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Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa

Positive periodic solutions of a periodic neutral delay logistic equation with impulsesI Yongkun Li Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, People’s Republic of China

article

info

a b s t r a c t

Article history: Received 4 April 2006 Received in revised form 1 March 2008 Accepted 25 March 2008

By using a fixed point theorem of strict-set-contraction, some criteria are established for the existence of positive periodic solutions for a periodic neutral functional differential equation with impulses of the form

dx(t )

Keywords: Positive periodic solution Neutral functional differential equation Impulse Strict-set-contraction

dt

" = x(t ) a(t ) −

m X

bi (t )x(t − τi (t )) −

i=1

x(tk ) − x(tk ) = −Ik (x(tk )),

k = 1, 2, . . . .

+

n X

# cj (t )x (t − σj (t )) , 0

j=1

© 2008 Elsevier Ltd. All rights reserved.

1. Introduction Recently, by using the continuation theory for k-set-contractions, Lu [1] studied the existence of positive periodic solutions of the following neutral logistic equation dN (t ) dt

" = N (t ) a(t ) − β(t )N (t ) −

m X

bi (t )N (t − τi (t )) −

n X

# cj (t )N (t − σj (t )) , 0

(1.1)

j=1

i=1

where a, β, bi , cj ∈ C (R, [0, +∞)), τi , σj ∈ C (R, [0, +∞)) (i = 1, 2, . . . , m, j = 1, 2, . . . , n) are ω-periodic functions. Also, Yang and Cao [2], by using Mawhin’s continuation theorem [3], investigated the existence of positive periodic solutions of (1.1) in the case of m = n. However, the main results obtained in [1,2] all required that cj ∈ C 1 , σj ∈ C 2 and σj0 < 1 (j = 1, 2, . . . , n). In this paper, we are concerned with the following neutral delay logistic equation with impulses

dx(t ) dt

" = x(t ) a(t ) −

m X i=1

bi (t )x(t − τi (t )) −

x(tk ) − x(tk ) = −Ik (x(tk )), +

k = 1, 2, . . . ,

n X j=1

# cj (t )x (t − σj (t )) , 0

(1.2)

where a, bi , cj ∈ C (R, R ) and τi , σj ∈ C (R, R), i = 1, 2, . . . , m, j = 1, 2, . . . , n are ω-periodic functions, Ik ∈ C (R+ , R+ ), there exists a positive integer p such that tk+p = tk + ω, Ik+p = Ik , k ∈ Z, [0, ω) ∩ {tk : k ∈ Z} = {t1 , t2 , . . . , tp }. For the ecological justification of (1.1), one can refer to [4–7]. The main purpose of this paper is to establish criteria to guarantee the existence of positive periodic solutions of (1.2) by using a fixed point theorem of strict-set-contraction [Theorems 3, 8]. +

I This work is supported by the National Natural Sciences Foundation of People’s Republic of China under Grant 10361006 and the Natural Sciences Foundation of Yunnan Province under Grant 2003A0001M. E-mail address: [email protected]

0898-1221/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2008.03.044

2190

Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196

For convenience, we introduce the notation R − 0ω a(s)ds

Θ := e

,

ω

Z

∆=

" m X

0

Π=

ω

Z

" m X

0

bi (s) +

i=1

n X

Θ bi (s) −

i=1

n X

# cj (s) ds,

j=1

# cj (s) ds,

f M = max {f (t )},

j=1

t ∈[0,ω]

f m = min {f (t )}, t ∈[0,ω]

where f is a continuous ω-periodic function. Throughout this paper, we assume that: Rω

− 0 a(s)ds (H1 ) Θ < 1. P := e P n (H2 ) m Θ bi (t ) − j=1 cj (t ) ≥ 0. i=1 Pm Pn 2 (H3 ) (1 + am ) 1Θ−Θ∆ ≥ maxt ∈[0,ω] i=1 bi (t ) + j=1 cj (t ) . Pm Pn M (H4 ) ΠΘ((a1−−Θ1)) ≤ mint ∈[0,ω] i=1 Θ bi (t ) − j=1 cj (t ) . P (H5 ) 1Θ−2Θ∆ nj=1 cjM < 1.

