Applied Mathematics and Computation 104 (1999) 207±215 www.elsevier.nl/locate/amc
A correction on the oscillatory behavior of solutions of certain second order nonlinear dierential equations Mingshu Peng
a,* ,
Weigao Ge a, Lihong Huang b, Qianli Xu c
a
c
Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, People's Republic of China b Department of Applied Mathematics, Hunan University Changsha, Hunan 410082, People's Republic of China Department of Mathematics,Yiyang Teachers College, Hunan 413049, People's Republic of China Received 23 September 1997
Abstract In this paper, we correct a result stated in a recent paper related to the oscillatory behavior of solutions of certain second order nonlinear dierential equations. Ó 1999 Elsevier Science Inc. All rights reserved. Keywords: Dierential equation; Oscillation; Nonlinearity
1. Introduction In a recent paper [1] the authors provided some sucient conditions for all solutions of the dierential equation r 0
a
t
y 0
t q
tf
y
t r
t
1
and the special case that r
t 0, i.e. r 0
a
t
y 0
t q
tf
y
t 0;
*
Corresponding author.
0096-3003/99/$ ± see front matter Ó 1999 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 9 8 ) 1 0 0 7 3 - 5
1a
208
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
where r is a positive quotient of odd integers (odd/odd) or even over odd integers (even/odd), a
t is an eventually positive function and f satis®es
i uf
u > 0 for u 6 0;
2
ii f 0
u P 0:
3
and In Theorem 3.5(b) of [1] an incorrect result about the oscillatory behavior of all solutions of Eq. (1a) is stated. The aim of this paper is to show this fact with two examples, as well as to prove the correct result. 2. On the oscillatory behavior of solutions of Eq. (1a) For notational simplicity, let r
w
t a
t
y 0
t ; the starting point of this paper is the following result proved in [1] as Theorem 3.5(b). Theorem 1 [1]. Suppose that Z 1 q
s ds 1
4
t0
holds. If r (even/odd), then all solutions of Eq. (1a) are oscillatory. The following examples show that Theorem 3.5(b) of [1] is not true. Example 1. Let us consider the dierential equation
t3
y 0
t2 0 y
t=t 0;
t > 0:
5
Obviously, r 2 and Eq. (4) holds. All conditions of Theorem 3.5(b) of [1] are satis®ed. Therefore Eq. (5) should oscillate. But this equation actually has a nonoscillatory solution given by y
t 1=t. Example 2. Consider the dierential equation
t
y 0
tr y
t=t 0;
t > 0;
6
where 0 < r p=q with p even and q odd integers. It is clear that Eq. (4) holds. Hence every condition of Theorem 3.5(b) is satis®ed. So Eq. (6) should be oscillatory. But in fact, this equation has a nonoscillatory solution given by y
t ÿt. The reason why Theorem 3.5(b) of [1] is incorrect is that in the proof of [1] we could not get a contradiction in case 2 and no contradiction could be
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
209
obtained in case 1 or 2 when y
t is eventually negative. The following results correct Theorem 3.5(b) of [1]. Theorem 2. Let r
even=odd and q(t) be an eventually positive function. If Eq. (4) holds and there exists T P t0 such that Z s 1=r Z 1 1 q
s ds ds 1;
7 1=r a
s T T then every bounded solution y
t of Eq. (1) is either oscillatory or monotonically converges to zero as t ! 1. Proof. Let y
t be a nonoscillatory bounded solution of Eq. (1a), say, y
t < 0 for t P t0 . We consider only this case because the proof for the case y
t > 0 for t P t0 is similar and is omitted. From Eq. (1a) we have the identity r 0 r1 a
t
y 0
t a
tf 0
y
ty 0
t ÿq
t ÿ :
8 f
y
t f
y
t2 Case 1. Suppose that y 0
t P 0 for t P t1 P t0 : Then from Eq. (8) we obtain 0 a
t
y 0
tr 6 ÿ q
t; t P t1
9 f
y
t which on integrating from t1 to t provides Z t a
ty 0
tr a
t1
y 0
t1 r 6 ÿ q
s ds: f
y
t f
y
t1 t1
10
Since y 0
t > 0 and y(t) is eventually negative, y
t ! M < 0 as t ! 1 and y
t 6 M for t P t1 . Consequently it follows from Eq. (3) that f
y
t 6 f
M < 0 for t P t1 : Hence from Eq. (9) we obtain Z t Z t r 0 a
ty
t P ÿ f
y
t q
s ds P ÿ f
M q
s ds: t1
Therefore we have 0
y
t P
ÿf
M1=r
Rt
t1
q
s ds1=r
t1 1=r
; a
t which on integrating from t1 to t provides Z s 1=r Z t 1 1=r q
s ds ds: y
t P y
t1
ÿf
M 1=r t1 a
s t1
11
By Eq. (7) the right side of Eq. (11) tends to 1 as t ! 1 whereas the left side of Eq. (11) is negative, which is a contradiction.
