA 2.6 ([14]). The equation
d3
y (t) + M y (t) = h (t) ; h 2 CT+ ;
dt3
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has a unique T -periodic solution
y (t) =
83
Z
T
G2 (t; s) h (s) ds;
0
84
where if 0
s
t
T;
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G2 (t; s)
=
3
88
and if 0
G2 (t; s)
t
s
=
3
89
+
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92
93
h
1 + exp( T )
exp
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2
1
T
2
2
2 exp
h
98
T
2
s
p
cos
T)
i sin
3 T
2
6
(t+T
2
!#
+
3
(t
2
s)
6
!
exp( (s t))
;
(1 exp( T ))
2
3
s)
p
1 + exp( T ) 2 exp 2T cos
"
!
p
3
sin
(t + T s)
exp
2
6
3 T
2
i
1
T
2
sin
p
3
(t
2
s)
6
!#
exp( (s t T ))
:
3 2 (1 exp( T ))
p
RT
LEMMA 2.7 ([14]). 0 G2 (t; s) ds = 1=M and if 3 T < 4 =3 holds, then
G2 (t; s) > 0 for all t 2 [0; T ] and s 2 [0; T ].
LEMMA 2.8 ([14]). The equation
d3
y (t) + a (t) y (t) = h (t) ; h 2 CT+ ;
dt3
has a unique positive T -periodic solution
(P2 h) (t) = (I
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2 exp
p
3
(t
2
p
T;
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sin
"
(t s)
2
2 exp
where
(T2 h) (t) =
Z
T2 B2 )
1
T2 h (t) ;
T
G2 (t; s) h (s) ds; (B2 y) (t) = [M
a (t)] y (t) :
0
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LEMMA 2.9 ([14]). If
p
3 T < 4 =3 holds, then P2 is completely continuous and
0 < (T2 h) (t)
(P2 h) (t)
M
kT2 hk ; h 2 CT+ :
m
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Existence of Positive Periodic Solutions
The following lemma is essential for our results on existence of positive periodic
solution of (2).
LEMMA 2.10. If x 2 CT then x is a solution of equation (2) if and only if
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x (t) = g (t; x (t
104
d3
[x (t) g (t; x (t
dt3
= [M a (t)] [x (t)
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= B2 [x (t)
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(t)))] + M [x (t)
g (t; x (t
g (t; x (t
x (t)
This yields
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(I
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Therefore,
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x (t)
g (t; x (t
T2 B2 ) (x (t)
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127
128
129
130
131
(t)))
(t))) = T2 B2 [x (t)
g (t; x (t
g (t; x (t
(t)))
g (t; x (t
(t)))) = T2 (f (t; x (t
(t))) = (I
T2 B2 )
= P2 (f (t; x (t
118
119
123
(t)))] + f (t; x (t
+ T2 (f (t; x (t
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122
g (t; x (t
(t)))] + f (t; x (t
112
113
121
a (t) g (t; x (t
(t)))) :
(4)
(t)))]
(t)))
a (t) g (t; x (t
a (t) g (t; x (t
(t)))
(t))) :
From Lemma 2.6, we have
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(t)))
PROOF. Let x 2 PT be a solution of (2). Rewrite (2) as
105
110
(t))) + P2 (f (t; x (t
1
T2 (f (t; x (t
(t)))
(t)))]
a (t) g (t; x (t
(t)))
(t)))
a (t) g (t; x (t
(t)))) :
a (t) g (t; x (t
a (t) g (t; x (t
(t)))) :
(t))))
(t)))) :
Obviously,
x (t) = g (t; x (t
(t))) + P2 (f (t; x (t
(t)))
a (t) g (t; x (t
(t)))) :
This completes the proof.
Lastly in this section, we state Krasnoselskii’s …xed point theorem which enables
us to prove the existence of positive periodic solutions to (1) and (2). For its proof we
refer the reader to ([15], p. 31).
