J. Mah. Anal. Appl. 321 (2006) 553 568 www.elsevier.com/locae/jmaa Necessary sufficien condiions for oscillaion of firs order nonlinear neural differenial equaions X.H. ang a,, Xiaoyan Lin b a School of Mahemaical Sciences Compuing echnology, Cenral Souh Universiy, Changsha, Hunan 410083, PR China b Deparmen of Mahemaics, Huaihua Universiy, Huaihua, Hunan 418008, PR China Received 17 March 2005 Available online 19 Sepember 2005 Submied by William F. Ames Absrac In his paper, we prove ha every soluion of he firs order nonlinear neural differenial equaion ] m x() px( τ) + q() x( σj ) βj sign x( σ 1 ) ] = 0,, oscillaes if only if ( m ) ] q(s)exp τ 1 ln p β j 1 s ds =, when ( m β j 1) ln p<0, q(s)ds =, when ( m β j 1) ln p>0, where p, τ>0, β j > 0, σ j 0, j = 1, 2,...,m, q C(, ), 0, )). 2005 Published by Elsevier Inc. his work is parially suppored by he NNSF (No. 10471153) of China. * Corresponding auhor. E-mail addresses: angxh@mail.csu.edu.cn (X.H. ang), xiaoyanlin98@homail.com (X. Lin). 0022-247X/$ see fron maer 2005 Published by Elsevier Inc. doi:10.1016/j.jmaa.2005.07.078
554 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 Keywords: Firs order neural differenial equaion; Superlinear; Sublinear; Oscillaion 1. Inroducion Consider he firs nonlinear neural delay differenial equaion ] m x() px( τ) + q() x( σj ) βj sign x( σ 1 ) ] = 0,, (1.1) where p,τ > 0, β j > 0, σ j 0, j = 1, 2,...,m, q C(, )), 0, )). When m β j = 1, he oscillaory behavior of soluions of Eq. (1.1) is similar o he linear neural delay differenial equaion, which has been sudied by many auhors, see 1,2,5,7,8] he references cied herein. When m β j 1, we only find hree papers 3,4,6] which deal wih he oscillaory behavior of soluions of some special forms of Eq. (1.1). he resuls obained here are he following. heorem 1.1. 3] Assume ha p = 0 m β j > 1. hen he following conclusions hold: (i) If here exiss λ>0 such ha m β j e λσ j < 1, (1.2) lim inf q()exp( e λ ) ] > 0, (1.3) hen every soluion of Eq. (1.1) oscillaes. (ii) If, for large, q(s) 0, s, + σ ], (1.4) where σ = max{σ 1,σ 2,...,σ m }, ha exiss μ>0 such ha m β j e μσ j > 1, (1.5) lim sup q()exp( e μ ) ] <, (1.6) hen Eq. (1.1) has an evenually posiive soluion. heorem 1.2. 4] Assume ha p = 1 m β j < 1. hen every soluion of Eq. (1.1) oscillaes if only if s β q(s)ds =, where β = m β j. (1.7)
X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 555 heorem 1.3. 4] Assume ha p = 1 m β j > 1. hen every soluion of Eq. (1.1) oscillaes if only if sq(s)ds =. (1.8) heorem 1.4. (6], see also 1, heorem 3.4.6]) Assume ha p<1 m β j < 1. hen every soluion of Eq. (1.1) oscillaes if only if q(s)ds =. Besides he several cases menioned in he above four heorems, we find no resuls in he lieraure on he oscillaion of soluions of Eq. (1.1) in he following wo cases: (i) p (0, 1) (1, ) m β j > 1; (ii) p (1, ) m