732 Caroline St. Port Angeles, WA Shimane University Matsue, Japan

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1 FIXED POINTS AND PROBLEMS IN STABILITY THEORY FOR ORDINARY AND FUNCTIONAL DIFFERENTIAL EQUATIONS T.A. Buron 1 and Tesuo Furumochi 2 1 Norhwes Research Insiue 732 Caroline S. Por Angeles, WA Deparmen of Mahemaics Shimane Universiy Masue, Japan ABSTRACT. In his paper we begin a sudy of sabiliy heory for ordinary and funcional differenial equaions by means of fixed poin heory. The paper is moivaed by a number of difficulies encounered in he sudy of sabiliy by means of Liapunov s direc mehod. We noice ha hese difficulies frequenly vanish when we apply fixed poin heory. This sudy is mainly resriced o conracion mappings. AMS (MOS) subjec classificaion. 34D2, 34K2, 47H1 1. INTRODUCTION Several problems in differenial equaions have come o ligh, mainly in he sudy of Liapunov s direc mehod. In a ypical problem we have a sysem of funcional differenial equaions (1.1) x = F(,x ), ogeher wih a Liapunov funcional V (,x ) and coninuous increasing funcions W i, saisfying a relaion (1.2) W 1 ( x() ) V (,x ) Typese by AMS-TEX

2 and he derivaive of V along a soluion of he differenial equaion saisfying (1.3) V (,x ) p()w 2 ( x() ) wih p(). The objecive is o inegrae V and show ha if he soluion does no go o zero, hen V ends o, a conradicion. Five fundamenal difficulies occur. Firs, i is an ar o consruc a Liapunov funcional saisfying (1.2) and (1.3). Second, condiions need o be placed on p() o accomplish he objecive and hese usually require ha p be posiive mos of he ime; hese are discussed in [4], [1], and [14], for example. Third, in so much of he work we need o ask ha he righ hand side of he differenial equaion be bounded when he space variable is bounded; his is discussed in deail in [1], [3; p. 231], [5], [6], [9] and [13]. Fourh, so ofen in Liapunov heory, and paricularly in Razumikhin heory, when we consider an equaion like x = a()x + b()x( h) here arises a poinwise comparison beween a and b; such discussions are found in [3; p. 265], [5], [6], [1], [18], for example. Ye differenial equaions have soluions expressed as inegral equaions and we invesigae if he comparison should no be averaged insead of being poinwise. This paper is dedicaed o a collecion of examples in which hese problems arise. We show ha all of hem can frequenly be avoided using fixed poin heory, paricularly in conjuncion wih Liapunov heory. In Par I we consider problems which are half linear and are sudied in par by a variaion of parameers formula. Tha formula enables us o se up proper fixed poin mappings in such a simple way. Par II considers fully nonlinear problems which require ad hoc echniques for he mappings. Liu [12] considers many nonlinear variaion of parameer problems which migh be very useful in ha connecion, bu we have no sudied ha. Sabiliy definiions may be found in [3], [7], and [17], for example. PART I: HALF-LINEAR EXAMPLES 2. AN ORDINARY DIFFERENTIAL EQUATION One of he enduring problems in sabiliy heory is o deermine sabiliy properies of he zero soluion of equaions sharing properies of (2.1) x = x 3 [/2 + sin n ]x =: f(,x) where n 1 is an odd posiive ineger. (i) The linear par is asympoically sable (AS), bu no UAS. Hence, AS of he zero soluion of (2.1) does no follow from ha of he linear par. (ii) Concerning Liapunov s direc mehod:

3 a) For each posiive ɛ < 1 here exis sequences n and s n, n < s n < n+1 and f( n,ɛ) <,f(s n,ɛ/2) >. b) f(,x) is no bounded for x bounded. These are he properies ha keep us from using Liapunov funcions which are no decrescen. We now show his may no be a difficuly when we use fixed poin echniques. By he variaion of parameers formula we wrie (2.1) as (2.2) x() = x e 2 /4 + Noice ha e (1/4)( 2 +s 2) [ sx(s)sin n s + x 3 (s)]ds. e (1/4)( 2 +s 2) sds = 2e (1/4)( 2) [e (1/4)2 1] 2. Hence, if < α < 1, we can find an n so large ha (2.3) e (1/4)( 2 +s 2) ssin n s ds α. Le a >,. be he supremum norm, and le (2.4) S = {φ : [, ) R φ() = x, φ a,φ C,φ() as }. The se S is complee. Firs, i is conained in he Banach space of bounded coninuous funcions wih he supremum norm; hus, any Cauchy sequence φ n has a coninuous limi φ. As he φ n () as we can prove ha φ() and φ a. Define P : S S by (2.5) (Pφ)() = x e (1/4)2 + e (1/4)( 2 +s 2) [ sφ(s)sin n s + φ 3 (s)]ds. THEOREM 2.1. If (2.3) holds hen he zero soluion of (2.1) is asympoically sable. Proof. We will give he deails for =, bu general offers no new difficulies. LEMMA 1. There exiss a > such ha P : S S; ha is, (Pφ)() is coninuous, (Pφ)() as, (Pφ)() a, (Pφ)() = x. Proof. (i) Clearly, (Pφ)() is coninuous.

