- Email: [email protected]
Counterexamples to the article “Oscillation and global attractivity in hematopoiesis model with delay time”
Applied Mathematics and Computation 168 (2005) 973–980
www.elsevier.com/locate/amc
Counterexamples to the article ‘‘Oscillation and global attractivity in hematopoiesis model with delay time’’ S.J. Yang, B. Shi *, D.C. Zhang, M.J. Gai Institute of Applied Mathematics, Naval Aeronautical Engineering Academy, Yantai, Shandong 264001, PR China
Abstract In this paper, we give some counterexamples to the article ‘‘oscillation and global attractivity in hematopoiesis model with delay time’’, [Appl. Math. Comput., 136 (2– 3) (2003) 241–250]. Ó 2004 Elsevier Inc. All rights reserved. Keywords: Counterexamples; Attractivity; Differential equations
1. Introduction In the recent paper of Saker [1], the author considered the differential equation _ ¼ pðtÞ
bpm ðt sÞ cpðtÞ 1 þ pn ðt sÞ
*
Corresponding author. E-mail address: [email protected] (B. Shi).
0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2004.09.023
ð1Þ
974
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
with initial value s 6 t 6 0; /ð0Þ > 0; / 2 C½½s; 0; Rþ ;
pðtÞ ¼ /ðtÞ;
ð2Þ
where s, b, c 2 (0, 1), m, n 2 N. Assume that Eq. (1) has a unique positive equilibrium p > 1 and satisfies the equation bpm ¼ cp: 1þ pn
ð3Þ
Then the author obtains the following main results: Theorem 2.1 (Saker [1]). Assume that (2) holds and n > m. Then other positive solutions of Eq. (1) which does not oscillate about p satisfies lim pðtÞ ¼ p:
ð4Þ
t!1
Theorem 2.5 (Saker [1]). Assume that (2) holds, n > m and bðn mÞð p þ p2 Þ
nþm1
bmð p þ p1 Þ
m1
ð1 þ ðp þ p1 Þn Þ2
p ecs s < : 2
ð5Þ
Then every positive solution of Eq. (1) satisfies lim pðtÞ ¼ p;
t!1
where p1 ¼ p csð p þ f bsÞ and p2 ¼ p þ f bs. First we sketch the proof of Theorem (2.1) which is given in Saker [1]. Proof. To prove this theorem we assume that pðtÞ > p for t sufficiently large (the proof when pðtÞ < p is similar and will be omitted). Set pðtÞ ¼ p þ zðtÞ: Then z(t) > 0 for t sufficiently large. The change of variables above reduces Eq. (1) to the equation m
z_ ðtÞ þ czðtÞ þ c p¼
bðp þ zðt sÞÞ n: 1 þ ð p þ zðt sÞÞ
But from (3) we have z_ ðtÞ þ czðtÞ þ
b pm bð p þ zðt sÞÞm ¼ 0: n 1þ p 1 þ ð p þ zðt sÞÞn
As pðtÞ > p, we have z(t) > 0. Then, for t sufficiently large, we have
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
z_ ðtÞ þ
975
m b pm ð1 þ pn Þð p þ zðt sÞÞ 1 60 n pm ð1 þ ð 1þ pn p þ zðt sÞÞ Þ
or m b pm ð1 þ pn Þð p þ zðt sÞÞ 1 m < 0: z_ ðtÞ 6 n 1 þ pn p þ zðt sÞÞ Þ p ð1 þ ð
ð6Þ
Then z(t) is decreasing. So, z(ts) > z(t). Also lim zðtÞ ¼ a; 2 ½0; 1Þ exists:
t!1
We prove that a = 0. Otherwise a > 0. Then there exist e 2 (0, a) and Te > 0 such that for t P Te, ae < z(t) < a + e. As z(t) is decreasing, we have a e < zðt sÞ: Then from (6) we have n
z_ ðtÞ þ
m
b pm ð1 þ ð p þ a eÞ Þ bð1 þ pn Þð p þ a eÞ 6 0; t P T e : n n ð1 þ p Þð1 þ ðp þ a eÞ Þ
ð7Þ
Integrating (7) from Te to 1, we obtain a contradiction. Then a = 0. So z(t) tends to zero. Therefore limt!1 pðtÞ ¼ p. The proof is complete. h In section 2, we shall give some counterexamples to show that Theorem (2.1) and Theorem (2.5) are not true. Consequently, the above proof is invalid.
