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For the catenoid profile $y(x)=E\cosh (x/E)$ spanning fixed-endpoint rings, the second variation of $F[y]=2\pi \int y\sqrt{1+y^{\prime 2}}dx$ isThis refers to axisymmetric graph variations, not all possible changes in surface topology. Strict positivity for every nonzero admissible variation gives a strict local minimum in the $C^1$ topology: the positive lowest Dirichlet eigenvalue of the regular Sturm-Liouville problem gives coercivity, and sufficiently small coefficient perturbations preserve it. The zero variation must be excluded from a strict-positivity hypothesis.

For $f\in C([-1,1])$, the equation $((1-x^2)y^{\prime }{)}^{\prime }=f$ has a bounded solution on $(-1,1)$ if and only if ${\int }_{-1}^{1}f=0$. All bounded solutions differ by a constant. To prove necessity, integrate the equation once. The quantity $(1-x^2)y^{\prime }$ has finite endpoint limits, and any nonzero endpoint limit would force logarithmic growth in $y$. Both limits must therefore vanish, yielding the zero-mean condition. For sufficiency set $F(x)={\int }_{-1}^{x}f(t)dt$ and $y(x)=C+{\int }_0^{x}F(s)/(1-s^2)ds$. Zero mean implies $|F(s)|\le \Vert f{\Vert }_{\infty }\min (1+s,1-s)$, so $y^{\prime }$ is bounded and $y$ extends continuously to the endpoints. The homogeneous solutions are a constant and $\frac{1}{2}\log ((1+x)/(1-x))$, proving uniqueness up to a constant. This is the Fredholm alternative for the zero eigenvalue of this singular Sturm-Liouville problem.

The eigenfunction expansion in Legendre polynomials is $y=C-{\sum }_{k\ge 1}{P}_k{\int }_{-1}^{1}f{P}_k/[k(k+1){\int }_{-1}^1{P}_k^2]$. It follows by projecting onto the eigenfunctions with eigenvalues $-k(k+1)$. Completeness follows because the Legendre polynomials span the polynomials, which are dense by the Weierstrass approximation theorem.

Continuity only in the open interval requires more care. Even if $f$ is integrable and has zero mean, the primitive of $F/(1-x^2)$ must still be bounded at both ends. For example, $F(x)=1/\log (e/(1-x^2))$ has zero endpoint limits and absolutely integrable derivative $f=F^{\prime }$, but $F/(1-x^2)$ has a divergent primitive. Thus zero mean alone is insufficient without endpoint control.

For the Sturm-Liouville problem $-(p{y}_n^{\prime }{)}^{\prime }={\lambda }_{n}w{y}_n$, assume that the boundary term $[p{y}_n^{\prime }{y}_m{]}_a^b$ vanishes. One integration by parts gives the displayed identity. Distinct eigenvalues make the right side zero by orthogonality of Sturm-Liouville eigenfunctions, so the derivatives are orthogonal with weight $p$, while the original functions are orthogonal with weight $w$. This conclusion uses the absence of the $qy$ term: for a general potential the extra integral $-\int q{y}_n{y}_m$ need not vanish. For the Chebyshev polynomial of the second kind of degree $n-1$, the two weights are $w=\sqrt{1-x^2}$ and $p=(1-x^2{)}^{3/2}$.

For $\ell >0$ and $k>0$, the Sturm-Liouville problem with one Dirichlet boundary condition and one Robin boundary condition has positive eigenvalues. Indeed, integration by parts givesfor every nonzero eigenfunction. Let $z=\ell \sqrt{\lambda }$ and $\kappa =k\ell$. The eigenfunctions are proportional to $\sin (zx/\ell )$, withFor each $n\ge 1$, there is exactly one root $z_n$ in $((n-1/2)\pi ,n\pi )$: on that interval $\tan z+z/\kappa$ increases strictly from $-\infty$ to a positive value. There are no other positive roots. The apparent root $z=0$ gives only the zero sine function and is not an eigenvalue.