2. Preliminaries The following lemma is fundamental to our discussion. Since the proof is similar to those in the literature [9,10], we omit the proof. Lemma 2.1. x(t ) is an ω-periodic solution of (1.2) is equivalent to x(t ) is an ω-periodic solution of the following: x(t ) =

Z

t +ω

G(t , s)x(s)

" m X

t

bi (s)x(s − τi (s)) +

i=1

n X

# cj (s)x (s − σj (s)) + 0

j=1

X

G(t , tk )Ik (x(tk )),

(2.1)

k:tk ∈[t ,t +ω)

where G(t , s) =

e−

Rs

t a(θ)dθ

1 − e−

Rω

0 a(θ)dθ

,

s ∈ [t , t + ω].

(2.2)

In order to obtain the existence of a periodic solution of system (2.1), we first make the following preparations: Let E be a Banach space and K be a cone in E. The semi-order induced by the cone K is denoted by ‘‘≤’’. That is, x ≤ y if and only if y − x ∈ K . In addition, for a bounded subset A ⊂ E, let αE (A) denote the (Kuratowski) measure of non-compactness defined by

( αE (A) = inf δ > 0 : there is a finite number of subsets Ai ⊂ A ) such that A =

[

Ai and diam(Ai ) ≤ δ

,

i

where diam(Ai ) denotes the diameter of the set Ai . Let E, F be two Banach spaces and D ⊂ E, a continuous and bounded map Φ : D → F is called k-set contractive if for any bounded set S ⊂ D we have

αF (Φ (S )) ≤ kαE (S ). Φ is called strict-set-contractive if it is k-set-contractive for some 0 ≤ k < 1. The following lemma cited from Ref. [8,11] is useful for the proof of our main results of this paper. Lemma 2.2 ([8,11]). Let K be a cone of the real Banach space X and Kr ,R = {x ∈ K : r ≤ kxk ≤ R} with R > r > 0. Suppose that Φ : Kr ,R → K is strict-set-contractive such that one of the following two conditions is satisfied: (i) Φ x ≤ 6 x, ∀x ∈ K , kxk = r and Φ x ≥ 6 x, ∀x ∈ K , kxk = R. (ii) Φ x ≥ 6 x, ∀x ∈ K , kxk = r and Φ x ≤ 6 x, ∀x ∈ K , kxk = R. Then Φ has at least one fixed point in Kr ,R .

Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196

2191

Define PC (R) = x : R → R, x|(tk ,tk+1 ) ∈ C ((tk , tk+1 ), R), ∃x(tk− ) = x(tk ), x(tk+ ), k ∈ Z

and

PC 1 (R) = x : R → R, x|(tk ,tk+1 ) ∈ C 1 ((tk , tk+1 ), R), ∃x0 (tk− ) = x0 (tk ), x(tk+ ), k ∈ Z . Set

X = {x : x ∈ PC (R), x(t + ω) = x(t ), t ∈ R} with the norm defined by |x|0 = supt ∈[0,ω] {|x(t )|}, and

Y = x : x ∈ PC 1 (R), x(t + ω) = x(t ), t ∈ R

with the norm defined by |x|1 = max{|x|0 , |x0 |0 }. Then X and Y are all Banach spaces. Define the cone K in Y by K = {x ∈ Y, x(t ) ≥ Θ |x|1 , t ∈ [0, ω]}.

(2.3)

Let the map Φ be defined by

(Φ x)(t ) =

Z

t +ω

G(t , s)x(s)

