Past exam of the options II course of the University of Oxford / 2004 / G / G2 / Solution

For a variation $y+\epsilon \eta$ with $\eta (0)=\eta (1)=0$, stationarity givesThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation $d{F}_{y^{\prime }}/dx-F_y=0$.

Differentiating the proposed first integral givesThus it is constant along an extremal when $F_x=0$, proving the Beltrami identity.

For the positive branch relevant to the endpoint data, $F=\sqrt{1+y^{\prime 2}}/y$. The identity becomesThe Euler-Lagrange equation itself gives $y{y}^{\prime \prime }=-(1+y^{\prime 2})$, so $(y^2{)}^{\prime \prime }=-2$. Integrating, or solving the first integral, yields a circle $y^2+(x-c{)}^2=R^2$. The equal endpoint values force $c=1/2$ and $R^2=5/4$. Hence the unique positive smooth extremal isIt stays positive on the entire interval and satisfies the second-order equation, including its turning point. Merely differentiating the first integral after dividing by $y^{\prime }$ would need extra care there. Geometrically this is an arc of a hyperbolic geodesic in the upper half-plane model.