Applied Mathematics and Computation 216 (2010) 1310–1315
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Existence of positive periodic solutions for neutral multi-delay logarithmic population model q Yan Luo *, Zhiguo Luo Department of Mathematics, Hunan Normal University, Changsha Hunan 410081, PR China
a r t i c l e
i n f o
Keywords: Neutral multi-delay logarithmic population model Existence k-Set contractive operator Positive periodic solution
a b s t r a c t In the paper, a new result is obtained for the existence of positive periodic solutions to a neutral multi-delay logarithmic population model. Our analysis mainly relies on an abstract continuous theorem of k-set contractive operator. We also give an example to illustrate the applicability of our results. Ó 2010 Elsevier Inc. All rights reserved.
1. Introduction Let x > 0 be a constant, C x ¼ fx : x 2 CðR; RÞ; xðt þ xÞ ¼ xðtÞg with the norm defined by jxj0 ¼ maxt2½0;x jxðtÞj, and C 1x ¼ fx : x 2 C 1 ðR; RÞ; xðt þ xÞ ¼ xðtÞg with the norm defined by kxk ¼ maxfjxj0 ; jx0 j0 g, then C x ; C 1x are both Banach spaces. ¼ 1 R x hðtÞdt; 8h 2 C x . Meanwhile, we denote h x 0 In the paper, we consider the following periodic neutral multi-delay logarithmic population model:
" # n m X X dN d ¼ NðtÞ rðtÞ aj ðtÞ ln Nðt rj ðtÞÞ bi ðtÞ ln Nðt si ðtÞÞ ; dt dt j¼1 i¼1
ð1:1Þ
where rðtÞ; aj ðtÞ; bi ðtÞ; rj ðtÞ; si ðtÞ are all in C x with r > 0; rj ðtÞ P 0 and si ðtÞ P 0; 8 t 2 ½0; x; 8 j 2 f1; . . . ; ng; 8 i 2 f1; . . . ; mg. Furthermore, rj ðtÞ 2 C 1 ðR; RÞ; si ðtÞ 2 C 2 ðR; RÞ, and s0i ðtÞ < 1; r0j ðtÞ < 1; 8 t 2 ½0; x; 8 j 2 f1; . . . ; ng; 8 i 2 f1; . . . ; mg. For the ecological justification of Eq. (1.1), see [2,4–6,10,11]. In [7], Lu and Ge studied Eq. (1.1), where rðtÞ; aj ðtÞ; bi ðtÞ; rj ðtÞ; si ðtÞ are defined as above. The purpose of this paper is to reestablish some criteria to guarantee the existence of positive periodic solutions of Eq. (1.1). By using an abstract continuation theorem for k-set contraction and some other analysis techniques, we obtain a new result on the existence of positive periodic solutions and give an example to illustrate the applicability of our results. Let NðtÞ ¼ exðtÞ , then Eq. (1.1) can be rewritten in the following form:
x0 ðtÞ ¼ rðtÞ
n X j¼1
aj ðtÞxðt rj ðtÞÞ
m X
ci ðtÞx0 ðt si ðtÞÞ;
ð1:2Þ
i¼1
where ci ðtÞ ¼ bi ðtÞð1 s0i ðtÞÞ; i ¼ 1; . . . ; m. It is easy to see that Eq. (1.1) has an x-positive periodic solution if and only if Eq. (1.2) possesses an x-periodic solution. q This paper was supported by the National Natural Sciences Foundation of China (10871063) and Scientific Research Fund of Hunan Provincial Education Department (07A038). * Corresponding author. E-mail address:
[email protected] (Y. Luo).
