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(1)

Oscillation by Impulses for a Second Order Neutral Delay Differential Equations

R. Sakthivel

PG and Research Department of Mathematics, Pachaiyappa’s College, Chennai, Tamilnadu, INDIA.

email:[email protected].

(Received on: September 28, 2018) ABSTRACT

In this paper, we consider the following second order impulsive neutral delay differential equation

{[𝑟(𝑡) (𝑥(𝑡) + 𝑝(𝑡)𝑥(𝜏(𝑡)))]+ 𝑞(𝑡)𝑥(𝜎(𝑡)) = 0, 𝑡 ≥ 𝑡0, 𝑡 ≠ 𝑡𝑘 ,

𝑥(𝑡𝑘) = 𝐼𝑘(𝑥(𝑡𝑘)), 𝑥(𝑡𝑘) = 𝐽𝑘(𝑥(𝑡𝑘)), 𝑘 = 1,2,3, … (𝐸) where 0 ≤ 𝑡0< 𝑡1< 𝑡2< ⋯ < 𝑡𝑘… with lim

k→∞𝑡𝑘= ∞, some interesting results are obtained, which illustrate the impulses play a very important role in giving rise to the oscillations of equations.

AMS Subject Classification: 34C10, 34A37.

Keywords: Oscillation, Impulsive, Neutral Delay Differential Equations.

1. INTRODUCTION

The oscillatory behavior of delay and neutral delay differential equation has been studied by many mathematicians, see for example

8,13,17,18

, and the references cited therein. In recent years, there has been an increasing interest on the oscillatory behavior of second order delay and neutral delay differential equations with impulse action, see for example

1,2,3,4,5,10,11, 12,15,16

and the references cited therein. In this paper some interesting results established, which shows that all solutions of equation oscillate when proper impulse controls are imposed.

In

4

, L. P. Gimenses and M. Federson studies the oscillatory behavior of the following

delay differential equation with impulses

(2)

{

𝑥

′′

(𝑡) + 𝑓(𝑡, 𝑥(𝑡), 𝑥

(𝑡)) + 𝑔(𝑡, 𝑥(𝑡), 𝑥(𝑡 − 𝜏)) = 0, 𝑡 ≥ 𝑡

0

, 𝑡 ≠ 𝑡

𝑘

𝑥(𝑡

𝑘

) = 𝐼

𝑘

(𝑥(𝑡

𝑘

)), 𝑥

(𝑡

𝑘

) = 𝐽

𝑘

(𝑥

(𝑡

𝑘

)), 𝑘 = 1,2,3, …, 𝑥(𝑡) = 𝜙(𝑡), 𝑡

0

− 𝜏 ≤ 𝑡 ≤ 𝑡

0

,

where 0 ≤ 𝑡

0

< 𝑡

1

< 𝑡

2

< ⋯ < 𝑡

𝑘

… with lim

k→∞

𝑡

𝑘

= ∞.

In

2

, E. M. Bonoto, L.P. Gimenses and M. Federson prove an oscillation theorem for the second order neutral differential equation with impulses

{

[𝑟(𝑡) (𝑥(𝑡) + 𝑝(𝑡)(𝑥(𝑡 − 𝜏)))

]

+ 𝑓(𝑡, 𝑥(𝑡), 𝑥(𝑡 − 𝛿)) = 0, 𝑡 ≥ 𝑡

0

, 𝑡 ≠ 𝑡

𝑘

, 𝑥(𝑡

𝑘

) = 𝐼

𝑘

(𝑥(𝑡

𝑘

)), 𝑥

(𝑡

𝑘

) = 𝐽

𝑘

(𝑥

(𝑡

𝑘

)), 𝑘 = 1,2,3, …, 𝑥(𝑡) = 𝜙(𝑡), 𝑡

0

− 𝜎 ≤ 𝑡 ≤ 𝑡

0

, where 𝛿, 𝜏 are positive constants, 0 ≤ 𝑡

0

< 𝑡

1

< 𝑡

2

< ⋯ < 𝑡

𝑘

… with lim

k→∞

𝑡

𝑘

= ∞ and 𝑡

𝑘+1

− 𝑡

𝑘

> 𝜎, where 𝜎 = max{𝜎, 𝜏}.

In fact, it is known that some particular cases of (E) oscillate without the presence of impulses. Our main results, namely Theorem 1 and Theorem 2, give conditions under which system (E) remains oscillatory. In order to obtain such results, we employ some ideas from

2

and

9

.

Our paper is organized as follows. In Section 2, we present a lemma, that plays an important role in the proof of main results in Section 3. In Section 4, an interesting example is also given.

2. PRELIMINARIES

Consider the second order neutral delay impulsive differential equation of the form { [𝑟(𝑡) (𝑥(𝑡) + 𝑝(𝑡)𝑥(𝜏(𝑡)))

]

+ 𝑞(𝑡)𝑥(𝜎(𝑡)) = 0, 𝑡 ≥ 𝑡

0

, 𝑡 ≠ 𝑡

𝑘

,

𝑥(𝑡

𝑘

) = 𝐼

𝑘

(𝑥(𝑡

𝑘

)), 𝑥

(𝑡

𝑘

) = 𝐽

𝑘

(𝑥

(𝑡

𝑘

)), 𝑘 = 1,2,3, … (1) where 0 ≤ 𝑡

0

< 𝑡

1

< 𝑡

2

< ⋯ < 𝑡

𝑘

… with lim

k→∞

𝑡

𝑘

= ∞.

