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Take the sky frame and disk frame to be right-handed orthonormal bases, with positive angular displacement measured from $\hat{X}$ toward $\hat{Y}$. In particular, positive $Z$ is toward the observer, so material crossing the sky plane with positive $\dot{Z}$ is at the ascending node. Write $c_d=\cos {\Omega }_d$, $s_d=\sin {\Omega }_d$, $C_d=\cos I_d$, and $S_d=\sin I_d$. The disk frame's first unit vector lies along its ascending node, and its second has positive $Z$ component. The orbital-frame rotation from inclination and node therefore givesEach column is a disk-frame unit vector in sky coordinates, so $\mathbf{X}=T\mathbf{x}$ and the inverse is $\mathbf{x}=T^{\mathsf{T}}\mathbf{X}$. At the disk's ascending node, the $Z$ component of its tangential velocity is positive, confirming the sign of the tilt.

Let $T_p=R_Z({\Omega }_p)R_X(I_p)$ define a provisional planet frame whose first axis is the planet's sky-plane ascending node. Its normal is ${\hat{z}}_p=(\sin {\Omega }_p\sin I_p,-\cos {\Omega }_p\sin I_p,\cos I_p)$. Expressing that normal in the disk orthonormal basis, with $\Delta \Omega ={\Omega }_p-{\Omega }_d$, givesThe last component is the dot product of the two orbital normals, hence the cosine of the mutual inclination. The mutual ascending node points along ${\hat{z}}_d\times {\hat{z}}_p$: motion there has positive disk-normal component. SetThen $\hat{x}\prime =(A\hat{x}+B\hat{y})/\sqrt{A^2+B^2}$, and the quadrant-correct longitude of ascending node is ${\Omega }_m=atan2(B,A)$. ThusThe tangent alone cannot fix the quadrant, and ${\Omega }_m$ is undefined for exactly coincident orbital planes.

For the two rotation routes, define $\psi$ as the oriented angle, within the planet's orbital plane, from its sky-plane ascending node to its disk-plane ascending node. If ${\hat{x}}_p,{\hat{y}}_p$ are the first two columns of $T_p$, take $\psi =atan2(\hat{x}\prime \cdot {\hat{y}}_p,\hat{x}\prime \cdot {\hat{x}}_p)$. Both routes describe the same planet orthonormal basis:The left route goes through the disk frame; the right goes through the sky-node planet frame. Comparing their third columns gives the two boxed relations, while comparison of first columns fixes the extra in-plane angle.

Put $I_d=\pi /2+{\delta }_d$ and $I_p=\pi /2+{\delta }_p$. A second-order Taylor expansion gives $\cos I_m=1-[({\delta }_p-{\delta }_d{)}^2+\Delta {\Omega }^2]/2+O({\delta }^4)$, and therefore the mutual inclination of nearly edge-on orbits satisfiesAll expansion angles are in radians. For plotting in degrees this is $I_m\simeq \sqrt{(I_p-87^{\circ }{)}^2+(4^{\circ }{)}^2}$. The given planet has exact $I_m\simeq 4.119^{\circ }$ and ${\Omega }_m\simeq 76.044^{\circ }$. The exact minimum occurs at $I_p=atan2(\sin I_d\cos \Delta \Omega ,\cos I_d)\simeq 86.993^{\circ }$, with $I_{m,\min }\simeq 3.995^{\circ }$. At $I_p=85^{\circ }$ and $95^{\circ }$, the second-order values are ${\sqrt{20}}^{\circ }\simeq 4.472^{\circ }$ and ${\sqrt{80}}^{\circ }\simeq 8.944^{\circ }$.

