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For fixed $\mu \ne 0$, a large positive root must lie close to a pole of the tangent function: away from its poles, $\tan x$ cannot balance the unbounded right side. Label the adjacent poles by $a_n=(n+1/2)\pi$ and write $x_n=a_n+{\delta }_n$. The Taylor series of $-\cot \delta$ givesSeek the asymptotic expansion $\delta =A/a_n+B/a_n^3+\cdots$. The coefficients of $a_n$ and $a_n^{-1}$ giveThus the large roots near tangent poles satisfyThe first correction already supplies the requested dependence on $\mu$. For $\mu >0$ the root approaches the pole from below; for $\mu <0$ it approaches from above. This labels roots by their nearby poles, avoiding an irrelevant finite shift in the enumeration of positive roots.

No positivity restriction on $\mu$ is required. The expansion requires $\mu \ne 0$ fixed and $|\mu |a_n\gg 1$, so the displacement is small. It is not uniform as $\mu \to 0$. At $\mu =0$ the roots are exactly $n\pi$, a different leading sequence; these cannot be recovered by setting $\mu =0$ in the pole expansion. If $\mu$ varies with $n$, its size must be checked against the small-displacement and successive-term conditions rather than using the fixed-parameter remainder blindly.

The method of multiple scales introduces independent variables for the fast oscillation and its slow modulation, for example $t_0=t$, $T=ϵt$, and, when needed, $T_2={ϵ}^{2}t$. The derivative becomes $d/dt={\partial }_{t_0}+ϵ{\partial }_T+{ϵ}^2{\partial }_{T_2}+\cdots$. One expands the solution in $ϵ$ while allowing its leading amplitudes and phases to depend on the slow times.

A solvability condition in the method of multiple scales removes resonant forcing of each fast homogeneous mode. Otherwise a correction contains a secular term, such as $ϵt\sin t$, and an ordinary asymptotic expansion fails on long times. The resulting slow amplitude equations incorporate that accumulated effect in the leading solution. One must still check that the amplitudes remain within the assumed weak-coupling regime and that omitted slower effects have not accumulated.

Assume $ϵ\to 0^{+}$. Since the drift $(1+x{)}^2$ is positive, the fast mode decays to the right; the matched asymptotic expansion has a left boundary layer at $x=0$ of width $O(ϵ)$, and the outer expansion uses the condition at $x=1$.

Write $y\sim y_0+ϵy_1$. The first two equations for the outer expansion areEnforcing $y_0(1)=1$ and $y_1(1)=0$, defineThen the outer asymptotic expansion isTo derive the correction, set $y_1=Eq$. Cancellation of the homogeneous terms leaves $q^{\prime }=-2/(1+x{)}^5-1/(1+x{)}^6$; integration and $q(1)=0$ give the expression above.

In the left boundary layer, set $\xi =x/ϵ$ and $y=Y_0(\xi )+ϵY_1(\xi )+\cdots$. The equation becomes $Y^{\prime \prime }+(1+ϵ\xi {)}^2{Y}^{\prime }+ϵY=0$. Put $A=e^{1/2}$ and $B=1-A$. The leading inner equation and matching give $Y_0^{\prime \prime }+Y_0^{\prime }=0$, $Y_0(0)=1$, $Y_0(\infty )=A$, hence $Y_0=A+B{e}^{-\xi }$. At the next order,Since $q(0)=53/80$, put $C=53A/80$. The matched inner correction satisfying $Y_1(0)=0$ isFor verification, the differential operator $d^2/d{\xi }^2+d/d\xi$ maps $e^{-\xi }P(\xi )$ to $e^{-\xi }(P^{\prime \prime }-P^{\prime })$. The large-$\xi$ common part is $A+ϵ(C-A\xi )$, precisely the small-$x$ outer expansion in the overlap $ϵ\ll x\ll 1$.

Adding the inner expansions and outer expansions and subtracting their common part gives the first-order composite expansion for a positive driftThis is uniformly correct through $O(ϵ)$ on $0\le x\le 1$. It satisfies the left boundary condition exactly; at the right boundary the remaining error is exponentially small. Polynomial growth in the inner correction is harmless because it is multiplied by the decaying exponential. For the smooth nonvanishing drift, the next matched terms are uniformly $O({ϵ}^2)$.