β j < 1. In his paper, we shall esablish some necessary sufficien condiions for oscillaion of soluions of Eq. (1.1) in he above case (i) case (ii), respecively. As is cusomary, a soluion x() of Eq. (1.1) is said o be oscillaory if i has arbirarily large zeros. Oherwise, i is said o be nonoscillaory. hroughou of his paper, we denoe β = m β j. 2. he superlinear case m β j > 1 heorem 2.1. Assume ha <p<1 m β j > 1. hen every soluion of Eq. (1.1) oscillaes if only if ( m ) ] q(s)exp τ 1 ln p β j 1 s ds =. (2.1) Proof. Sufficiency. Lex() be a nonoscillaory soluion of Eq. (1.1). We may wihou loss of generaliy assume ha x() > 0for 1 for some 1. Se z() = x() px( τ). (2.2) hen i follows from (1.1), (2.1) (2.2) ha m z () = q() x( σj ) βj 0 ( 0), 2 = 1 + ρ, (2.3) here in he sequel, ρ = max{τ,σ 1,...,σ m }. his shows ha z() is nonincreasing on 2, ). Hence, z() > 0, 2, (see 1, Lemma 5.1.4]), from (2.2) (2.3), we have n x() = p i z( iτ) + p n+1 x( nτ τ) i=0 (1.9)
556 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 = 1 τ n p i z( iτ) 1 τ i=0 +ρ (n+1)τ+ρ n iτ+ρ i=0 (i+1)τ+ρ p (+ρ s)/τ z(s) ds 1 τ 2 + nτ 2 + (n + 1)τ, n = 1, 2,..., which yields x( σ j ) 1 τ 1 τ +ρ σ j 2 +ρ 3 p (+ρ s)/τ z(s) ds +ρ 2 +ρ p (+ρ σ j s)/τ z(s) ds 1 τ p (+ρ s)/τ z(s) ds, p (+ρ s)/τ z(s) ds, 2 +ρ p (+ρ s)/τ z(s) ds 3 := 2 + 2ρ, j = 1, 2,...,m. (2.4) Subsiuing (2.4) ino (2.3), we obain z 1 ] β () q() p (+ρ s)/τ z(s) ds, τ 3. (2.5) 3 Se y() = z(s) ds, 3. (2.6) 3 hen i follows from (2.5) ha z () τ β p β(+ρ)/τ q() y() ] β, 3, (2.7) so z(s) τ β Subsiuing his ino (2.6), we have s y() τ β 3 τ β 3 p β(u+ρ)/τ q(u) y(u) ] β du, s 3. s p β(u+ρ)/τ q(u) y(u) ] β duds u p β(u+ρ)/τ ] β q(u) y(u) dsdu 3 = τ β+1 p β(s+ρ)/τ( p 3/τ ) ] β q(s) y(s) ds, ln p 3. (2.8) 3
X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 557 Se w() = p β(+ρ)/τ( p /τ p 3/τ ) q() y() ] β, 3. (2.9) hen (2.8) yields ( ) β ( τ β+1 ) β w() w(s)ds p β(+ρ)/τ( p /τ p 3/τ ) q(), ln p 3. (2.10) 3 Choose a 1 > 3 such ha 1 3 w(s)ds > 0. hen from (2.10), we have ( τ β+1 ln p 1 I follows ha ) β w() 1 ( 3 p β(+ρ)/τ( p /τ p 3/τ ) q()d 3 w(s)ds ) β ( d = 1 1 β 1 ( 1 1 ) 1 β w(s)ds, > 1. β 1 3 w(s)ds ) 1 β ( 3 w(s)ds ) 1 β ] 1 p β(+ρ)/τ( p /τ p 3/τ ) q()d <. (2.11) Noe ha p /τ p 3/τ p /τ 1 p ( 1 3 )/τ ], 1. hen (2.11) implies ha 1 p (β 1)s/τ q(s)ds <, (2.12) which conradics o (2.1) so he sufficiency is proved. Necessiy. We only need o prove ha he condiion q(s)exp τ 1 ln p(β 1)s ] ds < (2.13) implies ha Eq. (1.1) has an evenually posiive soluion. Noe ha s = τ ln p p βu/τ q(u)duds p βu/τ ( q(u) p u/τ p /τ ) du
