 Solution

Multiplication by an integrating factor givessince $p^{\prime }=ap$ and hence $(p{w}^{\prime }{)}^{\prime }=p(w^{\prime \prime }+a{w}^{\prime })$.

For the singular-endpoint equation, take $p=x^{-1/2}$, $r=x^{1/2}/2$, $q=0$. The indicial equation is $\alpha (2\alpha -3)=0$. The constant branch violates the zero endpoint value unless its coefficient vanishes; the nontrivial admissible branch therefore has $\boxed{\alpha =3/2}$. More precisely, an eigenfunction has $w=C{x}^{3/2}+O(x^{9/2})$ and $w^{\prime }=(3C/2)x^{1/2}+O(x^{7/2})$ near zero.

Integrate the self-adjoint equation after multiplying by $w$, initially on $[a,1]$ with $a>0$:At zero the boundary expression is $O(x^{3/2})$, the first integrand is $O(x^{1/2})$, and the second is $O(x^{7/2})$. Both integrals converge and the boundary term vanishes. Division by the positive weighted norm gives $\boxed{\lambda =I(w)}$.

The change of coordinate $z=x^{3/2}$, $w(x)=W(z)$ gives exactlyThus ${\lambda }_n=9n^2{\pi }^2/2$ and $w_n=\sin (n\pi x^{3/2})$. To prove a minimum over trial functions, rather than just identify the least eigenvalue, transform the quotient itself:The Dirichlet Poincaré inequality gives the last step, with equality precisely for a nonzero multiple of $W=\sin \pi z$. The trial class is the nonzero finite-energy class with zero Sobolev traces; endpoint values alone without an integrability requirement do not define a finite quotient. This is the Power-coordinate reduction of a singular Sturm-Liouville quotient.