Set $z=x^2/\epsilon$. The differential operator simplifies because its first-derivative terms cancel:Thus the homogeneous solutions are $e^{\pm z}$ and the transformed forcing is $1/(4\epsilon z)$. For the decaying-sector variation of parameters solution choose the contour directionsThe first integrand decays at positive-real infinity and the second at positive-imaginary infinity. Join each endpoint to $x$ within the first quadrant, avoiding zero. This also fixes the exponentially small homogeneous contributions, which cannot be determined from an algebraic asymptotic expansion alone.

After $s=t^2/\epsilon$, write $A(z)={\int }_{+\infty }^z{e}^{-s}ds/s$ and $B(z)={\int }_{-\infty +i0}^z{e}^{s}ds/s$. Their derivatives are $A^{\prime }=e^{-z}/z$ and $B^{\prime }=e^z/z$. Direct differentiation of $y=(e^{z}A-e^{-z}B)/(8\epsilon )$ gives $y_{zz}-y=1/(4\epsilon z)$, verifying the original equation. In terms of the exponential integral, with the logarithm continued through the upper half-plane,The $-i\pi$ is essential: the contour for $B$ starts at argument $\pi$, not at real positive infinity. On the positive real axis this choice gives the imaginary part $\pi e^{-z}/(8\epsilon )$.

Repeated endpoint integration by parts gives $e^z{E}_1(z)\sim z^{-1}(1-z^{-1}+2!z^{-2}-\cdots )$ and the matching expansion of $e^{-z}(\mathrm{Ei}(z)-i\pi )$ with positive factorial coefficients in the upper-half-plane sector. Their odd inverse-power corrections cancel. Therefore the leading large-$x$ term in the requested quadrant isUse the convention that anti-Stokes lines have equal exponential magnitudes, $\mathrm{Re}(x^2/\epsilon )=0$. They are $\arg x=\pi /4+m\pi /2$. The Stokes lines are $\arg x=m\pi /2$. Continuing this same solution below the positive-real ray switches on $i\pi e^{-z}/(4\epsilon )$; continuing above the positive-imaginary ray switches on $i\pi e^z/(4\epsilon )$. These exponentials remain decaying up to the neighbouring anti-Stokes lines. Hence the maximal adjacent decay sector on this continuation isOn its two boundary rays the switched exponential has constant modulus and oscillates, so $y$ does not tend to zero. Within the sector the leading algebraic term above persists. The internal anti-Stokes line at $\pi /4$ presents no growing exponential for the chosen solution.

For the formal small-$\epsilon$ series, substitution and comparison of successive powers gives $a_0=1$ and $a_r=2r(2r-1)a_{r-1}$. Thus all coefficients areThis is a factorially divergent asymptotic expansion. The ratio of consecutive term magnitudes is $(2r+2)(2r+1)/|z{|}^2$. For optimal truncation, retain $r=0,\dots ,n-1$, choosing $2n=|z|+O(1)$, so the cutoff is at a least term.

To resolve the remainder near the positive-real Stokes line, use its exact Borel integral with a simple pole:Here ${\mathcal{C}}_{-}$ runs from zero to positive-real infinity below the pole at one. Partial fractions reproduce the exponential-integral solution, including its positive imaginary part. The finite geometric identity for $(1-t^2{)}^{-1}$ gives the displayed exact remainder after integrating the first $n$ monomials.

Write $z=2n+i\mu \sqrt{n}+\nu$, with real bounded $\mu$ and bounded $\nu$. For a Gaussian saddle coalescing with a simple pole, put $t=1+u/\sqrt{n}$. The exponent becomes $-u^2-i\mu u+O(n^{-1/2})$ and $dt/(1-t^2)=-du/(2u)+O(n^{-1/2})du$. ThereforeThe integral is its Cauchy principal value plus $i\pi$. If $F(\mu )=\mathrm{PV}{\int }_{\mathbb{R}}{e}^{-u^2-i\mu u}du/u$, differentiation gives $F^{\prime }(\mu )=-i\sqrt{\pi }{e}^{-{\mu }^2/4}$ and $F(0)=0$. Consequently $F(\mu )=-i\pi \mathrm{erf}(\mu /2)$, and $J_n=-i\pi \mathrm{erfc}(\mu /2)/2+O(n^{-1/2})$.

