高阶非线性脉冲微分方程解的振动性

1、仲恺农业技术学院学报,18(1):4551,2005JournalofZhongkaiUniversityofAgricultureandTechnology文章编号:1006-0774(2005)01-0045-07高阶非线性脉冲微分方程解的振动性张超龙,胡小建(仲恺农业技术学院基础部,广东广州510225)摘要:对高阶非线性脉冲微分方程解的振动性态进行了研究,得到了解振动的充分条件,对解性态的关键性影响因素进行了较深入的探讨.关键词:高阶;脉冲;振动;微分方程;非线性中图分类号:O175.1文献标识码:AOscillationsofhigherorder2long,HUXiao2jian(

2、DepartmentScience,ZhongkaiUniversityofAgricultureandTechnology,Guangzhou510225,China)Abstract:Oscillationsofhigherordernonlinearordinarydifferentialequation(ODE)withimpulsesareinves2tigated,andsomesufficientconditionsaboutoscillationsareobtained,thenotableeffectfactorsuponthebe2haviorofsolutionsarec

3、onsideredfurther.Keywords:higherorder;impulse;oscillation;nonlinear;ordinarydifferentialequation(ODE)r(t)xxx(i)(i)(2n-1)(t)+f(t,x(t)=0,tt0,ttk,(1)(i)(t+(tk),i=0,1,2n-1,k=1,2,k)=gk(i)(x(i)(t+k)=x0,i=0,1,2n-1,这里n为正整数,0<t0<t1<t2<<tk<,k=1,2,limtk=+,x(0)(t)=x(t),下面3个条k件是经常用到的:(A):f(t,x(

4、t)在t0,+)×(-,+)上连续,()()()(x)ptx0;在区间t0,+上pt(x)>0(x0),(x)0,xR.0连续且不恒为零;x(i)i)(B):在(-,+)上gk(i)(u)连续,且存在正常数a(k,bk,i=0,1,2,2n-1,满足aki()()b(ki).x(C):0<a<r(t)<M对tt0,+),r(t)在t0,+)连续,对任意i=0,1,2,2n-1,有收稿日期:2004-09-18基金项目:广东省教育厅自然科学研究项目(Z03052).作者简介:张超龙(1974-),男,湖南泸溪人,助教,硕士.46仲恺农业技术学院学报第18卷r(t

5、)t0t1dt+b1(i)(i-1)r(t)t1t2dt+b1(i-1)(i)(i)b2(i-1)r(t)dt+t2t3+记x(i)b1(i-1)(i)(i)b2(i-1)(i)a(i-1)bmr(t)dt+=+,tmtm+1(2)(-tk)=lim(i-1)h-0(i-1)()(i-1)(-)()(i-1)(+)(i)+,x(tk)=limh+0hhi=1,2,2n-1,k=1,2,.)R(t0,>0)称为方程(1)的解,若它满足方程(1)的条件,且x(i)(t)在定义1函数x:t0,t0+)处左连续.tt0,t0+定义2方程(1)的解称为非振动的,如果这个解最终为正或最终为负,否则称

6、该解为振动的,如果(1)式的所有解为振动的,则称(1)式为振动的.由于高阶非线性脉冲微分方程可化为脉冲微分方程组,)献6.在下文中,我们假定(1)式的解在T0,+(B(C,2,2n-1,存在T引理1设x(t)为方程(1)的解,且条件(A)、t0,使当tT时,x(i)(t)>0(<0),x1)(t)1T使当tT1时,有x(i-1)(t)>0(<0).)证明:括号内情形类似可证).不妨设T=t0,当t(tk,tk+1(k=0,1,2,时,有x(i)(t)>0,x(i-1)(t)0.以下分两种情况讨论:(1):若对于一切tk>T,有x(i-1)(tk)<0.