For the disk evolution, neglect disk self-gravity, collisions that align the planes, and back-reaction on the planet. In the initial disk plane, use the complex inclination $w=I{e}^{i\Omega }$ for a ring and $w_p=I_m{e}^{i{\Omega }_m}$ for the planet. Linear Laplace-Lagrange secular theory has forced inclination $w_p$ and solution $w(a,t)=w_p[1-e^{-i\nu (a)t}]$ for initially flat rings. For an inner circular perturber, the Laplace coefficient formula is $\nu =n(M_p/M_{\star })\alpha b_{3/2}^{(1)}(\alpha )/4$, with $\alpha =a_p/a$. Its quadrupole approximation uses $b_{3/2}^{(1)}\simeq 3\alpha$, yieldingIn the $(I\cos \Omega ,I\sin \Omega )$ plot referenced to the initial disk, every ring runs clockwise on the same circle centered at $w_p$, starting at the origin. Inner rings advance more quickly. If the axes are instead referenced to the planet's plane, the relative tilt is $w-w_p=-w_p{e}^{-i\nu t}$: these circles are centered at the origin, and the mutual inclination is constant. Both conventions describe the same nodal precession; they should not be mixed.

At an intermediate epoch, rings with $\nu t\gtrsim 1$ have tilted substantially while outer rings with $\nu t\ll 1$ remain near the original disk plane. This radial variation is a planet-induced debris-disk warp. The characteristic affected radius grows like $t^{2/7}$. Conservative differential nodal precession preserves each ring's tilt relative to the planet, so it does not by itself align every orbit; an unresolved inner region can acquire a mean plane near the planetary plane through phase mixing. The secular interpretation is also described in Wyatt's planetary dynamics lectures.

Let $\mu =M_p/M_{\star }$, $q=\sqrt{1-\beta }$, and restrict first to $0\le \beta <1$. The radiation-pressure coefficient reduces the grain's central gravitational parameter to $G{M}_{\star }{q}^2$, whereas the planet's is approximately $G{M}_{\star }$. Their mean motions obey $n/n_p=q(1+ϵ{)}^{-3/2}$. Setting this ratio to $j/(j+1)$ gives the radiation-pressure shift of a mean-motion resonance:It lies interior to the planet exactly when $q(j+1)<j$, or $j>q/(1-q)$. Equality puts the frequency commensurability at the planet's radius. For $\beta =0$ every such commensurability is exterior.

There is a frame qualification in the printed encounter formula. The circular inertial speed of the grain at $a_p(1+{ϵ}_1)$ is $v_1=q{v}_p(1+{ϵ}_1{)}^{-1/2}$. Subtracting the planet's inertial velocity at conjunction gives the local planet-relative inertial speedThis is the velocity used in the question's two-body gravitational assist calculation. In a frame actually rotating about the star at $n_p$, the grain's tangential velocity is instead $v_1-n_p{a}_p(1+{ϵ}_1)$, with magnitude $v_p[1+{ϵ}_1-q(1-{ϵ}_1/2)]$. In particular its nonradiating shear is $3v_p{ϵ}_1/2$, rather than $v_p{ϵ}_1/2$. Thus the requested expression uses a local translating, inertial encounter convention despite the source's rotating-frame wording. The following derivation consistently uses that convention, and its coefficients are an encounter estimate rather than an exact rotating-frame result.

Take impact parameter $b\simeq a_p{ϵ}_1$ and a small gravitational scattering angle $\theta$. Expanding the supplied hyperbolic scattering relation gives $\theta \simeq 2G{M}_p/(b{v}_{\infty }^2)$. Initially the relative tangential velocity is $-v_{\infty }$; after the deflection it has tangential component $-v_{\infty }\cos \theta$. The relative speed is unchanged in the local hyperbolic Kepler orbit, but the stellar-frame kinetic energy changes. Taking the dot product with the planet's velocity givesThe specific orbital energy of the grain is $-G{M}_{\star }(1-\beta )/(2a)$. Comparing it immediately before and after the encounter at the same position givesConsequentlyThe encounter increases the grain's semi-major axis within this geometry. These steps assume ${ϵ}_1\ll 1$, $\theta \ll 1$, and an orbital kick small enough for the subsequent linearization.

Choose the initial conjunction longitude as zero. The synodic period is $2\pi /(n_p-n)$, and the planet's longitude accumulated by the next conjunction isThese are unwrapped longitudes, counting complete revolutions. The second uses the new mean orbital period while neglecting the encounter's immediate phase offset and the difference between true and mean conjunction. A comparison reduced modulo $2\pi$ could not implement the stated one-turn sensitivity test.

