\documentclass[11pt,a4paper,oneside,fleqn]{article} % usepackages \usepackage{latexsym} \usepackage{amstext,amsthm} \usepackage{amsmath,amssymb} \usepackage{amsfonts,german} % Einstellungen \setlength{\oddsidemargin}{0.46cm} \setlength{\textwidth}{16cm} \setlength{\mathindent}{1cm} \setlength{\topmargin}{-1cm} \setlength{\headheight}{0cm} \setlength{\headsep}{0cm} \setlength{\topskip}{0cm} \setlength{\textheight}{26cm} % Neue Kommandos \newcommand{\fa}{\qquad \forall \quad} \newcommand{\eps}{\varepsilon} \newcommand{\im}{{\rm Im}\,} \newcommand{\re}{{\rm Re}\,} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand {\field}[1]{\mathbb{#1}} \newcommand {\B}{\mathcal{B}} \newcommand {\C}{\field C} \newcommand {\D}{\mathcal{D}} \newcommand {\E}{\mathcal{E}} \newcommand {\F}{\field{F}} \newcommand {\G}{\mathcal{G}} \newcommand {\K}{\mathcal{K}} \newcommand {\Ll}{\mathcal{L}} \newcommand {\N}{\field N} \newcommand {\Oo}{\mathcal{O}} \newcommand {\PP}{\mathcal{P}} \newcommand {\Q}{\field Q} \newcommand {\R}{\field R} \newcommand {\Ss}{\mathcal{S}} \newcommand {\T}{\field T} \newcommand {\Z}{\field Z} % Umgebung fuer Saetze \newtheorem{satz}{Satz}%[section] \newtheorem{lemma}[satz]{Lemma} \newtheorem{hilfssatz}[satz]{Hilfssatz} \newtheorem{korollar}[satz]{Korollar} \newtheorem{prop}[satz]{Proposition} \newtheorem{defi}[satz]{Definition} \newtheorem{bem}[satz]{Bemerkung} \newtheorem{beis}[satz]{Beispiel} \newcommand{\Arsinh}{\operatorname{Arsinh}} \newcommand{\Arcosh}{\operatorname{Arcosh}} \newcommand{\Artanh}{\operatorname{Artanh}} \newcommand{\Arcoth}{\operatorname{Arcoth}} \newcommand{\dist}{\operatorname{dist}} \newcommand{\rot}{\operatorname{rot}} \newcommand{\Graph}{\operatorname{Graph}} \newcommand{\divergens}{\operatorname{div}} \begin{document} \pagestyle{empty} \noindent Mathematisches Institut LMU \hspace{7cm} 13.06.2005\\ Prof. Dr. H. Steinlein \\ \begin{center} \ % {\Large \bf {\"Ubungsblatt 9 zu MPIIA} } \end{center} \ % \vspace{1cm} \noindent{\bf Aufgabe 33: (4 Punkte)} \begin{itemize} \item[a)] Es seien \(f,g:\mathbb R^3\to\mathbb R\) und \(u:\mathbb R^3\to\mathbb R^3\) alle aus \(C^2(\mathbb R^3)\). Zeigen Sie: \begin{itemize} \item[1)] \(\divergens(\rot u)=0\). \item[2)] \(\nabla(\divergens u)-\rot(\rot u)=\Delta u\). \item[3)] \(\rot(fu)=\nabla f\times u+f(\rot u)\). \end{itemize} \item[b)] Zeigen Sie: Sind \(f,g:M\subset\mathbb R^n\to\mathbb R\) aus \(C^2(M)\) , so gilt \begin{align*} \Delta(fg)=(\Delta f)g+2\nabla f\cdot\nabla g+f\Delta g. \end{align*} \item[c)] Zeigen Sie: Ist \(F:I\to\mathbb R\), \(I\subset\,]0,\infty[\,\), eine \(C^2\)-Funktion und \(f(x):=F(r)\), wobei \(r:=|x|\), \(x\in\mathbb R^n\), so gilt \begin{align*} \Delta f(x)=F''(r)+\frac{n-1}{r}F'(r). \end{align*} \item[d)] Zeigen Sie: Ist \(f:\mathbb R^2\to\mathbb R\) aus \(C^2(\mathbb R^2)\) und \(F(r,\varphi):=f(r\cos\varphi, r\sin\varphi)\), so gilt \begin{align*} (\Delta