7、因为x(i+1)(t)0,故x(i)(t)>0且在(t1,t2上单调不减,+).从而当t(t1,t2时,有x(i)(t)x(i)(t1(i-1)(i-1)+(t2)x(i-1)(t1)+x(i)(t1)(t2-t1).+(t3)x(i-1)(t2)+x(i)(t2)(t3-t2).积分上式得同理可得xx(3)(4)+),由(3)、(4)式得注意到x(i)(t2)x(i)(t1+)+x(i)(t2)(t3-t2)x(i-1)(t3)x(i-1)(t2(i-1)(i-1)(i)(i)(t2)+a2b2xx(t2)(t3-t2)(i-1)()(i)(i)+)+x(i)(t1)(t2-t1)+a

8、2b2xi-1(t1x(t2)(t3-t2)b2用归纳法可得(i-1)x(i-1)(t1)+x+(i)+(t1)(t2-t1)+(i-1)x(i)(t1)(t3-t2).b2+(i)x(i-1)(i)(i-1)(i-1)(i-1)i-1)+(tm)b(t1)+x(i)(t1)(t2-t1)+b2xm-1b3(i)(i)(i)a(i-1)(t3-t2)+(i-1)(i-1)1)(tm-tm-1).b2b2b3b(mi-1(5)因为a(t1-t0)+(i)(i)(i-1)ab1a(t2-t1)+(i-1)(t3-t2)+(i-1)(i-1)(i-1)(tm-tm-1)b2b2b3bm-1dt+(i

9、-1)(i-1)r(t)b1b2(i)(i)(i)(i)r(t)dt.t0t1dt+(i-1)r(t)b1t1t2(i)(i)t2t3adt+(i-1)(i-1)1)r(t)b1b2b(mi-1(i)(i)(i)tmtm-1注意到a(ki)>0,b(ki-1)>0,由条件(C)知当m充分大时,(5)式的右端大于零.从而当m充分大时,有x(i-1)(tm)>0,与假设矛盾!所以对一切tk>T,情形(1)不可能出现.(2):若存在某个j,tj>T时,有x(i-1)(tj)0.因为x(i)(t)>0,所以x(i-1)(t)在(tj,tj+1上严格单调增加,从而当t

10、tj时,x(i-1)(t)>0.第1期张超龙,等:高阶非线性脉冲微分方程解的振动性47(B)、(C)成立,又设对某一i1,2,2n,存在Tt0,引理2设x(t)为方程(1)的解,且条件(A)、)上不恒为零,则当t充分大时,有x(i-1)当tT时,有x(t)>0,x(i)(t)0,且x(i)(t)在任何区间t,+(t)>0.证明:仅就括号外情形作证明(括号内情形类似可证)不妨设T=t0.下证对一切tkT,有x(i-1)(tk)>0.若不然,则存在某个tj,使x(i-1)(tj)0.由x(i)(t)0知当kj时,x(i-1)(t)在任何一个(tk,tk+1)上不恒为零.故存

11、在某个tltj使x(i)(t)在(tl,tl+1上单调不增,又由条件知x(i)(t)在任一区间t,+不恒为零,为方便起见,不妨设l=j,即x(i)(t)在(tj,tj+1上不恒为零,从而有x(i-1)(tj+1)<x(i-1)(t+j)a(i-1)(tj)0.(i-1)(i-1)(tj+1)<0.由归纳法可得知当t又当t(tj+1,tj+2时,有x(i-1)(t)x(i-1)(t+xj+1)aj+1(tj+m,tj+m+1时,有x(i-1)(t)<0.故在(tj+1,+)上有x(i-1)(t)<0,x(i)(t)0.由引理1可得当t充分大时,有x(i-2)(t)<

12、0.依次类推,反复应用引理1,可得当t充分大时,有xt0.这与x(t)>0(t(i-1)(tk)>0,再由x(i-1)(t,tk+知当t充分大时,有T)矛盾!故对一切的tk,有xx(i-1)(t)>0.引理2证毕.(,又设存在Tt0,当tT时,有x(t)>引理3设x(t)为方程(,(0.则存在TTl2n1,tT时,有x(i)(t)>0,i=0,1,li-1(-1)x(i)(t)>0,i=l+1,2n-1;(t)0.(6)r(t)x(2n-1)时,必有x(2n-1)(tk)>0.证明:(一)先当tk>T(k=1,2,(2n-1)(2n-1)(tj)

