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Let $q$ be the coordination number of a lattice, the number of nearest neighbours of each cell away from boundaries. Define an effective chemical potential incorporating the cell's translational entropy:Use $⟨\alpha {\alpha }^{\prime }⟩$ for an unordered nearest-neighbour bond, counted once. The occupation Hamiltonian including the chemical-potential term isUnder $n_{\alpha }=(1+{\sigma }_{\alpha })/2$, one bond contributesIn a regular lattice with periodic boundary conditions, the number of bonds is $Mq/2$, and each spin occurs in $q$ linear bond terms. ThereforeThe pair sum equals one half of the ordered sum used in the paper. Consequently the effective field isThus $\Xi ={\sum }_{\{\sigma \}}{e}^{-\beta {\mathcal{H}}_{\mathrm{eff}}}$ is the Ising model partition sum, including the displayed spin-independent constant if its absolute normalization is required. Dropping that constant changes $\Xi$ by a multiplicative factor but leaves spin probabilities and the phase-transition argument unchanged. For simple cubic cells in three dimensions, $q=6$ and the first term in $h$ is $6J$. Open boundaries replace $q$ by the local coordination near the surface; the stated uniform field is the bulk or periodic-lattice result.

Farther than a few edge lengths from the root, replace the radial coordinate $r$ by a physical coordinate $x=ar$ and smooth the master equation. It has the same constant-coefficient Fokker-Planck equation as in (b). Under the conventional first-time-derivative truncation,The radial drift $v=-C=a(z-2)/(z\tau )$ is outward. For $z=2$ the tree is the infinite line, its radial distance is an unbiased reflected walk, and $C=0$.

The root cannot absorb probability or allow negative radial distance. Write the equation as a continuity equation, using the Fokker-Planck probability currentThe appropriate continuum boundary condition is reflecting, or zero flux:It is not generally ${\partial }_{x}P(0,t)=0$: that simpler condition is correct only when $C=0$. With the finite-bias fixed-step Gaussian convention from (b), use $D=D_{\mathrm{fixed}}=2a^2(z-1)/(z^2\tau )$ instead; the drift sign and zero-current boundary condition are unchanged. A continuum boundary condition represents the microscopic root region after coarse-graining, rather than reproducing each discrete root probability at individual step times.

A uniform steady state satisfies $f(u^{\ast },v^{\ast })=g(u^{\ast },v^{\ast })=0$. Set $u=u^{\ast }+\tilde{u}$, $v=v^{\ast }+\tilde{v}$ and keep only first-order terms in the perturbations. Since the base state is constant, its Laplacian vanishes. Taylor expansion givesThe zero-flux boundary conditions linearize to homogeneous Neumann conditions, ${\partial }_n\tilde{u}={\partial }_n\tilde{v}=0$. The off-diagonal entries of the Jacobian describe the cross-effects of the two chemicals, while the diagonal entries describe local self-response.

A spherical bubble displaces liquid volume $4\pi a^3/3$. Neglect its own weight compared with the liquid buoyancy, as required by the stated gas/liquid approximation. Equating $4\pi a^3{\rho }_{l}g/3$ with the quadratic drag givesRetaining a nonnegligible gas density would add the factor $1-{\rho }_g/{\rho }_l$. Let $V_s=\sqrt{4ag/(3C_D)}>0$, so the upward slip is $u_1-u_2=V_s$.

With constant phase densities, adding the two volume balances gives ${\partial }_x[\alpha u_1+(1-\alpha )u_2]=0$. The closed lower end fixes the total volume flux to zero. Solving it together with the slip relation yieldsIndividual bubbles move upward for $0\le \alpha <1$, while small phase-fraction disturbances move at the characteristic speed $q^{\prime }(\alpha )=V_s(1-2\alpha )$. The latter is negative when $\alpha >1/2$: wave speed and bubble speed are different quantities.

A cleared region starts at the closed bottom, with ${\alpha }_L=0$ and ${\alpha }_R=3/4$. Its Rankine-Hugoniot condition speed is $[q]/[\alpha ]=V_s/4$. The intended clearing solution is thereforeIt is an entropy shock for the concave conservation law flux, since $q^{\prime }(0)=V_s>V_s/4>q^{\prime }(3/4)=-V_s/2$. Its speed also equals the bubble speed on its uncleared side, so it follows the last layer of bubbles. Under this continuation the last bubble reaches the outlet whenafter which $\alpha =0$ throughout the pipe.

