Oscillation criteria for second order linear matrix differential systems with damping

Journal of Computational and Applied Mathematics 229 (2009) 222–229

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Oscillation criteria for second order linear matrix differential systems with dampingI Haidong Liu ∗ , Fanwei Meng School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China

article

a b s t r a c t

info

Article history: Received 30 June 2008 Received in revised form 10 October 2008

Using a generalization of Sturm’s comparison theorem, some new oscillation criteria are established for the matrix differential system with damping

Keywords: Oscillation Comparison Matrix differential system Damping

under the hypothesis: P (t ) = P ∗ (t ) > 0, Q (t ) = Q ∗ (t ), Y (t ) are n × n matrices of real

(P (t )Y 0 )0 + R(t )Y 0 + Q (t )Y = 0 2

valued continuous functions on the interval [t0 , ∞), and R(t ) = R∗ (t ) ∈ C 1 ([t0 , ∞), Rn ). Our results are sharper than some previous results. Crown Copyright © 2008 Published by Elsevier B.V. All rights reserved.

1. Introduction Consider the second order linear matrix differential system with damped term

(P (t )Y 0 )0 + R(t )Y 0 + Q (t )Y = 0,

t > t0 ,

(1)

where P (t ) = P (t ) > 0 (i.e., P (t ) is positive definite), Q (t ) = Q (t ), Y (t ) are n × n matrices of real valued continuous ∗

∗

2

functions on the interval [t0 , ∞), and R(t ) = R∗ (t ) ∈ C 1 ([t0 , ∞), Rn ). By M ∗ we mean the transpose of the matrix M. A solution Y (t ) of (1) is said to be nontrivial solution if det Y (t ) 6= 0 for at least one point t ∈ [t0 , ∞). In this paper, we say that a nontrivial solution Y (t ) of (1) is prepared if for t ∈ [t0 , ∞), Y ∗ (t )P (t )Y 0 (t ) − (Y ∗ (t ))0 P (t )Y (t ) ≡ 0,

(2)

Y (t )R(t )Y (t ) − (Y (t )) R(t )Y (t ) ≡ 0,

(3)

∗

0

∗

0

i.e., Y (t )P (t )Y (t ) and Y (t )R(t )Y (t ) are symmetric. A prepared solution Y (t ) of (1) is called to be oscillatory, if det Y (t ) has arbitrarily large zeros on t ∈ [t0 , ∞). System (1) is said to be oscillatory on [t0 , ∞) if every nontrivial prepared solution is oscillatory. When R(t ) ≡ 0, system (1) reduces to the following matrix differential system: ∗

0

∗

(P (t )Y 0 )0 + Q (t )Y = 0,

0

t > t0 ,

(4)

which is the particular case of the matrix Hamiltonian system



U 0 (t ) = A(t )U (t ) + B(t )V (t ), V 0 (t ) = C (t )U (t ) − A∗ (t )V (t ),

I This research was supported by the NNSF of China (10771118).

∗

Corresponding author. E-mail address: [email protected] (H. Liu).

0377-0427/$ – see front matter Crown Copyright © 2008 Published by Elsevier B.V. All rights reserved. doi:10.1016/j.cam.2008.10.024

(5)

H. Liu, F. Meng / Journal of Computational and Applied Mathematics 229 (2009) 222–229

223

with A(t ) ≡ 0, B−1 (t ) = P (t ) and C (t ) = −Q (t ). When R(t ) 6≡ 0, system (1) in general cannot be rewritten to system (4) or system (5). Therefore, all existing oscillation results for systems (4) and (5) generally cannot be applied to system (1). By now, in [2–15,17–19], some criteria for oscillation of systems (4) and (5) have been established. Recently, Sun and Meng [16], Wang [20], Meng and Ma [21] have established some oscillation criteria of Kamenev type for system (1). Etgen and Pawlowski [4] showed that system (4) is oscillatory provided the scalar linear differential equation g [P (t )]y0


