Past exam of the options I course of the University of Oxford / 2005 / G / 1 / Solution

For fixed endpoint values, the Euler-Lagrange equation isIt follows by differentiating the functional along $y+\epsilon \eta$ with $\eta (0)=\eta (\pi )=0$, integrating the $F_{y^{\prime }}{\eta }^{\prime }$ term by parts and applying the fundamental lemma of the calculus of variations.

For this quadratic integrand it becomes $y^{\prime \prime }+ky=0$. If $k>0$, write $\nu =\sqrt{k}$; the condition at zero gives $y=B\sin (\nu x)$, and that at $\pi$ requires $B\sin (\nu \pi )=0$. Consequently $y=0$ is the only extremal when $\sqrt{k}$ is not an integer. If $\sqrt{k}=n$ is a positive integer, all extremals areThe integer case $n=0$ must be handled separately: $y^{\prime \prime }=0$ with both endpoints zero has only $y=0$. If negative real $k$ is allowed, the hyperbolic-function solutions also yield only zero under these endpoints. These statements concern stationary functions; no minimum is assumed.