Throughout this paper, we always assume that:

(i) 𝑟 ∈ 𝐶(𝑅, (0, +∞)), 𝑝 ∈ 𝑃𝐶

([𝑡

0

, ∞), 𝑅

+

), 𝑞 ∈ 𝐶([𝑡

0

, +∞), (0, +∞)), 𝑞(𝑡) > 0;

(ii) 𝜎, 𝜏 ∈ 𝑃𝐶

([𝑡

0

, ∞), 𝑅), 𝜏(𝑡) ≤ 𝑡, 𝜎(𝑡) ≤ 𝑡, 𝜏

(𝑡) = 𝜏

0

> 0, 𝜎

(𝑡) > 0,

t→∞

lim 𝜏(𝑡) = lim

t→∞

𝜎(𝑡) = ∞, 𝜏(𝜎(𝑡)) = 𝜎(𝜏(𝑡)) Where 𝜏

0

is a constant ;

(iii) 𝐼

𝑘

, 𝐽

𝑘

∈ 𝐶(𝑅, 𝑅) and there exist a positive constant 𝑎

𝑘

, 𝑎

𝑘

, 𝑏

𝑘

, 𝑐

𝑘

such that 𝑎

𝑘

𝐼𝑘𝑥(𝑥)

≤ 𝑎

𝑘

, 𝐽

𝑘

(𝑥) = 𝑐

𝑘

(𝑥) for all 𝑥 ≠ 0, 𝑘 = 1,2,3, … ;

(iv) 𝑝(𝑡) and 𝑝

(𝑡) are right continuous on (𝑡

𝑘

, 𝑡

𝑘+1

) with left lateral limits 𝑝(𝑡

𝑘

) =

1

𝑐𝑘

𝑝(𝑡

𝑘

) and 𝑝

(𝑡

𝑘

) =

1

𝑐𝑘

𝑝

(𝑡

𝑘

), for all 𝑘 ∈ 𝑁.

Let 𝐽 ⊂ 𝑅 be an interval, we define 𝑃𝐶(𝐽, 𝑅) = {𝑥 ∶ 𝐽 → 𝑅; x(t) is continuous everywhere

except some 𝑡

𝑘

’s at which 𝑥(𝑡

𝑘

) and 𝑥(𝑡

𝑘+

) exist and 𝑥(𝑡

𝑘

) = 𝑥(𝑡

𝑘

)};

(3)

𝑃𝐶

(𝐽, 𝑅) = {𝑥 ∈ 𝑃𝐶(𝐽, 𝑅): x(t) is continuously differentiable everywhere except some 𝑡

𝑘

at which 𝑥

(𝑡

𝑘

) and 𝑥

(𝑡

𝑘

) exist and 𝑥

(𝑡

𝑘

) = 𝑥

(𝑡

𝑘

)}.

We start by presenting a lemma which is borrowed from [10] replacing the left continuity by the right continuity of 𝑚(𝑡) and 𝑚

(𝑡) at 𝑡

𝑘

for all 𝑘 ∈ 𝑁.

Lemma 1. Suppose

(i) the sequence {𝑡

𝑘

}

𝑘∈𝑁

satisfies 0 ≤ 𝑡

0

< 𝑡

1

< 𝑡

2

< ⋯ < 𝑡

𝑘

… with lim

𝑘→∞

𝑡

𝑘

= ∞;

(ii) 𝑚, 𝑚

∶ 𝑅

+

→ 𝑅 are right continuous on 𝑅

+

\{𝑡

𝑘

∶ 𝑘 ∈ 𝑁}, there exist the lateral limits 𝑚(𝑡

𝑘

), 𝑚

(𝑡

𝑘

), 𝑚(𝑡

𝑘+

), 𝑚

(𝑡

𝑘+

) and 𝑚(𝑡

𝑘+

) = 𝑚(𝑡

𝑘

), 𝑘 = 1,2,3, … ;

(iii) for 𝑘 = 1,2,3, … and 𝑡 ≠ 𝑡

0

, we have

𝑚

(𝑡) ≤ 𝑝(𝑡)𝑚(𝑡) + 𝑞(𝑡), 𝑡 ≠ 𝑡

𝑘

, (2)𝑚(𝑡

𝑘+

) ≤ 𝛼

𝑘

𝑚(𝑡

𝑘

) + 𝛽

𝑘

, (3) where 𝑝, 𝑞 ∈ 𝐶(𝑅

+

, 𝑅), 𝛼

𝑘

and 𝛽

𝑘

are real constants with 𝛼

𝑘

≥ 0. Then the following inequality holds

𝑚(𝑡) ≤ 𝑚(𝑡

0

) ∏ 𝛼

𝑘

exp (∫ 𝑝(𝑠)𝑑𝑠

𝑡𝑡

0

)