210
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
Case 2. Suppose that y 0
t < 0 for t P t1 P t0 : Now integrating Eq. (1a) from t1 to t, we get Z t r r a
t
y 0
t a
t1
y 0
t1 ÿ q
sf
y
s ds t1
a
t1
y 0
t1 r ÿ f
y
t1 Z ÿ
t t1
Z
t t1
q
s ds
q
sf
y
s ÿ f
y
t1 ds:
12
Since y 0
t and y
t are both eventually negative, in view of Eq. (3), we have f
y
t 6 f
y
t1 < 0 for t P t1 . Hence from Eq. (12) we ®nd Z t r 0 a
t
y
t1 P ÿ f
y
t1 q
s ds: t1
0
Since y
t < 0 for t > t1 , we get R 1=r t q
s ds t1 y 0
t 6 ÿ
ÿf
y
t1 ; a
t1=r
13
which on integrating from t1 to t provides Z s 1=r Z t 1 1=r q
s ds ds: y
t 6 y
t1 ÿ
ÿf
y
t1 1=r t1 a
s t1
14
By Eq. (7) the right side of Eq. (14) is in®nite. However the left side is bounded. Case 3. Suppose that y 0
t oscillates. Then there exist a sequence fNk g such that y 0
Nk 0. Choose k suciently large such that Nk P t0 . Without loss of generality, we assume that y 0
t > 0 for t 2
Nk ; Nk1 . Further in view of Eq. (4) we have Z Nk1 q
s ds > 0:
15 Nk
Now from Eq. (8) we get Eq. (9) for t
Nk ; Nk1 . Integrating Eq. (9) from Nk to Nk1 provides Z Nk1 r a
sy 0
s Nk1 q
s ds P 0; ÿ f
y
s Nk Nk which contradicts Eq. (15). The proof is now complete. Example 3. Consider the dierential equation 2
0
5
y 0
t =t 5y
t =t 0;
t > 0:
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
211
It is easily veri®ed that Eqs. (4) and (7) hold. So every bounded solution of this equation is either oscillatory or monotonically converges to zero as t converges to in®nity. One such solution is y
t 1=t: Example 4. Let us consider the equation 2 0
t3
y 0
t 3ty
t 0;
t > 0:
One can verify that Eqs. (4) and (7) hold. But one can ®nd that this equation has an unbouded nonoscillatory solution by y
t ÿt. Theorem 3. Let r
even=odd and q(t) be an eventually positive function. If Eq. (7) holds and the following conditions are satisfactory: Z 1 q
s ds < 1;
16
i 0 < Z
ii lim
t!1
t
Z
1 1=r
a
s
t0
1 s
1=r q
s ds
ds 1;
17
then every bounded solution y
t of Eq. (1a) is either oscillatory or limt!1 y
t 0. Proof. The condition (4) is used only in the case that y
t is eventually positive in the proof of Theorem 2. Suppose y
t is a bounded nonoscillatory solution of Eq. (1a). We shall only consider the case y
t > 0 for t P t0 : Case 1. Suppose y 0
t P 0 for t P t1 P t0 : From Eq. (8) we can get Eq. (10). Hence we ®nd Z t r a
t1 y 0
t1 ; q
s ds 6 f
y
t1 t1 and therefore for t P t1 Z 1 r a
t
y 0
t ; q
s ds 6 f
y
t t Z
1 a
t1=r
1 t
1=r q
s ds
6
y 0
t f
y
t1=r
:
18
Integrating Eq. (18) from t1 to t we get Z
t t1
Z
1 1=r
a
s
1 s
1=r q
s ds
Z ds 6
t t1
y 0
s f
y
s
1=r
ds:
19
212
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
In view of Eqs. (1) and (3) we have f
y
t P f
y
t1 > 0 for t P t1 : Hence it follows from Eq. (19) that Z
t t1
Z
1 a
s
1=r
1=r
1
q
s ds
s
Z ds 6
t
y 0
s ds 6 1=r f
y
s
t1
y
t ÿ y
t1 f
y
t1
1=r
:
Z
t t1
y 0
s f 1=r
y
t
1
ds
20
By Eq. (17) the left side of Eq. (20) tends to in®nite as t ! 1. However the right side of Eq. (20) is bounded. R1 q
t dt > 0, there exists Case 2. Suppose y 0
t < 0 for t P t1 P t0 : Since t2 P t1 such that Z t q
s ds P 0:
21 t2
Now integrating Eq. (1a) from t2 to t and then using integration by parts we get Z t r r q
sf
y
s ds a
t
y 0
t a
t2
y 0
t2 ÿ t2
r
a
t2
y 0
t2 ÿ f
y
t Z
t2
t
0
0
y
sf
y
s
Z t2
s
Z t2
t
q
s ds
q
s ds ds:
22
Since y 0
t < 0 for t P t1 and f 0
y
t P 0, it follows from Eqs. (21) and (22) that Z t a
t
y 0
tr 6 a
t2
y 0
tr ÿ f
y
t q
s ds:
23 t2
0
Since y
t is eventually positive and y
t is eventually negative, y
t ! M > 0 as t ! 1 and y
t P M for t P t2 . Consequently it follows from Eq. (3) that f
y
t P f
M > 0 for t P t2 . Hence from Eq. (23) we have Z t r r 0 0 q
s ds:
24 a
t
y
t 6 a
t2
y
t2 ÿ f
M Therefore we obtain Z t r f
M q
s ds 6 a
t2
y 0
t2 : t2
Hence for t P t2 Z 1 f
M q
s ds 6 a
t
y 0
tr : t
