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In what follows you may assume that the following notation applies

$$y=y(x),\ y'=\ds\frac{dy}{dx}.$$

You may also assume that, unless otherwise stated, $y$ is a
sufficiently continuously differentiable function.

{\bf Question}

A particle moves along the $y$-axis and is at $y=0$ when $x=0$. It
reaches $y=a$ when $x=T$. The motion is governed by Hamilton's
principle which states that the particle moves so as to make

$$\ds\frac{1}{2}\ds\int_0^T \,dx ({y'}^2+\omega^2y^2)$$

stationary, where $\omega$ is a given constant. Show that the
motion is $y=a\ds\frac{\sinh \omega x}{\sinh \omega T}$. By also
computing the second variation, show that it is independent of the
path and also positive for all possible variations. Hence deduce
that this is a minimum extremal.


\medskip

{\bf Answer}

Let $I=\ds\frac{1}{2}\ds\int_0^T\,dx({y'}^2+\omega^2y^2)$

$F={y'}^2+\omega^2y^2$

$\ds\frac{\pl F}{\pl y}=2\omega^2y;\ \ds\frac{\pl F}{\pl y'}=2y'$

Therefore E-L is $2\omega^2y-\ds\frac{d}{dx}(2y')=0$

$\Rightarrow \omega^2-y-y''=0\ \ \ \omega=const$

Which can be solved easily to give

$y=A \cosh \omega x+B\sinh \omega x$

consts $A$ and $B$ from boundary conditions: $y(0)=0,\ y(T)=a$

$\Rightarrow A=0,\ B=\ds\frac{a}{\sinh \omega t}$

$\Rightarrow \un{y=\ds\frac{a \sinh \omega x}{\sinh \omega T}}$

\newpage
By computing second variation

\begin{eqnarray*} V_2 & = & \ds\int_0^T \left[{\eta'}^2
\ds\frac{\pl ^2 F}{\pl {y'}^2}+2\eta{\eta '} \ds\frac{\pl ^2
F}{\pl y\pl {y'}} +{\eta}^2\ds\frac{\pl ^2 F}{\pl {y}^2}\right]\\
& = & \ds\int_0^T[2{\eta '}^2+0+2\omega^2\eta ^2]\\ & = &
2\ds\int_0^T\,dx[{\eta '}^2+\omega^2\eta^2]
\end{eqnarray*}

This is independent of the path $y(x)$ and is also positive for
\un{all} possible $\eta(x) \ne 0$. Thus we have a weak minimum
(from lecture notes)
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