4 Nex, we show in wo seps ha (Pφ)() as. (ii) Noe ha for some M >. Hence e (1/4)( 2 +s 2) = e (1/4)( s)(+s) e (1/4)( s)+m e M e (1/4)( s) φ 3 (s)ds is he convoluion of an L 1 funcion wih a funcion ending o, so he inegral ends o zero. (iii) Now, for each ɛ (,1), for each a (ɛ,1), and for each φ S, here is a 1 > such ha: a) For 1 hen φ() < ɛ/4. b) And 2ae (3/4)2 1 < ɛ/2. Thus, if 2 1 we have e (1/4)2 1 e (1/4)s2 s φ(s) ds =e (1/4)2 [ e (1/4)s2 s φ(s) ds + e (1/4)s 2 s φ(s) ds] 1 2ae (1/4)2 e (1/4)2 1 + (ɛ/4) 2 s 2) sds 2ae (1/4)((2 1) ) + (1/2)ɛ <ɛ. Hence, (Pφ)() as. (iv) We now deermine a so ha Pφ a. We have for he α < 1 of (2.3). Pφ x + x + αa + 2a 3 1 e (1/4)( e (1/4)( 2 +s 2) [ ssin n s a + a 3 ]ds Hence, for a small enough and x small we have Pφ a. NOTICE. Once we find an a which works, we can obain smaller and smaller such a. In he ɛ δ definiion of sabiliy, for a given ɛ, we will find a < ɛ and hen find x small enough ha he mappings work. This value of x will deermine δ. Noice also ha for 1 e (1/4)( 2 +s 2) ds 1 e (1/4)( 2 +s 2) ds + 1 e (1/4)( 2 +s 2) sds

5 which is bounded (a condiion needed for sabiliy). LEMMA 2. If a is small enough, hen P is a conracion. Proof. We have (Pφ)() (Pψ)() e (1/4)( 2 +s 2) [ ssin n s φ ψ + φ ψ 2( φ 2 + ψ 2 )]ds α φ ψ + 2 π φ ψ ( φ 2 + ψ 2 ) β φ ψ for some β < 1 if a is small enough. Here, we used ables o evaluae he inegral and obain π. Now, o finish he proof, given ɛ > we can find a < ɛ and δ > so ha x < δ makes Lemma 1 rue. Then P has a unique fixed poin φ and ha is he soluion ending o zero. Hence, x = is asympoically sable. 3. INDEFINITE a AND UNBOUNDED r IN A DELAY EQUATION We now give a simple example which so quickly shows how o achieve some of our saed goals. In Hale [7; p. 18] is considered he classical problem x () = a()x() b()x( r) where a and b are bounded coninuous funcions, Using he Liapunov funcional a() δ >, b() θδ,θ < 1. V (,x ) = (1/2)(x 2 () + δ wih he riangle inequaliy we have r x 2 (s)ds) V (δ/2)(θ 1)(x 2 () + x 2 ( r)). I can now be argued ha his yields UAS. I is severe o ask ha a,b be bounded and ha b() is bounded by a all of he ime. Les ry o improve ha wih fixed poin heory. The half-linear equaion (3.1) x () = a()x() + b()g(x( r()))

6 presens severe challenges when we aemp o show ha soluions end o using Liapunov funcionals. We are ineresed in he case where a can be negaive some of he ime, a and b are relaed on average, boh a and r can be unbounded. The problem of boundedness of r is discussed in Buron-Havani [4], Knyazhishche- Shcheglov [1], and Yoshizawa [18], for example. Seifer [15] poins ou he need for r() o end o. Here, we ask ha a, b, and r be coninuous, ha (3.2) a(s)ds as, (3.3) e R s a(u)du b(s) ds α < 1,, (3.4) r(), r() as, here is an L > so ha if x, y L hen (3.5) g() = and g(x) g(y) x y. THEOREM 3.1. If (3.2)-(3.5) hold, hen every soluion of (3.1) wih small coninuous iniial funcion ends o as. Moreover, he zero soluion is sable a =. Proof. For he α and L, find δ > wih δ + αl L. Le ψ : (,] R be a given coninuous funcion wih ψ() < δ and le S = {φ : R R φ L,φ() = ψ() if,φ() as,φ C} where is he supremum norm. and Define P : S S by (Pφ)() = ψ() if (Pφ)() = e R a(s)ds ψ() + e R s a(u)du b(s)g(φ(s r(s)))ds,. Clearly, Pφ C. We now show ha (Pφ)() as. Le φ S and ɛ > be given. Then φ L, here exiss 1 > wih φ( r()) < ɛ if 1, and here exiss 2 > 1 such ha > 2 implies ha e R 1 a(u)du < ɛ/(lα). Then > 2 implies ha e R s a(u)du b(s)g(φ(s r(s)))ds