(p /τ y () ) = τ β p β( ρ τ)/τ q() y() ] β 0,. (2.20) 558 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 τ p (β 1)s/τ τ q(s)ds = ln p ln p Hence, from (2.13), we have Choose > + ρ such ha ( ) 2 β p βρ/τ τ s q(s)exp τ 1 ln p(β 1)s ] ds. p βu/τ q(u)duds <. (2.14) Define he sequence of funcions {y n ()} as follows: y 0 () = 2,, y n+1 () = 1 + τ β p βρ/τ s p β(u τ)/τ q(u)duds < 1. (2.15) s p β(u τ)/τ q(u) y n (u) ] β duds, (2.16), n= 1, 2,.... (2.17) By (2.15) (2.17) by inducion, i is easy o verify ha 1 y n+1 () y n () y 0 () = 2,, n= 1, 2,.... hen he limi lim n y n () = y() exiss for, ) 1 y() 2for, ). Applying Lebesgue s monoone convergence heorem o (2.17), we obain y() = 1 + τ β p βρ/τ I follows ha p /τ y () = τ β p βρ/τ s p β(u τ)/τ q(u) y(u) ] β duds,. (2.18) p β(s τ)/τ q(s) y(s) ] β ds,, (2.19) I follows from (2.15) ha here exiss a 1 > + τ such ha τ β p βρ/τ p 1/τ 1 p β(s τ)/τ ] β q(s) y(s) ds < τ 1. (2.21) If q() 0forlarge, hen x() = e ln p/τ is an evenually posiive soluion of Eq. (1.1). So, in he sequel, we only consider he case when q() 0 evenually, so p β(s τ)/τ q(s) y(s) ] β ds > 0,. (2.22)
X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 559 u(s) ds + y( 1 ), 1. (2.25) Se u() = p /τ y (),. (2.23) hen from (2.19) (2.23), we have 0 <u()<τ 1 p 1/τ, u () 0, 1, (2.24) y() = 1 Define a funcion v() as follows: τ 1 u( 1 + τ)( 1 ), 1 1 + τ, v() = p v( + τ) u( + τ)], < 1, (2.26) u() + pv( τ), 1 + iτ < 1 + (i + 1)τ, i = 1, 2,... I is easy o see ha v() coninues on, ) v() > 0for> 1 v() = u() + pv( τ), 1 + τ. (2.27) From (2.24) (2.26), we have v() u( 1 + τ) u( 1 )<τ 1 p 1/τ, 1 1 + τ. (2.28) hen from (2.24), (2.25), (2.27) (2.28) by using he fac y() 1for,wehave n 1 n 1 v() = p i u( iτ) + p n v( nτ) p i u( iτ) + p n τ 1 p 1/τ 1 τ = 1 τ i=0 n 1 iτ i=0 (i+1)τ nτ i=0 p ( τ s)/τ u(s) ds + τ 1 p n+ 1/τ p ( τ s)/τ u(s) ds + τ 1 p n+ 1/τ 1 p ( τ s)/τ u(s) ds + τ 1 p ( τ)/τ 1 = τ τ p( τ)/τ u(s) ds + 1 1 1 ] 1 τ p( τ)/τ u(s) ds + y( 1 ) = 1 τ p( τ)/τ y(), 1 1 + nτ 1 + (n + 1)τ, n = 1, 2,.... Noe ha y () 0for, i follows from he above ha v( σ j ) 1 τ p( τ σ j )/τ y( σ j ) 1 τ p( τ σ j )/τ y(), 1 + 2ρ, j = 1, 2,...,m. (2.29) ]
560 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 By (2.19), (2.23), (2.27) (2.29), we obain v() = pv( τ)+ u() = pv( τ)+ τ β p βρ/τ pv( τ)+ τ β p βρ/τ pv( τ)+ his shows ha he inequaliy v() pv( τ)+ q(s) q(s) p β(s τ)/τ q(s) p β(s τ)/τ q(s) y(s) ] β ds m τp ( s+τ+σ j )/τ v(s σ j ) ] β j ds m v(s σj ) ] β j ds, 1 + 2ρ. m v(s σ j ) βj sign v(s σ 1 ) ] ds, 1 + 2ρ, (2.30) has a posiive soluion v() on 1 + 2ρ, ). Similar o he proof of 1, Lemma 5.1.5], we can prove ha he corresponding equaion x() = px( τ)+ q(s) m x(s σ j ) βj sign x(s σ 1 ) ] ds, 1 + 2ρ, (2.31) has also a posiive soluion x() on 1 + 2ρ, ). Obviously, x() is also he evenually posiive soluion of Eq. (1.1), so he necessiy is proved. he proof is complee. heorem 2.2. Assume ha p>1 m β j > 1. hen every soluion of Eq. (1.1) oscillaes if only if q(s)ds =. (2.32) Proof. Sufficiency. Lex() be a nonoscillaory soluion of Eq. (1.1). We may wihou loss of generaliy assume ha x() > 0for 1 for some 1. Se z() as in (2.2). hen i follows from p>1 (2.32) ha z() < 0 z () 0for 2 for some 2 > 1, see 1, heorem 3.2.9]. Choose a posiive ineger n such ha nτ ρ. hen from (2.2), we have so x() > 1 p n z( + nτ), 2, (2.33) x( σ j )> 1 p n z( + nτ σ j ) 1 p n z(), 2 + ρ, j = 1, 2,...,m. Subsiuing his ino (2.3), we have ( z () q() 1 ) β p n z(), 2 + ρ. I follows ha