For $\beta =0$, $v_{\infty }\simeq v_p{ϵ}_1/2$, so $\delta ϵ\simeq 32{\mu }^2{ϵ}_1^{-5}$ and ${\lambda }_c\simeq 4\pi /(3ϵ)$. Conjunction-longitude sensitivity to an encounter givesRequiring this to exceed $2\pi$ gives ${ϵ}_1^7<64{\mu }^2/3$. At a high-index first-order mean-motion resonance, ${ϵ}_j=(1+1/j{)}^{2/3}-1\simeq 2/(3j)$, henceThe interpretation of the resonance-overlap encounter estimate is loss of coherent encounter phases: a small kick changes the next synodic encounter time by more than one planetary revolution. Resonant protection by repeatedly meeting at controlled orbital phases can then fail, allowing irregular kicks and diffusion. This sensitivity criterion motivates an unstable region; it is not a proof that every orbit or resonant phase in it must escape.

For a quantitative sketch with radiation, put $h=1-q(1-ϵ/2)$ and $d=1-q(1+ϵ{)}^{-3/2}$. Differentiating the conjunction formula and substituting the kick givesEvaluate this at the positive ${ϵ}_j$ and take $\mathcal{S}=1$ as the critical-index curve. At small $\beta$, increased encounter speed weakens planetary scattering, moving the threshold to larger $j$. As $\beta$ rises further, the radiation-pressure shift of a mean-motion resonance brings each commensurability closer to the planet, strengthening the close encounters. The curve turns over and approaches the exterior-limit boundary $j=q/(1-q)$. Larger planetary mass produces stronger kicks and a lower critical index. These competing effects are shown below; the curves retain the factors in $\mathcal{S}$, so their $\beta =0$ intercepts differ slightly from the leading high-$j$ formula. The boundary for disappearing exterior commensurabilities is geometric; divergence of the encounter approximation near $ϵ=0$ should not be read as a precise physical scattering rate.

Use perturbations $\mathbf{v},\mathbf{b},p,\vartheta$ for velocity, magnetic field, magnetohydrodynamic total pressure and buoyancy displacement variable. SetThis is advection by the background Keplerian shearing sheet. Subtracting the equilibrium before retaining first-order terms gives the nine linearized equations:The azimuthal coefficient $\Omega /2$ combines the Coriolis acceleration with perturbation advection of the background shear. The $-3\Omega b_x/2$ term is the winding of radial magnetic field by that shear. The tidal force cancels when the equations at a fixed position are subtracted; it does not add a separate Eulerian perturbation force. Magnetic pressure is already included in $p$, leaving only linear magnetic tension. In this normalization $\vartheta$ has dimensions of length because $D_0\vartheta =v_x$.

For an axisymmetric vertical mode of a shearing sheet with real $k\ne 0$, the two divergence constraints give $ik{v}_z=ik{b}_z=0$. The vertical momentum equation then yields $ikp=0$. Thus all three amplitudes vanish:Define the magnetic amplitudes $a_i=b_i/\sqrt{4\pi {\rho }_0}$ and the signed vertical Alfvén velocity $v_A=B_0/\sqrt{4\pi {\rho }_0}$. Write $A=k^2{v}_A^2$. The remaining equations for the normal mode areFor a growing or oscillatory mode with $\sigma \ne 0$, eliminating $a_x,a_y,\vartheta$ givesA nonzero velocity requires the determinant to vanish. After removing its factor $\sigma$, the dynamical dispersion relation isThis is the radially stratified magnetorotational dispersion relation, with $A=k^2{B}_0^2/(4\pi {\rho }_0)$. Keeping the original five-amplitude system instead gives characteristic polynomial $\sigma$ times this quartic. There is also a stationary balanced normal mode; division by $\sigma$ excludes it but loses no exponentially growing mode. At $k=0$ the divergence argument for vanishing vertical components does not apply, so that spatially uniform case must be treated separately.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.

