Past exam of the short options course of the University of Oxford / 2023 / 6 (Calculus of variations)

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.

Set $t=\sqrt{\lambda }x$ and $w(t)=y(t/\sqrt{\lambda })$. The chain rule turns the equation into the parameter-one Bessel differential equation. By the regular-at-zero solution property supplied in the paper,For a nonzero solution, $c\ne 0$, so $y(1)=0$ forces $J_k(\sqrt{\lambda })=0$. Conversely, every positive zero $z$ supplies the admissible eigenfunction $J_k(zx)$ with eigenvalue $z^2$.

Here is a direct justification of the infimum characterization that does not assume a minimizer exists. Let $u(x)=J_k(z_{k,1}x)$ and choose its sign so that $u>0$ on $(0,1)$. The first positive zero exists by the oscillatory large-argument behavior of the Bessel function of the first kind. At zero, its Frobenius series for Bessel functions gives $u(x)=C{x}^k(1+O(x^2))$. At one, its zero is simple by uniqueness for a regular second-order ordinary differential equation. Expanding $y=u(y/u)$ and using the equation for $u$ gives the Ground-state representation of a Sturm-Liouville energy:For completeness, apply the identity first on $[\epsilon ,1-\epsilon ]$. The discarded boundary term is $[x{u}^{\prime }{y}^2/u]$. It tends to zero at both ends: $x{u}^{\prime }/u\to k$ and $y=O(x)$ at zero, while $u^{\prime }/u=O((1-x{)}^{-1})$ and $y=O(1-x)$ at one. Taking $y=u$ achieves equality, soThe infimum is over the stated smooth admissible functions with finite energy; the equality also proves attainment by a multiple of $u$.

The trial function $y=x(1-x)$ givesIts Rayleigh quotient is $5k^2+10$, yielding the upper bound. Conversely, $1/x\ge x$ on $(0,1)$ and the derivative energy is nonnegative, so $I[y]\ge k^{2}J[y]$ for every admissible $y$. Part (b) therefore gives