0096-3003/$ - see front matter Ó 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2010.02.024
Y. Luo, Z. Luo / Applied Mathematics and Computation 216 (2010) 1310–1315
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2. Main lemmas In order to study Eq. (1.1), now we make some preliminary remarks and definitions. Definition 2.1. Let E be a Banach space, S E be a bounded subset, denote
aE ðSÞ ¼ inffd > 0 j there is a finite number of subsets Si S such that S ¼
[
Si and diam ðSi Þ 6 dg;
i¼1
then aE is called non-compactness measure of S or Kuratowski distance (see [1]), where diamðSi Þ denotes the diameter of set Si . Definition 2.2. Let E1 and E2 be Banach spaces, D E1 ; A : D ! E2 be a continuous and bounded operator. If there exists a constant k P 0 satisfying
aE2 ðAðSÞÞ 6 kaE1 ðSÞ for any bounded set S D, then A is called k-set contractive operator on D (see [1]). Assume that Ł : DomŁ E1 ! E2 is a Fredholm operator with index 0 (see [3]), from [1], we know that supfd > 0 j daE1 ðBÞ 6 aE2 ðŁðBÞÞg exists for any bounded set B DomŁ, so we can define
lðŁÞ :¼ supfd > 0 j daE ðBÞ 6 aY ðŁðBÞÞ; for any bounded set B DomŁg: Now let Ł : X ! Y be a Fredholm operator with index 0, X and Y be Banach Spaces, X X be an open and bounded set, and let N : X ! Y be a k-set contractive operator with k < lðŁÞ. By using the homotopy invariance of k-set contractive operator’s topological degree D½ðŁ; NÞ; X, [9] proved the following result. Lemma 2.1. Assume that Ł : X ! Y is a Fredholm operator with index 0, r 2 Y is a fixed point, N : X ! Y is a k-set contractive with k < lðŁÞ, where X X is bounded, open, and symmetric about 0 2 X. Furthermore, we also assume that
ðR1 Þ Łx–kNx þ kr;
8k 2 ð0; 1Þ;
8x 2 @ X \ DomŁ; 8x 2 @ X \ KerŁ;
ðR2 Þ ½QNðxÞ þ Qr; x ½QNðxÞ þ Qr; x < 0;
where ½; is a bilinear form on Y X, and Q is the projection of Y onto CokerŁ, where CokerŁ is the cokernel of the operator Ł. Then there exists a x 2 X satisfying Łx ¼ Nx þ r. In order to use Lemma 2.1 to study Eq. (1.2), we set Y ¼ C x , X ¼ C 1x ,
Łx ¼
dx dt
ð2:1Þ
and
Nx ¼
n X
aj ðtÞxðt rj ðtÞÞ
j¼1
m X
ci ðtÞx0 ðt si ðtÞÞ;
ð2:2Þ
i¼1
then Eq. (1.2) is equivalent to the equation
Łx ¼ Nx þ r;
ð2:3Þ
where r ¼ rðtÞ. Clearly, Eq. (1.2) has an x-periodic solution if and only if Eq. (2.3) has a solution x 2 C 1x . Lemma 2.2 [8, Lemma 3.2]. The differential operator Ł is a Fredholm operator with index 0, and satisfies lðŁÞ P 1. Lemma 2.3. If k ¼
Pm
i¼1 jc i j0 ,
then N : X ! C x is a k-set contractive operator.
As Lemma 2.3 can be proved in the same way as in the proof of Lemma 3.3 in [9], we omit it here. Lemma 2.4 [7, Lemma 4]. Suppose s 2 C 1x and l 2 CðR; RÞ with lða þ xÞ ¼ lðaÞ þ x; 8 a 2 R.