Put $q=e^{i\pi /6}$, so $q^3=i$. The bounded matching branch of the Airy function gives $u_Y=A\mathrm{Ai}(q(Y-C))$; the other independent Airy solution grows along the ray and is excluded. Integrating with the two wall conditions giveswhile the wall momentum equation gives the constant fluid pressureDefine the convergent Airy displacement integralsThe Airy function equation and its decaying derivative imply $I_1-C{I}_0=iq{\mathrm{Ai}}^{\prime }(-qC)$. Consequently, as $Y\to \infty$,Both differentiated lower momentum and the constant of integration are satisfied; retaining that constant is essential for the final matching.

For the middle region introduce a perturbation stream function $\Phi$ with $u={\Phi }^{\prime }$ and $v=-i\alpha \Phi$. The leading inviscid flow streamwise equation isFor a shear flow, this is the integrated form of the Rayleigh equation for inviscid shear flow. The leading matched solution for fixed $y=O(1)$ is $\Phi \sim BU$, so $u\sim B{U}^{\prime }$ and $v\sim -i\alpha BU$. In the overlap with the lower viscous boundary layer, the first constant correction iswhich agrees with the integrated lower solution. For a regular smooth profile, the normal momentum equation gives $p^{\prime }\sim -{\alpha }^2(U-c)\Phi$. Thus fluid pressure variation across a fixed-width middle region is $O({\alpha }^{2}B)$; in the final distinguished scaling this is smaller than its $O(\delta )$ value, so $p\sim \delta P_0$ throughout that region. Since $U\to 1$, the middle velocity tends to $\boxed{v\sim -i\alpha B}$, which does not decay and requires another region.

The fluid pressure equation becomes $p^{\prime \prime }-{\alpha }^{2}p=0$ when $U\sim 1$. Therefore the upper scale is $\Delta =1/\alpha$, with $z=\alpha y$. The inviscid decaying upper solution isMatching velocity as $z\downarrow 0$ gives $P_{\infty }\sim \alpha B$; matching fluid pressure to the middle region gives $\delta P_0\sim \alpha B$. This is three-layer long-wave shear-flow matching, expressed as a matched asymptotic expansion.

Under $\alpha =k{R}^{-1/4}$ with fixed positive $k$, $\delta =k^{-1/3}{R}^{-1/4}$, and $\alpha /\delta =k^{4/3}$. Eliminating the nonzero amplitude $A$ gives the Airy wall-layer matching determinant and dispersion relationEquivalently, with $\mathcal{T}(C)={\int }_{-qC}^{\infty e^{i\pi /6}}\mathrm{Ai}(\zeta )d\zeta =q{I}_0(C)$,The determinant form does not divide by a possibly vanishing $I_0$. The wavespeed is $c=\delta C$; temporal growth has sign $\mathrm{Im}C$ for positive $\alpha$.

There is a genuine conflict in the printed assumptions: fixed-$k$ $\alpha =k{R}^{-1/4}$ cannot satisfy $R^{-1/7}\ll \alpha$ as $R\to \infty$. For order-one lower matching amplitudes, constant leading middle fluid pressure requires ${\alpha }^2\ll \delta$, equivalently $\alpha \ll R^{-1/7}$; this is the reverse inequality and is satisfied by the final distinguished scaling. Thinness also requires $\alpha R\gg 1$. The dispersion relation above is therefore derived under the final distinguished limit, for which the complete consistent window is $R^{-1}\ll \alpha \ll R^{-1/7}$. Both printed limits cannot be imposed simultaneously.

One convention for the Papkovich–Neuber representation uses a harmonic vector potential $\mathbf{\Phi }$ and a harmonic scalar potential $\chi$:These are harmonic functions away from any singular force point. Since ${\nabla }^2(\mathbf{r}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, the representation gives $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$, the equations of homogeneous Stokes flow.

Place the point force at the origin. A velocity linear in $\mathbf{F}$, decaying as $r^{-1}$ and having the rotational symmetry of a point force is obtained from $\mathbf{\Phi }=c\mathbf{F}/r$, $\chi =0$. The vector components are harmonic functions for $r>0$; scalar dipole potentials would instead generate higher-order decaying singularities. The force normalization fixes $c=-1/(4\pi )$. Indeed, substitution gives the Stokeslet:To verify its strength, the Newtonian fluid stress tensor is ${\sigma }_{ij}=-3(\mathbf{F}\cdot \mathbf{r})r_i{r}_j/(4\pi r^5)$. Its outward traction integrated over any sphere surrounding the origin is $-\mathbf{F}$, because $\int \mathbf{nn}d\Omega =(4\pi /3)I$. Thus the localized force applied to the fluid is $\mathbf{F}$, as required. Translation of the origin gives the same Stokeslet centered at any prescribed force point.

