陳 頌,閆曉芳

(永城職業(yè)學院 基礎部,河南 永城 476600)

0 引言

p-Laplace方程是一類橢圓微分方程.含有p-Laplace算子的微分方程解的存在性問題受到了廣泛的關注,相關研究結果可參見文獻[1-3]等.p(t)-Laplace方程首先是從非線性彈性理論中引發(fā)出來的,p(t)-Laplace算子比p-Laplace算子有更復雜的非線性性質,相關結果可參見文獻[4-6]等.

本文討論了如下p(t)-Laplace方程解的存在性問題:

(1)

其中p∈C(,)是T-周期函數(shù),對于t∈有p(t)>1,f∈C(×N,N),且f(t,u)關于t是T-周期的.

1 定理

定理如果存在r>0和線性向量場V,使得對任意的u∈N,〈Vu,u〉≥0,當且僅當u=0時,〈Vu,u〉=0,且對所有的t∈[0,T],u∈N,當〈Vu,u〉=r時,都有〈f(t,u),Vu〉>0,〈f(t,u),V*u〉>0,則以上方程至少有一個解u滿足|u(t)|≤r,其中t∈[0,T].

令g∈C([0,T]×N,N).設X是Banach空間,定義映射A:X→X*:

定義映射G:X→X*:

引理[7]令p為定理中所定義的,A,g,G均為上文定義.若存在一個正常數(shù)M,使得

|g(t,u)|≤M,

(2)

?t∈[0,T],?u∈N,

則映射A+G:X→X*是滿的.特別地,存在u∈X使得A(u)+G(u)=0,且u是以下問題的解:

(3)

定義映射Q=Qr:N→N:

令g(t,u)=f(t,Q(u))-Q(u).則g∈C(×N,N).

則由引理,映射A+G:X→X*是滿的.特別地,存在u∈X使得A(u)+G(u)=0,且u是以下問題的解:

(4)

現(xiàn)在證明u滿足|u(t)|≤r,?t∈[0,T].首先證明以下論斷.

論斷存在t∈[0,T],使得|u(t)|≤r.

證明反證法.設對所有的t∈[0,T]都有|u(t)|>r.考慮函數(shù)

φ(t)=〈|u′(t)|p(t)-2u′(t),Vu(t)〉,

t∈[0,T],

故

φ′(t)=|u′(t)|p(t)+

〈f(t,Qu(t)),Vu(t)〉+

〈u(t)-Qu(t)〉,t∈[0,T].

因為|Qu(t)|=r,有

〈f(t,Qu(t)),Vu(t)〉≥0,

又因

故

〈u(t)-Qu(t),Vu(t)〉=

(|u(t)|-r)|u(t)|>0.

故

φ′(t)>0, ?t∈[0,T].

(5)

這表明φ′(t)在[0,T]上嚴格增,這和φ(0)=φ(T)矛盾.故論斷成立.

2 定理的證明

假設u不滿足|u(t)|≤r,?t∈[0,T],則存在t∈[0,T],使得|u(t)|>r.假設u在上定義,且為T-周期的,則u∈C1(,N).設

由假設,可以找到σ<τ使得

h(σ)=maxh(t),

|u(τ)|=r,t∈,

且當t∈[σ,τ]時,|u(t)|>r.故|u(σ)|>r且h′(σ)=0.

令

ψ(t)=〈u(t),Vu(t)〉,

ψ1(t)=〈u′(t),Vu(t)〉,

ψ2(t)=〈u(t),Vu′(t)〉.

類似于式(5)的證明,有

將V換成V*,同理可得

〈u′(t),V*u(t)〉′>0, ?t∈[σ,τ],

則

〈V′u(t),u(t)〉′>0, ?t∈[σ,τ],

故

故

ψ″(t)=〈u(t),Vu(t)〉″=

(〈u′(t),Vu(t)〉′+

〈u(t),Vu′(t)〉′=

因此,ψ′(t)在[σ,τ]上嚴格增.由于

因此,對t∈[σ,τ],有h′(t)

故u滿足

|u(t)|≤r,?t∈[0,T].

故可推出Qu(t)=u(t).因此u是問題(4)的解,故u是問題(1)的解.定理得證.

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