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Choose homogeneous Dirichlet boundary conditions, $y(a)=y(b)=0$, and nonzero admissible functions with finite $F$ and positive $G$. Variations obey the same zero endpoint values. At a stationary Rayleigh quotient, set $\lambda =F[y]/G[y]$. ThenThe boundary term vanishes, and the fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Thus stationary functions are eigenfunctions. Conversely, multiply that equation by $y$ and integrate by parts to obtain $F=\lambda G$; its first variation then vanishes by the same calculation. This proves that stationary values are the corresponding eigenvalues. Homogeneous Neumann boundary conditions also make the displayed boundary terms vanish; general Robin boundary conditions can require a boundary contribution to the quadratic form.

For $p=1$, $q=0$, $w=x$ on $(0,\pi )$, the polynomial trial function $y=x(\pi -x)$ satisfies the endpoint conditions. Its Rayleigh quotient usesso the polynomial estimate is $\boxed{{\lambda }_{\mathrm{poly}}=20/{\pi }^3\simeq 0.64503}$. The trigonometric trial function $y=\sin x$ givesand hence $\boxed{{\lambda }_{\mathrm{trig}}=2/\pi \simeq 0.63662}$. Both are upper bounds on the lowest eigenvalue, by the Rayleigh-Ritz method; neither trial is an exact eigenfunction, so the inequalities are strict.

Their closeness has a useful explanation beyond their similar single-arch shapes. Both trial functions are symmetric under $x\mapsto \pi -x$, forcing ${\int }_0^{\pi }x{y}^{2}dx=(\pi /2){\int }_0^{\pi }{y}^{2}dx$. The reflection-symmetric trials for a linearly weighted Rayleigh quotient therefore reduce this problem to a constant multiple of the ordinary Dirichlet energy quotient. The sine is the best trial within this symmetric class, and the polynomial's value is only a factor $10/{\pi }^2\simeq 1.0132$ larger. The unrestricted lowest eigenfunction can be asymmetric because the weight $x$ is asymmetric.