\documentclass[12pt]{article} \usepackage[latin1]{inputenc} \usepackage{ngerman} \usepackage{bbm} \usepackage{amssymb, amsmath} \usepackage{dsfont} \usepackage{fancyhdr} \usepackage{a4wide} \pagestyle{fancy} \lhead{10. November 2004} \chead{} \rhead{\bfseries Vorlesung 7} \newcounter{Satz} \newcounter{Lemma} \newcounter{Proposition} \newcounter{Korollar} \newcounter{Beispiel} % 1 Art der Aussage/Counter 2 Name 3 Verweisname \newenvironment{aussage}[3][~]{ \bigskip \par \vskip0pt plus 30pt \penalty -300 \vskip0pt plus -30pt \refstepcounter{#1} \label{#1:#3} %{\bf \underline{#1 \arabic{#1}}} #2: \\} {\medskip \par} {\bf \underline{#1 \arabic{#1}}} #2: \begin{quote} \vspace*{-0.25cm}} {\end{quote} \medskip \par} % 1 Art der Aussage 2 Name 3 Verweisname \newenvironment{sonst}[3][~]{\bigskip \par \vskip0pt plus 30pt \penalty -300 \vskip0pt plus -30pt \label{#1:#3} {\bf \underline{#1}} #2: \begin{quote}} {\end{quote} \medskip \par} \newcommand{\verweis}[2]{#1 \ref{#1:#2}} % 1 Art der Aussage 2 Verweisname \newcommand{\RR}{\mathds{R}} \newcommand{\QQ}{\mathds{Q}} \newcommand{\ZZ}{\mathds{Z}} \newcommand{\NN}{\mathds{N}} \begin{document} \begin{aussage}[Satz]{(Integralsatz von Green)}{} Sei $M$$\subset$$\RR^2$ offene Menge, die sich in endl. viele Standardbereiche zerlegen l"asst. $ V = ( V_1, V_2 )$ stetig diffb. Vektorfeld in Umgebung von $\overline{M} = M \setminus \delta M$ $$ \Rightarrow \int_{M} \left( \frac{\partial V_1}{\partial y} - \frac{\partial V_2}{\partial x} \right) d(x, y) = - \int_{\delta M} V_{1}dx + V_{2}dy $$ Hierbei soll der Rand so durchlaufen werden, dass das Innere immer links liegt. \\ $ $ \\ Beweis: \\ Beweise Satz f"ur jeden Standardbereich. Allg. Satz folgt durch Aufsummierung. \\ Also o.B.d.A. $M$ ist Standardbereich. \\ W"ahle folgende explizite Parameterdarstellung von $\delta M$ $$ \left\{ \begin{array}{c c c} [0,4] & \leftarrow & \RR^2 \\ t & \mapsto & ( x(t), y(t) ) \end{array} \right . $$ mit $$ x(t) = \left \{ \begin{array}{c c} a+t(b-a) & 0 \leq t \leq 1 \\ b & 1\leq t \leq 2 \\ b-(t-2)(b-a) & 2\leq t \leq 3 \\ a & 3 \leq t \leq 4 \end{array} \right . $$ $$ y(t) = \left \{ \begin{array}{c c} \varphi(a+t(b-a)) & 0 \leq t \leq 1 \\ \varphi(b)+(t-1)(\psi(b)-\varphi(b)) & 1\leq t \leq 2 \\ \varphi(b-(t-2)(b-a)) & 2\leq t \leq 3 \\ \varphi(a)-(t-3)(\psi(a)-\varphi(a)) & 3 \leq t \leq 4 \end{array} \right . $$ \begin{eqnarray*} \int_M\frac{\partial V_1}{\partial y} d(x,y) & \stackrel{Satz 1}{=} & \int_a^b(\int_{\varphi(x)}^{\psi(x)}\frac{\partial V_1}{\partial y}dy)dx \\ & = & \int_a^b(V_1(x,\psi(x))-V_1(x,\varphi(x)))dx \\ & = & - \int_b^aV_1(x,\psi(x))dx - \int_a^bV_1(x, \varphi(x))dx \\ & = & - \int_2^3V_1(x(t),y(t))x'(t)dt - \int_0^1V_1(x(t),y(t))x'(t)dt \\ & = & - \int_0^4V_1(x(t),y(t))x'(t)dt \qquad (\hbox{ denn } \int_1^2V_1(x(t),y(t))x'(t)=0) \\ & = & - \int_{\delta S}V_1(x,y)dx \end{eqnarray*} Analog ist zu zeigen: $ - \int_M \frac{\partial V_2}{\partial x}d(x,y) = - \int_{\delta S}V_2dy $ \phantom{hallo}\hfill$\square$ \end{aussage} \begin{aussage}[Beispiel]{}{} $ M = \{ ( x, y) \in \RR^2 \mid 0 \leq x^2 + y^2 < 1\} $ \\ $ V_{1}= 2xy^3 \quad V_{2}= 3x^2y^2 $ $$\int_{\delta M} 2xy^3dx + 3x^2y^2dy= -\int_{M} (6xy^2 - 6xy^2) d( x, y) = 0 $$ \end{aussage} \begin{aussage}[Beispiel]{}{} $M$ wie oben $V_{1}=x^{4} - y^3 \quad V_{2}=x^3 - y^{4}$ \begin{eqnarray*} \int_{\delta M} (x^{4} - y^3)dx + (x^3 - y^{4})dy & = & - \int_{M} (-3y^2 - x^2)d(x,y) \\ & = & 3\int_M (x^2+y^2) d(x,y) \quad ( \hbox{ Polarkoord.: } x= r cos\varphi \hbox{, } y= r sin\varphi ) \\ & = & 3 \int_{0}^{1}(\int_{0}^{2\pi} r^2r dr)d\varphi = 6\pi \int_{0}^{1} r^3 dr = \frac{6\pi}{4}= \frac{3}{2}\pi \end{eqnarray*} \end{aussage} \begin{aussage}[Satz]{(Gaußscher Integralsatz)}{} Sei $M \subset \RR^3$ Standardbereich $V=(V_{1}, V_{2}, V_{3})$ stetig dif\/f'bares Vektorfeld auf Umgebung von $\overline{M}$ \\ Sei $N$ "au"serer Normalenvektor auf $\delta M$ \\ $$ \Rightarrow \int_{M} div V \cdot d(x, y, z) = \int_{\delta M} (V\cdot N) d\sigma $$ $ div V = \frac{\partial V_{1}}{\partial x}+\frac{\partial V_{2}}{\partial y}+\frac{\partial V_{3}}{\partial z} $ \\ $ M= \bigcup_{(x, y) \in M_{z}} \{(x, y, z) \in \RR^3 \mid \varphi(x, y) \leq z \leq \psi (x, y)\} $ \\ $$ \begin{array}{c c c c c c c} \delta M = & \Gamma_{\varphi} & \cup & \Gamma_{\psi} & \cup & \hbox{Teil parallel z-Achse} & \\ = & \delta M_1 & = & \delta M_2 & = & \delta M_3 & \end{array} $$ F"ur $p = (x, y, z) \in \Gamma_{\psi}$ ist $N (p) = \frac{(-\psi_{x}, -\psi_{y}, 1)}{\sqrt{1+\psi_{x}^2+\psi_{y}^2}}$ \\ F"ur $p = (x, y, z) \in \Gamma_{\varphi}$ ist $N(p) = \frac{(\varphi_{x}, \varphi_{y}, -1)}{\sqrt{1+\varphi_{x}^2+\varphi_{y}^2}}$ \\ \newpage Beweis: Setze $N (p) = (N_{1}, N_{2}, N_{3}) (p)$ \\ \begin{eqnarray*} \int_{M} \frac{\partial V_{3}}{\partial z} (x, y, z) d(x, y, z) & \stackrel{Satz 1}{=} & \int_{M_{z}} (\int_{\varphi(x, y)}^{\psi(x, y)} \frac{\partial V_{3}}{\partial z} (x, y, z)dz)d(x, y) \\ & \stackrel{H.D.I.