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\begin{document}

PHZ3113: Solution for Homework 11. \medskip

1. For small oscillation we have derived the Euler-Lagrange equations,
which read in matrix notation
$$ \left(\matrix{2&1\cr 1&1\cr}\right)\,\left(\matrix{\ddot{\phi}\cr
   \ddot{\psi}\cr}\right) + \left(\matrix{2g/l&0\cr 0&g/l\cr}\right)\,
   \left(\matrix{\phi\cr\psi\cr}\right)\ =\ 0\ . $$
This is solved by the exponential ansatz (physical is the real part
of the solution):
$$ \Phi(t) = \left(\matrix{\phi\cr\psi\cr}\right)\ =\ e^{i\,\omega\,t}
   \left(\matrix{\phi_0\cr\psi_0\cr}\right)~~\Rightarrow~~
   \left(\matrix{\ddot{\phi}\cr\ddot{\psi}\cr}\right)\ =\
   - \omega^2\,e^{i\,\omega\,t} \left(\matrix{\phi_0\cr\psi_0\cr}
   \right) $$
$$ \left(\matrix{-2\omega^2+2g/l&-\omega^2\cr -\omega^2&-\omega^2+g/l
   \cr}\right)\,\left(\matrix{\phi\cr\psi\cr}\right)\ =\ 0\ . $$
$$ 0\ =\ {\rm det}\,\left|\matrix{-2\omega^2+2g/l&-\omega^2\cr -
   \omega^2&-\omega^2+g/l\cr}\right|\ = \omega^4 - \frac{4g}{l}\, 
   \omega^2 + \frac{2g^2}{l^2} $$
with eigenmodes
$$ \omega_{\pm}\ =\ \sqrt{\frac{g}{l}}\,\sqrt{2\pm \sqrt{2}}\ . $$

2. For the $\omega_+$ eigenvalue we have
$$ 0 = \frac{g}{l}\, \left(
   \matrix{\left(-2-2\sqrt{2}\right)&\left(-2-\sqrt{2}\right)\cr
           \left(-2-\sqrt{2}\right) &\left(-1-\sqrt{2}\right)}\right)
   \,\left(\matrix{\phi_+\cr\psi_+\cr} \right)\ \Rightarrow\
   \frac{\psi_+}{\phi_+}=\frac{-2-2\sqrt{2}}{2+\sqrt{2}}=-\sqrt{2}$$
and for $\omega_-$
$$ 0 = \frac{g}{l}\, \left(
   \matrix{\left(-2+2\sqrt{2}\right)&\left(-2+\sqrt{2}\right)\cr
           \left(-2+\sqrt{2}\right) &\left(-1+\sqrt{2}\right)}\right)
   \,\left(\matrix{\phi_+\cr\psi_+\cr} \right)\ \Rightarrow\
   \frac{\psi_+}{\phi_+}=\frac{-2+2\sqrt{2}}{2-\sqrt{2}}=\sqrt{2}\,.$$
Therefore, the solutions (real part) can be written
\begin{eqnarray*}
  \phi(t)   &=& \phi_+(t) + \phi_-(t)\,,\\
  \psi(t) &=& -\sqrt{2}\,\phi_+(t) + \sqrt{2}\,\phi_-(t)\,.
\end{eqnarray*}
with
\begin{eqnarray*}
  \phi_+(t) &=& A_+\cos(\omega_+t) + B_+\sin(\omega_+t)\,,\\
  \phi_-(t) &=& A_-\cos(\omega_-t) + B_-\sin(\omega_-t)\,,\\
\end{eqnarray*}

3. 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}

\end{document}

