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Normalize the second-order equation as $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$. If every point other than zero and infinity is ordinary and both exceptions are at worst regular singular points, then $zP$ and $z^{2}Q$ extend holomorphically across zero and stay bounded at infinity. Liouville's theorem makes them constants, yielding the displayed Euler-Cauchy equation. If both exceptions must be genuinely singular, exclude $(A,B)=(0,0)$ and $(2,0)$, which make zero and infinity respectively ordinary.

For $y^{\prime \prime }+\lambda x^{-2}y=0$ on $0<a<x<b$ with zero endpoint values, write $t=\log (x/a)$ and $y=\sqrt{x}w(t)$. The Euler-Cauchy equation becomes $w^{\prime \prime }+(\lambda -1/4)w=0$. Thus $y_n=\sqrt{x}\sin [n\pi \log (x/a)/\log (b/a)]$ and the exact spectrum is displayed above. Leading WKB quantization omits the constant $1/4$, while producing the same leading eigenfunction shape after quantization.

In a flat matter-dominated universe with $a\propto t^{2/3}$ and $P\propto {\rho }^{4/3}$, the squared sound speed scales as $t^{-2/3}$. Each comoving Fourier mode then satisfies an Euler-Cauchy equation, with gravitational growth at long wavelengths and logarithmic-time acoustic oscillations at short wavelengths.

During matter domination in a flat FLRW metric, the cosmological continuity equation gives $\rho \propto a^{-3}$, and the Friedmann equation gives $a\propto {\tau }^2$ and $8\pi G\rho a^2=12/{\tau }^2$. The CDM density equation in a matter-radiation universe, with radiation neglected, is therefore ${\delta }^{\prime \prime }+2{\delta }^{\prime }/\tau -6\delta /{\tau }^2=0$. Its Euler-Cauchy equation powers satisfy $(p-2)(p+3)=0$, proving the displayed general solution. The growing density contrast is proportional to the scale factor, while the decaying mode is proportional to $a^{-3/2}$. This all-scale form uses the CDM-comoving synchronous gauge; other density slices require their corresponding gauge transformation.

In deep radiation domination, $a\propto \tau$ and $a^{\prime }/a=1/\tau$. Neglect ${\Omega }_c{\delta }_c$ against the radiation gravity source, so the supplied CDM density equation in a matter-radiation universe becomesOn superhorizon scales omit the $k^2{\delta }_r/3$ pressure-gradient term. Under adiabatic initial conditions, ${\delta }_r=(4/3){\delta }_c$ then givesThe Euler-Cauchy equation characteristic polynomial is $p^2-4$, soThese are the general growing and decaying adiabatic solutions in this approximation. A regular primordial growing mode discards the divergent ${\tau }^{-2}$ solution. Although its conformal-time growth is again quadratic, now ${\delta }_c\propto a^2$, rather than ${\delta }_c\propto a$ in matter domination.

If “general” is taken without imposing adiabaticity, both independent relative-entropy integration constants must also be retained. The gradient-free radiation equation givesSubstitution into the matter equation producesThis is the general solution of the leading supplied coupled system, described by the superhorizon radiation-era entropy integration constants. Setting both $S_0$ and $S_1$ to zero recovers the adiabatic result. Regularity and the full initial Einstein field equations constraints can further restrict which of these leading solutions represents a prescribed physical initial mode.

On scales well inside the horizon, radiation pressure drives oscillations at frequency approximately $k/\sqrt{3}$. Their amplitude is bounded at leading order, and their gravitational forcing of a slowly varying matter contrast can be averaged away. With this approximation, as well as negligible matter self-gravity,hence the logarithmic CDM growth after radiation-era horizon entry isThis is the Mészáros effect. It does not mean that ${\delta }_r$ itself vanishes. For a rapidly oscillating source $R_{\gamma }\cos (\omega \tau +\phi )$, the matter equation has a particular responseIts leading second derivative reproduces $3{\delta }_r/{\tau }^2$; the remaining terms are smaller by powers of $(k\tau {)}^{-1}$. Thus the suppressed cold-matter forcing by rapid radiation oscillations is only order ${\delta }_r/(k\tau {)}^2$ and does not change the secular logarithm. Radiation forcing near horizon entry determines $C_0,C_{\log }$ and cannot be averaged away there. Neither this smoothing approximation nor the neglect of matter gravity remains uniformly valid near equality.

