ASYMPTOTIC FORMS OF WEAKLY INCREASING POSITIVE SOLUTIONS FOR QUASILINEAR ORDINARY DIFFERENTIAL EQUATIONS

Transcription

1 Elecronic Journal of Differenial Equaions, Vol. 2007(2007), No. 126, pp ISSN: URL: hp://ejde.mah.xsae.edu or hp://ejde.mah.un.edu fp ejde.mah.xsae.edu (login: fp) ASYMPTOTIC FORMS OF WEAKLY INCREASING POSITIVE SOLUTIONS FOR QUASILINEAR ORDINARY DIFFERENTIAL EQUATIONS KEN-ICHI KAMO, HIROYUKI USAMI Absrac. Asympoic forms are deermined for posiive soluions which are called weakly increasing soluions o quasilinear ordinary differenial equaions. 1. Inroducion In his paper we consider he equaion under he following condiions: ( u α 1 u ) + p() u β 1 u = 0 (1.1) (A1) α and β are posiive consans saisfying α β; (A2) p() is a C 1 funcion defined near + saisfying he asympoic condiion p() σ for some σ R as. By condiion (A2) equaion (1.1) can be rewrien in he form ( u α 1 u ) + σ (1 + ε()) u β 1 u = 0, (1.2) where ε() = σ p() 1 saisfies lim ε() = 0. Of course, here and in wha follows he symbol f() g() as means ha lim f()/g() = 1. Some preparaory resuls for equaion (1.1) are sill valid for more general equaions han (1.1); so i is convenien o consider he auxiliary equaion ( u α 1 u ) + q() u β 1 u = 0, (1.3) where we assume ha α and β saisfy condiion (A1) and q C([ 0, ); (0, )). A funcion u is defined o be a soluion of equaion (1.3) if u C 1 [ 1, ) and u α 1 u C 1 [ 1, ) and i saisfies equaion (1.3) on [ 1, ) for sufficienly large 1. I is easily seen ha all posiive soluions u() of (1.3) are classified ino he following hree ypes according as heir asympoic behavior as : (I) Asympoically linear soluion: u() c 1 for some consan c 1 > 0; (II) Weakly increasing soluion: u () 0 and u() ; (III) Asympoically consan soluion: u() c 1 for some consan c 1 > Mahemaics Subjec Classificaion. 34E10, 34C41. Key words and phrases. Quasilinear ordinary differenial equaion; posiive soluion. c 2007 Texas Sae Universiy - San Marcos. Submied May 24, Published Sepember 28,

2 2 K. KAMO, H. USAMI EJDE-2007/126 Concerning qualiaive properies of posiive soluions, he sudy of asympoic behavior of asympoically linear soluions and asympoically consan soluions are raher easy, because heir firs approximaions are given by definiion. On he oher hand, we can no easily know how he weakly increasing posiive soluions behave excep in he case α = 1; see [1, 3]. In [3, Secion 20], equaion (1.1) wih α = 1 has been considered sysemaically, and asympoic forms of weakly increasing posiive soluions are given by means of he parameers β and σ. When α 1, as far as he auhors are aware, here are no works in which asympoic forms of weakly increasing posiive soluions are sudied sysemaically. Moivaed by hese facs in his paper we make an aemp o find ou asympoic forms of weakly increasing posiive soluions of (1.1) for he general case α > 0. Furhermore we will also esablish more han obained in [3] in he case α > β. In fac, some of our resuls are new even hough α = 1. To gain an insigh ino our problem, we consider he ypical equaion ( u α 1 u ) + σ u β 1 u = 0, (1.4) where σ R is a consan. Noe ha equaion (1.1) can be regarded as a perurbaion of (1.4). Equaion (1.4) has a weakly increasing posiive soluion of he form c ρ, (c > 0, 0 < ρ < 1) if and only if min{α, β} + 1 < σ < max{α, β} + 1. This soluion is uniquely given by u 0 () = Ĉk, (1.5) where k = α σ + 1 (0, 1), Ĉ = {α(1 k)k α } 1 β α. (1.6) α β From his simple observaion we can see ha asympoic forms of weakly increasing posiive soluions of (1.3) may be srongly affeced by ha of he coefficien funcion q(). Furhermore we conjecure ha weakly increasing posiive soluions u of (1.1) behave like u 0 () given by (1.5) and (1.6) if ε() is sufficienly small a. We will show ha he above conjecure is rue in many cases. In fac, we can obain he following heorems which are he main resuls of he paper: Theorem 1.1. Le α > β. (i) Suppose ha β + 1 < σ < α + 1. Then, every weakly increasing posiive soluion u of (1.1) has he asympoic form u() u 0 () as, where u 0 is given by (1.5) and (1.6). (ii) Suppose ha σ = α + 1; namely p() α 1 as. Then, every weakly increasing posiive soluion u of (1.1) has he asympoic form u() α 1 ( α α β ) α (log ) α as. Theorem 1.2. Le α < β. Suppose ha α 1 and 1/2 < k < 1 ( (α+β +2)/2 < σ < β + 1). Suppose furhermore ha eiher ε() 2 d < (1.7) or ε () d < (1.8)