f)(r\cos\varphi,r\sin\varphi)=\Big(\frac{\partial^2 F}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 F}{\partial \varphi^2}+\frac{1}{r}\frac{\partial F}{\partial r}\Big)(r,\varphi). \end{align*} \end{itemize} \noindent{\it L\"osungsvorschlag:} a) 1) \begin{align*} \divergens(\rot u)&=\frac{\partial}{\partial x}\big((\rot u)_x\big)+ \frac{\partial}{\partial y}\big((\rot u)_y\big)+\frac{\partial}{\partial z}\big((\rot u)_z\big) \\&=\frac{\partial}{\partial x}\big(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z}\big)+\frac{\partial}{\partial y}\big(\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x}\big) +\frac{\partial}{\partial z}\big(\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\big) =0.\\ &\big[\text{ Weil } \frac{\partial^2 u_3}{\partial x\partial y}=\frac{\partial^2 u_3}{\partial y\partial x},\ \frac{\partial^2 u_2}{\partial x\partial z}=\frac{\partial^2 u_2}{\partial z\partial x},\ \frac{\partial^2 u_1}{\partial y\partial z}=\frac{\partial^2 u_1}{\partial z\partial y} \end{align*} weil \(u\in C^2(\mathbb R^3)\), und alle partielle Ableitungen 2. Ordnung damit stetig sind.\(\ \big]\) 2) \begin{align*} \nabla&(\divergens u)-\rot(\rot u) =\nabla\Big(\frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}+\frac{\partial u_3}{\partial z}\Big) -\rot\big( (\rot u)_x, (\rot u)_y, (\rot u)_z\big) \\& =\Big(\frac{\partial^2 u_1}{\partial x^2}+\frac{\partial^2 u_2}{\partial x\partial y}+\frac{\partial^2 u_3}{\partial x\partial z},\frac{\partial^2 u_1}{\partial y\partial x}+ \frac{\partial^2 u_2}{\partial y^2}+\frac{\partial^2 u_3}{\partial y\partial z}, \frac{\partial^2 u_1}{\partial z\partial x}+\frac{\partial^2 u_2}{\partial z\partial y}+\frac{\partial^2 u_3}{\partial z^2}\Big) \\&\quad -\Big(\frac{\partial}{\partial y}\big(\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\big)-\frac{\partial}{\partial z}\big(\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x}\big), \frac{\partial}{\partial z}\big(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z}\big)-\frac{\partial}{\partial x}\big(\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\big), \\&\qquad\qquad\qquad\frac{\partial}{\partial x}\big(\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x}\big)-\frac{\partial}{\partial y}\big(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z}\big)\Big) \\&= \Big(\frac{\partial^2 u_1}{\partial x^2}+\frac{\partial^2 u_1}{\partial y^2}+\frac{\partial^2 u_1}{\partial z^2}, \frac{\partial^2 u_2}{\partial x^2}+\frac{\partial^2 u_2}{\partial y^2}+\frac{\partial^2 u_2}{\partial z^2},\frac{\partial^2 u_3}{\partial x^2}+\frac{\partial^2 u_3}{\partial y^2}+\frac{\partial^2 u_3}{\partial z^2}\Big) \\&=\big(\Delta u_1,\Delta u_2,\Delta u_3\big)=\Delta u. \end{align*} \(\big[\ \) Weil \begin{align*} &\frac{\partial^2 u_1}{\partial z\partial x}=\frac{\partial^2 u_1}{\partial x\partial z}; \ \frac{\partial^2 u_2}{\partial y\partial x}=\frac{\partial^2 