THEOREM 2.1 (Krasnoselskii). Let D be a closed convex nonempty subset of a
Banach space (B; k:k) : Suppose that A and B map D into B such that
(i) x; y 2 D; implies Ax + By 2 D;
(ii) A is completely continuous,
(iii) B is a contraction mapping.
Then there exists z 2 D with z = Az + Bz:
Ardjouni et al.
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Positive Periodic Solutions for (1)
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To apply Theorem 2.1, we need to de…ne a Banach space B, a closed convex subset D
of B and construct two mappings, one is a contraction and the other is a completely
continuous. So, we let (B; k:k) = (CT ; k:k) and D = f' 2 B : L ' Kg, where L is
non-negative constant and K is positive constant. We express equation (3) as
137
' (t) = (B1 ') (t) + (A1 ') (t) := (H1 ') (t) ;
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where A1 ; B1 : D ! B are de…ned by
(A1 ') (t) = P1 ( f (t; ' (t
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140
143
144
145
(t)))) ;
(5)
and
(B1 ') (t) = g (t; ' (t
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142
(t))) + a (t) g (t; ' (t
(t))) :
(6)
In this section we obtain the existence of a positive periodic solution of (1) by
considering the three cases; g (t; x) > 0, g (t; x) = 0 and g (t; x) < 0 for all t 2 R, x 2 D.
We assume that function g (t; x) is locally Lipschitz continuous in x. That is, there
exists a positive constant k such that
jg (t; x)
146
g (t; y)j
k kx
yk ; for all t 2 [0; T ] ; x; y 2 D:
(7)
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In the case g (t; x) > 0, we assume that there exist positive constants k1 and k2
such that
k1 x g (t; x) k2 x; for all t 2 [0; T ] ; x 2 D;
(8)
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k2 < 1;
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153
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and for all t 2 [0; T ] ; x 2 D;
k1 m
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M:
(10)
LEMMA 3.1. Suppose that (7) holds. If B1 is given by (6) with
k < 1;
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F (t; x)
(9)
(11)
then B1 : D ! B is a contraction.
PROOF. Let B1 be de…ned by (6). Obviously, B1 ' is continuous and it is easy to
show that (B1 ') (t + T ) = (B1 ') (t). So, for any '; 2 D, we have
j(B1 ') (t)
Then kB1 '
(B1 ) (t)j
B1 k
k k'
jg (t; ' (t
(t)))
g (t;
(t
(t)))j
k k'
k:
k. Thus B1 : D ! B is a contraction by (11).
Besides, by the complete continuity of P1 , it is easy to verify the following lemma.
p
LEMMA 3.2. Suppose that 3 T < 4 =3 and the conditions (8)-(10) hold. Then
A1 : D ! B is completely continuous.
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Existence of Positive Periodic Solutions
p
THEOREM 3.1. Suppose that 3 T < 4 =3 and the conditions (7)-(11) hold with
k1 m
M
L =
and K =
. Then equation (1) has a positive T -periodic
(1 k1 ) M
(1 k2 ) m
solution x in the subset
k1 m
M
D= '2B:
'
:
(1 k1 ) M
(1 k2 ) m
PROOF. By Lemma 3.1, the operator B1 : D ! B is a contraction. Also, from
Lemma 3.2, the operator A1 : D ! B is completely continuous. Moreover, we claim
