ON THE GLOBAL STABILITY OF A NEUTRAL DIFFERENTIAL EQUATION WITH VARIABLE TIME-LAGS

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1 Bulletin of athematical Analysis and Applications ISSN: , URL: Volume 9 Issue 4(2017), Pages ON THE GLOBAL STABILITY OF A NEUTRAL DIFFERENTIAL EQUATION WITH VARIABLE TIE-LAGS YENER ALTUN, CEIL TUNÇ Abstract. In this work, we get assumptions that guaranteeing the global exponential stability (GES) of the zero solution of a neutral differential equation (NDE) with time-lags. By help of the Liapunov-Krasovskii functional (LKF) approach, we obtain a new result related (GES) of the zero solution of the studied (NDE). An example is given to illustrate the applicability and correctness of the obtained result by ATLAB-Simulink. The obtained result includes and improves the results found in the literature. 1. Introduction In 2014, Keadnarmol and Rojsiraphisal [10] considered the first order neutral differential equation (NDE) with two variable time-lags, d [x+px()] = ax+btanhx(t σ(t)). (1.1) dt Using Lyapunov functionals, the authors established some sufficient conditions for the (GES) of solutions of (NDE) (1.1). By this work, the authors [10] established an improved criterion for the (GES) of solutions of (NDE) (1.1). In this paper, instead of (NDE) (1.1), we consider the first order non-linear (NDE) with two variable time-lags: d [x+p(t)x(] = a(t)h(x) b(t)g(x()) dt +c(t)tanhx(t σ(t)),t 0, (1.2) where represents d dt,a,b,c,p : [t 0, ) [0, ),t 0 0, and g,h : R R are continuous functions with g(0) = 0,h(0) = 0; the function c is continuous and differentiable, and p(t) p 0 < 1 ( p 0 -constant). The variable time-lags τ and σ are continuous and differentiable, defined by τ(t) : [0, ) [0,τ 0 ] and σ(t) : [0, ) [0,σ 0 ] satisfying where τ 0 > 0,σ 0 > 0,δ 1 > 0,δ 2 (> 0) R. 0 τ(t) τ 0, 0 σ(t) σ 0, τ (t) δ 1, σ (t) δ 2 < 1, (1.3) 2000 athematics Subject Classification. 34K20, 34K40, 47N20. Key words and phrases. (NDE); (GES); (LKF); matrix inequality; multiple time-lags. c 2017 Universiteti i Prishtinës, Prishtinë, Kosovë. Submitted September 30, Published October 30, Communicated by Tongxing Li. 31

2 32 Y. ALTUN, C. TUNÇ Throughout the paper, we assume that assumptions given by (1.3) hold when we need x shows x(t). For (NDE) (1.2), we assume the existence initial condition x 0 (θ) = φ(θ),θ [ r,0], where r = max{τ 0,σ 0 }, φ C([ r,0];r). Define h 1 (x) = and g 1 (x) = h(x) x,x 0 dh(0) dt,x = 0 g(x) x,x 0 dg(0) dt,x = 0. It is seen from (NDE) (1.2) and (1.4), (1.5) that (1.4) (1.5) d dt [x+p(t)x()] = a(t)h 1(x)x b(t)g 1 (x())x(t τ(t)) +c(t)tanhx(t σ(t)). (1.6) In this paper, we discuss the (GES) of the zero solution of (NDE) (1.2). eanwhile, itiswellknownthat(ndes)withoutorwithtime-lagsoftenoccurinmanyscientific areas such as engineering techniques fields, physics, medicine and etc. (see [1-29] and the references therein). Therefore, it is worth investigating the (GES) of (NDE) (1.2). In the relevant literature, the most of researchers have focused on the qualitative properties of special case of (NDEs) (1.1) and (1.2) with constant coefficients and constant time-lags like d [x+px(t τ)] = ax+btanhx(t σ) (1.7) dt or its different models. During the investigations, the authors benefited from different methods such as the Liapunovs function (direct) method, Liapunov-Krasovskii functional(lkf) method, integral inequalities, LI, perturbation techniques, model transformations, etc., to obtain specific conditions on the various qualitative properties of(ndes)(see[1-29]). It is also worth mentioning that this paper especially motivated by the results of Keadnarmol and Rojsiraphisal [10] and those can be found in the references of this paper. When we consider (NDEs) (1.1), (1.2) and (1.7) and compare our equation, (NDE) (1.2) with that discussed by Keadnarmol and Rojsiraphisal [10], (NDE) (1.1), it follows that (NDE) (1.2) includes and improves (NDE) (1.1). In fact, if we choose p(t) = p is a constant, a(t)h(x) = ax,a > 0, a R is a positive constant, b(t) = 0, g(.) = 0 and c(t) = 1, then (NDE) (1.2) reduces to (NDE) (1.1) and includes (NDE) (1.7). This fact clearly shows how this work improves the results of [10] and do a contribution to the relevant literature. In addition, giving an example and using ATLAB-Simulink show the other novelty of this paper. These are the originality of this paper.