𝑡0<𝑡𝑘<𝑡

+ ∫ ∏

𝑡𝑡 𝑠<𝑡𝑘<𝑡

exp (∫ 𝑝(𝑢)𝑑𝑢

𝑠𝑡

)

0

𝑞(𝑠)𝑑𝑠 +

∑ ∏ 𝛼

𝑗

exp (∫ 𝑝(𝑠)𝑑𝑠

𝑡𝑡

𝑘

)

𝑡𝑘<𝑡𝑗<𝑡

𝑡0<𝑡𝑘<𝑡

𝛽

𝑘

, 𝑡 ≥ 𝑡

0

. (4)

For notation convenience, let

𝑦(𝑡) = 𝑥(𝑡) + 𝑝(𝑡)𝑥(𝜏(𝑡)). (5) Lemma 2. Let 𝑥(𝑡) be a solution of equation (1). Suppose that there exists some 𝑇 ≥ 𝑡

0

such that 𝑥(𝑡) > 0 for 𝑡 ≥ 𝑇. If 𝑐

𝑘

≥ 1 and

(𝐻

1

) lim

t→∞

∫ 1

𝑟(𝑠) [ ∏ 𝑐

𝑘

max {𝑎

𝑘

, 𝑐

𝑘

}

𝑡𝑗0<𝑡𝑘≤𝑠

]

𝑡

𝑡𝑗

𝑑𝑠 = +∞,

then 𝑦

(𝑡

𝑘

) ≥ 0 and 𝑦

(𝑡) = [𝑥(𝑡) + 𝑝(𝑡)𝑥(𝜏(𝑡))] ≥ 0 for 𝑡 ∈ [𝑡

𝑘

, 𝑡

𝑘+1

) where 𝑡

𝑘

≥ 𝑇 and 𝑘 ∈ 𝑁, 𝑦(𝑡) is defined by (5).

Proof. Suppose 𝑥(𝑡) > 0 for 𝑡 ≥ 𝑇. At first we shall prove that 𝑦

(𝑡

𝑘

) ≥ 0 for any 𝑡

𝑘

≥ 𝑇 and 𝑘 ∈ 𝑁. If it is not true, then there exist some 𝑗

0

such that 𝑡

𝑗0

≥ 𝑇, 𝑦

(𝑡

𝑗0

) < 0. Let 𝑦

(𝑡

𝑗0

) = −𝛼 where 𝛼 > 0.

From equation (5) and (iii), we have

𝑦

(𝑡

𝑘

) = 𝑥

(𝑡

𝑘

) + 𝑝

(𝑡

𝑘

)𝑥(𝜏(𝑡

𝑘

)) + 𝑝(𝑡

𝑘

)𝑥

(𝜏(𝑡

𝑘

))𝜏

(𝑡

𝑘

)

= 𝐽

𝑘

(𝑥

(𝑡

𝑘

)) + 𝑐

𝑘

𝑝

(𝑡

𝑘

)𝑥(𝜏(𝑡

𝑘

))

+ 𝑐

𝑘

𝑝(𝑡

𝑘

)𝑥

(𝜏(𝑡

𝑘

))𝜏

(𝑡

𝑘

) = 𝑐

𝑘

𝑥

(𝑡

𝑘

) + 𝑐

𝑘

𝑝

(𝑡

𝑘

)𝑥(𝜏(𝑡

𝑘

)) + 𝑐

𝑘

𝑝(𝑡

𝑘

)𝑥

(𝜏(𝑡

𝑘

))𝜏

(𝑡

𝑘

) = 𝑐

𝑘

𝑦

(𝑡

𝑘

),

for all 𝑘 ∈ 𝑁.

Since 𝑥(𝑡) > 0 for 𝑡 ≥ 𝑇, without loss of generality we assume that 𝑥(𝜎(𝑡)) > 0 and 𝑥(𝜏(𝑡)) > 0, for all 𝑡 ≥ 𝑇.

𝑦(𝑡) = 𝑥(𝑡) + 𝑝(𝑡)𝑥(𝜏(𝑡)) > 0 for 𝑡 ≥ 𝑇.

From (1), for 𝑡 ∈ [𝑡

𝑘

, 𝑡

𝑘+1

), 𝑘 ∈ 𝑁, 𝑡

𝑘

≥ 𝑇 we have

(4)

(𝑟(𝑡)𝑦

(𝑡))

= −𝑞(𝑡)𝑥(𝜎(𝑡)) ≤ 0. (6) Hence 𝑟(𝑡)𝑦

(𝑡) is non increasing function on each interval [𝑡

𝑘

, 𝑡

𝑘+1

), 𝑘 ∈ 𝑁 such that 𝑡

𝑘

> 𝑇.