13、.令令S(t)=r(t)x(2n-1)(t)若不然,存在自然数j,满足tj>T,因此,x(2n-1)(t+xj)aj(2n-1)(2n-1)(t+(t),由(1)式知当t(tj+i-1,tj+i,i=1,2,r(t+j)xjj)=-(>0).由于S(t)=r(t)x有(t)=(r(t)x(2n-1)(t)-f(t,x(t)-p(t)(x(t)0.S()I)若对一切tk,有x2n-2(tk)>0,这时显然有(2n-2)x+S(tj)=-<0,t(tj,tj+1.即有(t)>0(tt0).(7)(t)0.所以S(t)在区间(tj+i-1,tj+i(i=1,2,)上单调

14、不增.故有S(t)由于S(t)=r(t)x(2n-1)(t),Sx(2n-1)(t)-r(t),t(tj,tj+1(8)(2n-1)同理有S(tj+2)S(t+S(tj+1)-<0,即j+1)aj+1S(t)-aj+1(2n-1)(2n-1)<0,t(tj+1,tj+2.(9)(t)-,t(tj+1,tj+2.故xr(t)当t(tj,tj+1时,对(8)式两边从tj到tj+1积分有:x(2n-1)(2n-1)(tj+1)x(2n-1)(t+j)-r(t)dt.tjtj+1(10)同样对(9)式从tj+1到tj+2积分得x(2n-1)(2n-1)(tj+2)x(2n-2)(t+j+1

15、)-aj+1tj+248(2n-2)仲恺农业技术学院学报第18卷(2n-2)bj+1xx(t+j)-(+tj)tjtj+1tjtj+1r(t)dt-dt-(2n-1)aj+1r(t)dttj+1tj+2tj+1tj+2(2n-2)bj+1(2n-2)-r(t)(2n-1)(2n-2)bj+1r(t)dttj+mtj+mtj+m-1.由数学归纳法可得:x(2n-1)(2n-1)(2n-1)(2n-1)(tj+m)x(2n-2)(t+j+m-1)-aj+m-1aj+m-2aj+1r(t)dtdt(2n-2)bj+m-1x(2n-2)(tj+m-1)-(2n-1)aj+m-1(2n-1)aj+m-2

16、(2n-1)aj+1r(t)tj+m-1(2n-2)bj+1(2n-2)bj+2(2n-2)bj+m-1(2n-2)(+)-r(t)tjtj+1dt-(2n-1)(2n-2)bj+1r(t)dttj+1tj+2a-(n-2)(2n-2)n-2)bj2bj+2b(j2+1+m-1(2n-1)(2n-1)(2n-1)tj+mr(t)tj+m-1dt.由条件(C)知,当m充分大时,x(2n-2)(tj+m)<0,故当t,x(22t<0.x(2n-2)(t)<0矛盾!2n-2)若有某个tj,使x(i-2)tj),x(n-2)a(>0知,当t(tj,tj+1时,j有x(2n-2)

17、(t)<x(-2)(.x(2n-2)(tj+1)<0,x(2n-2)(2n-2)(2n-2)(t+(tj+1)<0.xj+1)aj+1(2n-2)(tj+2)<0,当t(tj+1,tj+2时,有x(2n-2)(t)<x(2n-2)(t+j+1)<0.特别地有xx(2n-2)(2n-2)(2n-2)(t+(tj+2)<0.xj+2)aj+2(2n-2)(tj+m)<0,当t(tj+m-1,tj+m时,有x(2n-2)(t)<x(2n-2)(t+j+m-1)<0.特别地有xx(2n-2)(2n-2)(2n-2)(t+(tj+m)<0

18、.xj+m)aj+m从而由单调性知t充分大时,x(2n-2)(t)<0.当x(2n-1)(t)<0,x(2n-2)(t)<0,由引理1知,t充分大时x(2n-3)(t)<0,反复应用引理知当t充分大时x(t)<0产生矛盾!故假设x(2n-1)(tj)不成立.(t)0,S(t)>0,当t(tk,tk+1,有r(t)x(2n-1)(t)从而证明了tk>T时,x(2n-1)(tk)>0.又因为Sr(tk+1)x(2n-1)(tk+1),当t(tk+1,tk+2,有r(t)x(2n-1)(t)r(tk+2)x(2n-1)(tk+2),依次类似推理可得当t

