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For real smooth coefficients with $p,w>0$, the Rayleigh quotient $\Lambda [y]=\int [p(y^{\prime }{)}^2+q{y}^2]dx/\int w{y}^{2}dx$ has the displayed first variation. Its stationary points solve the Sturm-Liouville equation on an admissible class whose boundary variations make $[p{y}^{\prime }\eta {]}_a^b=0$. Homogeneous Dirichlet boundary conditions do this by setting $\eta =0$ at the ends; homogeneous Neumann boundary conditions do it through $y^{\prime }=0$. More general self-adjoint boundary conditions can require a boundary contribution to the quadratic form.

For an eigenfunction, the weaker identity $[py{y}^{\prime }{]}_a^b=0$ suffices to give $\lambda =\Lambda [y]$ by integration by parts, but it does not by itself imply the condition on every variation. An explicit counterexample uses $p=q=w=1$ on $[0,1]$ and $y=\cos \pi x+\sin \pi x$. This solves $-y^{\prime \prime }+y=(1+{\pi }^2)y$ and has $[y{y}^{\prime }{]}_0^1=0$. Define a nearby familyIt keeps $y_{\epsilon }(0)=1$, $y_{\epsilon }^{\prime }(0)=\pi$, $y_{\epsilon }(1)=-1+\epsilon$, and $y_{\epsilon }^{\prime }(1)=\pi /(-1+\epsilon )$, so $[y_{\epsilon }{y}_{\epsilon }^{\prime }{]}_0^1=0$ for small $\epsilon$. Nevertheless, $\eta ={\partial }_{\epsilon }{y}_{\epsilon }{|}_0$ has $\eta (0)=0$, $\eta (1)=1$, and ${\int }_0^1{y}^{2}dx=1$, giving $\delta \Lambda =2[y^{\prime }\eta {]}_0^1=-2\pi \ne 0$. Boundary-energy cancellation alone therefore cannot replace a specified variational domain.

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.

For a nonzero real function, write $\Lambda =F/G$ with $G={\int }_a^{b}w{y}^{2}dx>0$. Multiplying the Sturm-Liouville equation by $y$ and integrating givesThus the stated vanishing boundary product proves $\lambda =\Lambda [y]$.

For the variational statement, the first variation isWith admissible boundary conditions that also make $[p{y}^{\prime }\eta {]}_a^b=0$ for every allowed variation, the fundamental lemma of the calculus of variations identifies stationary functions with the eigenfunctions and stationary values with their eigenvalues. Zero Dirichlet boundary conditions, zero Neumann boundary conditions, and matched periodic data give familiar examples. The single printed condition $[py{y}^{\prime }{]}_a^b=0$ proves the quotient identity but, by itself, does not specify the admissible variations needed for stationarity; the boundary conditions for variational Rayleigh quotients make this distinction explicit.

For the half-line equation, multiplication by $x^{n-1}$ gives $p=w=x^{n-1}$ and $q=x^{n+1}$. Use the regular radial boundary condition at zero and square-integrable decay at infinity, under which the proposed Gaussian is admissible. With $y=e^{a{x}^2}$, $a<0$, $y^{\prime }=2axy$, soThe ratio of the two Gaussian integrals is $n/(-4a)$, from a change of variables and the Gamma function recurrence. Differentiating gives ${\Lambda }^{\prime }(a)=-n[1-1/(4a^2)]$, so the minimum on $a<0$ occurs at $a=-1/2$. The Gaussian variational bound for a radial harmonic oscillator therefore givesHere the trial actually satisfies the differential equation with $\lambda =n$. Moreover, on the same regular decaying domain, integration by parts gives $F[y]-nG[y]={\int }_0^{\infty }{x}^{n-1}(y^{\prime }+xy{)}^{2}dx\ge 0$. Hence the estimate is exact, with $\boxed{{\lambda }_0=n}$. For $n=1$, regular radial data mean the Neumann boundary condition $y^{\prime }(0)=0$; imposing a zero Dirichlet boundary condition there would exclude this trial and give a different problem.

For nonzero $y$ with $G[y]>0$, the first variation of the Rayleigh quotient under $y\mapsto y+\epsilon h$ isChoose fixed admissible homogeneous boundary conditions that make the variation boundary term vanish, such as zero endpoint values or zero endpoint derivatives. Compactly supported $h$ then show that a stationary function satisfies the Sturm-Liouville equation with $\lambda =\Lambda [y]$. Conversely an eigenfunction satisfying those admissible conditions makes the variation zero. Multiplying its differential equation by $y$ and integrating by parts givesso the stipulated energy boundary term yields $\Lambda [y]=\lambda$. The displayed energy condition alone is not a full specification of the admissible variations; it must be paired with boundary conditions for this variational equivalence.

For the final Dirichlet problem take the trial function $y=\sin x$, with $p=1,q=0,w=x$. ThenThe Rayleigh-Ritz variational principle makes this an upper bound for the lowest eigenvalue. It is strictly above the true value: equality would require $\sin x$ itself to be a lowest eigenfunction, but $y^{\prime \prime }+\lambda xy=(-1+\lambda x)\sin x$ cannot vanish throughout $(0,\pi )$ for a constant $\lambda$. ThereforeAlthough the weight vanishes at the left endpoint, it is positive in the interior and its weighted quadratic norm is positive for every nonzero admissible continuous function; this does not obstruct the stated Dirichlet variational comparison.

Introduce a Lagrange multiplier for the constraint $G=1$ and vary $F-\lambda G$ with $y\mapsto y+ϵ\eta$, where $\eta (a)=\eta (b)=0$. The first variation isThe fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Multiplying by $y$, integrating by parts and using the endpoint conditions gives $F=\lambda G$. Therefore the normalization implies $\lambda =F$. Assume the coefficient and function regularity needed for these variations and integrations; the condition $w>0$ makes the constraint derivative nonzero at a normalized function.

For the radial problem, set $\sigma =r\psi$. A direct differentiation gives ${\sigma }^{\prime \prime }=r{\psi }^{\prime \prime }+2{\psi }^{\prime }$, soFinite $\psi (0)$ supplies the first boundary value. The Rayleigh quotient is $R[\sigma ]={\int }_0^1{\sigma }^{\prime 2}dr/{\int }_0^1{\sigma }^{2}dr$. For the two trials, ${\sigma }_1=r(1-r)$ and ${\sigma }_2=r(1-r^2)$, the numerator-denominator pairs are $(1/3,1/30)$ and $(4/5,8/105)$, respectively. HenceThe exact eigenvalues are $n^2{\pi }^2$, so $E_{\min }={\pi }^2\simeq 9.86960$. Both trials give upper bounds by the Rayleigh-Ritz variational principle, and the first estimate is more accurate, since it is smaller. Trial admissibility uses the Dirichlet energy form for $\sigma$; the first trial's nonzero radial derivative at the origin does not invalidate this variational calculation.

To improve, add independent trial functions and minimize the quotient over their span. Here merely combining the two supplied trials gives no improvement: their two-dimensional stiffness and mass matrices areThe lowest Ritz value is still ten. Instead add the symmetric quartic $\tau =r^2(1-r{)}^2$. On the span of ${\sigma }_1,\tau$, the generalized eigenvalue problem hasIts determinant is $(E^2-112E+1008)/529200$, so the least root givesThis is strictly better than both original estimates. The symmetry obstruction in a polynomial Rayleigh-Ritz trial space explains why the cubic addition failed while the quartic helps. It is also the shifted-interval version of the polynomial Rayleigh-Ritz trial space for a one-dimensional box.