" m X

t

n X

bi (s)x(s − τi (s)) +

# G(t , tk )Ik (x(tk )),

(2.4)

k:tk ∈[t ,t +ω)

j=1

i=1

X

cj (s)x (s − σj (s)) + 0

where x ∈ K , t ∈ R, and G(t , s) is defined by (2.2). Definition 2.1 ([12]). The set F is said to be quasi-equicontinuous in [0, ω] if for any > 0 there exists a δ > 0 such that if x ∈ F , k ∈ Z, t 0 , t 00 ∈ (tk−1 , tk ] ∩ [0, ω], and |t 0 − t 00 | < d, then |x(t 0 ) − x(t 00 )| < . Lemma 2.3 (Compactness criterion [12]). The set F ⊂ X is relatively compact if and only if: (a) F is bounded, that is, kxk ≤ M for each x ∈ F and some M > 0; (b) F is quasi-equicontinuous in [0, ω]. In what follows, we will give some lemmas concerning K and Φ defined by (2.3) and (2.4), respectively. Lemma 2.4. Assume that (H1 )–(H3 ) hold. (i) If aM ≤ 1, then Φ : K → K is well defined. (ii) If (H4 ) holds and aM > 1, then Φ : K → K is well defined. Proof. For any x ∈ K , it is clear that Φ x ∈ Y. In view of (H2 ), for x ∈ K , t ∈ [0, ω], we have m X

bi (t )x(t − τi (t )) +

n X

cj (t )x0 (t − σj (t )) ≥

bi (t )x(t − τi (t )) −

≥

m X

Θ bi (t )|x|1 −

i=1

=

n X

cj (t )|x0 (t − σj (t ))|

j=1

i=1

j=1

i=1

m X

n X

cj (t )|x|1

j=1

" m X

Θ bi (t ) −

i=1

n X

# cj (t ) |x|1 ≥ 0.

(2.5)

j=1

Therefore, for x ∈ K , t ∈ [0, ω], we find

|Φ x|0 ≤

1 1−Θ

ω

Z

x(s)

" m X

0

bi (s)x(s − τi (s)) +

i=1

#

n X

cj (s)x (s − σj (s)) ds +

1

0

j=1

p X

1 − Θ k=1

Ik (x(tk ))

and

Θ (Φ x)(t ) ≥ 1−Θ

Z

Θ = 1−Θ

Z

t +ω

x(s)

t

≥ Θ |Φ x|0 .

0

" m X

bi (s)x(s − τi (s)) +

n X

i=1

ω

x(s)

" m X i=1

# cj (s)x (s − σj (s)) ds + 0

j=1

#

n

bi (s)x(s − τi (s)) +

X j=1

cj (s)x (s − σj (s)) ds + 0

p Θ X Ik (x(tk )) 1 − Θ k=1

p Θ X Ik (x(tk )) 1 − Θ k=1

(2.6)

2192

Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196

Now, we show that (Φ x)(t ) ≥ Θ |(Φ x)|1 , t ∈ [0, ω]. From (2.4), we have

(Φ x) (t ) = G(t , t + ω)x(t + ω) 0

" m X

bi (t + ω)x(t + ω − τi (t + ω)) +

i=1

− G(t , t )x(t )

" m X

n X

# cj (t + ω)x (t + ω − σj (t + ω)) 0

j=1

bi (t )x(t − τi (t )) +

n X

# cj (t )x (t − σj (t )) + a(t )(Φ x)(t ) 0

j=1

i=1

" m X

= a(t )(Φ x)(t ) − x(t )

bi (t )x(t − τi (t )) +

i=1

n X

# cj (t )x (t − σj (t )) . 0

(2.7)

j=1

It follows from (2.7) that if (Φ x)0 (t ) ≥ 0, then

(Φ x)0 (t ) ≤ a(t )(Φ x)(t ) ≤ aM (Φ x)(t ) ≤ (Φ x)(t ).

(2.8)

On the other hand, from (2.5), (2.7) and (H3 ), if (Φ x)0 (t ) < 0, then

−(Φ x) (t ) = x(t ) 0

" m X

ai (t )x(t − τi (t )) +

≤ |x|

" m X

# cj (t )x (t − σj (t )) − a(t )(Φ x)(t ) 0

j=1

i=1 2 1

n X

bi (t ) +

n X

# cj (t ) − am (Φ x)(t )

j=1

i=1

# Z ω "X n m X Θ2 2 cj (s) ds − am (Φ x)(t ) |x|1 Θ bi (s) − ≤ (1 + a ) 1−Θ 0 j=1 i=1 # " Z ω n m X X Θ m |x|1 cj (s) ds − am (Φ x)(t ) Θ |x|1 bi (s) − = (1 + a ) Θ |x|1 1−Θ 0 j=1 i=1 " # Z t +ω n m X X 0 m cj (s)|x (s − σj (s))| ds − am (Φ x)(t ) ≤ (1 + a ) bi (s)x(s − τi (s)) − G(t , s)x(s) m

