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 \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}