3 ON THE GLOBAL STABILITY OF A NEUTRAL DIFFERENTIAL Preliminaries For convenience, let D 1 (t) = x + p(t)x(t τ(t)). Hence, (NDE) (1.6) can be written as D 1 = d dt [x+p(t)x()] = a(t)h 1(x)x b(t)g 1 (x())x(t τ(t))+c(t)tanhx(t σ(t)). Therefore, we have { D 1 = a(t)h 1 (x)x b(t)g 1 (x())x(t τ(t))+c(t)tanhx(t σ(t)) 0 = D 1 +x+p(t)x(). (2.1) Definition 1. The solution x = 0 of (NDE)(1.6) is (ES) if x Kexp( λt) sup x(s) = Kexp( λt) x 0 s, (2.2) r s 0 where K(> 0) R, λ(> 0) R, and x t s = sup r s 0 x(t+s). Lemma 2. Let N R n n be any symmetric and positive definite matrix and x,y R n. Then ±2x T y x T Nx+y T N 1 y. Proposition 3. Let > 0,µ > 0, p(t) p 0 < 1, and 0 τ(t) τ 0. If x : [ τ 0, ) R satisfies and x sup x(s) = x 0 s,t [ τ 0,0] s [ τ 0,0] x p 0 x() + exp( µt), then there are positive constants ε,m [0, lnp0 τ 0 ] such that and x x 0 s exp( mt)+ p 0 exp(ετ 0 ) < 1 exp( εt) N exp( ϑt), (2.3) 1 p 0 exp(ετ 0 ) 1 p 0 exp(ετ 0) where t 0,N = x 0 s + and ϑ = min{m,ε}. Proof. In view of the assumptions p(t) p 0 < 1 and 0 τ(t) τ 0 one can find sufficient small positive constant ε,m [0, lnp0 τ 0 ] such that p 0 exp(ετ 0 ) < 1 and p 0 exp(mτ 0 ) < 1. We verify that inequality (2.3) is true. If µ ε, we can choose µ = ε; else if µ > ε, we have exp( µt) exp( εt). Let t = 0. Hence we have x(0) p 0 x( τ(0)) + p 0 < x 0 s + sup τ 0 s 0 1 p 0 exp(ετ 0 ) N. x(s) + Therefore, estimate (2.3) is true when t = 0. Now, let t > 0. Assume that inequality (2.3) fails. Then, there is t > 0 such that x(t ) > x 0 s exp( mt )+ 1 p 0 exp(ετ 0 ) exp( εt ) N exp( ϑt ) (2.4)

4 34 Y. ALTUN, C. TUNÇ and x(t) x 0 s exp( mt)+ 1 p 0 exp(ετ 0 ) exp( εt) N exp( ϑt)forallt [0,t ). I. Let t > τ(t ) > 0. Then, x(t ) p 0 x(t τ(t )) + exp( µt ) p 0 { x 0 s exp( m(t τ(t )))+ + exp( εt ) 1 p 0 exp(ετ 0 ) exp( ε(t τ(t )))} p 0 exp(mτ 0 ) x 0 s exp( mt )+ p 0exp(ετ 0 ) 1 p 0 exp(ετ 0 ) exp( εt ) + exp( εt ) x 0 s exp( mt )+ II. Let τ 0 < 0 < t < τ(t ). Then and hence, it follows that 1 p 0 exp(ετ 0 ) exp( εt ) N exp( ϑt ). x(t τ(t )) x 0 s = sup x(s), s [ τ 0,0] x(t ) p 0 x(t τ(t ) + exp( µt ) x 0 s exp( mt )+ N exp( ϑt ). 1 p 0 exp(ετ 0 ) exp( εt ) Thus, for both the cases I and II, we have a contradiction to inequality (2.4). Therefore, inequality (2.3) is true for all t Exponential stability We assume that there exist nonnegative a i,b i,m i,n i and positive constants c i,(i = 1,2), such that for t t 0, a 1 a(t) a 2, b 1 b(t) b 2, c 1 c(t) c 2, c (t) 0, (3.1) m 1 g 1 (x) m 2, n 1 h 1 (x) n 2. (3.2) Theorem 4. Let a i,b i,m i,n i be nonnegative constants and estimate (1.3) holds. Thentrivialsolutionof(NDE)(1.6)is(GES)iftheoperatorD 1 isstable(i.e. p(t) p 0 < 1) and there exist positive constants c 1,c 2,q 1,α,k and constants q i,(i = 2,3,...,6), such that Ω = 2kq 1 2q 2 (1,2) (1,3) q 1 c 2 q 3 (2,2) (2,3) 0 q 5 +2q 6 (3,3) 0 2q 6 c 1 (1 δ 2 ) 0 2q 6 < 0, (3.3) where (1,2) = q 1 a 1 n 1 + q 2 q 3 q 4, (1,3) = q 1 b 1 m 1 + q 2 p 0 + q 3, (2,2) = 2q 4 2q 5 2q 6 +αe 2kτ0 +c 2 e 2kσ0, (2,3) = q 4 p 0 +q 5 +2q 6, (3,3) = 2q 6 α(1 δ 1 ) and the symbols indicates the elements below the main diagonal of the symmetric