19、充分大时,x(2n-1)(t)>0.(二)为方便起见,不妨设tt0时,x(2n-1)(t)>0.从而有x(2n-2)(t)在(tk,tk+1上单调增加.若对一切tk,有x(2n-2)(tk)<0,这时显然有x(2n-2)(t)<0(tt0).若有某个tj,使x(2n-2)(tj)0,则由x(2n-2)(t)的单调性及a(k2n-2)>0知当t>tj时,有x(2n-2)(t)>0.由此知存在T1T,使下面两情形之一成立:(A1)x(2n-1)(t)>0,x(2n-2)(t)>0,tT1;(B1)x(2n-1)(t)>0,x(2n-2)(

20、t)<0,tT1;当(A1)成立时,利用引理1可知当t充分大时,有x(2n-3)(t)>0.反复利用引理1,最终可得当t充分大时,有x(2n-1)(t)>0,x(2n-2)(t)>0,x(t)>0,x(t)>0.当(B1)成立时,类似前面的讨论,利用引理2可知当t充分大时,有x(2n-3)(t)>0,且进一步可推知存在T2T1,当tT2时,有以下两种情形之一成立:(A2)x(2n-3)(t)>0,x(2n-4)(t)>0,tT2;(B2)x(2n-3)(t)>0,x(2n-4)(t)<0,tT2.重复上述讨论,最终可得:存在TT

21、及l1,3,2n-1,使当tT时,有第1期张超龙,等:高阶非线性脉冲微分方程解的振动性49x(i)(t)>0,i=0,1,l;i-1(-1)x(i)(t)>0,i=l+1,l+2,2n-1;(t)0.(2n-r(t)x1)引理3证毕.注:方程(1)存在最终负解的情形,也有类似引理2、引理3的结论.(B)和(C)成立,a(k0)1,b(k0),b(ki)1,i=1,2,2n-1,k=1,2,且定理1设条件(A)、t0+t2n-1p(t)dt=+,则方程(1)每一个有界解是振动的.证明:假设方程(1)有一个非振动的有界解x(t),不失一般性,设x(t)>0,tt0.由引理3把(6

22、)式分为两种情形:(i)若l=1即有x(t)>0,x(t)>0,x(t)<0,x󰂪(t)>0,x(4)<0,x(2n-1)(t)>0.(ii)若l(t)>0,x(t)>0,x󰂪(t)>0,xl(t)>0,xl+1(t)<0,x(2n-1)(t)>0.3即有x(t)>0,x(0)(t)>0,t(tk,tk+1,k=1,2,以上两种情况都有x.因此x(t)在t(tk,tk1)单调递增.即当ak)是单调递增,即当tt0时,x(t)x(0).(1)1,x(t)在t0,+(r(t)x(2n

23、-t1)(s)=-f(s,x(s)-p(s)(x(0)-cp()tk,tk+1.t(11)其中c=(x(t0)>0.对(11)n-,tkk-(rn-ds-cstk2n-1p(s)ds,t(tk,tk+1.(12):(a):设情形(i)成立,当t(tk,tk+1有,tkts2n-1(r(s)x(2n-1(s)ds=tkts2n-1d(r(s)x(2n-1)(s)t=t2n-1(2n-1)(2n-1)2n-1(t+r(t)x-tkr(tk)xk)-(2n-1)t2n-1(2n-1)(2n-1)2n-1(t+r(t)x-tkr(tk)xk)-(2n-1)Mstks2n-2r(s)xx(2n-1

24、)(s)dst2n-2(2n-1)tk(s)ds=2n-2=t2n-1r(t)x(2n-1)-tk2n-1r(tk)x(2n-1)(+tk)+Mi=0(-1)i+1i(i)tx(t)+i!2n-2Mi=0(-1)ii(i)+tkx(tk).i!特别,对任意自然数k有tktk+1s2n-1(r(s)x2n-2(2n-1)()(s)dstk+1r(tk+1)x2n-1(tk+1)-tk2n-22n-12n-1r(tk)x(2n-1)(t+k)+Mi=0(-1)i+1i(i)tk+1x(tk+1)+Mi!i=0(-1)ii(i)+tkx(tk).i!不论i为奇数或偶数都有i()()+)-x(i)(t