t2
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
So we have 0
y
t 6 ÿ f
M
1=r
1 a
t
Z t
1
213
1=r q
s ds
:
Integrating Eq. (25) from t2 to t, we get Z 1 1=r Z t 1 1=r q
s ds ds: y
t 6 y
t2 ÿ f
M 1=r t2 a
s s
25
26
By Eq. (17) the right side of Eq. (26) tends to ÿ1 at t ! 1 however the left side of Eq. (26) is positive. R1 q
s ds > 0 we have Eq. (15) Case 3. Suppose y 0
t is oscillatory. Since and the contradiction follows as in Theorem 2 (case 3). The proof is complete. Example 5. Consider the dierential equation 0 1 0 6 2
y
t y 5
t 0: t2 t2 It is easily veri®ed that Eqs. (7), (16) and (17) hold. Therefore according to Theorem 3, every solution y
t of this equation is either oscillatory or monotonically converges to zero as t ! 1. In fact this equation has such a solution given by y
t 1=t. If the properties of f are replaced by:
i f 0
u P 0
for all u 2
0; 1;
27
ii f 0
u 6 0
for u 2
ÿ1; 0;
28
iii uf
u 6 0
u 6 0
and
f
0 P 0:
29
Then the following result can be obtained. Theorem 4. Let r
even=odd. If Eqs. (27), (28) and (29) hold. Then every solution y
t of Eq. (1a) is either oscillatory or monotonically converges to zero as t ! 1. Proof. Suppose on the contrary that y
t is a nonoscillatory solution of Eq. (1a) and y
t do not monotonically converge to zero as t ! 1. Without loss of generality, say, y
t < 0 for t P t0 . We shall consider only this case because the proof for the case y
t > 0 for t P t0 is similar. From Eqs. (27) and (28) we can get f
u > 0
for all u 6 0:
Case 1. Suppose that y 0
t 6 0 for t P t1 P t0 : Then from Eq. (8) we have Eq. (10). By Eq. (4) the right side of Eq. (10) tends to ÿ1 as t ! 1. However the left side of Eq. (10) is nonnegative.
214
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
Case 2. Suppose that y 0
t > 0 for t P t1 P 0 . Condition (4) implies (21). Now integrating Eq. (1a) from t2 to t and then using integration by parts we get Eq. (22). Since y 0
t > 0 for t P t1 and f 0
y
t 6 0; it follows from Eqs. (21) and (22) that Eq. (23) holds. Since y(t) is eventually negative and y 0
t > 0 for t P t1 ; y
t ! M > 0 as t ! 1 and y
t P M for t > t1 . Consequently it follows from Eq. (28) that f
y
t P f
M > 0 for t > t1 . Hence from Eq. (23) we have Eq. (24). By Eq. (4) the right side of Eq. (24) tends to ÿ1 as t ! 1 whereas the left side of Eq. (24) is positive. R1 q
s ds > 0 (by Eq. (4)) we Case 3. Suppose that y 0
t oscillates. Since have Eq. (15) and the contradiction follows as in Theorem 2 (case 3). Example 6. Consider the dierential equation 2 0
t2
y 0 2y 2 =t 0;
t > 0:
31
Obviously, r 2 and Eq. (4) holds. It is clear that all conditions of Theorem 4 are satis®ed. So every solution y
t of Eq. (31) is either oscillatory or limt!1 jy
tj 0. In fact, this equation has two solutions given by y
t 1=t: Example 7. Consider the dierential equation
y 0 r 0 ÿ re
rÿ2t y 2 0;
t > 0;
32
where r is any positive quotient of even over odd intergers. However condition Eq. (4) does not hold because Z 1 Z 1 q
s ds
ÿre
rÿ2t ds < 1: t0
t0
Hence the conditions of Theorem 4 are violated. Indeed Eq. (32) has a nonoscillatory solution given by y
t et . Example 8. Consider the dierential equation r
0
y 0 =t y 4
t=
1 sin 2 t 0;
t > 0;
33
where r (even/odd). Since Z t0
1
Z q
s ds
1 t0
ds > 1 sin2 s
Z
1 t0
ds 1 2
Eq. (4) holds. It follows from Theorem 3 that every solution y
t of Eq. (33) is either oscillatory or satis®es limt!1 jy
tj 0.
M. Peng et al. / Appl. Math. Comput. 104 (1999) 207±215
215
Acknowledgements This work was supported by the National Natural Science Fund of China and the Fund of Doctoral Program Research of the Education Ministry of China. References [1] P.J.Y. Wong, R.P. Agarwal, Oscillatory behavior of solutions of certain second order nonlinear dierential equations, J. Math. Anal. Appl. 198 (1996) 337±354.