Oscillation by Impulses for a Second-Order Delay Differential Equation

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1 PERGAMON Computers and Mathematics with Applications 0 ( Oscillation by Impulses for a Second-Order Delay Differential Equation L. P. Gimenes and M. Federson Departamento de Matemática Instituto de Ciências Matemáticas e de Computação Universidade de São Paulo-Campus de São Carlos CP 668, , São Carlos, SP, Brazil <parron><federson>@icmc.usp.br (Received June 2005; revised and accepted June 2006 Abstract We consider a certain second-order nonlinear delay differential equation and prove that the all solutions oscillate when proper impulse controls are imposed. An example is given. c 2006 Elsevier Science Ltd. All rights reserved. Keywords Delay differential equations, Second-order, Nonlinear, Oscillation, Impulses.. INTRODUCTION In recent years, there has been an increasing interest on the oscillatory behavior of second-order nonlinear delay differential equation. For example, see the recent papers 6. However, there are only a few papers on second-order nonlinear delay differential equations with impulses. See, for instance, 7,8. For the general theory of impulsive ordinary differential equations, the reader is referred to the boo 9 and to some results on the oscillatory behavior of some second-order nonlinear impulsive ordinary differential equations, please see 0 2. Some nonimpulsive delay differential equations are nonoscillatory, but they may become oscillatory if some proper impulse controls are added to them. The purpose of this paper is then to study the oscillatory behavior of solutions of a second-order nonlinear delay differential equations with impulses. In 2, He and Ge study the oscillatory behavior of the following second-order nonlinear impulsive ordinary differential equation: ( r(t(x (t σ f(t, x(t=0, t t0, t, x ( t = I (x(, x ( t = J (x (, =,2..., ( *Supported by CAPES, Brazil. Author to whom all correspondence should be addressed /06/$ - see front matter c 2006 Elsevier Science Ltd. All rights reserved. Typeset by AMS-TEX PII:00

2 2 L. P. Gimenes and M. Federson where 0 < < < < with = and σ is any quotient of positive odd integers. In 7, Peng and Ge prove an oscillation theorem for the second-order delay differential equation with impulses (r(t(x (t σ f(t, x(t,x(t τ=0, t, t, x ( t = I (x(, x (t =J (x (, =,2..., x(t=φ(t, τ t, (2 where τ>0, 0 < < < < with = and >τ. In this paper, we adapt the techniques applied by the authors in 7 and 2 to prove that the equation x (tf(t, x(t,x (t g(t, x(t,x(t τ=0, t, t, x( =I (x(t, x ( =J (x (t, =,2..., (3 x(t=φ(t, τ t, oscillates, where τ>0, 0 < < < < with = and >τ. While in 7 and 2 the authors prove their results provided a solution exists, we assume that f and g are dominated by continuous functions (see (H and (H 2 below in order to guarantee the existence of a global (forward solution of problem (3. The other assumptions are similar to theirs. Our paper is organized as follows. In Section 2, we present a lemma that plays an important role in the proof of the main result. In Section 3, we obtain the oscillatory behavior of (3 through impulse controls. An example is given in Section 4. Consider the impulsive differential equation 2. PRELIMINARIES x (tf(t, x(t,x (t g(t, x(t,x(t τ=0, t, t, x( =I (x(t, x ( =J (x (t, =,2..., (4 satisfying the initial value condition x(t =φ(t, τ t, (5 where φ, φ : τ, R have at most a finite number of discontinuities of firsind and are right continuous at these points. We assume that (H f : τ, R R R is continuous, nonnegative and f(t, u, v z(t, for all u, v R, where z(t is continuous in τ, ; (H 2 g : τ, R R R is continuous, ug(t, u, v > 0, for all uv > 0 and g(t, u, v ϕ(v p(t and g(t, u, v v q(t, for all v 0, where p(t and q(t are continuous in τ,, p(t 0, xϕ(x > 0, for all x 0 and ϕ (x 0; (H 3 I, J : R R are continuous, I (0 = J (0=0, Nand there exist positive numbers a, b, c and such that a I (x x b, c J (x x, x 0, =,2,...;