There is a boundary qualification for this dense initial mixture. The characteristics in the uncleared state point into the pipe from the top. To continue the displayed solution all the way to the stated clearance time, its incoming top trace must remain $\alpha (1^{-},t)=3/4$ until clearing. Initial characteristics alone cover the shock only until $t=4/(3V_s)$, when its backward right-state characteristic first meets the top boundary. Atmospheric pressure by itself is not a boundary condition for this scalar phase-fraction equation. Thus the displayed formula and printed clearance time give the intended maintained-trace branch; a different top phase-fraction closure could change the later evolution. This is the incoming boundary condition in dense bubble clearing, and is separate from the local derivation of the conservation law.

The two wall boundary conditions areDefine the capillary parameter for a falling filmRetain $S=O(1)$ in the long-wave approximation. Thus surface tension is retained in the leading normal stress balance even though geometric slope corrections are small. For $h_x=O(1)$ in the rescaled coordinates, the dimensional surface slope is $\epsilon h_x$.

The three free surface boundary conditions, accurate through $O(\epsilon )$, areThe first is the kinematic boundary condition; the second is the zero tangential traction condition. For the third, the dimensional liquid-side pressure is, to the required order,Division by $\rho g{h}_0\sin \alpha$ gives $p=-3S{h}_{xx}/2+\epsilon w_z$ in dimensionless variables. Subtracting $\bar{p}(h)=(1-h)\cot \alpha$ gives the stated perturbation condition. The additional slope-times-shear contribution to normal stress vanishes at this order because the tangential traction is zero. The omitted geometric and tangential-stress corrections are $O({\epsilon }^2)$ under this ordering.

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.

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.

The condition $f(0)=0$ means the river maintains the water table at the impermeable base level: $h(0,t)=0$. The liquid is exposed to atmospheric pressure there. It does not imply zero drainage, because $h_x$ may be singular at the outlet.

Far from the river, lateral drainage has not significantly affected storage. Rainfall simply fills pore volume, so the dimensional height is $h\sim Rt/\varphi$. In the dimensionless variables this is $h\sim t$, giving $f(\infty )=1$.

Although the physical aquifer is finite, these boundary conditions describe its early similarity solution before the drainage-affected region reaches $x=L$. In dimensional variables its horizontal extent is of order $\sqrt{KR}t/\varphi$, so the semi-infinite approximation requires $t\ll T=\varphi L/\sqrt{KR}$.

Use the dimensionless equation from part (a). For the similarity ansatz $h=t^{a}f(\eta )$, $\eta =x/t^b$, its three terms have time factorsA nontrivial similarity solution retaining storage, lateral drainage and nonzero rainfall requires the exponents to agree:Unlike a source-free diffusion equation, the nonzero prescribed rainfall fixes an amplitude-versus-time scale directly, so no additional conserved quantity or boundary condition is required to determine these exponents. Degenerate spatially uniform solutions are an exception to this all-terms balance. The rain-fed aquifer similarity solution satisfiesIf the dimensional equation is used directly without scaling out coefficients, its similarity profile instead obeys $\varphi (f-\eta f^{\prime })=K(f{f}^{\prime }{)}^{\prime }+R$. The normalization above is the one in which the subsequent condition $f(\infty )=1$ applies.

With $R=0$, substitute $h=X(x)T(t)$ into the dimensional Boussinesq equation for an unconfined aquifer:For a positive draining profile, take $\lambda >0$. Normalize $T(t^{\ast })=1$, so $X$ has the dimensions of height. The temporal ordinary differential equation integrates toThe spatial ordinary differential equation and boundary conditions areHere $\lambda$ has units of inverse length with this normalization. The scale freedom between $X$ and $T$ is fixed by the chosen value $T(t^{\ast })=1$; the amplitude of $X$, and hence $\lambda$, must be matched to the available water.

This separable aquifer drawdown is exact if the inherited height profile has the separated spatial shape. For a general rainfall-generated initial condition, it describes the late drought regime; it cannot reproduce an arbitrary profile at the instant rainfall stops merely by choosing one temporal constant.

