f**n
发帖数: 401 | 1 现有如下一个非线性微分方程:
dx(t)/dt=\lambda(t)-p(x)=\lambda(t)-p(x(t))
其中\lambda(t)为t地函数,与x地取值无关.
p(x)为x的单调增函数.p(x)>0
\lambda(t),p(x)都是很一般的函数,连续,可导...反正需要用什么条件就用什么吧
这个方程的初始条件为:
x(0)=\theta,
\lambda()和p()都与\theta无关.
另外有个条件不知道有没有用:
\lambda(0)-p(x(0))=\lambda(0)-p(\theta)=0.
由于x(t)的初值与\theta有关,x(t)也一般的与\theta有关.
我的问题是:
当t=0时,\frac{\partial x(t)}{\partial\theta}=? 就是说x(t)对\theta的偏导在
t=0时的值是多少?
我的猜想是,当t=0时,\frac{\partial x(t)}{\partial\theta}=1?Of course this is
only a guess...
Thanks and bow! | x******n
发帖数: 24 | 2 You may proof it in this way:
Assume you have the solution x(t, \theta) for your initial problem,
then integral your equation from [0,t] , you get
x(t,\theta)-\theta=\int_0^t \lambda(s)ds - \int_0^t p(x(s,\theta)) ds
then differential by \theta and evluate at t=0
you get
\frac{\partial x(t)}{\partial\theta}=1 at t=0
since integral domain of the second integral is [0,0], it is 0
that's it!
在 fern (无嗔~~左手) 的大作中提到: 】 |
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