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\begin{document}
 \parindent=0cm
  \section*{Local flow theorem.}
In the proof of Fundamental theorem,
$K=Lipschitz\,\,constant\,\,of\,\,f,$


$M=upper\,\,bound\,\,for\,\,f\,\,on\,\,\overline{B_b}(x_0)$.
$\Rightarrow\exists !$ integral curve $\alp_{x_0}$ defined at
least on $J=[-a,a]$ if $a<min\{\frac{b}{M}\frac{1}{K}\},$ and
$\alp_{x_0}(J)\subset\overline{B}$.

따라서 $\overline{B_{\frac{b}{2}}}(x_0)$안의 모든 점 $x$에 대해
$\overline{B_{\frac{b}{2}}}(x)\subset\overline{B_b}(x_0)$ 가
성립한다. 또한 $\eps<min\{\frac{b/2}{M},\frac{1}{K}\}$ 에 대해
$\exists !$ integral curve $\alp_x$ defined at least on
$[-\eps,\eps]$ , and
$\alp_x[-\eps,\eps]\subset\overline{B_{\frac{b}{2}}}(x)\subset\overline{B_b}(x_0)$.

Now define
$\alp:[-\eps,\eps]\times\overline{B_{\frac{b}{2}}}(x_0)\rightarrow\overline{B_b}(x_0)\subset
W $ by $\alp (t,x)=\alp_x(t).$

{\it $\alp$ is called a {\bf local flow} of $f$.}\\

$\hs{3em}Claim\,:\,\alp \,\,is\,\, continuous.$

\begin{thm}(Continuous dependency on initial conditions)

$f:W(\subset\rb^{n+1})\rightarrow\rb^n$, Lipschitz. Let
$y(t),z(t)$ be solutions of $x'=f(t,x)$ on $[t_0,t_1]$. Then
$|y(t)-z(t)|\leq|y(t_0)-z(t_0)|e^{K(t-t_0)}$, $\forall t
\in[t_0,t_1]$.


(K=Lipschitz constant)
\end{thm}

위 정리의 증명을 위해 다음 보조정리를 먼저 보이자.

\begin{lem}(Gronwall's inequality) Let $u:[t_0,t_1]\rightarrow\rb$
be continuous, non-negative function. Suppose $u(t)\leq
C+\int_{t_0}^tKu(s)ds,\forall t\in[t_0,t_1]$ for some constants
$C, K\geq 0$.Then $u(t)\leq Ce^{K(t-t_0)}$.

\end{lem}

\begin{proof}
1.( $C>0$ ) $v(t):=C+\int_{t_0}^tKu(s)ds$라 놓자. $u(t)\leq v(t)$
이고

$v'(t)=Ku(t)$, $\frac{d}{dt}\log
v(t)=\frac{v'}{v}=K\frac{u}{v}\leq K$ 이므로 양변을 적분하면,

$\log v(t)\leq K(t-t_0)+\log C\Rightarrow $ $v(t)\leq
Ce^{K(t-t_0)}$.

$\therefore u(t)\leq Ce^{K(t-t_0)}$.

2. ( $C=0$ )  Choose $C_i>0$ such that $C_i\rightarrow 0$ and
apply $1$ to $C_i$. Then

$u(t)\leq C_i e^{K(t-t_0)}$ and take limit.


\end{proof}

{\bf (proof of the theorem 1.)}

Let $u(t)=|y(t)-z(t)|$.

$y(t)-z(t)=y(t_0)-z(t_0)+\int_{t_0}^tf(s,y(s))-f(s,z(s))ds$.

$u(t)=|y(t)-z(t)|\leq u(t_0)+\int_{t_0}^t|f(s,y(s))-f(s,z(s))|ds$.

$u(t)\leq u(t_0)+\int_{t_0}^tKu(s)ds\,\,\,\,\,\,$  (Lipschitz
condition.)

$u(t)\leq u(t_0)e^{K(t-t_0)}\,\,\,\,\,\,\,$ by the {\it Gronwall's
inequality. }

$\therefore\,|y(t)-z(t)|\leq|y(t_0)-z(t_0)|e^{K(t-t_0)}.\hs{4em}\square$\\

{\bf proof of the claim : $\alp$ is continuous.}

$|\alp(t,x)-\alp(t_1,x_1)|\leq|\alp(t,x)-\alp(t,x_1)|+|\alp(t,x_1)-\alp(t_1,x_1)|$

$\hs{8.5em}\leq|\alp_x(t)-\alp_{x_1}(t)|+|\alp_{x_1}(t)-\alp_{x_1}(t_1)|$\\

$\hs{2em}\alp(t,x_1)=x_1+\int_{0}^tf(\alp(s,x_1))ds.$

$\hs{2em}\alp(t_1,x_1)=x_1+\int_{0}^{t_1}f(\alp(s,x_1))ds.$

$\hs{2em} and\,\, by\,\,theorem\,1. $\\

$\hs{8.5em}\leq|\alp_x(0)-\alp_{x_1}(0)|e^{K|t|}\,+\,|\int_{t_1}^tf(x(s))ds|$


$\hs{8.5em}\leq|x-x_1|e^{K|t|}\,+\,|\int_{t_1}^tf(x(s))ds|$

$\hs{8.5em}\leq|x-x_1|e^{K\eps}+M|t-t_1|\rightarrow 0$ as
$(t,x)\rightarrow (t_1,x_1)$\\

$\hs{4em}\therefore\alp$  is  continuous.$\hs{8em}\square$\\

{\bf Fact.} $\alp  $ is $\ccn$ if $f$ is $\ccn.$\\

{\bf Reference.} Hirsch and Smale. Differential equations,
Dynamical systems and Linear algebra. Ch. 15.

  \end{document}