2. Counterexamples Consider the following differential equation x_ ðtÞ ¼
bxm ðt sÞ cxðtÞ 1 þ xmþ1 ðt sÞ
ð8Þ
with initial value xðtÞ ¼ /ðtÞ;
s 6 t 6 0; /ð0Þ > 0; / 2 C½½s; 0; Rþ ;
ð9Þ
where s, b, c 2 (0, 1) and m P 4. Then the initial value problem of (8) and (9) is an example of the initial value problem of (1) and (2). First, we list some lemmas which will be needed in the following proofs. Lemma 1 (Saker [1]). Assume that n P m and p(t) is the solution of the initial value problem (1) and (2) in (0,1), Then 0 < P ðtÞ < A;
976
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
where 8 > < max /ð0Þ; bc ; A¼ xm > 1 : max /ð0Þ; bc 1þx n ;
if n ¼ m; if n > m;
1
where x1 ¼
1n
m nm
.
Lemma 2 (Guo et al. [2]). Assume that #ðtÞ; xðtÞ 2 C 1 ½J ; R are lower and upper solutions of the initial value problem u_ ¼ f ðt; uÞ;
uð0Þ ¼ u0 ;
respectively, and there exists a positive constant L such that for every t2J,x P y, f ðt; xÞ f ðt; yÞ 6 Lðx yÞ: Then for every t 2 J, we have #ðtÞ 6 xðtÞ; where f ðt; uÞ 2 C½J R; R. Remark 1. By Lemma 2, the solution of inequality x_ ðtÞ þ cxðtÞ 6 bm ;
xð0Þ 6 /ð0Þ;
satisfies
b b xðtÞ 6 m þ /ð0Þ m ect ; c c
ð10Þ
where is a positive constant. Lemma 3. Assume that sffiffiffi!mþ1 rffiffiffiffiffiffiffiffiffiffiffiffi!m1 b ðm þ 1Þ mþ1 m P 4 and ¼ : c 2 m1
ð11Þ
Then Eq. (8) has a unique positive equilibrium x and satisfies x > 1. Proof. Consider equation bxm ¼ cx: 1 þ xmþ1 For x > 0, we have 1 þ xmþ1 ¼
b m1 x : c
ð12Þ
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
From (11), it is easy to see that x ¼ Let
qffiffiffiffiffiffiffiffiffiffiffi
bðm1Þ cðmþ1Þ
977
is a positive solution of Eq. (12).
b f ðxÞ ¼ 1 þ xmþ1 xm1 : c Then f ðxÞ ¼ 0; and ðm 1Þb m2 x f_ ðxÞ ¼ ðm þ 1Þxm c bðm 1Þ m2 2 ¼ ðm þ 1Þx x cðm þ 1Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! bðm 1Þ bðm 1Þ ¼ ðm þ 1Þxm2 x þ x : cðm þ 1Þ cðm þ 1Þ So, if 0 < x < x, we have f_ ðxÞ < 0; if x > x, q weffiffiffiffiffiffiffiffiffiffi haveffi f_ ðxÞ > 0. That is, equation bðm1Þ f(x) = 0 has a unique positive solution x ¼ cðmþ1Þ . Also, Eq. (8) has a unique positive equilibrium x. From (11), it is easy to see that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi bðm 1Þ x ¼ > 1: cðm þ 1Þ The proof is complete.
h
Remark 2. From Lemma 3, we see that if (11) holds, then Eq. (8) satisfies (3). Theorem 1. Assume that (11) and 1 m1 c 0 < /ðtÞ 6 6 < 1; s 6 t 6 0; /ð0Þ ¼ b
ð13Þ
hold. Then the solutions of the initial value problem (8) and (9) satisfies 0 < xðtÞ 6 : Proof. From (13), we have /ð0Þ ¼ P
b m : c
From (8) and Lemma 1, we have if 0 6 t 6 s, x_ ðtÞ 6 bxm ðt sÞ cxðtÞ 6 bm cxðtÞ:
ð14Þ
978
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
That is, x_ ðtÞ þ cxðtÞ 6 bm : By Remark 1 and (14), we have b b xðtÞ 6 m þ /ð0Þ m ect 6 /ð0Þ 6 : c c By induction, we can prove for t 2 ½0; ks; k 2 N 0 < xðtÞ 6 : The proof is complete.
