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\begin{document}
 \parindent=0cm
  \section*{Fundamental theorem of ordinary differential equation.}
  \begin{thm}(Fundamental theorem of o.d.e.)

$W\subseteq\rb^n:open,$ $f:W\rightarrow\rb^n$: "locally
Lipschitz".

Given a system of o.d.e with initial condition
\[ (*) \left\{
         \begin{array}{cc}
               x'=& f(x) \\
             x(0)=& x_0
          \end{array} \right. \]

$\exists \,a>0$ and a solution $x:(-a,a)\rightarrow W$ of (*). If
$x,y$ are solutions of (*), defined on $I(=open\,\,interval$)
containing 0 then $x\equiv y$ on $I$.






\end{thm}


\begin{defn}{\it (Lipschitz condition)}
{\it $f:W\rightarrow\rb^n$ is {\bf Lipschitz }on $W$ if
$\exists\,K>0$ such that} {\it $|f(y)-f(x)|\leq
K|y-x|\,\,\forall\,x,y\in W$. Here $K$ is called a Lipschitz
constant. $f:W\rightarrow\rb^n$ is locally Lipschitz on $W$ if
each $x\in W$
has a neighborhood on which f is Lipschitz. }\\
\end{defn}

{\bf Note.} $f:W\rightarrow\rb^n$, $\mathcal{C}^1\Rightarrow f$ is
locally Lipschitz.\\

(����) Let $\phi(t)=f(x+t(y-x))$. Then
$f(y)-f(x)=\phi(1)-\phi(0)=\int_0^1\phi'(t)dt$.

$\therefore f(y)-f(x)=\int_0^1Df(x+ t(y-x))(y-x)dt$.

$|Df(x+t(y-x))||y-x|\leq K|y-x|\,\,\,\,$  where $\,\,K=max||Df||$
on the neighborhood considered.
$\therefore|f(y)-f(x)|\leq K|y-x|$.\\

{\bf (proof of the fundamental theorem)}

Key observation $(*)\Leftrightarrow x(t)=x_0+\int_0^t f(x(s))ds$.

$f$ is Lipschitz on $B=\{x\in W\,\,|\,\,\,|x-x_0|\leq b\}\subset
W$ with Lipschitz constant $K$ and let $|f(x)|\leq M$ on
$\overline{B}$.

Choose $a>0$ such that $a<min\{\frac{b}{M},\frac{1}{K}\}$ and let
$J=[-a,a]$. Construct approximation solution
$u_0,u_1,\cdot\cdot\cdot,$ as follows :

$\hspace{3em}u_0(t)=x_0,u_{k+1}(t)=x_0+\int_0^tf(u_k(s))ds$, $t\in
J.$\\


$\hspace{0em}$(1) {\it $u_k(t)\in B$ : use induction.}

      $|u_{k+1}(t)-x_0|\leq\int_0^t|f(u_k(s))|ds\leq\int_0^tMds=Mt<Ma<b$.\\

$\hspace{0em}$(2) {\it $\exists L\geq 0$ such that
$|u_{k+1}(t)-u_k(t)|\leq(Ka)^k\,\,L,\,\,\,\forall\, k$} :

$|t|<a$�� ��� $t$�� ���� $L=max|u_1(t)-u_0(t)|$ �� ����.
Lipschitz  ���ǰ� induction�� ����

$|u_{k+1}(t)-u_k(t)|\leq\int_0^t|f(u_k(s))-f(u_{k-1}(s))|ds$

$\hspace{7.2em}\leq\int_0^tK|u_k(s)-u_{k-1}(s)|ds$

$\hspace{7.3em}\leq
K(Ka)^{k-1}La=(Ka)^kL$.\\



$\hspace{0em}$(3) {\it $\{u_k\}$ is a Cauchy sequence (with
respect to uniform norm)} :

For $r,s>N$,
$|u_r(t)-u_s(t)|\leq\dis{\sum_{k=N}^\infty|u_{k+1}(t)-u_k(t)|\leq\sum_{k=N}^\infty(Ka)^kL}$\\

$a$�� ���ǿ� ���� $Ka<1$ �̰� ���� $N\rightarrow\infty$�̸� ��
���� 0�� ����. ���� $u_k$�� uniform limit $u$�� ������ �̰���
�ٷ� �츮�� ã�� ���̴�. ��,

$u_{k+1}(t)=x_0+\int_0^tf(u_k(s))ds$ ���� �纯��
$\underset{k\rightarrow\infty}{\lim}$ �� ���ϸ�\\

$u(t)=x_0+\underset{k\rightarrow\infty}{\lim}\int_o^tf(u(s))ds$


$\hspace{2em}=x_0+\int_0^t\underset{k\rightarrow\infty}{\lim}
f(u_k(s))ds$ $\hspace{2em}by \,\,Lebesgue's
\,\,dominated\,\,convergence\,\, theorem$

$\hs{2em}=x_0+\int_0^tf(u(s))ds$$\hspace{4em}
by\,\,Lipschitz\,\,condition$\\

���� $u$�� ���й�����, ���� �̺й������� �����ϰ� $f$�� �����̹Ƿ� $u$�� $\mathcal{C}^1$�̴�.\\

(4) {\it Uniqueness} :

���� ���� �ٹ濡���� $x\equiv y$�� �ǰ� �� �� ������ ������.

