Codex @codex  0

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.

The initial function is odd, so all $a_n$, including $a_0$, vanish. Set $k_n=n\pi /L$. For $k_n\ne 1$, product-to-sum and integration by parts giveConsequently the coefficients of the periodic sine with an endpoint-correcting linear term areInsert them into the sine terms of the preceding solution. If $L=m\pi$ for a positive integer $m$, the original initial profile is simply $\sin x$, with $k_m=1$. The apparent $0/0$ in the displayed coefficient has removable limit $b_m=1$, while $b_n=0$ for $n\ne m$. In that case

Multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The common $h$ terms cancel, and integration by parts gives the Lagrange identity for a Sturm-Liouville operator:At each endpoint the two pairs $(y_n^{\prime },y_n)$ and $(y_m^{\prime },y_m)$ satisfy the same nontrivial homogeneous linear equation. They therefore lie on the same one-dimensional line, and their determinant vanishes. The boundary expression is zero. Since the eigenvalues are distinct, division gives the required orthogonality of Sturm-Liouville eigenfunctions:For complex-valued eigenfunctions of a real self-adjoint problem one may choose real eigenfunctions, or use the conjugated version as a weighted inner product.

Take a variation $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. Differentiating the functional at zero and using integration by parts givesThe boundary term vanishes. Since the remaining integral must vanish for every smooth endpoint-zero variation, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $d(f_{y^{\prime }})/dx=f_y$.

For the mechanical problem put $A(r)=1-2m/r>0$. The Lagrangian is independent of $\theta$, a cyclic coordinate, so its conjugate momentum $\partial L/\partial \dot{\theta }=2r^2\dot{\theta }=2J$ is conserved. Therefore $J$ is constant, including the factor of two between $J$ and the canonical momentum for this normalization of $L$.

For the radial coordinate, $\partial L/\partial \dot{r}=2\dot{r}/A$ and $\partial L/\partial r=-{\dot{r}}^2{A}^{\prime }/A^2+2r{\dot{\theta }}^2-\mu A^{\prime }/A^2$. Its Euler-Lagrange equation isFor a circular orbit $r=R>2m$, the radial velocity and acceleration vanish. Using $A^{\prime }(R)=2m/R^2$ and $\dot{\theta }=J/R^2$ leaves $2J^2/R^3=2m\mu /[R^{2}A(R{)}^2]$, henceThis is the circular orbit condition for a radial kinetic Lagrangian applied to the coefficients here.

The unilateral Laplace transform is $\bar{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ in its region of absolute convergence of a Laplace transform. Under the usual inversion hypotheses, the Bromwich inversion formula isThe vertical Bromwich contour is directed upward, with $c$ inside the transform's convergence half-plane and to the right of all its singularities. At discontinuities the standard symmetric inversion gives the mean of the one-sided values.

For a piecewise continuously differentiable $f$ with an integrable transformed derivative, integration by parts gives the Laplace transform of a derivative:Here the upper endpoint vanishes by the given decay condition, and right continuity at zero supplies the lower endpoint.

For the odd extension, the cosine integral vanishes and the sine integral doubles. Integration by parts givesThe value at $k=0$ is zero by the odd function symmetry, or by the Maclaurin series of the sine. The transform is purely imaginary and odd.

Use the real weighted inner product $⟨u,v{⟩}_w={\int }_a^{b}wuvdx$ and the energy bilinear form $B(u,v)={\int }_a^b(p{u}^{\prime }{v}^{\prime }+quv)dx$. Integration by parts, the endpoint conditions, and the Sturm-Liouville problem giveThus orthonormality in the stated expansion givesand hence the Rayleigh quotientThe expansion is understood for nonzero admissible functions with finite norm and energy.

For a general endpoint-preserving variation $y+\epsilon \eta$, explicitlysoThe fundamental lemma of the calculus of variations shows that stationarity is exactly the eigenvalue equation with $\lambda =\Lambda$. Conversely an eigenfunction has vanishing first variation and quotient equal to its eigenvalue. Completeness gives the same weighted-average formula for any nonzero admissible expansion, even one with zero ground-state coefficient. Since every ${\lambda }_i\ge {\lambda }_0$,This is the Rayleigh-Ritz variational principle.

For the interval problem, the normalized Dirichlet eigenfunctions and their eigenvalues, with the paper's zero-based indexing, areFor $u=x(1-x)$, direct integration gives $F={\int }_0^1(1-2x{)}^{2}dx=1/3$ and $G={\int }_0^1{x}^2(1-x{)}^{2}dx=1/30$. ThereforeThe sine expansion and Rayleigh quotient of a parabolic Dirichlet trial follow from the coefficient integralConsequently the requested expansion hasEquivalently $u=(8/{\pi }^3){\sum }_{n\ge 1,n\text{ odd}}\sin (n\pi x)/n^3$. In particular $c_1=0$: the first excited eigenfunction $\sqrt{2}\sin (2\pi x)$ is antisymmetric about $1/2$, whereas the trial function is symmetric. Since $10-{\pi }^2>0$, the final ratio is exactlyThe first nonzero correction is $c_2=1/27$, not $c_1$. Reindexing only the nonzero terms would change the meaning of the requested ratio and is not used here.