h
Remark 3. From Theorem 1, it is easy to see that Saker [1, Theorem 2.1] is not true. Consider equation x_ ðtÞ ¼
bxm ðt sÞ cxðtÞ 1 þ xmþ3 ðt sÞ
ð15Þ
with initial value (9), where s, b, c 2 (0,1) and m P 6. Lemma 4. Assume that mþ3 m1 b 4 mþ3 mþ3 4 m P 6 and ¼ : c 4 m1 Then Eq. (15) has a unique positive equilibrium x ¼
ð16Þ
bðm1Þ cðmþ3Þ
14
and satisfies x > 1.
Proof. Similar to the that in the proof of Lemma 3, it is easy to be shown and we will omit it. h Theorem 2. Assume that (13) and (16) hold. Then the solutions of the initial value problem (15) and (9) satisfies 0 < xðtÞ 6 : Proof. Similar to the proof of Theorem 1, it is easy to be proved and we will omit it. h Remark 4. From Theorem 2, it is easy to see that Saker [1, Theorem 2.1], is not true either. Consider equation x_ ðtÞ ¼
bxm ðt sÞ cxðtÞ 1 þ xmþ5 ðt sÞ
with initial value (9), where s, b, c 2 (0,1) and m P 8.
ð17Þ
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
979
Lemma 5. Assume that m P 8 and
mþ5 m1 b 6 mþ5 mþ5 6 ¼ : c 6 m1
ð18Þ 1
Þ6 and satisfies x > 1. Then Eq. (17) has a unique positive equilibrium x ¼ ðbðm1Þ cðmþ5Þ Proof. Similar to that in the proof of Lemma 3, it is easy to be shown and we will omit it. h Theorem 3. Assume that (13) and (18) hold. Then the solutions of the initial value problem (17) and (9) satisfies 0 < xðtÞ 6 : Proof. Similar to that in the proof of Theorem 1, it is easy to be shown and we will omit it. h Lemma 6. Assume that (9) holds and m P 8. Then there exists a s0 > 0 such that 5bðx þ x2 Þ
2mþ4
bmðx þ x1 Þ
m1
mþ5 2
p ecs0 s0 < ; 2
ð1 þ ðx þ x1 Þ Þ where x1 ¼ x csðx þ f bsÞ; x2 ¼ x þ f bs.
ð19Þ
Proof. Let
h m1 i ct 2mþ4 b 5ð2x þ f btÞ m 2x ctðx þ f btÞ e t gðtÞ ¼ : h i mþ5 2 1 þ 2x ctðx þ f btÞ
It is obvious that lim gðtÞ ¼ 0: t!0
So, there exists a s0 > 0 such that p gðs0 Þ < : 2 Then it is easy to see that (19) is true. Now we consider equation x_ ðtÞ ¼
bxm ðt s0 Þ cxðtÞ 1 þ xmþ5 ðt s0 Þ
ð20Þ
with initial value xðtÞ ¼ /ðtÞ;
s0 6 t 6 0; /ð0Þ > 0; / 2 C½½s0 ; 0; Rþ ;
ð21Þ
980
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
where b,c2(0,1) and m P 8. From Lemma 6, we have (19). Then by Saker [1, Theorem 2.5], every solution of the initial problem (20) and (21) satisfies lim xðtÞ ¼ x:
t!1
But from Theorem 3, the solutions of the initial problem (20) and (21) satisfies 0 < xðtÞ 6 < x. That is, Saker [1, Theorem 2.5], is not true.
Acknowledgment This work is supported by the Distinguished Expert Science Foundation of the Naval Aeronautical Engineering Academy.
