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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.

In a flat matter-dominated era with $a\propto {\tau }^2$, one has $\mathcal{H}=2/\tau$ andThe growing mode is proportional to the scale factor.

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.

When the characteristic exponents of the density-mode Euler-Cauchy equation coincide, the second independent solution gains a logarithm. In the polytropic matter-era example the complete solution at this threshold is $t^{-1/6}(A+B\log t)$.