\documentclass[12pt ]{article} \setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true cm} \usepackage{graphicx} \usepackage{hangul} \usepackage{amscd,amsmath} \usepackage{amsfonts} \usepackage{amssymb,theorem} \usepackage{longtable} \newcommand{\hs}{\hspace} \newcommand{\inv}{^{-1}} \newcommand{\vphi}{\varphi} \newcommand{\paxi}{\frac{\partial}{\partial x_i}} \newcommand{\payj}{\frac{\partial}{\partial y_j}} \newcommand{\paxj}{\frac{\partial}{\partial x_j}} \newcommand{\payi}{\frac{\partial}{\partial y_i}} \newcommand{\dis}{\displaystyle} \newcommand{\disi}{\displaystyle{\sum_{i=1}^n}} \newcommand{\disj}{\displaystyle{\sum_{j=1}^n}} \newcommand{\pa}{\partial} \newcommand{\Aff}{\mbox{\it Aff}} \newcommand{\aff}{\mbox{\it aff}} \newcommand{\cc}{\mathcal{C}} \newcommand{\ccn}{\mathcal{C}^{\infty}} \newcommand{\sbb}{\mathbb{S}} \newcommand{\rb}{\mathbb{R}} \newcommand{\rc}{\mathcal{R}} \newcommand{\alp}{\alpha} \newcommand{\bet}{\beta} \newcommand{\del}{\delta} \newcommand{\gam}{\gamma} \newcommand{\vep}{\varepsilon} \newcommand{\eps}{\epsilon} \newcommand{\lam}{\lambda} \newcommand{\kap}{\kappa} \newcommand{\sig}{\sigma} \newcommand{\ome}{\omega} \newcommand{\Gam}{\Gamma} \newcommand{\Ome}{\Omega} \newcommand{\Sig}{\Sigma} \newcommand{\Del}{\Delta} \newcommand{\Lam}{\Lambda} \newtheorem{thm}{정리} \newtheorem{cor}[thm]{따름정리} \newtheorem{lem}[thm]{보조정리} \newtheorem{prop}[thm]{명제} \newtheorem{cl}{Claim} {\theorembodyfont{\rm} \newtheorem{ex}{예} \newtheorem{que}{질문} \newtheorem{notation}{Notation}[section] \newtheorem{defn}{정의} \newtheorem{rem}{주} \newtheorem{note}{Note} } \renewcommand{\thenote}{} \renewcommand{\therem}{} \newenvironment{proof}{{\bf 증명}}{\hfill\framebox[2mm]{}} \newenvironment{proof1}{{\bf 정리증명}}{\hfill\framebox[2mm]{}} \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.\\ (증명) 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 $aN$, $|u_r(t)-u_s(t)|\leq\dis{\sum_{k=N}^\infty|u_{k+1}(t)-u_k(t)|\leq\sum_{k=N}^\infty(Ka)^kL}$\\ $a$의 정의에 의해 $Ka<1$ 이고 따라서 $N\rightarrow\infty$이면 위 식은 0로 간다. 따라서 $u_k$는 uniform limit $u$를 가지고 이것이 바로 우리가 찾던 해이다. 즉, $u_{k+1}(t)=x_0+\int_0^tf(u_k(s))ds$ 에서 양변에 $\underset{k\rightarrow\infty}{\lim}$ 를 취하면\\ $u(t)=x_0+\underset{k\rightarrow\infty}{\lim}\int_o^tf(u(s))ds$ $\hspace{2em}=x_0+\int_0^t\underset{k\rightarrow\infty}{\lim} f(u_k(s))ds$ $\hspace{2em}by \,\,Lebesgue's \,\,dominated\,\,convergence\,\, theorem$ $\hs{2em}=x_0+\int_0^tf(u(s))ds$$\hspace{4em} by\,\,Lipschitz\,\,condition$\\ 따라서 $u$는 적분방정식, 따라서 미분방정식을 만족하고 $f$가 연속이므로 $u$는 $\mathcal{C}^1$이다.\\ (4) {\it Uniqueness} : 먼저 작은 근방에서는 $x\equiv y$가 되게 할 수 있음을 보이자. $\exists\,\epsilon>0\,\,such\,\,that\,\,x\equiv y\,\,on\,\,(-\eps,\eps)$ : 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 $t_1\in J$. $\hs{2.9em}=|\int_0^{t_1}x'(s)-y'(s)ds|$ $\hs{2.9em}\leq\int_0^{t_1}|x'(s)-y'(s)|ds$ $\hs{2.9em}\leq\int_0^{t_1}|f(x(s))-f(y(s))|ds$ $\hs{2.9em}\leq K\int_0^{t_1}|x(s)-y(s)|ds$ $\hs{2.9em}\leq KQ\eps0$ such that $\widetilde{x}\equiv\widetilde{y}$ on $(-\eps,\eps)$. 따라서 $x(t)\equiv y(t)$ on $(t_0-\eps,t_0+\eps)$이다. 즉 $A$에 들어가는 $t_0$의 근방을 잡았으므로 $A$는 open이다.${\hs{4em}}\square$\\ {\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 theorem으로부터 unique local existence 는 얻었고, 다음의 $\mathcal{I}$를 생각하자. $\mathcal{I}=\{I=(a,b)\,|\,\exists\sigma:(a,b)\rightarrow M\,:\,integal\,\,curve,\,\sigma(0)=p\}$. 만일 두 개의 curve $\sigma:I\rightarrow M,\tau:J\rightarrow M$가 (1),(2)를 만족한다면, uniqueness에 의해 $I\cap J$에서는 $\sigma\equiv\tau$여야 한다. 즉 겹치는 곳에서는 항상 같으므로 잘 붙여나갈 수 있다. 이제 $(a(p),b(p))=\dis{\bigcup_{I\in\mathcal{I}}}I$ 라 두고 $\sigma$를 "union" of integral curves라 두면 된다. \end{proof} \end{document}