We now consider the impulsive differential inequality

(𝑟(𝑡)𝑦(𝑡))≤ 0, 𝑡 > 𝑡𝑗0, 𝑡 ≠ 𝑡𝑘, 𝑘 = 𝑗0+ 1, 𝑗0+ 2, … 𝑦(𝑡𝑘) = 𝑐𝑘𝑦(𝑡𝑘), 𝑘 = 𝑗0+ 1, 𝑗0+ 2 , …

Let 𝑚(𝑡) = 𝑟(𝑡)𝑦

(𝑡), then

𝑚(𝑡) ≤ 0, 𝑡 > 𝑡𝑗0, 𝑡 ≠ 𝑡𝑘, 𝑘 = 𝑗0+ 1, 𝑗0+ 2, … 𝑚(𝑡𝑘) = 𝑐𝑘𝑚(𝑡𝑘), 𝑘 = 𝑗0+ 1, 𝑗0+ 2 , …

By Lemma 1, we have

𝑚(𝑡) ≤ 𝑚(𝑡

𝑗0

) ∏ 𝑐

𝑘

𝑡𝑗0<𝑡𝑘<𝑡

𝑖. 𝑒., 𝑦

(𝑡) ≤ 𝑟(𝑡

𝑗0

) 𝑦

(𝑡

𝑗0

)

𝑟(𝑡) ∏ 𝑐

𝑘

.

𝑡𝑗0<𝑡𝑘<𝑡

Then, using the fact that

𝑦(𝑡

𝑘

) = 𝑥(𝑡

𝑘

) + 𝑝(𝑡

𝑘

)𝑥(𝜏(𝑡

𝑘

)) = 𝐼

𝑘

(𝑥(𝑡

𝑘

)) + 𝑐

𝑘

𝑝(𝑡

𝑘

)𝑥(𝜏(𝑡

𝑘

))

≤ 𝑎

𝑘

𝑥(𝑡

𝑘

) + 𝑐

𝑘

𝑝(𝑡

𝑘

)𝑥(𝜏(𝑡

𝑘

))

≤ max{𝑎

𝑘

, 𝑐

𝑘

}𝑦(𝑡

𝑘

), 𝑘 = 𝑗

0

+ 1, 𝑗

0

+ 2, … Applying Lemma 1, we have

𝑦(𝑡) ≤ 𝑦(𝑡

𝑗0

) ∏ max{𝑎

𝑘

, 𝑐

𝑘

} + ∫ ∏ max {𝑎

𝑘

, 𝑐

𝑘

} [ 𝑟(𝑡

𝑗0

)𝑦

(𝑡

𝑗0

)

𝑟(𝑠) ∏ 𝑐

𝑘

𝑡𝑗0<𝑡𝑘<𝑠

]

𝑠<𝑡𝑘<𝑡 𝑡 𝑡𝑗0 𝑡𝑗0<𝑡𝑘<𝑡

𝑑𝑠

≤ ∏ max(𝑎

𝑘

, 𝑐

𝑘

) [𝑦(𝑡

𝑗0

) − 𝛼𝑟(𝑡

𝑗0

) ∫ 1

𝑟(𝑠) ∏ 𝑐

𝑘

max{𝑎

𝑘

, 𝑐

𝑘

} 𝑑𝑠

𝑡𝑗0<𝑡𝑘≤𝑠 𝑡

𝑡𝑗0

]

𝑡𝑗0<𝑡𝑘<𝑡

, and since 𝑦(𝑡) > 0 for 𝑡 ≥ 𝑇, the last inequality contradicts (𝐻

1

). Therefore 𝑦

(𝑡

𝑘

) ≥ 0 for all 𝑡

𝑘

, 𝑡

𝑘

≥ 𝑇. Since 𝑟(𝑡)𝑦

(𝑡) is non-increasing on [𝑡

𝑘

, 𝑡

𝑘+1

), it is clear that

𝑦

(𝑡) ≥ 𝑟(𝑡

𝑘+1

)𝑦

(𝑡

𝑘+1

)

𝑟(𝑡) ≥ 0 for 𝑡 ∈ [𝑡

𝑘

, 𝑡

𝑘+1

), 𝑡 ≥ 𝑇.

Thus the proof of Lemma 2 is complete.

3. MAIN RESULTS

In this section, we present some sufficient conditions for the oscillation of all solutions of equations (1).

Theorem 1. Assume that condition (𝐻

1

) hold. If 𝑐

𝑘

≥ 1 and

t→∞

lim ∫ ∏ 𝑐

𝑘

𝑞(𝑠)[1 − 𝑝(𝜎(𝑠))]𝑑𝑠 = ∞. (7)

𝑠<𝑡𝑘<𝑡 𝑡 𝑡𝑗0

then all solutions of equation (1) oscillates.

(5)

Proof. Let 𝑥(𝑡) be a non-oscillatory solution of equation (1). Without loss of generality we may assume that 𝑥(𝑡) > 0, 𝑡 ≥ 𝑇. By Lemma 2, 𝑦(𝑡) ≥ 0, 𝑦

(𝑡) ≥ 0 for 𝑡 ∈ [𝑡

𝑘

, 𝑡

𝑘+1

), where 𝑡

𝑘

≥ 𝑇 and 𝑘 ∈ 𝑁.

By equation (6), we have

(𝑟(𝑡)𝑦

(𝑡))

+ 𝑞(𝑡)𝑥(𝜎(𝑡)) ≤ 0.