u_2}{\partial x\partial y}; \ \frac{\partial^2 u_3}{\partial x\partial z}=\frac{\partial^2 u_3}{\partial z\partial x};\ \\& \frac{\partial^2 u_3}{\partial y\partial z}=\frac{\partial^2 u_3}{\partial z\partial y};\ \frac{\partial^2 u_3}{\partial y\partial z}=\frac{\partial^2 u_2}{\partial z\partial y};\ \frac{\partial^2 u_1}{\partial y\partial x}=\frac{\partial^2 u_1}{\partial x\partial y} \end{align*} wie oben. \(\ \big]\) 3) \begin{align*} u&=(u_1,u_2,u_3);\ fu=(fu_1,fu_2,fu_3);\\ \rot(fu)&=\Big(\frac{\partial}{\partial y}(fu_3)-\frac{\partial}{\partial z}(fu_2), \frac{\partial}{\partial z}(fu_1)-\frac{\partial}{\partial x}(fu_3), \frac{\partial}{\partial x}(fu_2)-\frac{\partial}{\partial y}(fu_1)\Big) \\&=\Big(\big[u_3 \frac{\partial f}{\partial y}+f \frac{\partial u_3}{\partial y}\big]-\big[ u_2 \frac{\partial f}{\partial z}+f \frac{\partial u_2}{\partial z}\big], \big[ u_1\frac{\partial f}{\partial z}+f \frac{\partial u_1}{\partial z}\big]-\big[ u_3 \frac{\partial f}{\partial x} +f \frac{\partial u_3}{\partial x}\big], \\&\qquad\qquad\qquad \big[ u_2 \frac{\partial f}{\partial x}+f \frac{\partial u_2}{\partial x}\big]-\big[ u_1 \frac{\partial f}{\partial y}+f\frac{\partial u_1}{\partial y}\big]\Big) \\& =f\Big(\frac{\partial u_3}{\partial y} - \frac{\partial u_2}{\partial z}, \frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x}, \frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\Big) \\&\quad+\Big(\big(\nabla f\big)_2 u_3-\big(\nabla f\big)_3u_2, \big(\nabla f\big)_3u_1-\big(\nabla f\big)_1u_3, \big(\nabla f\big)_1u_2-\big(\nabla f\big)_2u_1\Big) \\&= f\rot(u)+\nabla f\times u. \end{align*} b) \begin{align*} \Delta(fg)&=\sum_{j=1}^{n}\frac{\partial^2}\partial x_{j}^{2}(fg) =\sum_{j=1}^{n}\frac{\partial}{\partial x_j}\big(f\frac{\partial g}{\partial x_j}+g\frac{\partial f}{\partial x_j}\big) \\&=\sum_{j=1}^{n}\big(\frac{\partial f}{\partial x_j}\,\frac{\partial g}{\partial x_j}+f\frac{\partial^2 g}{\partial x_j^2}+\frac{\partial g}{\partial x_j}\,\frac{\partial f}{\partial x_j}+g\frac{\partial^2 f}{\partial x_j^2}\big)\\& =g\big(\sum_{j=1}^{n}\frac{\partial^2 f}{\partial x_j}\big)+2\sum_{j=1}^{n}\big(\frac{\partial f}{\partial x_j}\,\frac{\partial g}{\partial x_j}\big) +f\big(\sum_{j=1}^{n}\frac{\partial^2 g}{\partial x_j}\big)\\& =g\Delta f+2\nabla f\cdot\nabla g+f\Delta g. \end{align*} c) Man verwendet die Kettenregel (mehrmals): \begin{align*} g(x)&=|x|;\quad f=F\circ g;\ f:\mathbb R^n\to\mathbb R;\\ &\frac{\partial g}{\partial x_j}(x)=\frac{x_j}{|x|}; \ \frac{\partial^2 g}{\partial x_j^2}(x)=\frac{\partial}{\partial x_j}\big(\frac{x_j}{|x|}\big) =\frac{|x|\frac{\partial}{\partial x_j}(x_j)-x_j\frac{\partial}{\partial x_j}(|x|)}{|x|^2} =\frac{|x|-x_j\frac{x_j}{|x|}}{|x|^2}=\frac{|x|^2-x_j^2}{|x|^3}; \\& \frac{\partial f}{\partial x_j}(x)=\Big(\frac{\partial}{\partial x_j}\big(F\circ g)\Big)(x) =F'(g(x))\,\frac{\partial g}{\partial