5 ON THE GLOBAL STABILITY OF A NEUTRAL DIFFERENTIAL 35 matrix Ω. Proof. We define a (LKF) V = V 1 +V 2 +V 3 = V 1 (t)+v 2 (t)+v 3 (t) by V 1 (t) = e 2kt [D 1,x,0] q q 2 q 3 0 D 1 x = e 2kt q 1 D1, q 4 q 5 q 6 0 V 2 (t) = α e 2k(s+τ0) x 2 (s)ds+c(t) e 2k(s+σ0) tanh 2 x(s)ds, V 3 (t) = ηe 2kt D 2 1, t σ(t) where D 1 = x + p(t)x(t τ(t)),q 1 > 0,α > 0,c(t) > 0,q i R,(i = 2,...,6), and η(> 0) R, we determine it later. Differentiating V 1 and V 2 along system (2.1), we get 1 (t) = e2kt (2kq 1 D q 1D 1 D 1 ) = 2e 2kt kq 1 D e 2kt [D 1,x,0] q 1 q 2 q 3 0 q 4 q q 6 Benefited from the formula, x x() = x (s)ds, we obtain 1(t) = 2e 2kt kq 1 D e 2kt [D 1,x, x (s)ds+x x()] q 1 q 2 q 3 0 q 4 q q 6 D a(t)h 1 (x)x b(t)g 1 (x())x(t τ(t))+c(t)tanhx(t σ(t)) D 1 +x+p(t)x() x (s)ds x+x() = 2e 2kt kq 1 D e 2kt [D 1 q 1,D 1 q 2 +q 4 x,d 1 q 3 +q 5 x q 6. x (s)ds+q 6 x q 6 x()] a(t)h 1 (x)x b(t)g 1 (x())x(t τ(t))+c(t)tanhx(t σ(t)) D 1 +x+p(t)x() x (s)ds x+x() = 2e 2kt kq 1 D e2kt { D 1 q 1 a(t)h 1 (x)x D 1 q 1 b(t)g 1 (x())x(t τ(t))+d 1 q 1 c(t)tanhx(t σ(t)) D 2 1q 2 +D 1 q 2 x+d 1 q 2 p(t)x() D 1 q 4 x+q 4 x 2 +q 4 p(t)xx() +D 1 q 3 +q 5 x q 6 [ x (s)ds D 1 q 3 x+d 1 q 3 x() x (s)ds q 5 x 2 +q 5 xx() x (s)ds] 2 +q 6 x x (s)ds q 6 x() x (s)ds