that B1 + A1 ' 2 D for all '; 2 D. Since F (t; x)
k1 m > 0 which implies
f (t; x) + a (t) g (t; x) < 0, then for any '; 2 D, by Lemma 2.2 and Lemma 2.4, we
have
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(B1 ) (t) + (A1 ') (t)
174
= g (t;
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176
177
178
179
180
181
(t
(t))) + P1 ( f (t; ' (t
(t))) + a (t) g (t; ' (t
(t))))
M
kT1 ( f (t; ' (t
(t))) + a (t) g (t; ' (t
(t))))k
k2 (t
(t)) +
m
Z T
M
k2 M
max
+
G1 (t; s) (f (s; ' (s
(s))) a (s) g (s; ' (s
(s)))) ds
(1 k2 )m
m t2[0;T ] 0
Z T
M
k2 M
+
max
G1 (t; s) (f (s; ' (s
(s))) a (s) g (s; ' (s
(s)))) ds
(1 k2 )m
m t2[0;T ] 0
Z
M T
k2 M
+
G1 (t; s) M ds
(1 k2 ) m
m 0
M
1
M
k2 M
+
M
=
:
=
(1 k2 ) m
m M
(1 k2 ) m
On the other hand, by Lemma 2.2 and Lemma 2.4,
182
(B1 ) (t) + (A1 ') (t)
183
= g (t;
184
k1 (t
(t
(t))) + P1 ( f (t; ' (t
(t))) + a (t) g (t; ' (t
(t))))
Z T
(t)) +
G1 (t; s) (f (s; ' (s
(s))) a (s) g (s; ' (s
(s)))) ds
0
185
186
187
188
189
190
191
192
193
Z T
k12 m
G1 (t; s) k1 mds
+
(1 k1 ) M
0
1
k1 m
k12 m
=
+ k1 m
=
:
(1 k1 ) M
M
(1 k1 ) M
Then B1 + A1 ' 2 D for all '; 2 D. Clearly, all the hypotheses of the Krasnoselskii
theorem are satis…ed. Thus there exists a …xed point x 2 D such that x = A1 x + B1 x.
By Lemma 2.5 this …xed point is a solution of (1) and the proof is complete.
EXAMPLE 3.1. Consider the following third-order nonlinear neutral di¤erential
equation with variable delay
d3
[x (t)
dt3
g (t; x (t
(t)))] = a (t) x (t)
f (t; x (t
(t))) ;
(12)
Ardjouni et al.
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195
196
197
198
199
where T = ;
93
(t) = sin2 (t) ; a (t) =
f (t; x) =
202
203
204
sin2 (t) + 0:8; g (t; x) = 0:6 sin
x
2
; and
x
sin2 (t)
x
+ 0:48 sin2
+ 0:2:
+ 0:12 sin2 (t) sin
2
x + 1:6
2
2
Then Equation (12) has a positive -periodic solution x satisfying 0:2
see this, a simple calculation yields
x
2:5. To
k = 0:3; m = 0:8; M = 1; k1 = 0:2; k2 = 0:5; L = 0:2; K = 2:5:
De…ne the set D = f' 2 B : 0:2
F (t; x) =
200
201
1
5
'
2:5g. Then for x 2 [0:2; 2:5] we have
sin2 (t)
+ 0:2
x2 + 1:6
0:81 < 1 = M:
On the other hand,
F (t; x) =
sin2 (t)
+ 0:2
x2 + 1:6
0:2 > 0:16 = k1 m:
By Theorem 3.1, Equation (12) has a positive
x 2:5.
-periodic solution x such that 0:2
205
REMARK 3.1. When g (t; x) = cx, Theorem 3.1 reduces to Theorem 6.2 of [14].
206
In the case g (t; x) = 0, we have the following theorem.
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209
210
211
212
213
214
215
216
217
218
219
220
p
THEOREM 3.2 ([14]). If 3 T < 4 =3 holds, k2 = 0 and 0 < F (t; x)
equation (1) has a positive T -periodic solution x in the subset
D1 =
'2B:0<'
M
m
M , then
:
In the case g (t; x) < 0, we substitute conditions (8)-(10) with the following
ditions respectively. We assume that there exist negative constants k3 and k4
that
k3 x g (t; x) k4 x; for all t 2 [0; T ] ; x 2 D;
m
k3 <
;
M
and for all t 2 [0; T ] ; x 2 D
consuch
k3 M < F (t; x)
(15)
m:
(13)
(14)
p
THEOREM 3.3. Suppose that 3 T < 4 =3, (7) and (11)-(15) hold with L = 0
and K = 1. Then equation (1) has a positive T -periodic solution x in the subset
D2 = f' 2 B : 0 < ' 1g.
