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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.

Combine the logarithms first and factor out the limiting constant:Apply the Taylor expansion $\ln (1+u)=u-u^2/2+O(u^3)$ with $u=1/(2x)+1/(2x^2)$. The first three nonzero terms, including the constant, areThis is an asymptotic expansion in reciprocal powers of $x$.

Apply the Poincaré-Lindstedt method to the truncated system from part (b). Since $y=\dot{x}$, its scalar equation isIntroduce the stretched time $\tau =\omega t$, with asymptotic expansionsPrimes in this part mean derivatives with respect to $\tau$. At first order,The trigonometric identitiesreduce this toA periodic $X_1$ cannot have a secular term, so both fundamental-harmonic coefficients must vanish. Taking $A>0$ givesIn particular $\mu >0$ is necessary. Once these conditions hold, a bounded particular correction isadding fundamental sine and cosine terms adjusts the phase and the first correction to the amplitude. Consequently the required leading periodic orbit isHere $A$ is the leading amplitude, not an assertion that the exact finite-$ϵ$ amplitude is independent of $ϵ$.

To justify existence beyond formal cancellation, the leading radial displacement in a Poincaré return map is $2\pi ϵB(r)$, whereAt the positive root $A$, $B^{\prime }(A)=-5\mu \ne 0$. The implicit function theorem, applied after dividing the displacement by $ϵ$, continues this root to a periodic orbit for sufficiently small positive $ϵ$. Its radial Floquet multiplier is $1-10\pi \mu ϵ+O({ϵ}^2)$, so it is attracting. These statements apply to the truncated system with fixed $\mu >0$; its interpretation as a local approximation to the original three-dimensional system also requires $\mu$ small, so that $A=O(\sqrt{\mu })$ stays near the Hopf bifurcation.

Write $E(h)$ for the supplied approximation. The asymptotic expansion iswhere $O$ is Big O notation. At the refined parameter $h/3$,Multiply this identity by nine and subtract the first one. The unknown leading truncation error coefficient cancels:Thus the Richardson extrapolation estimate isThe two supplied values giveThe correction to the finer approximation is $(3-2)/8=1/8$. Neglecting the higher-order remainder would also give $c=9/8$.

The conclusion is asymptotic: the leading error order improves from two to at least three when both parameter values are in the range of the stated asymptotic expansion. With no bound on the remainder constant, the two finite observations alone do not guarantee that $25/8$ is closer to $Y$. Indeed, for any proposed value $L$, the polynomial modelsatisfies $E(1)=2$, $E(1/3)=3$ and the required asymptotic expansion with true limit $L$. This is why finite observations do not certify Richardson extrapolation accuracy: asymptotic cancellation provides an improved-order estimator, while a numerical error certificate needs quantitative remainder information.

For every fixed truncation index $N$, the remainder must be smaller than the last retained term:This is an asymptotic expansion relative to the sequence in part (i). The number of retained terms is fixed before taking the limit. The associated series need not converge for any nonzero parameter; its role is to give successively smaller truncation errors. If $a_n$ is an asymptotic sequence, the same condition can be stated as a remainder $O(a_{N+1})$ when the next expansion condition is available, but the displayed little o notation formulation is the defining one.

Let $b=\pi /4$ and introduceDeform the contour upward to the steepest descent contour $t=e^{i\pi /4}r$ from zero and the endpoint path $t=(b^2+is{)}^{1/2}$ from $b$. More precisely the original integral is the first path traversed forward minus the endpoint path traversed forward. Connect the truncated paths where $\mathrm{Im}(t^2)=R$ and let $R\to \infty$. That connector has exponentially small factor $e^{-xR}$, and no pole of $\tan t$ is crossed: all its poles are real and outside the original segment.

On the first path ${\tan }^{2}t=t^2+O(t^4)$, so its leading contribution isThe last equality follows by $u=x{r}^2$, $\Gamma (3/2)=\sqrt{\pi }/2$.