For fixed $k,\ell$, write $z_n=(n-1/2)\pi +{\epsilon }_n$. The root equation becomes $\kappa \cot {\epsilon }_n=(n-1/2)\pi +{\epsilon }_n$, henceAn optional unit normalization uses ${\int }_0^{\ell }{\sin }^2(\sqrt{{\lambda }_n}x)dx=\ell /2+k/[2(k^2+{\lambda }_n)]$. The positive sign of the Robin boundary term matters: reversing it can introduce nonpositive modes, as in nonpositive modes in a Dirichlet-Robin interval expansion.

For positive diffusivity $D(x)$, a diffusion equation with potential-driven drift has diffusive flux $J=-D(c_x+\beta V^{\prime }c)$. On a finite interval with zero flux at both endpoints, the total amount is conserved. The substitution $c=e^{-\lambda t}{e}^{-\beta V}y$ converts the decay eigenvalue problem to the Sturm-Liouville problemThe Rayleigh quotient is $\lambda =\int p|y^{\prime }{|}^2/\int w|y{|}^2\ge 0$. Its zero eigenfunction is constant, giving the stationary solution $c_{\mathrm{eq}}=C{e}^{-\beta V}$. This is a spatial Boltzmann distribution; $D$ changes relaxation rates but cancels out of the zero-flux equilibrium profile. For a sufficiently regular positive $D$ and bounded $V$, the constant eigenfunction is the unique zero mode up to scale.

For $\omega >0$ on $x>0$, use the regular finite branch at zero, decay at infinity, and a finite quadratic-form energy. For trial functions with $[x{y}^2{]}_0^{\infty }=0$, integration by parts givesThe Rayleigh quotient is at least $\omega$, with equality exactly for a nonzero multiple of $e^{-\omega x}$. Direct substitution verifies that this is an eigenfunction with eigenvalue $\omega$, so the lower bound is attained. The zero coefficient $p(0)=0$ makes this a singular-endpoint Sturm-Liouville problem; regularity replaces a prescribed zero endpoint value. Continuous piecewise differentiable trials of finite form energy are allowed in the Rayleigh-Ritz method, even if their derivative has a jump away from the endpoint.

For the Volterra integration operator $Af(x)={\int }_0^{x}f(t)dt$, its adjoint is $A^{\ast }g(x)={\int }_x^{1}g(t)dt$. With $a_n=(n-\frac{1}{2})\pi$, the input singular vectors are $\sqrt{2}\cos (a_{n}x)$ and the output singular vectors $\sqrt{2}\sin (a_{n}x)$, with singular values $1/a_n$. The associated Sturm-Liouville problem has Neumann condition at zero and Dirichlet condition at one for the cosine basis; the sine basis has the reversed conditions. These complete orthonormal bases give a singular value system for an injective compact operator with dense range.

Let $L=w^{-1}[-(d/dx)(pd/dx)+q]$ be the regular self-adjoint operator of a Sturm-Liouville problem, with eigenfunctions orthonormal for the weighted inner product $\int w\overline{u}v$. For $\mu$ outside its spectrum, the displayed kernel satisfies $(L_x-\mu )G_{\mu }=\delta (x-\xi )$ with an ordinary, unweighted Dirac delta. Indeed weighted completeness gives $w(\xi ){\sum }_n{y}_n(x)y_n(\xi {)}^{\ast }=\delta (x-\xi )$. Thus $u(x)=\int G_{\mu }(x,\xi )f(\xi )d\xi$ solves $(L-\mu )u=f$. A kernel integrated against $w(\xi )d\xi$ instead omits the prefactor. For real coefficients, the weighted reciprocity relation is $G_{\mu }(x,\xi )/w(\xi )=G_{\mu }(\xi ,x)/w(x)$ when the spectral parameter is real and outside the spectrum. A zero eigenvalue prevents inversion at $\mu =0$ without a forcing compatibility condition or a projected inverse.

For a regular self-adjoint Sturm-Liouville problem with positive weight $w$ and self-adjoint boundary conditions, eigenfunctions with distinct eigenvalues are orthogonal in the weighted inner product $\int w\overline{u}v$. Integration by parts proves this by equating $\int \overline{u}Lv$ and $\int \overline{Lu}v$.