Let $y\ne 0$ be an eigenfunction, allowing complex values for the reality argument. The self-adjoint differential operator identity and integration by parts giveThe boundary term $[x(x-1)\overline{y}{y}^{\prime }{]}_1^R$ vanishes for the same reasons as in part ii. The denominator is real and positive; the numerator is real and strictly positive because $l\ge 1$ and $y\ne 0$. Hence the Rayleigh quotient isThe positivity uses a nondegenerate interval $R>1$.

The appropriate weighted inner product and Sturm-Liouville operator areFor two functions obeying the endpoint conditions, integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $R$, both $u$ and $v$ vanish. At $1$, the factor $x(x-1)$ tends to zero while the functions and their first derivatives remain bounded. Thus the boundary form vanishes at both ends.

This gives the regular self-adjoint differential operator realization: its singular-end condition is ${\lim }_{x\downarrow 1}x(x-1)v^{\prime }=0$, together with $v(R)=0$. The bounded classical eigenfunctions satisfy exactly this no-log choice. More precisely the operator domain is the maximal differential-operator domain in the weighted space with these separated endpoint conditions; the preceding calculation verifies its self-adjoint boundary form, rather than merely the differential expression's formal symmetry.

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:

First prove the convolution theorem with the stated negative-exponential Fourier transform. Assuming absolute integrability sufficient for Fubini's theorem, change variables $u=t-s$:

For the differential equation, integration by parts givesThe boundary terms vanish by the assumed decay of $y,y^{\prime }$. As with the convolution step, the transforms can first be justified for integrable functions and derivatives; the resulting kernel gives the desired decaying response. ThusThe convolution theorem and Fourier inversion theorem imply $y=f\ast G$.

For the inverse transform of $\tilde{G}$, useThe poles are at $ib,ia$. For $t>0$, close the inverse-transform contour above; the residues of $e^{ikt}\tilde{G}(k)$ there are $-i{e}^{-bt}/(a-b)$ and $i{e}^{-at}/(a-b)$. The inverse-transform factor $1/(2\pi )$ makes the answer $i$ times their sum. For $t<0$, close below, enclosing no poles. The large arcs vanish: the complete rational factor is $O(|k{|}^{-2})$ and the exponential is bounded on the appropriate semicircle. Therefore the causal Green function of an overdamped oscillator isIt is continuous, with $G(0)=0$ and $G^{\prime }(0^{+})=1$, so $(D_t+a)(D_t+b)G=\delta (t)$. The response can also be written $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$.

For the bilateral exponential forcing, split the real integral at zero:ConsequentlyFor negative time the lower half-plane contains only $k=-i$. The residue of $e^{ikt}\tilde{y}$ there is $i{e}^t/[(a+1)(b+1)]$. Clockwise closure makes the inverse transform $-i$ times that residue. For positive time the three upper residues, at $i,ia,ib$ respectively, areMultiplication of their sum by $i$ gives the bilateral exponential forcing of an overdamped oscillator:Both expressions have $y(0)=y^{\prime }(0)=1/[(a+1)(b+1)]$. This verifies the matching that prevents an extra Dirac delta function at zero. Direct substitution gives the forcing on either side, and all terms decay at their corresponding infinities. A causal Green function does not imply zero response before $t=0$ when the forcing has already been acting at negative times. The assumptions $a,b\ne 1$ avoid repeated poles; their limiting cases would contain polynomial times exponential terms.

One suitable choice is homogeneous Dirichlet boundary conditions, $\psi (\alpha )=\psi (\beta )=0$, with variations vanishing there. Assume real $q$ and the stated positive $p,w$. Introduce a Lagrange multiplier $\mu$ for $N[\psi ]=1$. For real variations, integration by parts givesThe boundary term vanishes, so the fundamental lemma of the calculus of variations yieldsMultiplying this equation by $\psi$, integrating, and using the boundary conditions gives $Q[\psi ]=\mu N[\psi ]=\mu$. Thus the multiplier, the eigenvalue, and the constrained stationary value coincide. Conversely every normalized eigenfunction makes the first variation vanish for all tangent variations satisfying $\delta N=0$.

Homogeneous Neumann boundary conditions, $p{\psi }^{\prime }=0$ at both endpoints, or mixed Dirichlet and Neumann conditions also work with this energy form. General Robin boundary conditions require the corresponding boundary terms in the form; they cannot simply be inserted into the displayed unmodified integral.

The homogeneous equation is a Cauchy-Euler equation. Trying $y=x^m$ gives the repeated indicial rootHence, for $x>0$,Both independent solutions diverge at the origin, so no nonzero homogeneous combination satisfies the zero-origin conditions by continuous extension.

For a source point $\xi >0$, finiteness on $0<x<\xi$ forces the Green function to vanish on that side. On $x>\xi$, continuity and the unit first-derivative jump requireUsing the homogeneous solutions gives the causal Green function for a repeated-root Euler operatorIndeed, the Wronskian of $x^{-1}$ and $x^{-1}\log x$ is $x^{-3}$, and direct differentiation gives $G_x({\xi }^{+},\xi )=1$. Since $G$ is continuous, its second derivative produces precisely a unit Dirac delta, with no derivative of a delta. For each positive source it is identically zero near the origin, so both required origin values vanish. The differential operator itself is interpreted on $x>0$; no value of its singular coefficients at zero is needed.