0

+ g [Q (t )] y = 0

is oscillatory, where g is a positive linear functional. So the very large number of well-known oscillation criteria for the above scalar linear differential equation can be used to determine associated oscillation criteria for system (4). Motivated by the work of [4], in this paper, we derive some new comparison theorems and use it to obtain some new oscillation theorems for system (1). We now recall for the sake of convenience of reference the following definition from earlier literature. Definition. Let R be the linear space of n × n matrices with real entries, ℘ ⊂ R be the subspace of n × n symmetric matrices, and g be a linear functional on R. The linear functional g is said to be positive if (i) g [M ] > 0 whenever M ∈ ℘ and M > 0, (ii) g [N ] ≥ 0 whenever N ∈ ℘ and N ≥ 0 (i.e., N is positive semi-definite). Equivalently, the linear functional g is a positive functional if g [N ] ≥ 0 for all N ∈ ℘ such that N ≥ 0; g [M ] > 0 for all M ∈ ℘ such that M > 0. From [1] we get each positive functional g on R is bounded (i.e., continuous) with kg k = g (In ) (In denotes the identity matrix in R). 2. Main results Before presenting our main theorem, we state the following lemmas which are needed for proving the theorem. Lemma 1 ([7]). Let g be a positive linear functional on R, then for all A, B ∈ R, g 2 [A∗ B] 6 g [A∗ A]g [B∗ B]. Lemma 2. Given the matrix differential system (1) and

[F (t )X 0 ]0 + G(t )X = 0,

t > t0 ,

(6)

where F (t ) > 0 (i.e., F (t ) is positive definite), G(t ) and X (t ) are n × n matrices of real valued continuous functions on the interval [t0 , ∞). Let Y (t ) be a prepared solution of (1) such that Y (t ) is nonsingular on some interval [a, b] ⊂ [t0 , ∞). For all f (t ) ∈ C 1 [a, b], let

 Z t  ρ(t ) = exp −2 f (s)ds ,



S (t ) = ρ(t ) (PY 0 Y −1 )(t ) + f (t )P (t ) +

  R 2

(t )



on [a, b], If U (t ) is a solution of (6), then b [(U ∗ FU 0 − U ∗ SU )(t )]tt = =a =

b

Z

U ∗ (F − ρ P )U 0 (t )dt + 0




a

Z

b

a



U ∗ (t ) M − G −

ρ RP −1 R



4

(t )U (t )dt

b

Z

(U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t )dt ,

+

(7)

a

where



M (t ) = ρ(t ) Q (t ) + (f 2 P )(t ) − (fP )0 (t ) −

 0 R 2

 (t ) .

Proof. From (2), P (t ) = P ∗ (t ) and R(t ) = R∗ (t ) we have that S (t ) is symmetric for [a, b]. From (3), we get

(RP −1 S )(t ) = (ρ RY 0 Y −1 )(t ) + (ρ fR)(t ) +

ρ RP −1 R 2

= (ρ(Y −1 )∗ (Y 0 )∗ R)(t ) + (ρ fR)(t ) + = (S P

∗ −1

R)(t ) = (SP

−1

R)(t ).

(t )

ρ RP −1 R 2

(t ) (8)

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H. Liu, F. Meng / Journal of Computational and Applied Mathematics 229 (2009) 222–229

From (1) and (8) we have



S 0 (t ) = −2(fS )(t ) + ρ(t ) −RY 0 Y −1 − Q − P (Y 0 Y −1 )2 + (fP )0 +



 = −2(fS )(t ) + ρ(t ) −RP −1 ρ −1 S − fP −    R − ρ −1 S − fP − P −1 ρ −1 S − fP − 2

R

R 2

(t )



2  R 2

 0 

−Q + (fP )0 +

 0  R 2

(t )

1

= −(ρ −1 SP −1 S )(t ) + (ρ RP −1 R)(t ) − M (t ),

(9)

4

where



M (t ) = ρ(t ) Q (t ) + (f 2 P )(t ) − (fP )0 (t ) −

 0 R 2

 (t ) .