3 Oscillation by Impulses 3 (H 4 t t < <s c b ds =. Now we define a solution of the impulsive problem (4,(5. Definition 2.. A function x(t : τ, R is a solution of problem (4,(5 if (i x(t and x (t are continuous on, \{ ; N}, there exist lateral its x(t, x (t, x(t, x (t with x(t =x(and x (t =x (, N; (ii x(t fulfills (4,(5; (iii x( and x ( fulfill (4, for each N. By PC(a, b, R n we mean the Banach space of piecewise right continuous functions ψ :a, b R n with the usual supremum norm. If x PC( τ,σ,r n, where R, σ, then for each t,σ we define x t PC( τ,0, R n byx t (s=x(ts for τ s 0. We denote by C(a, b, R n the subspace of PC(a, b, R n of continuous functions with the induced norm. Remar 2.. By using the transformation y(t =x (t, the nonimpulsive equation in (4 can be transformed into the following system: x (t =y(t, y (t= f(t, x(t,y(t g(t, x(t,x(t τ, t. (6 Consider the function F :, R 3 Rgiven by F (t, x 0,x,x 2 =f(t, x 0,x 2 g(t, x 0,x. If ψ PC( τ,0, R 2, ψ =(ψ,ψ 2, we define h(t, ψ =(ψ 2 (0, F (t, ψ (0,ψ ( τ,ψ 2 (0. Then system (6 with the impulsive conditions can be reduced to the system z (t =h(t, z t, t, t, z( =H (z(t, (7 where z(t =(x(t,y(t, z t =(x t,y t and H (z(t =(I (x(t,j (x (t. In this way, under Hypotheses (H to(h 3, in particular the dominance of f and g, imply the global existence of solutions of (7 by 3, Theorem 3.. Therefore, we can guarantee that there is a solution of (4 in,. Now we define an oscillatory solution of the impulsive problem (4,(5. Definition 2.2. A solution of (4,(5 is said to be nonoscillatory if it is eventually positive or eventually negative. Otherwise, it is called oscillatory. Now we present a lemma which is a version of Theorem.4. in 9 replacing the left continuity by the right continuity of m(t and m (t at, N. Lemma 2.. Suppose (i the sequence { } N satisfies 0 < < < < with =, (ii m, m : R R are continuous on R \{ ; N}, there exist the lateral its m(t, m (t, m(t, m (t and m(t =m(,=,2,..., (iii for =,2,... and t,wehave m (t p(tm(tq(t, t, (8 m( m(t b, (9

4 4 L. P. Gimenes and M. Federson where p, q C(R, R, and b are real constants with 0. Then the following inequality holds: ( t t ( t m(t m( exp p(s ds exp p(u du q(s ds < <t s< <t s ( t (0 d j exp p(s ds b, t. < <t <t j<t Remar 2.2. If inequalities (8 and (9 are reversed, then inequality (0 is also reversed. 3. MAIN RESULT In this section, we will show that every solution of (4,(5 is oscillatory under hypotheses (H to (H 4. In the sequel, let x(t be a solution of (4,(5. Lemma 3.. Suppose (H to(h 4 are fulfilled and there exists T such that x(t > 0 for t T τ. Then x ( 0 and x (t 0 for t,, where T. Proof. Suppose x(t > 0, for t T τ. Then x(t τ > 0, t T. At first, we prove that x (t 0, T. If otherwise, there exists some t j T such that x (t j < 0. From (H 3 and (4, we obtain x (t j =J j (x (t j c jx (t j < 0. Let x (t j = α,α>0. By (H and (H 2, given t t j,t j, we have x (t g(t, x(t,x(t τ p(tϕ(x(t τ 0. Then x (t is nondecreasing in t t j,t j. Moreover, x (t j x (t j = α<0, x (t j2 x (t j =J j (x (t j c jx (t j c j( α < 0, x (t j3 x (t j2 =J j2 (x (t j2 c j2x (t j2 c j2c j ( α < 0, and by induction one can prove that n x (t jn i= Hence, x (t is decreasing in t j,. We now consider the impulsive differential inequalities c ji α<0. ( x (t 0, t>t j, t, = j,j2,..., x ( c x (t, = j,j2,... By Lemma 2. with m(t =x (t, we have m(t m(t j c, t j< <t that is, x (t x (t j c. (2 t j< <t