In deep matter domination, the scale factor is $a\propto {\tau }^2$, so $a^{\prime }/a=2/\tau$. Neglect radiation in the gravity source, including its backreaction, to obtainFor a power ${\tau }^p$ the Euler-Cauchy equation becomes $p(p-1)+2p-6=(p-2)(p+3)=0$. Its two independent matter-era growing and decaying density modes give the full approximate solutionThere is no $k$-dependent pressure term for cold dark matter. Thus this leading equation is valid both outside and inside the horizon in the stated comoving synchronous gauge, as long as the perturbations remain linear and the omitted radiation gravity is negligible. A different time slicing would give a different density variable.

For $k>0$, put $\omega =k/\sqrt{3}$, $C_g=A/{\tau }_i^2$ and $C_d=B{\tau }_i^3$. The remaining radiation equation is a forced harmonic oscillator:The matter-era forced radiation response can be written exactly within the matter-background approximation asThe last term is a Green's function particular solution; changing its lower limit only changes the two oscillatory constants. For the decaying source, a slowly varying particular solution obeys ${\delta }_{r,p}\simeq (4/3){\delta }_c^{\prime \prime }/{\omega }^2$, since a second time derivative is smaller than ${\omega }^2{\delta }_{r,p}$ by order $(k\tau {)}^{-2}$. Consequently the subhorizon asymptotic form isFor a pure growing mode, the constant offset and acoustic terms are an exact solution of this reduced radiation equation. In particular, the late radiation contrast does not follow the ever-growing ${\delta }_c$.

Before crossing, a regular growing adiabatic cosmological perturbation has both contrasts proportional to ${\tau }^2$, with ${\delta }_r=(4/3){\delta }_c$ to leading order. Afterwards cold dark matter keeps this quadratic growth, whereas radiation undergoes subhorizon radiation acoustic oscillations about a constant forced offset. To make the requested sketch concrete, one regular leading matter-era solution on both sides isIts small-$k\tau$ expansion gives ${\delta }_r=(4/3){\delta }_c+O((k\tau {)}^4)$, and its oscillation frequency is the radiation sound frequency. The sketch plots amplitudes in units of $\epsilon$; $\epsilon$ can be chosen small enough that the whole plotted interval remains linear. The line at ${\tau }_h$ uses the wavelength-to-particle-horizon convention in the question, rather than asserting that the smooth transition occurs at a sharp surface.

During radiation domination, the cosmological continuity equation gives ${\rho }_R\propto a^{-4}$. The flat Friedmann equation in conformal time is $(a^{\prime }/a{)}^2=8\pi G{a}^2{\rho }_R/3$. Hence $a^{\prime 2}$ is constant, and choosing the Big Bang at $\tau =0$ yieldsFor the CDM density equation in a matter-radiation universe, neglect ${\rho }_C{\delta }_C$ while retaining ${\delta }_C$ itself. The adiabatic initial conditions give ${\delta }_R=4{\delta }_C/3$, soThis Euler-Cauchy equation has powers ${\tau }^p$ with $p(p-1)+p-4=0$, namely $p=2,-2$. Selecting the growing superhorizon adiabatic CDM mode in radiation domination gives

During matter domination, ${\rho }_C\propto a^{-3}$. The flat Friedmann equation in conformal time implies $a^{\prime 2}\propto a$, and henceWith radiation neglected, the CDM density equation in a matter-radiation universe reduces toFor this Euler-Cauchy equation, ${\delta }_C\propto {\tau }^p$ requires $p(p-1)+2p-6=(p-2)(p+3)=0$. Thus the matter-era growing and decaying density modes are ${\tau }^2$ and ${\tau }^{-3}$, respectively, andThere is no explicit $k$ in this pressureless equation, so the result holds on all scales within the stated CDM-comoving synchronous gauge description. The amplitude can still depend on $k$ through its earlier evolution; the result does not equate density contrasts on different time slicings.