Since 𝑦

(𝑡) ≥ 0 for 𝑡 ∈ [𝑡

𝑘

, 𝑡

𝑘+1

]. Hence we have

𝑟(𝑡)𝑦

(𝑡) ≤ 𝑟(𝜎(𝑡))𝑦

(𝜎(𝑡)) 𝑦

(𝜎(𝑡))

𝑦

(𝑡) ≥ 𝑟(𝑡) 𝑟(𝜎(𝑡)) Now

𝑥(𝑡) = 𝑦(𝑡) − 𝑝(𝑡)𝑥(𝜏(𝑡))𝑥(𝜎(𝑡)) = 𝑦(𝜎(𝑡)) − 𝑝(𝜎(𝑡))𝑥 (𝜏(𝜎(𝑡)))

≥ 𝑦(𝜎(𝑡)) − 𝑝(𝜎(𝑡))𝑦 (𝜏(𝜎(𝑡))) ≥ [1 − 𝑝(𝜎(𝑡))]𝑦(𝜎(𝑡)) which together with equation (6), leads to

(𝑟(𝑡)𝑦

(𝑡))

≤ −𝑞(𝑡)𝑥(𝜎(𝑡)) ≤ −𝑞(𝑡)[1 − 𝑝(𝜎(𝑡))]𝑦(𝜎(𝑡)) (𝑟(𝑡)𝑦

(𝑡))

𝑦(𝜎(𝑡))

≤ −𝑞(𝑡)[1 − 𝑝(𝜎(𝑡))]

Define

𝑤(𝑡) = 𝑟(𝑡)𝑦

(𝑡) 𝑦(𝜎(𝑡))

Then 𝑤(𝑡

𝑘+

) ≥ 0 (𝑘 = 1,2,3, … . ), 𝑤(𝑡) ≥ 0 for 𝑡 ≥ 𝑡

0

. If 𝑡 = 𝑡

𝑘

𝑤

(𝑡) = [𝑟(𝑡)𝑦

(𝑡)]

𝑦(𝜎(𝑡)) − 𝑟(𝑡)𝑦

(𝑡)𝑦

(𝜎(𝑡))𝜎

(𝑡) [𝑦(𝜎(𝑡))]

2

≤ −𝑞(𝑡)[1 − 𝑝(𝜎(𝑡))] − 𝑤(𝑡)𝑦

(𝜎(𝑡))𝜎

(𝑡) 𝑦(𝜎(𝑡))

≤ −𝑞(𝑡)[1 − 𝑝(𝜎(𝑡))] − 𝑤

2

(𝑡)𝜎

(𝑡)

𝑟(𝜎(𝑡)) . ≤ −𝑞(𝑡)[1 − 𝑝(𝜎(𝑡))].

At 𝑡 = 𝑡

𝑘

,

𝑤(𝑡

𝑘

) = 𝑟(𝑡

𝑘

)𝑦

(𝑡

𝑘

)

𝑦(𝜎(𝑡

𝑘

)) = 𝑟(𝑡

𝑘

)𝑐

𝑘

𝑦

(𝑡

𝑘

)

𝑦(𝜎(𝑡

𝑘

)) 𝑤(𝑡

𝑘

) = 𝑐

𝑘

𝑤(𝑡

𝑘

).

Therefore

𝑤

(𝑡) ≤ −𝑞(𝑡)[1 − 𝑝(𝜎(𝑡))], 𝑡 ≠ 𝑡

𝑘

, 𝑘 = 𝑗

0

+ 1, 𝑗

0

+ 2, … . . 𝑤(𝑡

𝑘

) = 𝑐

𝑘

𝑤(𝑡

𝑘

), 𝑡 ≠ 𝑡

𝑘

. Therefore by Lemma 1, we have

𝑤(𝑡) ≤ 𝑤(𝑡

𝑗0

) ∏ 𝑐

𝑘

− ∫ ∏ 𝑐

𝑘

𝑞(𝑠)[1 − 𝑝(𝜎(𝑠))]𝑑𝑠.

𝑠<𝑡𝑘<𝑡 𝑡

𝑡𝑗0 𝑡𝑗0<𝑡𝑘<𝑡

Since 𝑤(𝑡) > 0 for 𝑡 > 𝑇, the last inequality contradicts with (7). This completes the proof.

(6)

Theorem 2. Assume that 𝜎(𝑡) ≤ 𝜏(𝑡) and (𝐻

1

) holds. If there exists ℎ ∈ 𝑃𝐶

([𝑡

0

, ∞), 𝑅

+

) such that

t→∞

lim ∫ ∏ 𝑐

𝑘

[𝑄(𝑠)ℎ(𝑠) − 1

4 (1 + 1

𝜏

02

) [ℎ

(𝑠)]

2

𝑟(𝜎(𝑠)) ℎ(𝑠)𝜎

(𝑠) ]

𝑠<𝑡𝑘<𝑡 𝑡

𝑡𝑗0

𝑑𝑠 = ∞, (8) where 𝑄(𝑡) = min{𝑞(𝑡), 𝑞(𝜏(𝑡))} holds, then every solution 𝑥 of (1) is oscillatory.

Proof. Let x be a non oscillatory solution of equation (1). Without loss of generality we can assume 𝑥(𝑡) > 0 for some 𝑇 ≥ 𝑡

0

.

