\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{\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*{Vector field.} Let $\pi:TM\rightarrow M$ be a canonical projection. $\hspace{3.8em}X_p\mapsto p$ \begin{defn}{\it A {\bf vector field} X on $A\subset M$ is a "section" of $\pi:TM\rightarrow M$ over $A$, i.e., A vector field is a map $X:A\rightarrow TM$ with $\pi\circ X=id.$}\\ \end{defn} *그림 21*\\ 즉 vector field $X$는 각 점 $p$에 대해 $X_p$를 대응시키는 map이다.\\ \begin{thm} Let $X$ be a vector field. Then the followings are equivalent; (1) $X$ is $\ccn$. (2) $ (U,x)$is a coordinate chart on $M$ and $X(p)=\sum a_i(p)\frac{\pa}{\pa x_i}(p)$ on $U$ $\Rightarrow a_i\in \ccn(U,\rb)$. (3) $V$ is open in $M$, $f\in\ccn(V,\rb)\Rightarrow Xf\in\ccn(V,\rb).$ \end{thm} \begin{proof} $(1)\Rightarrow(2)$ $a_i(p)=Xx_i(p)=dx_i(X_p)$ and $dx_i$ is a coordinate function on $\pi\inv(U)=\widehat{U}$ $\therefore dx_i$ is $\ccn\Rightarrow a_i=dx_i\circ X\in\ccn(U,\rb)$. $(2)\Rightarrow(3)$ $Xf$가 locally $\ccn$임을 보이면 된다. $Xf=\sum a_i\frac{\pa f}{\pa x_i }$ 인데 $a_i$와 $f$ 즉 $\frac{\pa f}{\pa x_i}$가 $\ccn$이므로 $Xf$는 locally $\ccn$이다. $(3)\Rightarrow(1)$ $\hspace{1em}X:U\hspace{1.5em}\rightarrow\hspace{1.5em} \widehat{U}=\pi\inv(U)$ $\hspace{6.5em}\vphi\downarrow\hspace{6em}\downarrow\hat{\vphi}$ $\hspace{6em}\vphi(U)\overset{induced\,\,by\,\,X}{\dashrightarrow}\vphi(U)\times\rb^n$ $\hspace{6.5em}x\hspace{2.5em}\mapsto\hspace{1.5em}(x,a_1,\cdot\cdot\cdot,a_n)$ 각 coordinate function이 $id,a_1,\cdot\cdot\cdot,a_n$인데 $a_i=Xx_i$이고 $x_i$는 $\ccn$이므로 (3)에 의해 $\forall a_i$는 $\ccn$이다. 따라서 각 coordinate function이 $\ccn$이므로 $X$은 $\ccn$이 된다. \end{proof} \begin{defn}{\it Let $X$ be a $\ccn$ vector field on $M$. A $\ccn$ curve $\sigma$ is an {\bf integral curve} of $X$, if $\frac{d\sigma}{dt}(t)=X(\sigma(t))$ for $\forall t\in dom(\sigma)$.}\\ \end{defn} $X$는 local하게는 open subset of $\rb^n$에서의 $\ccn$ vector field로 볼 수 있다. $X=(f_1(x),\cdot\cdot\cdot,f_n(x))$(or $X=\sum f_i(x)\paxi$)라 두면 integral curve를 찾는다는 것은 $\sigma(t)=(x_1(t),\cdot\cdot\cdot,x_n(t))$ such that $\frac{d\sigma}{dt}=X(\sigma(t))$ 를 만족하는 $\sigma$를 찾는 문제이고 곧 이는 아래의 system of o.d.e. $(*)$를 푸는 것과 같다. \[ (*) \left\{ \begin{array}{cc} \frac{dx_1}{dt}=& f_1(x) \\ \,\,\, \cdot \\ \,\,\,\cdot \\ \,\,\,\cdot \\ \frac{dx_n}{dt}=& f_n(x) \end{array} \right. \]\\ {\bf 예 1.} $X=y\frac{\pa}{\pa x}+\frac{\pa}{\pa y}=(y,1)$으로 주어졌을 때 integral curve $\sigma$는? \[ (*) \left\{ \begin{array}{cc} \frac{dx}{dt}=& y\\ \frac{dy}{dt}=& 1 \end{array} \right. \]\\ $\therefore y=t+c,x=\frac{1}{2}t^2+ct+d.$\\ {\bf 예 2.} $X=y\frac{\pa}{\pa x}+x\frac{\pa}{\pa y}=(y,x)$ with initial condition $(x(0),y(0))=(1,0)$. 미분방정식을 풀면 $-\infty