}{=} & \int_{M_{z}} [V_{3} (x, y, \psi(x, y)) - V_{3} (x, y, \varphi(x, y))] d(x, y) \\ & = & \int_{M_{z}} V_{3} (x, y, \psi(x, y)) \frac{\sqrt{\psi_{x}^2+\psi_{y}^2+1}}{\sqrt{\psi_{x}^2+\psi_{y}^2+1}}d(x,y) \\ & + & \int_{M_{z}} V_{3} (x, y, \varphi(x, y)) \frac{\sqrt{\varphi_{x}^2+\varphi_{y}^2+1}}{\sqrt{\varphi_{x}^2+\varphi_{y}^2+1}}d(x,y) \\ & & \hbox{Auf dem oberen Rand } \delta M_{1} \hbox{ ist } N_{3}(p) = \frac{1}{\sqrt{\psi_{x}^2+\varphi_{y}^2+1}} \\ & & \hbox{Auf dem unteren Rand } \delta M_{2} \hbox{ ist } N_{3}(p) = \frac{-1}{\sqrt{\psi_{x}^2+\varphi_{y}^2+1}} \\ & & \hbox{Zur z-Achse paralleler Teil } \delta M_{3} \hbox{ ist } N_{3}(p) = 0 \\ & = & \int_{M_{z}}V_{3}\cdot N_{3}\cdot\sqrt{\psi_{x}^2+\psi_{y}^2+1} \, d(x,y)\\ & + & \int_{M_{z}} V_{3}\cdot N_{3}\cdot\sqrt{\varphi_{x}^2+\varphi_{y}^2+1} \, d(x,y) \\ & = & \int_{\delta M_{1}} V_{3} N_{3} d\sigma + \int_{\delta M_{2}} V_{3} N_{3} d\sigma + \int_{\delta M_{3}} V_{3} N_{3} d\sigma \\ & = & \int_{\delta M} V_{3}\cdot N_{3}\cdot d\sigma \end{eqnarray*} Analog (wegen M x-projezierbar) $$ \int_{M} \frac{\partial V_{1}}{\partial x} d(x, y, z) = \int_{\delta M} V_{1} \cdot N_{1} d\sigma $$ $$ \int_{M} \frac{\partial V_{2}}{\partial x} d(x, y, z) = \int_{\delta M} V_{2} \cdot N_{2} d\sigma $$ \begin{eqnarray*} \Rightarrow \int_{M} div V d(x, y, z)& = & \int_{M} \left( \frac{\partial V_{1}}{\partial x}+ \frac{\partial V_{2}}{\partial y} + \frac{\partial V_{3}}{\partial z} \right) d(x, y, z) \\ & = & \int_{\delta M} (V_{1}N_{1} + V_{2}N_{2} + V_{3}N_{3}) d\sigma = \int_{\delta M} V\cdot N d\sigma \end{eqnarray*} \end{aussage} \begin{sonst}[Bemerkung]{}{} $ \int_{\delta M} V_{i}N_{i} d\sigma$ ist f"ur parametrisierte Fl"achen definiert. \\ Man hat f"ur die 3 Summanden verschiedene Parametriesierungen. $\Rightarrow$ Beweis, dass das Fl"achenintegral unabh"angig von der Parametrisierung ist kommt sp"ater. \end{sonst} \begin{aussage}[Beispiel]{}{} Sei $M =\{ (x, y, z) \in \RR^3 \mid 0 \leq x^2+y^2 < 1 \hbox{, } 0 < z < \sqrt{1 - (x^2 + y^2)} \} $ \\ $V = (V_{1}, V_{2}, V_{3})$ mit $V_{1} = xz^2$, $ V_{2} = x^2y - z^3$, $V_{3} = 2xy + y^2z$ \begin{eqnarray*} \int_{\delta M} V\cdot N d\sigma & = & \int_{M} div V d(x, y, z) \\ & = & \int_{M} (z^2 + x^2 + y^2) d(x, y, z) \\ & = & \int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}} r^2r^2 cos \theta d\theta dy dr \quad \hbox{ (mit Polarkoord.)} \\ & = & \cdots \end{eqnarray*} \end{aussage} \begin{aussage}[Satz]{(Integralsatz von Stokes)}{} Sei $D \subset \RR^3$ offen und $V = ( V_1, V_2, V_3 ): D \rightarrow \RR^3$ zweimal stetig diff'bares Vektorfeld. \\ Sei $M$$\subset$$\RR^2$ offene Menge, die sich in endl. viele Standardbereiche zerlegen l"asst. \\ Sei $ F:M \rightarrow F \subset \overline{F} \subset D \subset \RR^3$ eine parametrisierte Fl"ache $$ \Rightarrow \int_{F}(rot{V} \cdot N)d\sigma = \int_{\partial F} ( V_1dx + V_2dy + V_3dz ) $$ Bew.: \\ $$ \hbox{Sei } \stackrel{\sim}{k}: \left \{ \begin{array}{l} [a,b] \rightarrow \partial M \\ t \mapsto (\stackrel{\sim}{u}(t),\stackrel{\sim}{v}(t)) \end{array} \right. \hbox{ parametrisierung von } \partial F $$ Sei $ f(u,v) = (x(u,v), y(u,v), z(u,v))$ \\ Setze zur Abk"urzung $\stackrel{\sim}{V_1}(u,v) = V_1(x(u,v), y(u,v), z(u,v))$ \begin{eqnarray*} \Rightarrow \int_{\partial F}V_1dx & = & \intop_{a}^{b}\stackrel{\sim}{V_1}(\stackrel{\sim}{u}(t)\stackrel{\sim}{v}(t))\frac{d}{dt} (x(\stackrel{\sim}{u}(t), \stackrel{\sim}{v}(t)))dt \\ & \stackrel{Kettenregel}{=} & \intop_{a}^{b}\stackrel{\sim}{V_1}(\stackrel{\sim}{u}(t)\stackrel{\sim}{v}(t)) \{\frac{\partial x}{\partial u}(\stackrel{\sim}{u}(t)\stackrel{\sim}{v}(t))\stackrel{\sim}{u}'(t)+ \frac{\partial x}{\partial v}(\stackrel{\sim}{u}(t) \stackrel{\sim}{v}(t))\stackrel{\sim}{v}'(t)\}dt \\ & \stackrel{Def. Kurvenint.}{=} & \int_{\partial F} \{ \stackrel{\sim}{V_1}(u,v) \frac{\partial x}{\partial u}(u,v)du +\stackrel{\sim}{V_1}(u,v) \frac{\partial x}{\partial v}(u,v)dv\} \\ & \stackrel{Green}{=} & \int_{M}\{\frac{\partial}{\partial u}(\stackrel{\sim}{V_1}(u,v)\frac{\partial x}{\partial v}(u,v)) - \frac{\partial}{\partial v}(\stackrel{\sim}{V_1}(u,v)\frac{\partial x}{\partial u}(u,v))\}d(u,v) \\ & = & \int_{M}\{\frac{\partial}{\partial u}V_1(x(u,v),y(u,v),z(u,v)) \frac{\partial x}{\partial v}(u,v)) - \\ & & \phantom{\int_{M}}\frac{\partial x}{\partial v}V_1(x(u,v),y(u,v),z(u,v)) \frac{\partial x}{\partial u}(u,v))\}d(u,v) \\ \end{eqnarray*} \begin{eqnarray*} & \stackrel{Produktregel}{=} & \int_{M} \{(\frac{\partial V_1}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial V_1}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial V_1}{\partial z}\frac{\partial z}{\partial u} ) \frac{\partial x}{\partial v} + V_1\frac{\partial^2x}{\partial u \partial v } - \\ & & \phantom{\int_{M}} (\frac{\partial V_1}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial V_1}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial V_1}{\partial z}\frac{\partial z}{\partial v} \frac{\partial x}{\partial u} - V_1\frac{\partial^2x}{\partial u \partial v } \} d(u,v) - \\ & & \phantom{\int_{M}} ( \frac{\partial x \partial y}{\partial u \partial v} - \frac{\partial x \partial y}{\partial v \partial u} - \frac{\partial x \partial y}{\partial u \partial v}) \frac{\partial V_1}{\partial y} + (\frac{\partial x \partial z}{\partial v \partial u} - \frac{\partial x \partial z}{\partial u \partial v}) \frac{\partial V_1}{\partial z}\}d(u,v) = (\ast) \end{eqnarray*} Es war vergleiche $ \mathsection 5$: $$ N(u,v)=\frac{(D_x, D_y, D_z)}{\sqrt{{D_x}^2+{D_y}^2+{D_z}^2}}(u,v) \hbox{ mit } D_y = \left| \begin{array}{c c} \frac{\partial z}{\partial u} & \frac{\partial x}{ \partial u} \\ \frac{\partial z }{\partial