Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / G / G2

At a regular constrained stationary curve, choose a constant multiplier $\lambda$ and make the augmented functional $I-\lambda J$ stationary under endpoint-fixed variations. Its integrand is $H=F-\lambda G$ and the equation is $d{H}_{y^{\prime }}/dx-H_y=0$. This normal multiplier formulation requires the first variation of the constraint not to vanish identically; the usual abnormal case would require two multipliers. When $H_x=0$,so $\boxed{y^{\prime }{H}_{y^{\prime }}-H=c_0}$ is constant, proving the requested first integral.

For the chain, $ds=\sqrt{1+(y^{\prime }{)}^2}dx$, so the fixed physical length is exactly $J[y]$. Its gravitational potential, apart from the constant factor $\rho g$, is $I[y]={\int }_{-a}^{a}y\sqrt{1+(y^{\prime }{)}^2}dx$. Thus $H=(y-\lambda )\sqrt{1+(y^{\prime }{)}^2}$, and the Euler–Lagrange equation isThe first integral becomes $(y-\lambda )/\sqrt{1+(y^{\prime }{)}^2}=c$, after renaming the constant with the opposite sign. The differential equation also gives $(y-\lambda )y^{\prime \prime }=1+(y^{\prime }{)}^2$, so the nondegenerate solution integrates to $y=\lambda +c\cosh ((x-x_0)/c)$. Equal endpoint heights force $x_0=0$: the two cosh values agree only when their arguments are equal or opposite, and $a>0$ excludes the former. HenceFor the hanging energy minimum take $c>0$. Since $\sqrt{1+(y^{\prime }{)}^2}=\cosh (x/c)$,For $L>2a$ the length equation has exactly one positive solution: its right side decreases from infinity to $2a$ as $c$ increases from zero to infinity. For $L=2a$ the only feasible shape is the straight segment, obtained as a limiting case; for $L<2a$ no chain can span the endpoints. A negative $c$ gives the inverted stationary catenary, rather than the physically hanging minimum.