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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.

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.

Introduce a constant Lagrange multiplier $\mu$ for this isoperimetric constraint and make $A-\mu (C-B)$ stationary. Its integrand is $(1-\mu )y-\mu y^{\prime 2}$, so the Euler-Lagrange equation givesThe endpoint conditions therefore force a parabola $y=Kx(1-x)$. Its area and budget areSolving the quadratic and choosing positive area givesThis requires $B>0$; no positive-area barrier is possible for a nonpositive budget. The corresponding multiplier is $\mu =1/(1+4K)>0$.

To verify a global maximum rather than just stationarity, use the sharp area-cost bound for a barrier with fixed endpoints. For $w=x(1-x)$, integration by parts gives $2A={\int }_0^1{y}^{\prime }{w}^{\prime }dx$, and ${\int }_0^1{w}^{\prime 2}dx=1/3$. The Cauchy-Schwarz inequality impliesFor $A>0$ the last expression increases strictly, yielding exactly the boxed upper bound. Equality requires $y^{\prime }$ proportional to $w^{\prime }$, hence $y=Kw$ after imposing the endpoints. Thus the displayed optimum is unique and lies above the baseline throughout the interior.

Multiply the Sturm-Liouville form for $y_n$ by $y_m$ and subtract the equation for $y_m$ multiplied by $y_n$. Integration givesThe boundary expression is zero because $p\to 0$ and the functions and first derivatives remain finite. Since $n,m$ are distinct positive integers, this provesFor the derivatives, apply integration by parts once to the same Sturm-Liouville form:Thus the new weight isThis is derivative orthogonality when the Sturm-Liouville potential vanishes. For $n=1$ the derivative is identically zero, which is consistent with the identity.

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.

Take $\gamma >0$ and set $A=\pi /2$. The Rayleigh quotient of the trial for the quantum harmonic oscillator isIts denominator is $\pi /(2\gamma )$. The kinetic numerator is ${\gamma }^2$ times that denominator. For the remaining integral, put $t=\gamma x$ and useobtained from ${\cos }^{2}t=(1+\cos 2t)/2$ and integration by parts. Thus the compact cosine trial bound for the quantum harmonic oscillator isDifferentiation gives $R^{\prime }=2\gamma -({\pi }^2-6)/(6{\gamma }^3)$, so the minimizing parameter satisfiesThe quotient tends to infinity at either end of $\gamma >0$, and the two positive terms are equal at this stationary point, proving it is the global minimum. ThereforeFor comparison, the exact ground state has value $1$. The trial is admissible for the energy quadratic form: it is continuous, has compact support, and its weak derivative is square-integrable. Its first derivative jumps at the support endpoints, so it need not belong to the smaller twice-differentiable operator domain; this does not invalidate the Rayleigh-Ritz variational principle bound.

For $t<0$, all forcing points satisfy $\tau >t$, so the Green's function formula becomesAn integration by parts gives $\int \tau e^{-2\tau }d\tau =-(\tau /2+1/4)e^{-2\tau }$. ConsequentlyThe exponential has the required decay as $t\to -\infty$.

Let $G(k)=\tilde{g}(k)$. Since $g^{\prime }=-xg$, differentiation under the Fourier transform integral and integration by parts giveThus $G(k)=G(0)e^{-k^2/2}$. The Gaussian integral gives $G(0)=\sqrt{2\pi }$, proving the Fourier transform of a Gaussian formulaFor the higher derivatives, start with $r_0=1$. If $g^{(m)}=r_{m}g$, the product rule givesIf $r_m$ has degree $m$ and leading coefficient $(-1{)}^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1{)}^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1{)}^m{\mathrm{He}}_m$, where ${\mathrm{He}}_m$ is the Probabilists' Hermite polynomial.

Assume the necessary derivatives decay enough for all boundary terms to vanish. One integration by parts givesRepeating proves the Fourier transform of a derivative rule $\boxed{\widetilde{f^{(m)}}=(ik{)}^m\tilde{f}}$. On the other hand, differentiation under the integral givesand therefore $\boxed{\widetilde{x^{m}f}=i^m{\tilde{f}}^{(m)}}$.

Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$, with $a_n\ne 0$. Applying the second rule and part (a) giveswhereOnly the $j=n$ term can contribute degree $n$. Its leading coefficient is $\sqrt{2\pi }{a}_n(-i{)}^n\ne 0$, so $q_n$ has exactly the same degree $n$, with no possible leading-term cancellation. The coefficients of this transformed polynomial can be complex even when those of $p_n$ are real.

First expand the variational integrand:Its needed partial derivatives are ${\mathcal{L}}_f=3f^{\prime }+4f$ and ${\mathcal{L}}_{f^{\prime }}=2f^{\prime }+3f-x$. The Euler-Lagrange equation is thereforeThe general solution is $f=A\cosh (\sqrt{2}x)+B\sinh (\sqrt{2}x)-1/4$. Subtracting the two endpoint equations gives $B=0$; adding them gives $A=1/[4\cosh \sqrt{2}]$. HenceThis is the unique stationary solution. It is also the unique global minimizer: after discarding the fixed boundary terms, integration by parts rewrites $I$ as ${\int }_{-1}^1[f^{\prime 2}+2f^2+f-2x^2]dx$. For an endpoint-zero variation $\eta$, the linear part vanishes at the displayed solution andwith equality only for $\eta =0$.

Write the endpoint condition in its linear Robin boundary condition form $y^{\prime }(1)=y(1)/\alpha$. For a positive eigenvalue $\lambda =q^2$, the left Dirichlet boundary condition gives $y=C\sin (qx)$, and the right condition givesFor each positive root $q_n$, the squared norm isConsequently the sine coefficients in any complete expansion areWhen $\alpha >1$, these are all modes and the Sturm-Liouville eigenfunction expansion gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all eigenvalues follows from integration by parts and the Cauchy-Schwarz inequality:for a nonzero admissible $y$, since $y(1)={\int }_0^1{y}^{\prime }dx$.

However, the original PDF states only $\alpha >0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha =1$, $\lambda =0$ admits $y=x$, so the correct expansion isFor $0<\alpha <1$, a negative eigenvalue $\lambda =-{\kappa }^2$ admits $y=\sinh (\kappa x)$, withThere is exactly one positive root: $\tanh \kappa /\kappa$ decreases strictly from $1$ to zero. For example, its derivative has the sign of $\kappa {\mathrm{sech}}^2\kappa -\tanh \kappa <0$, because $\sinh \kappa \cosh \kappa >\kappa$. Thus the corrected expansion isThese are the nonpositive modes in a Dirichlet-Robin interval expansion. As an explicit counterexample to the printed claim, at $\alpha =1$ take $f(x)=x$: it satisfies both boundary conditions, and each positive root obeysbecause $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne 0$. Likewise $f=\sinh (\kappa x)$ is a counterexample for $\alpha <1$, by orthogonality of Sturm-Liouville eigenfunctions with distinct eigenvalues. The formulas above give the full solution without silently assuming away the omitted modes.

Put $s_n(x)=\sin (q_{n}x)$ and $s_m(x)=\sin (q_{m}x)$. Their differential equations are $-s_n^{\prime \prime }=q_n^2{s}_n$ and $-s_m^{\prime \prime }=q_m^2{s}_m$. Subtract the two cross-multiplied equations and use integration by parts:The lower boundary term is zero because both functions vanish there. At the upper endpoint, the Robin boundary condition gives $s_j^{\prime }(1)=s_j(1)/\alpha$, soSince distinct positive roots have $q_n^2\ne q_m^2$, we concludeThis orthogonality holds for every $\alpha >0$, even when the sine family requires the additional nonpositive mode for completeness as explained in part (c).

For real $z>0$, apply integration by parts to the gamma integral with $u=t^z$ and $dv=e^{-t}dt$:The boundary term vanishes: $t^z\to 0$ at zero, and exponential decay dominates the power at infinity. This proves the Gamma function recurrence

For fixed initial and final coordinates, the two Euler-Lagrange equations areIndeed the first variation of the functional isIntegration by parts moves the time derivatives from the variations; their endpoint values are zero. The fundamental lemma of the calculus of variations then gives the two equations above.