t

≤ (1 + a ) m

Z

t +ω

G(t , s)x(s)

i=1

j=1

" m X

n X

t

bi (s)x(s − τi (s)) +

i=1

# cj (s)x (s − σj (s)) ds − am (Φ x)(t ) 0

j=1

= (1 + am )(Φ x)(t ) − am (Φ x)(t ) = (Φ x)(t ).

(2.9)

It follows from (2.8) and (2.9) that |(Φ x)0 |0 ≤ |Φ x|0 . So |Φ x|1 = |Φ x|0 . By (2.6) we have

(Φ x)(t ) ≥ Θ |Φ x|1 . Hence, Φ x ∈ K . The proof of (i) is complete. (ii) In view of the proof of (i), we only need to prove that (Φ x)0 (t ) ≥ 0 implies

(Φ x)0 (t ) ≤ (Φ x)(t ). From (2.5), (2.7), (H2 ) and (H4 ), we obtain

(Φ x) (t ) ≤ a(t )(Φ x)(t ) − Θ |x|1 0

" m X

bi (t )x(t − τi (t )) −

i=1

" ≤ a(t )(Φ x)(t ) − Θ |x|

2 1

Θ

m X

# cj (t )|x (t − σj (t ))| 0

j=1

bi (t ) −

i=1

aM − 1

n X

n X

# cj (t )

j=1

Z

ω

" m X

n X

#

cj (s) ds bi (s) + Θ (1 − Θ ) 0 j=1 i=1 " # Z t +ω m n X X 1 M M ≤ a (Φ x)(t ) − (a − 1) |x|1 bi (s)|x|1 + cj (s)|x|1 ds 1−Θ t i=1 j=1

≤ a (Φ x)(t ) − Θ |x| M

2 1

Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196

≤ aM (Φ x)(t ) − (aM − 1)

Z

t +ω

G(t , s)x(s)

" m X

t

≤ a (Φ x)(t ) − (a − 1) M

M

Z

bi (s)x(s − τj (s)) +

i=1 t +ω

G(t , s)x(s)

" m X

t

n X

2193

# cj (s)|x0 (s − σj (s))| ds

j=1

bi (s)x(s − τj (s, x(s))) +

i=1

n X

# cj (s)x (s − σj (s)) ds 0

j=1

= aM (Φ x)(t ) − (aM − 1)(Φ x)(t ) = (Φ x)(t ). The proof of (ii) is complete.

Lemma 2.5. Assume that (H1 )–(H3 ) hold and R

Pn

j=1

cjM < 1.

¯ R → K is strict-set-contractive, (i) If aM ≤ 1, then Φ : K Ω T ¯ R → K is strict-set-contractive, (ii) If (H4 ) holds and aM > 1, then Φ : K Ω T

where ΩR = {x ∈ Y : |x|1 < R}. Proof. We only need to prove P (i), sincethe proof of (ii) is similar. It is easy to see that Φ is continuous and bounded. Now n ¯ R . Let η = αY (S ). Then, for any positive number we prove that αY (Φ (S )) ≤ R j=1 cjM αY (S ) for any bounded set S ⊂ Ω

ε< R

Pn

cjM η, there is a finite family of subsets {Si } satisfying S =

j=1

S

i

Si with diam(Si ) ≤ η + ε . Therefore

|x − y|1 ≤ η + ε for any x, y ∈ Si .

(2.10)

As S and Si are precompact in X, it follows that there is a finite family of subsets {Sij } of Si such that Si =

S

j

Sij and

|x − y|0 ≤ ε for any x, y ∈ Sij .