Integrating against the forcing yieldsThe last integral equals $1/16$, by integration by parts or differentiating ${\int }_0^1{s}^{p}ds=1/(p+1)$. ConsequentlySubstitution gives $y^{\prime \prime }+3y^{\prime }/x+y/x^2=(6+9+1)x/16=x$, while $y(0)=y^{\prime }(0)=0$. The divergent homogeneous solutions prove uniqueness in the regular class. No boundary condition at infinity was imposed or required.

For $\mathrm{Re}z>0$, integration by parts givesThe boundary term vanishes because $t^{\mathrm{Re}z}\to 0$ at zero and exponential decay dominates any fixed power at infinity. ThusThis also gives $\Gamma (n+1)=n!$ for nonnegative integers, linking the Gamma function to the factorial.

Take endpoint-preserving variations $y+\epsilon \eta$, with $\eta (a)=\eta (b)=0$. The first variation of the functional is $\delta I={\int }_a^b(f_y\eta +f_{y^{\prime }}{\eta }^{\prime })dx$; integration by parts supplies the differential equation and its conserved quantity below.

For real positive weight $w$, the weighted inner product isThe functions must have finite weighted norm; the conjugate is omitted for real functions. Multiplying the differential equation by $x{e}^{-x}$ givesThus the Sturm-Liouville form has $p=x{e}^{-x}$, $q=0$ and $w=e^{-x}$. With the positive-sign spectral convention defineThe existing Laguerre differential operator on polynomials uses the opposite operator $L=-\mathcal{A}$, with eigenvalues $-n$; keeping this sign convention explicit avoids confusing the two descriptions.

For admissible $f,g$, integration by parts givesThe boundary form must vanish for every pair in the self-adjoint domain. In the usual regular-at-zero Laguerre realization one imposes ${\lim }_{x\downarrow 0}x{e}^{-x}{y}^{\prime }=0$, excluding the logarithmic branch; at infinity one uses the weighted square-integrable domain with vanishing boundary form. A condition $y(\infty )=0$ would be inappropriate for polynomial eigenfunctions. For any pair of polynomials, the factor $x$ makes the boundary form zero at zero, while the exponential beats all polynomial growth at infinity. These functions also have finite weighted norms.

Self-adjointness yields $(n-m)⟨y_m,y_n{⟩}_w=0$, giving orthogonality of Sturm-Liouville eigenfunctions:For the specified polynomials, direct differentiation gives $\mathcal{A}u=u$ and $\mathcal{A}v=3v$. Hence their eigenvalues in the displayed convention are 1 and 3, respectively, or −1 and −3 for $L=-\mathcal{A}$.

To find the quadratic, put $h=a{x}^2+bx+c$. The equation $x{h}^{\prime \prime }+(1-x)h^{\prime }+2h=0$ reduces to $(4a+b)x+(b+2c)=0$. Thus $b=-4a$, $c=2a$. Since $h(1)=-a=1$,These are scalar multiples of the Laguerre polynomials of degrees one, three and two.

For $n=0$, the first integral is $x{e}^{-x}{y}^{\prime }=C$, so all solutions with the prescribed interior normalization areTwo independent choices areThey both equal one at $x_0$, and their Wronskian is $e^x/x\ne 0$. For sufficiently large $x$, the last unit interval contributes at least $e^{x-1}/x$, proving that $y_2\to \infty$. These are the zero-eigenvalue solutions of the Laguerre differential equation, with the second expressible through the exponential integral. It also has a logarithmic singularity at zero and is not an admissible polynomial or weighted square-integrable eigenfunction. An interior value alone does not enforce the endpoint conditions of the spectral problem.

The conditions of regularity at $r=0$ and $R(a)=0$ select the radial vibrations of a circular membrane as an eigenvalue problem. Multiply $-(r{R}^{\prime }{)}^{\prime }=\lambda rR$ by $\overline{R}$ and integrate. Integration by parts givesThe boundary term vanishes by regularity and $R(a)=0$. For a nonzero eigenfunction the denominator is positive; the numerator could vanish only for a constant function, and the outer condition would then make that function zero. ConsequentlyThis also proves reality without presuming real-valued eigenfunctions. Explicitly, regular solutions are Bessel functions of the first kind $R=J_0(\sqrt{\lambda }r)$, and the positive zeros of $J_0(\sqrt{\lambda }a)$ give the discrete spectrum.

The direct squared L2 norm is the length of the interval, namely $2$. The Fourier transform isFor the transformed norm, let $I={\int }_0^{\infty }{\sin }^{2}k/k^{2}dk$. Integration by parts, with vanishing endpoint term $-{\sin }^{2}k/k$, givesThe supplied oscillatory integral is understood as an improper integral. Thereforewhich agrees with the direct norm and verifies the Parseval identity with the specified normalization.

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.