x_j}(x) =F'(r)\,\frac{x_j}{|x|}; \\& \frac{\partial^2 f}{\partial x_j^2}(x) =\frac{\partial}{\partial x_j}\big(F'(g(x))\,\frac{x_j}{|x|}\big) =\frac{\partial}{\partial x_j}\big(F'(g(x))\big) \, \frac{x_j}{|x|} +F'(g(x)) \frac{\partial}{\partial x_j}\big(\frac{x_j}{|x|}\big)\\& =\frac{\partial}{\partial x_j}\big((F'\circ g)(x)\big)\frac{x_j}{|x|} +F'(r)\,\frac{|x|^2-x_j^2}{|x|^3} =\big(F''(g(x)) \frac{\partial g}{\partial x_j}\big)\frac{x_j}{|x|} +F'(r)\,\frac{|x|^2-x_j^2}{|x|^3}; \\ & \Delta f(x)=\sum_{j=1}^n\frac{\partial^2 f}{\partial x_j^2}(x) =\sum_{j=1}^n\Big\{ F''(g(x))\frac{x_j}{|x|}\frac{x_j}{|x|}+F'(r)\,\frac{|x|^2-x_j^2}{|x|^3}\Big\} \\&=F''(r)\Big(\frac{\sum_{j=1}^n x_j^2}{|x|^2}\Big)+F'(r)\frac{1}{|x|^3}\sum_{j=1}^n\big(|x|^2-x_j^2\big) =F''(r)\cdot1+F'(r)\frac{1}{|x|^3}\big(n|x|^2-|x|^2\big)\\&=F''(r)+\frac{n-1}{r}F'(r). \end{align*} d) Man verwendet wieder die Kettenregel. \begin{align*} F&=f\circ g;\ g(r,\varphi)=(r\cos\varphi, r\sin\varphi)=(g_1(r,\varphi), g_1(r,\varphi));\\ \frac{\partial F}{\partial r}(r,\varphi)&=\frac{\partial f}{\partial x_1}(g(r,\varphi))\,\frac{\partial g_1}{\partial r}(r,\varphi) +\frac{\partial f}{\partial x_2}(g(r,\varphi))\,\frac{\partial g_2}{\partial r}(r,\varphi) \\&=\frac{\partial f}{\partial x_1}(g(r,\varphi)) \cos\varphi +\frac{\partial f}{\partial x_2}(g(r,\varphi))\sin\varphi ; \\ \frac{\partial^2 F}{\partial r^2}(r,\varphi)&=\frac{\partial}{\partial r}\Big[ \frac{\partial f}{\partial x_1}(g(r,\varphi)) \cos\varphi +\frac{\partial f}{\partial x_2}(g(r,\varphi))\sin\varphi\Big] \\&=\Big[\frac{\partial^2 f}{\partial x_1\partial x_1}(g(r,\varphi))\frac{\partial g_1}{\partial r}(r,\varphi) +\frac{\partial^2 f}{\partial x_2\partial x_1}(g(r,\varphi))\frac{\partial g_2}{\partial r}(r,\varphi) \Big]\cos\varphi \\&\ +\Big[\frac{\partial^2 f}{\partial x_1\partial x_2}(g(r,\varphi))\frac{\partial g_1}{\partial r}(r,\varphi) +\frac{\partial^2 f}{\partial x_2\partial x_2}(g(r,\varphi))\frac{\partial g_2}{\partial r}(r,\varphi) \Big]\sin\varphi\\ &=\frac{\partial^2 f}{\partial x_1^2}(g(r,\varphi))\cos^2\varphi +\frac{\partial^2 f}{\partial x_2^2}(g(r,\varphi))\sin^2\varphi +2\frac{\partial^2 f}{\partial x_1\partial x_2}(g(r,\varphi))\sin\varphi\cos\varphi;\\ \frac{\partial F}{\partial \varphi}(r,\varphi)&=\frac{\partial f}{\partial x_1}(g(r,\varphi))\,\frac{\partial g_1}{\partial \varphi}(r,\varphi) +\frac{\partial f}{\partial x_2}(g(r,\varphi))\,\frac{\partial g_2}{\partial \varphi}(r,\varphi) \\&=\frac{\partial f}{\partial x_1}(g(r,\varphi)) (-r\sin\varphi) +\frac{\partial f}{\partial x_2}(g(r,\varphi))(r\cos\varphi) ; \\ \frac{\partial^2 F}{\partial \varphi^2}(r,\varphi) &=\frac{\partial}{\partial\varphi}\Big[\frac{\partial f}{\partial x_1}(g(r,\varphi)) (-r\sin\varphi) +\frac{\partial f}{\partial x_2}(g(r,\varphi))(r\cos\varphi) \Big] \\&=\frac{\partial}{\partial\varphi}\Big[\frac{\partial f}{\partial x_1}(g(r,\varphi))\Big](-r\sin\varphi)+\frac{\partial