Therefore, from (6) and (9) we obtain 0 ∗ 0 ∗0 0 ∗ ∗0 ∗ 0 ∗ 0 (U ∗ FU  − U SU ) (t ) = (U FU − U GU − U SU − ρU S U − U SU )(t )  0 0 = U ∗ FU 0 − U ∗ GU − U ∗ SU + U ∗ ρ −1 SP −1 SU − U ∗ RP −1 RU + U ∗ MU − U ∗ SU 0 (t ) 4 i h ρ ∗ = U ∗ 0 FU 0 − U ∗ GU − U ∗ RP −1 RU + U ∗ MU + (U 0 − ρ −1 P −1 SU )∗ ρ P (U 0 − ρ −1 P −1 SU ) − U 0 ρ PU 0 (t ) 4    ∗0  ρ RP −1 R 0 ∗ (t )U (t ) + (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t ). = U (F − ρ P )U (t ) + U (t ) M − G −

4

So we get

[(U FU − U SU )(t )] ∗

0

∗

t =b t =a

Z

b

U (F − ρ P )U ∗0

=

0




(t )dt +

a

b

Z



U (t ) M − G − ∗

ρ RP −1 R

a

Z +



4

(t )U (t )dt

b

(U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t )dt .

a

The proof of Lemma 2 is complete.



Lemma 3 (Comparison Theorem I). Suppose for any [a, b] ⊂ [t0 , ∞), there exist a functional f (t ) ∈ C 1 [a, b], a positive linear functional g ∈ R and a solution U (t ) of (6) such that :

    ρ RP −1 R (i) g U ∗ (t ) M − G − (t )U (t ) ≥ 0, t ∈ [a, b], 4  ∗0  0 (ii) g U (t )(F − ρ P )(t )U (t ) ≥ 0, t ∈ [a, b],  ∗   ∗  (iii) g U (a)U (a) = g U (b)U (b) = 0, t ∈ [a, b],     (iv) for any c ∈ [a, b], g U ∗ (c )U (c ) = 0 implies that g U ∗ 0 (c )P (c )U 0 (c ) > 0, where ρ(t ), M (t ) are the same as in Lemma 2. If Y (t ) is a prepared solution of (1), then Y (t ) is singular for at least one t ∈ [a, b]. Proof. Suppose to the contrary that Y (t ) is nonsingular on [a, b]. Then S (t ) = ρ(t ) (PY 0 Y −1 )(t ) + f (t )P (t ) + 12 R(t ) exists on [a, b] and (7) holds. Applying the positive linear functional g to both sides of (7), and using the linearity and continuity of g, we obtain





g [U ∗ FU 0 ](b) − g [U ∗ SU ](b) − g [U ∗ FU 0 ](a) + g [U ∗ SU ](a) b

Z

g U ∗ (t )(F − ρ P )(t )U 0 (t ) dt + 0



= a



b

Z





g U ∗ (t ) M − G −

ρ RP −1 R

a

Z + a

4



 (t )U (t ) dt

b

g (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t ) dt .





(10)

H. Liu, F. Meng / Journal of Computational and Applied Mathematics 229 (2009) 222–229

225

From Lemma 1 we have g 2 [(U ∗ FU 0 )(t )] 6 g [U ∗ (t )U (t )] · g [(FU 0 )∗ (t )(FU 0 )(t )] and g 2 [(U ∗ SU )(t )] 6 g [U ∗ (t )U (t )] · g [(SU )∗ (t )(SU )(t )]. By the assumption (iii), g [(U ∗ FU 0 )(t )] = g [(U ∗ SU )(t )] = 0 when t = a or t = b. So (10) reduces to b

Z

0



b

Z

g U ∗ (t )(F − ρ P )(t )U 0 (t ) dt +







g U ∗ (t ) M − G −

ρ RP −1 R 4

a

a



 (t )U (t ) dt

b

Z

g (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t ) dt = 0.



+



(11)

a

On the other hand g (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t )





        = g U ∗ 0 (t )ρ(t )P (t )U 0 (t ) − g U ∗ 0 (t )S (t )U (t ) − g (P −1 SU )∗ (t )P (t )U 0 (t ) + g (ρ −1 P −1 SU )∗ (t )S (t )U (t ) . By evaluating this expression at 1 in the    x = a, and by using Lemma   manner suggested above  and the assumption (iii), we get g U ∗ 0 (a)S (a)U (a) = 0, g (P −1 SU )∗ (a)P (a)U 0 (a) = 0 and g (ρ −1 P −1 SU )∗ (a)S (a)U (a) = 0. So we have g (U 0 − ρ −1 P −1 SU )∗ (a)(ρ P )(a)(U 0 − ρ −1 P −1 SU )(a) = g U ∗ (a)ρ(a)P (a)U 0 (a) .