3 EJDE-2007/126 ASYMPTOTIC FORMS 3 holds. Then, every weakly increasing posiive soluion u of (1.1) has he asympoic form u() u 0 () as, where u 0 is given by (1.5) and (1.6). Theorem 1.3. Le α < β. Suppose ha α 1 and 0 < k < 1/2( α + 1 < σ < (α + β + 2)/2). Suppose furhermore ha eiher (1.7) or (1.8) holds. Then, he same conclusion as in Theorem 1.2 holds. Remark 1.4. (i) In Theorem 1.1 he differeniabiliy of p is acually unnecessary. Similarly, in Theorems 1.2 and 1.3, he differeniabiliy of p is unnecessary when (1.7) is assumed. (ii) When α = 1 and ε() 0, Theorems 1.1, 1.2 and 1.3 were obained by [1] and [3, Corollaries 20.2, 20.3]. Remark 1.5. Le α < β. We can no prove wheher or no our conjecure is rue in he following cases: (i) α 1 and 0 < k < 1/2; (ii) α 1 and 1/2 < k < 1; and (iii) k = 1/2. We noe ha exisence resuls of weakly increasing posiive soluions o (1.3) and (1.1) are known for he case α > β. In fac, equaion (1.1) has a weakly increasing posiive soluion if and only if β + 1 < σ α + 1; see Remark 3.4 in Secion 3. In conras, i seems ha here are no such useful resuls for he case α < β. Bu we can show many concree examples of hose equaions ha have weakly increasing posiive soluions. The paper is organized as follows. In Secion 2 we give preparaory lemmas employed laer. In Secion 3 we consider equaion (1.1), as well as (1.3), under he sub-homogeneiy condiion α > β. When q() saisfie < lim inf q()/ σ lim sup q()/ σ < for some σ R, we can obain a resul which may be called as asympoic equivalence heorem for equaion (1.3); see Corollary 3.3. Theorem 1.1 is a direc consequence of his corollary. In Secion 4 we consider only equaion (1.1) under he super-homogeneiy condiion α < β, and we prove Theorems 1.2 and 1.3 here. Oher relaed resuls are found in [2, 4, 5, 6]. 2. Preparaory lemmas Lemma 2.1. Le w C 1 [ 0, ), w () = O(1) as, and w L λ [ 0, ) for some λ > 0. Then, lim w() = 0. Proof. We have w() λ w() = w( 0 ) λ w( 0 ) + = w( 0 ) λ w( 0 ) + (λ + 1) 0 ( w(s) λ w(s)) ds 0 w(s) λ w (s)ds. By our assumpions he above inegral converges. Hence lim w() λ w() exiss in R. Since w L λ [ 0, ), he limi mus be 0. The proof is complee. Lemma 2.2. Le u C 1 [ 0, ), u() > 0 and u () 0 as. Then, u () u() for sufficienly large, and he funcion u()/ is decreasing near.