f}{\partial x_1}(g(r,\varphi))\frac{\partial}{\partial\varphi}\big[(-r\sin\varphi\big] \\& \ +\frac{\partial}{\partial\varphi}\Big[\frac{\partial f}{\partial x_2} (g(r,\varphi))\Big](r\cos\varphi)+\frac{\partial f}{\partial x_2}(g(r,\varphi))\frac{\partial}{\partial\varphi}\big[r\cos\varphi)\big] \\&=\Big[\frac{\partial^2 f}{\partial x_1\partial x_1}(g(r,\varphi))\frac{\partial g_1}{\partial \varphi}(r,\varphi)+\frac{\partial^2 f}{\partial x_1\partial x_2}(g(r,\varphi))\frac{\partial g_2}{\partial \varphi}(r,\varphi)\Big] (-r\sin\varphi) \\&\quad+\Big[\frac{\partial f}{\partial x_1}(g(r,\varphi))\Big](-r\cos\varphi) \\&\ + \Big[\frac{\partial^2 f}{\partial x_1\partial x_2}(g(r,\varphi))\frac{\partial g_1}{\partial \varphi}(r,\varphi)+\frac{\partial^2 f}{\partial x_2\partial x_2}(g(r,\varphi))\frac{\partial g_2}{\partial \varphi}(r,\varphi)\Big](r\cos\varphi) \\&\quad+\Big[\frac{\partial f}{\partial x_2}(g(r,\varphi))\Big](-r\sin\varphi) \\&= \Big[\frac{\partial^2 f}{\partial x_1^2}(g(r,\varphi))\Big]r^2\sin^2\varphi +\Big[\frac{\partial^2 f}{\partial x_2^2}(g(r,\varphi))\Big]r^2\cos^2\varphi \\&\ -(2r\sin\varphi\cos\varphi)\Big[\frac{\partial^2f}{\partial x_1\partial x_2}(g(r,\varphi))\Big] \\&\ -\Big[(r\cos\varphi)\frac{\partial f}{\partial x_1}(g(r,\varphi))+(r\sin\varphi)\frac{\partial f}{\partial x_2}(g(r,\varphi))\Big] \end{align*} \begin{align*} \Big(\frac{\partial^2 F}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 F}{\partial \varphi^2}+\frac{1}{r}\frac{\partial F}{\partial r}\Big)(r,\varphi) &=\Big[\frac{\partial^2 f}{\partial x_1^2}(g(r,\varphi))\cos^2\varphi +\frac{\partial^2 f}{\partial x_2^2}(g(r,\varphi))\sin^2\varphi \\&\qquad+2\frac{\partial^2 f}{\partial x_1\partial x_2}(g(r,\varphi))\sin\varphi\cos\varphi\Big] \\&+\frac{1}{r^2}\Big\{\Big[\frac{\partial^2 f}{\partial x_1^2}(g(r,\varphi))\Big]r^2\sin^2\varphi +\Big[\frac{\partial^2 f}{\partial x_2^2}(g(r,\varphi))\Big]r^2\cos^2\varphi \\&\qquad\ -(2r\sin\varphi\cos\varphi)\Big[\frac{\partial^2f}{\partial x_1\partial x_2}(g(r,\varphi))\Big] \\&\qquad\ -\Big[(r\cos\varphi)\frac{\partial f}{\partial x_1}(g(r,\varphi))+(r\sin\varphi)\frac{\partial f}{\partial x_2}(g(r,\varphi))\Big]\Big\} \\&+ \frac{1}{r}\Big\{\frac{\partial f}{\partial x_1}(g(r,\varphi)) \cos\varphi +\frac{\partial f}{\partial x_2}(g(r,\varphi))\sin\varphi\Big\} \\&=\frac{\partial^2 f}{\partial x_1^2}(g(r,\varphi))\big[\cos^2\varphi+\sin^2\varphi\big] \\&\quad + \frac{\partial^2 f}{\partial x_2^2}(g(r,\varphi)) \big[\sin^2\varphi+\cos^2\varphi\big] \\&=\Big(\frac{\partial^2 f}{\partial x_1^2}+\frac{\partial^2 f}{\partial x_2^2}\Big)(g(r,\varphi)) =(\Delta f)(r\cos\varphi,r\sin\varphi). \end{align*} \(\big[\ \) Verwendet wurde (zweimal), da\ss\ (weil \(f\) aus \(C^2\) ist) \begin{align*} \frac{\partial^2 f}{\partial x_1\partial x_2}(g(r,\varphi))=\frac{\partial^2 f}{\partial x_2\partial x_1}(g(r,\varphi)).