Write the operator in divergence form:The Legendre differential equation therefore givesThese are the eigenvalues of $\mathcal{L}$. With the common positive-sign convention for a Sturm-Liouville problem, use $-\mathcal{L}$ instead:This is a singular Sturm-Liouville problem, because $p$ vanishes at the endpoints. Bounded eigenfunctions select the nonlogarithmic branches and satisfy $p{P}_n^{\prime }\to 0$ there. The Lagrange identity for a Sturm-Liouville operator givesThe eigenvalues are distinct, proving the orthogonality conditionThe weight is one. No choice of normalization of the Legendre polynomials is needed for this conclusion.

Use the weighted inner product $⟨u,v{⟩}_r={\int }_a^{b}r(x)u(x)v(x)dx$. For real eigenfunctions, integration by parts with the zero endpoint values givesThus $({\lambda }_n-{\lambda }_m)⟨y_m,y_n{⟩}_r=0$. The regular separated Sturm-Liouville problem has simple eigenvalues: for a fixed eigenvalue, prescribing $y(a)=0$ leaves only the initial flux $p(a)y^{\prime }(a)$ free, so uniqueness of the associated first-order system makes any two eigenfunctions proportional. Distinct eigenfunctions are therefore orthogonal, and choosing their signs and scales so that $\int r{y}_n^2=1$ givesThe positive energy form $\int (p{y}^{\prime 2}+q{y}^2)$ realizes $r^{-1}\mathcal{L}$ as a positive self-adjoint operator in this weighted space, with compact inverse on the bounded interval. The spectral theorem for compact self-adjoint operators supplies a complete eigenbasis. This is the standard regular Sturm-Liouville eigenfunction expansion, also valid with continuous $p$ by working with the continuous flux $p{y}^{\prime }$.

For a source $F$, expand a zero-endpoint solution of $(\mathcal{L}-\mu r)y=F$ in that basis. Multiplication by $y_n$ and unweighted integration giveSince $\mu$ is not an eigenvalue, every coefficient is determined. The spectral Green function of a regular Sturm-Liouville problem is consequentlyWeighted completeness means $r(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$ as a distribution, so applying $\mathcal{L}-\mu r$ to this series produces the required delta source. The source integration is with $d\xi$, explaining the absence of an additional factor $r(\xi )$ in the stated kernel.

For the subsequent operator $L=d^2/d{x}^2$ the sign is explicitly reversed: $y_n=\sqrt{2/\pi }\sin (nx)$ and ${\lambda }_n=-n^2$. To cover every real non-eigenvalue $\mu$, define the homogeneous solutionIt satisfies $S_{\mu }(0)=0$, $S_{\mu }^{\prime }(0)=1$ and $S_{\mu }^{\prime \prime }=\mu S_{\mu }$. Matching the two endpoint solutions continuously at $x=\xi$ with derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ yieldsThe denominator is nonzero precisely off the Dirichlet eigenvalues $\mu =-n^2$. In particular $G_0=-x_{<}(\pi -x_{>})/\pi$. At $x=\xi =\pi /2$, its spectral expansion givesThis odd reciprocal-square sum from a Dirichlet Green function proves $\boxed{\sum _{j=0}^{\infty }(2j+1{)}^{-2}={\pi }^2/8}$.

Work at fixed enclosed mass $m$, so the perturbations are fluid Lagrangian perturbations. The equilibrium acceleration vanishes. Varying the factor $r^{-4}$ in the gravitational force contributes $4Gm\delta r/(4\pi r^5)$, while varying the factor multiplying the acceleration gives no first-order term because the background acceleration is zero. With $\delta r=r\xi e^{-i\omega t}$, the linearized momentum equation isConservation of the mass of each displaced shell gives its fractional density change as minus the fractional volume change. ThusThe adiabatic relation is $\delta p=\gamma p\delta \rho /\rho =-\gamma p\chi$. Use $dm/dr=4\pi r^2\rho$ and the stellar hydrostatic equation $p^{\prime }=-Gm\rho /r^2$ to turn the mass-coordinate momentum equation into $(\delta p{)}^{\prime }-{\omega }^2\rho r\xi +4p^{\prime }\xi =0$. It follows that the radial stellar pulsation equation isThe stellar adiabatic exponent may vary with radius, so it must remain inside the derivative.