On the endpoint path $dt=ids/(2\sqrt{b^2+is})$, and ${\tan }^{2}b=1$. Its negative orientation gives the leading endpoint termTaking real parts and inserting $b=\pi /4$ givesThe oscillatory first coefficient can vanish at particular values of $x$; the stated remainder still holds as an additive asymptotic expansion.

Put $\delta =\sqrt{ϵ}$ and write the outer asymptotic expansion as $y_0+\delta y_{1/2}+{\delta }^2{y}_1+\cdots$. Equating powers in the differential equation givesThe first two equations give $y_0=A{e}^{x^2/2}$ and $y_{1/2}=B{e}^{x^2/2}$. Since $y_0^{(3)}=A(x^3+3x)e^{x^2/2}$, an integrating factor yieldsHere $A,B,C$ are matching constants; boundary conditions have not yet been imposed.

Let $g_x(t)=(1+xt{)}^{-1}$. Repeated integration by parts using the factor $e^{-t}$ givesBoundary terms at infinity vanish, and $g_x^{(n)}(t)=(-1{)}^{n}n!x^n(1+xt{)}^{-n-1}$. ThereforeFor every fixed $N$, this is $o(x^N)$ as $x\downarrow 0$, provingFor each fixed $x>0$ the displayed infinite series diverges, since successive term magnitudes have ratio $(n+1)x\to \infty$. Its usefulness is precisely that of an asymptotic expansion, not a convergent power series.

For an asymptotic sequence $a_{n+1}(ϵ)=o(a_n(ϵ))$, the asymptotic expansion condition is that every fixed truncation has remainder smaller than its last retained term:This limit is taken as $ϵ\to 0$ in the specified direction or sector. The number $N$ stays fixed while taking that limit. The infinite series need not converge for any nonzero parameter.

For the forward ray of i, the ultrarelativistic longitudinal Doppler factor follows from the exact expression in terms of the Lorentz factor:Indeed, $\gamma \beta =\sqrt{{\gamma }^2-1}$ for a positive boost. If $\beta =1-ϵ$, the exact formulas and their consistent asymptotic expansions areEquivalently,The printed equality is not exact. For a concrete ultrarelativistic example, $\gamma =10$ gives the correct $D=10+\sqrt{99}\simeq 19.949874$, whereas $\sqrt{4{\gamma }^2-1}=\sqrt{399}\simeq 19.974984$. The latter expands as $2\gamma -1/(4\gamma )+O({\gamma }^{-3})$, with the wrong first correction.

One can obtain the printed expression by using $D^2=2/ϵ-1$ but replacing ${\gamma }^2$ by only its leading approximation $1/(2ϵ)$. This retains a constant term in the first equation while discarding a same-order term in the second. Thus $D\approx \sqrt{4{\gamma }^2-1}$ is admissible only as a leading relative-accuracy estimate, of the same order as $D\sim 2\gamma$; it is not the stated identity or a correct next-order approximation.

A spherical three-momentum ultraviolet cutoff givesFor $m>0$ and $\Lambda \gg m$, the asymptotic expansion isIn particular, the leading ultraviolet divergence is quartic. For $m=0$, $\mathcal{E}={\Lambda }^4/(16{\pi }^2)$ exactly.

Write $\phi =⟨\psi ⟩$. Under the proposed identification, the scalar potential from a field-dependent momentum cutoff is $V_{\mathrm{ind}}(\phi )=\mathcal{E}(|\phi |)$. It is even and nonnegative, and its derivative with respect to $|\phi |$ is $|\phi {|}^2\sqrt{{\phi }^2+m^2}/(4{\pi }^2)>0$ away from zero. Consequently its only global minimum is at $\phi =0$: this contribution alone does not cause spontaneous symmetry breaking. At large $|\phi |$ it grows quartically. At small $|\phi |$ with fixed positive $m$, however,Thus the quartic sketch is an asymptotic shape, not the exact small-field behavior for a massive original field. Identifying a regulator with a new field is an additional modeling prescription; an actual effective potential also depends on the new field's dynamics and any other allowed terms.

