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To obtain Sturm-Liouville form, multiply the equation by $(1-x^2{)}^{-1/2}$. The derivative identitygivesThe weight determines the weighted inner product on $(-1,1)$. It is integrable but singular at the endpoints, where $p$ also vanishes. Thus this is a singular-endpoint Sturm-Liouville problem, not a regular problem with strictly positive endpoint values of $p$. The transformation in part (ii) specifies its admissible Dirichlet realization; these are the Dirichlet eigenfunctions of the Chebyshev differential operator.

Use the ordinary real inner product $⟨u,v⟩={\int }_a^{b}uvdx$ on the domain of functions obeying the two separated boundary conditions. Integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $a$, the two boundary-value vectors $(u(a),u^{\prime }(a))$ and $(v(a),v^{\prime }(a))$ satisfy the same homogeneous linear relation. If $A_2=0$, both function values vanish; if $A_2\ne 0$, their derivatives are the same constant multiple $-A_1/A_2$ of their values. Either way $v{u}^{\prime }-u{v}^{\prime }=0$. The same argument applies at $b$. HenceThis proves symmetry, and the regular separated Sturm-Liouville problem has the same adjoint domain, so $L$ is self-adjoint. To see the domain point, require the boundary form to vanish against every admissible $u$: the independently variable endpoint data force an adjoint-domain function to obey those same endpoint relations.

As usual, each endpoint pair $(A_1,A_2)$ and $(B_1,B_2)$ must be nonzero, and the coefficients must have the regularity needed for these operations. An identically zero pair would impose no condition at that endpoint and would not give the claimed self-adjoint operator. The related operator ${\omega }^{-1}L$ is self-adjoint in the weighted inner product ${\int }_a^b\omega uvdx$.

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 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.

Projecting the Sturm-Liouville eigenfunction expansion onto $y_m$ using the weighted inner product givesFor the partial sum $f_N={\sum }_{n=0}^N{a}_n{y}_n$, orthonormality therefore givesFor $f$ of finite weighted norm, the given completeness means that $f_N\to f$ in that norm. The triangle inequality implies $|\Vert f_N{\Vert }_w-\Vert f{\Vert }_w|\le \Vert f_N-f{\Vert }_w\to 0$. Taking the limit of the finite identity proves Parseval's identity:This argument justifies the limiting step without assuming that a double infinite series can be integrated term by term. As usual, the expansion is understood for square-integrable functions with respect to the weighted measure; the boundary and smoothness assumptions of the regular Sturm-Liouville problem provide the usual stronger interpretations when needed.

Substitute $T=e^{-\nu t}R(r)$ into the axisymmetric heat equation and cancel the exponential:Thus the radial ordinary differential equation and its Sturm-Liouville form areThe weight in this Sturm-Liouville problem is therefore $\boxed{w=r}$. The outer Dirichlet boundary condition is $R(1)=0$, and the physical solution is regular at the centre. The factor $r$ in the weighted inner product is also the radial part of the polar area element.

Multiply the separated Sturm-Liouville problem by real $R$ and apply integration by parts:The outer endpoint term vanishes because $R(1)=0$. At zero, regularity gives vanishing radial flux. This can be obtained from the equation without assuming a bounded derivative: let $F=r\kappa R^{\prime }$. Since $F^{\prime }=-\nu rR$ and $R$ is bounded, $F$ has a finite limit at zero. If that limit were nonzero, positive bounded thermal diffusivity would make $R^{\prime }$ have fixed sign and magnitude at least a constant times $1/r$ sufficiently near zero, contradicting bounded $R$. Hence $F(0)=0$, and in fact $F(r)=-\nu {\int }_0^{r}sR(s)ds=O(r^2)$. This also forces $RF\to 0$ for bounded $R$.

The boundary terms therefore vanish, provingAs usual for diffusion, $\kappa$ is nonnegative and positive for $0<r\le 1$; the later power-law case may vanish at the centre. The argument applies in that case as well.

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.