s0 ðtÞ < 1; 8 t 2 ½0; x. Then the function t sðtÞ has a inverse lðtÞ satisfying
Remark 2.1. By using Lemma 2.4, we see that if g 2 C x ; s 2 C 1x and s0 ðtÞ < 1; 8 t 2 ½0; x, then gðlðt þ xÞÞ ¼ gðlðtÞ þ xÞ ¼ gðlðtÞÞ; 8 t 2 R, where lðtÞ is the inverse function of t sðtÞ, which together with l 2 CðR; RÞ implies that gðlðtÞÞ 2 C x . 3. Main results Since s0i ðtÞ < 1; r0j ðtÞ < 1; 8 t 2 ½0; x, we see that either t rj ðtÞ or t si ðtÞ has a unique inverse. Now we set lj ðtÞ to represent the inverse of t rj ðtÞ; j ¼ 1; . . . ; n, and set ci ðtÞ to represent the inverse of t si ðtÞ; i ¼ 1; . . . ; m. Further, we denote
CðtÞ :¼
n X j¼1
0 m X aj ðlj ðtÞÞ bi ðci ðtÞÞ : 0 1 rj ðlj ðtÞÞ i¼1 1 s0i ðci ðtÞÞ
ð3:1Þ
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Theorem 3.1. Assume the following conditions hold: (H1) There exists a constant h > 0 such that jCðtÞj P h; 8 t 2 ½0; x, where CðtÞ is defined by (3.1), Pn Pm P 0 1=2 0 < 1 and m (H2) j¼1 jaj j0 x þ i¼1 jbi j0 j1 si j0 i¼1 jbi j0 j1 si j0 < 1. Then Eq. (1.1) has at least an x-positive periodic solution. Remark 3.1. A similar result was given in [7] (see Theorem in [7]), where the condition ðH2 Þ
0 11=2 !1=2 X n n pffiffiffi m X X 1 1=2 0 0 @ a A xþ jbi j0 j1 si j0 < 1 and 2jaj j0 lj þ 2 j¼1 j j¼1 i¼1 0
m X
jbi j0 j1 s0i j0 < 1
i¼1
was used, where lj ¼ jrj mj xj0 ; mj is determined by rj ðtÞ such that t j 2 ½mj x x=2; mj x þ x=2Þ, where tj satisfy rj ðtj Þ ¼ tj . The condition ðH2 Þ of Theorem 3.1 in our result is simpler than the one in the theorem in [7], so our result is more practical. Proof. Suppose that xðtÞ is an arbitrary x-periodic solution of the following operator equation
Łx ¼ kNx þ kr;
k 2 ð0; 1Þ;
where Ł and N are defined by (2.1) and (2.2), respectively. Then xðtÞ satisfies
" 0
x ðtÞ ¼ k rðtÞ
n X
aj ðtÞxðt rj ðtÞÞ
m X
j¼1
# 0
ci ðtÞx ðt si ðtÞÞ :
ð3:2Þ
i¼1
Integrating both sides of (3.2) over [0, x], we have
Z
x
" rðtÞ
0
n X
aj ðtÞxðt rj ðtÞÞ þ
m X
j¼1
# 0
bi ðtÞxðt si ðtÞÞ dt ¼ 0;
i¼1
i.e.,
Z
x
0
n X
aj ðtÞxðt rj ðtÞÞdt
Z
m X
x
0
j¼1
0
bi ðtÞxðt si ðtÞÞdt ¼ r x:
ð3:3Þ
i¼1
Let t rj ðtÞ ¼ s, i.e., t ¼ lj ðsÞ, then
Z
x
aj ðtÞxðt rj ðtÞÞdt ¼
Z
xrj ðxÞ
rj ð0Þ
0
aj ðlj ðsÞÞ xðsÞds: 1 r0j ðlj ðsÞÞ
