= Solution
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,
$$
y(x)=cJ_k(\sqrt\lambda x).
$$
For a nonzero solution, $c\ne0$, 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 $u(x)=Cx^k(1+O(x^2))$. 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>:
$$
I[y]-z_{k,1}^2J[y]=\int_0^1xu(x)^2\left(\frac d{dx}\frac{y(x)}{u(x)}\right)^2dx\geq0.
$$
For completeness, apply the identity first on $[\varepsilon,1-\varepsilon]$. The discarded boundary term is $[xu'y^2/u]$. It tends to zero at both ends: $xu'/u\to k$ and $y=O(x)$ at zero, while $u'/u=O((1-x)^{-1})$ and $y=O(1-x)$ at one. Taking $y=u$ achieves equality, so
$$
\boxed{z_{k,1}^2=\inf_{y\ne0,\ y(0)=y(1)=0}\frac{I[y]}{J[y]}}.
$$
The infimum is over the stated smooth admissible functions with finite energy; the equality also proves attainment by a multiple of $u$.
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