5 Oscillation by Impulses 5 Now considering (2 annowing that x( =I (x(t b x(t, = j,j2,...,by Lemma 2., we conclude that x(t t j< <t b t x(t j x (t j t j t j< <s c ds. (3 b By (H 4 and taing j sufficiently large, we find x(t 0. But this is a contradiction, since x(t > 0, for t T τ. Therefore, x (t 0, T. It follows from (H 3 that x ( c x (t 0 for any T. Because x (t is decreasing in,, then x (t x ( 0, t,, T and the proof is complete. Remar 3.. When x(t is eventually negative, under hypotheses (H to(h 4, one can prove in a similar way that x ( 0 and x (t 0 for t,, where T. Theorem 3.. Suppose (H to(h 4 are fulfilled and there exists a positive integer 0 such that a, for all 0.If t p(u du =, (4 < <u then all solutions of (4,(5 oscillate. Proof. We suppose, without loss of generality, tha 0 =. Let x(t be a nonoscillatory solution of (4,(5. We can assume that x(t > 0, t. By Lemma 3., x (t 0 and x ( 0, t,, where. By (H 3 and the fact that a, =,2,..., we obtain x( <x(t x( x(t 2. It follows that x(t is nondecreasing in,. Now let x (t m(t = ϕ(x(t τ. (5 Then m( 0 and m(t 0, t.by(h and equation (4, we have m (t = f(t, x(t,x (t g(t, x(t,x(t τ ϕ(x(t τ p(t, t, t, τ. It follows from (H 3, equation (4, a and ϕ (x 0 that m( = x (tϕ (x(t τx (t τ ϕ 2 (x(t τ x ( ϕ(x( τ x (t ϕ(x(t τ = m(t (6 and m( τ = x ( τ ϕ(x( x (t τ ϕ(a x(t x (t τ ϕ(x(t = m(t τ. (7 Then using (6 and (7, by Lemma 2., we obtain m(t m(s s< <t t s u< <t p(u du, s t. (8

6 6 L. P. Gimenes and M. Federson Let s and t t. It follows from (6 and (8 that t t m( d m(t d m( p(u du = d m( d p(u du. Similarly, nowing that t 2 >τ and using (7 and the above inequality, we get m(t 2 d 2 m(t 2 d 2 m( τ p(u du By induction, we obtain τ d 2 m(t τ τ d 2 m( p(u du d 2 d m( d 2 d t p(u du p(u du d 2 t m(t n d d 2 d n m( d d 2 d n p(udu d 2 d n = d n d n p(udu d n < <t n t n 2 n m( t < <u p(u du p(u du p(u du.. p(udu Then in view of (4 and m(t n 0, we find a contradiction as n, and the proof is finished. With the next corollaries, we intend to show that inequality (4 is fulfilled. We use Theorem 3. to conclude the results. Corollary 3.. Suppose (H to(h 4 are fulfilled and there exists a positive integer 0 such that a and, for all 0.If < <u p(udu =, then all solutions of (4,(5 oscillate. Proof. Suppose, without loss of generality, tha 0 =. Since /, we have ( t n t p(u du = p(u du = ( t ( t = p(u du < <u d d 2 d n p(udu p(u du p(udu ( t n d p(u du p(u du p(u du p(udu =. t n p(u du d d 2 t 2 d d 2 d n p(udu t 2 p(u du

7 Oscillation by Impulses 7 Then condition (4 is satisfied. Hence, by Theorem 3., all solutions of the impulsive system (4,(5 oscillate. Corollary 3.2. Suppose (H to(h 4 are fulfilled and there exist a positive integer 0 and a constant α>0such that a and / t α, for all 0.If t α p(tdt =, then all solutions of (4,(5 oscillate. Proof. Suppose, without loss of generality, tha 0 = and. Since / t α, for all 0, we obtain < < < < and Thus, t < <u p(u du = d d d 2 = d t α 2, t α 2 t α 3 t α 3,..., d 2 t α 2 t α 3 t α n t α d n,... n ( n t ( t ( t p(u du < <u d d 2 d n p(udu p(u du t α np(udu ( = t n u α p(u du u α p(udu ( p(u du d p(u du t n p(u du d d 2 t 2 d d 2 d n p(udu t α 2 p(u du t α 3 p(u du t 2 t n t α np(u du u α p(u du = u α p(u du t 2 u α p(u du u α p(u du =. Hence, condition (4 is satisfied and Theorem 3. implies all solutions of (4,(5 oscillate. Theorem 3.2. Suppose (H to(h 4 are fulfilled and ϕ(ab ϕ(aϕ(b, for any ab 0.If t then all solutions of (4,(5 oscillate. < <u ϕ(a p(u du =, (9