According to Remark 2.1 on Lemma 2.4, we have
aj ðlj ðsÞÞ xðsÞ 2 C x : 1 r0j ðlj ðsÞÞ Thus,
Z
x
aj ðtÞxðt rj ðtÞÞdt ¼
Z
0
x
0
aj ðlj ðsÞÞ xðsÞds; 1 r0j ðlj ðsÞÞ
j ¼ f1; . . . ; ng:
ð3:4Þ
i ¼ f1; . . . ; mg:
ð3:5Þ
Similarly,
Z
x
0
0
bi ðtÞxðt si ðtÞÞdt ¼
Z 0
x
0
bi ðci ðtÞÞ xðtÞdt; 1 s0i ðci ðtÞÞ
Substituting (3.4) and (3.5) into (3.3), we get
rx ¼
Z
x
CðtÞxðtÞdt:
0
Considering assumption ðH1 Þ, we know jCðtÞj P h > 0, and it follows from the integral mean value theorem that there exists g 2 ½0; x satisfying
jxðgÞj ¼
r r 6 : jCðgÞj h
In view of
jxðtÞj ¼ jxðgÞ þ
Z g
t
r x0 ðtÞdtj 6 þ h
Z
x
jx0 ðtÞjdt; 0
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Y. Luo, Z. Luo / Applied Mathematics and Computation 216 (2010) 1310–1315
we obtain
r jxj0 6 þ h
Z
x
jx0 ðtÞjdt:
ð3:6Þ
0
Multiplying both sides of (3.2) by x0 ðtÞ and integrating them over [0, x], we have
Z
x
Z Z xX Z xX x n m 0 0 0 0 jx ðtÞj dt ¼ k rðtÞx ðtÞdt aj ðtÞxðt rj ðtÞÞx ðtÞdt ci ðtÞx ðt si ðtÞÞx ðtÞdt 0 0 0 j¼1 i¼1 Z x Z Z n m x x X X 6 jrj0 jx0 ðtÞjdt þ jaj j0 jxj0 jx0 ðtÞjdt þ jci ðtÞx0 ðt si ðtÞÞjjx0 ðtÞjdt: 2
0
0
0
0
j¼1
0
i¼1
By using Cauchy–Schwarz inequality, we have
Z
x
jx0 ðtÞj2 dt 6
jrj0 þ
0
n X
jaj j0 jxj0
!Z
þ
x
jx0 ðtÞj2 dt
1=2
jci ðtÞx0 ðt si ðtÞÞj2 dt
1=2 Z
0
i¼1
x1=2
0
j¼1 m Z X
x
x
jx0 ðtÞj2 dt
1=2 ð3:7Þ
:
0
Meanwhile, we see that
Z
x
jci ðtÞx0 ðt si ðtÞÞj2 dt
1=2
Z
¼
0
x
1 1s c
0 i ð i ðtÞÞ
0
Z
¼
x
0
jci ðci ðtÞÞx0 ðtÞj2 dt
1=2
ð1 s0i ðci ðtÞÞÞjbi ðci ðtÞÞx0 ðtÞj2 dt
1=2
1=2
6 j1 s0i j0 jbi j0
Z
x
jx0 ðtÞj2 dt
1=2 :
ð3:8Þ
0
Substituting (3.8) into (3.7), we can find
Z
x
2
0
jx ðtÞj dt 6
jrj0 þ
0
n X
jaj j0 jxj0
!Z
x
jx0 ðtÞj2 dt
1=2
x1=2 þ
0
j¼1
m X
1=2
j1 s0i j0 jbi j0
Z
i¼1
x
jx0 ðtÞj2 dt;
0
which gives
Z
x 0
2
jx ðtÞj dt
1=2 jrj0 þ
6
0
n X
! jaj j0 jxj0 x1=2 þ
j¼1
m X
j1 s0i j1=2 0 jbi j0
Z
i¼1
x
jx0 ðtÞj2 dt
1=2 :
0
Substituting (3.6) into the above formula, we get
Z
x 0
2
jx ðtÞj dt
1=2
6 jrj0 x
0
n X
þ
j¼1
Pn
From the assumption
Z
1=2
x
jx0 ðtÞj2 dt
j¼1 jaj j0
1=2
xþ
Pm
" #Z 1=2 n m x X X r 1=2 0 1=2 jaj j0 x þ jaj j0 x þ j1 si j0 jbi j0 jx0 ðtÞj2 dt : h 0 j¼1 i¼1
i¼1 j1
ð3:9Þ
s0i j1=2 0 jbi j0 < 1, it follows from (3.9) that there exists constant M > 0 such that