Let $\mathbf{n}=\mathbf{X}/R$, $\zeta =6\pi \mu a$, and letUse the method of reflections for Stokes flow. At the fixed sphere, the first sphere produces the incident Stokeslet velocity ${\mathbf{u}}_{\infty }(0)=\mathsf{A}\mathbf{V}$ to leading order. In Faxén translation law, set the second sphere's translational velocity to zero. The applied holding force is thereforeThe first sphere's finite-radius potential dipole adds $O(V{a}^3/R^3)$ to the incident velocity, as does the Laplacian term in Faxén translation law at the fixed sphere. Multiplying by $\zeta$ gives the stated next correction. There is no intermediate $O(\mu V{a}^3/R^2)$ term. This is the holding force and torque for a sphere in a distant Stokeslet.

The vorticity of a Stokeslet at displacement $\mathbf{r}$ is $\mathbf{F}\times \mathbf{r}/(4\pi \mu r^3)$. At the fixed sphere, $\mathbf{r}=-\mathbf{X}$, soSet its angular velocity to zero in Faxén rotation law. The applied holding couple isThe displayed sign is the external couple needed to oppose the ambient rotation, rather than the hydrodynamic couple on the sphere.

The leading reflected flow at the first sphere is $\mathsf{G}(\mathbf{X}){\mathbf{F}}_2=-{\mathsf{A}}^2\mathbf{V}$. Apply Faxén translation law there with its prescribed force $\mathbf{F}=\zeta \mathbf{V}$. Since $(I+\mathbf{nn}{)}^2=I+3\mathbf{nn}$, the mobility correction from a fixed distant sphere isThe next correction comes from finite-radius terms in the incident/reflected flow and in Faxén translation law, together with the fixed sphere's induced stresslet and holding-couple rotlet. Each gives $O(V{a}^4/R^4)$ at the first sphere. The symbol $\mathbf{V}$ is its isolated-sphere velocity scale, not its actual velocity in the two-sphere problem.

The first sphere has no applied couple. Its angular velocity follows from Faxén rotation law and the reflected Stokeslet:The component of ${\mathbf{F}}_2$ parallel to $\mathbf{X}$ contributes no vorticity. The next reflected rotlet and stresslet give the stated error order.

Take $z$ positive upwards and let $f$ denote axial body force per unit volume. With inertia and surface tension omitted, the extensional equations for a slender Newtonian column areMass conservation gives the first equation. The second uses the Trouton ratio three: the leading radial velocity is $u_r=-r{w}_z/2$, the radial normal traction gives $p=p_{\mathrm{ext}}-\mu w_z$, and the axial stress is $-p_{\mathrm{ext}}+3\mu w_z$. An axial control volume balance then gives the displayed equation. The approximation requires a slender column and slowly varying radius.

Let $h=R-a$. Since the container base is closed, the total volume flux at every section is zero. The column carries flux $\pi a^{2}w$, so the annulus carries $-\pi a^{2}w$ and has mean speed of order $a|w|/h$. Its interfacial shear stress is therefore of order $\lambda \mu a|w|/h^2$. Transmitting this stress through the core produces an axial velocity variation of order $\lambda a^2|w|/h^2$. The plug-flow criterion for a column in a low-viscosity annulus is consequently

Use modified pressure $P=p_{\mathrm{ext}}+\rho gz$, removing the annular fluid's hydrostatic pressure. In a local planar gap, with $y=0$ at the column and $y=h$ at the container wall, lubrication theory givesThe required return flux per unit circumference is $q_g=-aw/2$ to leading order in $h/a$. Its pressure-driven component is larger than the Couette component by $a/h$, so the annular return flow around a slumping column givesThe annular shear contribution to the integrated axial force is smaller than this pressure-gradient term by $h/a$. Retaining the leading terms and substituting $f=-(\rho +\Delta \rho )g$ gives the dimensional equationsThe boundary conditions are $w(0,t)=0$ at the base and $w_z(L(t),t)=0$ at the free upper end; the latter expresses zero excess axial extensional stress.