0





So, by assumption (iv) and using the continuity of g and noting the fact that (U 0 −ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 −ρ −1 P −1 SU )(t ) is nonnegative, we have there is a subinterval [a, a0 ) of [a, b] on which g (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t ) > 0





(The same reason shows that there is also a subinterval (b0 , b] of [a, b] on which g [(U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t )] > 0). So b

Z

g (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t ) dt





a a0

Z

g (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t ) dt > 0.



>



(12)

a

Thus from assumption (i), (ii) and (12) we have b

Z

g U ∗ (t )(F − ρ P )(t )U 0 (t ) dt + 0





b

Z

a





g U ∗ (t ) M − G −

ρ RP −1 R



4

a

 (t )U (t ) dt

b

Z

g (U 0 − ρ −1 P −1 SU )∗ (t )(ρ P )(t )(U 0 − ρ −1 P −1 SU )(t ) dt > 0.



+



a

which contradicts (11). The proof of Lemma 3 is complete.



In the following theorem, we need second order scalar equations of the form

(p(t )y0 )0 + q(t )y = 0,

t ∈ [t0 , ∞),

(13)

where p, q ∈ C ([t0 , ∞), R) and p(t ) > 0. The definitions concerning the oscillation of solutions of (13) are the same as usual. Theorem 1. Suppose Eq. (13) is oscillatory. If there exist a functional f (t ) ∈ C 1 [t0 , ∞) and a positive linear functional g ∈ R such that

    ρ RP −1 R (i) g M (t ) − q(t )In − (t ) > 0, 4

(ii) g [p(t )In − ρ P (t )] > 0, on [t1 , ∞) for some t1 > t0 , where ρ(t ), M (t ) are the same as in Lemma 2, then system (1) is oscillatory.

226

H. Liu, F. Meng / Journal of Computational and Applied Mathematics 229 (2009) 222–229

Proof. Let Y (t ) be a nontrivial prepared solution of (1), and let u(t ) be a nontrivial solution of (13). Since (13) is oscillatory, for each T > t1 , there are points a, b ∈ R, T 6 a 6 b, such that u(a) = u(b) = 0. If we define F (t ) = p(t )In and G(t ) = q(t )In , then U (t ) = u(t )In is a solution of the system

[F (t )X 0 ]0 + G(t )X = 0.

(14)

From the assumptions of this theorem, it is easy to verify that the functional f (t ) and the positive linear functional g and the solution U (t ) of the system (14) satisfy the hypotheses of Lemma 3 on [a, b]. So Y (t ) is singular for at least one t ∈ [a, b] and then system (1) is oscillatory. The proof of Theorem 1 is complete.  Theorem 2. If there exist a functional f (t ) ∈ C 1 [t0 , ∞) and a positive linear functional g ∈ R such that the scalar equation




0 0

+ g M (t ) −

ρ(t )g [P (t )]y



ρ RP −1 R



4

 (t ) y = 0

(15)

is oscillatory, where ρ(t ), M (t ) are the same as in Lemma 2, then system (1) is oscillatory. Proof. Suppose g satisfies the assumptions of the theorem. We can assume kg k = g (In ) = 1. Because if g (In ) 6= 1, then we can choose e g = g /g (In ) such that e g (In ) = 1, and


0 0

ρ(t )e g [P (t )]y

    ρ RP −1 R e + g M (t ) − (t ) y = 0 4

is oscillatory if and only if (15) is. Let



p(t ) = ρ(t )g [P (t )],

q(t ) = g M (t ) −



ρ RP −1 R



4

 (t ) ,

then



g M (t ) − q(t )In −



ρ RP −1 R 4



 (t ) = q(t )[1 − g (In )] = 0,

g [p(t )In − ρ(t )P (t )] = p(t )[g (In ) − 1] = 0. So from Theorem 1 we have that system (1) is oscillatory. The proof of Theorem 2 is complete.