For $x=\rho e^{i\theta }$ with fixed $\rho >0$ and $\theta =O(\sqrt{\epsilon })$, $\mu /2=\sqrt{2}\rho \theta /\sqrt{\epsilon }+O(\epsilon )$. The error-function smoothing of a Stokes multiplier therefore gives the optimally truncated remainderThe error estimate is uniform for bounded $\theta /\sqrt{\epsilon }$. On the Stokes line the multiplier is half its switched-on value; above it the exponentially small term switches off, and below it switches on. This term is invisible at every algebraic order. Near the line the remainder can exceed the least retained term by a square-root factor because a pole meets the saddle point.

The printed auxiliary Gaussian hint has inconsistent constants. Direct saddle point expansion of its nonsingular integral gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$ under its printed scaling, rather than $\frac{1}{2}\sqrt{\pi /(2n)}{e}^{-{\mu }^2/8}$. The derivation above uses the correct saddle point normalization and independently fixes both the smoothing width and the contour sign.

Take the logarithmic derivative of the proposed asymptotic expansion. Through order $z^{-1}$, $u^{\prime }/u=\lambda +\mu /z+O(z^{-2})$ and $u^{\prime \prime }/u={\lambda }^2+2\lambda \mu /z+O(z^{-2})$. Matching the constant and inverse-power terms givesThe second formula assumes a simple root, $2\lambda +f_0\ne 0$. At a repeated root the ansatz can fail: the next equation first requires $f_1\lambda +g_1=0$, and if that holds higher orders determine the remaining behaviour.

For the Bessel differential equation, $\lambda =\pm i$ and $\mu =-1/2$. Set $u=e^{\lambda z}{z}^{-1/2}v(z)$. Direct substitution leaves $v^{\prime \prime }+2\lambda v^{\prime }+(1/4-{\nu }^2)z^{-2}v=0$. With $v=1+a/z+O(z^{-2})$, matching $z^{-2}$ gives $a=(1/4-{\nu }^2)/(2\lambda )$. ThusTwo constant multiples of the Hankel functions normalize their leading coefficients to one and realize these solutions in their appropriate sectors.

For generic $\nu$, these are sectorial expansions, not unchanged expansions for a single branch over arbitrarily continued $\arg z$. The factors $z^{-1/2}$ require a branch, and crossing sector boundaries can introduce the other exponential by Stokes phenomenon and analytic continuation. More precisely, the standard normalized first Hankel solution has its expansion for $-\pi +\delta \le \arg z\le 2\pi -\delta$, and the second for $-2\pi +\delta \le \arg z\le \pi -\delta$, with any fixed positive $\delta$ inside these sectors. In particular both are valid in the common slit sector $-\pi <\arg z<\pi$, away from its boundaries. These sector statements are recorded in NIST's Hankel asymptotics. Exceptional half-integer orders have terminating elementary expressions, so a blanket assertion of nonzero Stokes mixing for every order would be too strong.

Let $b=e^{i\theta }$ and use the method of steepest descents with phase $\Phi (z)=-z^2$. For $z=x+iy$, $\mathrm{Re}\Phi =y^2-x^2$ and $\mathrm{Im}\Phi =-2xy$. The simple saddle point is $z=0$: the real axis is its descending path and the imaginary axis its ascending path. The quadratic complex endpoint descent path from $b$ satisfies $\mathrm{Im}(z^2)=\sin 2\theta$ and can be parametrized bywith the square-root branch chosen continuously from $b$. Increasing $s$ reduces the real part of the exponent by $\lambda s$.

For $0\le \theta <\pi /2$, that endpoint path goes to positive real infinity. Deform the original contour into the real path from $-d$ through the saddle point towards $+\infty$, followed backwards along the endpoint descent path to $b$. The integrand is an entire function, so finite contour deformations are legitimate; connecting arcs in the far decaying sector vanish. If $T_b$ denotes the oriented integral from $b$ to that infinity, this givesThe last expression follows by setting $z^2=b^2+s$: $dz/ds=1/(2z)$, so the endpoint integral is $e^{-\lambda b^2}{\int }_0^{\infty }{e}^{-\lambda s}/[2z(s)]ds$. This derives its phase and orientation directly.