\quad\big] \end{align*} \vspace{1cm} \noindent{\bf Aufgabe 34: (4 Punkte)} Es sei \(f:\mathbb R^2\setminus\{0\}\to\mathbb R\) gegeben durch \begin{align*} f(x,y):=x^2y^2\ln(x^2+y^2). \end{align*} Es ist \((1,1,\ln 2)\in\Graph(f):=\{(x,y,f(x,y))\,|\ (x,y)\in\mathbb R^2\setminus\{0\}\}\). \begin{itemize} \item[a)] Stellen die Tangentialebene an \(\Graph(f)\) in diesem Punkt als Graph einer Funktion \(\ell:\mathbb R^2\to\mathbb R\) dar. \item[b)] Approximieren Sie \(\Graph(f)\) in der N\"ahe von \((1,1,\ln 2)\) durch den Graph eines Polynoms 2. Grades in \(x\) und \(y\). (Hinweis: Taylor). \end{itemize} \noindent{\it L\"osungsvorschlag:} a) Man braucht das Taylorpolynom 1. Ordnung: \begin{align*} \ell(x,y)&=f(1,1)+\frac{\partial f}{\partial x}(1,1)(x-1)+\frac{\partial f}{\partial x}(1,1)(y-1);\\ \frac{\partial f}{\partial x}(x,y)&=2xy^2\ln(x^2+y^2)+x^2y^2\frac{2x}{x^2+y^2};\ \frac{\partial f}{\partial x}(1,1)=2\ln 2+1;\\ \frac{\partial f}{\partial y}(x,y)&=2yx^2\ln(x^2+y^2)+x^2y^2\frac{2y}{x^2+y^2};\ \frac{\partial f}{\partial y}(1,1)=2\ln 2+1;\\ \ell(x,y)&=\ln 2+(2\ln 2+1)(x-1)+(2\ln 2+1)(y-1) \\&=-3\ln 2-2+(2\ln 2+1)(x+y). \end{align*} \([\ \ \ell\) ist eine Polynom 2. Grad in \((x,y)\), also ist \(\Graph(\ell)\) eine Ebene; \(\ell(1,1)=f(1,1);\) \( \frac{\partial \ell}{\partial x}(1,1)= \frac{\partial f}{\partial x}(1,1);\) \( \frac{\partial \ell}{\partial y}(1,1)= \frac{\partial f}{\partial y}(1,1)\), und damit ist \(\Graph(\ell)\) die Tangentialebene am \(\Graph(f)\) im Punkte \((1,1,\ln 2).\ ]\) b) Man braucht das Taylorpolynom 2. Ordnung: \begin{align*} (T_2(f))(x,y)&=f(1,1)+\frac{\partial f}{\partial x}(1,1)(x-1)+\frac{\partial f}{\partial x}(1,1)(y-1)\\&\ +\frac12\Big[\frac{\partial^2 f}{\partial x^2}(1,1)\Big](x-1)^2 +\frac12\Big[\frac{\partial^2 f}{\partial y^2}(1,1)\Big](y-1)^2 \\&\qquad+\Big[\frac{\partial^2 f}{\partial x\partial y}(1,1)\Big](x-1)(y-1) \\&=\ell(x,y)+\frac12\Big[\frac{\partial^2 f}{\partial x^2}(1,1)\Big](x-1)^2 +\frac12\Big[\frac{\partial^2 f}{\partial y^2}(1,1)\Big](y-1)^2 \\&\qquad +\Big[\frac{\partial^2 f}{\partial x\partial y}(1,1)\Big](x-1)(y-1); \\ \frac{\partial^2 f}{\partial x^2}(x,y)&= \frac{\partial}{\partial x}\Big[2xy^2\ln(x^2+y^2)+x^2y^2\frac{2x}{x^2+y^2}\Big] \\&=2y^2\ln(x^2+y^2)+\frac{10x^2y^2}{x^2+y^2}-\frac{4x^4y^2}{(x^2+y^2)^2}; \quad \frac{\partial^2 f}{\partial x^2}(1,1)=2\ln2+4; \\ \frac{\partial^2 f}{\partial y^2}(x,y)&= \frac{\partial}{\partial y}\Big[2x^2y\ln(x^2+y^2)+x^2y^2\frac{2y}{x^2+y^2}\Big] \\&=2x^2\ln(x^2+y^2)+\frac{10x^2y^2}{x^2+y^2}-\frac{4x^2y^4}{(x^2+y^2)^2}; \quad \frac{\partial^2 f}{\partial y^2}(1,1)=2\ln2+4; \\& \frac{\partial^2 f}{\partial y\partial x}(x,y) =\frac{\partial}{\partial y}\Big[2xy^2\ln(x^2+y^2)+x^2y^2\frac{2x}{x^2+y^2}\Big] \\&=4xy\ln(x^2+y^2)+\frac{4(xy^3+x^3y)}{x^2+y^2}-\frac{4x^3y^3}{(x^2+y^2)^2}; \\\frac{\partial^2 f}{\partial y\partial x}(1,1)&=4\ln2+3; \\ (T_2(f))(x,y)&=-3\ln 2-2+(2\ln 2+1)(x+y)+\frac12(2\ln2+4) (x-1)^2\\&\quad +\frac12(2\ln2+4) (y-1)^2+(4\ln2+3)(x-1)(y-1) \\& =(2+\ln2)(x^2+y^2)+(4\ln2+3)xy-(4\ln2+6)(x+y)+(3\ln2+5) \end{align*} \(\big[\ \) Verwendet wurde, da\ss\ (weil \(f\) aus \(C^2\) ist) \begin{align*} \frac{\partial^2 f}{\partial x\partial y}(x,y)=\frac{\partial^2 f}{\partial y\partial x}(x,y). \end{align*} Man sieht dieses auch durch direktes Ausrechnen.