For $\lambda =0$ and initially constant $a=a_0$, put $G=\Delta \rho g/(3\mu )$. Then $w_{zz}=G$, so

For $\lambda >0$, defineThis function records the leading thin-annulus resistance; differences from curvature and the Couette term enter at the already neglected relative order $h/a$. Choose the extensional screening length of a slumping column and the associated scales asHere $h_0=R-a_0$ denotes the initial annular gap in this question. With $Z=z/\hat{z}$, $T=t/\hat{t}$ and $W=w/\hat{w}$, the dimensional equations reduce to

Initially $\varphi ={\varphi }_0$ and ${\Lambda }_0=L_0/\hat{z}$, so $W_{ZZ}=1+W$, with $W(0)=0$ and $W_Z({\Lambda }_0)=0$. The initial velocity profile of an annularly confined column isThe column thickens everywhere below its stress-free top, while that initial thickening rate is zero at the top.

For a tall column, ${\Lambda }_0\gg 1$, the limiting forms away from exponentially small end corrections areThe bulk falls at approximately $-\hat{w}$, where excess weight balances annular hydraulic resistance. Extensional stresses adjust this velocity to zero in a basal boundary layer of thickness $\hat{z}$; almost all thickening occurs there.

For a short column on this scale, ${\Lambda }_0\ll 1$, while still slender, the limits areThe annular-resistance term $W$ is small, so excess weight is balanced predominantly by extensional viscous stress. The top speed is $-\hat{w}{\Lambda }_0^2/2$, recovering the zero-annular-viscosity result. Thus $\hat{z}$ is the vertical distance over which extensional stress communicates the basal constraint before annular resistance screens it. The sketches below use separate natural normalizations for the two limits.

At leading order in the lubrication approximation, surface arclength is horizontal distance. An insoluble surfactant is transported by the surface velocity, without bulk exchange or diffusion. Conservation of insoluble surfactant on a moving interface therefore becomesThe local hydrostatic pressure and capillary normal stress giveSolve $\mu u_{zz}=p_x$ with no-slip boundary condition at $z=0$ and Marangoni stress $\mu u_z(h)={\gamma }_x=-A{C}_x$. The velocity, surface velocity and volume flux per unit width areMass conservation, $h_t+q_x=0$, yields the thin-film equations with insoluble surfactant:and the corresponding surfactant equation isThe hydrostatic and capillary terms thus affect both the film flux and the surface transport, with different coefficients.

For finite-mass Marangoni spreading on a liquid film, first neglect both pressure-gradient terms. If the spread has size $\ell$, then $C\sim M/\ell$ and $C_x\sim M/{\ell }^2$, giving surface speed $u_s\sim AM{h}_0/(\mu {\ell }^2)$. Equating $\ell /t$ to this speed givesSet $\ell =(AM{h}_{0}t/\mu {)}^{1/3}$ exactly as a similarity scale, and writeOn the positive half of the pool, substitution produces two ordinary differential equations:Symmetry fixes the flux integration constant in the second equation to zero, giving $H{\Gamma }^{\prime }=-\eta /3$ wherever $\Gamma >0$. Substitution into the first equation gives $\eta H^{\prime }=H$. Thus the linear similarity profiles for surfactant spreading have the formThe concentration vanishes at the moving front. The two integral constraints are total surfactant mass and conservation of the fluid volume relative to the undisturbed layer:The first gives $k={\eta }_N^2/3$; the second gives $k{\eta }_N=2$. Therefore ${\eta }_N^3=6$. In physical variables, the solution isOutside the pool the leading outer solution has $h=h_0$, $C=0$. In particular, $h(x_{N-},t)=2h_0$: the subscript means the one-sided limit just behind the front. As a consistency check, the surface speed there is $x_N/(3t)={\dot{x}}_N$ and the fluid jump satisfies the moving-front mass conservation condition. This ideal outer solution has a height jump from $2h_0$ to $h_0$.