Recall that 𝑦(𝑡) = 𝑥(𝑡) + 𝑝(𝑡)𝑥(𝜏(𝑡)). By Lemma 2, 𝑦(𝑡) > 0, 𝑦

(𝑡) > 0 for 𝑡 ∈ [𝑡

𝑘

, 𝑡

𝑘+1

), where 𝑡

𝑘

≥ 𝑇 and 𝑘 ∈ 𝑁.

Define a Riccati transformation,

𝑤(𝑡) = ℎ(𝑡) 𝑟(𝑡)𝑦

(𝑡) 𝑦(𝜎(𝑡)) . Then 𝑤(𝑡

𝑘+

) ≥ 0 (𝑘 = 1,2,3, … ), 𝑤(𝑡) ≥ 0 for 𝑡 ≥ 𝑡

0

. 𝑤

(𝑡) = ℎ

(𝑡)(𝑟(𝑡)𝑦

(𝑡))

𝑦(𝜎(𝑡)) + ℎ(𝑡)(𝑟(𝑡)𝑦

(𝑡))

𝑦(𝜎(𝑡)) − ℎ(𝑡)𝑟(𝑡)𝑦

(𝑡)𝑦

(𝜎(𝑡))𝜎

(𝑡) [𝑦(𝜎(𝑡))]

2

= ℎ

(𝑡)

ℎ(𝑡) 𝑤(𝑡) + ℎ(𝑡) (𝑟(𝑡)𝑦

(𝑡))

𝑦(𝜎(𝑡)) − 𝑤(𝑡) 𝑦

(𝜎(𝑡))𝜎

(𝑡) [𝑦(𝜎(𝑡))] . Since 𝑦

(𝜎(𝑡))𝑟(𝜎(𝑡)) ≥ 𝑟(𝑡)𝑦

(𝑡), and

𝑤

(𝑡) ≤ ℎ

(𝑡)

ℎ(𝑡) 𝑤(𝑡) + ℎ(𝑡) (𝑟(𝑡)𝑦

(𝑡))

𝑦(𝜎(𝑡)) − 𝑤(𝑡) 𝑟(𝑡)𝑦

(𝑡)𝜎

(𝑡) 𝑟(𝜎(𝑡))𝑦(𝜎(𝑡))

≤ ℎ

(𝑡)

ℎ(𝑡) 𝑤(𝑡) + ℎ(𝑡) (𝑟(𝑡)𝑦

(𝑡))

𝑦(𝜎(𝑡)) − 𝑤

2

(𝑡) 𝜎

(𝑡) ℎ(𝑡)𝑟(𝜎(𝑡)) . Similarly, we introduce another Riccati transformation,

𝑣(𝑡) = ℎ(𝑡) 𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡)) 𝑦(𝜎(𝑡)) . Then 𝑣(𝑡

𝑘+

) ≥ 0 (𝑘 = 1,2,3, … ), 𝑣(𝑡) ≥ 0 for 𝑡 ≥ 𝑡

0

. 𝑣

(𝑡) = ℎ

(𝑡) 𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡))

𝑦(𝜎(𝑡)) + ℎ(𝑡) [𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡))]

𝑦(𝜎(𝑡))

− ℎ(𝑡) 𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡))𝑦

(𝜎(𝑡))𝜎

(𝑡) [𝑦(𝜎(𝑡))]

2

= ℎ

(𝑡)

ℎ(𝑡) 𝑣(𝑡) + ℎ(𝑡) (𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡)))

𝑦(𝜎(𝑡)) − 𝑣(𝑡) 𝑦

(𝜎(𝑡))𝜎

(𝑡)

𝑦(𝜎(𝑡)) .

Since 𝑟(𝑡)𝑦

(𝑡) ≤ 0, 𝑦

(𝑡) > 0, 𝜎(𝑡) ≤ 𝜏(𝑡) and 𝑦

(𝜎(𝑡))𝑟(𝜎(𝑡)) ≥ 𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡)).

(7)

𝑣(𝑡) ≤ℎ(𝑡)

ℎ(𝑡)𝑣(𝑡) +ℎ(𝑡) (𝑟(𝜏(𝑡))𝑦(𝜏(𝑡)))

𝑦(𝜎(𝑡)) − 𝑣(𝑡)𝑟(𝜏(𝑡))𝑦(𝜏(𝑡))𝜎(𝑡) 𝑟(𝜎(𝑡))𝑦(𝜎(𝑡))

≤ℎ(𝑡)

ℎ(𝑡)𝑣(𝑡) +ℎ(𝑡) (𝑟(𝜏(𝑡))𝑦(𝜏(𝑡)))

𝑦(𝜎(𝑡)) − 𝑣2(𝑡) 𝜎(𝑡) ℎ(𝑡)𝑟(𝜎(𝑡)).

Now

𝑤(𝑡) + 1

𝜏02𝑣(𝑡) ≤ℎ(𝑡)

ℎ(𝑡)𝑤(𝑡) +ℎ(𝑡)(𝑟(𝑡)𝑦(𝑡))

𝑦(𝜎(𝑡)) − 𝑤2(𝑡) 𝜎(𝑡)

ℎ(𝑡)𝑟(𝜎(𝑡))+ 1 𝜏02

(𝑡) ℎ(𝑡) 𝑣(𝑡) + 1

𝜏02

ℎ(𝑡) (𝑟(𝜏(𝑡))𝑦(𝜏(𝑡)))

𝑦(𝜎(𝑡)) − 1

𝜏02𝑣2(𝑡) 𝜎(𝑡) ℎ(𝑡)𝑟(𝜎(𝑡)).