X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 561 2 +ρ q(s)ds p nβ 2 +ρ ( z(s) ) βz (s) ds = p nβ (β 1) 1( z( 2 + ρ) ) 1 β ( z() ) 1 β ] <p nβ (β 1) 1 z( 2 + ρ) ] 1 β, 2 + ρ, so 2 +ρ q(s)ds <, which conradics o (2.32) so he sufficiency is proved. Necessiy. We only need o prove ha he condiion q(s)ds < (2.34) implies ha Eq. (1.1) has an evenually posiive soluion. By (2.34), we can choose > +τ +σ such ha q(s)ds p 1. 2 (2.35) Define he sequence of funcions {x n ()} as follows: x 0 () = 1,, p 1 { p 1 2 + x n ( + τ) x n+1 () = + +τ q(s) m x n (s σ j )] β j ds},, x n+1 ( ), <. (2.36) (2.37) n = 1, 2,... By (2.35) (2.37) by inducion, i is easy o verify ha p 1 2p x n+1() x n () x 0 () = 1,, n= 1, 2,.... hen he limi lim n x n () = x() exiss for, ) (p 1)/2p x() 1for, ). Applying Lebesgue s monoone convergence heorem o (2.37), we obain { +τ x() = p 1 p 1 + x( + τ)+ q(s) 2 m x(s σj ) ] } β j ds,. (2.38) I is easy o see ha x() is also he evenually posiive soluion of Eq. (1.1), so he necessiy is proved. he proof is complee.
562 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 3. he sublinear case m β j < 1 heorem 3.1. Assume ha p>1 m β j < 1. hen every soluion of Eq. (1.1) oscillaes if only if ( m ) ] q(s)exp τ 1 ln p β j 1 s ds =. (3.1) Proof. Sufficiency. Lex() be a nonoscillaory soluion of Eq. (1.1). We may wihou loss of generaliy assume ha x() > 0 for all 1. Se z() = px( τ) x(). hen i follows from (1.1) (3.1) ha m z () = q() x( σ j ) βj 0 ( 0), 2 = 1 + ρ. (3.3) his shows ha z() is nondecreasing on 2, ). Hence, z() > 0, 2, (see 1, heorem 3.2.9]), from (3.2) (3.3), we have x() = 1 τ which yields p i z( + iτ) 1 τ i=1 i=1 p i +iτ +(i 1)τ p (s+τ )/τ z(s) ds = 1 τ p( τ)/τ x( σ j ) 1 τ p( τ σ j )/τ σ j z(s) ds z(s) ds 1 τ +iτ i=1 +(i 1)τ z(s) ds, 2, 1 τ p( τ ρ)/τ (3.2) p (s+τ )/τ z(s) ds z(s) ds, 2 + ρ, j = 1, 2,...,m. (3.4) Subsiuing (3.4) ino (3.3), we obain z 1 β () q() τ p( τ ρ)/τ z(s) ds], 3 = 2 + ρ. (3.5) Se y() = hen i follows from (3.5) ha z(s) ds, 3. (3.6) z () τ β p β( τ ρ)/τ q() y() ] β, 3, (3.7)
X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 563 so s z(s) τ β 3 Subsiuing his ino (3.6), we have p β(u τ ρ)/τ ] β q(u) y(u) du, s 3. Se y() τ β τ β s 3 p β(u τ ρ)/τ ] β q(u) y(u) duds p β(u τ ρ)/τ q(u) y(u) ] β = τ β+1 p β(τ+ρ)/τ ln p u dsdu p (β 1)s/τ q(s) y(s) ] β ds, 3. (3.8) w() = p (β 1)/τ q() y() ] β, 3. (3.9) hen (3.8) yields w() ( β ( τ w(s)ds) β+1 p β(τ+ρ)/τ ) β p (β 1)/τ q(), 3. (3.10) ln p Inegraing (3.10) from 3 o,wehave ( τ β+1 p β(τ+ρ)/τ I follows ha ln p ) β 3 p (β 1)/τ q()d 3 w() = 1 1 β ( ( 3 w(s)ds) β d w(s)ds ) 1 β. 3 p (β 1)/τ q()d <, (3.11) which conradics o (3.1) so he sufficiency is proved. Necessiy. We only need o prove ha he condiion q(s)exp τ 1 ln p(β 1)s ] ds < (3.12) implies ha Eq. (1.1) has an evenually posiive soluion. Noe ha
