四阶微分差分方程的非线性边值问题的存在性

JOURNAL󰀁OF󰀁DALIAN󰀁JIAOTONG󰀁UNIVERSITYVo.l31󰀁No.2

󰀁Apr.2010󰀁

󰀁文章编号:1673󰀁9590(2010)02󰀁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:A

0󰀁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]

H(!),满足

,但对于带时滞项的微分差

0

∀+%

!d!=%,使得当0!t!1,

H(!)

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

2

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)

[-󰀁,1]&

=f(t,x,x󰀂,x ,x (t-󰀁),x󰀂)

x (t)=󰀂(t),-󰀁!t!0(1)(2)(3)

C[-󰀁,0]&C[0,1],使得当-󰀁!t!1时,∀(t)!#(t),∀ (t)∋f(t,∀(t),∀(t-󰀁),∀󰀂(t)),# (t)!f(t,#(t),#(t-󰀁),#󰀂(t)),

1

则对任意的函数󰀂(t)∃C[-󰀁,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󰀂)u(t)=󰀂(t),-󰀁!t!0

u(1)=A

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(2)h(∃,%)∃C(R),且h(∃,%)对固定的∃关于%单调不减

1

则对任意的函数󰀂(t)∃C[-󰀁,0],当∀(t)!󰀂(t)!#(t),-󰀁!t!0,且h(∀(1),∀󰀂(1))!0,h(#(1),#󰀂(1))∋0时,边值问题

u =f(t,u,u(t-󰀁),u󰀂)u(t)=󰀂(t),-󰀁!t!0h(u(1),u󰀂(1))=0

(10)(11)(12)

2

x(0)=B,x(1)=C,h(x (1),x󰀂(1))=0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.

(7)(8)(9)

1󰀁预备定理

本节考虑时滞边值问题

u=f(t,u,u(t-󰀁),u󰀂,T1u,T2u)u(t)=󰀂(t),-󰀁!

t!0

(4)(5)

有解x(t),使得∀(t)!u(t)!#(t),0!t!1,

h(u(1),u󰀂(1))=0(6)其中,󰀁>0与A及󰀂(t)同上,Tiu= i(t)+K(t,s)u(s)ds,K(t,s)于[0,1]∀

0

i

i

1

#[0,1]上

连续, ,1]上连续.i(t)于[0

下面将讨论边值问题(4)~(6)解得存在性.定理1󰀁假设

3

(1)f(t,u,v,w)∃C([0,1]#R),对任意的r>0存在[0,+%)上的正值连续函数

*

[5]

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󰀁mai:lwanggc@d.lcn.

󰀁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)=󰀂(t),-󰀁!t!0,u(1)=∀(1)

有解,任取其一并记为∀0(t),显然∀(t)!

∀0(t)

=

%

12[#0(1)-∀0(1)]2

%

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)#n(1)-∀n(1)=

(13)

10(1)-∀0(1)](14)n[#2

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(1))<0h(#n(1),#󰀂n(1))>0#nj(t))x0(t),#󰀂nj(t))x󰀂0(t),

-󰀁!

t!1,j)%

∀ 󰀂 󰀂ni(t))x0(t),∀ni(t))x0(t),

-󰀁!t!1,i)%

即u0(t),!u0(t)都是满足式(10),且u0(t)=󰀂(t),-󰀁!t!0,h(u0(1),u󰀂,u0(t)0(1))∋0=󰀂(t),-󰀁!

t!0,h(u0(1),u󰀂,由0(1))∋0

式(11),(12)可知u0(t)!u0(t),-󰀁!t!1,u0(1)=u0(1),于是有u󰀂0(1)∋u󰀂0(1),这样从条件(2)得0!u󰀂0(1))!

h(u0(1),u󰀂0(1))!

h(u0(1),h(u0(1),

0,即h(u0(1),u󰀂0(1))!

