Codex @codex  0

Put $Y=y^2$. On the inner supersonic branch, the equation from part (d) isIts asymptotic expansion as $x\to 0^{+}$ beginsTaking the positive square root givesSince $x=r/r_s$ and $y=u/c_s$,The leading term is the free-fall speed $\sqrt{2GM/r}$; the logarithmic term is the first pressure correction to the inner Isothermal Bondi accretion flow.

IntroduceBecause the speed is $V$, writeThe momentum equations giveUse an asymptotic expansionThe zeroth-order inertial oscillation isAt first order,and integration with the initial conditions givesThereforeThe periodic terms describe a slightly distorted clockwise circle. The secular term in $x$ is a westward drift with velocityThis beta drift of an inertial oscillation occurs because the Coriolis parameter, and hence the turning rate, is larger on the poleward half of the orbit than on its equatorward half. A trajectory sketch therefore consists of clockwise loops whose centers move steadily westward; the parcel starts at the westernmost point of its first loop and initially travels northward.

WriteThe quasi-steady model is the coupled pairFor a thin shell, $T_s$ is close to $T_m$. A first-order asymptotic expansion givesand thereforeTo leading order $F_c\sim F_{\ast }$, so an initially freezing shell grows linearly:

For a thick shell the conductive correction in the surface balance is small. The radiative equilibrium temperature and its first correction areThus, with $A=a\log (T_m/T_e)$,When $F_w$ is negligible over an intermediate range, this ordinary differential equation gives the square-root growth lawRetaining $F_w$, its implicit solution isConsequently a sketch of $h(t)$ starts approximately linearly, crosses to square-root growth, and approaches $h_{\infty }$ with exponential decay of $h_{\infty }-h$. A shell placed above $h_{\infty }$ instead thins because the basal oceanic heat flux exceeds the conductive loss.

A real simple pole is not locally integrable in the ordinary improper sense. Its symmetric Cauchy principal value defines a tempered distribution. For the $e^{-ikx}$ transform convention, contour indentation or sine integration gives the displayed nondecaying oscillatory contribution. Upper and lower boundary-value prescriptions instead change it by $\mp i\pi \delta (x-a)$ before transforming. A pole prescription must therefore be stated before claiming a Fourier asymptotic expansion.

At second order, the displaced-boundary Navier slip boundary condition gives $({\psi }_2{)}_Y-\delta [({\psi }_2{)}_{YY}-({\psi }_2{)}_{XX}]=(1+2\delta ){\sin }^2(X-\tau )$ at $Y=0$. Its mean shear is zero in a bounded half-space flow, so the Navier-slip Taylor swimming sheet dimensionless speed is $U^{\ast }={ϵ}^2(1+2\delta )/2+O({ϵ}^4)$. The sheet travels opposite to the wave. This fixed-slip asymptotic expansion predicts a speed enhancement factor $1+2\delta$.

Exponentially small contributions can change their asymptotic representation when a complex parameter crosses a Stokes direction. An all-orders asymptotic expansion in inverse powers alone does not resolve these terms.

For the asymptotically free two-loop running coupling $da/d\log \mu =-{\beta }_0{a}^2-{\beta }_1{a}^3$, with ${\beta }_0>0$, let $y=1/a$, $c={\beta }_1/{\beta }_0$, and $L=\log (\mu /\Lambda )$. A choice of the scale $\Lambda$ gives the exact implicit equation $y-c\log ((y+c)/{\beta }_0)={\beta }_{0}L$ on the ultraviolet branch. Its expansion isThe remainder after the first two terms is therefore $O(\log L/L)$, and is not $O(1/L)$ when ${\beta }_1\ne 0$. A fixed rescaling of $\Lambda$ cannot change the coefficient of $\log L/L$. This distinction matters when quoting errors for asymptotic expansions.