(2.11)

In addition, for any x ∈ S and t ∈ [0, ω], we have

|(Φ x)(t )| =

t +ω

Z

G(t , s)x(s)

t

≤

" m X

bi (s)x(s − τi (s)) +

1−Θ

Z 0

ω

" m X

# cj (s)x (s − σj (s)) ds 0

j=1

i=1

R2

n X

bi (s) +

i=1

n X

# cj (s) ds := H

j=1

and

# " n m X X 0 cj (t )x (t − σj (t )) bi (t )x(t − τi (t )) + |(Φ x) (t )| = a(t )(Φ x)(t ) − x(t ) j=1 j=1 ! n m X X ≤ aM H + R2 cjM . bM + i 0

i=1

j=1

Applying Lemma 2.3, we know that Φ (S ) is precompact in X. Then, there is a finite family of subsets {Sijk } of Sij such that S Sij = k Sijk and

|Φ x − Φ y|0 ≤ ε for any x, y ∈ Sijk . From (2.7), (2.10)–(2.12) and (H2 ), for any x, y ∈ Sijk , we obtain

( " # m n X X 0 |(Φ x) − (Φ y) |0 = max a(t )(Φ x)(t ) − a(t )(Φ y)(t ) − x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] j=1 j=1 " # ) m n X X 0 + y(t ) bi (t )y(t − τi (t )) + cj (t )y (t − σj (t )) i=1 j=1 ( " # m n X X 0 ≤ max {|a(t )[(Φ x)(t ) − (Φ y)(t )]|} + max x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] t ∈[0,ω] i=1 j=1 " # ) m n X X 0 − y(t ) bi (t )y(t − τj (t )) + cj (t )y (t − σj (t )) j=1 j=1 0

0

(2.12)

2194

Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196

( " ! m n X X 0 ≤ a |(Φ x) − (Φ y)|0 + max x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] j=1 j=1 !# ) n m X X 0 bi (t )y(t − τi (t )) + cj (t )y (t − σj (t )) − i=1 j=1 ) ( " # n m X X 0 + max y(t ) cj (t )y (t − σj (t )) [x(t ) − y(t )] bi (t )y(t − τi (t )) + t ∈[0,ω] j=1 i=1 ( ) q m X X M 0 0 ≤ a ε + R max bi (t )|x(t − τi (t )) − y(t − τi (t ))| + cj (t )|x (t − σj (t )) − y (t − σj (t ))| M

t ∈[0,ω]

( + ε max

t ∈[0,ω]

i=1

j=1

n

q

X

X

bi (t )y(t − τi (t , y(t ))) +

i=1

! + R(η + ε)

BM i

n X

i=1

= Rη

n X

cj (t )|y (t − σj (t ))|

j=1

m X

≤ aM ε + Rε

) 0

! cjM

m X

+ Rε

j=1

bM i +

i=1

n X

! cjM

j=1

! + Hˆ ε,

cjM

(2.13)

j=1

Pm

ˆ = aM + 2R i=1 bM where H i + 2R From (2.12) and (2.13) we have |Φ x − Φ y|1 ≤

R

n X

Pn

j=1

cjM .

! η + Hˆ ε for any x, y ∈ Sijk .

cjM

j=1

As ε is arbitrary small, it follows that

αY (Φ (S )) ≤ R

n X

! αY (S ).

cjM

j=1

Therefore, Φ is strict-set-contractive. The proof of Lemma 2.5 is complete.

3. Main result Our main result of this paper is as follows: Theorem 3.1. Assume that (H1 )–(H3 ), (H5 ) hold. (i) If aM ≤ 1, then system (1.2) has at least one positive ω-periodic solution. (ii) If (H4 ) holds and aM > 1, then system (1.2) has at least one positive ω-periodic solution. Θ (1−Θ )

Proof. We only need to prove (i), since the proof of (ii) is similar. Let R = 1Θ−2Θ and 0 < r < . Then we have 0 < r < R. Π ∆ From Lemmas 2.4 and 2.5, we know that Φ is strict-set-contractive on Kr ,R . In view of (2.7), we see that if there exists x∗ ∈ K such that Φ x∗ = x∗ , then x∗ is one positive ω-periodic solution of system (1.2). Now, we shall prove that condition (ii) of Lemma 2.2 holds. First, we prove that Φ x 6≥ x, ∀x ∈ K , |x|1 = r. Otherwise, there exists x ∈ K , |x|1 = r such that Φ x ≥ x. So |x| > 0 and Φ x − x ∈ K , which imply that

(Φ x)(t ) − x(t ) ≥ Θ |Φ x − x|1 ≥ 0 for any t ∈ [0, ω].

