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\Blatt{1}{WS 2009/2010, 26.10.2009}
         {Mechanik der Kontinua}
         {Prof. J.~L.~van~Hemmen}

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\begin{deflist}{20.}

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\item[1.]
{\bf Kreuzprodukt}
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Zeigen Sie dass es zu jedem $\mbox{\boldmath{$\omega$}} \in {\Bbb
  R}^3$ eine eindeutige antisymmetrische Matrix $\Omega$ gibt, f"ur
die f"ur alle ${\bf v} \in {\Bbb R}^3$ gilt
\[
\Omega \, {\bf v} =  \mbox{\boldmath{$\omega$}} \times  {\bf v}
\]

\item[2.]
{\bf Drehung}
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Sei $\Omega$ eine antisymmetrische $3 \times 3$ Matrix und
$\varepsilon>0$ klein.

\begin{deflist}{(0)}
\item[(i)] 
Interpretieren Sie geometrisch die Abbildung
\[
D(\varepsilon)\, {\bf v} = (\one + \varepsilon\, \Omega) \, {\bf v}\ .
\]
Benutzen Sie dazu die Darstellung von $\Omega$ aus Aufgabe 1 und
rechnen Sie in einem Koordinatensystem, in dem gilt {\boldmath{$\omega$}}$=(0,0,1)^T$.

\item[(ii)] Berechnen Sie die n-te Iteration $D(\varepsilon)^n\, {\bf
    v}$.
\item[(iii)] Berechnen und interpretieren Sie $\lim_{n\to\infty}
  D(\varphi/n)^n$.
\end{deflist}


\item[3.]
{\bf Gradient in gedrehten Koordinaten}
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Gegeben seien zwei gegeneinander verdrehte Koordinatensysteme
(ungestrichene und gestrichene Koordinaten), also
\begin{displaymath}
  {\bf x'} = O {\bf x} \,,
\end{displaymath}
wobei $O$ eine orthogonale Matrix ist. Zeigen Sie dass für die
Ableitung nach gestrichenen Koordinaten
\begin{displaymath}
  \nabla' = \left( \begin{array}{c}
             \partial x' \\
             \partial y' \\
             \partial z'
            \end{array} \right)
\end{displaymath}
gilt
\begin{displaymath}
  \nabla' = O \nabla \,.
\end{displaymath}

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\item[4.]
{\bf Ableitungen eines Vektorfelds}
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Zeigen Sie, dass die Ableitung $\nabla_i u_j = \frac{\partial
  u_j(x_1,x_2,x_3)}{\partial x_i}$ des Vektorfelds $(x_1,x_2,x_3)
\mapsto {\bf u}(x_1,x_2,x_3) \in {\Bbb R}^3$ ein Tensor ist. Hinweis: Benutzen
Sie das Resultat aus Aufgabe 3.


\item[5.]
{\bf Gau{\3}'scher Integralsatz}
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Beweisen Sie den Gau{\3}'schen Integralsatz f"ur
\begin{enumerate}
\item den Spezialfall des Geschwindigkeitsfelds in einer zylindrischen
  Röhre (Achse ist die x-Achse), die von einer Flüssigkeit mit
  Geschwindigkeit $\mathbf{v}(\mathbf{x}) = \mathbf{v}(x,y,z) =
  (v(x),0,0)$ durchstömt wird.  Das Integrationsvolumen sei der
  Zylinder. 
  
  Warum ist bei einer \emph{inkompressiblen} Flüssigkeit in obigem
  Spezialfall $\div \mathbf{v}(\mathbf{x})=0$?
  
  Warum gilt für die Dichte $\varrho$ einer \emph{kompressiblen} Flüssigkeit
  in obigem Spezialfall
  \begin{displaymath}
    \frac{\partial
      \varrho(\mathbf{x},t)}{\partial t}=-\div [ \varrho(\mathbf{x},t)
    \mathbf{v}(\mathbf{x},t) ] \,?
  \end{displaymath}
\item ein beliebiges Vektorfeld $\mathbf{v}(\mathbf{x})$ und einen
  achsparallelen Quader als Integrationsvolumen.
\end{enumerate}




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\footnotesize
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Besprechung der Übungen Dienstags 10-12 HS3, Mittwochs 8-10 HS2 und
Donnerstags 10-12 HS3. \\
Übungsleitung: Dr. Moritz Franosch, mail@Franosch.org,
http://www.t35.ph.tum.de.

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