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For real eigenfunctions $y_i,y_j$, multiply their Sturm-Liouville problems crosswise and subtract. Integration givesAt $R$, both eigenfunctions vanish, while at $1$ their bounded values and bounded first derivatives multiply a factor tending to zero. The right side is zero. Distinct eigenvalues therefore give the orthogonality of Sturm-Liouville eigenfunctionsFor complex eigenfunctions, replace $y_i$ by $\overline{y_i}$, in accordance with the weighted inner product.

Use a Lagrange multiplier $\mu$ for the norm constraint. For real variations $u+\delta \eta$ with $\eta (0)=\eta (\pi )=0$, integration by parts givesThus stationarity is equivalent to the Sturm-Liouville problemMultiplying by $u$ and integrating shows $F[u]=\mu G[u]=\mu$, so the constrained stationary values are the eigenvalues, and the normalized stationary functions are their eigenfunctions. Conversely the eigenvalue equation makes the constrained first variation vanish. The self-adjoint operator has weight one.

For $ϵ=0$, zero and negative eigenvalues yield no nonzero solution with both Dirichlet boundary conditions. Positive eigenvalues have solutions $u=A\sin (\sqrt{\lambda }x)$ after the left condition, and the right condition gives $\sqrt{\lambda }=n$, $n\ge 1$. Hence the lowest eigenpair iswhere the amplitude follows from ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$.

The Rayleigh-Ritz variational principle bounds the full lowest eigenvalue by $F[Y_1]$. Its derivative contribution is one. For the linear potential, reflection about $\pi /2$ givesThus the sine trial bound for a linear potential on an interval isFor $ϵ>0$ the bound is strict, since $-Y_1^{\prime \prime }+ϵx{Y}_1=(1+ϵx)Y_1$ is not a constant multiple of $Y_1$. Also ${\lambda }_{ϵ}\ge 1$ by the Poincaré inequality and the nonnegative potential. Equality in the upper bound holds at $ϵ=0$.

For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The $q$ terms cancel. The resulting Lagrange identity for a Sturm-Liouville operator isIntegrating first over $[ϵ,R]$ and then taking the endpoint limits givesThe needed assumptions are that the improper integral exists and that the displayed boundary expression tends to zero at both endpoints. Bounded $y_m,y_n$ and their first derivatives are sufficient because $p\to 0$, but are stronger than necessary. Merely knowing that $p\to 0$ does not control arbitrarily growing eigenfunctions. With the vanishing boundary form and distinct eigenvalues, division by ${\lambda }_n-{\lambda }_m$ proves orthogonality of Sturm-Liouville eigenfunctions:For complex eigenfunctions in a real self-adjoint differential operator problem, use $\overline{y_m}$ instead, giving the corresponding weighted inner product.

Multiplying the particular equation by $e^{-x}$ puts it in Sturm-Liouville form:ConsequentlyIts polynomial eigenfunctions are the Laguerre polynomials. Products of these polynomials, with or without one derivative, grow only algebraically at infinity. The exponential in $p$ therefore makes the boundary form vanish there; its factor $x$ makes it vanish at zero. Their orthogonality is thus

For $\lambda =1$, substitution of $Ax+B$ gives $A+B=0$. For $\lambda =2$, substitution of $A{x}^2+Bx+C$ gives $4A+B=0$ and $B+2C=0$. Choosing $y_n(0)=1$, which fixes the otherwise arbitrary nonzero multiplicative constants, yieldsTheir product is $1-3x+\frac{5}{2}{x}^2-\frac{1}{2}{x}^3$. Using the Gamma function identity ${\int }_0^{\infty }{x}^k{e}^{-x}dx=k!$, also obtainable by repeated integration by parts, verifies the required orthogonality directly:

Use the real eigenfunctions, or insert a complex conjugate on the first factor for complex functions. Multiply $-(p{y}_n^{\prime }{)}^{\prime }=n^{2}w{y}_n$ by $y_m$, do the same with $m,n$ interchanged, and subtract. Integration by parts givesThe boundary term vanishes because both functions and derivatives are bounded while $p\to 0$. For distinct squared indices, orthogonality of Sturm-Liouville eigenfunctions therefore yieldsA second integration by parts, using the equation for $y_n$, givesThus derivative orthogonality when the Sturm-Liouville potential vanishes givesThe index qualification matters. The source initially allows integer indices, so $n\ne m$ alone does not imply distinct eigenvalues: $m=-n$ gives the same $n^2$. Taking the conventional nonnegative labels $n,m\ge 0$ makes the stated conclusion valid. For the polynomial solutions below, $y_{-n}=y_n$, so opposite indices are an explicit counterexample to an unqualified orthogonality claim.

