Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 4 / 16A (Variational Principles)

For a variation $y+\epsilon \eta$ preserving both endpoint values and slopes, $\eta ={\eta }^{\prime }=0$ at the endpoints. The first variation isIntegrate the last term twice and the middle term once. All boundary terms vanish, givingThe fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equationFor the present functional this becomes $1+y-y^{(4)}=0$. Its general solution is $y=-1+A\cosh x+B\sinh x+C\cos x+D\sin x$. The left endpoint gives $C=-A$ and $D=-B$. The right endpoint then givesThe coefficient determinant is $2(1+\cosh \pi )>0$, so $A=1$, $B=0$. The stationary function is $y_{\ast }(x)=-1+\cosh x-\cos x$.

For any admissible perturbation $\eta$, exact quadratic expansion, followed by the stationarity equation and integration by parts, givesApply the supplied Poincare inequality first to $\eta$, then to ${\eta }^{\prime }$, whose endpoint values also vanish:Thus $J[y_{\ast }+\eta ]\le J[y_{\ast }]$ for every admissible perturbation, proving a global maximum through the clamped second-variation inequality.