(3.1)

Moreover, for t ∈ [0, ω], we have

(Φ x)(t ) =

Z

t +ω

G(t , s)x(s)

t

≤

" m X

bi (s)x(s − τi (s)) +

j=1

i=1

1 1−Θ

Z r |x|0 0

ω

" m X i=1

n X

bi (s) +

n X j=1

# cj (s) ds

# cj (s)x (s − σj (s)) ds 0

Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196

1

=

1−Θ

2195

Π |x|0

< Θ |x|0 .

(3.2)

In view of (3.1) and (3.2), we have

|x|0 ≤ |Φ x| < Θ |x|0 < |x|0 , which is a contradiction. Finally, we prove that Φ x 6≤ x, ∀x ∈ K , |x|1 = R also hold. For this case, we only need to prove that

Φ x 6< x,

x ∈ K , |x|1 = R.

Suppose, for the sake of contradiction, that there exist x ∈ K and |x|1 = R such that Φ x < x. Thus x − Φ x ∈ K \ {0}. Furthermore, for any t ∈ [0, ω], we have x(t ) − (Φ x)(t ) ≥ Θ |x − Φ x|1 > 0.

(3.3)

In addition, for any t ∈ [0, ω], we find

(Φ x)(t ) =

t +ω

Z

G(t , s)x(s)

" m X

t

bi (s)x(s − τi (s)) +

i=1

Θ ≥ Θ |x|21 1−Θ

Z

ω

" m X

0

#

n X

cj (s)x (s − σj (s)) ds 0

j=1

Θ bi (s) −

n X

# cj (s) ds

j=1

i=1

Θ2 1R2 1−Θ = R.

=

(3.4)

From (3.3) and (3.4), we obtain

|x| > |Φ x|0 ≥ R, which is a contradiction. Therefore, conditions (i) and (ii) of Lemma 2.2 hold. By Lemma 2.2, we see that Φ has at least one non-zero fixed point in K . Thus, system (2.1) has at least one positive ω-periodic solution. Therefore, it follows from Lemma 2.1 that system (1.2) has a positive ω-periodic solution. The proof of Theorem 3.1 is complete. Remark 3.1. Since we do not restrict σj to be non-negative, in fact, (1.2) is a neutral functional differential equation of mixed type. An example Consider the following non-impulsive equation x0 (t ) = x(t )

1 + cos t 4π

− (2 + sin t )x(t − τ1 (t )) −

1 − sin t 0 x (t − σ1 (t )) , 20

(3.5)

where τ , σ ∈ C (R, R) are 2π -periodic functions. Obviously, a(t ) =

1 + cos t 4π

,

b1 (t ) = 2 + sin t ,

c1 (t ) =

1 − sin t 20

.

Furthermore, we have 1

Θ = e− 2 , ∆=

Z

2π

max {Θ b1 (t ) − c1 (t )} > 3 > 0,

0≤t ≤2π

1

[Θ b1 (s) − c1 (s)]ds = 16π e− 2 −

0

(1 + aM )

∆=

Θ2 ∆ > 59 and 1−Θ

1−Θ

Θ 2∆

1 10

99 10

1

π + 6π e− 2 ,

max {b1 (t ) + c1 (t )} < 4,

0≤t ≤2π

π,

c1M < 1.68 × 10−3 < 1.

Hence, (H1 )–(H3 ) and (H5 ) hold and aM ≤ 1. According to Theorem 3.1, system (3.5) has at least one positive 2π -periodic solution. Remark 3.2. Since we do not require σ10 (t ) < 1, the results in [1,2] cannot be used to obtain the existence of positive 2π -periodic solutions of (3.5).

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Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196

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