Torres[1]讨论了一类线性周期边值问题的格林函数, 在一定条件下证明了其格林函数是不变号的.Cabada等[2]假设格林函数是正的情形时, 证明了非线性周期边值问题正解的存在性.其他一些相关的工作可参见文献[3-8].Graef等[9]和Ma[10]分别讨论了在格林函数非负和变号情形下非线性周期边值问题正解的存在性.Zhong等[11]讨论了当
(1) |
实际上, 当
(2) |
式中,
(3) |
其中格林函数的变号性是由参数不同取值导致的, 结论的证明方法来自文献[12].
C[0, 2π]表示[0, 2π]上连续函数的Banach空间, 范数为
首先假设:
H1:α≠β, α≠0, β≠0, α≠±1, β≠±1;
H2:f∈C[0, ∞)是连续的,且f(0)>0;
H3:g∈L1[0, 2π],对于任意区间不是几乎处处等于零.
将f(x)再进行如下连续延拓:当x < 0时,规定f(x)=f(0),仍记为f(x).定义算子
式中,
是边值问题(3)的格林函数.利用常规方法可以证明下面的引理.
引理1 设H1,H2和H3满足.边值问题式(3)存在解y∈C[0, 2π],等价于y是算子S在空间C[0, 2π]的不动点.
定理1 设H1满足,则格林函数G(t, s)在[0, 2π]×[0, 2π]变号.
证明 根据参数α与β的关系,可划分为如下20个参数区域Ⅰ-XX,见图 1.
仅给出对于区域0 < β < α的讨论,其余区域的讨论方法类似.
1) π>t>s>0情形:因为
故恒有G(t, s)>0.事实上,
2)π>s>t>0情形:因为
故恒有G(t, s)>0.事实上,
3) 2π>s>π>t>0情形:因为G(t, s)与情形2)相同, 所以
由于
4) 2π>t>π>s>0情形:因为G(t, s)与情形1)相同, 所以
由于
5) 2π>t>s>π情形:因为G(t, s)与情形1)相同, 所以
由于
且当t→π时,s→π;当t→2π时,s→2π.故由单调性与凹凸性可知其符号.
6) 2π>s>t>π情形:因为G(t, s)与情形2)相同, 所以
令
且当t→π时,s→π;当t→2π时,s→2π.故由单调性与凹凸性可知其符号.
综上可知,对于区域0 < β < α来说,格林函数符号情况可见图 2.
对于t∈[0, 2π], 记
再假设
H4:存在ε>0,使得对于t∈[0, 2π],
引理2 如果, H1,H3和H4满足,令
证明 如果p(t)≡0,则∀t∈[0, 2π],有[G(t, s)g(s)]+=0, a.e.s∈[0, 2π].由H4可知,∀t∈[0, 2π],[G(t, s)g(s)]-=0, a.e.s∈[0, 2π].故G(t, s)g(s)=0, a.e.s∈[0, 2π].从而由定理1可知,存在使g几乎处处等于零的区间,这与H3矛盾.证毕.
定义算子T为
(Ty)(t)=
引理3 如果H1~H4满足,令0 < δ < 1,则存在正数λ, 使得当λ∈(0, λ)时,方程y(t)=(Ty)(t)有一个正解
定理2 如果H1~H4满足,则存在λ0>0,使得当λ∈(0, λ0)时,式(3)有一个正解.
证明 令
|f(s)|≤ξf(0)(1+ε).所以
(4) |
如果q(t)=0, 式(4)成立.
设δ∈(ξ, 1), 由引理3可知
所以存在λ1>0, 使得当λ∈(0, λ1)时,
(5) |
当x, y∈[-k, k], 且|x-y|≤λ0δf(0)‖p‖0时,
(6) |
当λ∈(0, λ0)时,定义算子H如下:
设θ∈(0, 1), y∈C[0, 2π],若y=θHy,下证‖y‖0≠λδf(0)‖p‖0.否则‖y‖0=λδf(0)‖p‖0, 则由式(5)和式(6)得
(7) |
(8) |
由式(4),式(7)和式(8)可知:
(9) |
故与
可见yλ是式(3)的一个正解.
3 例子例1 考虑边值问题:
可以验证定理2的条件满足, 所以存在λ0>0,当0 < λ < λ0时有正解.
例2 考虑边值问题:
可以验证定理2的条件满足,所以存在λ0>0,当0 < λ < λ0时有正解.
这里含导数部分边值的系数分别取正数和负数,并且
本文主要研究的是由边值条件中系数α, β不同取值导致格林函数变号的二阶非线性边值问题,其中α≠β.如果H1~H4满足,则存在λ0>0,使得当λ∈(0, λ0)时,式(3)有一个正解.当α=β=0时,其格林函数不变号.而当α=β=1时,实际上是周期边值问题,其格林函数也是不变号的.
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