7 1 e R s a(u)du b(s) Lds + e R s a(u)du b(s) ɛds 1 e R 1 a(u)du 1 e R 1 s αle R 1 a(u)du + αɛ ɛ + αɛ. a(u)du b(s) Lds + αɛ To see ha P is a conracion under he supremum norm, if φ,η S, hen (Pφ)() (Pη)() wih α < 1 by (3.3). e R s a(u)du b(s) φ η ds α φ η Hence, for each such iniial funcion, P has a unique fixed poin in S which solves (3.1) and ends o. To ge sabiliy for soluions saring a =, le ɛ > be given and do he above work for L = ɛ. 4. RELATIONS BETWEEN a, b, AND r Consider he scalar equaion (4.1) x = a()x + where r() r. We suppose ha r() b(, s)g(x(s))ds (4.2) here exiss α < 1 wih e R s a(u)du s s r(s) b(s,u) duds α. Also, for each ɛ > here exis 1 > and T > such ha 2 1 and 2 + T imply ha (4.3) e R 2 a(s)ds < ɛ and e R a(s)ds as. Finally, here is an L > such ha x, y L imply ha (4.4) g(x) g(y) x y and g() =. THEOREM 4.1. If (4.2) and (4.3) hold hen he zero soluion of (4.1) is asympoically sable a =.

8 Proof. For he L and α find δ > so ha δ + αl L. Le ψ : [ r,] R be a given coninuous iniial funcion wih ψ() < δ and define S = {φ : [ r, ) R φ L,φ C,φ = ψ,φ() as } where denoes he supremum norm. and Define P : S S by (Pφ)() = e R a(s)ds ψ() + (Pφ)() = ψ() if r e R s a(u)du s s r(s) b(s, u)g(φ(u))duds,. Now, le ɛ > and φ S be given. Then φ L and here exiss 1 > such ha 1 r implies ha φ() < ɛ, while 2 1 and 2 + T imply ha e R a(s)ds 2 < ɛ. Thus, by he second par of (4.3) we may suppose 1 so large ha e R 1 a(s)ds < ɛ. Hence, 2 + T implies ha (Pφ)() 2 δɛ + e R 2 s a(u)du e R s a(u)du 2 b(s, u) Lduds s r(s) + e R s s a(u)du b(s, u) ɛds 2 s r(s) δɛ + ɛαl + ɛα. Thus, (Pφ)() as. Also, φ L implies Pφ L by choice of δ. Finally, (Pφ)() (Pη)() Hence, P has a unique fixed poin in S. e R s a(u)du s α φ η. s r(s) b(s, u) duds φ η 5. A SCALAR NEUTRAL EQUATION We now consider problems in which he ode par is no necessarily sable and he delay par will be used o sabilize i. Such work wih Liapunov funcions is found, for example, in Buron [2; p. 139]. The idea is o conver o a neural equaion. Here, he size of he delay will cerainly play a role.

9 The reader can confirm ha he work we do here also carries hrough on a nonlinear equaion x () = a()x() + b()x( r) + h()g(x( r)) where g saisfies a local Lipschiz condiion, as in previous secions. Bu in order o no obscure he main par, we consider he scalar equaion x () = a()x() + b()x( r) = [ a() + b( + r)]x() (d/d) b(s + r)x(s)ds r (5.1) =: A()x() (d/d) b(s + r)x(s)ds. r I is assumed ha (5.2) [ a(s) + b(s + r)]ds as, (5.3) r b(u + r) du + A(s)e R s s A(u)du b(u + r) duds α < 1, s r and ha whenever φ() as, hen (5.4) A(s)e R s A(u)du s s r b(u + r)φ(u)duds as. EXAMPLE 5.1. Le a() =,b() = b < a consan. Then (5.2) and (5.4) are immediaely saisfied, while (5.3) needs 2br < 1. THEOREM 5.1. If (5.2)-(5.4) hold, hen for every coninuous ψ : [ r,] R, he soluion x(,,ψ) as. Proof. Le S = {φ : [ r, ) R φ C,φ = ψ,φ() as }. We can inegrae by pars and wrie he soluion of (6.1) as x() = e R A(s)ds ψ() e R s s A(u)du (d/ds) b(u + r)x(u)duds s r = e R A(s)ds ψ() e R s A(u)du s s r b(u + r)x(u)du

10 A(s)e R s s A(u)du b(u + r)x(u)duds. s r = e R A(s)ds ψ() b(u + r)x(u)du + e R A(u)du b(u + r)ψ(u)du r r A(s)e R s s A(u)du b(u + r)x(u)duds. s r Now, define P : S S by (Pφ)() = ψ() if r and for hen (Pφ)() = e R A(s)ds ψ() b(u + r)φ(u)du + e R A(u)du b(u + r)ψ(u)du r r A(s)e R s s A(u)du b(u + r)φ(u)duds. s r By (5.2) and (5.4), (Pφ)() as. To see ha P is a conracion, if φ,η S hen (Pφ)() (Pη)() + r b(u + r) φ η du A(s)e R s s A(u)du b(u + r) duds φ η s r α φ η by (5.3). Hence, P is a conracion and here is a unique soluion ending o zero. 6. AN ENTIRELY DELAYED EQUATION Here, we consider (6.1) x = b()g(x( h()) where b(), h() r, and 1 h () > so ha he funcion defined by h() is sricly increasing and has an inverse r(). We can wrie (6.1) as (6.2) or x = [b(r())/(1 h (r()))]g(x) + (d/d) (6.3) x = c()g(x) + (d/d) h() h() [b(r(s))g(x(s))/(1 h (r(s))]ds c(s)g(x(s))ds. We suppose ha (6.4) g(x) = x + G(x), lim x [G(x)/x] =,c().