4 4 K. KAMO, H. USAMI EJDE-2007/126 Proof. Since u() = u( 0 ) + u (s)ds u( 0 ) + u ()( 0 ), 0 we have u () u() 0 u () u( 0 ). Noing he assumpion u ( ) = 0 we find ha u () u() < 0 near. Since (u()/) = (u () u())/ 2, he proof is compleed. 3. Sub-homogeneous case: α > β Throughou he secion we assume ha α > β. As a firs sep we give he growh esimaes for weakly increasing posiive soluions of (1.3): Lemma 3.1. Le u be a weakly increasing posiive soluion of (1.3). following esimaes hold near : (α β ) α { ( ) 1/αds } α q(r)dr α 1 u() ( α β ) α { α s 1 s β/α where 1 is a sufficienly large number. ( s ) 1/αds } α r β q(r)dr Then he (3.1) Noe ha s β q(s)ds < if equaion (1.3) has a weakly increasing posiive soluion; see (ii) of Remark 3.4. Proof of Lemma 3.1. We may assume ha u, u > 0 for 1. Since u saisfies for large, and u is increasing, we have ha is u () α = u () α u() β q(s)ds, q(s)u(s) β ds, (3.2) ( 1/α. u ()u() β/α q(s)ds) (3.3) An inegraion of his inequaliy on he inerval [ 1, ] gives α } ( {u() 1/αds, α u(1 ) α q(r)dr) α β which proves he firs inequaliy of (3.1). On he oher hand, by he decreasing naure of u()/ shown in Lemma 2.2, we find from (3.2) ha Accordingly, u () α ( u() 1 ) β s β q(s)ds. u ()u() β/α β/α( 1/α. s q(s)ds) β As before we can ge he second inequaliy in (3.1). The proof is complee. s

5 EJDE-2007/126 ASYMPTOTIC FORMS 5 To give he main resul in his secion, le us consider he wo equaions, of he same ype, ( u α 1 u ) + q 1 () u β 1 u = 0, (3.4) ( u α 1 u ) + q 2 () u β 1 u = 0. (3.5) Here, we assume ha 0 < α < β and q 1, q 2 C([ 0, ); (0, )). Theorem 3.2. Suppose ha q 1 () q 2 () as, (3.6) ( ) 1/αds ( 1/αds C s β/α r β q 1 (r)dr q 1 (r)dr) (3.7) 0 s 0 s hold for some consan C > 0. If u 1 and u 2 are weakly increasing posiive soluions of equaions (3.4) and (3.5), respecively, hen u 1 () u 2 () as. Corollary 3.3. Suppose ha q 1 and q 2 saisfy (3.6) and 0 < lim inf q 1()/ σ lim sup q 1 ()/ σ < for some σ (β + 1, α + 1]. If u 1 and u 2 are weakly increasing posiive soluions of equaions (3.4) and (3.5), respecively, hen u 1 () u 2 () as. Theorem 1.1 is an immediae consequence of Corollary 3.3. Indeed, o see (ii) of Theorem 1.1, i suffices o noice he fac ha he equaion ( u α 1 u ) + α 1( β ) 1 u β 1 u = 0 (α β) log has a weakly increasing posiive soluion given explicily by α ( 1 α ) α (log ) α. α β Proof of Theorem 3.2. Pu z() = u 1 ()/u 2 (), 0, 0 being sufficienly large. Then, z saisfies he equaion z + 2u 2() u 2 () z + u 2() β 1 [ (u2 ()z + u α 2()z) 1 α q 1 ()z β q 2 ()u 2() 1 α z ] = 0. If z (T ) = 0 for some T, hen ( z (T ) = α 1 q 1 (T )u 2 (T ) β 1 u 2(T ) 1 α q2 (T ) z(t ) q 1 (T ) z(t )β α). Thus, if z = 0 in he region z > (q 1 ()/q 2 ()) 1/(), hen z aains a local minimum here; while if z = 0 in he region 0 < z < (q 1 ()/q 2 ()) 1/(), hen z aains a local maximum here. Noe ha by our assumpion lim (q 1 ()/q 2 ()) 1/() = 1. These simple observaions are used below. Since Lemma 3.1 and condiions (3.6) and (3.7) imply ha z() is bounded and bounded from 0, we can pu 0 < l = lim inf z() lim sup z() = L <. We claim ha l = L; ha is lim z() (0, ) exiss. Suppose for conradicion ha l L. We rea he following four cases separaely: (a)l 1 > l; (b)l > 1 l; (c)l > l 1; (d)1 L > l. Suppose ha case (a) occurs. We can find wo sequences {T n } and { n } saisfying lim n T n = lim n n = (3.8)

