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非線性二階差分方程三點(diǎn)邊值問(wèn)題的研究

2021-09-09 22:37:34魏文英紀(jì)玉德郭彥平

魏文英 紀(jì)玉德 郭彥平

摘 要:為了拓展非線性離散邊值問(wèn)題的基本理論,研究了一類(lèi)非線性二階差分方程三點(diǎn)邊值問(wèn)題正解存在性的充分條件。首先,給出了相應(yīng)的二階差分方程三點(diǎn)邊值問(wèn)題解的表達(dá)式并證明其性質(zhì);其次,在Banach空間中構(gòu)造合適的錐和積分算子,運(yùn)用錐上的Krasnoselskiis不動(dòng)點(diǎn)定理,在非線性項(xiàng)允許變號(hào)的條件下,獲得非線性二階差分方程三點(diǎn)邊值問(wèn)題正解存在性的充分條件;最后,通過(guò)2個(gè)例子證明主要定理和結(jié)果的有效性。結(jié)果表明,定理?xiàng)l件得證且離散邊值問(wèn)題滿足正解的存在性。所研究的方法在二階離散邊值問(wèn)題理論證明方面效果良好,對(duì)探究非線性高階多點(diǎn)離散邊值問(wèn)題具有一定的借鑒意義。

關(guān)鍵詞:差分方程;離散邊值問(wèn)題;不動(dòng)點(diǎn)定理;錐;正解;存在性

中圖分類(lèi)號(hào):O175?? 文獻(xiàn)標(biāo)識(shí)碼:A

doi:10.7535/hbkd.2021yx04006

收稿日期:2021-05-11;修回日期:2021-06-09;責(zé)任編輯:張士瑩

基金項(xiàng)目:河北省自然科學(xué)基金(A2015208051);河北省高等學(xué)校科學(xué)技術(shù)研究指導(dǎo)項(xiàng)目(Z2019027)

第一作者簡(jiǎn)介:魏文英(1982—),女,河北邯鄲人,講師,碩士,主要從事常微分方程及差分方程方面的研究。

通訊作者:紀(jì)玉德副教授。E-mail:ji_yude@163.com

魏文英,紀(jì)玉德,郭彥平.非線性二階差分方程三點(diǎn)邊值問(wèn)題的研究[J].河北科技大學(xué)學(xué)報(bào),2021,42(4):360-368.WEI Wenying,JI Yude,GUO Yanping.Research of the three-point boundary value problems for nonlinear second-order difference equation[J].Journal of Hebei University of Science and Technology,2021,42(4):360-368.

Research of the three-point boundary value problems for nonlinear second-order difference equation

WEI Wenying1,JI Yude2,GUO Yanping2

(1.School of Software,Hebei Polytechnic Institute,Shijiazhuang,Hebei 050091,China;2.School of Science,Hebei University of Science and Technology,Shijiazhuang,Hebei 050018,China)

Abstract:In order to extend the basic theory of nonlinear discrete boundary value problems,this paper studied the sufficient conditions for the existence of positive solutions for a class of nonlinear second-order difference equations with three-point boundary value problems.Firstly,the expressions of the solutions for the corresponding three-point boundary value problems for second-order difference equations were given and their properties were proved; Secondly,by constructing suitable cone and integral operator in Banach space and utilizing Krasnoselskii's fixed point theorem in cones,the sufficient conditions for the existence of positive solutions of three-point boundary value problems for nonlinear second-order difference equations were obtained under the condition that the nonlinear term was allowed to change sign.Finally,two examples were given to illustrate the validity of the main theorems and results.The results show that the conditions of the theorem are proved and the discrete boundary value problems satisfies the existence condition of positive solutions.The method is effective in the theoretical proof of the second-order discrete boundary value problem,and has reference for the study of the nonlinear high-order multi-point discrete boundary value problems.