11 (Wih considerable care, we can reduce c().) For a given iniial funcion ψ : [ r,] R, he soluion is given by x() = e R c(s)ds ψ() + e R s s c(u)du [ c(s)g(x(s)) + (d/ds) c(u)g(x(u))du]ds. s h(s) Inegraing ha derivaive by pars will yield x() = e R c(s)ds ψ() e R s c(u)du c(s)g(x(s))ds + c(u)g(x(u))du e R c(u)du c(u)g(x(u))du h() h() c(s)e R s c(u)du s s h(s) c(u)g(x(u))duds. We suppose ha here is a δ > and γ > so ha if (6.5) g = max x γ g(x),g = max x γ G(x) (his is no needed in he linear case) hen δ + G + g h() c(u)du + g h() c(u)du (6.6) +g (This, oo, is no needed in he linear case.) e R s s c(u)du c(s) c(u)duds < γ. s h(s) EXAMPLE 6.1. If g(x) = x hen g = γ and G =. If h = 1, hen he inverse of 1 is r() = + 1. We hen need δ + γ 1 c(u)du + γ 1 c(u)du + γ If we ake c o be a posiive consan, his reduces o c < 1/3. e R s s c(u)du c(s) c(u)duds < γ. s 1 Two more condiions are needed. Firs, here are λ, µ > such ha (6.7) x, y γ imply g(x) g(y) µ x y, G(x) G(y) λ x y.

12 Also, (6.8) λ + µ c(u)du + µ e R s s c(u)du c(s) c(u)duds α < 1. h() s h(s) EXAMPLE 6.2. If c is consan, g(x) = x, and h = 1 hen λ =,µ = 1, and (6.8) is saisfied if c < 1/2. THEOREM 6.1. Le (6.4)-(6.8) hold and c(s)ds as. Then ψ < δ implies ha x(,,ψ) as. Proof. For he given ψ wih ψ < δ define S = {φ : [ r, ) R φ C,φ(),φ = ψ, φ γ}. Then define P : S S by φ S implies ha (Pφ)() = ψ() if r and for hen (Pφ)() = e R c(s)ds ψ() e R s c(u)du c(s)g(φ(s))ds + c(u)g(φ(u))du e R c(u)du c(u)g(ψ(u))du h() h() e R s s c(u)du c(s) c(u)g(φ(u))duds. s h(s) I is easy o argue ha if φ() as and (6.6) holds hen (Pφ)() as. Also, (6.6) shows ha φ S implies ha Pφ γ. To see ha P is a conracion, (Pφ)() (Pη)() λ φ η + µ c(u)du φ η h() by (6.8). +µ φ η e R s s c(u)du c(s) c(u)duds s h(s) α φ η REMARK 6.1. This problem has been widely discussed and much is known abou i. Noe firs ha our condiions can hold even when b() is zero over arbirarily long ime periods, so long as i has an infinie inegral. By conras, sandard resuls usually ask ha b be inegrally posiive in some sense. On he oher hand, some of he bes known resuls ask ha b s := sup h +h b(u + h) du < 3/2.

13 These are discussed in some deail in Knyazhishche-Shcheglov [1] which uses sophisicaed Liapunov heory. Their resul asks ha b s ln 4, which is much beer han ours, bu also requires a very complicaed inegrally posiive condiion which requires a separae heorem o verify. Oher reamen of (6.1) by Liapunov heory is found in Buron-Havani [4; pp. 68 and 79] where i is required ha b be inegrally posiive. Furher resuls are found in Yoneyama [16], for example. 7. A d-dimensional NEUTRAL VOLTERRA EQUATION We acually wan soluions o be L 1 [, ), bu he se S which we have been consrucing would no be complee if we asked for such funcions. However, we will go abou maers in a very indirec manner here. Consider he d-dimensional sysem (7.1) x = Ax + where (7.2) C, D, We le (7.3) Q = A + Then C( s)x(s)ds + D( s)x(s)ds C(v)dv L 1 [, ) and are coninuous. C(v)dv and assume ha all roos of Q have negaive real pars. (7.4) x = Qx (d/d) C(v)dvx(s)ds + D( s)x(s)ds s and s x() = e Q x e Q( s) (d/ds) C(v)dvx(u)duds s u s (7.5) + e Q( s) D(s u)x(u)duds. We define S as before (wihou a norm resricion since (7.1) is linear) and define P : S S hrough (7.5) bu inegrae by pars o ge (Pφ)() = e Q x e Q( s) s s u C(v)dvφ(u)du