References [1] S.H. Saker, Oscillation and global attractivity in hematopoiesis model with delay time, Applied Mathematics and Computation 136 (2–3) (2003) 241–250. [2] D.J. Guo, J.X. Sun, Z.L. Liu, Functional Methods of Nonlinear Ordinary Differential Equations, Shandong Science and Technology Press, Jinan, 1995, in Chinese.
www.elsevier.com/locate/amc
Counterexamples to the article ‘‘Oscillation and global attractivity in hematopoiesis model with delay time’’ S.J. Yang, B. Shi *, D.C. Zhang, M.J. Gai Institute of Applied Mathematics, Naval Aeronautical Engineering Academy, Yantai, Shandong 264001, PR China
Abstract In this paper, we give some counterexamples to the article ‘‘oscillation and global attractivity in hematopoiesis model with delay time’’, [Appl. Math. Comput., 136 (2– 3) (2003) 241–250]. Ó 2004 Elsevier Inc. All rights reserved. Keywords: Counterexamples; Attractivity; Differential equations
1. Introduction In the recent paper of Saker [1], the author considered the differential equation _ ¼ pðtÞ
bpm ðt sÞ cpðtÞ 1 þ pn ðt sÞ
*
Corresponding author. E-mail address: [email protected] (B. Shi).
0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2004.09.023
ð1Þ
974
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
with initial value s 6 t 6 0; /ð0Þ > 0; / 2 C½½s; 0; Rþ ;
pðtÞ ¼ /ðtÞ;
ð2Þ
where s, b, c 2 (0, 1), m, n 2 N. Assume that Eq. (1) has a unique positive equilibrium p > 1 and satisfies the equation bpm ¼ cp: 1þ pn
ð3Þ
Then the author obtains the following main results: Theorem 2.1 (Saker [1]). Assume that (2) holds and n > m. Then other positive solutions of Eq. (1) which does not oscillate about p satisfies lim pðtÞ ¼ p:
ð4Þ
t!1
Theorem 2.5 (Saker [1]). Assume that (2) holds, n > m and bðn mÞð p þ p2 Þ
nþm1
bmð p þ p1 Þ
m1
ð1 þ ðp þ p1 Þn Þ2
p ecs s < : 2
ð5Þ
Then every positive solution of Eq. (1) satisfies lim pðtÞ ¼ p;
t!1
where p1 ¼ p csð p þ f bsÞ and p2 ¼ p þ f bs. First we sketch the proof of Theorem (2.1) which is given in Saker [1]. Proof. To prove this theorem we assume that pðtÞ > p for t sufficiently large (the proof when pðtÞ < p is similar and will be omitted). Set pðtÞ ¼ p þ zðtÞ: Then z(t) > 0 for t sufficiently large. The change of variables above reduces Eq. (1) to the equation m
z_ ðtÞ þ czðtÞ þ c p¼
bðp þ zðt sÞÞ n: 1 þ ð p þ zðt sÞÞ
But from (3) we have z_ ðtÞ þ czðtÞ þ
b pm bð p þ zðt sÞÞm ¼ 0: n 1þ p 1 þ ð p þ zðt sÞÞn
As pðtÞ > p, we have z(t) > 0. Then, for t sufficiently large, we have
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
z_ ðtÞ þ
975
m b pm ð1 þ pn Þð p þ zðt sÞÞ 1 60 n pm ð1 þ ð 1þ pn p þ zðt sÞÞ Þ
or m b pm ð1 þ pn Þð p þ zðt sÞÞ 1 m < 0: z_ ðtÞ 6 n 1 þ pn p þ zðt sÞÞ Þ p ð1 þ ð
ð6Þ
Then z(t) is decreasing. So, z(ts) > z(t). Also lim zðtÞ ¼ a; 2 ½0; 1Þ exists:
t!1
We prove that a = 0. Otherwise a > 0. Then there exist e 2 (0, a) and Te > 0 such that for t P Te, ae < z(t) < a + e. As z(t) is decreasing, we have a e < zðt sÞ: Then from (6) we have n
z_ ðtÞ þ
m
b pm ð1 þ ð p þ a eÞ Þ bð1 þ pn Þð p þ a eÞ 6 0; t P T e : n n ð1 þ p Þð1 þ ðp þ a eÞ Þ
ð7Þ
Integrating (7) from Te to 1, we obtain a contradiction. Then a = 0. So z(t) tends to zero. Therefore limt!1 pðtÞ ¼ p. The proof is complete. h In section 2, we shall give some counterexamples to show that Theorem (2.1) and Theorem (2.5) are not true. Consequently, the above proof is invalid.
