 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2015/6/a/solution

Use the chain rule along a sufficiently smooth extremal of the functional. For

E = y^{'} F_{y^{'}} - F

\frac{d E}{d x} = y^{''} F_{y^{'}} + y^{'} \frac{d}{d x} F_{y^{'}} - F_{x} - F_{y} y^{'} - F_{y^{'}} y^{''} = y^{'} (\frac{d}{d x} F_{y^{'}} - F_{y}) - F_{x} .

F

has no explicit

x

dependence,

F_{x} = 0

y^{'} F_{y^{'}} - F = constant along the extremal .

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