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\begin{center}
{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf HOME AND CLASS WORK -- SET 4}\\%
\end{center}
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Solution for assignment 17: Double pendulum eigenvectors and 
explicit orbits. \bigskip

$$ \frac{g}{l}\,\left(\matrix{-2\,(2\pm\sqrt{2})+2&-(2\pm\sqrt{2})\cr 
   -(2\pm\sqrt{2})&-(2\pm\sqrt{2})+1}\right)
   \left(\matrix{\phi_0\cr\psi_0\cr}\right)\ =\
   \frac{g}{l}\,\left(\matrix{a_\pm&b_\pm\cr b_\pm&c_\pm}\right)
   \left(\matrix{\phi_0\cr\psi_0\cr}\right)\ =\ 0\ $$
with $a_+=-2-2\sqrt{2}$, $b_+=-2-\sqrt{2}$, $c_+=-1-\sqrt{2}$ 
and  $a_-=-2+2\sqrt{2}$, $b_-=-2+\sqrt{2}$, $c_-=-1+\sqrt{2}$.
With the ansatz given in the problem we find: 
$$ \left(\matrix{a_\pm&b_\pm\cr b_\pm&c_\pm}\right)
   \left(\matrix{1\cr\psi_{0\pm}\cr}\right) = 0~~\Rightarrow~~
   \psi_{0\pm} = -\frac{a_\pm}{b_\pm} = -\frac{b_\pm}{c_\pm}\,,
   ~~\psi_{0+} = - \sqrt{2}~~{\rm and}~~\psi_{0-} = \sqrt{2}  $$
and, therefore, the eigenvectors
$$ \Phi_{0+} = \left(\matrix{1\cr-\sqrt{2}\cr}\right)~~~{\rm and}~~~
   \Phi_{0-} = \left(\matrix{1\cr \sqrt{2}\cr}\right)\ .             $$
They are {\bf not} orthogonal, because the previously solved eigenvalue
problem is of the form
$$ V\,\Phi_0 = \omega^2\,T\,\Phi_0~~~{\rm with}~~~
   V = \left(\matrix{2g/l&0\cr 0&g/l\cr}\right)
   ~~{\rm and}~~T = \left(\matrix{2&1\cr 1&1\cr}\right)\ . $$
Multiplying with $T^{-1}$ from the left does not help, because the 
matrix $T^{-1}V$ will not be symmetric. Instead a generalization
of the usual orthogonality definition works, see M\&T (12.55)
or  Goldstein Mechanics. % p.357.
The general solution for the double pendulum problem is given 
by the real part of our ansatz, which can be written
$$\Phi(t)=\left(A_+\cos\omega_+t + B_+\sin\omega_+t\right)
          \left(\matrix{1\cr -\sqrt{2}\cr}\right) +
          \left(A_-\cos\omega_-t + B_-\sin\omega_-t\right)
          \left(\matrix{1\cr \sqrt{2}\cr}\right)\ .        $$
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}
and the time dependence of the angles is
\begin{eqnarray} \nonumber
  \phi(t) &=&\ + A_+\cos\omega_+t + B_+\sin\omega_+t +
   A_-\cos\omega_-t + B_-\sin\omega_-t\,,\\ \nonumber
  \psi(t) &=& ( - A_+\cos\omega_+t - B_+\sin\omega_+t\, +
   A_-\cos\omega_-t + B_-\sin\omega_-t)\,\sqrt{2}\,.
\end{eqnarray}
For the initial conditions
$$ \phi_0=0\,,~~~\dot{\phi}_0=1\,,~~~\psi_0=0\,,~~~\dot{\psi}_0=-1$$
at time $t=0$ see the figure for a plot up to $t=50\,\sqrt{l/g}$ (next
page).

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

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