Theorem 3 (Comparison Theorem II). Suppose f (t ) ∈ C 1 [t0 , ∞) and a(t ) = exp −2

Rt





f (s)ds . Then system (1) is oscillatory

if and only if

(P˜ (t )X 0 )0 + R˜ (t )X 0 + Q˜ (t )X = 0,

t > t0 ,

(16)

is oscillatory, where P˜ (t ) = a(t )P (t ),

R˜ (t ) = a(t )R(t ),

Q˜ (t ) = a(t ) Q (t ) + (f 2 P )(t ) − (fP )0 (t ) − (fR)(t ) .





1 Proof. Suppose Y (t ) is a solution of system (1), we define X (t ) = a− 2 (t )Y (t ), t ∈ [t0 , ∞), then 1 1 X 0 (t ) = a− 2 (t )Y 0 (t ) + a− 2 (t )f (t )Y (t ),

1

Y 0 (t ) = a 2 (t )X 0 (t ) − f (t )Y (t ).

(17)

So we get 1

a(t )P (t )X 0 (t ) = a 2 (t ) P (t )Y 0 (t ) + f (t )P (t )Y (t ) .





From (1), (17) and (18), we have

  1 (a(t )P (t )X 0 (t ))0 = −a 2 (t )f (t ) P (t )Y 0 (t ) + f (t )P (t )Y (t )   1 + a 2 (t ) −R(t )Y 0 (t ) − Q (t )Y (t ) + (fP )0 (t )Y (t ) + f (t )P (t )Y 0 (t ) .   1 1 = a 2 (t )Y (t ) (fP )0 (t ) − Q (t ) − (f 2 P )(t ) − a 2 (t )R(t )Y 0 (t )   1 = a 2 (t )Y (t ) (fP )0 (t ) − Q (t ) − (f 2 P )(t ) + (fR)(t ) − a(t )R(t )X 0 (t ).   = a(t )X (t ) (fP )0 (t ) − Q (t ) − (f 2 P )(t ) + (fR)(t ) − a(t )R(t )X 0 (t ).

(18)

H. Liu, F. Meng / Journal of Computational and Applied Mathematics 229 (2009) 222–229

227

So we get

(P˜ (t )X 0 (t ))0 + R˜ (t )X 0 (t ) + Q˜ (t )X (t ) = 0, where P˜ (t ) = a(t )P (t ),

R˜ (t ) = a(t )R(t ),

The proof of Theorem 3 is complete.

Q˜ (t ) = a(t ) Q (t ) + (f 2 P )(t ) − (fP )0 (t ) − (fR)(t ) .







Theorem 4 (Comparison Theorem III). If there exists f (t ) ∈ C 1 [t0 , ∞) such that f (t )P (t ) = −R(t ). Then system (1) is oscillatory if and only if

(P (t )X 0 )0 + Q˜ (t )X = 0,

t > t0 ,

(19)

is oscillatory, where

  (fP )0 (t ) (f 2 P )(t ) ˜ Q (t ) = Q (t ) + − . 2

4

1

 R t

Proof. Suppose Y (t ) is a solution of system (1), we define X (t ) = a 2 (t )Y (t ), where a(t ) = exp −



f (s)ds , t ∈ [t0 , ∞),

then X 0 (t ) = −

1 2

1

1

· a 2 (t )f (t )Y (t ) + a 2 (t )Y 0 (t ).

(20)

So we get 1



1

P (t )X 0 (t ) = a 2 (t ) P (t )Y 0 (t ) −

2

 · f (t )P (t )Y (t ) .

(21)

From (1) and (21), we have 1



1 2

1

 · f (t )P (t )Y (t )

(P (t )X (t )) = − · a (t )f (t ) P (t )Y (t ) − 2 2   1 1 1 + a 2 (t ) −R(t )Y 0 (t ) − Q (t )Y (t ) − · (fP )0 (t )Y (t ) − · f (t )P (t )Y 0 (t ) 2 2  2  0 (fP ) (t ) 1 (f P )(t ) 1 = −a 2 (t )(fP + R)(t )Y 0 (t ) + − Q (t ) − a 2 (t )Y (t ) 4 2  2  (f P )(t ) (fP )0 (t ) = − Q (t ) − X (t ). 0

0

0

4

2

So we have

(P (t )X 0 (t ))0 + Q˜ (t )X (t ) = 0, where

  (fP )0 (t ) (f 2 P )(t ) ˜ Q (t ) = Q (t ) + − . 2

4

The proof of Theorem 4 is complete.