564 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 s p βu/τ q(u)duds = Hence, from (3.12), we have s τ ln p = τ ln p p (β 1)s/τ q(s)ds q(s)exp τ 1 ln p(β 1)s ] ds. p βu/τ q(u)duds <. (3.13) Choose > + ρ such ha ( ) 2 β s p β(u+ρ)/τ q(u)duds < 1. (3.14) τ If q() 0for, hen x() = e ln p/τ is an evenually posiive soluion of Eq. (1.1). So, in he sequel, we only consider he case when q() 0for. Define he sequence of funcions {y n ()} as follows: y 0 () = 2,, y n+1 () = 1 + τ β, n= 1, 2,.... s p β(u+ρ)/τ q(u) y n (u) ] β duds, By (3.14) (3.16) by inducion, i is easy o verify ha 1 y n+1 () y n () y 0 () = 2,, n= 1, 2,.... (3.15) (3.16) hen he limi lim n y n () = y() exiss for, ) 1 y() 2for, ). Applying Lebesgue s monoone convergence heorem o (3.16), we obain y() = 1 + τ β s p β(u+ρ)/τ q(u) y(u) ] β duds,. (3.17) I follows ha p /τ y () = τ β p β(s+ρ)/τ q(s) y(s) ] β ds,, (3.18) Se ( p /τ y ()) = τ β p β(+ρ)/τ q() y() ] β 0,. (3.19) u() = p /τ y (),. (3.20)
X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 565 hen from (3.17), (3.18), (3.19) (3.20), we have u() 0, u () 0,, (3.21) y() = u(s) ds + 1, 1. (3.22) From (3.21) (3.22), we find p i u( + iτ) 1 τ i=1 = 1 τ +ρ+iτ i=1 +ρ+(i 1)τ +ρ = 1 τ p(+ρ)/τ p (+ρ s)/τ u(s) ds +ρ 1 τ p(+ρ)/τ p (+ρ s)/τ u(s) ds u(s) ds +ρ u(s) ds + 1 = 1 τ p(+ρ)/τ y( + ρ),. (3.23) Noe ha q() 0for, so we can choose 1 > + ρ such ha u( 1 ) = τ β 1 I follows from (3.21) ha p β(s+ρ)/τ q(s) y(s) ] β ds > 0. (3.24) u() > 0, u () 0, 1. (3.25) Define a funcion v() as follows: v() = p i u( + iτ),. (3.26) i=1 By (3.23) (3.25), i is easy o see ha v() coninues on, ) v() > 0for, v() = 1 p u( + τ)+ v( + τ) ],, (3.27) v() 1 τ p(+ρ)/τ y( + ρ),. (3.28) I follows from (3.26) he fac ha y () 0for ha ]
566 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 v( σ j ) 1 τ p(+ρ σ j )/τ y( + ρ σ j ) 1 τ p(+ρ σ j )/τ y(), 1,j= 1, 2,...,m. (3.29) hen, by (3.18), (3.20), (3.24), (3.27) (3.29), we obain v() = 1 ] v( + τ)+ u( + τ) p = 1 +τ v( + τ)+ τ β p β(s+ρ)/τ q(s) y(s) ] ] β ds p = 1 u( 1 ) + v( + τ)+ τ β p +τ 1 +τ 1 u( 1 ) + v( + τ)+ τ β p 1 u( 1 ) + v( + τ)+ p his shows ha he inequaliy { v() 1 u( 1 ) + v( + τ)+ p +τ 1 +τ 1 1 q(s) q(s) ] p β(s+ρ)/τ ] β q(s) y(s) ds p β(s+ρ)/τ q(s) m τp ( s ρ+σ j )/τ v(s σ j ) ] ] β j ds m v(s σj ) ] ] β j ds, 1. m v(s σj ) βj sign v(s σ 1 ) ] } ds, 1, (3.30) has a posiive soluion v() on 1, ). Similar o he proof 1, Lemma 5.1.5], we can prove ha he corresponding equaion { x() = 1 +τ m u( 1 ) + x( + τ)+ q(s) x(s σj ) βj sign x(s σ 1 ) ] } ds, p 1 1, (3.31) has also a posiive soluion x() on 1, ). Obviously, x() is also he evenually posiive soluion of Eq. (1.1), so he necessiy is proved. he proof is complee. 4. Remarks Combining heorems 1.4, 2.1, 2.2 3.1, we have he following corollaries. Corollary 4.1. Assume ha ( m β j 1) ln p<0. hen every soluion of Eq. (1.1) oscillaes if only if ( m ) ] q(s)exp τ 1 ln p β j 1 s ds =. (4.1)