94
221
222
223
224
Existence of Positive Periodic Solutions
PROOF. By Lemma 3.1, the operator B1 : D ! B is a contraction. Also, from
Lemma 3.2, the operator A1 : D ! B is completely continuous. Moreover, we claim
that B1 + A1 ' 2 D for all '; 2 D. In fact, for any '; 2 D, by Lemma 2.2 and
Lemma 2.4, we have
225
(B1 ) (t) + (A1 ') (t)
226
= g (t;
(t))) + P1 ( f (t; ' (t
(t))) + a (t) g (t; ' (t
(t))))
M
kT1 ( f (t; ' (t
(t))) + a (t) g (t; ' (t
(t))))k
k4 (t
(t)) +
m
Z T
M
G1 (t; s) (f (s; ' (s
(s))) a (s) g (s; ' (s
(s)))) ds
max
m t2[0;T ] 0
Z T
M
G1 (t; s) (f (s; ' (s
(s))) a (s) g (s; ' (s
(s)))) ds
max
m t2[0;T ] 0
Z
M
1
M T
G1 (t; s) mds =
m
= 1:
m 0
m M
227
228
229
230
231
232
(t
On the other hand, by Lemma 2.2 and Lemma 2.4,
233
(B1 ) (t) + (A1 ') (t)
234
= g (t;
(t
k3 (t
235
k3 +
236
Z
(t))) + P1 ( f (t; ' (t
(t))) + a (t) g (t; ' (t
(t))))
Z T
(t)) +
G1 (t; s) (f (s; ' (s
(s))) a (s) g (s; ' (s
(s)))) ds
0
T
G1 (t; s) ( k3 M ) ds
0
= k3 + ( k3 M )
237
238
239
240
241
242
1
= 0:
M
Then B1 + A1 ' 2 D for all '; 2 D. Clearly, all the hypotheses of the Krasnoselskii
theorem are satis…ed. Thus there exists a …xed point x 2 D such that x = A1 x + B1 x.
Since F (t; x) > k3 M , it is clear that x (t) > 0, hence x 2 D2 . By Lemma 2.5 this
…xed point is a solution of (1) and the proof is complete.
REMARK 3.2. When g (t; x) = cx, Theorem 3.3 reduces to Theorem 6.6 of [14].
243
244
4
245
We express equation (4) as
246
247
248
Positive Periodic Solutions for (2)
' (t) = (B2 ') (t) + (A2 ') (t) := (H2 ') (t) ;
where A2 ; B2 : D ! B are de…ned by
(A2 ') (t) = P2 (f (t; ' (t
(t)))
a (t) g (t; ' (t
(t)))) ;
(16)
Ardjouni et al.
249
and
(B2 ') (t) = g (t; ' (t
250
251
252
253
254
255
256
257
258
261
264
(17)
p
LEMMA 4.1. Suppose that 3 T < 4 =3 and the conditions (8)-(10) hold. Then
A2 : D ! B is completely continuous.
REMARK 4.1. Notice that B2 in this section is de…ned exactly the same as that in
Section 3. Hence Lemma 3.1 still holds true.
Similar to the results in Section 3, we have
THEOREM 4.1. Assume that the hypotheses of Theorem 3.1 hold, then equation
(2) has a positive T -periodic solution x in the subset
D= '2B:
k2
M
'
1
m
:
THEOREM 4.2. Assume that the hypotheses of Theorem 3.2 hold, then equation
(2) has a positive T -periodic solution x in the subset
D1 = ' 2 B : 0 < '
262
263
(t))) :
Moreover, by the complete continuity of P2 , it is easy to verify
259
260
95
1
m
:
THEOREM 4.3. Assume that the hypotheses of Theorem 3.3 hold, then equation
(2) has a positive T -periodic solution x in the subset
D2 = f' 2 B : 0 < '
265
1g :
267
Acknowledgement. The authors would like to thank the anonymous referee for
his/her valuable comments and good advice.
268
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266
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277
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