6 6 K. KAMO, H. USAMI EJDE-2007/126 and lim z(t n) = L, n lim z( n) = l, n < T n < n+1 for n = 1, 2,.... (3.9) n Since lim (q 1 ()/q 2 ()) 1/() = 1, we may assume ha z( n ) < (q 1 ( n )/q 2 ( n )) 1/(). For sufficienly large n N he minimum of z() on he inerval [T n 1, T n ] mus be aained a an inerior poin, say (T n 1, T n ). Obviously, z ( ) = 0 and z ( ) 0. However, since z( ) z( n ) for sufficienly large n, we ge a conradicion o he above observaion. Hence case (a) doeo occur. Nex suppose ha case (c) occurs. As in case (a) we can find wo sequences {T n } and { n } saisfying (3.8) and (3.9). For sufficienly lage n N he maximum of z() on he inerval [ n, n+1 ] mus be aained a an inerior poin, say ( n, n+1 ). Obviously, z ( ) = 0 and z ( ) 0. Since z( ) z(t n ) and z(t n ) > (q 1 (T n )/q 2 (T n )) 1/() for sufficienly large n, we ge a conradicion as before. Hence case (c) doeo occur. Similarly we can show ha he oher cases can no occur. Therefore lim z() = lim u 1 ()/u 2 () = m (0, ) exiss. Finally, by L Hospial s rule we have u 1 () ( m = lim u 2 () = u lim 1() α ) 1/α u 2 ( ()α [u = lim 1() α ] ) 1/α [u 2 ()α ] ( = lim q 1 ()u 1 () β q 2 ()u 2 () β ) 1/α = m β/α ; ha is m = m β/α. Since α > β, we have m = 1. This complees he proof. Remark 3.4. Concerning he exisence properies of weakly increasing posiive soluions, we know he following resuls: (i) If β q()d < ; and ( 1/αd q(s)ds) =, (3.10) Then, equaion (1.3) has weakly increasing posiive soluions [2, Example 1]. (ii) Conversely, if equaion (1.3) has a weakly increasing posiive soluion, hen we can show ha β q()d < ; and β/α( 1/αd s q(s)ds) β =. (3.11) In fac, he firs condiion in (3.11) follows from [2, Example 2]; while he second one is an immediae consequence of he esimaes in Lemma 3.1. In paricular, when α = 1, we find ha condiions (3.10) and (3.11) are he same; ha is, equaion (1.3) (wih α = 1) has a weakly increasing posiive soluion if and only if (3.10) (wih α = 1) holds.

7 EJDE-2007/126 ASYMPTOTIC FORMS 7 4. Super-homogeneous case: α < β Throughou his secion we assume ha α < β. In his case he siuaion seems o be more complicaed han in he previous case. The main purpose of he secion is o give he proofs of Theorems 1.2 and 1.3. To his end we need several lemmas. Lemma 4.1. Le 0 < lim inf q()/ σ lim sup q()/ σ < for some σ (α+1, β +1). Then every weakly increasing posiive soluion u of (1.1) saisfies u() = O(u 0 ()) and u () = O(u 0()) as, where u 0 is he exac soluion of (1.4) given by (1.5) and (1.6). Proof. As in he proof of Lemma 3.1 we obain (3.3). An inegraion of (3.3) on he inerval [, ) for large will give { ( 1/αds } α/(β α) u() C 1 q(r)dr) C2 u 0 (), s where C 1 and C 2 are posiive consan. Furhermore, by (3.2) we find ha ( ) 1/α ( 1/α u () = q(s)u(s) β ds C3 s ds) σ+kβ = C4 k 1 = O(u 0()) as, where C 3 and C 4 are posiive consans. This complees he proof. Lemma 4.2. Le σ (α + 1, β + 1), and u be a weakly increasing posiive soluion of equaion (1.1). Pu s = log u 0 () and v = u/u 0. Then (i) v, v = O(1) as s, and v + v > 0 near, where = d/ds; (ii) v(s) saisfieear he equaion v + a v bv + b( v + v) 1 α v β + bδ(s)( v + v) 1 α v β = 0, (4.1) where a = 2 1 k 0, b = 1 k k > 0, and δ(s) = ε(). Proof. We will prove only (i), because (ii) can be proved by direc compuaions. We noe ha u, u > 0. Since u = u 0 v, he boundedness of v follows from Lemma 4.1. Noing du/d = Ĉkk 1 (v + v), we have v + v > 0. On he oher hand, since d/ds = /k, we have v = d ( u ) d d u 0 ds This complees he proof. = u u 0 u 0u u 2 0 k C k 1 k 2k = O(1) as s. Lemma 4.3. Le he assumpion eiher of Theorem 1.2 or Theorem 1.3 hold, and v be as in Lemma 4.2. Then v L 2 [, ) for sufficienly large. Proof. We noe ha condiions (1.7) and (1.8), respecively, are equivalen o δ(s) 2 ds < (4.2) and δ(s) ds <. (4.3) We firsly consider he case where assumpions of Theorem 1.2 hold. In his case, he consan a appearing in (4.1) is posiive. We muliply he boh sides of (4.1) by v. Since α 1, we have (1 + δ(s))( v + v) 1 α v (1 + δ(s))v 1 α v; and so we obain v v + a v 2 bv v + bv 1 α+β v + bδ(s)v 1 α+β v 0. (4.4)