Let $w=c_s^2$ be constant and small, and write $\theta =i\mathbf{k}\cdot \mathbf{v}$. The supplied synchronous gauge equations implyDifferentiating the first and eliminating $\theta$ givesFor the dominant single component, the Einstein equation gives $h^{\prime \prime }+\mathcal{H}{h}^{\prime }=-3{\mathcal{H}}^2(1+3w)\delta$. Therefore the exact algebraic intermediate relation isThe requested nonrelativistic density equation in synchronous gauge follows to leading nonrelativistic order: discard pressure corrections to the background, Hubble damping, and metric tensor terms while retaining $c_s^2{k}^2\delta$, which can be comparable to gravity at the Jeans wavenumber. Since $w\ll 1$, that scale lies well inside the Hubble radius, so keeping its enhanced gradient term is consistent. The resulting equation iswhere the Friedmann equation gave $3{\mathcal{H}}^2/2=4\pi G{a}^2{\bar{\rho }}_m$. Now ${\delta }^{\prime }=a\dot{\delta }$ and ${\delta }^{\prime \prime }=a^2(\ddot{\delta }+H\dot{\delta })$, soThis is the specified approximation, not an exact finite-$w$ relativistic identity. The pressure-gradient term stabilizes short wavelengths, while gravity drives the long-wavelength Jeans instability.

Now apply the stipulated equation to $P=K{\rho }^{4/3}$ with $K>0$. In the pressure-negligible matter background,Thus $c_s^2{k}^2/a^2$ is proportional to $t^{-2}$, exactly like the gravitational term. DefineThe comoving Jeans wavenumber and $q$ are constant in this era. The equation becomes the Euler-Cauchy equationSetting $\delta \propto t^p$ gives $p^2+p/3+2(q-1)/3=0$. For distinct real roots the full pair of modes isThese are the Jeans modes of a four-thirds polytropic cosmological fluid. The physical Jeans length isHere ${\lambda }_J\propto a$, so its comoving value is constant. For $\lambda >{\lambda }_J$, $q<1$ and $p_{+}>0$, giving genuine growth; the other solution decays. In the long-wavelength pressure-free limit, $p_{+}=2/3$ and $p_{-}=-1$, the familiar matter-era modes. At $\lambda ={\lambda }_J$, the two powers are $0$ and $-1/3$, so the leading mode is constant rather than growing.

For $1<q<25/24$ both powers are negative: pressure prevents growth, although the modes still decay without oscillation. At $q=25/24$ they coincide and the complete solution isFor $q>25/24$ they become a complex-conjugate pair; a real basis isThese short-wavelength acoustic oscillations have a decaying envelope. The onset of instability is $q=1$, while the threshold for oscillatory time dependence is slightly larger, $q=25/24$; expansion damping explains why these two thresholds are not identical.

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.

For $c\ne 0$ and on an interval where $x\ne 0$, the logarithmic derivative substitution givesThe quadratic terms cancel. Multiplying by $cx$ gives the linearization of a Riccati equation:Multiplying $x$ by a nonzero constant leaves $y$ unchanged. If $c=0$, that substitution is undefined, but the original equation is already a first-order linear ordinary differential equation, solvable by an integrating factor.

For the particular equation, $c=1$ and the linear equation is $x^{\prime \prime }+x^{\prime }/t-{\lambda }^{2}x/t^2=0$. Use the PDF's logarithmic coordinate $\tau =\log t$ and write $X(\tau )=x(e^{\tau })$. The chain rule gives $x^{\prime }=X^{\prime }/t$ and $x^{\prime \prime }=(X^{\prime \prime }-X^{\prime })/t^2$, leaving $X^{\prime \prime }-{\lambda }^{2}X=0$. Thus $x=A{t}^{\lambda }+B{t}^{-\lambda }$, as also follows from the Euler-Cauchy equation. Recovering the Riccati equation solution yieldsThe initial value imposes $A=-3B$; choose $A=3,B=-1$. ThereforeThe denominator vanishes only at $t_{\ast }>0$, and its numerator is then nonzero. This is a simple pole: $y(t)\sim 1/(t-t_{\ast })$. The maximal interval containing $t=1$ is $(t_{\ast },\infty )$; the same expression on $(0,t_{\ast })$ is a separate solution branch. These are poles of a Riccati solution from zeros of its linearizing solution.