2. Counterexamples Consider the following differential equation x_ ðtÞ ¼
bxm ðt sÞ cxðtÞ 1 þ xmþ1 ðt sÞ
ð8Þ
with initial value xðtÞ ¼ /ðtÞ;
s 6 t 6 0; /ð0Þ > 0; / 2 C½½s; 0; Rþ ;
ð9Þ
where s, b, c 2 (0, 1) and m P 4. Then the initial value problem of (8) and (9) is an example of the initial value problem of (1) and (2). First, we list some lemmas which will be needed in the following proofs. Lemma 1 (Saker [1]). Assume that n P m and p(t) is the solution of the initial value problem (1) and (2) in (0,1), Then 0 < P ðtÞ < A;
976
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
where 8 > < max /ð0Þ; bc ; A¼ xm > 1 : max /ð0Þ; bc 1þx n ;
if n ¼ m; if n > m;
1
where x1 ¼
1n
m nm
.
Lemma 2 (Guo et al. [2]). Assume that #ðtÞ; xðtÞ 2 C 1 ½J ; R are lower and upper solutions of the initial value problem u_ ¼ f ðt; uÞ;
uð0Þ ¼ u0 ;
respectively, and there exists a positive constant L such that for every t2J,x P y, f ðt; xÞ f ðt; yÞ 6 Lðx yÞ: Then for every t 2 J, we have #ðtÞ 6 xðtÞ; where f ðt; uÞ 2 C½J R; R. Remark 1. By Lemma 2, the solution of inequality x_ ðtÞ þ cxðtÞ 6 bm ;
xð0Þ 6 /ð0Þ;
satisfies
b b xðtÞ 6 m þ /ð0Þ m ect ; c c
ð10Þ
where is a positive constant. Lemma 3. Assume that sffiffiffi!mþ1 rffiffiffiffiffiffiffiffiffiffiffiffi!m1 b ðm þ 1Þ mþ1 m P 4 and ¼ : c 2 m1
ð11Þ
Then Eq. (8) has a unique positive equilibrium x and satisfies x > 1. Proof. Consider equation bxm ¼ cx: 1 þ xmþ1 For x > 0, we have 1 þ xmþ1 ¼
b m1 x : c
ð12Þ
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
From (11), it is easy to see that x ¼ Let
qffiffiffiffiffiffiffiffiffiffiffi
bðm1Þ cðmþ1Þ
977
is a positive solution of Eq. (12).
b f ðxÞ ¼ 1 þ xmþ1 xm1 : c Then f ðxÞ ¼ 0; and ðm 1Þb m2 x f_ ðxÞ ¼ ðm þ 1Þxm c bðm 1Þ m2 2 ¼ ðm þ 1Þx x cðm þ 1Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! bðm 1Þ bðm 1Þ ¼ ðm þ 1Þxm2 x þ x : cðm þ 1Þ cðm þ 1Þ So, if 0 < x < x, we have f_ ðxÞ < 0; if x > x, q weffiffiffiffiffiffiffiffiffiffi haveffi f_ ðxÞ > 0. That is, equation bðm1Þ f(x) = 0 has a unique positive solution x ¼ cðmþ1Þ . Also, Eq. (8) has a unique positive equilibrium x. From (11), it is easy to see that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi bðm 1Þ x ¼ > 1: cðm þ 1Þ The proof is complete.
h
Remark 2. From Lemma 3, we see that if (11) holds, then Eq. (8) satisfies (3). Theorem 1. Assume that (11) and 1 m1 c 0 < /ðtÞ 6 6 < 1; s 6 t 6 0; /ð0Þ ¼ b
ð13Þ
hold. Then the solutions of the initial value problem (8) and (9) satisfies 0 < xðtÞ 6 : Proof. From (13), we have /ð0Þ ¼ P
b m : c
From (8) and Lemma 1, we have if 0 6 t 6 s, x_ ðtÞ 6 bxm ðt sÞ cxðtÞ 6 bm cxðtÞ:
ð14Þ
978
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
That is, x_ ðtÞ þ cxðtÞ 6 bm : By Remark 1 and (14), we have b b xðtÞ 6 m þ /ð0Þ m ect 6 /ð0Þ 6 : c c By induction, we can prove for t 2 ½0; ks; k 2 N 0 < xðtÞ 6 : The proof is complete.