v} & \frac{\partial x}{\partial v} \end{array} \right | \hbox{, } D_z = \left| \begin{array}{c c} \frac{\partial y}{\partial u} & \frac{\partial y}{ \partial u} \\ \frac{\partial x }{\partial v} & \frac{\partial y}{\partial v} \end{array} \right | \hbox{ und } $$ $ D = \sqrt{{D_x}^2+{D_y}^2+{D_z}^2} = \|f_u \times f_v\| $ \\ \begin{eqnarray*} (\ast) & = & \int_{M}\{-D_z \frac{\partial V_1}{\partial y} + D_y \frac{\partial V_1}{\partial z}\}d(u,v) \\ & = & \int_{M}(-\frac{\partial V_1}{\partial y}\frac{D_z}{D} + \frac{\partial V_1}{\partial z}\frac{D_y}{D})\|f_u \times f_v\|d(u,v) \\ & = & \int_{F}(-\frac{\partial V_1}{\partial y}\frac{D_z}{D} + \frac{\partial V_1}{\partial z}\frac{D_y}{D})d\sigma \end{eqnarray*} Analog erh"alt man: \begin{eqnarray*} \int_{\partial F} V_2dy & = & \int_{F}(-\frac{\partial V_2}{\partial z}\frac{D_x}{D} + \frac{\partial V_2}{\partial x}\frac{D_z}{D})d\sigma \\ \int_{\partial F} V_3dy & = & \int_{F}(-\frac{\partial V_3}{\partial x}\frac{D_y}{D} + \frac{\partial V_3}{\partial y}\frac{D_x}{D})d\sigma \end{eqnarray*} Summation ergibt: \begin{eqnarray*} \int_{\partial F} V_1dx + V_2dy + V_dz & = & \int_{F}((\frac{\partial V_3}{\partial y} - \frac{\partial V_2}{\partial z})\frac{D_x}{D}+(\frac{\partial V_1}{\partial z} - \frac{\partial V_3}{\partial x}) \frac{D_y}{D} + (\frac{\partial V_2}{\partial x } - \frac{\partial V_1}{\partial y})\frac{D_x}{D})d \sigma \\ & = & \int_{F}rot{V} \cdot Nd \sigma \end{eqnarray*} \end{aussage} \begin{sonst}[Beispiel]{}{} $ D = \RR^3, \, V(x,y,z) = (-y,x,1)$ \\ $F=$ obere Halbkugel vom Radius $2$ \\ $M = \{(u,v) \in \RR^2 \, | \, u^2+v^2 \le 4 \} $ \\ $$ f : \left\{ \begin{array}{l} M \rightarrow F \subset \RR^3 \\ (u,v) \mapsto (x(u,v),y(u,v),z(u,v)) \end{array} \right. \hbox{ mit } \left\{ \begin{array}{l l} x(u,v) = u \\ y(u,v) = v \\ z(u,v) = \sqrt{4-u^2-v^2} \end{array} \right. $$ \begin{eqnarray*} rot{V} & = & (\frac{\partial V_3}{\partial y} - \frac{\partial V_2}{\partial z}, \frac{\partial V_1}{\partial z} - \frac{\partial V_3}{\partial x}, \frac{\partial V_2}{\partial x } - \frac{\partial V_1}{\partial y}) = (0,0,2) \\ N & = & \frac{(-\frac{\partial z }{\partial u},-\frac{\partial z }{\partial v},1)} {\sqrt{1+(\frac{\partial z}{\partial u})^2 +(\frac{\partial z}{\partial v})^2}} \\ rot{V} \cdot N & = & \frac{2}{\sqrt{1+(\frac{\partial z}{\partial u})^2 +(\frac{\partial z}{\partial v})^2}} = \frac{2}{\|f_u \times f_v \|} \\ \int_{F}rot{V} \cdot N d\sigma & = & \int_{M}\frac{2}{\|f_u \times f_v \|} \|f_u \times f_v \| d(u,v) = \int_{M}2d(u,v)= 8\pi \end{eqnarray*} Satz von Stokes: \\ $$ \int_{\partial F} (-ydx+xdy+dz) = 8\pi $$ \end{sonst} \end{document}