Impose $y=x$ before varying. Writing $\delta ={\delta }_1+{\delta }_2$, the restricted integrand isIts middle term is the total derivative of $\alpha ({\gamma }_1+{\gamma }_2)(x+\beta x^2/2)$, so it can be dropped for fixed endpoints by part (b). The remaining Euler-Lagrange equation is $\delta \ddot{x}=\alpha \beta$. Integrating twice and imposing $x(0)=0$, $x(T)=S$ givesThis is the unique global minimizer among paths on the specified line. To prove it, add any admissible variation $\eta$ with $\eta (0)=\eta (T)=0$. The change in fuel, after dropping the fixed boundary term, isThe first integral vanishes by integration by parts and the equation for $x$. Equality forces $\eta =0$. No monotonicity or restriction to the segment between endpoints was imposed in the question; the minimizing coordinate can initially decrease for sufficiently large $T$.

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.

For an admissible nonzero real trial function, integration by parts and the zero-flux boundary conditions reduce the Rayleigh quotient toA nonzero constant trial function attains zero. Positivity of $p$ also shows that equality requires $y^{\prime }=0$. HenceIf the conserved total amount is $M={\int }_{-L}^{L}cdx$, the normalized stationary solution isIts diffusive flux vanishes everywhere. The Boltzmann distribution fixes the spatial profile, while the diffusivity affects the relaxation toward it. Boundedness of $V$ makes the normalization finite on this finite interval.

Take nonzero real trial functions $y$ with $y,y^{\prime }$ square-integrable and sufficient decay for integration by parts. For a compactly supported variation $y\mapsto y+\epsilon \eta$, the first variations areThe quotient rule for the functional then yieldsBy the fundamental lemma of the calculus of variations, stationarity for all such $\eta$ is equivalent toConversely, multiplying a decaying eigenfunction equation by $y$ and using integration by parts gives $F[y]=EG[y]$, so its Rayleigh quotient equals $E$ and its first variation vanishes. Thus the stationary values of the functional are the energy eigenvalues. The word "extrema" in the question includes stationary points here: higher eigenfunctions can be saddle points of the unrestricted Rayleigh quotient.

The Rayleigh-Ritz variational principle givesChoose an admissible trial family, calculate its Rayleigh quotient, and minimize over its parameters to obtain the best upper bound within that family. Equality holds for the exact ground state; a restricted family generally gives only an upper bound.

Multiplication by $x^2$ gives the Sturm-Liouville formso the weight is $w(x)=x^2$. Use the weighted inner product $⟨f,g{⟩}_w={\int }_0^{\pi /2}{x}^2\overline{f}gdx$ and the Sturm-Liouville operator $L=-x^{-2}(x^{2}d/dx{)}^{\prime }$. Integration by parts gives its boundary form:The upper endpoint term vanishes by the Dirichlet boundary conditions, and the lower one vanishes for the regular solutions. More precisely, the radial Dirichlet spectrum in three dimensions identifies $L$ under the unitary transformation $v=xu$ with $-v^{\prime \prime }$ on $(0,\pi /2)$ with zero endpoint values; this specifies the self-adjoint realization at the singular endpoint as well as proving the formal identity.

For eigenfunctions $u_a,u_b$ with different eigenvalues, the same identity gives $({\lambda }_b-{\lambda }_a)⟨u_a,u_b{⟩}_w=0$. Taking real eigenfunctions proves the required orthogonality of Sturm-Liouville eigenfunctions.

Set $v=xu$. Direct differentiation gives $u^{\prime \prime }+2u^{\prime }/x=v^{\prime \prime }/x$, so $-v^{\prime \prime }=\lambda v$. Finiteness of $u$ at zero requires $v(0)=0$, and the other endpoint gives $v(\pi /2)=0$. The Dirichlet eigenvalues are positive: integration by parts gives $\lambda \int |v{|}^2=\int |v^{\prime }{|}^2$, and a nonzero $v$ cannot have zero right-hand side. Thus $v=C\sin (\sqrt{\lambda }x)$, and the second endpoint givesDefine $u_n(0)=2n$ by its limit. This is the radial Dirichlet spectrum in three dimensions with radius $\pi /2$. Its weighted squared norm is ${\int }_0^{\pi /2}{x}^2{u}_n^{2}dx=\pi /4$.