Expand the derivative and multiply by $r^3$. The self-adjoint differential equation becomesHence $f(r)=\gamma p{r}^4$ and $g(r)=r^3[(3\gamma -4)p{]}^{\prime }+{\omega }^2\rho r^4$ in the requested form. Multiplying by a real eigenfunction and integrating gives the boundary term $[\gamma p{r}^4\xi {\xi }^{\prime }{]}_0^R$. It vanishes for a regular centre and bounded admissible surface displacement and derivative, since $p(R)=0$. Therefore the weighted stellar pulsation Rayleigh quotient isReal coefficients and the symmetric endpoint conditions give the Sturm-Liouville problem; the Rayleigh-Ritz variational principle identifies its minimum with the lowest squared frequency. For complex amplitudes the corresponding quadratic forms use modulus squares, and the real eigenfunctions may be chosen for this calculation.

Integrate the second term of $K$ by parts. Its boundary term $-[(3\gamma -4)p{r}^3{\xi }^2{]}_0^R$ again vanishes. The remaining integrand isSince $\chi =a+3\xi$, one has $3{\xi }^2+2a\xi =({\chi }^2-a^2)/3$. Substitution proves the positive-square radial pulsation energy identityIf $\gamma \ge 4/3$ throughout the positive-pressure interior, both terms are nonnegative for every admissible displacement. No negative squared frequency is then possible. Hence radial dynamical instability requires $\gamma <4/3$ somewhere inside the star.

For the opposite necessary condition, test the quotient with a nonzero constant $\xi =C$. It has $\chi =3C$ andIf $\gamma \le 4/3$ everywhere and is strictly below $4/3$ on a positive-pressure interval, this is negative. The lowest eigenvalue is at most this negative trial quotient, proving instability. Strict dynamical stability requires the stronger weighted conditionand in particular $\gamma >4/3$ somewhere. This is the pressure-weighted radial instability criterion viewed as a necessary test for strict stability.

The word “stability” needs a marginal qualification: if $\gamma =4/3$ throughout, $\xi =C$ solves the pulsation equation with $\omega =0$. The star is marginally stable in this radial sense, so the strict $\gamma >4/3$ requirement applies to positive fundamental squared frequency, not to a definition that includes a zero-frequency neutral mode. There is no exponentially growing mode in this equality case, but the zero restoring force also permits a secular solution linear in time.

The conditions are useful as rapid tests, but local threshold crossings are not a complete criterion for a mixed exponent profile. If $\gamma \ge 4/3$ everywhere and exceeds it on a positive-pressure interval, the square identity forces $K>0$ for every nonzero eigenfunction: zero energy would first force constant $\xi$ and then force that constant to vanish. If $\gamma \le 4/3$ everywhere with strict inequality on an interval, the constant trial proves instability. When both types of region occur, $\gamma <4/3$ somewhere does not alone prove instability, and $\gamma >4/3$ somewhere does not alone prove stability. A negative pressure-weighted average proves instability, while a positive one tests only the homologous displacement; another radial shape may still have negative energy. Determining the mixed case requires minimizing the full weighted stellar pulsation Rayleigh quotient or solving the radial stellar pulsation equation. These are radial adiabatic statements and do not rule out nonradial, convective or nonadiabatic instabilities.

For the real Sturm-Liouville problem, multiply the equations for $y_n$ and $y_m$ by $y_m$ and $y_n$, respectively, and subtract. Integration by parts givesThe boundary term vanishes because both functions are zero at both endpoints. Distinct eigenvalues therefore imply weighted orthogonality. For equal indices the integral is the squared weighted norm, so

Let $c_n={\int }_0^{1}w{y}_n^{2}dx>0$. For $y=\sum a_n{y}_n$ in the operator domain, with convergence sufficient to evaluate its quadratic energy, weighted orthogonality givesFor finite sums these identities follow immediately; the corresponding convergent energy expansion follows by approximation in the operator or quadratic-form domain. For nonzero $y$, the Rayleigh quotient is a weighted average of the eigenvalues, soEquality occurs precisely when only a lowest-eigenvalue component is present.

