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Year 2 of the mathematics course of the University of Oxford

Short options course of the University of Oxford

Past exam of the short options course of the University of Oxford

2015

Use the <chain rule> along a sufficiently smooth <extremal> of the <functional>. For $E=y'F_{y'}-F$,
$$
\begin{aligned}
\frac{dE}{dx}
&=y''F_{y'}+y'\frac d{dx}F_{y'}
-F_x-F_yy'-F_{y'}y''\\
&=y'\left(\frac d{dx}F_{y'}-F_y\right)-F_x.
\end{aligned}
$$
The bracket vanishes by the <Euler-Lagrange equation>. When $F$ has no explicit $x$ dependence, $F_x=0$, giving the <Beltrami identity>:
$$
\boxed{y'F_{y'}-F=\text{constant along the extremal}.}
$$

Solution

Past exam of the short options course of the University of Oxford / 2015 / 6 / a / Solution

Past exam of the short options course of the University of Oxford / 2015 / 6 / a / Solution

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