14 or + Qe Q( s) s s u C(v)dvφ(u)duds s e Q( s) D(s u)φ(u)duds (Pφ)() = e Q x u C(v)dvφ(u)du (7.6) Qe Q( s) s s u C(v)dvφ(u)duds + s e Q( s) D(s u)φ(u)duds. Now P will be a conracion if (Pφ)() (Pη)() [ C(v) dvdu u s + Qe Q( s) C(v)dv duds s u s + e Q( s) D(s u) duds] φ η (7.7) α φ η,α < 1. THEOREM 7.1. If (7.2), (7.3), and (7.7) hold hen every soluion x(,,x ) of (1) ends o as. 8. A SEARCH FOR NORMS In his secion we approach Secion 7 differenly. Consider he sysem z = Az + B( s)z(s)ds, where A is consan, B(s) ds <, and Z() is he n n marix of soluions wih Z() = I. I is known (cf. Buron [2; pp ]) ha z = is UAS if and only if Z L 1 [, ). Quesion: Wha hen is he appropriae norm for soluions? 1. When soluions are US, he supremum norm seems righ. 2. When soluions end o, his should be refleced in he norm or in he space.

15 3. When Z L 1, he norm should include z(s) ds. Thus, we now le wih norm S = {φ : [, ) R d φ C, φ(u) du <,φ() } φ = sup φ() + φ(u) du =: φ + φ 1. < This space is complee. We now reurn o he work from he previous problem and sudy i from a differen poin of view. Consider he d-dimensional sysem (8.1) x = Ax + where (8.2) C, D, We le (8.3) Q = A + C( s)x(s)ds + D( s)x(s)ds C(v)dv L 1 [, ) and are coninuous. C(v)dv and assume ha all roos of Q have negaive real pars. Then exacly as in he las secion we wrie (8.1) as an inegral equaion, inegrae by pars, and define P : S S by (Pφ)() = e Q x u C(v)dvφ(u)du (8.6) Qe Q( s) s s u C(v)dvφ(u)duds + s e Q( s) D(s u)φ(u)duds. REMARK 8.2. Noe ha if φ() as, hen each inegral is he convoluion of an L 1 - funcion wih a funcion ending o. (Do he middle inegral wice.) Hence, each inegral ends o. In exacly he same way, if φ L 1 hen we have Pφ L 1. Now P : S S independen of any norm. Under he sup norm, S is complee and P will be a conracion if (Pφ)() (Pη)() [ C(v) dvdu u

16 s + Qe Q( s) C(v)dv duds s u s + e Q( s) D(s u) duds] φ η (8.7) α φ η,α < 1. And if i is a conracion, hen i has a unique fixed poin in S and ha fixed poin is a soluion of (8.1) which is inegrable. THEOREM 8.1. If (8.2), (8.3), and (8.7) hold hen every soluion x(,,x ) of (8.1) is in L 1 [, ) and so he zero soluion is UAS. 9. ULTIMATE BOUNDEDNESS Ulimae boundedness using Liapunov funcionals has been a very elusive problem. The reader is referred o Yoshizawa [17] for definiions of uniform boundedness (UB) and uniform ulimae boundedness (UUB). Many ypical deails are found in Buron [3; pp ]. Very special condiions on he Liapunov funcionals are required and Hale [7; p. 139] declines o discuss hem because of severe resricions on he size of he delay. If we consider (9.1) x = a()x + b()g(x( h)) + f() wih a() b( + h) and f() m, g(x) r x, and a() α >, hen he naural Liapunov funcional will no work unless a, b, f are bounded. The usual argumen goes like his. Define V (,x ) = x 2 () + b(s + h) g 2 (x(s))ds h so ha V = 2a()x 2 + 2b()xg(x( h)) + 2f()x + b( + h) g 2 (x) b() g 2 (x( h)) and we need [ 2a() + b() + m 2 + r 2 b( + h) ]x 2 + f 2 ()/m 2 (C) 2a() b() + r 2 b( + h) + m 2 + γ where m, γ are posiive consans. Because of he assumed boundedness of b and f we arrive a he sysem W 1 ( x() ) V (,x ) W 2 ( x() + h W 3 ( x(s) )ds)

17 V W 3 ( x() ) + K for appropriae K > and wedges W i. Because of he very special relaion wih W 3 in boh he upper bound and in he derivaive, his will readily yield UB and UUB (Buron [3; p. 278]). We now use a conracion o relax (C). In (9.1), suppose ha here are posiive consans M, L, K, µ wih (9.2) e R s a(u)du f(s) ds M, (Thus, on average f is dominaed by a.) (9.3) e R s a(u)du b(s) ds L, (And on average b is also dominaed by a.) (9.4) g(x) g(y) µ x y,g() =,µlk + M + 1 < K, (This implies ha µl < 1.) and for each 1 > and ɛ > here exiss 2 > 1 such ha > 2 implies ha (9.5) e R a(s)ds < ɛ and e R 1 a(u)du < ɛ. (Thus, a is posiive on average.) THEOREM 9.1. If (9.2)-(9.5) hold, hen every soluion of (9.1) saisfies x(,,ψ) < K for all large. Proof. Le ψ : [ h,] R be a given coninuous iniial funcion and define S = {φ : [ h, ) R φ = ψ,φ C, φ() < K for large }. Nex define P : S S by (Pφ)() = ψ() if < (Pφ)() = ψ()e R a(s)ds + e R s a(u)du [b(s)g(φ(s h)) + f(s)]ds,. Le φ S be given. Then here exiss J such ha φ J and here exiss 1 such ha 1 h implies ha φ() K. Selec ɛ > so ha φ() ɛ + LµJɛ < 1/2. Nex, find 2 and le 2 > 1 so ha we will have (Pφ)() φ() e R a(s)ds + M 1 + e R 1 s a(u)du e R a(u)du 1 b(s) µjds + e R s a(u)du b(s) Kµds 1