(15)

故存在一致收敛的子序列{#nj(t)},{∀ni(t)}使得

!#(t),-󰀁!t!1,于是∀󰀂(1)∋∀󰀂0(1),由条件(2)得

h(∀󰀂󰀂(1))!00(1),∀0(1))!h(∀(1),∀󰀁󰀁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#0(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),#󰀂󰀂(1))∋00(1))∋h(#(1),#同理,如果上式等式成立,则定理得证,否则取d1=1[#0(1)+∀0(1)],并考虑边值问题2

u =f(t,u,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󰀂(1))>0,则取∀1(t)=∀0(t),#1(t)=u(t);如果h(u(1),u󰀂(1))<0,则取∀1(t)=u(t),#1(t)=#0(t).于是#1(1)-∀1(1)=-∀0(1)],再置d2=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∀∀#2(t)!#1(t),-󰀁!t!11(t)!2(t)!

1#[#1(1)-∀1(1)]2(1)-∀2(1)=

21[#0(1)2

u󰀂于是h(u0(1),u󰀂h(u0(1),0(1)),0(1))=u󰀂,u0(t)=u0(t)=󰀂(t),-󰀁!t!0.0(1))=0󰀁󰀁定理3󰀁假设

(1)f(t,u,v,w,p,q)∃C([0,1]#R),对任意的r>0存在[0,+%)上的正值连续函数

+%

!

H(!),满足0d!=%,使得当0!t!1,

H(!)

5

∀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.

1

(4)存在函数∀(t),#(t)∃[-󰀁,1]&C[-󰀁,0]&C[0,1],使得当-󰀁!t!1时,∀(t)!#(t),且对任给&(t)∃[-󰀁,1]&C[-󰀁,0]&C[0,1],满足∀(t)!&(t)!#(t),

∀ (t)∋f(t,∀(t),∀(t-󰀁),∀󰀂(t),

[T1&](t),[T2&](t)),2

1

2

1[#1(1)+∀1(1)],并考虑2

󰀁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󰀂(t)∃C[-󰀁,0]和实数A,当∀(t)!

󰀂(t)!#(t),-󰀁!

t!

0,且h(∀(1),

∀󰀂(1))!0,h(#(1),#󰀂(1))∋0时,边值问题(4)~(6)有解x(t),使得∀(t)!u(t)!#(t),-󰀁!t!

1.

21

(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!# (t-∀(1),∀ (t-󰀁)!

!

󰀂(t)!

󰀁),h(∀ (1),∀󰀂(1))

0,h(# (1),

#󰀂(1))∋0.

4

则边值问题(1)~(3)有解x(t)∃C[0,1]&C[-󰀁,0]&C[-󰀁,1],使得∀(t)!x(t)!#(t),-󰀁!

1

3

2

证明󰀁对于&(t)∃C[0,1],定义范数

2

∗u∗=0max,则C[0,1]是具有∗+∗&(t)!t!1的Banach空间,让

2

∋={&(t):&(t)∃C[0,1],∀(t)!

&(t)!#(t),&(t)=^󰀂(t),-󰀁!t!

0,&(1)=A^}

2

t!1

K(t,s)u(s)ds,i=1,2,∀

0

i

证明󰀁令x󰀂=u,并定义积分算子如下:Tiu= i(t)+

其中

, 1(t)=B+(C-B)t2(t)=C-BK1(t,s)=K2(t,s)=

(t-1)s,0!s!

t!1

(s-1)t,0!t!s!1s,0!s!t!1

s-1,0!t!s!1

,

则∋是Banach空间C[0,1]中的一个有界闭凸

子集.

由定理1,对任给的&(t)∃C[0,1],边值问题

u=f(t,u,u(t-󰀁),u󰀂,T1&,T2&)u(t)=󰀂^(t),-󰀁!t!0

h(u(1),u󰀂(1))=0

2

有解u(t)∃C[0,1]满足不等式

∀(t)!u(t)!#(t),-󰀁!

t!1

(4)󰀂(5)󰀂(6)󰀂

2

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 )

u(t)=󰀂(t),-󰀁!

t!0

(1)󰀂(2)󰀂

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)∃

*

*

*

*

h(u(1),u󰀂(1))=0(3)󰀂

若∀ =∀ =##*,#*,显然有∀*!*,且对任何&(t)∃C[0,1],#(t)!&(t)!∀(t),∀ (t)!&(t)!