6 36 Y. ALTUN, C. TUNÇ +q 6 x x (s)ds q 6 x 2 +q 6 xx() q 6 x() x (s)ds+q 6 xx() q 6 x 2 ()} = e 2kt {(2kq 1 2q 2 )D [ q 1 a(t)h 1 (x)+q 2 q 3 q 4 ]xd 1 +(2q 4 2q 5 2q 6 )x 2 +2D 1 q 1 c(t)tanhx(t σ(t)) +2[ q 1 b(t)g 1 (x()) +q 2 p(t)+q 3 ]x()d 1 +2(q 4 p(t)+q 5 +2q 6 )xx()+2d 1 q 3 2q 6 x 2 () 4q 6 x() +2(q 5 +2q 6 )x x (s)ds 2q 6 [ x (s)ds Using conditions (3.1), (3.2) and p(t) p 0 < 1, we have x (s)ds] 2 }. x (s)ds 1(t) e 2kt {(2kq 1 2q 2 )D ( q 1 a 1 n 1 +q 2 q 3 q 4 )xd 1 +2(q 4 q 5 q 6 )x 2 +2q 1 c 2 D 1 tanhx(t σ(t)) +2( q 1 b 1 m 1 +q 2 p 0 +q 3 )D 1 x() +2(q 4 p 0 +q 5 +2q 6 )xx() +2q 3 D 1 +2(q 5 +2q 6 )x x (s)ds 2q 6 x 2 () 4q 6 x() x (s)ds x (s)ds 2q 6 [ x (s)ds] 2 }, (3.4) 2(t) αe 2k(t+τ0) x 2 αe 2k(t+τ0 τ(t)) (1 τ (t))x 2 (() +c (t) t σ(t) e 2k(s+σ0) tanh 2 x(s)ds+c(t)e 2k(t+σ0) tanh 2 x c(t)e 2k(t+σ0 σ(t)) (1 σ (t))tanh 2 x(t σ(t)). Using conditions (1.3), (3.1) and applying the estimate tanh 2 x x 2, we obtain 2 (t) e2kt {(αe 2kτ0 +c 2 e 2kσ0 )x 2 α(1 δ 1 )x 2 () c 1 (1 δ 2 )tanh 2 x(t σ(t))}. (3.5) Combining equations (3.4) and (3.5), we have 1(t)+ 2(t) e 2kt {(2kq 1 2q 2 )D ( q 1 a 1 n 1 +q 2 q 3 q 4 )xd 1 +(2q 4 2q 5 2q 6 +αe 2kτ0 +c 2 e 2kσ0 )x 2 +2q 1 c 2 D 1 tanhx(t σ(t))

7 ON THE GLOBAL STABILITY OF A NEUTRAL DIFFERENTIAL 37 +2( q 1 b 1 m 1 +q 2 p 0 +q 3 )D 1 x() +2(q 4 p 0 +q 5 +2q 6 )xx() +2q 3 D 1 x (s)ds+2(q 5 +2q 6 )x +[ 2q 6 α(1 δ 1 )]x 2 () 4q 6 x() x (s)ds c 1 (1 δ 2 )tanh 2 x(t σ(t)) 2q 6 [ x (s)ds] 2 } = e 2kt l T (t)ωl(t) x (s)ds where l(t) = [D 1,x,x(t τ(t)),tanhx(t σ(t)), x (s)ds] T and Ω is defined by (3.3). aking use of assumption Ω < 0, we have 1(t)+ 2(t) e 2kt l T (t)ωl(t) < 0. Therefore, there is a constant λ,λ > 0, such that ( V 1(t)+ 2(t) λe 2kt D x 2 + x( 2 ) + tanhx(t σ(t)) 2 + x (s)ds 2 λe 2kt x(t) 2. Calculating the derivate of V 3 along system (2.1), we have 3(t) = 2e 2kt η(d 1 D 1 +kd 2 1) = 2e 2kt η{[x+p(t)x()] [ a(t)h 1 (x)x b(t)g 1 (x())x(t τ(t))+c(t)tanhx(t σ(t))] +k[x+p(t)x()] 2 } = 2e 2kt η{ a(t)h 1 (x)x 2 b(t)g 1 (x())xx(t τ(t)) +c(t)xtanhx(t σ(t)) a(t)h 1 (x)p(t)xx(t τ(t)) b(t)g 1 (x())p(t)x 2 () +c(t)p(t)x()tanhx(t σ(t)) +kx 2 +2kp(t)xx()+kp 2 (t)x 2 ()}. Utilizing conditions (3.1), (3.2) and p(t) p 0 < 1, we have 3(t) e 2kt η{ 2a 1 n 1 x 2 2b 1 m 1 xx() +2c 2 xtanhx(t σ(t)) 2a 1 n 1 p 0 xx() 2b 1 m 1 p 0 x 2 ()+2c 2 p 0 x()tanhx(t σ(t)) +2kx 2 +4kp 0 xx()+2p 2 0 kx2 ()}. By means of Lemma 2, we find 3(t) e 2kt η{(2k 2a 1 n 1 + b 1 m p 0 k 2 + c a 1 n 1 p 0 2 )x 2 +( 2b 1 m 1 p 0 +2p 2 0k + c 2 p )x 2 ()+2tanh 2 x(t σ(t))}.