Choose homogeneous Dirichlet boundary conditions $\psi (a)=\psi (b)=0$, and take a nonzero real admissible trial $\psi$. Write the numerator as $F[\psi ]$ and denominator as $G[\psi ]>0$. For an endpoint-zero variation $\eta$, integration by parts givesThus the first variation of the Rayleigh quotient isBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiplying an eigenfunction equation by $\psi$ and integrating gives $F=\lambda G$, so its first variation vanishes. This establishes the equivalence for the chosen Sturm-Liouville problem. Homogeneous Neumann boundary conditions, or a mixed Dirichlet-Neumann pair, also cancel the boundary term; general Robin boundary conditions require adding the corresponding boundary contribution to the numerator.

Use separation of variables in the form $u=X(x)T(t)$, with $T^{\prime }/T=X^{\prime \prime }/X=-\lambda$. Then $T=e^{-\lambda t}$ and the spatial Sturm-Liouville problem isThese real separated Robin boundary conditions give a self-adjoint differential operator, so the eigenvalues are real. All three sign cases must be considered.

If $\lambda ={\mu }^2>0$, write $X=A\cos \mu x+B\sin \mu x$. The first condition gives $B=A/\mu$, while the second reduces toThus $\mu =n\pi$, and scaling the eigenfunction gives $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$.

If $\lambda =0$, then $X=A+Bx$. The first condition gives $B=A$, and the second gives $B=A+B$, so $A=B=0$; there is no zero eigenvalue.

If $\lambda =-{\kappa }^2<0$, write $X=A\cosh \kappa x+B\sinh \kappa x$. The first condition gives $B=A/\kappa$, and the second becomesHence $\kappa =1$, giving the single negative eigenvalue $-1$ and spatial eigenfunction $e^x$. Completeness of this regular Sturm-Liouville problem now givesThis is Robin diffusion with a growing mode. The growth is compatible with the stated conditions: they prescribe the same coordinate derivative at both ends, not identical dissipative outward-normal conditions.

The original PDF has a typographical omission: its displayed equation has $\lambda w(x)$ rather than $\lambda w(x)y(x)$ on the right. Taken literally, it is an inhomogeneous boundary value problem, not the stated Sturm-Liouville problem. The quotient and the later eigenfunction expansion identify the intended equation, which the following derivation produces explicitly.

Choose homogeneous Dirichlet boundary conditions $y(a)=y(b)=0$ and a nonzero real admissible $y$. Then $G[y]>0$ because $w>0$. For an endpoint-zero variation $\eta$, integration by parts givesThe first variation of the Rayleigh quotient is thusBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiply this corrected eigenvalue equation by $y$ and integrate. The same boundary cancellation gives $F[y]=\lambda G[y]$, so the first variation vanishes. This proves the requested equivalence for the intended equation. Homogeneous Neumann boundary conditions, or a Dirichlet condition at one end and a Neumann condition at the other, also eliminate the boundary term. General Robin boundary conditions would require the corresponding boundary contribution in the numerator.

Use the normalized eigenfunctions with the stated weighted inner product. Orthogonality gives $G[\hat{y}]={\sum }_n|c_n{|}^2$. The corrected Sturm-Liouville problem and the vanishing boundary term giveFor a complex trial the numerator means $\int (p|{\hat{y}}^{\prime }{|}^2+q|\hat{y}{|}^2)dx$, agreeing with the given formula for real trials. ThereforeIt is a convex combination of the eigenvalues, with equality only for a trial proportional to the lowest eigenfunction. The Rayleigh-Ritz variational principle consequently estimates ${\lambda }_0$ by choosing a nonzero admissible trial family and minimizing this quotient over its parameters:The result is an upper bound. A more flexible trial space can lower it, while equality requires that the exact lowest eigenfunction belong to the trial space. One need not know the expansion coefficients to calculate the quotient directly from the trial function.

Use the standard regular Sturm-Liouville problem hypotheses: positive weight $w$, real coefficients, and real self-adjoint boundary conditions. The positive weight and boundary-domain assumptions are implicit in the requested real orthonormal basis; real differential coefficients alone would not guarantee them for arbitrary complex boundary conditions.

