# 差分方程建模 Modelling w/Differential Eqtns MATH140001

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Example. Consider $f(x)=\sqrt{|x|}$. Then $\left(x_{1}, x_{2}\right)=(0, \infty)$,
$$F(x)=2\left(\sqrt{x}-\sqrt{x_{0}}\right) \text {. }$$
and
$$\varphi(t)=\left(\sqrt{x_{0}}+\frac{t}{2}\right)^{2}, \quad-2 \sqrt{x_{0}}<t<\infty$$

Then $\phi \in C^{1}\left(\left(T_{-}, T_{+}\right)\right)$and
$$\lim {t \uparrow T{+}} \phi(t)=x_{2}, \quad \text { respectively } \quad \lim {t \downarrow T{-}} \phi(t)=x_{1} .$$
In particular, $\phi$ exists for all $t>0$ (resp. $t<0$ ) if and only if
$$T_{+}=\int_{x_{0}}^{x_{2}} \frac{d y}{f(y)}=+\infty$$

## MATH140001COURSE NOTES ：

$$s=\sigma(t), \quad y=\eta(t, x)$$
(which map the fibers $t=$ const to the fibers $s=$ const). Denoting the inverse transform by
$$t=\tau(s), \quad x=\xi(s, y),$$
a straightforward application of the chain rule shows that $\phi(t)$ satisfies
$$\dot{x}=f(t, x)$$
if and only if $\psi(s)=\eta(\tau(s), \phi(\tau(s)))$ satisfies
$$\dot{y}=\dot{\tau}\left(\frac{\partial \eta}{\partial t}(\tau, \xi)+\frac{\partial \eta}{\partial x}(\tau, \xi) f(\tau, \xi)\right)$$