8 38 Y. ALTUN, C. TUNÇ Let us choose the constant η as { λ 2 η = min{1 ξ, 1 2 } if ψ 0, λ 2 min{1 ξ, 1 1 ψ 2 } if ψ > 0, where ξ = 2b 1 m 1 p 0 +2p 2 0 k+ c 2p and ψ = 2k 2a 1 n 1 + b 1 m p 0 k 2 + c a 1 n 1 p 0 2. Hence, we can obtain 1(t)+ 2(t)+ 3(t) λ 2 e2kt x(t) 2 < 0. Since (t) is negative definite and 0 τ(t) τ 0, 0 σ(t) σ 0 then, V(x) V(x(0)) for all t 0, with V(x(0)) = V 1 (x(0))+v 2 (x(0))+v 3 (x(0)) = q 1 [x(0)+p(0)x( τ(0))] 2 +α 0 τ(0) e 2k(s+τ0) x 2 (s)ds+c(0) +η[x(0)+p(0)x( τ(0))] 2 q 1 (1+p 0 ) 2 x 0 2 s +α 0 +c 2 0 σ(0) τ(0) 0 σ(0) e 2k(s+σ0) tanh 2 x(s)ds e 2k(s+τ0) ( sup x(s) ) 2 ds r s 0 e 2k(s+σ0) ( sup x(s) ) 2 ds+η(1+p 0 ) 2 x 0 2 s r s 0 q 1 (1+p 0 ) 2 x 0 2 s +αe 2kτ0 τ 0 x 0 2 s +c 2 e 2kσ0 σ 0 x 0 2 s +η((1+p 0 ) 2 x 0 2 s = x 0 2 s where = q 1 (1+p 0 ) 2 +αe 2kτ0 τ 0 +c 2 e 2kσ0 σ 0 +η((1+p 0 ) 2. From ηe 2kt D 1 2 V(x) x 0 2 s, we obtain D 1 e kt, where = η x 0 s. Because of D 1 = x+p(t)x(), we have x = D 1 p(t)x() D 1 + p(t)x() e kt +p 0 x(). Since p(t) p 0 < 1 and 0 τ(t) τ 0, we can choose sufficiently small positive constant ϑ = k < lnp0 τ 0 so that p 0 e ϑτ0 < 1. Utilizing Proposition 3, we have Choosing γ = max{ x 0 s, x ( x 0 s + 1 p 0e }, we obtain ϑτ 0 1 p 0 e ϑτ0)e ϑt,t 0. x 2γe ϑt. This implies that the zero solution of (NDE) (1.6) is (ES). By radially unboundedness, it is also (GES) with rate of convergence k = ϑ > 0. Remark 5 If k = 0 one can easily see that the zero solution of (NDE) (1.6) is

9 ON THE GLOBAL STABILITY OF A NEUTRAL DIFFERENTIAL 39 uniformly asymptotically stable when the following criterion holds: Ω = 2q 2 (1,2) (1,3) q 1 c 2 q 3 (2,2) (2,3) 0 q 5 +2q 6 (3,3) 0 2q 6 c 1 (1 δ 2 ) 0 2q 6 < 0, (3.6) where (1,2) = q 1 a 1 n 1 + q 2 q 3 q 4, (1,3) = q 1 b 1 m 1 + q 2 p 0 + q 3, (2,2) = 2q 4 2q 5 2q 6 +α+c 2, (2,3) = q 4 p 0 +q 5 +2q 6 and (3,3) = 2q 6 α(1 δ 1 ). Example 1 As a special case of (NDE) (1.2), we consider the following nonlinear (NDE) with variable time- lags, Here, [ d x+ 1 ] dt 8+t 2x() [ = (2+exp( t)) )[ ( 1 2 +exp( t) x+ x ] 1+x 2 x()+ x() 1+x 2 () + 1 tanhx(t σ(t)),t 0. (3.7) 3 D 1 (t) = x+ 1 8+t 2x(),p(t) = 1 8+t 2 a(t) = 2+exp( t),b(t) = 1 2 +exp( t),c(t) = 1 3 ] τ(t) = σ(t) = sin2 t 20,τ (t) = sin2t 20 h(x) = x+ x 1+x 2,h 1(x) = g(x) = x+ x 1+x 2,g 1(x) = < 1 { x,x 0 2 h (0),x = 0 { x 2,x 0 g (0),x = 0. Then, we have h(0) = 0,n 1 = 1 h 1 (x) 2 = n 2 g(0) = 0,m 1 = 1 g 1 (x) 2 = m 2 a 1 = 2 a(t) = 2+exp( t) 3 = a 2 b 1 = 1 2 b(t) = 1 2 +exp( t) 3 2 = b 2 p(t) = 1 8+t = p 0 < 1 c 1 = c 2 = 1 3,δ 1 = 1 5,δ 2 = 1 10,α = 5 8,k = 1 4 q 1 = q 2 = 3 2,q 3 = q 5 = q 6 = 0,q 4 = 1,

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