Using the definition of 𝑦(𝑡) and applying (1) we get,

(𝑟(𝑡)𝑦

(𝑡))

+ 𝑞(𝑡)𝑥(𝜎(𝑡)) + 𝑞(𝜏(𝑡))𝑥 (𝜎(𝜏(𝑡))) + 1

𝜏

02

(𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡)))

= 0, for 𝑡 ≠ 𝑡

𝑘

, 𝑘 ∈ ℕ and thus

(𝑟(𝑡)𝑦

(𝑡))

+ 𝑄(𝑡)𝑦(𝜎(𝑡)) + 1

𝜏

02

(𝑟(𝜏(𝑡))𝑦

(𝜏(𝑡)))

≤ 0, 𝑡 ≠ 𝑡

𝑘

, 𝑘 ∈ ℕ

𝑤(𝑡) + 1

𝜏02𝑣(𝑡) ≤ℎ(𝑡)

ℎ(𝑡)𝑤(𝑡) − 𝑤2(𝑡) 𝜎(𝑡)

ℎ(𝑡)𝑟(𝜎(𝑡))+ 1 𝜏02

(𝑡)

ℎ(𝑡)𝑣(𝑡) − 1 𝜏02

𝑣2(𝑡)𝜎(𝑡)

ℎ(𝑡)𝑟(𝜎(𝑡))− 𝑄(𝑡)ℎ(𝑡)

≤[ℎ(𝑡)]2𝑟(𝜎(𝑡)) 4ℎ(𝑡)𝜎(𝑡) + 1

4𝜏02

[ℎ(𝑡)]2𝑟(𝜎(𝑡))

ℎ(𝑡)𝜎(𝑡) − 𝑄(𝑡)ℎ(𝑡) [𝑤(𝑡) + 1 𝜏02𝑣(𝑡)]

≤ − [𝑄(𝑡)ℎ(𝑡) −1 4(1 + 1

𝜏02) [ℎ(𝑡)]2𝑟(𝜎(𝑡)) ℎ(𝑡)𝜎(𝑡) ].

Now, 𝑤(𝑡

𝑘

) + 1

𝜏

02

𝑣(𝑡

𝑘

) = ℎ(𝑡

𝑘

)𝑟(𝑡

𝑘

)𝑦

(𝑡

𝑘

) 𝑦(𝜎(𝑡

𝑘

)) + 1

𝜏

02

ℎ(𝑡

𝑘

)𝑟(𝜏(𝑡

𝑘

))𝑦

(𝜏(𝑡

𝑘

)) 𝑦(𝜎(𝑡

𝑘

))

= ℎ(𝑡

𝑘

)𝑟(𝑡

𝑘

)𝑐

𝑘

𝑦

(𝑡

𝑘

) 𝑦 (𝜎(𝑡

𝑘

)) + 1

𝜏

02

ℎ(𝑡

𝑘

)𝑟(𝜏(𝑡

𝑘

))𝑐

𝑘

𝑦

(𝜏(𝑡

𝑘

)) 𝑦 (𝜎(𝑡

𝑘

))

= 𝑐

𝑘

[𝑤(𝑡

𝑘

) + 1

𝜏

02

𝑣(𝑡

𝑘

)] , 𝑘 = 𝑗

0

+ 1, 𝑗

0

+ 2, ….

By Lemma 1, 𝑤(𝑡) + 1

𝜏

02

𝑣(𝑡) ≤ [𝑤(𝑡

𝑗0

) + 1

𝜏

02

𝑣(𝑡

𝑗0

)] ∏ 𝑐

𝑘

𝑡𝑗0<𝑡𝑘<𝑡

− ∫ ∏ 𝑐

𝑘

[𝑄(𝑠)ℎ(𝑠) − 1

4 (1 + 1

𝜏

02

) [ [ℎ

(𝑠)]

2

𝑟(𝜎(𝑠)) ℎ(𝑠)𝜎

(𝑠) ] 𝑑𝑠

𝑠<𝑡𝑘<𝑡 𝑡

𝑡0

which contradicts with (8). This completes the proof.

(8)

Corollary 3. Suppose that (𝐻

1

) holds. If ℎ(𝑡) = 1 in Theorem 2, then

t→∞

lim ∫ ∏ 𝑐

𝑘

𝑄(𝑠)𝑑𝑠 = ∞, (9)

𝑠<𝑡𝑘<𝑡 𝑡

𝑡𝑗0

where 𝑄 is defined as in Theorem 2, then every solution of (1) oscillates.

Corollary 4. Suppose that (𝐻

1

) holds. If ℎ(𝑡) = 𝑡 in Theorem 2 and

t→∞

lim ∫ ∏ 𝑐

𝑘

[𝑠𝑄(𝑠) − 1

4 (1 + 1

𝜏

02

) 𝑟(𝜎(𝑠))

𝑠𝜎

(𝑠) ] 𝑑𝑠 = ∞, (10)

𝑠<𝑡𝑘<𝑡 𝑡

𝑡𝑗0

where 𝑄 is defined as in Theorem 2, then every solution of (1) oscillates.