< M:
ð3:10Þ
0
Substituting (3.10) into (3.6), we have
r jxj0 6 þ Mx1=2 :¼ M 1 : h
ð3:11Þ
Again from (3.2), we get
jx0 j0 6 jrj0 þ
n X
jaj j0 jxj0 þ
j¼1
From condition
jx0 j0 6
i¼1 jci j0
0
x
6
Pm
i¼1 j1
s0i j0 jbi j0 < 1, it is easy to see that
Pn
1
j¼1 jaj j0 M 1
Pm
i¼1 jc i j0
By Lemma 2.4, we see
Z
jci j0 jx0 j0 :
i¼1
Pm
jrj0 þ
m X
:¼ M 2 :
ð3:12Þ
lj ðxÞ ¼ lj ð0Þ þ x; 8 j 2 f1; . . . ; ng. So
Z lj ðxÞ Z x aj ðtÞð1 r0j ðtÞÞ aj ðlj ðtÞÞ j ; aj ðtÞdt ¼ xa dt ¼ dt ¼ 0 0 1 rj ðlj ðtÞÞ 1 rj ðtÞ lj ð0Þ 0
j ¼ f1; . . . ; ng:
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Y. Luo, Z. Luo / Applied Mathematics and Computation 216 (2010) 1310–1315
Similarly,
Z
x
0
0
bi ðci ðtÞÞ dt ¼ 1 s0i ðci ðtÞÞ
Z
x 0
0
bi ðtÞdt ¼ 0;
i ¼ f1; . . . ; mg:
Thus,
C¼
1
x
Z
x
CðtÞdt ¼
0
" n Z 1 X
x
j¼1
x
0
m X aj ðlj ðtÞÞ dt 0 1 rj ðlj ðtÞÞ i¼1
Z
x
0
# 0 n X bi ðci ðtÞÞ j : a dt ¼ 0 1 si ðci ðtÞÞ j¼1
From jCðtÞj P h; 8 t 2 ½0; x, we have
Z 1 Z x X n 1 x j j ¼ jC ¼ a CðtÞdt ¼ jCðtÞjdt P h > 0: x 0 8 j¼1 x 0 9 Now, we take M 3 > max
= n o P M1 ; M 2 ; Pnr and X ¼ x : x 2 C 1x ; kxk < M3 . Then k ¼ m i¼1 jc i j0 < 1 6 lðŁÞ. So by (3.11) : j¼1 aj ; <
and (3.12), we can find that all conditions of Lemma 2.1 except ðR2 Þ hold. Next we will prove that the condition ðR2 Þ of Lemma 2.1 is also satisfied. In order to do this, we define a bounded bilinear form ½; on C x C 1x as follows:
½y; x ¼
Z
x
yðtÞxðtÞdt:
0
Also we define Q : Y ! Coker Ł by Qy ¼ x1
Rx 0
yðtÞdt. Obviously,
fxjx 2 ker Ł \ @ Xg ¼ fxjx ¼ M3 ; or x ¼ M 3 g: Without loss of generality, we may assume that x ¼ M 3 . Thus
½QNðxÞ þ Qr;x ½QNðxÞ þ Qr;x ¼ M 23
"Z
x
rðtÞdt M 3
0
n Z X j¼1
# "Z
x
x
aj ðtÞdt
0
rðtÞdt þ M3
0
n Z X j¼1
#
x
aj ðtÞdt 0
# " # X X m m 2 2 2 2 aj r þ M 3 aj ¼ x M3 r M 3 aj r þ M 3 aj < 0: ¼ x M3 r M3 j¼1 j¼1 j¼1 j¼1 "
m X
# "
m X
#
"
Therefore, by Lemma 2.1, we obtain Eq. (1.1) has at least an x-positive periodic solution. The proof of Theorem 3.1 is complete. P Pm 0 0 1=2 Since j1 s0i j0 6 1; 8 i 2 f1; . . . ; mg, it follows that m i¼1 jbi j0 1 si 0 6 i¼1 jbi j0 1 si 0 . So from Theorem 3.1, we have the following result. h Corollary 3.2. Assume the following conditions hold: 0 exists a