Define the aspect ratio and the thickness-based Reynolds number byUse the nondimensionalizationThe small transverse velocity scale follows from incompressibility. Dropping stars, the base fields become $\bar{u}=2z-z^2$ and $\bar{p}=(1-z)\cot \alpha$. Before subtracting the base state, the dimensionless Navier-Stokes equation readsFor example, division of the downslope momentum equation by $\mu U_0/h_0^2$ makes the inertial coefficient $\rho U_0{h}_0^2/(\mu L)=\mathrm{Re}\epsilon$ and the pressure gradient coefficient $2\epsilon$. The transverse equation has been divided instead by $\mu U_0/h_0^2$ after using its smaller velocity scale; this gives the displayed $\mathrm{Re}{\epsilon }^2$ coefficient.

Now substitute $u=\bar{u}+\hat{u}$, $w=\hat{w}$, $p=\bar{p}+\hat{p}$. Since ${\bar{u}}_{zz}+2=0$, ${\bar{p}}_z=-\cot \alpha$, and the base fields have no $x$ or $t$ dependence, subtraction givesThese are nonlinear equations for the deviations; small amplitude has not been assumed. The later asymptotic expansion takes $\mathrm{Re}=O(1)$ as $\epsilon \to 0$, or more generally requires the retained inertial correction to remain small.

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.

Integrating the total velocity across the film givesIndeed the three inertial polynomial terms integrate to $\mathrm{Re}{h}^6{h}_x(1/30-1/6+2/3)=8\mathrm{Re}{h}^6{h}_x/15$. Integrating the continuity equation and using the kinematic boundary condition gives $h_t+q_x=0$. Hence the Benney equation isNormal gravity smooths height variations on the upper side of the slope, while the inertial correction can amplify them. Surface tension suppresses sufficiently short wavelengths.

Perturbation theory develops approximate solutions in a small parameter around a simpler reference problem. A power-series asymptotic expansion gives equations for successive correction coefficients; a matched asymptotic expansion handles cases where one expansion is not uniformly valid.

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.

A regular perturbation admits an asymptotic expansion about the limiting problem without changing its relevant scale or losing solutions or conditions. A regular perturbation of a simple algebraic root follows from the implicit function theorem. A full problem can still be a singular perturbation even when some individual solution branches are regular.

Richardson extrapolation improves an approximation by combining two resolutions so that their leading truncation errors cancel. The useful surprise is that the size of that leading error need not be known: its scaling under refinement is enough. This lets an existing finite difference method or quadrature rule produce a more accurate answer from two runs, without first deriving a new method from scratch.

Suppose $A(h)$ approximates a fixed limit $L$ as the positive resolution parameter $h$ tends to zero, with an asymptotic expansionHere $p$ is known, $c$ is independent of $h$, and $O$ denotes Big O notation. The same asymptotic expansion, with the same $c$, must describe both $h$ and $h/2$. Nested grids make it natural to reuse samples, although nesting itself does not establish this error model. Define the linear combinationThe weights sum to one, preserving $L$, while $2^{p}c(h/2{)}^p-c{h}^p=0$. Thus Richardson extrapolation gives $R(h)=L+O(h^{p+q})$: the next available term in the asymptotic expansion determines the improved order. Justin Solomon's numerical analysis text develops this cancellation for finite differences and applies it to numerical integration.^[ref]

Geometrically, regard the approximation as a function of $t=h^p$. Its first two terms are a straight line $L+ct$. The two computed points lie at positive $t$; extending the line through them to $t=0$ gives $R(h)$. This is why the operation is extrapolation: it estimates a value outside the sampled parameter interval. The remaining curvature corresponds to the uncancelled terms of the asymptotic expansion.