$\exists\,\epsilon>0\,\,such\,\,that\,\,x\equiv
y\,\,on\,\,(-\eps,\eps)$ :

Choose $\eps>0$ such that $x(J),y(J)\in
B,\,\,J=[-\eps,\eps]\subseteq[-a,a]$ and $\eps<\frac{1}{K}$.

Let $Q=\underset{J}{max}|x(t)-y(t)|$  $\,\,\,\,\,(J$:compact and
$x,y$:continuous$\Rightarrow\exists\,\,max.$)

$\hs{2.9em}=|x(t_1)-y(t_1)|$ for some $t_1\in J$.

$\hs{2.9em}=|\int_0^{t_1}x'(s)-y'(s)ds|$

$\hs{2.9em}\leq\int_0^{t_1}|x'(s)-y'(s)|ds$

$\hs{2.9em}\leq\int_0^{t_1}|f(x(s))-f(y(s))|ds$

$\hs{2.9em}\leq K\int_0^{t_1}|x(s)-y(s)|ds$

$\hs{2.9em}\leq
KQ\eps<Q\hs{2em}:\hs{2em}contradiction\,\,unless\,\,Q=0.$

���� $Q=0$ �� �Ǵ� $\eps$�� ��Ұ� $\,\,\eps$�ٹ濡���� $x\equiv
y$�̴�.

���� global�ϰԵ� $x\equiv y$�� �������� ���̱� ���� $A=\{t\in
I\,|\,x(t)=y(t)\}$ �� ��������. $I$�� connected���� �̿��ؼ�
$A\subset I$�� open�̰� closed���� ���̸� $A=I$�̴�.
$(A\neq\emptyset$�̹Ƿ�)

���� $A=\{t\in I\,|\,x(t)=y(t)\}$���̹Ƿ� Hausdorff ���ǿ��� �̴�
�翬�� closed�̴�. $A$�� open���� ���̱� ���� $t_0\in A$ ��
��������. �� $x(t_0)=y(t_0)=p$. �׸���  $x,y$�� ���� $(*)$�� �ذ�
�ȴ�.

\[ (*) \left\{
         \begin{array}{cc}
               x'=& f(x) \\
             x(t_0)=& p
          \end{array} \right. \]

{\bf Note.} $x(t)$�� $(*)$�� ���̸� $\widetilde{x}(t)=x(t+t_0)$
���� $(*)$�� �ذ� �ǰ� �� �� initial condition��
$\widetilde{x}(0)=x(t_0)=p$�̴�.

$(\because)
\widetilde{x}'=x'(t+t_0)=f(x(t+t_0))=f(\widetilde{x}(t))$.\\

���� $\widetilde{x}(t)=x(t+t_0),\widetilde{y}(t)=y(t+t_0)$ ��
�θ�, �̵��� �� �� $(*)$�� �ذ� �Ǹ鼭 ���� initial condition
$(t,x)=(0,p)$�� ������. {\it uniqueness} ������ �պκп���
$0$�ٹ濡���� $\widetilde{x}\equiv\widetilde{y}$�� �ǰ� �� ��
�ִٰ� �����Ƿ� $\exists\eps>0$ such that
$\widetilde{x}\equiv\widetilde{y}$ on $(-\eps,\eps)$. ����
$x(t)\equiv y(t)$ on $(t_0-\eps,t_0+\eps)$�̴�. �� $A$�� ����
$t_0$�� �ٹ��� ������Ƿ� $A$�� open�̴�.${\hs{4em}}\square$\\

{\bf Unique maximal integral curve on $M$.}

\begin{thm}
Let $X$ be $\ccn$ vector field on Hausdorff $M$. $\forall p\in M,
\exists ! $ maximal integral curve $\sigma=\sigma_p$ with
$\sigma(0)=p$. i.e., $\exists$ open interval
$(a(p),b(p))\subset\rb$ and

$\sigma:(a(p),b(p))\rightarrow M$ such that

(1) $0\in(a(p),b(p))$ and $\sigma(0)=p$.

(2) $\sigma$ is an integral curve.

(3) If $\gamma:(c,d)\rightarrow M$ is a smooth curve with (1) and
(2), then $\gamma=\sigma|_{(c,d)}$.


\end{thm}

\begin{proof}
Fundamental theorem���κ��� unique local existence �� �����,
������ $\mathcal{I}$�� ��������.

$\mathcal{I}=\{I=(a,b)\,|\,\exists\sigma:(a,b)\rightarrow
M\,:\,integal\,\,curve,\,\sigma(0)=p\}$.

���� �� ���� curve $\sigma:I\rightarrow M,\tau:J\rightarrow M$��
(1),(2)�� �����Ѵٸ�, uniqueness�� ���� $I\cap J$������
$\sigma\equiv\tau$���� �Ѵ�. �� ��ġ�� �������� �׻� �����Ƿ� ��
�ٿ����� �� �ִ�. ����
$(a(p),b(p))=\dis{\bigcup_{I\in\mathcal{I}}}I$ �� �ΰ� $\sigma$��
"union" of integral curves�� �θ� �ȴ�.

\end{proof}






  \end{document}