Let $L{y}_n={\lambda }_n{y}_n$. By self-adjointness,Taking $m=n$ first shows that each eigenvalue is real: $⟨y_n,y_n{⟩}_w>0$ for a nonzero eigenfunction, since $w>0$. ThereforeFor distinct eigenvalues this proves orthogonality of Sturm-Liouville eigenfunctions:The weight is essential: this need not be orthogonality under the unweighted integral.

Expand the forcing and the solution in the normalized eigenfunctions: $1={\sum }_n{b}_n{y}_n$ and $y={\sum }_n{c}_n{y}_n$. Their coefficients areSince all ${\lambda }_n=n^2{\pi }^2-1$ are nonzero, the equation gives $c_n=b_n/{\lambda }_n$. Hence the required Sturm-Liouville eigenfunction expansion isThis is constant forcing in a logarithmic Dirichlet Sturm-Liouville problem. The coefficients are $O(n^{-3})$, so the solution series converges uniformly and has zero endpoint values. Applying $L$ is justified in the weighted square-integrable sense; the forcing sine series equals one at interior points, while its endpoint sums are zero. The equation is imposed in the interior, so this does not contradict the boundary conditions.

As an independent check, solving $-Y^{\prime \prime }-Y=1$ directly givesIt has both zero endpoint values and satisfies $Ly=1$. The invertibility of the Dirichlet operator makes it the same solution as the expansion.

Let $R[y]=F[y]/G[y]$, where $G[y]>0$ for a nonzero admissible real function. Writing $\lambda =R[y]$, its first variation isChoose boundary conditions making the boundary term vanish, for example homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions, or a corresponding mixed pair. Arbitrary interior variations then give the Sturm-Liouville problemMultiplying this equation by $y$ and integrating by parts, with $[py{y}^{\prime }{]}_{\alpha }^{\beta }=0$, yieldsThus eigenfunctions are stationary directions of the quotient. The lowest eigenvalue is its global minimum; higher eigenfunctions need not be unrestricted maxima or minima. General Robin boundary conditions require including their boundary energy in the quotient unless the displayed endpoint products also vanish.

The Rayleigh-Ritz method chooses admissible trial functions depending on finitely many coefficients and minimizes this quotient over them. Expansion in a complete orthonormal basis with weight $w$ shows $R[y]\ge {\lambda }_0$, so the trial minimum is an upper bound on the true lowest eigenvalue. In a finite basis, the minimization becomes a generalized eigenvalue problem for the stiffness and weighted mass matrices. Enlarging the trial space can only lower or preserve this upper bound.

On its specified domain of admissible functions, a self-adjoint operator equals its adjoint, including the adjoint domain. Its integration identity on the interval isFor a differential operator, the boundary conditions must make the boundary form from integration by parts vanish. Formal symmetry of the differential expression alone does not specify a self-adjoint boundary-value problem.

Apply this identity to two eigenfunctions. Since the weight is real,First take $m=n$: positivity of ${\int }_a^{b}w|y_n{|}^{2}dx$ shows ${\lambda }_n={\lambda }_n^{\ast }$, so the eigenvalues are real. For distinct eigenvalues, subtraction therefore givesThis is orthogonality of Sturm-Liouville eigenfunctions in the weighted inner product, rather than necessarily in the unweighted inner product.

Using the normalized eigenfunctions and the same change of variables, their weighted inner product isFor distinct positive integers $m,n$, the product-to-sum identity givesbecause each integrated sine has integer-multiple-of-$\pi$ endpoints. For $m=n$, the integral is one by the normalization already calculated. HenceThis verifies orthonormality explicitly, with ${\delta }_{mn}$ the Kronecker delta.

For an eigenfunction in the operator domain, $H{\psi }_n=E_n{\psi }_n$. Substitution into the Rayleigh quotient givesThe cancellation only needs a finite nonzero norm; orthonormalization makes both denominator and norm equal to one.

Set $y(x,t)=X(x)T(t)$ in the wave equation. A nonzero product obtained by separation of variables obeysThe fixed-end Dirichlet boundary conditions require $X(0)=X(L)=0$. The zero and negative spatial eigenvalue cases produce only the trivial solution; the nontrivial eigenfunctions areFor each normal mode, $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$. By superposition and completeness of the Fourier sine basis, the general finite-energy solution has the expansionIf $f(x)=y(x,0)$ and $g(x)=y_t(x,0)$, orthogonality gives $A_n=(2/L){\int }_0^{L}f(x)\sin (k_{n}x)dx$ and $B_n=2{\int }_0^{L}g(x)\sin (k_{n}x)dx/(L{\omega }_n)$. Individual products are separable; their sum is the general solution.