For $\pi /2<\theta \le \pi$, the descent path from $b$ goes to negative real infinity. Deform from $-d$ towards that infinity and then backwards along this endpoint path to $b$. There is no saddle point segment, anduses the same formula with the negative square-root branch. At $\theta =\pi /2$ the contour can run from $-d$ to zero and then up the imaginary axis. It gives a half-Gaussian contribution and the dominant imaginary endpoint term $i{e}^{\lambda }/(2\lambda )$. The change in the subdominant saddle point contribution is a Stokes phenomenon; it does not change the leading endpoint term on either side of that angle.

The Gaussian saddle and endpoint dominance by angle follows by comparing exponential sizes and gives all fixed-angle leading behaviours:At $\theta =\pi /4$, the endpoint term has magnitude $O({\lambda }^{-1})$, so the saddle point term of order ${\lambda }^{-1/2}$ still dominates. In the middle interval the endpoint term grows exponentially for $\pi /4<\theta <3\pi /4$, is oscillatory of order ${\lambda }^{-1}$ at $3\pi /4$, and decays exponentially for $3\pi /4<\theta <5\pi /6$. On the latter interval it nevertheless decays more slowly than the fixed-endpoint term. Equality of their exponential rates means $\cos 2\theta =1/2$ on the left half-plane, hence $\theta =5\pi /6$. At that angle both terms are required; their coefficient magnitudes are unequal, so their sum has no complete destructive cancellation.

These fixed-angle asymptotic expansions are not uniform if $\theta$ is allowed to vary with $\lambda$ near a dominance transition. The endpoint/saddle balance near $\pi /4$ involves $\lambda (\theta -\pi /4)=O(\log \lambda )$; the two-endpoint competition near $5\pi /6$ has an angular scale $O({\lambda }^{-1})$.

The following sketches show the contour orientations, the saddle point and both endpoints. The hyperbolic endpoint paths are followed backwards when returning from infinity to $b$; the dotted far-end connectors indicate the vanishing connecting arc.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

Use the characteristic polynomial in the convention $p(\lambda )=\det (\lambda I-A)$:For fixed $\alpha >0$ with $\alpha \ne 1$, all three limiting eigenvalues are simple. Expanding this polynomial at each root, or applying first-order perturbation of a simple eigenvalue, givesThese are fixed-$\alpha$ asymptotic expansions, not uniform through $\alpha =1$. At $\alpha =1/3$ the displayed correction to ${\lambda }_{\alpha }$ vanishes because $\lambda =1/3$ is an exact root for every $ϵ$; there is no second nonzero term for that branch.

The two roots near $1$ interact when the correction $ϵ/(\alpha -1)$ is comparable to their unperturbed gap $\alpha -1$. Thus the distinguished limit is $\alpha =1+d\sqrt{ϵ}$, with bounded real $d$. Set $h=\sqrt{ϵ}$ and $\lambda =1+hL+h^{2}N+\cdots$. The polynomial becomesAt orders $h^2$ and $h^3$,With $D=\sqrt{d^2+4}$, the two interacting eigenvalues areThe $h^2$ correction to the pair is included to resolve the next perturbative term as well as the leading splitting. In particular the previously excluded fixed case $\alpha =1$ is $d=0$:The half-integer powers reflect square-root splitting of a defective double eigenvalue: the limiting matrix has a nontrivial Jordan block at $1$.

For matching, write $\Delta =\alpha -1$ and take the overlap $\sqrt{ϵ}\ll |\Delta |\ll 1$. For $d\to +\infty$, $L_{+}=d+d^{-1}+O(d^{-3})$, $N_{+}=1+O(d^{-2})$, while $L_{-}=-d^{-1}+O(d^{-3})$, $N_{-}=O(d^{-2})$. Consequently the distinguished branches reduce toThe regular coefficient satisfies $(3\alpha -1)/[(\alpha +1)(\alpha -1)]=1/\Delta +2/(2+\Delta )=1/\Delta +1+O(\Delta )$, proving agreement. For $d\to -\infty$ the labels $+$ and $-$ exchange roles and the same matching holds. The isolated negative branch also matches because $-2/(\alpha +1)=-1+O(\Delta )$. As a spectral check, positive diagonal symmetrization of a matrix transforms this tridiagonal matrix to a real symmetric matrix with off-diagonal entries $\sqrt{2ϵ}$ and $\sqrt{ϵ}$; its eigenvalues are real for positive $ϵ$.