8 8 K. KAMO, H. USAMI EJDE-2007/126 An inegraion gives v a v 2 dr b 2 v2 + bv2 α+β 2 α + β + b δ(r)v 1 α+β vdr cons; (4.5) ha is a v 2 dr + b δ(r)v 1 α+β vdr O(1) as s. Here we have employed (i) of Lemma 4.2. Le he inegral condiion (1.7) hold; ha is, le (4.2) hold. By he Schwarz inequaliy we have ( ) 1/2 ( ) 1/2 a v 2 dr C 1 δ(r) 2 dr v 2 dr O(1) for some consan C 1 > 0. Therefore v L 2 [, ). Nex le (1.8) hold. Using inegral by pars, we obain from (4.5) v a v 2 dr b b[1 + 2 v2 δ(r)]v2 α+β b s + δ(r)v 2 α+β dr cons. 2 α + β 2 α + β Noing (i) of Lemma 4.2, we find ha v L 2 [, ). Secondly we consider he case where assumpions of Theorem 1.3 hold. As above, we muliply boh he sides of (4.1) by v. Since α 1, we have (1+δ(s))(v+ v) 1 α v (1 + δ(s))v 1 α v, and so we obain a v 2 v v bv v + b(1 + δ(s))v 1 α+β v. An inegraion on he inerval [, s] gives a v 2 dr v2 2 bv2 2 + bv2 α+β 2 α + β + b δ(r)v 1 α+β vdr + cons. As before, we will obain v L 2 [, ). This complees he proof. Proof of Theorem 1.2. To his end i suffices o show ha lim s v(s) = 1, where v(s) is he funcion inroduced in Lemma 4.2. The proof is divided ino hree seps. Sep 1. We claim ha lim inf s v(s) > 0; or equivalenly lim inf u()/u 0 () > 0. The proof is done by conradicion. Suppose ha lim inf s v(s) = 0. Firsly, we suppose ha v(s) decrease o 0 as s. This means ha u()/u 0 () decreases o 0 as. Accordingly we have u () α = = ( u() u 0 () p(r)u(r) β dr p(r)u 0 (r) β( u(r) ) βdr u 0 (r) ) β p(r)u 0 (r) β dr = C 1 1 σ u() β, where C 1 > 0 is a consan. Consequenly we obain he differenial inequaliy u C 2 (1 σ)/α u β/α for some consan C 2 > 0 near. Bu his differenial inequaliy implies ha u()/u 0 () v(s) C 3 > 0 for some consan C 3 > 0. This is an obvious conradicion. Nex suppose ha lim inf s v(s) = 0 and v(s) changes he sign in any neighborhood of. We noice ha, if v = 0 in