\(\ \big]\) F\"ur Grafen von \(\ell\) und \(T_2(f)\), siehe PS-Datei auf der Homepage. \ % \vspace{0.5cm} \noindent{\bf Aufgabe 35: (4 Punkte)} Eine Funktion \(u:\mathbb R^2\supset U\to\mathbb R\) hei\ss t {\it harmonisch}, falls \(u\) aus \(C^2(U)\) ist und die Gleichung \(\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0\) erf\"ullt ist. Sei \(f=u+iv:\mathbb C\supset U\to\mathbb C\) {\it komplex} differenzierbar, und \(u,v\in C^2(U,\mathbb R)\) (hier interpretieren wir \(u(x+iy)=u(x,y), v(x+iy)=v(x,y)\)). Zeigen Sie, da\ss\ \(u\) und \(v\) harmonisch sind. \noindent{\it L\"osungsvorschlag:} Man hat, da\ss \begin{align*} \lim_{z\to z_0}&\frac{f(z)-f(z_0)}{z-z_0} \\&=\lim_{(s,t)\to(0,0)}\frac{\big[u(x_0+s,y_0+t)+i v(x_0+s,y_0+t)\big]-\big[u(x_0,y_0)+iv(x_0,y_0)\big]}{s+it} \end{align*} existiert f\"ur alle \(z_0=x_0+iy_0\in U\) (Analysis I). Insbesonders muss (\((s,t)=(r,0)\)) \begin{align*} \lim_{r\to0}&\frac{\big[u(x_0+r,y_0)+i v(x_0+r,y_0)\big]-\big[u(x_0,y_0)+iv(x_0,y_0)\big]}{r} \\&=\lim_{r\to0}\frac{u(x_0+r,y_0)-u(x_0,y_0)}{r}+i\lim_{r\to0}\frac{v(x_0+r,y_0)-v(x_0,y_0)}{r} \\&=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0) \end{align*} gleich (\((s,t)=(0,\rho)\)) \begin{align*} \lim_{\rho\to0}&\frac{\big[u(x_0,y_0+\rho)+i v(x_0,y_0+\rho)\big]-\big[u(x_0,y_0)+iv(x_0,y_0)\big]}{i\rho} \\&=\frac{1}{i}\lim_{\rho\to0}\frac{u(x_0,y_0+\rho)-u(x_0,y_0)}{r} +\lim_{\rho\to0}\frac{v(x_0,y_0+\rho)-v(x_0,y_0)}{\rho} \\&=-i\frac{\partial u}{\partial y}(x_0,y_0)+\frac{\partial v}{\partial y}(x_0,y_0) \end{align*} sein. (Alle Grenzwerten existieren, weil \(u,v\) aus \(C^2\) sind). Weil \(u,v\) reellwertige sind, folgt \begin{align} \label{eq1} \frac{\partial u}{\partial x}(x_0,y_0)&=\frac{\partial y}{\partial y}(x_0,y_0),\\ \label{eq2} \frac{\partial u}{\partial y}(x_0,y_0)&=-\frac{\partial v}{\partial x}(x_0,y_0). \end{align} Aus \eqref{eq1} (und da\ss\ \(u,v\) aus \(C^2\) sind) folgt, da\ss\ \begin{align*} \frac{\partial^2 u}{\partial y\partial x}(x_0,y_0)= \frac{\partial^2 v}{\partial y^2}(x_0,y_0) \end{align*} und aus \eqref{eq2} \begin{align*} \frac{\partial^2 u}{\partial x\partial y}(x_0,y_0)= -\frac{\partial^2 v}{\partial x^2}(x_0,y_0) . \end{align*} Aber weil \(u,v\) aus \(C^2\) sind, gilt \(\frac{\partial^2 u}{\partial x\partial y}(x_0,y_0)=\frac{\partial^2 u}{\partial y\partial x}(x_0,y_0)\), und damit, da\ss\ \begin{align*} \frac{\partial^2 v}{\partial y^2}(x_0,y_0)=-\frac{\partial^2 v}{\partial x^2}(x_0,y_0) \end{align*} d.h. \begin{align*} (\Delta v)(x_0,y_0)= \frac{\partial^2 