# (t),由& (t),可得&(t)=*(t)=&

T1&󰀂(t)=T2&*(t),&*(t),于是

∀ ,T1&󰀂*∋f(t*,T2&*,∀*,∀*(t-󰀁),∀*),# ∋f(t,T1&󰀂**,T2&*,#*,#*(t-󰀁),#*)∀󰀂(t)!#*(t)!*(t),-󰀁!#󰀂*(1))

故∀'(3)的下、'上解,由*(t),#*(t)是问题(1)~定理2,边值问题(1)~'(3)有解'u(t),且∀*(t)!u(t)!#t!1,注意到关系式x (t)*(t),0!=u(t),易见,x(t)=T1u= 1(t)+∀(t),-󰀁!参考文献:

[1]BERNFELDSR,LASHMIKANTHANV.Anintroduction

tononlinearboundaryvalueproblems[M].NewYork:AcademicPress,1974.

0

4

∋,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),#(t)∃C[0,1]32

&C[-󰀁,0]&C[-󰀁,1],使得

∀(t)∋f(t,&(t),&󰀂(t),#

(4)

(4)

4

t!0,

h(∀󰀂0!h(#*(1),∀*(1))!*(1),

∀ (t),∀ (t-󰀁),∀(t))

(t)!f(t,&(t),&󰀂(t),# (t),

# (t-󰀁),#(t))

󰀂

󰀂

K(t,∀

1

1

则称∀(t),#(t)分别为式(1)的下解与上解,其4

中&(t)∃C[0,1],#(t)!&(t)!∀(t),∀ (t)!&(t)!# (t).

定理4󰀁假设

󰀂

(1)f(t,x,x󰀂,x ,x (t-󰀁),x)∃C([0,1]#R),满足Nagumo条件,且关于x (t-󰀁)单调不增.

5

s)u(s)ds是(1)~(3)的解,且#(t)!x(t)!

t!1.

󰀁88大连交通大学学报第31卷

[2]AFTABIZADEHAR.Existenceanduniquenesstheo󰀁

remsfourthorderboundaryvalueproblems[J].Math.A󰀁nalApp.l,1986,6:415󰀁426.

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[4]王国灿.四阶微分方程非线性边值问题的可解性[J].大连铁道学院学报,1998(4):5󰀁8.

[5]苗树梅,周钦德.奇摄动微分差分方程边值问题[J].吉林大学自然科学学报,1987(3):1󰀁7.

ExistenceofSolutionsforNonlinearBoundaryValueProblemsof

FourthOrderDifferential󰀁DifferenceEquation

WANGGuo󰀁can

(SchoolofMathematicsandPhysics,DalianJiaotongUniversity,Dalian116028,China)

Abstract:Nonlinearboundaryvalueproblemsoffourth󰀁orderdifferential󰀁differenceequationisstudiedbymeansofdifferentialinequalitytheories.Basedonsuitcondition,theexistenceofsolutionswasestablished.Theresultshowsthatitisseemsnewtoapplythesetechniquetosolveotherboundaryvalueproblems.Keywords:fourthorderdifferential󰀁differenceequation;nonlinearboundaryvalueproblem;differentialine󰀁

quality

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1

2

2

(1.SchoolofElectrical&InformationEngineering,DalianJiaotongUniversity,Dalian116028,China;2.SchoolofElectronic

Abstract:Basedontheconventionallumpedparametercircuitmode,lpiezoelectrictransformerelectricexpres󰀁sioncanbededucedfromadmittancematrixYparametersoftwo󰀁portnetwork,andthecharacteristicsofdisk󰀁typestep󰀁downpiezoelectricandinverseconnectedRosentypetransformersaresimulatedandanalysed.TheresultsindicatethattheelectriccharacteristicsoftheinverseconnectedRosentypeisbasicallyconsistantwiththatofsomeotherstep󰀁downpiezoelectrictransformerswhentheloadismatched.Keywords:equivalentcircui;tYparameter;disk󰀁typeofpiezoelectrictransformer;voltagegain;step󰀁down