8 8 L. P. Gimenes and M. Federson Proof. Let x(t be a nonoscillatory solution of (4,(5. We can assume that x(t > 0, t. By Lemma 3., x (t 0 and x ( 0, t,, where. Now let m(t be defined by (5. Then m( 0 and m(t 0, t.by(h and equation (4, we have m (t p(t, t, t, τ. It follows from (H 3, equation (4, ϕ(ab ϕ(aϕ(b and ϕ (x 0 that m( = x ( ϕ(x( τ x (t ϕ(x(t τ = m(t (20 and m( τ = x ( τ ϕ(x( x (t τ ϕ(a x(t x (t τ ϕ(a ϕ(x(t = ϕ(a m(t τ. (2 Then using (20 and (2, by Lemma 2., we obtain m(t m(s s< <t t s u< <t Let s and t t. It follows from (2 and (22 that m( d m(t d p(u du, s t. (22 t t m( p(u du = d m( d p(u du. Similarly, nowing that t 2 >τ and using (2 and the above inequality, we get m(t 2 d 2 m(t 2 d 2 m( τ p(u du τ t2 d 2 ϕ(a m(t τ p(u du τ t2 d 2 ϕ(a m( p(u du Then by induction, we obtain m(t n d 2d ϕ(a m( d 2d ϕ(a t p(u du d 2 d d 2 d t n m( p(udu ϕ(a ϕ(a ϕ(a 2 ϕ(a n d p(u du p(u du. ϕ(a ϕ(a 2 ϕ(a n 2 p(udu ϕ(a ϕ(a 2 ϕ(a n d d 2 d n 2 t n 2 d d 2 d n n d d 2 d n t ϕ(a = m( p(u du. ϕ(a ϕ(a 2 ϕ(a n < <u p(udu But in view of (9 and m(t n 0, we find a contradiction as n, and the proof is finished.

9 Oscillation by Impulses 9 Corollary 3.3. Suppose (H to(h 4 are fulfilled and there exist a positive integer 0 and a constant α>0such that ϕ(a / t α, for all 0.If t α p(tdt =, then all solutions of (4,(5 oscillate. The proof of Corollary 3.3 is omitted, since it can be deduced from Theorem 3.2 and it is similar to that of Corollary AN EXAMPLE Consider the impulsive delay differential equation x (tx(t τ arctan x (t =0, t 0, t, ( x( = x(t, x ( =x (t, =,2,..., x(t=φ(t, τ t 0, (23 where >τ,=,2,... and φ, φ : τ,0 R are continuous. Since ϕ(v =v,p(t=,a =b =(/ and c = =,=,2,..., hypotheses (H to (H 3 are satisfied. Notice that t t < <s c ds = b = t < <s < <s t 2 < <s ds ds ds < <s ds =( 2 (t 2 3 (t 3 t 2 = =. Thus, (H 4 is also satisfied. Le 0 =. Then a and = for all. And since p(u du = 0 du =, it follows from Corollary 3. that all solutions x(t of (23 oscillate. REFERENCES. X. Lin, Oscillation of second-order nonlinear neutral differential equations, J. Math. Anal. Appl. 309, , ( Q. Meng and J. Yan, Bounded oscillation for second order non-linear neutral delay differential equations in critical and non-critical cases, Nonlinear Anal. 64, , ( Y.G. Sun and S.H. Saer, Oscillation for second-order nonlinear neutral delay difference equations, Appl. Math. Comput. 63, , ( B.G. Zhang and S. Zhu, Oscillation of second-order nonlinear delay dynamic equations on time scales, Computers Math. Applic. 49 (4, , ( Q. Yang, L. Yang and S. Zhu, Interval criteria for oscillation of second-order nonlinear neutral differential equations, Computers Math. Applic. 46 (5/6, , (2003.

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