Keywords:

difference equation;discrete boundary value problem;fixed point theorem;cone;positive solution;existence

多點(diǎn)邊值問(wèn)題的研究起源于應(yīng)用數(shù)學(xué)和物理領(lǐng)域。例如,彈力穩(wěn)定性理論中的很多問(wèn)題,都可以用多點(diǎn)邊值問(wèn)題的方法進(jìn)行處理。因此,多點(diǎn)邊值問(wèn)題受到了很多研究者的關(guān)注,參見(jiàn)文獻(xiàn)[1]、文獻(xiàn)[2]及其參考文獻(xiàn)。

在文獻(xiàn)[3]和文獻(xiàn)[4]中,研究人員應(yīng)用不動(dòng)點(diǎn)定理,證明了如下三點(diǎn)邊值問(wèn)題

u″(t)+a(t)f(u(t))=0, 0

多個(gè)正解的存在性,其中非線性項(xiàng)f是超線性的或次線性的,0<η<1,0<α<1η,f∈C([0,+.SymboleB@),(0,+.SymboleB@)),a∈C0,1,0,+.SymboleB@,且存在t0∈0,1,使得at0>0。而后,XU等[5-6]證明了上述三點(diǎn)邊值問(wèn)題至少存在1個(gè)和3個(gè)解的存在性,其中0<α,η<1,f∈C(R,R)。

LI等[7]利用Krasnoselskiis不動(dòng)點(diǎn)定理研究了如下三點(diǎn)邊值問(wèn)題

u″(t)+a(t)u′(t)+λf(t,u(t))=0, 0

至少有1個(gè)正解的存在性,其中0<α,η<1,λ>0,a∈C([0,1],(-.SymboleB@,0)),且非線性項(xiàng)f∈C([0,1]×R,R)。

對(duì)于二階差分方程滿足局部(混合周期)和非局部邊界條件解的存在性和多解性,人們進(jìn)行了廣泛研究[8-18]。從文獻(xiàn)來(lái)看,關(guān)于非線性差分方程多點(diǎn)邊值問(wèn)題解的存在性的研究成果還很少。受上述文獻(xiàn)中研究方法的啟發(fā),本文利用錐上的Krasnoselskiis不動(dòng)點(diǎn)定理,研究非線性二階三點(diǎn)離散邊值問(wèn)題

Δ2u(t)+a(t)Δu(t)+λf(t,u(t))=0, t∈{0,1,…,n-2},(1)

Δu(0)=0,u(n)=αu(η)(2)

正解存在性的充分條件,其中Δu(t)=u(t+1)-u(t),且有∑n-1t=t0Δut=un-ut0和Δ∑n-1t=t0ut=un成立。

假設(shè)條件:

H1λ>0,0<α<1,η∈1,2,…,n-1和a(t)<0,t∈0,1,…,n;

H2f∈C({0,1,…,n}×R,R),且當(dāng)t,u∈0,1,…,n×R時(shí),存在M>0,使得ft,u>-M。

利用錐上的Krasnoselskiis不動(dòng)點(diǎn)定理,獲得離散邊值問(wèn)題(1)和問(wèn)題(2)至少存在1個(gè)正解的充分條件。

1 預(yù)備知識(shí)

引理1[19](Krasnoselskiis不動(dòng)點(diǎn)定理) 設(shè)E是Banach空間,K是E上的1個(gè)錐,并且Ω1和Ω2是E的有界開(kāi)集,滿足0∈Ω1Ω1Ω2,假設(shè)算子A:K∩Ω2\Ω1→K是全連續(xù)算子,且滿足如下2個(gè)條件之一:

1)‖Au‖≤‖u‖, u∈K∩Ω1,且‖Au‖≥‖u‖, u∈K∩Ω2,

2)‖Au‖≥‖u‖, u∈K∩Ω1,且‖Au‖≤‖u‖, u∈K∩Ω2,

那么,算子A在K∩Ω2\Ω1中至少存在1個(gè)不動(dòng)點(diǎn)。

對(duì)于u∈u0,u1,…,un,令‖u‖=maxt∈0,1,…,nut,易知X=u0,u1,…,unui∈R,i=0,1,…,n關(guān)于范數(shù)‖·‖構(gòu)成Banach空間。

引理2 假設(shè)條件H1成立,則對(duì)任意y∈X,二階離散三點(diǎn)邊值問(wèn)題

Δ2ut+atΔut+yt=0, t∈0,1,…,n-2,(3)

Δu(0)=0,u(n)=αu(η)(4)

有唯一解,

ut=-∑t-1j=11pj∑j-1i=0pi+1yi+11-α∑n-1j=11pj∑j-1i=0pi+1yi-

α1-α∑η-1j=11pjpi+1yi,(5)

其中pt=∑t-1i=011-ai。

證明 假設(shè)ut滿足邊值問(wèn)題(3)和問(wèn)題(4)。因?yàn)閜t=∑t-1i=011-ai,有pt≥0,p0=1。用pt乘以式(3)的兩邊,得到ΔptΔut+pt+1yt=0,