Assume completeness as permitted and take a sufficiently regular solution of the diffusion equation. It belongs to the weighted expansion space at each positive time; its modal coefficients satisfy ${\dot{a}}_n=-{\lambda }_n{a}_n$. Alternatively differentiate its energy directly and use the preceding inequality:Thus the squared weighted norm is at most its initial value times $e^{-2{\lambda }_{1}t}$. This describes decay when ${\lambda }_1>0$; positivity of $p,w$ alone does not require ${\lambda }_1>0$ if the potential $q$ can be negative. The displayed inequality remains valid regardless of that sign.

Let $m(r)$ be the background enclosed mass, so $g=Gm/r^2$ and $m^{\prime }=4\pi r^2\rho$. The background hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use a Lagrangian fluid displacement $\xi ={\xi }_r$ of each spherical mass shell; $\Delta$ denotes its Lagrangian perturbation. Conservation of shell mass givesand the adiabatic pressure law gives the Lagrangian pressure perturbationThis retains perturbed self-gravity: a shell keeps its enclosed mass while its gravitational acceleration changes with its radius.

In mass coordinates, the exact radial shell equation is $\ddot{r}=-4\pi r^2\partial p/\partial m-Gm/r^2$. Its first-order perturbation isHydrostatic equilibrium gives ${\partial }_{m}p=-g/(4\pi r^2)$, and ${\partial }_m\Delta p=(\Delta p{)}^{\prime }/(4\pi r^2\rho )$. Thus the perturbation of the pressure-force geometry contributes another $2g\xi /r$, in addition to the gravity term:Put $\eta =\xi /r$, so ${\xi }^{\prime }+2\xi /r=r{\eta }^{\prime }+3\eta$. Expanding the derivative, using constant $\gamma$ and $p^{\prime }=-\rho g$, yields the radial stellar pulsation equation:The coefficient $4$ includes both geometric pressure and self-gravity effects; dropping the gravitational perturbation would not give this equation.

For a normal mode $\xi (r,t)=\xi (r)e^{-i\omega t}$, multiply by $\rho r^2{\xi }^{\ast }$ and integrate. Since ${\xi }^{\ast }/r={\eta }^{\ast }$, integration by parts givesRegularity at the center requires $\xi =O(r)$ with finite $\eta$, so the lower endpoint vanishes. For an isolated star take $p(R)=0$ and the free-surface condition $\Delta p=0$, with a regular finite-energy displacement; these eliminate the upper boundary term. HenceThese boundary conditions give the self-adjoint radial Sturm-Liouville problem behind the weighted stellar pulsation Rayleigh quotient.

Use the homologous trial fluid displacement ${\xi }_r=r$, or $\eta =1$. It is regular at the center, and $\Delta p=-3\gamma p$ vanishes at the free surface. The derivative term in the Rayleigh quotient is zero, leavingThe ratio is a positive weighted average of $g/r$. Since the central mass density is its maximum,For $\gamma >4/3$ multiplication preserves this inequality, proving the central-density upper bound for a radial stellar frequency:The full positive quadratic form also shows radial stability for constant $\gamma >4/3$ under these assumptions. A uniform-density equilibrium has constant $g/r$, making the homologous displacement an exact mode and saturating the bound.

If $\gamma <4/3$, the same trial value is strictly negative. Therefore ${\omega }_{\min }^2<0$: the star has an exponentially growing radial instability, with growth rate $\sqrt{-{\omega }_{\min }^2}$. One must use this negative trial quotient, rather than carrying the central-density inequality across the sign change unchanged. At the limiting value $\gamma =4/3$, the homologous displacement has zero restoring force and is a neutral radial mode in this constant-exponent model.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

Let $X=ϵx$ and suppose $k^2(X)$ stays positive and sufficiently smooth in the region considered. Write $y=\exp [{ϵ}^{-1}{S}_0(X)+S_1(X)+\cdots ]$ in ${ϵ}^2{y}_{XX}+k^2(X)y=0$. The leading and first transport equations of the WKB approximation areWith $K=\sqrt{k^2}>0$, they give $S_0^{\prime }=\pm iK$ and $S_1=-\frac{1}{2}\log K+\text{constant}$. Thus the leading WKB approximation for a slowly varying oscillator isThe amplitude is fixed by the transport equation, not just by replacing the constant wave number in a sinusoid. The local validity condition is that variation of $K$ over a wavelength is small, for example $|dK/dx|/K^2\ll 1$.