Multiply the eigenvalue equation of the Sturm-Liouville problem by $y$ and integrate. Integration by parts givesSince $y$ is a nonzero eigenfunction and $\omega >0$, the denominator is strictly positive. Thus the Rayleigh quotient isThe endpoint term must be retained for a general Robin boundary condition; it vanishes for homogeneous Dirichlet boundary conditions but need not vanish in the present general problem.

The left boundary condition is $y(0)=0$, while the right one is $y^{\prime }(\ell )=-ky(\ell )$. Hence $[y{y}^{\prime }{]}_0^{\ell }=-ky(\ell {)}^2$. The Rayleigh quotient becomesEvery term on the right is nonnegative. If the whole expression were zero, in particular ${\int }_0^{\ell }(y^{\prime }{)}^{2}dx=0$, so $y$ would be constant. Its zero value at the left endpoint would force $y\equiv 0$, contrary to its being an eigenfunction. ThereforeThis is the positivity mechanism for Dirichlet-Robin eigenvalues on a finite interval. The sign of the boundary contribution is positive for the convention $y^{\prime }+ky=0$ with $k>0$.

Self-adjoint form. Grouping the two derivative terms gives $[(1+ϵx)u^{\prime }{]}^{\prime }-ϵu=-\lambda u$. Thus a regular Sturm-Liouville operator isThe boundary conditions are mixed Neumann boundary conditions at zero and Dirichlet boundary conditions at $\pi$. For $ϵ\ge 0$, $p>0$, $q\ge 0$ and $w>0$, as required. The boundary term $[p(f{g}^{\prime }-f^{\prime }g){]}_0^{\pi }$ vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.

Constrained variation. Introduce a Lagrange multiplier $\Lambda$ for the normalization and varyIts first variation isThe endpoint term vanishes because $v^{\prime }(0)=0$ and $\delta v(\pi )=0$. The Euler-Lagrange equation is thereforewhich is exactly the original eigenvalue problem for the boxed choices of $p,q,w$. Multiply by $v$ and integrate by parts. The same endpoint conditions giveThus the stationary functional values are the eigenvalues, with $\Lambda =\lambda$. More generally, before normalization the relevant Rayleigh quotient is

Unperturbed ground mode. At $ϵ=0$, the equation is $-u^{\prime \prime }=\lambda u$. The quadratic form is nonnegative, and zero eigenvalue would force a constant function incompatible with the endpoint at $\pi$. For positive $\lambda$, $u=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. The derivative condition forces $B=0$, while the other endpoint gives $\cos (\pi \sqrt{\lambda })=0$. HenceThe lowest eigenvalue and normalized eigenfunction are

Trial-function bound. The Rayleigh-Ritz method says that the minimum eigenvalue is no larger than the quotient of any admissible trial function. The same $U_0$ still satisfies both boundary conditions and retains unit norm. Its derivative is $U_0^{\prime }=-\sqrt{2/\pi }\sin (x/2)/2$, givingUsing ${\sin }^2(x/2)=(1-\cos x)/2$,Therefore the affine-parameter Sturm-Liouville eigenvalue bound isThis is an upper bound for every $ϵ\ge 0$, not merely a small-parameter approximation. As a lower-bound check, $p\ge 1$ and the unperturbed variational minimum imply ${\mathcal{R}}_{ϵ}[v]\ge 1/4+ϵ$ for every admissible normalized $v$. Thus

The relevant weighted inner product is $⟨u,v{⟩}_w={\int }_0^{1}w(x)u(x)v(x)dx$. The normalized eigenfunctions of the Sturm-Liouville problem have the orthonormality propertyHere ${\delta }_{nm}$ is the Kronecker delta: the integral is one for $n=m$ and zero otherwise. The weight is $w$, rather than $p$ or the unweighted measure.

For a nonzero admissible radial profile define the radial diffusion Rayleigh quotientMinimizing the numerator subject to ${\int }_0^{1}r{R}^{2}dr=1$ introduces a multiplier $\nu$. The Euler-Lagrange equation for $r\kappa R^{\prime 2}-\nu r{R}^2$ isThe allowed variations vanish at $r=1$, and the natural condition at the centre is zero radial flux. Thus the minimum, when attained, is the smallest decay eigenvalue ${\nu }_0$. Equivalently, expansion in weighted-orthogonal eigenfunctions makes $\mathcal{R}[R]$ a weighted average of their decay eigenvalues. The Rayleigh-Ritz variational principle therefore givesAdmissible trial functions must obey the outer boundary condition, be nonzero and have finite weighted norm and energy; they need not solve the differential equation. If the diffusion operator has no lowest eigenfunction, the same variational argument bounds the infimum of its spectrum. Existence of a fundamental diffusion mode is an additional assumption for strongly vanishing diffusivity, not a consequence of the identity alone.

