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{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf HOME AND CLASS WORK -- SET 2}\\%
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Solution for assignment 10. \bigskip

\noindent 1. The Lagrangian is
$$ L = \frac{1}{2}\,\dot{x}^2 - \frac{1}{2}\,x^2\ . $$
Thus, the Euler-Lagrange equation becomes
$$ 0 = \frac{\partial L}{\partial x} - \frac{d\ }{dt}\,
       \frac{\partial L}{\partial\dot{x}} = -x - \ddot{x} $$
and the general solution is given by $x(t)=A\,\sin(t)+B\,\cos(t)$.
The integration constants are determined by the initial conditions
$$ x(0) = B = 0~~{\rm and}~~x(\pi/2)=A=1 $$
so that the exact path is
$$ x(t) = \sin(t)\ . $$
For this path the action is given by
$$ S\left[x(t)\right] = \frac{1}{2}\int_0^{\pi/2} dt\,\left[
   \cos^2(t) - \sin^2(t)\right] =
   \frac{1}{2}\int_0^{\pi/2} dt\,\cos(2t) = 0\ .  $$
\bigskip

\noindent 2. For a linear path $x(t)=a+b\,t$ subject to the boundary 
conditions:
$$ x(0) = a = 0~~{\rm and}~~x(\pi/2)=b\,\frac{\pi}{2}=1~\Rightarrow\ 
   b=\frac{2}{\pi} $$
so that this path is
$$ x(t) = \frac{2}{\pi}\,t\ . $$
For this path the action is given by
\begin{eqnarray} \nonumber
  S\left[x(t)\right] &=& \frac{1}{2}\int_0^{\pi/2} dt\,\left[
  \left(\frac{2}{\pi}\right)^2 - \left(\frac{2}{\pi}\right)^2\,t^2
  \right] = \frac{2}{\pi^2}\int_0^{\pi/2} dt\,\left(1-t^2\right)
  \\ \nonumber &=& \frac{2}{\pi^2}\left[\frac{\pi}{2}-\frac{1}{3}
   \left(\frac{2}{\pi}\right)^3 \right] = \frac{1}{\pi}\left(1-
   \frac{\pi^2}{12}\right)=\frac{1}{\pi}-\frac{\pi}{12}=0.0565\dots\ .  
\end{eqnarray}
\bigskip

\noindent 3. Assuming that the previous result is in $J\cdot s$,
we find for the large number 
$$ S/\hbar\approx 5.37\times 10^{32}\ .$$
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