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\begin{center}
{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf HOME AND CLASS WORK -- SET 6}\\%
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Solution for assignment 26: 
\medskip

Double pendulum solution and plot (continuation of 25).
\medskip

Let us take minors with respect to the first row of the determinant.
For the $\omega_+$ frequency the ratio of the two minors is
$$ \frac{\triangle_{1+}}{\triangle_{2+}} = \frac{-1-\sqrt{2}}{2+\sqrt{2}} 
  = \frac{(-1-\sqrt{2})}{(2+\sqrt{2})}\,\frac{(1-\sqrt{2})}{(1-\sqrt{2})} 
  = \frac{1}{-\sqrt{2}} $$
and for $\omega_-$ it is
$$ \frac{\triangle_{1-}}{\triangle_{2-}} = \frac{-1+\sqrt{2}}{2-\sqrt{2}} 
  = \frac{(-1+\sqrt{2})}{(2-\sqrt{2})}\,\frac{(1+\sqrt{2})}{(1+\sqrt{2})} 
  = \frac{1}{\sqrt{2}}\,. $$
Therefore, the solutions (real part) can be written
\begin{eqnarray*}
  \phi_+(t) &=& A_+\cos(\omega_+t) + B_+\sin(\omega_+t)\,,\\
  \phi_-(t) &=& A_-\cos(\omega_+t) + B_-\sin(\omega_+t)\,,\\
  \phi(t)   &=& \phi_+(t) + \phi_-(t)\,,\\
  \psi(t) &=& -\sqrt{2}\,\phi_+(t) + \sqrt{2}\,\phi_-(t)\,.
\end{eqnarray*}
The four constants are determined by the four initial value, e.g., 
$\phi_0,\,\dot{\phi_0},\,\psi_0,\,\dot{\psi_0}$ at time $t=0$:
\begin{eqnarray} \nonumber
  \phi_0 &=& A_+ + A_-\,,~~~~~~~~~~~
  \psi_0\ =\ \sqrt{2}\ (-A_+ + A_-)\,,\\ \nonumber
  \dot{\phi}_0 &=& \omega_+ B_+ + \omega_- B_-\,,~~~~
  \dot{\psi}_0\ =\ \sqrt{2}\ (-\omega_+ B_+ + \omega_- B_-)\,,
\end{eqnarray}
which gives
\begin{eqnarray} \nonumber
  A_+ &=& \frac{\phi_0}{2} - \frac{\psi_0}{2\sqrt{2}}\,,~~~~~~~~~~~
  A_-\ =\ \frac{\phi_0}{2} + \frac{\psi_0}{2\sqrt{2}}\,,\\ \nonumber
  B_+ &=& \frac{\dot{\phi}_0}{2\omega_+} - 
          \frac{\dot{\psi}_0}{2\sqrt{2}\,\omega_+}\,,~~~~~
  B_-\ =\ \frac{\dot{\phi}_0}{2\omega_-} + 
          \frac{\dot{\psi}_0}{2\sqrt{2}\,\omega_-}\,.
\end{eqnarray}
For the initial conditions
$$ \phi_0=0\,,~~~\dot{\phi}_0=1\,,~~~\psi_0=0\,,~~~\dot{\psi}_0=-1$$
at time $t=0$ the figure gives a plot up to $t=50\,\sqrt{l/g}$ (next
page).

\begin{figure}[tb]
\vspace{5pc}
 \centerline{\hbox{ \psfig{figure=DoublePendulum.pdf,width=18.0cm} }}
% \caption{Ladder for assignment~14.} \label{fig_lader}
\end{figure}

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