Take real admissible functions with finite $F,G$, positive $G$, and a self-adjoint domain for the Sturm-Liouville operator. Compactly supported variations $\eta$ avoid any boundary contribution at infinity. Formal self-adjointness givesFor $\Lambda =F/G$, the quotient rule therefore givesVanishing for arbitrary variations requiresConversely an eigenfunction in the self-adjoint domain makes this variation zero. Multiplying its equation by $y$ and integrating also shows directly that its stationary quotient is $\boxed{\Lambda [y_n]={\lambda }_n}$.

Thus the stationary values are the admissible eigenvalues. Under the usual positive-weight, semibounded hypotheses the lowest is the global minimum, while higher eigenfunctions generally give saddle points. The assumption $y\to 0$ alone is not a complete domain specification for arbitrary $\rho ,\sigma$: the required integrals and boundary form must exist. The rapidly decaying oscillator trials in part (b) meet these conditions.

Normalize the Gaussian trial first. The Gaussian integral givesIts second and fourth moments are $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Since ${\psi }^{\prime }=-\alpha x\psi$, the kinetic term is $\int ({\psi }^{\prime }{)}^2={\alpha }^2⟨x^2⟩=\alpha /2$. At the specified coupling, the energy is thereforeThe stationary condition isThe quadratic factor is positive for real $\alpha$, so the unique positive stationary width is $\alpha =2$. Also $E^{\prime \prime }={\alpha }^{-3}+9{\alpha }^{-4}>0$ and $E$ diverges at both ends of $\alpha >0$, proving a global minimum within this trial family. Its value isThe associated Sturm-Liouville problem isHere $\rho =1$, $\sigma =x^2+2x^4$ and $w=1$. The Rayleigh-Ritz variational principle makes the exact lowest eigenvalue the minimum over all normalized admissible states. Restricting to Gaussians can only raise this minimum, hence $\boxed{E_0\le 13/8}$. The inequality is strict because $H{\psi }_{\ast }/{\psi }_{\ast }=2-3x^2+2x^4$ is not constant: the trial is not an exact eigenfunction. This is the Gaussian trial for a quadratic-plus-quartic oscillator.

Fix one self-adjoint choice of separated homogeneous Dirichlet boundary conditions or Neumann boundary conditions at each endpoint. The stated product condition then makes the boundary term $[awy{y}^{\prime }{]}_{\alpha }^{\beta }$ vanish. Multiplication of $Ly=\lambda wy$ by $y$ and integration by parts giveHence define the quadratic functionalsand $\Lambda [y]=F[y]/G[y]$ for nonzero admissible $y$. The first expression for $F$ is linear in the operator $L$, and $G$ is linear in the weight $w$.

To prove the minimum statement, expand an admissible function in a complete weighted orthonormal system of eigenfunctions of the Sturm-Liouville problem, $y={\sum }_n{c}_n{y}_n$, with ordered eigenvalues ${\lambda }_0\le {\lambda }_1\le \cdots$. Self-adjointness and orthogonality of Sturm-Liouville eigenfunctions giveEquality is attained by a lowest eigenfunction, soThis is the Rayleigh-Ritz variational principle. The endpoint product condition is shorthand for fixed choices of boundary type: allowing each trial function to choose different endpoint types would not define one fixed self-adjoint eigenvalue problem.

Move the differential terms to the left in the sign convention of part (a). Then $\tilde{a}=1+x$, $\tilde{c}=0$, and $\tilde{b}=(3+2x)-{\tilde{a}}^{\prime }=2(1+x)$. The integrating factor is therefore $w=e^{2x}$, up to a positive constant. The Sturm-Liouville problem isThe trial $y=x{e}^{-x}$ is admissible because $y(0)=0$ and $y^{\prime }=e^{-x}(1-x)$ vanishes at one. Its Rayleigh quotient has numeratorand denominatorConsequently the Rayleigh-Ritz variational principle givesThe numerator is positive for every nonzero admissible function, so the lowest eigenvalue is positive. The trial is not an exact eigenfunction: substituting it would require $x^2+2x-1=\lambda x$ for every $x$. Thus the upper inequality is actually strict. This example underlies the exponential-polynomial trial space for a mixed Sturm-Liouville problem.