由差分的性質(zhì)可推出:

p1Δu1-p0Δu0=-p1y0,

p2Δu2-p1Δu1=-p2y1,

ptΔut-pt-1Δut-1=-ptyt-1,

ptΔut-p0Δu0=-∑t-1i=0pi+1yi。

由邊界條件Δu0=0,得ptΔut=-∑t-1i=0pi+1yi,

所以

Δut=-1pt∑t-1i=0pi+1yi。(6)

一方面,可得

Δu1=u2-u1=-1p1∑0i=0pi+1yi,

Δu2=u3-u2=-1p2∑1i=0pi+1yi,

Δut-1=ut-ut-1=-1pt-1∑t-2i=0pi+1yi。

將上述方程兩邊相加,得

ut-u0=-∑t-1j=11pj∑j-1i=0pi+1yi。(7)

另一方面,有

Δun-1=un-un-1=-1pn-1∑n-2i=0pi+1yi,

Δun-2=un-1-un-2=-1pn-2∑n-3i=0pi+1yi,

Δut=ut+1-ut=-1pt∑t-1i=0pi+1yi。

類(lèi)似的,將上述方程兩邊相加,得

un-ut=-∑n-1j=t1pj∑j-1i=0pi+1yi。(8)

將式(7)和式(8)兩邊相加,得到

u0=un+∑n-1j=t1pj∑j-1i=0pi+1yi+∑t-1j=11pj∑j-1i=0pi+1yi。

由邊界條件un=αuη,可得

u0=αuη+∑n-1j=11pj∑j-1i=0pi+1yi=

αu0-∑η-1j=11pj∑j-1i=0pi+1yi+∑n-1j=11pj∑j-1i=0pi+1yi,

u0=11-α∑n-1j=11pj∑j-1i=0pi+1yi-α1-α∑η-1j=11pj∑j-1i=0pi+1yi。(9)

因此,合并式(7)和式(9),可得到式(5)。反之,假設(shè)ut是由式(5)給出的,帶入式(3)和式(4)成立。

證畢。

引理3 假設(shè)條件H1成立。如果y∈X且yt≥0, t∈0,1,…,n,則離散邊值問(wèn)題(3)和問(wèn)題(4)的解ut滿足

mint∈0,1,…,nut≥γ‖u‖,(10)

其中γ=αn-ηn-αη。

證明 分2步進(jìn)行證明。

第1步,對(duì)任意t∈0,1,…,n,證明ut≥0。

由式(6)和yt≥0,可知對(duì)任意t∈0,1,…,n,Δut≤0,因此ut是單調(diào)遞減函數(shù)。如果un≥0,由于ut是單調(diào)遞減函數(shù),因而很容易得到上述結(jié)論;如果un<0,由邊界條件un=αuη,有uη

第2步,證明式(10)成立。

由ut的單調(diào)性,可知u0=maxt∈0,1,…,nut,un=mint∈0,1,…,nut。

因?yàn)棣?ut=-atΔut-yt≤0,所以可得到ut在0,1,…,n上的離散點(diǎn)圖形是凹的。由于ut是單調(diào)遞減函數(shù),因而有Δut≤0,根據(jù)差分性質(zhì)可知

Δun-1+…+Δuηn-η=uη-unn-η≥Δun-1+…+Δu0n-0=u0-unn,

u0≤un+nuη-nnn-η=un+nun-αuηαn-η=unn-αηαn-η=1γun,

可以得出un≥γu0,即mint∈0,1,…,nut≥γ‖u‖。

證畢。

注1 關(guān)于引理3,若α>1和un>0,由un=αuη,得到un>uη,這與單調(diào)性矛盾;若un=0,由函數(shù)的單調(diào)性得到ut≡0, t∈η,n,但需要yt≡0, t∈0,1,…,n。因此,本文要求0<α<1。

引理4 假設(shè)條件H1成立,w-t是如下邊值問(wèn)題的解:

Δ2ut+atΔut+1=0, t∈0,1,…,n-2,

Δu0=0,un=αuη,

則當(dāng)t∈0,1,…,n時(shí),存在S>0,使得0≤w-t≤S。

證明 由引理3的證明,可知w-0=maxt∈0,1,…,nw-t,且w-t≥0, t∈0,1,…,n。由式(9)可得

w-0=11-α∑n-1j=11pj∑j-1i=0pi+1-α1-α∑η-1j=11pj∑j-1i=0pi+1=

∑η-1j=11pj∑j-1i=0pi+1+11-α∑n-1j=η1pj∑j-1i=0pi+1。

因此,令

S=w-0=∑η-1j=11pj∑j-1i=0pi+1+11-α∑n-1j=η1pj∑j-1i=0pi+1。

則當(dāng)t∈0,1,…,n時(shí),0≤w-t≤S。

證畢。

設(shè)wt=λMw-t,則wt≤λMS(λ和M來(lái)自條件H1和H2),若ut是邊值問(wèn)題(1)和問(wèn)題(2)的解,則

Δ2ut+wt+atΔut+wt=

Δ2ut+atΔut+λMΔ2w-t+atΔw-t=

-λft,ut-λM=

-λft,ut+M。

令gt,u=ft,u+M,對(duì)任意t,u∈0,1,…,n×R, gt,u>0。

因此可得到引理5。

引理5 函數(shù)ut是邊值問(wèn)題(1)和問(wèn)題(2)的解,當(dāng)且僅當(dāng)u~t=ut+wt是如下邊值問(wèn)題的1個(gè)解,

Δ2u~t+atΔu~t+λgt,u~t-wt=0, t∈0,1,…,n-2,(11)

Δu~0=0, u~n=αu~η,(12)

其中u~t>wt, t∈0,1,…,n。

2 主要結(jié)果

定理1 假設(shè)條件H1和H2成立,且函數(shù)ft,u滿足

limu→+.SymboleB@inft∈η,…,nft,uu=+.SymboleB@,(13)

則存在一個(gè)常數(shù)λ>0,對(duì)任意λ∈0,λ,使得離散邊值問(wèn)題(1)和問(wèn)題(2)至少有1個(gè)正解。

證明 由引理2,易知邊值問(wèn)題(11)和問(wèn)題(12)有一個(gè)正解u~=u~t,當(dāng)且僅當(dāng)u~是方程算子ut=Aut的1個(gè)正解,其中:

Aut=-λ∑t-1j=11pj∑j-1i=0pi+1gi,ui-wi+

λ1-α∑n-1j=11pj∑j-1i=0pi+1gi,ui-wi-

αλ1-α∑η-1j=11pj∑j-1i=0pi+1gi,ui-wi。(14)

令K=uu∈X,u≥0,mint∈0,1,…,nut≥γ‖u‖。顯然K是X上的錐,并且依據(jù)引理3,AKK。根據(jù)文獻(xiàn)[20],容易看出A:K→K是全連續(xù)的。證明A滿足引理1的條件。

首先,設(shè)g~r1=supt∈0,1,…,n,0≤u≤r1gt,u,其中r1>0。由式(13)可得limr1→+.SymboleB@r1g~r1=0,則存在1個(gè)常數(shù)R1>0,使得

R1g~R1=maxr1>0r1g~r1。(15)

記G=g~R1,設(shè)

λ=minγR1MS, 1-αR1GS,(16)

并且對(duì)任意λ<λ,令Ω1=u∈X:‖u‖

R1≥us-ws≥γR1-λMS>γR1-λMS≥0, 0<γ<1(17)

gs,us-ws≤g~R1=G, s∈0,1,…,n。

因此,由式(14)和引理4可得:

Aut≤-λ1-α∑n-1j=11pj∑j-1i=0pi+1gi,ui-wi≤

λG1-α∑n-1j=11pj∑j-1i=0pi+1≤

λG1-α∑η-1j=11pj∑j-1i=0pi+1+11-α∑n-1j=η1pj∑j-1i=0pi+1=

λGS1-α≤R1,

對(duì)任意u∈K∩Ω1,得

‖Au‖≤‖u‖。(18)

另一方面,由式(13),有

limu→+.SymboleB@inft∈η,…,ngt,uu=limu→+.SymboleB@inft∈η,…,nft,u+Mu=+.SymboleB@。

因此,令R2=nR1(n是大于1的正整數(shù)),使得對(duì)任意t∈η,…,n,u≥γn-1nR2,有g(shù)t,u≥ξu,其中ξ>0滿足

αγλξn-1n1-α∑n-1j=η1pj∑j-1i=0pi+1≥1。(19)