X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 567 Corollary 4.2. Assume ha ( m β j 1) ln p>0. hen every soluion of Eq. (1.1) oscillaes if only if q(s)ds =. (4.2) Now we consider he firs nonlinear neural delay differenial equaion ] m x() p()x( τ) + q() x( σj ) βj sign x( σ 1 ) ] = 0,, (4.3) where p C(, ), 0, )), τ, β j > 0 q() are he same as in Eq. (1.1). In view of he proof of heorems 2.1 2.3, we have he following heorems. heorem 4.1. Assume ha m β j > 1. hen he following conclusions hold: (i) If here exiss a p 1 (0, 1) such ha p 1 p() 1,, (4.4) ( m ) ] q(s)exp τ 1 ln p 1 β j 1 s ds =, (4.5) hen every soluion of Eq. (4.3) oscillaes; (ii) If here exiss a p 2 (0, 1) such ha 0 p() p 2,, (4.6) ( m ) ] q(s)exp τ 1 ln p 2 β j 1 s ds <, (4.7) hen Eq. (4.3) has an evenually posiive soluion; (iii) If here exis p 3,p 4 (1, ) such ha p 3 p() p 4,, (4.8) hen every soluion of Eq. (4.3) oscillaes if only if q(s)ds =. (4.9) heorem 4.2. Assume ha m β j < 1. hen he following conclusions hold: (i) If here exiss p 1 (0, 1) such ha 0 p() p 1,, (4.10)
568 X.H. ang, X. Lin / J. Mah. Anal. Appl. 321 (2006) 553 568 hen every soluion of Eq. (4.3) oscillaes if only if q(s)ds = ; (4.11) (ii) If here exiss a p 2 (1, ) such ha 1 p() p 2,, (4.12) ( m ) ] q(s)exp τ 1 ln p 2 β j 1 s ds =, (4.13) hen every soluion of Eq. (4.3) oscillaes; (iii) If here exiss a p 3 (1, ) such ha p() p 3,, (4.14) ( m ) ] q(s)exp τ 1 ln p 3 β j 1 s ds <, (4.15) hen Eq. (4.3) has an evenually posiive soluion. References 1] L.H. Erbe, Q. Kong, B.G. Zhang, Oscillaion heory for Funcional Differenial Equaions, Dekker, New York, 1995. 2] I. Gyori, G. Ladas, Oscillaion heory of Delay Differenial Equaions wih Applicaions, Clarendon Press, Oxford, 1991. 3] X.H. ang, Oscillaion for firs order nonlinear delay differenial equaions, J. Mah. Anal. Appl. 264 (2001) 510 521. 4] X.H. ang, X.Q. Li, Necessary sufficien condiions for oscillaion of nonlinear neural differenial equaions, Hunan Ann. Mah. 17 (1997) 57 60 (in Chinese). 5] X.H. ang, J.S. Yu, Firs order nonlinear differenial inequaliies wih deviaing argumens, Appl. Mah. J. Chinese Univ. Ser. B 15 (2000) 21 27. 6] L.W. Wang, Oscillaion of firs order nonlinear neural differenial equaions, Aca Mah. Appl. Sinica 14 (1991) 348 359. 7] J.S. Yu, Firs order nonlinear differenial inequaliies wih deviaing argumens, Aca Mah. Sinica 33 (1990) 152 159. 8] B.G. Zhang, J.S. Yu, Oscillaion nonoscillaion for neural differenial equaions, J. Mah. Anal. Appl. 172 (1993) 11 23.