18 φ() ɛ + M + LµJɛ + KµL < K (1/2). Thus, P : S S. In fac, i is mapped ino a closed subse of S. We will show in a momen ha P is a conracion. Now S may no be complee because a Cauchy sequence may have a limi φ wih φ() K for large. However, S is mapped ino a complee subse of iself. To see his, refer o a sandard proof of he conracion mapping heorem; if φ S, consider he sequence {P k φ} which we have jus shown o be in S and (P k φ)() < K (1/2) for large. This is a Cauchy sequence and i is in he Banach space of bounded coninuous funcions wih he supremum norm so i has a limi η and η() < K (1/4) for large. Tha limi is he fixed poin. Now (Pφ)() (Pη)() e R s a(u)du b(s) µds φ η µl φ η, a conracion since µl < 1. This example shows ha for his half-linear equaion, he classical resricion (C) can be significanly improved using fixed poin heory. PART II: NONLINEAR EXAMPLES The variaion of parameers formula does wo hings for us. I allows us o define a mapping so ha he image of a funcion has he correc iniial condiion and we can prove ha he image ends o zero as. One of he classical fully nonlinear examples in Liapunov heory may be found in Hale [7; pp ] and i concerns (E) x () = a()x 3 () + b()x 3 ( h) wih a and b bounded coninuous funcions saisfying An appropriae Liapunov funcional is a() δ <, b() < qδ, < q < 1. V (,x ) = (1/4)x 4 () + (δ/2) x 6 (s)ds h so ha V (,x ) (δ/2)(q 1)(x 6 () + x 6 ( h)). From hese relaions and Marachkoff s argumen one can prove ha he zero soluion is uniformly asympoically sable. We will now perurb (E) in several differen ways and see wha can be proved using fixed poin heory.

19 1. A NONLINEAR ODE Consider he scalar equaion (1.1) x () = a()x 3 + g(,x 2,x) where and a(), a()d =,g(,y,) =, (1.2) g(,y,x) g(,y,w) b() y x w. Now we come o a recurring problem. We require ha for each bounded coninuous funcion z 2 () wih z 2 () c for some c >, here is an α < 1 wih e R s a(u)z2 (u)du b(s)z 2 (s)ds α,. An obvious sufficien condiion is ha a() kb() and a() for all and some k > 1. Bu ha is much oo severe. We would like for a o be zero on long inervals when b is nonzero. Bu here we come o a real difficuly. In hese fully nonlinear problems we will use he unknown exac soluion as par of he mapping. Thus, we need o rely on addiional informaion o ensure ha soluions exis on [, ). An example of (1.1) is x = a()x 3 + b()x 3. If a() < b() and b() > on any inerval [, 1 ], however shor, here are soluions wih finie escape ime. Hence, i will be necessary o work wih paricular iniial imes for which ( ) [ a(s) + b(s)]ds for all. Obviously, if a() b() for all, his would hold for any. LEMMA 3. Le (*) hold and le x R. Then x(,,x ) is defined for all. Proof. Le x() = x(,,x ) be a soluion of (1.1) wih maximal inerval of definiion [, 1 ). I is known ha 1 = or lim 1 x() =. Then define a Liapunov funcion V (x) = x so ha along he soluion wih have V (x()) a() x() 3 + b() x() 3 = [ a() + b()]v 3.

20 If we separae variables and inegrae we obain V () 2 + V ( ) 2 2 [ a(s) + b(s)]ds so ha x() 3 x( ) 3, a conradicion o he finie escape ime. We now assume ha for a given x and a given c < x, here is an α < 1 such ha if z 2 () is coninuous and x 2 z2 () c, hen (1.3) e R s a(u)z2 (u)du b(s)z 2 (s)ds α for. LEMMA 4. Suppose ha if n and if ( ) K n = sup [ a(s) + b(s)]ds, hen lim K n =. n n n If x() is a soluion of (1.1) on [, ) eiher x() or here is a c > wih x() c. Proof. Suppose here is a sequence n, a sequence s n > n, and a c > wih x( n ) and x(s n ) = c. Rename indices so ha x( n ) < c/2 for all n. Using he V as in he proof of Lemma 3, and aking k n = sn we have, upon inegraion of V, he relaion n [ a(s) + b(s)]ds V 2 (s n )) + V 2 ( n )) 2k n. If some k n, hen V (s n ) V ( n ), a conradicion. Hence, k n for all n and k n K n as n so a conradicion since V ( n ). V 2 ( n )) 2k n c 2, In paricular, from his resul, for each x here is a soluion x(,,x ) =: z() defined on [, ). If we srenghen he Lipschiz condiion in (1.2) i is unique. We will now ouline he mehod o be used on hese problems. 1. We shall assume or prove ha here is a and for each x here is a unique soluion x(,,x ) =: z() on [, ), When (*) and (**) hold we have shown in he lemmas how his migh be done in a paricular problem. 2. Hence, z() is he unique soluion of (1.4) x = a()x 2 ()x + g(,z 2 (),x),x( ) = x.