令Ω2=u∈X:‖u‖

us-ws≥γR2-λMS>γR2-γR1≥γn-1nR2,

g(s,u(s)-w(s))≥ξ(u(s)-w(s))≥γξ(n-1)nR2, s∈{0,1,…,n}。

由式(14)和式(19),可得

Aun=-λ∑n-1j=11pj∑j-1i=0pi+1gi,ui-wi+

λ1-α∑n-1j=11pj∑j-1i=0pi+1gi,ui-wi-

αλ1-α∑η-1j=11pj∑j-1i=0pi+1gi,ui-wi=

αλ1-α∑n-1j=11pj∑j-1i=0pi+1gi,ui-wi-

αλ1-α∑η-1j=11pj∑j-1i=0pi+1gi,ui-wi=

αλ1-α∑n-1j=η1pj∑j-1i=0pi+1gi,ui-wi,

所以

Aun=αλ1-α∑n-1j=η1pj∑j-1i=0pi+1gi,ui-wi,(20)

Aun≥αγλξn-1R2n1-α∑n-1j=η1pj∑j-1i=0pi+1≥R2>‖u‖。

即對(duì)u∈K∩Ω2時(shí),有

‖Au‖≥‖u‖。(21)

因此由式(18)、式(21)和引理1,可得到算子A在K∩Ω2\Ω1上有1個(gè)不動(dòng)點(diǎn)使R1≤‖u~‖≤R2。根據(jù)式(17),得到u~t>wt,因此u~t是邊值問(wèn)題(1)和問(wèn)題(2)的1個(gè)正解。

證畢。

定理2 假設(shè)條件H1和H2成立,并且函數(shù)ft,u滿足

limu→+.SymboleB@inft∈η,…,nft,u=+.SymboleB@和limu→+.SymboleB@supt∈0,1,,…,nft,uu=0,(22)

則存在1個(gè)常數(shù)λ>0,對(duì)任意λ∈λ,+.SymboleB@,使得離散邊值問(wèn)題(1)和(2)至少有1個(gè)正解。

證明 顯然A:K→K是全連續(xù)算子,下面的證明與定理1的證明過(guò)程類(lèi)似。

證明算子A滿足引理1的條件。

ζ=2MS1-ααγ∑n-1j=η1pj∑j-1i=0pi+1-1。(23)

一方面,由式(22)的第1個(gè)極限,有

limu→+.SymboleB@inft∈η,…,ngt,u=limu→+.SymboleB@inft∈η,…,nft,u+M=+.SymboleB@,

則存在N>0,t∈η,…,n和u≥N,使得

gt,u≥ζ。

設(shè)λ=NMS和R1=2λMSγ,則對(duì)任意λ≥λ,u∈K和‖u‖=R1,有

us-ws≥γR2-λMS=λMS≥λMS≥N(24)

和gs,us-ws≥ζ成立。

由式(20)和式(23),可知

Aun=αλ1-α∑n-1j=η1pj∑j-1i=0pi+1gi,ui-wi≥

αλζ1-α∑n-1j=η1pj∑j-1i=0pi+1=

αλζ1-α×2MS1-ααγζ=2λMSγ=R1=‖u‖。(25)

令Ω1=u∈X:‖u‖

‖Au‖≥‖u‖。(26)

另一方面,由式(22)的第2個(gè)極限,可知

limu→+.SymboleB@supt∈0,1,…,ngt,uu=limu→+.SymboleB@supt∈0,1,,…,nft,u+Mu=0。

因此,對(duì)任意ε>0,存在R~2>R1,t∈0,1,…,n和u≥R~2,使得gt,u≤εu。這里ε滿足

ελ1-α∑n-1j=η1pj∑j-1i=0pi+1≤1。(27)

因?yàn)間t,u非負(fù)連續(xù)函數(shù),存在σ>0,使得

0≤gt,u≤εσ, t∈0,1,…,n, N≤u≤R~2。

令R2=maxR~2,σ和Ω2=u∈X:‖u‖

由式(14)和式(27),可得

Aun≤λ1-α∑n-1j=11pj∑j-1i=0pi+1gi,ui-wi≤

εR2λ1-α∑n-1j=11pj∑j-1i=0pi+1≤R2,

即,當(dāng)u∈K∩Ω2時(shí),

‖Au‖≤‖u‖。(28)

因此,由式(26)、式(28)和引理1,得到A在K∩Ω2\Ω1上有1個(gè)不動(dòng)點(diǎn),使得R1≤‖u~‖≤R2。根據(jù)式(24),得到u~t>wt,因此u~t是邊值問(wèn)題(1)和問(wèn)題(2)的1個(gè)正解。證畢。

