Past exam of the options II course of the University of Oxford / 2005 / G / 2

Write $I[y]={\int }_a^{b}G(x,y,y^{\prime })dx$. For endpoint-zero variations $\eta$, the first variations areFor a normal constrained extremal, choose a variation ${\eta }_0$ with $\delta I({\eta }_0)\ne 0$. Given any $\eta$, use the implicit function theorem to choose a small correction in the ${\eta }_0$ direction so that $y+t\eta +s(t){\eta }_0$ preserves the constraint. Differentiating it and the stationarity condition at zero givesIntegrate the derivative-of-variation term by parts. Since the variations vanish at both endpoints, the boundary term is zero. The fundamental lemma of the calculus of variations then givesThe regularity condition on the constraint is what permits the coefficient of $F$ to be normalized to 1. In a degenerate case, the general multiplier statement uses ${\lambda }_{0}F-{\lambda }_{1}G$ with the two coefficients not both zero; the source's displayed form is the normal case.

For a string described as a function graph $r(x)=(x,y(x))$, its arc-length element is $|r^{\prime }(x)|dx=\sqrt{1+y^{\prime }(x{)}^2}dx$. Inextensibility therefore givesThis also forces $l\ge 2a$, with equality only for the straight function graph.

For the upper function graph, the vertical cross-section of the enclosed region has length $y(x)$, so its area is $J[y]={\int }_0^{2a}y(x)dx$. A maximizing nonnegative arch therefore uses $F=y$ and $G=\sqrt{1+y^{\prime 2}}$ in the normal Euler-Lagrange equation. It becomesIntegrating once gives $y^{\prime }/\sqrt{1+y^{\prime 2}}=(c-x)/\lambda$. The concave upper branch has $\lambda >0$ and integrates to $y=\sqrt{{\lambda }^2-(x-c{)}^2}+d$. The equal zero endpoint heights force $c=a$ and $d=-\sqrt{{\lambda }^2-a^2}$. HenceIt is a circular arc. To find its parameter, substitute into the length integral: $\sqrt{1+y^{\prime 2}}=\lambda /\sqrt{{\lambda }^2-(x-a{)}^2}$, givingThis scalar equation determines $\lambda$ numerically when needed; no explicit solution is required.

The arch is a global maximum in its smooth function graph regime. For $\lambda >0$, the integrand $y-\lambda \sqrt{1+y^{\prime 2}}$ is concave and strictly concave in its slope. Its supporting-tangent inequality, integrated against any competing function graph with the same endpoint values, has first-order term zero by the Euler-Lagrange equation. Thus $J[y]-\lambda I[y]\le J[y_{\ast }]-\lambda I[y_{\ast }]$. On the fixed-length class this is $J[y]\le J[y_{\ast }]$; strict slope concavity also gives uniqueness.

The source leaves the length range implicit. A twice continuously differentiable function graph of this form needs $\lambda >a$, giving $2a<l<\pi a$. At $l=2a$ the straight function graph is the degenerate solution. The semicircle limit $l=\pi a$ has vertical endpoint tangents and is outside that smooth function graph class. For $l>\pi a$, no arc in the claimed smooth function graph family exists, so the stated formula cannot describe a twice continuously differentiable maximizing function graph under those unrestricted data. Allowing overhanging curves changes the admissible class.