9 EJDE-2007/126 ASYMPTOTIC FORMS 9 he region 0 < v < [1 + δ(s)] 1/(β α), hen v > 0; while if v = 0 in he region v > [1 + δ(s)] 1/(β α), hen v < 0. Therefore, in his case he curve v = v(s) mus cross he curve v = [1 + δ(s)] 1/(β α) infiniely many imes as s. Therefore, we can find ou wo sequences {ξ n } and {η n } saisfying and ξ n < η n < ξ n+1, n = 1, 2,... ; lim n ξ n = lim n η n = v(η n ) 0 a, v(ξ n ) = [1 + δ(ξ n )] 1 β α 1 a. An inegraion of (4.4) on [ξ n, η n ] yields 1 2 { v(η n) 2 v(ξ n ) 2 } + a + ηn ξ n v 2 dr b 2 {v(η n) 2 v(ξ n ) 2 } b 2 α + β {v(η n) 2 α+β v(ξ n ) 2 α+β } + b ηn ξ n δ(r)v 1 α+β vdr 0. (4.6) From equaion (4.1) and (i) of Lemma 4.2 we know ha v = O(1) as s. This fac and he fac ha v L 2 [, ) imply ha lim s v(s) = 0 by Lemma 2.1. Accordingly (4.6) is equivalen o o(1) + o(1) b b (o(1) 1) + (o(1) 1) + o(1) 0 as s. 2 2 α + β Leing n, we ge b/2 b/(2 α + β) 0 a conradicion o he assumpion β > α. Therefore, lim inf s v(s) > 0. Sep 2. We claim ha here is a limi lim s v(s) (0, ). To see his, we inegrae (4.1) muliplied by v: v a + b v 2 dr b 2 v2 + b ( v + v) 1 α v β vdr δ(r)( v + v) 1 α v β vdr = cons. (4.7) Suppose ha condiion (1.7) holds; or equivalenly (4.2) holds. Since v L 2 [, ), he firs, and he hird inegrals in he lef hand side of (4.7) converge as s. The mean value heorem shows ha (v + v) 1 α = ( 1 + v v ) 1 αv 1 α = v 1 α + (1 α) (v + θ(r) v) α v, (4.8) where θ(r) is a quaniy saisfying 0 < θ(r) < 1. Therefore, ( v + v) 1 α v β vdr = {v 1 α+β v + (1 α)(v + θ(r) v) α v β v 2 }dr = v(s)2+β α 2 + β α + O(1) v 2 dr + cons. So we find ha he funcion v 2 /2+v 2+β α /(2+β α) has a finie limi. This fac shows ha lim s v(s) = m (0, ) exiss. Suppose ha (1.8) or equivalenly

10 10 K. KAMO, H. USAMI EJDE-2007/126 (4.3) holds. By (4.8), we have = δ(r)( v + v) 1 α v β vdr {δ(r)v 1 α+β v + δ(r)(1 α)(v + θ(r) v) α v β v 2 }dr = δ(s)v2+β α 2 + β α β α δ(r)v 2+β α dr + cons + O(1) v 2 dr as s. Hence, as before we know ha he funcion v 2 /2 + v 2+β α /(2 + β α) has a finie limi. Therefore m = lim s v(s) (0, ) exiss. Sep 3. Finally, we le s in equaion (4.1). Then, we have lim s v(s) = b(m m 1+β α ). Since v = o(1), we mus have lim s v(s) = 0, implying m = 1. The proof of Theorem 1.2 is complee. Proof of Theorem 1.3. Firsly we show ha lim inf s v(s) > 0. The proof is done by conradicion. Suppose ha lim inf s v(s) = 0. We may assume ha v changes he sign in any neighborhood of, because he case in which v decreases o 0 can be reaed as in he proof of Theorem 1.2 (Sep 1). Since v(s) akes local maxima in he region v (1+δ(s)) 1/(β α), here are he following sequences { } and { } saisfying v( ) = v( ) = 0, < < +1, lim = lim =, n n lim v( ) = 0, v() (1 + δ( )) 1/(β α). n Now, we decompose α in he form α = m ρ, where m N and ρ > 0. Alhough here are infiniely many such choices of decomposiion for α, we fix one choice for a momen. We rewrie equaion (4.1) as v a v bv + b( v + v) m+1+ρ v β + bδ(s)( v + v) m+1+ρ v β = 0. We muliply boh he sides by (v + v) m v and hen inegrae he resuling equaion on he inerval [, ] o obain + b v v(v + v) m dr a (v + v) 1+ρ vv β dr + b (v + v) m v 2 dr b v v(v + v) m dr δ(r)(v + v) 1+ρ vv β dr = 0. (4.9) The binomial expansion implies m c k v v k+1 v m k dr a k=0 + b (v + v) 1+ρ vv β dr + b (v + v) m v 2 dr b k=0 δ(r)(v + v) 1+ρ vv β dr = 0, m c k v m k+1 v k+1 dr