constant h > 0 such that jCðtÞj P h; 8 t 2 ½0; x, where CðtÞ is defined by (3.1), H10 There Pm Pn 0 1=2 < 1. H2 j¼1 jaj j0 x þ i¼1 jbi j0 j1 si j0 Then Eq. (1.1) has at least an x-positive periodic solution. P P 0 1=2 0 Since j1 s0i j0 > 1; 8 i 2 f1; . . . ; mg, it follows that m 6 m i¼1 jbi j0 j1 si j0 i¼1 jbi j0 j1 si j0 . So by Theorem 3.1, we have the following result. Corollary 3.3. Assume the following conditions hold: 0 exists a constant h > 0 such that jCðtÞj P h; 8 t 2 ½0; x, where CðtÞ is defined by (3.1), H10 There Pm 0 H2 i¼1 jbi j0 j1 si j0 < 1. Pm 1=2 1 jb j j1s0i j0 i¼1 i 0 . Then Eq. (1.1) has at least an x-positive periodic solution for d, where d ¼ max jaj j0 < nx j2f1;...;ng
4. Example Now, we give an example to demonstrate our results. Example 4.1. Let us consider the following equation:
dN 1 1 d ¼ NðtÞ rðtÞ ðcos2 t þ 1Þ ln Nðt pÞ ð3 cos tÞ ln Nðt pÞ ; dt 32 64 dt
ð4:1Þ
Y. Luo, Z. Luo / Applied Mathematics and Computation 216 (2010) 1310–1315
1315
1 1 where rðtÞ ¼ cos t 32 ðcos2 t þ 1Þ sin t 64 ð3 cos tÞ cos t. 1 1 ðcos2 t þ 1Þ; b1 ðtÞ ¼ 64 ð3 cos tÞ; r1 ðtÞ ¼ s1 ðtÞ ¼ p. So r ¼ Corresponding to Eq. (1.1), we have n ¼ m ¼ 1; a1 ðtÞ ¼ 32 p > 0; r0 ¼ s0 ¼ 0; l ðtÞ ¼ c ðtÞ ¼ p þ t. Let x ¼ 2p. Thus 1 1 1 1 64
1 1 1 ðcos2 t þ 1Þ þ sin t P ; 32 64 64 2p þ 1 ¼ < 1: 16
CðtÞ ¼ a1 ðl1 ðtÞÞ b01 ðc1 ðtÞÞ ¼ ja1 j0 x þ jb1 j0 j1 s01 j01=2
Hence, the conditions H01 and H02 in the Corollary 3.2 hold. So from Corollary 3.2, we obtain that Eq. (4.1) has a 2p-positive periodic solution. In fact, it is easy to see that NðtÞ ¼ esin t is a 2p-positive periodic solution of Eq. (4.1). However, the condition ðH2 Þ of Theorem in [7] is not satisfied. Since
pffiffiffi pffiffiffi 1=2
pffiffiffi 1=2 1 1 1=2 1 0 p 1 1=2 p ð 2p þ 1Þ1=2 x þ 2ja1 j0 l1 þ jb1 j0 j1 s01 j1=2 ¼ 2 p þ 2 ¼ ja1 j0 þ þ > 1; 0 2 2 32 16 16 4 4 where we set l1 ¼ jr1 m1 xj0 ¼ jp 2pj0 ¼ p. Thus, the result of [7] is not applicable to the example. Acknowledgements The authors are grateful to the referees for their valuable comments which have led to improvement of the presentation. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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