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\lhead{Technische Universität Berlin\\Numerische Mathematik I - WS 2019/2020}
\rhead{Prof.~V.~Mehrmann\\M.~Froidevaux}
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\newcommand{\CurrentWeekNb}{4}
\newcounter{taskNb}
\stepcounter{taskNb}

\begin{document}
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%{\LARGE Numerische Mathematik}\\[1.5ex]
%%{\large \CurrentWeekNb. \"Ubungsblatt}\\ 
{\Large Sketches of solutions for the exercise sheet \#\CurrentWeekNb }\\ 
{$\hookrightarrow$ Homeworks must be submitted by Thursday 14.11.19.}
\end{center}
%
%%\def\exerciseList{001, 002, 003}
\begin{enumerate}
\item[] \textbf{Solution \arabic{taskNb}} \stepcounter{taskNb}
Wir schreiben $\theta(t):=[\theta_1(t), \theta_2(t)]^T$. Mit dem Ansatz $\theta(t)=e^{i\omega t}[v_1, v_2]^T$ bekommt man
\[
\ddot{\theta}(t)=(i\omega)^2e^{i\omega t}[v_1, v_2]^T = -\omega^2 \theta(t).
\]
Die vereinfachte Bewegungsgleichung gibt dann
\[
A {\theta}(t) = \ddot{\theta}(t) = -\omega^2 \theta(t)
\]
mit $A=\frac{g}{L}\begin{bmatrix}
-2&1\\2&-2
\end{bmatrix}$. Man kann die Faktoren $e^{i\omega t}$ auf beiden Seiten der Gleichung kürzen, und man bekommt die Gleichung $A [v_1, v_2]^T = -\omega^2 [v_1, v_2]^T$, die offensichtlich als Lösung die Eigenpaaren $(\lambda, v)$ hat, wobei $\lambda:=-\omega^2$ und $v=[v_1, v_2]^T$.

Man berechnet die Eigenwerte von $A$ und findet 
\[
\lambda_{\pm} = -\big(\frac{g}{L} (2\pm \sqrt{2})\big).
\]
Die Eigenvektoren dazu sind
\[
v_{\mp} = \begin{bmatrix} 1 \\ \mp \sqrt{2} \end{bmatrix}.
\]
Deswegen kann die allgemeine Lösung als 
\[
\theta(t)= C_1 e^{i\omega_+ t} v_{-} + C_2 e^{i\omega_- t} v_{+}
\]
geschrieben werden, wobei $\omega_{\pm}=\sqrt{-\lambda_{\pm}}$, und $C_1, C_2$ Konstante sind, die anhand der Anfangswerte bestimmen werden müssen.

Da $\theta(0)=C_1 \begin{bmatrix} 1 \\ - \sqrt{2} \end{bmatrix} +C_2 \begin{bmatrix} 1 \\ + \sqrt{2} \end{bmatrix} \stackrel{!}{=} \begin{bmatrix} 1 \\ + \sqrt{2} \end{bmatrix} 1\text{e-}8$, findet man $C_1=0$, $C_2=1\text{e-}8$.

Schließlich ist die Lösung wie folgt:
\[
\begin{bmatrix} \theta_1(t) \\ \theta_2(t) \end{bmatrix} =  e^{i\sqrt{\frac{g}{L}(2-\sqrt{2})} t} \begin{bmatrix} 1\text{e-}8 \\ + \sqrt{2}\text{e-}8 \end{bmatrix}.
\]

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\item[] \textbf{Solution \arabic{taskNb}} \stepcounter{taskNb} 

Die Bewegungsgleichungen werden mit dem Zustandsvektor 
\[
y(t) :=[y_1(t), y_2(t), y_3(t), y_4(t)]^T :=[\theta_1(t), \theta_2(t), \dot{\theta}_1(t), \dot{\theta}_2(t) ]^T
\]
wie folgt umgeschrieben:
\[
\label{eq:Bewegung}
\dot{y}=
\begin{bmatrix}
y_3\\
y_4\\
\frac{1}{D} \left[ -\sin(\delta) \big(y_4^2+y_3^2\cos(\delta)\big)-\frac{g}{L}\big(2\sin(y_1) - \sin(y_2)\cos(\delta)\big)\right],\\
\frac{1}{D}\left[\sin (\delta)\big(2y_3^2+y_4^2\cos(\delta)\big)-\frac{2g}{L}\big(\sin(y_2)-\cos(\delta) \sin (y_1)\big)\right]
\end{bmatrix}
\]
wobei $\delta=\delta(y(t)):=y_1(t)-y_2(t)$ und $D=D(y(t)):= 2-\cos^2(\delta(y(t)))$.



\end{enumerate}
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