3. Examples In this section, we will work out two examples to illustrate our results. Example 1. Consider the following matrix differential system:

(P (t )Y 0 )0 + R(t )Y 0 + Q (t )Y = 0,

t > 1,

(22)

where P , R, Q are n × n symmetric matrices and P > 0 with

 P =

t

∗

 ∗ , ∗

 R=

β ∗

 ∗ , ∗

Q =

 −1 αt ∗

 ∗ , ∗

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H. Liu, F. Meng / Journal of Computational and Applied Mathematics 229 (2009) 222–229

.

where α 6= 0 and β 6= 0 are constants. Choose f (t ) = 0, g [A] = a11 , where A = (aij ) is a n × n matrix, then

ρ(t ) = 1,



g [P (t )] = t ,

g M (t ) −



ρ RP −1 R 4



 4α − β 2 (t ) = . 4t

So we get the equation

(ty0 )0 +

4α − β 2 4t

y = 0.

(23)

By the main result of Wintner [22] as follows: Theorem A. (p(t )u0 )0 + q(t )u = 0, (p(t ) > 0, and q(t ) ∈ C ([t0 , ∞), R)) is oscillatory if

Z

∞

Z

p−1 (t )dt = ∞,

∞

q(t )dt = ∞.

We have that (23) is oscillatory for α > β 2 /4 > 0. Then by Theorem 2, we have that system (22) is oscillatory for α > β 2 /4 > 0. Example 2. Consider the following matrix differential system: l m Y 00 + Y 0 + 2 Y = 0, t t

t > 1,

(24)

m I , l, m are constants. t2 n . Let f (t ) = 0, g [A] = a11 , where A = (aij ) is a n × n matrix, then     ρ RP −1 R m − l2 /4 + l/2 ρ(t ) = 1, g [P (t )] = 1, g M (t ) − (t ) = . 2

where P (t ) = In , R(t ) = tl In and Q (t ) =

4

t

So we get the equation y00 +

m − l2 /4 + l/2 t2

y = 0.

(25)

By the main result of Kong [23] as follows: Theorem B. Equation y00 + q(t )y = 0 is oscillatory provided that for each r > t0 , there exists α > 1 such that 1 lim sup α−1 t →∞ t

Z

1 lim sup α−1 t t →∞

Z

t

(s − r )α q(s)ds >

α2 4(α − 1)

(26)

(t − s)α q(s)ds >

α2 . 4(α − 1)

(27)

r

and t r

We will prove that (25) is oscillatory when (l − 1)2 < 4m. In fact, for any constant α > 1 and for each r > 1, the left-hand side of (26) takes the form 1 lim sup α−1 t →∞ t

t

Z

(s − r )α

m − l2 /4 + l/2 s2

r

Since (l − 1)2 < 4m, that is, m − l2 /4 + l/2 > and hence m − l2 /4 + l/2

>

α−1

ds =

1 , 4

m − l2 /4 + l/2

α−1

lim sup t →∞

(t − r )α tα

=

m − l2 /4 + l/2

α−1

.

2 we can choose an appropriate constant α > 1 such that m − l2 /4 + l/2 > α4 ,

α2 . 4(α − 1)

Thus, (26) holds. On the other hand 1 lim sup α−1 t →∞ t

t

Z r

(t − s)α

m − l2 /4 + l/2 s2

1 ds > lim sup α−1 t →∞ t

Z r

t

(s − r )α

m − l2 /4 + l/2 s2

ds.

Hence (27) holds. By Theorem B, we have that (25) is oscillatory when (l − 1)2 < 4m. So by Theorem 2 we have that system (24) is oscillatory for (l − 1)2 < 4m.

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