4 Examples

Example:1 Consider the following second order impulsive type neutral delay differential equation

{

[𝑥(𝑡) + (1 − 1

𝑡 ) 𝑥(𝑡 − 𝜋)]

′′

+ 𝑡𝑥(𝑡) = 0, 𝑡 ≥ 1, 𝑡 ≠ 2

𝑘

, 𝑘 = 1,2,3, … 𝑥(2

𝑘

) = ( 𝑘 + 1

𝑘 ) 𝑥((2

𝑘

)

), 𝑥

(2

𝑘

) = ( 𝑘 + 1

𝑘 ) 𝑥

((2

𝑘

)

), 𝑘 = 1,2,3, … .

(11)

we have 𝑟(𝑡) = 1, 𝑎

𝑘

= 𝑐

𝑘

=

𝑘+1𝑘

, 𝑝(𝑡) = 1 −

1𝑡

, 𝑞(𝑡) = 𝑡, 𝑡

𝑘

= 2

𝑘

, 𝑡

0

= 1, 𝜏(𝑡) = 𝑡 − 𝜋 and 𝜎(𝑡) = 𝑡.

Obviously,

t→∞lim∫ 1

𝑟(𝑠)[ ∏ 𝑐𝑘

max (𝑎𝑘, 𝑐𝑘)

𝑡𝑗0<𝑡𝑘<𝑠

] 𝑑𝑠

𝑡 𝑡𝑗0

= lim

t→∞∫ 1

𝑟(𝑠)[ ∏ 𝑐𝑘

max (𝑐𝑘, 𝑐𝑘)

𝑡𝑗0<𝑡𝑘<𝑠

]

𝑡

𝑡𝑗0 𝑑𝑠

= lim

t→∞∫ 𝑑𝑠

𝑡

𝑡𝑗0 = +∞.

Thus hypotheses (𝐻

1

) and (𝐻

2

) of Lemma 2 satisfied. Now

t→∞lim∫ ∏ 𝑐𝑘𝑞(𝑠)[1 − 𝑝(𝜎(𝑠))]𝑑𝑠

𝑡0<𝑡𝑘<𝑡 𝑡 𝑡0

= lim

t→∞∫ ∏ 𝑘 + 1 𝑘 𝑑𝑠

𝑡0<𝑡𝑘<𝑡 𝑡 1

= lim

t→∞∫ ∏ 𝑘 + 1

𝑡0<𝑡𝑘<𝑡 𝑘

𝑡1

1 𝑑𝑠 + ∫ ∏ 𝑘 + 1

𝑡0<𝑡𝑘<𝑡 𝑘

𝑡2

𝑡1 𝑑𝑠 + ⋯ = ∞.

(9)

Therefore, by Theorem 1, every solution of equation (11) oscillates.

Example: 2 Consider the following second order impulsive type neutral delay differential equation

{

𝑡𝑥(𝑡) + 7𝑥(𝑡 − 1) + 𝑡𝑥 ( 𝑡

2 ) = 0, 𝑡 ≥ 1, 𝑡 ≠ 2

𝑘

, 𝑘 = 1,2,3, … 𝑥(2

𝑘

) = ( 𝑘 + 1

𝑘 ) 𝑥((2

𝑘

)

), 𝑥

(2

𝑘

) = ( 𝑘 + 1

𝑘 ) 𝑥

((2

𝑘

)

), 𝑘 = 1,2,3, … .

(12)

We have,

𝑟(𝑡) = 𝑡, 𝑎𝑘= 𝑐𝑘=𝑘+1𝑘 , 𝑝(𝑡) = 1, 𝑞(𝑡) = 𝑡, 𝑡𝑘= 2𝑘, 𝑡0= 1, 𝜏(𝑡) = 𝑡

and 𝜎(𝑡) =

𝑡

2

. Obviously,

t→∞

lim ∫ 1

𝑟(𝑠) [ ∏ 𝑐

𝑘

max (𝑎

𝑘

, 𝑐

𝑘

)

𝑡𝑗0<𝑡𝑘<𝑠

]

𝑡 𝑡𝑗0

𝑑𝑠 = lim

t→∞

∫ 1

𝑠 [ ∏ 𝑐

𝑘

max (𝑐

𝑘

, 𝑐

𝑘

)

𝑡𝑗0<𝑡𝑘<𝑠

]

𝑡 𝑡𝑗0

𝑑𝑠

= lim

t→∞

∫ 1

𝑠 𝑑𝑠

𝑡

𝑡𝑗0

= +∞.

Thus hypotheses (𝐻

1

) and (𝐻

2

) of Lemma 2 satisfied.

Now,

t→∞

lim ∫ ∏ 𝑐

𝑘

𝑄(𝑠)𝑑𝑠 = ∞.

𝑡𝑗0<𝑡𝑘<𝑠 𝑡

𝑡𝑗0

It follows from Corollary 3 that every solution of (12) is oscillatory.

REFERENCES

1. L. Berenzansky and E. Braverman, On oscillation of a second order impulsive delay differential equation, J. Math. Anal. Appl. 39, 217-225 (2000).

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