21 3. PROBLEM. In wha space does z() lie? We wan o show ha i lies in (1.5) S = {φ : [, ) R φ( ) = x,φ() as,φ C}. Here, will denoe he supremum norm. The unique soluion of (1.4) is (1.6) x() = x e R a(s)z 2 (s)ds + e R s a(u)z2 (u)du g(s,z 2 (s),x(s))ds. 4. Define P : S S by (1.7) (Pφ)() = x e R a(s)z 2 (s)ds + e R s a(u)z2 (u)du g(s,z 2 (s),φ(s))ds. 5. If P has a fixed poin, i is z, and so z S which means ha z(). 6. In his example, when (*) and (**) hold we know from he lemmas ha eiher: a) z(), so here is nohing o prove, or b) z() c > so a()z 2 ()d =. Thus, we assume ha b) holds. 7. Clearly, (Pφ)( ) = x and Pφ C. We now show ha (Pφ)() as and we ake = for breviy. Le ɛ > and φ S be given and le c > be found. Find 1 so ha φ() < ɛ/2 if 1. Then using (1.3) we obain 1 e R s a(u)z2 (u)du g(s,z 2 (s),φ(s)) ds e R s a(u)z2 (u)du b(s)z 2 (s) φ(s) ds + e R s a(u)z2 (u)du b(s)z 2 (s) φ(s) ds 1 e R 1 a(u)z 2 (u)du 1 e R 1 s a(u)z2 (u)du b(s)z 2 (s) φ ds +(ɛ/2) e R s a(u)z2 (u)du b(s)z 2 (s)ds α φ e R 1 a(u)z 2 (u)du + (ɛ/2)α.

22 The firs erm ends o zero as and he second erm can be made as small as we please. 8. To see ha we have a conracion, for φ,ψ S we have (Pφ)() (Pψ)() e e R s a(u)z2 (u)du g(s,z 2 (s),φ(s)) g(s,z 2 (s),ψ(s)) ds R s a(u)z2 (u)du b(s)z 2 (s) φ(s) ψ(s) ds α φ ψ. Thus, P does have a fixed poin and i is in S. 11. A FIRST NONLINEAR DELAY EQUATION Le a and b be coninuous, h >, and consider he equaion (11.1) x () = a()x 3 () + b()x( h)x 2 () This is a difficul problem even if we ask ha a() > b() since b()x( h)x 2 () can dominae a()x 3 () along a soluion in a neighborhood of zero. Bu under ha condiion i is possible o find a mildly unbounded Liapunov funcion which will show ha all soluions can be coninued for all fuure ime, and ha is crucial for he fixed poin mehod we use. In his problem any will work and we simply le =. Here are he deails. LEMMA 5. Le ψ : [ h,] R be a given coninuous iniial funcion and le z() be he unique soluion deermined by ψ. Then z exiss on [, ). Proof. The condiions of he example ensure ha here is a unique local soluion. I is well known ha he only way in which a soluion can fail o be defined for all fuure ime is for here o exis T > wih lim sup T z() = +. In paricular, hen z( h) is a bounded coninuous funcion on [,T]. Now a() > so here is A > wih a() > A on [,T] and here is a B > wih B > b() z( h) on ha same inerval. If we define V (,x) = (1 + x )e K hen he derivaive of V along any soluion x() of (11.1) saisfies V e K [ a() x 3 + b()x( h) x 2 K K x ] e K [ A x 3 + Bx 2 K] for large enough K. Thus, V (,x()) is bounded on [,T] and so i can no happen ha limsup T x() =.

23 There will, of course, be problems in which we will be unable o find such a Liapunov funcion. In such cases we mus sae ha soluions defined for all fuure ime end o zero. We are no going o do he work of Lemma 4 here. Insead, for breviy we ask ha (11.2) b() ka(),k < 1. THEOREM Le (11.2) hold and le a() c >. For a given iniial funcion η : [ h, ] R, eiher: a) x 2 (,,η) L 1 [, ) or b) x(,,η) as. Proof. We le z() be he unique soluion of (11.1) and consider (11.3) x = a()z 2 ()x + b()z 2 ()x( h) so ha he soluion can be wrien as x() = η()e R a(s)z2 (s)ds + Then we define our se S by e R s a(u)z2 (u)du b(s)z 2 (s)x(s h)ds. S = {φ : [ h, ) R φ() = η() on [ h,],φ() as,φ C} and he mapping P (which will map S S if a) fails) by (Pφ)() = η() if h and (Pφ)() = η()e R a(s)z2 (s)ds + e R s a(u)z2 (u)du b(s)z 2 (s)dsφ(s h)ds for. Our condiion (11.2) readily yields ha his is a conracion. Now, if z 2 is no in L 1 [, ), hen a()z 2 ()d = and we can verify ha P : S S. Hence, here is a unique fixed poin in S, which is he unique soluion z, and i goes o zero. This complees he proof. REMARK For compleeness of our discussion of (E), consider (11.4) x () = a()x 3 () + b()x 2 ( h)x. We assume ha (11.5) a() c > and ha (11.6) a( h) k b(),k > 1.