If $k^2=-{\kappa }^2<0$, the same calculation gives the evanescent formAt a classical turning point where $k^2$ vanishes, these amplitudes diverge and the separated WKB descriptions cease to be uniform. For a simple zero, an Airy turning-point connection formula replaces them in an $O({ϵ}^{2/3})$ interval in $X$; it connects oscillation to exponential behavior with the usual quarter-phase shift. A higher-order zero needs a different local canonical problem, so the Airy statement assumes a simple turning point.

For a smooth positive weight $r$ on the closed interval, let $T={\int }_0^{\pi }\sqrt{r(x)}dx$. The local wave number in the Sturm-Liouville problem is $\sqrt{\lambda }\sqrt{r}$, and the condition at zero selects a sine. Imposing the condition at $\pi$ gives the WKB quantization conditionFor weighted normalization ${\int }_0^{\pi }r|y_n{|}^{2}dx=1$, rapidly oscillating averaging gives $C_n\sim \sqrt{2/T}$. There is no turning-point phase correction here because $r$ stays positive at both endpoints with Dirichlet boundary conditions.

In the specified example $r(x)=(x+1{)}^{-2}$, so $T=\log (1+\pi )=L$ and ${\lambda }_n^{\mathrm{WKB}}=(n\pi /L{)}^2$. To solve exactly, set $t=\log (x+1)$ and $y=\sqrt{x+1}w(t)$. Direct differentiation reduces the Euler-Cauchy equation to $w^{\prime \prime }+(\lambda -1/4)w=0$ on $0<t<L$, with zero endpoint values. Hence the inverse-square weighted Dirichlet spectrum isValues $\lambda \le 1/4$ give no nonzero solution satisfying both endpoint conditions. The WKB eigenfunction shape after leading quantization is actually identical to this exact shape; the leading eigenvalue omits $1/4$.

Using the exact eigenvalue as the reference, its relative error isThus $n\ge 3$ is sufficient and necessary among positive integers. Referencing the approximate value instead gives $n>5L/\pi$, and the same integer answer.

Use primes for differentiation with respect to $X$. Since $X=ϵx$, substituting the amplitude-phase representation into the differential equation and separating real and imaginary parts givesThe second equation is exactly $(\sigma A^2{)}^{\prime }=0$. Thus, with a real constant $C$ and $A\ne 0$,This exact amplitude-phase equation is the Ermakov-Pinney equation for the oscillator. Conversely, any real nonvanishing $A$ satisfying it and $\sigma =C/A^2$ gives an exact complex solution. Existence is not an asymptotic assumption: choose independent real solutions $u,v$ of the original equation, form $y=u+iv$, and set $A=(u^2+v^2{)}^{1/2}$. Their nonzero constant Wronskian prevents simultaneous zeros. The phase derivative satisfies $\sigma A^2=u{v}_x-u_{x}v$, so it is constant. Complex conjugation gives a second branch, and linear combinations recover real solutions.

For the WKB approximation, work on an interval where $K=|k|>0$ is smooth and bounded away from zero. Choose the positive phase branch and an $ϵ$-independent positive $C$. The equations naturally admit even-power expansionsThere is no order-$ϵ$ phase correction in this convention. Leading order gives ${\sigma }_0=K$ and $A_0=(C/K{)}^{1/2}$. Comparing order ${ϵ}^2$ in the real equation yieldsConsequentlyThe constants absorb the normalization. The opposite phase branch changes the sign of $\sigma$ and its correction. A zero of $k$ is a classical turning point where this particular expansion fails; the exact amplitude-phase equation itself remains valid for a suitable complex solution.