Multiply the eigenfunction equation by $\overline{y_n}$ and integrate. By self-adjointness, the number ${\int }_a^b\overline{y_n}\mathcal{L}{y}_{n}dx$ is real, whereasThe second integral is strictly positive, so $\boxed{{\lambda }_n\in \mathbb{R}}$. Because the coefficients and boundary conditions are real, the real and imaginary parts of any complex eigenfunction separately obey the same equation and conditions. At least one is nonzero, so a real eigenfunction can be chosen. More generally, each eigenspace has a real basis, which can be orthonormalized using the weighted inner product.

The orthogonality of Sturm-Liouville eigenfunctions, with normalization chosen to one, isFor the real choice, omit the complex conjugate.

The product rule makes the drift term cancel:Consequently the transformed eigenvalue problem isBoth $p$ and the weight $w$ are positive, as required for this Sturm-Liouville problem.

For sufficiently smooth functions satisfying the same boundary conditions, integration by parts gives the boundary formThe stated endpoint relation annihilates this expression. This proves the required self-adjoint differential operator identity; for the usual regular Sturm-Liouville problem, separated real boundary conditions give a self-adjoint realization on its domain.

In this problem the transformed diffusive flux is $J=-e^{-\lambda t}p{y}^{\prime }$. Thus the physical boundary conditions areequivalently $y^{\prime }(\pm L)=0$ when the endpoint values of $p$ are positive. In particular $p{f}^{\prime }=p{g}^{\prime }=0$ at the endpoints, which makes the boundary form vanish directly. The inner product used here is the ordinary integral; the weight $w$ enters the generalized eigenvalue problem, not this boundary-form calculation.

Apply the identity from (c) to the two real eigenfunctions. The equations of their Sturm-Liouville problem giveSince the eigenvalues differ,This is orthogonality in the weighted inner product.

Separation of variables replaces a compatible partial differential equation by ordinary differential eigenvalue equations. For example, with a thermally diffusive rod, $u_t=D{u}_{xx}$, try $u=X(x)T(t)$. Wherever division is valid,The two sides depend on independent variables, so each must be constant. Homogeneous end temperatures $u(0,t)=u(L,t)=0$ give $X_n=\sin (n\pi x/L)$ and ${\lambda }_n=(n\pi /L{)}^2$. The general initial-temperature profile is represented by superposing these separated modes:Each spatial mode decays at its own rate. For a fixed-end wave equation the same spatial eigenfunctions instead have oscillatory time factors $A_n\cos (cn\pi t/L)+B_n\sin (cn\pi t/L)$, with coefficients fixed by initial displacement and velocity. Similar constructions separate Laplace's equation in suitable coordinate systems.

A single product is usually only one mode, not the most general solution. Linear superposition and completeness of the spatial eigenfunctions are what extend the method to general allowed data. Nonhomogeneous boundary values can first be removed by a known stationary or particular solution. Separation is effective when the geometry, coefficients and boundary conditions admit a common separated eigenproblem; an arbitrary nonlinear equation or inseparable geometry need not do so. Self-adjoint Sturm-Liouville problems explain the weighted orthogonality and completeness in many physical applications.

A bounded spherically symmetric solution of the Helmholtz equation in a ball with zero value at radius $a$ satisfies $-(r^2{\psi }^{\prime }{)}^{\prime }=k^2{r}^2\psi$. The substitution $u=r\psi$ gives $u^{\prime \prime }+k^{2}u=0$ with $u(0)=u(a)=0$, so the displayed modes have $n\ge 1$; the apparent singularity at zero is removable. Their Rayleigh quotient is ${\int }_0^a{r}^2{\psi }^{\prime 2}dr/{\int }_0^a{r}^2{\psi }^{2}dr$. The trial profile $1-r^2/a^2$ gives the upper bound $21/(2a^2)$, compared with the exact lowest value ${\pi }^2/a^2$. Boundedness at zero eliminates the $1/r$ solution and makes the endpoint term in the Sturm-Liouville problem vanish.