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

Use the chain rule along a sufficiently smooth extremal of the functional. For $E=y^{\prime }{F}_{y^{\prime }}-F$,The bracket vanishes by the Euler-Lagrange equation. When $F$ has no explicit $x$ dependence, $F_x=0$, giving the Beltrami identity:

For the specified functional,Since $y>0$, the Beltrami identity gives $y\sqrt{1+y^{\prime 2}}=R$ for some $R>0$. This suggests a circular arc, but it is worth deriving its equation without dividing by $y^{\prime }$: such division would lose the point where the arc is horizontal.

Substitute the displayed partial derivative and $F_y=-\sqrt{1+y^{\prime 2}}/y^2$ in the full Euler-Lagrange equation. Combining terms over the common denominator givesThus $(y{y}^{\prime }{)}^{\prime }=-1$, so $y{y}^{\prime }=h-x$ and a second integration givesThis establishes the semicircular extremals of the hyperbolic length functional and excludes constant $y$, which would satisfy the first-order Beltrami identity for a suitable $R$ but not the full Euler-Lagrange equation.

The two prescribed endpoints giveSubtracting yields $4h=4(d^2-c^2)$. ThereforeIn particular the centre is $\boxed{(d^2-c^2,0)}$. Admissibility requires $c,d>0$, as the prescribed heights must lie in $y>0$. Under that condition the radicand is positive throughout $[-1,1]$: it is a concave quadratic polynomial whose two endpoint values are $4c^2$ and $4d^2$. The positive root consequently gives one smooth admissible arc, uniquely determined by the endpoints.

When an endpoint's horizontal coordinate is fixed but its height is free, the natural boundary conditions for a free endpoint require $F_{y^{\prime }}=0$ there. In this problem,Every admissible extremal is an upper circular arc with centre on $y=0$, by part (b). Its tangent is horizontal precisely above its centre. Thus imposing the natural boundary conditions for a free endpoint at $x=1$ forces $h=1$. The fixed height at $x=-1$ gives $R^2=4+4c^2$, soIts free endpoint height is $y(1)=2\sqrt{1+c^2}$. Positivity on the whole interval follows because the radicand is at least $4c^2$.

There can be no solution satisfying the natural boundary conditions for a free endpoint at both ends: an upper circular arc has only one point with a horizontal tangent, whereas the two endpoints have distinct horizontal coordinates. Therefore