Away from zero the Dirac delta distribution is absent. On either half-line, use $u=1+k|z|$ and the power-law ansatz $\psi =u^{\alpha }$. The Euler-Cauchy equation becomesThe growing term is excluded by boundedness at either end. Thus the independently bounded half-line solutions are

The flat matter-dominated Friedmann equation gives $\rho =1/(6\pi G{t}^2)$ when $a=t^{2/3}$. With $P=C{\rho }^{4/3}$,For linear density modes with polytropic pressure, substitution gives the Euler-Cauchy equationA power law has exponent satisfying $n(n-1)+4n/3+{\bar{c}}_s^2{k}^2-2/3=0$. Thus, for distinct roots,For $k\ll k_J$, the two modes approach $t^{2/3}$ and $t^{-1}$: gravity produces a growing linear cosmological density perturbation and a decaying mode. For $k\gg k_J$, put $\nu =\sqrt{{\bar{c}}_s^2{k}^2-25/36}$; the real solution isThese are pressure-supported oscillations with a decaying envelope. For the repeated-root cosmological density mode $k=k_J$, the complete solution is $t^{-1/6}(A_k+B_k\log t)$. The specified $k_J=5/(6{\bar{c}}_s)$ separates real exponents from logarithmic oscillations; the growing exponent actually crosses zero at $k=\sqrt{2/3}/{\bar{c}}_s$, a slightly smaller value.

Each broad face supplies surface tension $\gamma$ pulling the rounded hole edge into the remaining sheet. Their resultant is $2\gamma$ per unit circumference. At the inner boundary the fluid's outward normal is $-{\mathbf{e}}_r$, so the boundary traction is outward in the radial direction when $h{\sigma }_{rr}=-2\gamma$. The right panel of the preceding diagram shows these two capillary pulls.

With uniform thickness and $p_{\mathrm{ext}}=0$, the axisymmetric viscous-sheet stretching equations reduce toThis Euler-Cauchy equation gives $u=Ar+B/r$. The fixed outer rim imposes $u(R_0)=0$, hence $B=-A{R}_0^2$. Because $(ru{)}_r/r=2A$, conservation of mass gives $\dot{h}=-2Ah$, independent of $r$. Thus a uniform sheet remains uniform.

The radial stress is ${\sigma }_{rr}=6\mu A-2\mu B/r^2$. Its value at the hole edge determinesThe edge is material, so $\dot{R}=u(R)$ andNeglecting the initially tiny hole's volume, conservation of mass gives $\pi h(R_0^2-R^2)=\pi h_0{R}_0^2$, or $h=h_0/(1-x^2)$ with $x=R/R_0$. Therefore the capillary growth of a hole in a viscous sheet obeysFor a finite initial hole $x_i$, replace $h_0$ in the denominator by $h_0(1-x_i^2)$. A nonzero seed is needed: the exact initial condition $x(0)=0$ gives the stationary solution of this differential equation. For a small positive seed, $x$ initially grows exponentially at rate $\gamma /(\mu h_0)$.

Here $q=\kappa /R$, so the equation from part (b) becomes the Euler-Cauchy equationThe power-law ansatz $B_z=R^s$ gives the indicial equationFor $0<\kappa <1/2$, the general field is thereforeThe polynomial discriminant changes sign atFor $\kappa >{\kappa }_c$, define $\mu =\sqrt{{\kappa }^2-1/4}$. A real form of the solution isThus the field has a log-periodic oscillation: its phase is periodic in $\ln R$, so it oscillates as the radius changes geometrically.

