Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 9

For a variation $y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$, the first variation of the functional isThe boundary term vanishes by the fixed endpoints. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation

A missing explicit dependence produces a first integral of the Euler-Lagrange equation. The two cases follow by differentiating the candidate conserved quantity, as shown in the child solutions.

When $F_y=0$, the Euler-Lagrange equation reduces immediately to $d{F}_{y^{\prime }}/dx=0$. ThereforeThis is conservation of the generalized momentum associated with a cyclic coordinate.

Along a differentiable curve, the chain rule givesby the Euler-Lagrange equation. Thus when $F_x=0$ the Beltrami identity is

Here $F$ has no explicit $x$ dependence, so use the Beltrami identity. With $p=y^{\prime }$, one has $F_{y^{\prime }}=yp/\sqrt{1+p^2}$ and thereforeThis first integral reduces the second-order Euler-Lagrange equation to a first-order relation.

Choose the tangent angle $\psi =\arctan y^{\prime }$ in $(-\pi /2,\pi /2)$, so $1/\sqrt{1+(y^{\prime }{)}^2}=\cos \psi$. At the maximum, the first integral has value $C=y_0(1-\lambda )$. HenceA nonconstant regular branch below its positive maximum has $y>0$ and $\psi \ne 0$, giving $\lambda >1$. For completeness, strictness follows from the differential equation rather than assuming it at every point: the Euler-Lagrange equation simplifies toAt the maximum, $y_0{y}^{\prime \prime }=1-\lambda$ implies $\lambda \ge 1$. If $\lambda =1$, the positive-height initial data $y=y_0$, $y^{\prime }=0$ give the constant solution uniquely, contradicting the existence of the lower-height branch. Thus $\lambda =1$ is only the degenerate horizontal solution. The first integral with $\lambda >1$ also excludes $y=0$ on this regular branch.

Rearrange the first integral from part (iv) and use the half-angle identities:Eliminating $\lambda$ givesSquaring removes the sign of the slope; the rising and falling sides of the same profile have opposite tangent angles but obey the same height relation. Heights and angles must remain on the regular branch with $\cos \psi >0$.