v}{\partial x^2}(x_0,y_0)+\frac{\partial^2 v}{\partial y^2}(x_0,y_0) =0. \end{align*} \"Ahnlich folgt, da\ss\ \((\Delta u)(x_0,y_0)=0\). \ % \vspace{1cm} \noindent{\bf Aufgabe 36: (4 Punkte)} Sei \(f:\mathbb R^m\to\mathbb R^n\) aus \(C^1(\mathbb R^m)\). \begin{itemize} \item[a)] Zeigen Sie, da\ss\ \begin{align*} f(x)-f(y)=\int_0^1 \big[f'(x+t(y-x))\big](y-x)\,dt \end{align*} f\"ur alle \(x,y\in \mathbb R^m\). (Hinweis: Kettenregel). \item[b)] Sei \(\|A\|_{2}:=\sqrt{\sum_{i=1}^n\sum_{j=1}^m |a_{ij}|^2}\), \(A=(a_{ij})\in\mathbb R^{n\times m}\). Zeigen Sie, da\ss\ \begin{align*} |Ax|\leq \|A\|_2|x| \end{align*} f\"ur all \(x\in \mathbb R^m\). \item[c)] Zeigen Sie, da\ss\ \begin{align*} \big|f(x)-f(y)\big|\leq C|x-y| \ \ \text{mit }\ C:=\sup_{t\in[0,1]}\|f'(x+t(y-x))\|_2 \end{align*} f\"ur alle \(x,y\in \mathbb R^m\). \end{itemize} \noindent{\it L\"osungsvorschlag:} a) Die Funktion \(g:\mathbb R\to\mathbb R^n\) definiert durch \(g(t)=f(x+t(y-x))\) ist aus \(C^1(\mathbb R\) (weil \(h(t)=t(y-x)\) aus \(C^1(\mathbb R)\) ist, und \(g=f\circ h\); Kettenregel). Damit ist \begin{align*} g(1)-g(0)=\int_0^1g'(t)\,dt. \end{align*} Nach der Kettenregel ist \(g'(t)=f'(h(t))\circ h'(t)\). Es gilt \(h'(t)=y-x\) und damit \begin{align*} g'(t)=f'(h(t))\circ h'(t) =f'(x+t(y-x))\cdot (y-x). \end{align*} Ausserdem ist \(g(1)=f(h(1))=f(x+1\cdot(y-x))=f(y), g(0)=f(h(0))=f(x+0\cdot(y-x))=f(x)\), d.h., \begin{align*} f(y)-f(x)=g(1)-g(0)=\int_0^1 g'(t)\,dt = \int_0^1 f'(x+t(y-x)) (y-x)\,dt. \end{align*} \(\big[\ \) Man bemerke den Druckfehler in der Aufgabeformulierung. \(\ \big]\) b) \begin{align*} |Ax|^2&=\Big|\big(\sum_{j=1}^m a_{1j}x_j,\ldots,\sum_{j=1}^ma_{nj}x_j\big)\Big|^2= \sum_{i=1}^n\big(\sum_{j=1}^m a_{ij}x_j\big)^2 \\&\leq \sum_{i=1}^n\big(\sum_{j=1}^m|a_{ij}|^2\big)\big(\sum_{j=1}^m x_j{}^2\big) \\\Big[\text{ weil (Cauchy-Schwarz) } &\big(\sum_{j=1}^m a_{ij}x_j\big)^2 =\big|\langle a_{i\bullet}, x\rangle\big|^2 \leq |a_{i\bullet}|^2|x|^2= \big(\sum_{j=1}^m |a_{ij}|^2\big)\big(\sum_{j=1}^m x_j^2\big) \ \ \Big] \\&=\big(\sum_{j=1}^m x_j{}^2\big)\big(\sum_{i=1}^n\sum_{j=1}^m |a_{ij}|^2\big) =\|A\|_2|x|^2. \end{align*} c) \begin{align*} |f(x)-f(y)|&=|f(y)-f(x)| =\big|\int_0^1 f'( x+t(y-x))(y-x)\, dt\big| \\&\leq \int_0^1 \big|f'(x+t(y-x))(y-x)\big|dt \\&\leq \int_0^1 \|f'(x+t(y-x))\|_2|y-x| dt \\&\leq |y-x|\int_0^1 \sup_{t\in[0,1]}\|f'(x+t(y-x))\|_2 \,dt \\&=|y-x|\ \sup_{t\in[0,1]}\|f'(x+t(y-x))\|_2 \int_0^11\, dt \\&=C|x-y|\ \ \text{mit }\ C:=\sup_{t\in[0,1]}\|f'(x+t(y-x))\|_2. \end{align*} \vspace{1cm} {\bf \noindent Abgabe bis Montag 20.06.2005, 11.15 Uhr in den MPIIA \"Ubungskasten im 1.~Stock vor der Bibliothek. \\ Unter {\tt http://www.mathematik.uni-muenchen.de/$\sim$sorensen} sind die Bl\"atter im Internet abrufbar. \\ Sprechstunden: H. Steinlein: \hspace{1cm} Mo 10-11, Zimmer 318\\ \hspace*{2.95cm} T. S\o rensen: \hspace{1.02cm} Mi 14-15, Zimmer 335} \end{document}