h
Remark 3. From Theorem 1, it is easy to see that Saker [1, Theorem 2.1] is not true. Consider equation x_ ðtÞ ¼
bxm ðt sÞ cxðtÞ 1 þ xmþ3 ðt sÞ
ð15Þ
with initial value (9), where s, b, c 2 (0,1) and m P 6. Lemma 4. Assume that mþ3 m1 b 4 mþ3 mþ3 4 m P 6 and ¼ : c 4 m1 Then Eq. (15) has a unique positive equilibrium x ¼
ð16Þ
bðm1Þ cðmþ3Þ
14
and satisfies x > 1.
Proof. Similar to the that in the proof of Lemma 3, it is easy to be shown and we will omit it. h Theorem 2. Assume that (13) and (16) hold. Then the solutions of the initial value problem (15) and (9) satisfies 0 < xðtÞ 6 : Proof. Similar to the proof of Theorem 1, it is easy to be proved and we will omit it. h Remark 4. From Theorem 2, it is easy to see that Saker [1, Theorem 2.1], is not true either. Consider equation x_ ðtÞ ¼
bxm ðt sÞ cxðtÞ 1 þ xmþ5 ðt sÞ
with initial value (9), where s, b, c 2 (0,1) and m P 8.
ð17Þ
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
979
Lemma 5. Assume that m P 8 and
mþ5 m1 b 6 mþ5 mþ5 6 ¼ : c 6 m1
ð18Þ 1
Þ6 and satisfies x > 1. Then Eq. (17) has a unique positive equilibrium x ¼ ðbðm1Þ cðmþ5Þ Proof. Similar to that in the proof of Lemma 3, it is easy to be shown and we will omit it. h Theorem 3. Assume that (13) and (18) hold. Then the solutions of the initial value problem (17) and (9) satisfies 0 < xðtÞ 6 : Proof. Similar to that in the proof of Theorem 1, it is easy to be shown and we will omit it. h Lemma 6. Assume that (9) holds and m P 8. Then there exists a s0 > 0 such that 5bðx þ x2 Þ
2mþ4
bmðx þ x1 Þ
m1
mþ5 2
p ecs0 s0 < ; 2
ð1 þ ðx þ x1 Þ Þ where x1 ¼ x csðx þ f bsÞ; x2 ¼ x þ f bs.
ð19Þ
Proof. Let
h m1 i ct 2mþ4 b 5ð2x þ f btÞ m 2x ctðx þ f btÞ e t gðtÞ ¼ : h i mþ5 2 1 þ 2x ctðx þ f btÞ
It is obvious that lim gðtÞ ¼ 0: t!0
So, there exists a s0 > 0 such that p gðs0 Þ < : 2 Then it is easy to see that (19) is true. Now we consider equation x_ ðtÞ ¼
bxm ðt s0 Þ cxðtÞ 1 þ xmþ5 ðt s0 Þ
ð20Þ
with initial value xðtÞ ¼ /ðtÞ;
s0 6 t 6 0; /ð0Þ > 0; / 2 C½½s0 ; 0; Rþ ;
ð21Þ
980
S.J. Yang et al. / Appl. Math. Comput. 168 (2005) 973–980
where b,c2(0,1) and m P 8. From Lemma 6, we have (19). Then by Saker [1, Theorem 2.5], every solution of the initial problem (20) and (21) satisfies lim xðtÞ ¼ x:
t!1
But from Theorem 3, the solutions of the initial problem (20) and (21) satisfies 0 < xðtÞ 6 < x. That is, Saker [1, Theorem 2.5], is not true.
Acknowledgment This work is supported by the Distinguished Expert Science Foundation of the Naval Aeronautical Engineering Academy.
References [1] S.H. Saker, Oscillation and global attractivity in hematopoiesis model with delay time, Applied Mathematics and Computation 136 (2–3) (2003) 241–250. [2] D.J. Guo, J.X. Sun, Z.L. Liu, Functional Methods of Nonlinear Ordinary Differential Equations, Shandong Science and Technology Press, Jinan, 1995, in Chinese.