注2 若at≡0, t∈0,1,…,n,則定理1和定理2仍然成立,文獻(xiàn)[16]所研究的問(wèn)題就是該特殊情況。

注3 在定理1和定理2中,非線性項(xiàng)f的有界性假設(shè)是至關(guān)重要的。因?yàn)榇_定gt,u=ft,u+M是根據(jù)這個(gè)假設(shè),然后可得出引理5的結(jié)論,最后利用Krasnoselskiis不動(dòng)點(diǎn)定理,獲得2個(gè)重要的結(jié)果。

3 例 證

為了說(shuō)明本文所獲得的結(jié)果,舉出如下2個(gè)例子。

例1 考慮二階離散邊值問(wèn)題

Δ2u-Δu+λu2-5etcosu=0, t∈0,1,…,n-2,(29)

Δu0=0,un=45un2。(30)

易知α=45,η=n2,a(t)=-1,f(t,u)=u2-5etcosu。

令M=5en,則對(duì)于t∈0,1,…,n,f(t,u)>-5en,且limu→+.SymboleB@inft∈η,…,nft,u=+.SymboleB@,

因此滿足定理1的條件。再令γ=12,可得

S=∑n-1j=11pj∑j-1i=0pi+1-11-α∑η-1j=11pj∑j-1i=0pi+1=n2-9n4+12。

令R1=(5en-5)12,則式(15)成立,且G=g~(R1)=10(en-1)。

因此,根據(jù)定理1,對(duì)于任何λ<λ*=R1(1-α)GS=(5en-5)1250(en-1)(n2-9n4+12),邊值問(wèn)題(29)和問(wèn)題(30)至少存在1個(gè)正解u~(t),并且滿足‖u~(t)‖≥(5en-5)12。

例2 考慮二階離散邊值問(wèn)題

Δ2u-Δu+λu23-5t2cosu=0, t∈0,1,…,n-2,(31)

Δu0=0,un=45un2。(32)

易知α=45,η=n2,a(t)=-1,f(t,u)=u23-5t2cosu。

令M=5n2,則對(duì)于t∈0,1,…,n,f(t,u)>-5n2,且

limu→+.SymboleB@inft∈η,…,nft,u=+.SymboleB@和limu→+.SymboleB@supt∈0,1,…,nft,u=0,

因此滿足定理2的條件。再令γ=12,可得

S=n2-9n4+12,ζ=2MS(1-α)αγ∑n-1j=η1pj∑j-1i=0pi+1-1=40n2n2-9n4+1232n2-3n。

對(duì)任意N≥(ζ)12,有g(shù)(t,u)≥ζ成立。

因此,根據(jù)定理2,對(duì)于任何λ≥λ*=NMS=2105n2n2-9n4+1232n2-3n-12,邊值問(wèn)題(31)和問(wèn)題(32)至少存在1個(gè)正解u~(t),并且滿足‖u~(t)‖≥2λMSγ=20n2λ(n2-9n4+12)。

4 結(jié) 語(yǔ)

1)利用錐上的Krasnoselskiis不動(dòng)點(diǎn)定理,在非線性項(xiàng)允許變號(hào)的情形下,研究了非線性二階三點(diǎn)離散邊值問(wèn)題正解存在性的充分條件,通過(guò)2個(gè)數(shù)值例子驗(yàn)證了所獲得理論結(jié)果的正確性。

2)結(jié)果表明,離散邊值問(wèn)題滿足定理正解的存在性條件。所提出的研究方法在二階離散邊值問(wèn)題理論證明方面效果良好,對(duì)探究非線性高階多點(diǎn)離散邊值問(wèn)題具有一定的借鑒意義。

3)本研究?jī)H考慮了非線性二階三點(diǎn)離散邊值問(wèn)題,未能完全體現(xiàn)出彈力穩(wěn)定性理論中的更多問(wèn)題,且僅考慮了非線性項(xiàng)在有界情形下正解的存在性問(wèn)題;此外,所使用的錐上不動(dòng)點(diǎn)定理僅能獲得至少1個(gè)正解的存在性,對(duì)于非線性項(xiàng)更一般化的條件以及多解的存在性還有待于進(jìn)一步研究。下一步計(jì)劃對(duì)非線性高階多點(diǎn)離散邊值問(wèn)題以及分?jǐn)?shù)階q-差分邊值問(wèn)題解的存在性進(jìn)行深入探討,并在多種非線性項(xiàng)情形下進(jìn)行數(shù)值驗(yàn)證。

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