25、k),i=1,2,2n-1.0(-1)i(bk-1)xi(tk)(-1)i(xi(tk()从而对任意自然数m和t(tm,tm+1)有tkmtk+1s2n-1(r(s)x(2n-1)(s)dsti+12n-1r(t)x2n-2(2n-1)(t)-t1i2n-1r(t1)x(2n-1)(t+k)+2n-2MMi=0(-1)tk2n-1i(i)tx(t)+Mi!(+tk)i=0(-1)mi(i)+t1x(tk)-i!(-1)i2n-2x(2n-1)-x(2n-1)(tk)+Mk=2k=2i=0i(i)(i)+tk(x(tk)-x(tk)i!502n-1t1仲恺农业技术学院学报2n-2第18卷-(2n

26、-1)!Mx(t)-mr(t1)x(2n-1)(t1)+m+i=0(-i2n-21)i+1i(i)+t1x(t1)-i!Mk=2tk2n-1(2n-1)bk2n-2-1)x(2n-1)(tk)+M(2n-1)k=2i=0(-1)i(i)(i)tk(bk-1)x(tk)i!(-1)i+1-(2n-1)!Mx(t)-联合上述不等式和(12)式有2n-1t1r(t1)x(t1)+M+i=0i(i)+t1x(t1).i!t2n-2-(2n-1)!Mx(t)-2n-1t1r(t1)x(2n-1)(t1)+M+i=0(-1)i+1i(i)+t1x(t1)-ci!st12n-1p(s)ds.当t+时,x(t

27、)+,与x(t)有界矛盾!(b):设情形(ii)成立.).因此,对任意自然数m,有由b(k0)1,x(t)是非负严格单调递增,tt1,+x(t)=x(+tm)t+(s)ds,t(txtmm,tm+1),x(tm)=x(+tm-1)+(sxttt2=(t1)+m+(s)ds,xt1m-1t2x(t)=k=2(x(+tk)-x(tk)+x(t1)+k=1(s)ds+xx(s)ds.tktmk+1t(13)(t)>0,t(tk,tk+1),k由x1有,(1)(t)>x(t1+)a1(t1),t(t1,t2).xx(1)(1)(1)(1)(t2)>a2(t1),x(t)>x(t

28、2+)a2(t2)>a2(t1),t(t2,t3.特别有xxxa1x由数学归纳法知,对任意自然数k,有111(t)>x(t+(t1),t(tk,tk+1.xk)akak-1a1x()()()结合(13)式和a(k0)1,有m-1(t1)x(t)>xk=1a1a11a11()k()k-()()(tk+1-tk),t(tm,tm+1.由条件(C)和b(k0)1,有m-1k=1a1a11a11()k()k-(tk+1-tk)+,(m).故当t+时,x(t)+,这与x(t)是有界矛盾.因此方程(1)的一切有界解是振动的.证毕!(B)和(C)成立,若对任意正整数m有ak0>b&g

29、t;0,又bk2n-11,且对任意定理2设条件(A)、k=1m()()>0有则方程(1)的一切解是振动的.上非减.+|x|<+t0inff(t,x(t)dt=+,(14)(t)证明:设方程(1)有非振动解x(t),不妨设x(t)>0,tt0,由引理3知,x0,tt0,x(t)在t0,+x(t1)x(t0+),x(t2)x(t1+)a10x(t1)a10x(t0+),x(t3)x(t2+)a20x(t2)a20a10x(t0+).()()()()()由数学归纳法得x(tm)x(tm-1)am-1x(t-1)a1a2+(0)(0)(0)0)+a(m-1x(t0)>bx(t0).第1期张超龙,等:高阶非线性脉冲微分方程解的振动性51从而可保证x(t)有正的下界.且t0tf(s,x(s)ds(2n-1)t1t0把(1)式由t0到t1积分,得r(t0)xinff(s,x(s)ds.(t)+f(s,x(s)ds=r(t)x|x|+t0t10(2n-1)(t0+).类似地把(1)式由tk-1到tk积分(k为自然数)得r(tk)x(2n-1