Substitution of $y=x^p$ into the homogeneous Euler-Cauchy equation gives $(p(p-1)+p-1)x^{p-2}=(p^2-1)x^{p-2}$, so $\boxed{p=1,-1}$. Use the solution $u(x)=x$ that vanishes at zero and $v(x)=1/x$ that decays at infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-2/x$. The half-line Green function for the first cylindrical harmonic is thereforeIt is continuous at $x=s$, its first derivative has jump $1/2-(-1/2)=1$, and it solves the homogeneous equation elsewhere. Thus $L_{x}G=\delta (x-s)$ for integration with ordinary measure $ds$. The boundary conditions hold for each fixed $s>0$.

Integrating against the forcing givesSince the two integrals are $2-(x^2+2x+2)e^{-x}$ and $e^{-x}$, respectively,A Taylor expansion gives $y=-x/2+O(x^2)$ near zero, while $y\sim -1/x$ at infinity. This verifies the two endpoint limits; direct differentiation verifies $Ly=e^{-x}$ on $x>0$. The differential expression itself is singular at zero, so the endpoint condition is imposed by continuity. No nonzero combination of $x$ and $1/x$ satisfies both homogeneous endpoint conditions, proving uniqueness.

For separation of variables, write $B=F(r)\Theta (\theta )$. The exterior Laplace equation separates intoWith $\mu =\cos \theta$, angular regularity at the poles selects $\lambda =\ell (\ell +1)$ and the Legendre polynomial $P_{\ell }(\mu )$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }-\ell (\ell +1)F=0$, an Euler-Cauchy equation with solutions $r^{\ell },r^{-\ell -1}$. Decay at infinity excludes the first branch for every $\ell$, including its constant $\ell =0$ case. ThereforeThe expansion uses axisymmetric spherical harmonics. The written scalar boundary-value problem permits the $\ell =0$ term; any additional physical exclusion of that sector must be stated separately.

Use separation of variables, writing $\Psi =F(r)\Theta (\theta )$. After multiplication by $r^2/(F\Theta )$, the radial and angular terms are functions of independent variables, so setWith $x=\cos \theta$, the angular equation isRegularity at both poles selects $\lambda =\ell (\ell +1)$, $\ell =0,1,\dots$, and the Legendre polynomials $P_{\ell }(x)$; the other angular branch is singular at a pole. The radial Euler-Cauchy equation is $r^2{F}^{\prime \prime }+2r{F}^{\prime }-\ell (\ell +1)F=0$. A power $r^s$ gives $(s-\ell )(s+\ell +1)=0$, hence $F=a_{\ell }{r}^{\ell }+b_{\ell }{r}^{-\ell -1}$.

Superposing these axisymmetric spherical harmonics givesThe coefficients must give convergence appropriate to the domain. In a region containing the origin, regularity removes the negative powers; in an exterior region with zero potential at infinity, it removes the growing powers and the constant. No such radial restriction is needed for a general annular region.

Put $\Phi (r,\theta )=R(r)\Theta (\theta )$ for separation of variables. Multiplication by $r^2/(R\Theta )$ givesSingle-valuedness requires a $2\pi$-periodic angular function. The nonzero possibilities are $\lambda =n^2$ for integers $n\ge 0$. For $n\ge 1$, $\Theta$ is a combination of $\cos n\theta$ and $\sin n\theta$, while the radial Euler-Cauchy equation has solutions $r^n,r^{-n}$. At $n=0$, periodicity excludes an angular term proportional to $\theta$, and $(r{R}^{\prime }{)}^{\prime }=0$ gives $R=A_0+B_0\ln r$.

Superposing the separated harmonic functions yieldsTo obtain the general solution, expand any sufficiently regular harmonic function in a Fourier series on circles; each Fourier coefficient satisfies the corresponding radial equation. The coefficients must give the required convergence on compact annuli. Because no condition at zero or infinity is imposed in this part, neither radial branch is discarded.

For the Euler-Cauchy equation, substitute $y=x^n$ on an interval not containing zero. The coefficient of $x^{n-2}$ isThe distinct roots giveThese solutions are linearly independent since their Wronskian is $2x^3\ne 0$ away from zero. Both nevertheless approach zero as $x\to 0$, which will make the endpoint conditions in the next part insufficient for uniqueness.