定义2如果函数F:Ω→n满足下列条件,则称函数F属于F(Ω,h,ω):
(H1) 对任意的(x,t1),(x,t2)∈Ω,‖F(x,t2)-F(x,t1)‖≤|h(t2)-h(t1)|;
(H2) 对任意的(x,t1),(x,t2),(y,t1),(y,t2)∈Ω,
‖F(x,t2)-F(x,t1)-F(y,t2)+F(y,t1)‖≤ω(‖x-y‖)|h(t2)-h(t1)|.
引理5[2]假设F:Ω→n满足条件(H1),若[α,β]⊂[a,b],x:[α,β]→n是方程(2)的解,则对任意的t1,t2∈[α,β],有
‖x(t2)-x(t1)‖≤|h(t2)-h(t1)|.
引理6[2]假设给定F∈F(Ω,h,ω),x:[α,β]→n,[α,β]⊂[a,b],xk:[α,β]→n,如果x是函数列{xk}的极限函数,且对每个k∈,s∈[α,β],(xk(s),s)∈Ω,(x(s),s)∈Ω,DF(xk(τ),t)存在,则DF(x(τ),t)存在,且
引理7[2]给定F∈F(Ω,h,ω),且x:[α,β]→n,[α,β]⊂[a,b],xk:[α,β]→n是一列函数,如果x是{xk}的极限函数,且对任意的s∈[α,β],(x(s),s)∈Ω,则DF(x(τ),t)存在.
考虑微分方程(3),其中x′表示x∈n的分布导数.
假设[a,b]⊂,且分布f满足如下假设条件:
(H3) 在[a,b]上对任意给定的x∈n,f(x,·)是DHK可积的;
(H4) 在[a,b]上存在f-,f+∈DHK,使得对任意的x∈n满足f-⪯f(x,·)⪯f+;
(H5) 存在连续的增函数ω:[0,∞)→,ω(0)=0和分布l∈DHK,使得对任意的t∈[a,b],x,y∈n,有
|f(x,t)-f(y,t)|⪯ω(‖x-y‖)|l(t)|.
![](https://img.fx361.cc/images/2023/0219/7beb70bd1ab2e19f4301a2b88a857c3858a3bcea.webp)
![](https://img.fx361.cc/images/2023/0219/b84a174e23bbb1e08e4d7deb48fe8fce77abf5fe.webp)
(6)
其中右侧积分表示DHK积分.
![](https://img.fx361.cc/images/2023/0219/7beb70bd1ab2e19f4301a2b88a857c3858a3bcea.webp)
(7)
![](https://img.fx361.cc/images/2023/0219/3369e2d8ff60194fad9789d8f27c293bbebc14c8.webp)
![](https://img.fx361.cc/images/2023/0219/19703bdc34182f0f78336ed4af2a0bc35d4ddec3.webp)
(8)
证明:由(H3),(H4)和式(5)有
∀I0⊂[a,b],
因此
(10)
同理可证,当t1≥t2时,有|F(x,t2)-F(x,t1)|≤h(t1)-h(t2),因此(H1)成立.
![](https://img.fx361.cc/images/2023/0219/54d76da12cb97caddfc48f616b34a19956690814.webp)
当t1≥t2时,类似可证.因此,(H2)成立,从而结论成立.
![](https://img.fx361.cc/images/2023/0219/2efb747751da0cb249169c258f3205f7492a3640.webp)
![](https://img.fx361.cc/images/2023/0219/3e66a5a8ee3a3c56d063976cd1833f20fab93f43.webp)
则由(H5)可知,
![](https://img.fx361.cc/images/2023/0219/485a82cb1671239b34f6093556b07e3225fbade6.webp)
定理1函数x:[α,β]→n,[α,β]⊂[a,b]是方程(3)在[α,β]上的解当且仅当x是广义常微分方程(2)的解,其中F由式(7)给出.
证明:假设x:[α,β]→n是方程(3)的解,由命题1,DF(x(τ),t)存在,且对任意的t1,t2∈[α,β],有
![](https://img.fx361.cc/images/2023/0219/b84a174e23bbb1e08e4d7deb48fe8fce77abf5fe.webp)
所以,x是方程(2)的解.反之,若x是方程(2)的解,则由命题1知,x:[α,β]→n满足式(6).又由引理5知,对任意的t1,t2∈[α,β],满足
‖x(t2)-x(t1)‖≤|h(t2)-h(t1)|.
所以x是连续的.因此x是方程(3)的解.
引理9[2]假设F:Ω→n,F∈F(Ω,h,ω),取满足
则存在Δ-,Δ+>0,使得在区间[t0-Δ-,t0+Δ+]上存在广义常微分方程(2)的解x:[t0-Δ-,t0+Δ+]→n,且
![](https://img.fx361.cc/images/2023/0219/5f5a7dd04e31066ba3466640bd62349b09d0780e.webp)
则存在Δ-,Δ+>0,使得在区间[t0-Δ-,t0+Δ+]上存在方程(3)的解x:[t0-Δ-,t0+Δ+]→n,且满足
![](https://img.fx361.cc/images/2023/0219/3afa3b6cce57dc0f51e3253751f0838e7d0eea8b.webp)
定义4若方程(2)的每个解y:[t0,t0+σ]→n,y(t0)=x(t0),都存在η1>0,使得对∀t∈[t0,t0+η]∩[t0,t0+σ]∩[t0,t0+η1],满足x(t)=y(t),则方程(2)的解x:[t0,t0+η]→n称为在未来是局部唯一的.
![](https://img.fx361.cc/images/2023/0219/2e9061451f3d6a9dd24d4a758351fcfcb1b78735.webp)
对于方程(3)可以定义相同的概念.
引理10[2]设F∈F(Ω,h1,ω1),其中h1是左连续的增函数,ω1:[0,∞)→是增函数,且当r>0时,ω1(r)>0,ω1(0)=0,对任意的u>0,
(12)
定理3(唯一性) 设分布f满足条件(H3)~(H5),ω(r)>0,r>0,且对任意的u>0,
(13)
![](https://img.fx361.cc/images/2023/0219/3afa3b6cce57dc0f51e3253751f0838e7d0eea8b.webp)
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