Past exam of the short options course of the University of Oxford / 2023 / 6 / a / Solution

Let $y$ be a smooth constrained minimizer. For any smooth variation $h$ vanishing at the endpoints, use the normalized admissible curve $y_{\epsilon }=(y+\epsilon h)/\sqrt{J[y+\epsilon h]}$. Since $J[y]=1$, differentiating at zero givesIn particular this holds for every compactly supported test function $h$ in $(0,1)$. Integration by parts and the fundamental lemma of the calculus of variations givewhich is the required Bessel differential equation with $\boxed{\lambda =I[y]}$. The lemma used here says that a continuous function whose integral against every compactly supported smooth test function is zero must itself vanish. No multiplier-existence theorem is needed: the explicit normalization produces the variation directly.

The endpoint $0$ causes no divergent energy for these variations: smoothness and $y(0)=h(0)=0$ imply $y,h=O(x)$. Also $J[v]>0$ for every nonzero admissible $v$, and $v/\sqrt{J[v]}$ has unit $J$ and energy $I[v]/J[v]$. Thus minimizing $I$ with $J=1$ is equivalent to minimizing the scale-invariant Rayleigh quotient $I/J$. The zero function is excluded whenever this quotient is written.