24 From he argumen in he previous secion one readily concludes ha for each coninuous iniial funcion, a unique soluion exiss on [, ). The nex resul is proved direcly from he variaion of parameers formula. THEOREM Le (11.5) and (11.6) hold. For a given coninuous η : [ h,] R eiher: a) x 2 (,,η) L 1 [, ) or b) x(,,η) as. 12. EQUATION (E), ITSELF We reurn o (E) which we wrie as (12.1) x = a()x 3 () + b()x 3 ( h) wih a view o showing ha soluions end o wihou requiring ha a() be bounded. For breviy we ask ha (12.2) b( + h) γa(),a() >,γ < 1. (Wih more work we could ask ha here is a wih [ a(s)+ b(s+h) ]ds.) Under his assumpion we can use he Liapunov funcion V (,x ) = x + h b(s + h) x 3 (s) ds and obain V [ a() + b( + h) ] x 3 which yields sabiliy of he zero soluion. This means ha if z() is he unique soluion of (12.1) wih coninuous iniial funcion η : [ h,] R, we can make z as small as we please by aking η small. Thus, we ask ha here exiss M > wih (12.3) h b(u + h) du M, hen find β > wih (12.4) 2β + γ =: α < 1, and finally ake η so small ha (12.5) h b(u + h) x 2 (u,,η)du β.

25 While his does place an inegral bound on b, here is no bound on a. THEOREM Le (12.2) o (12.5) hold wih η seleced and le a() c >. Eiher: a) x 2 (,,η) L 1 [, ) or b) x(,,η) as. Proof. Le z() = x(,,η) which is he unique soluion of (12.6) x = a()z 2 ()x + b()z 2 ( h)x( h),x = η. Define S = {φ : [ h, ) R φ() = η() on [ h,],φ C,φ() as } and define (Pφ)() = η() for h and for define (Pφ)() = η()e R a(s)z2 (s)ds + e R s a(u)z2 (u)du b(s)z 2 (s h)φ(s h)ds. If a) fails, hen a(s)z2 (s)ds and we can show ha P : S S. Now i seems o require a rick from Liapunov heory o show ha we have a conracion. Noe ha b(s)z 2 (s h)φ(s h) Hence, = b(s + h)z 2 (s)φ(s) b(s + h)z 2 (s)φ(s) + b(s)z 2 (s h)φ(s h) = b(s + h)z 2 (s)φ(s) (d/ds) s s h b(u + h)z 2 (u)φ(u)du. (Pφ)() = η()e R a(s)z2 (s)ds + e R s a(u)z2 (u)du b(s + h)z 2 (s)φ(s)ds e R s s a(u)z2 (u)du (d/ds) b(u + h)z 2 (u)φ(u)du. s h We inegrae ha las erm by pars o obain e R s s a(u)z2 (u)du s h b(u + h)z 2 (u)φ(u)du + e R s s a(u)z2 (u)du a(s)z 2 (s) b(u + h)z 2 (u)φ(u)duds s h

26 = b(u + h)z 2 (u)φ(u)du + e R a(u)z2 (u)du b(u + h)z 2 (u)η(u)du h h + e R s s a(u)z2 (u)du a(s)z 2 (s) b(u + h)z 2 (u)φ(u)duds. s h Now, i follows readily ha we have a conracion. 13. TWO QUICK EXAMPLES We wan o cover he obvious perurbaions of (E) so we look a wo of hose which are very simple. Firs, le a and b be coninuous and consider (13.1) x () = a()x 2 ( h)x + b()x 3 ( h). I is readily argued ha soluions can be defined for all fuure ime, regardless of he signs of he funcions. Le η : [ h,] R be a given coninuous iniial funcion and le z() = x(,,η) which is he unique soluion of (13.2) x = a()z 2 ( h)x + b()z 2 ( h)x( h),x = η. Define S as in earlier secions and define (Pφ)() = η()e R a(s)z2 (s h)ds + e R s a(u)z2 (u h)du b(s)z 2 (s h)φ(s h)ds. We shall ask ha (13.3) b() k a(),k < 1. THEOREM If (13.3) holds and if a() c >, eiher x 2 (,,η) L 1 [, ) or x(,,η) as. Finally, le a() > be a coninuous funcion and consider x = a()x( h)x 2 (). If a() c >, hen soluions can have finie escape ime. Thus, le η : [ h, ] (, ) and le z() = x(,,η). Noice ha x = [a()z( h)z()]x() is linear wih negaive coefficien so he soluion decreases monoonically. I is never zero by he uniqueness heorem. We have x() = η()e R a(s)z(s h)z(s)ds,

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