四阶微分差分方程的非线性边值问题的存在性

1、第31卷 第2期2010年4月大连交通大学学报J OURNA L O F DAL IAN JI AOTONG UN I V ERS I TYV o.l31 N o.2A pr.2010文章编号:1673 9590(201002 0085 04四阶微分差分方程的非线性边值问题的存在性王国灿(大连交通大学理学院,辽宁大连116028*摘 要:利用微分不等式技巧研究了某一类四阶微分差分方程的非线性边值问题,在上下解存在的条件下,得到了解的存在性定理.结果表明:这种技巧为其它边值问题的研究提出了崭新的思路.关键词:四阶微分差分方程;非线性边值问题;微分不等式中图分类号:O175.8 文献标识码:A0

2、引言四阶微分方程的各种边值问题已经受到了广泛而深入的研究1 4,但对于带时滞项的微分差分方程的研究相对较少,本文利用微分不等式技巧,考虑一般的四阶微分差分方程的非线性边值问题x(4=f(t,x,x ,x,x(t- ,x (1 x(t= (t,- !t!0(2 x(0=B,x(1=C,h(x(1,x (1=0(3其中, 0与A,B,C为给定的常数, (t为- ,0上的一阶连续可微函数,在上下解存在的条件下,得到了解的存在性和唯一性定理.1 预备定理本节考虑时滞边值问题u=f(t,u,u(t- ,u ,T1u,T2u(4 u(t= (t,- !t!0(5h(u(1,u (1=0(6其中, 0与A及

3、 (t同上,T i u=i(t+10K i(t,su(sd s,K i(t,s于0,1#0,1上连续,i(t于0,1上连续.下面将讨论边值问题(4(6解得存在性.定理15 假设(1f(t,u,v,wC(0,1#R3,对任意的r0存在0,+%上的正值连续函数H(!,满足+%0!H(!d!=%,使得当0!t!1, u!r,v!r时,f(t,u,v,w!H(w;(2对固定的t,u,w,f(t,u,v,w关于v单调不增;(3存在函数(t,#(t- ,1& C1- ,0&C20,1,使得当- !t!1时,(t!#(t,(tf(t,(t,(t- , (t, #(t!f(t,#(t,#(t- ,# (t,则

4、对任意的函数 (tC1- ,0和实数A,当(t! (t!#(t,- !t!0,且(1!A !#(1时,边值问题u=f(t,u,u(t- ,u (7u(t= (t,- !t!0(8u(1=A(9有解x(t,使得(t!u(t!#(t,0!t!1,其中 0与A及 (t同上.定理2 假设(1定理1的条件(1,(2,(3成立(2h(,%C(R2,且h(,%对固定的关于%单调不减则对任意的函数 (tC1- ,0,当(t ! (t!#(t,- !t!0,且h(1, (1 !0,h(#(1,# (10时,边值问题u=f(t,u,u(t- ,u (10u(t= (t,- !t!0(11h(u(1,u (1=0(

5、12*收稿日期:2008 08 24作者简介:王国灿(1963-,男,教授,硕士,主要从事常微方程边值问题的研究E m ai:l w anggcd.l cn.86大连交通大学学报第31卷有解u(t,使得(t!u(t!#(t,- !t!1.证明 首先当(1=#(1时,由(t!#(t可得 (1# (1,又据条件(1知# (1 (1,即有 (1=# (1,则边值问题u=f(t,u,u(t- ,u ,u(t= (t,- !t!0,u(1=(1在- ,1上满足(t!u(t!#(t,- !t!1的解u(t就是所求的解.其次考虑(1!#(1时的情形.据定理1知边值问题u=f(t,u,u(t- ,u ,u(t

6、= (t,- !t!0,u(1=(1有解,任取其一并记为0(t,显然(t!0(t!#(t,- !t!1,于是 (1 0(1,由条件(2得h(0(1, 0(1!h(1, (1!0如果上式等式成立,则0(t便是式(10(12的解,否则考虑边值问题u=f(t,u,u(t- ,u ,u(t= (t,- !t!0,u(1=#(1又由定理1,此问题有解,并任取其一记为#(t,亦有0(t!#0(t!#(t,0!t!1,显然#0 (1# (1,再以h(,%的单调性得h(#0(1,# 0(1h(#(1,# (10同理,如果上式等式成立,则定理得证,否则取d1=12#0(1+0(1,并考虑边值问题u=f(t,u,

7、u(t- ,u ,u(t= (t,- !t!0,u(1=d1从定理1可得其解存在性,任取其一记为u(t,则0(t!u(t!#0(t,- !t!1.如果h(u(1,u (1=0,则定理为真;如果h(u(1,u (10,则取1(t=0(t,#1(t=u(t;如果h(u(1,u (10,则取1(t=u(t,#1(t=#0(t.于是#1(1-1(1=12#0(1-0(1,再置d2=12#1(1+1(1,并考虑边值问题u=f(t,u,u(t- ,u ,u(t= (t,- !t!0,u(1=d2显然问题有解,任取其一记为u(t,于是与1(t,#1(t的类似选取可得2(t,#2(t满足1(t!2(t!#2(

8、t!#1(t,- !t!1#2(1-2(1=12#1(1-1(1=122#0(1-0(1于是利用数学归纳法可得两串序列n(t%1,#n(t%1满足0(t!1(t!(!n(t!(!#n(t!(!#1(t!#0(t(13#n(1-n(1=12n#0(1-0(1(14此外,n(t,#n(t, n(t,# n(t还于- !t!1上一致有界,同等连续.又由n(t,#n(t的选取可知h(n(1, n(10(15故存在一致收敛的子序列#n j(t,ni(t使得#nj(tx0(t,# nj(tx 0(t,- !t!1,j%ni(tx0(t, ni(tx 0(t,- !t!1,i%即u0(t,!u0(t都是满足

