四階時(shí)滯微分方程邊值問(wèn)題的正解

汪媛媛, 李永祥

(西北師范大學(xué)數(shù)學(xué)與統(tǒng)計(jì)學(xué)院,甘肅蘭州730070)

1 引言及預(yù)備知識(shí)

記C=C([-τ,0],R),則其按范數(shù)構(gòu)成 Banach 空間.令C+={φ∈C|φ(θ)≥0,θ∈[-τ,0]}.考慮四階時(shí)滯微分方程邊值問(wèn)題

正解的存在性,其中,f:I×C+?[0,+∞)連續(xù),I=[0,1],?(t)∈C([-τ,0],[0,+∞)),?(0)=0,對(duì)?t∈I,ut(θ)=u(t+θ),θ∈[-τ,0],

本文始終假設(shè):

兩端簡(jiǎn)單支撐的彎曲彈性梁的平衡狀態(tài)可用四階邊值問(wèn)題

來(lái)描述[1-2],其中,f:[0,1]×R×R?R連續(xù),關(guān)于邊值問(wèn)題(2)以及更廣泛的常微分方程邊值問(wèn)題解的存在性,已有許多研究工作[3-11].近年來(lái),伴隨著時(shí)滯微分方程理論的發(fā)展以及其在物理學(xué)、自動(dòng)控制理論、生物學(xué)、經(jīng)濟(jì)學(xué)、人口理論等多門學(xué)科中的廣泛應(yīng)用,時(shí)滯微分方程邊值問(wèn)題已逐漸成為一個(gè)研究的熱點(diǎn)[12-25].

對(duì)不含時(shí)滯的情形,即τ=0時(shí),問(wèn)題(1)退化為下面的常微分方程邊值問(wèn)題

其中,f:I×[0,+∞)?[0,+∞)連續(xù).問(wèn)題(3)已被許多作者研究[3-6],其中,文獻(xiàn)[3]給出了其正解的存在性定理.

對(duì)含有時(shí)滯項(xiàng)的情形,即τ≠0時(shí),宋利梅等在文獻(xiàn)[12]中討論了邊值問(wèn)題

正解的存在性,其中,f是定義在I×C+上的非負(fù)連續(xù)函數(shù),p(t)是定義在I上的非負(fù)可測(cè)函數(shù).他們運(yùn)用錐拉升與錐壓縮不動(dòng)點(diǎn)定理證明了問(wèn)題(4)正解的存在性.

本文考慮更一般的四階時(shí)滯微分方程邊值問(wèn)題(1).通過(guò)對(duì)不動(dòng)點(diǎn)指數(shù)的精確計(jì)算,證明了只要f0適當(dāng)小,f∞適當(dāng)大,或者f0適當(dāng)大,f∞適當(dāng)小時(shí),問(wèn)題(1)至少存在一個(gè)正解.

稱u(t)∈C4[-τ,1]為問(wèn)題(1)的一個(gè)解,如果u(t)滿足下面的條件:

定義線性算子T:C[-τ,1]?C[-τ,1]為

則算子T是方程(9)的解算子,且T把C[-τ,1]中的有界集映為C[-τ,1]中的有界集.由Gelfand公式得

其中,r(T)是算子T的譜半徑,顯然r(T)>0.設(shè)L=π4-aπ2-b,易見(jiàn)L是線性邊值問(wèn)題(9)對(duì)應(yīng)的最小特征值,因此r(T)=1/L.

引理3T是全連續(xù)算子.

證明T的連續(xù)性顯然,只需證明T的等度連續(xù)性即可.

由于G1和G2在[0,1]×[0,1]一致連續(xù),即對(duì)?ε>0,?η>0,?t1,t2∈[0,1],當(dāng)|t1-t2|<η時(shí)有

聯(lián)立以上兩式可得

又因?yàn)楫?dāng)t∈[-τ,0]時(shí),Au(t)=0;當(dāng)t∈(0,1)時(shí),Au(t)>0,從而可得A(K)?K.由T的全連續(xù)性可知A是全連續(xù)的.證畢.

引理 5[9]?γ∈(0,1),使得對(duì)?x∈K,當(dāng)t∈時(shí),有‖xt‖C≥γ‖x‖.

引理6[3]設(shè)A:K?K全連續(xù),如果μAu≠u,?u∈?Kr1,且0<μ≤1時(shí),i(A,Kr1,K)=1,其中,Kr1={u∈K|‖u‖C

引理7[3]設(shè)A:K?K全連續(xù),如果下列條件滿足:

則i(A,KR,K)=0,其中,KR={u∈K|‖u‖C

2 主要結(jié)果

定理1f是定義在I×C+上的非負(fù)連續(xù)函數(shù),且f把I×C+中的有界集映為[0,+∞)中的有界集,如果下列條件之一滿足:

則邊值問(wèn)題(1)至少存在一個(gè)正解.

證明假設(shè)條件(H1)成立,則由?(t)≡0,得vt=0,t∈[0,1].

由(H1)的第一個(gè)不等式,因?yàn)閒0<δL,則存在適當(dāng)?shù)膔0,滿足0

下證當(dāng)0<μ≤1時(shí),對(duì)?x∈?Kr0,有μAx≠x.反設(shè)?x0∈?Kr0,μ0∈(0,1],使得μ0Ax0=x0,則由A的定義,x0滿足

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