Worked example: a centered derivative of sine. For the smooth function $f(x)=\sin x$, the exact derivative at $x=0$ is $L=\cos 0=1$. The centered difference quotient, a central finite difference, isThe Taylor series of sine gives the signed truncation errorThus $p=2$, $q=2$ and $c=-1/6$. Halving $h$ divides the leading truncation error by four. Combining the two centered difference quotients with Richardson extrapolation givesIndeed, the $h^4$ coefficient becomes $(4/16-1)/(3\cdot 120)=-1/480$; the $h^2$ coefficient vanishes. This is the even-power case of Richardson extrapolation discussed in Persson's authored lecture notes.^[ref]

For $h=0.4$ and $h/2=0.2$, the actual computations areThe absolute error drops from about $2.6454\times 10^{-2}$ for $A(0.4)$ to $5.3080\times 10^{-5}$ for $R(0.4)$, an improvement by roughly a factor of 498. Refining again gives $R(0.2)\approx 0.999996670632607$, whose absolute error is about $3.3294\times 10^{-6}$. The refined truncation errors decrease by nearly four for $A$ and nearly sixteen for $R$, reflecting their second and fourth orders.

The logarithmic scale on each axis turns power laws into straight lines. The dashed reference curves are $h^2/6$ and $h^4/480$, the magnitudes of the leading truncation errors. The plotted errors are computed with 70-digit decimal arithmetic, rather than clipped to a display floor; this isolates the asymptotic expansion from roundoff error.

Worked example: integrating an exponential. Approximate the definite integral of the exponential function on $[0,1]$:For numerical integration, the composite trapezoidal rule adds the areas of trapezoids between successive samples; this is the quadrature use of the endpoint average that also occurs in the ODE trapezoidal rule. On $n$ equal intervals, with $h=1/n$, defineSumming the geometric series gives $T_n=(e-1)\frac{h}{2}\frac{e^h+1}{e^h-1}$. Expanding the exponential function near zero yieldsThe leading truncation error again has $p=2$, $q=2$. On the uniform nested grids with $n$ and $2n$ intervals, Richardson extrapolation therefore givesThis linear combination is exactly composite Simpson's rule on the finer grid, whose spacing is $h/2$. On each coarse panel, an old endpoint receives weight $h/6$ and the new midpoint weight $2h/3$; shared old endpoints receive weight $h/3$. These are the usual Simpson's rule weights $1,4,2,\dots ,4,1$ multiplied by $h/6$. The equivalence is explained directly in LeMesurier's university notes.^[ref]

For two coarse panels and four fine panels,The absolute errors are about $0.0356493$, $0.00894008$ and $0.0000370135$, respectively. Reusing the fine-grid samples makes Richardson extrapolation especially attractive here: the resulting Simpson's rule needs no samples beyond those already used by $T_{2n}$.

When the improvement is justified. Richardson extrapolation requires the stated asymptotic expansion and the correct positive $p$. If the actual leading exponent is $r\ne p$, its coefficient after combination is $c(2^{p-r}-1)/(2^p-1)$, so the leading truncation error generally remains. Merely knowing $A(h)-L=O(h^p)$ does not supply a common coefficient $c$ or a higher-order remainder. A discontinuity of a function, a singular endpoint, or unrelated changes between grids can invalidate the assumed asymptotic expansion; refinement alone does not guarantee improvement for every numerical problem.

Roundoff error and errors in the sampled values survive the cancellation of truncation error. If the coarse and fine values have absolute perturbations bounded by $\epsilon$, the perturbation in $R(h)$ is bounded by $(2^p+1)\epsilon /(2^p-1)$; for $p=2$ this is $5\epsilon /3$. In floating-point arithmetic, a centered difference quotient can also amplify sample errors through division by $h$, so an indefinitely smaller $h$ need not be more accurate. Persson's notes describe this balance between truncation error and roundoff error.^[ref]

The same cancellation idea appears in step-doubling error estimation. For an order-$p$ one-step method for an ordinary differential equation, two half steps each contribute local truncation error of order $h^{p+1}$, so their accumulated leading local error has ratio $2^{-p}$ to one full step. This explains the denominator $2^p-1$ there; the appropriate exponent follows from that local error model, rather than automatically from a generic use of Richardson extrapolation.

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.