 Solution

Apply the Euler-Lagrange equation to the integrand of $F-\lambda G$, namely $p{y}^{\prime 2}-q{y}^2-\lambda r{y}^2$. With fixed zero endpoint values, it gives $(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0$. Multiplying by $y$ and using integration by parts gives $F=\lambda G$, so each stationary Rayleigh quotient $F/G$ equals its eigenvalue.

In the half-line problem, $p=x$, $q=-x$, $r=1$, so the quotient is $R[y]={\int }_0^{\infty }x(y^{\prime 2}+y^2)dx/{\int }_0^{\infty }{y}^{2}dx$. For $y=e^{-\beta x}$, $\beta >0$, its numerator is $({\beta }^2+1)/(4{\beta }^2)$ and denominator $1/(2\beta )$. Hence $R=(\beta +{\beta }^{-1})/2$, minimized at $\beta =1$:For the compactly supported linear trial, $A>0$, the numerator is $A^2/2+A^4/12$ and denominator $A^3/3$. Thus $R=3/(2A)+A/4$, minimized at $A=\sqrt{6}$, givingThe derivative jump at the cutoff is allowed in the quadratic-form trial space of the Rayleigh-Ritz method: the trial is continuous and has finite form energy.

The exponential ground state of a linearly weighted Sturm-Liouville problem proves which estimate is best. Vanishing endpoint terms giveEquality occurs for $y=C{e}^{-x}$, which also satisfies the differential equation with $\lambda =1$. Thus the exponential trial gives the exact smallest eigenvalue, whereas the linear trial gives a larger upper bound. Finiteness at the singular endpoint supplies the regular branch with $xy{y}^{\prime }\to 0$; all the displayed trials have the required vanishing boundary terms.