9、式(10,且u0(t=(t,- !t!0,h(u0(1,u 0(10,u0(t= (t,- !t!0,h(u0(1,u 0(10,由式(11,(12可知u0(t!u0(t,- !t!1,u0(1=u0(1,于是有u 0(1u 0(1,这样从条件(2得0!h(u0(1,u 0(1!h(u0(1,u 0(1!0,即h(u0(1,u 0(1!h(u0(1,u 0(1,于是h(u0(1,u 0(1=h(u0(1,u 0(1=0,u0(t=u0(t= (t,- !t!0.定理3 假设(1f(t,u,v,w,p,qC(0,1#R5,对任意的r0存在0,+%上的正值连续函数H(!,满足+%0!H(!d!=%

10、,使得当0!t!1,u!r,v!r,p!r,q!r时,f(t,u,v,w,p,q!H(w.(2对固定的t,u,v,f(t,u,v,w,p,q关于v单调不减.(3定理2的条件(2成立.(4存在函数(t,#(t- ,1&C1-,0&C20,1,使得当- !t!1时,(t!#(t,且对任给&(t- ,1&C1- ,0&C20,1,满足(t!&(t!#(t,(tf(t,(t,(t- , (t,T1&(t,T2&(t,第2期王国灿:四阶微分差分方程的非线性边值问题的存在性87#(t!f(t,#(t,#(t- ,# (t,T1&(t,T2&(t,则对任意的函数 (tC1- ,0和实数A,当(t! (t!#

11、(t,- !t!0,且h(1, (1!0,h(#(1,# (10时,边值问题(4(6有解x(t,使得(t!u(t!#(t, - !t!1.证明 对于&(tC20,1,定义范数u=m ax0!t!1&(t,则C20,1是具有+的Banach空间,让=&(t:&(tC20,1,(t!&(t!#(t,&(t= (t,- !t!0,&(1=A则是Banach空间C20,1中的一个有界闭凸子集.由定理1,对任给的&(tC20,1,边值问题u=f(t,u,u(t- ,u ,T1&,T2&(4 u(t= (t,- !t!0(5h(u(1,u (1=0(6 有解u(tC20,1满足不等式(t!u(t!#(t,

12、- !t!1所以,u(t,于是定义映射F:.不难验证F在上是完全连续映射,于是根据Schauder不动点定理存在u*(t,使得Fu*(t=u*(t,即u*(t是边值问题(4 (6的解,从而定理得证.2 主要结果本节利用微分算子研究边值问题(1(3的存在性.若两个函数存在函数(t,#(tC40,1 &C3- ,0&C2- ,1,使得(4(tf(t,&(t,& (t,(t,(t- , (t#(4(t!f(t,&(t,& (t,#(t,#(t- ,# (t则称(t,#(t分别为式(1的下解与上解,其中&(tC40,1,#(t!&(t!(t,(t !&(t!#(t.定理4 假设(1f(t,x,x ,x

13、,x(t- ,x C(0,1 #R5,满足N agum o条件,且关于x(t- 单调不增.(2定理2的条件(2成立.(3存在上下解#(t和(t,使得#(t!(t,(t!#(t,- !t!1,且#(0!B!(0,#(1!C!(1,(t- ! (t! #(t- ,h(1, (1!0,h(#(1, # (10.则边值问题(1(3有解x(tC40,1 &C3- ,0&C2- ,1,使得(t!x(t! #(t,- !t!1证明 令x =u,并定义积分算子如下:T i u =i(t+10K i(t,su(sd s,i=1,2,其中1(t=B+(C-Bt,2(t=C-BK1(t,s=(t-1s,0!s!t!

14、1(s-1t,0!t!s!1, K2(t,s=s,0!s!t!1s-1,0!t!s!1则x=T1u,x =T2u,且边值问题(1(3转化为下述问题:u=f(t,T1u,T2u,u,u (t- ,u(1 u(t= (t,- !t!0(2h(u(1,u (1=0(3 若=*,#=#*,显然有*!#*,且对任何&(tC40,1,#(t!&(t!(t,(t! &(t!#(t,由&*(t=&(t,可得&(t= T1&*(t,& (t=T2&*(t,于是*f(t,T1&*,T2&*,*,*(t- , *, #*f(t,T1&*,T2&*,#*,#*(t- ,# *(t! (t!#*(t,- !t!0,h(

15、*(1, *(1!0!h(#*(1,# *(1故*(t,#*(t是问题(1(3的下、上解,由定理2,边值问题(1(3有解u(t,且*(t !u(t!#*(t,0!t!1,注意到关系式x(t =u(t,易见,x(t=T1u=1(t+10K1(t, su(sd s是(1(3的解,且#(t!x(t!(t,- !t! 1.参考文献:1BERNFELD S R,LAS HM I KANTHAN V.A n i ntroducti onto nonli near boundary va l ue prob le m sM.N e w Y ork:A cade m i c Press,1974.88大连交通

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18、equati o n is stud ied by m eans of d ifferenti a l i n equality theories.Based on suit cond ition,the ex i s tence of solutions w as established. The resu lt sho w s t h at it is see m s ne w to app l y t h ese techn ique to so lve other boundary va l u e proble m s.K ey w ords:fourth order d iffer

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22、 type P iezoelectric T ransf o r m er for AC/DCC.36th I EEE Powe r E lectron ics Spec i a li sts Con ference2005,P iscataway:Instit u te o fE l ec trica l and E lectron ics Eng i neers Inc.,2005:624 629.5L I M K J,KANG S H,K I M H H,e t a.l D esi gn and perfor m ance of m i n i aturized piez oe lect

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