11 EJDE-2007/126 ASYMPTOTIC FORMS 11 where c k = ( ) m k are he binomial coefficiens. Now, we evaluae each erm in he lef hand side. For k {0, 1,..., m 1} we obain v v k+1 v m k dr = = m k k + 2 d ( v k+2 ) v m k dr dr k + 2 v k+3 v m k 1 dr = o(1) a. For k = m obviously we have v v k+1 dr = 0. Hence he firs erm of he lef hand side of (4.9) ends o 0 a. The second erm is dominaed by cons v 2 dr, and hence i ends o zero a. Nex, we compue he hird erm. For k {1, 2,..., m} we have have v m+1 vdr = 1 m + 2 Therefore, he hird erm is equal o v m k+1 v k+1 dr cons v 2 dr. For k = 0, we ( v(sn ) m+2 v( ) m+2) = v() m+2 + o(1) a. m + 2 o(1) bv() m+2 a. m + 2 To evaluae he fourh erm we employ he mean value heorem o obain (v + v) 1+ρ = v 1+ρ + (1 + ρ) (v + θ(r) v) ρ v, where θ(r) is a quaniy beween 0 and 1. Hence we can compue = (v + v) 1+ρ vv β dr v 1+ρ+β vdr + (1 + ρ) = v() 2+ρ+β v( ) 2+ρ+β 2 + ρ + β (v + θ(r) v) ρ v 2 v β dr + (1 + ρ) O(1) v 2 dr = v() 2+ρ+β + o(1) a. 2 + ρ + β Finally by Schwarz s inequaliy we find ha he las erm is dominaed by he quaniy ( ) 1/2 ( ) 1/2 cons δ(r) 2 dr v 2 dr = o(1) a. (Noe ha (1.7) or equivalenly (4.2) is assumed.) Consequenly, from (4.9) we obain he formula o(1) b m + 2 v() m+2 b ρ + β v() 2+ρ+β + o(1) = 0 a. This implies ha lim n v( ) = [(m β α)/(m + 2)] 1/(β α). Since m can be moved arbirarily, his is an obvious conradicion. Therefore lim inf s v > 0. We are now in a posiion o show lim s v(s) = 1. Since lim inf s v(s) > 0, we find ha lim inf u()/u 0 () > 0. Inegraing equaion (3.2) (wih q replaced by p), we furher find ha lim inf u ()/u 0() > 0. Since v + v = u ()/u 0(), we

12 12 K. KAMO, H. USAMI EJDE-2007/126 obain lim inf s (v + v) > 0. Recalling equaion (4.1), we find ha v(s) = O(1) as s. Since we have already shown ha v L 2 [, ), Lemma 2.1 shows ha lim s v = 0. Consequenly, as in he proof of Theorem 1.2 we can prove ha lim s v(s) = 1. This complees he proof of Theorem 1.3. References [1] R. Bellman, Sabiliy heory of differenial equaions, MaGraw-Hill, New York, (Reprin: Dover Publicaions, Inc., 1969) [2] Á. Elber and T. Kusano, Oscillaion and non-oscillaion heorems for a class of second order quasilinear differenial equaions, Aca Mah. Hung. 56 (1990), [3] I. T. Kiguradze and T. A. Chanuria, Asympoic properies of soluions of nonauonomous ordinary differenial equaions, Kluwer Academic Publisher, Dordrech, [4] M. Kiano and T. Kusano, On a class of second order quasilinear ordinary differenial equaions, Hiroshima Mah. J. 25 (1995), [5] R. A. Moore and Z. Nehari, Nonoscillaion heorems for a class of nonlinear differenial equaions, Trans. Amer. Mah. Soc. 93 (1959), [6] K. Nishihara, Asympoic behaviors of soluions of second order differenial equaions, J. Mah. Anal. Appl. 189 (1995), Ken-ichi Kamo Division of Mahemaics, School of Medicine, Liberal Ars and Sciences, Sapporo Medical Universiy, S1W17, Chuoku, Sapporo , Japan address: kamo@sapmed.ac.jp Hiroyuki Usami Deparmen of Mahemaics, Graduae School of Science, Hiroshima Universiy, Higashi - Hiroshima, , Japan address: usami@mis.hiroshima-u.ac.jp