Return to dimensional variables. The river at base level imposes the drained boundary condition, while the right-hand end is impermeable:As long as the right-hand end remains wet, the latter is $h_x(L,t)=0$. The initial condition for the drought is the height profile inherited at $t=t^{\ast }$; stopping rainfall changes the forcing but does not instantaneously remove stored water.

For the approximate similarity profile, $F^{\prime }=A\mathrm{erf}(\xi /\sqrt{2})$, and thereforeFor the prescribed undisturbed horizontal streams, a suitable choice is $D=0$, which gives zero transverse velocity at both far-field limits. Then $V=-\Delta U\sqrt{CS}B{e}^{-{\xi }^2/2}$ inside the layer. The nonzero internal crossflow is consistent with the streamwise widening of the velocity transition.

Another possible convention is to make the line $y=0$ a streamline, imposing $V(0)=0$. This requires $D=-B$, equivalently $F(0)=0$, and gives $V(\pm \infty )=\Delta U\sqrt{CS}B$. It therefore requires a uniform outer crossflow relative to the coordinates used for the centred error function profile. A single constant cannot impose both zero far-field crossflow and $V(0)=0$ unless $\Delta U=0$. The physical lateral boundary conditions, or the chosen material centreline, determine which convention is appropriate. Since $B,D=O(r)$, these transverse velocities are of order $\Delta U^2/U_c$ for fixed $C,S$.

The two external stream velocities imposeLet $r=\Delta U/U_c\ll 1$ and seek $F=O(r)$ over $\xi =O(1)$. The nonlinear term $F{F}^{\prime \prime }$ is then $O(r^2)$, so the leading ordinary differential equation isFor $G=F^{\prime \prime }$, this is the first-order linear differential equation $G^{\prime }+\xi G=0$. Its integrating factor is $e^{{\xi }^2/2}$, giving $F^{\prime \prime }=M{e}^{-{\xi }^2/2}$. An integration using the error function givesThe two far-field boundary conditions require $E=0$ and $M=r/\sqrt{2\pi }$. A second integration usesConsequentlyThe relation between $A$ and $B$ cancels the extra $\xi e^{-{\xi }^2/2}$ term when differentiating. It gives the familiar error function mean velocity profileto first order in $r$. The additive constant $D=O(r)$ does not affect this velocity profile at the retained order; its interpretation comes from the transverse velocity.

In a thin boundary layer, take streamwise velocity and distance to be order one, set $z=\delta \zeta$, and use incompressibility to set $w=\delta W$. The streamwise convective acceleration is order one, whereas its dominant viscous contribution after division by the Reynolds number is ${\mathrm{Re}}^{-1}{\delta }^{-2}{u}_{\zeta \zeta }$. Their dominant balance selectsWith this choice, the exact rescaled Navier-Stokes equations areThus the leading pressure is independent of $\zeta$. Matching it to the outer solution from part (a), evaluated near $z=0$, gives $p=p_{\ast }-x^2/2$, hence $-p_x=x$. The leading boundary-layer equation is thereforeThe exact surface shear stress condition rescales to $u_{\zeta }+{\delta }^2{W}_x=\delta \beta x=\gamma x$. Since $W(x,0)=0$ for all $x$, its tangential derivative also vanishes there. The leading surface boundary conditions areMatching to the outer solution requiresThe second matching condition specifies the leading growth of $W$, allowing a constant displacement correction. Requiring $W+\zeta \to 0$ would impose an extra condition not supplied by the leading outer solution. This is the stress-driven stagnation-point boundary layer.

The held-flat interface imposes the no-penetration boundary condition, while its specified tangential traction gives a shear stress condition:A differentiable boundary value $w(x,0)\equiv 0$ implies $w_x(x,0)=0$, so the second boundary condition is also $u_z(x,0)=\beta x$. There is no imposed no-slip boundary condition at this interface; tangential velocity is determined by the shear stress and the fluid equations.

The solution in part (a) has zero tangential traction, and so remains valid for the whole interface if and only if $\beta =0$. At the single point $x=0$ the imposed shear stress vanishes for any $\beta$, but this does not satisfy a boundary condition that must hold for every $x$.

Set $g=f_1^{\prime }$. The boundary value problem becomesFor $y_1=1+{\zeta }^2$, direct substitution gives $y_1^{\prime \prime }+\zeta y_1^{\prime }-2y_1=2+2{\zeta }^2-2(1+{\zeta }^2)=0$. This growing polynomial solution is useful for reduction of order.

Write $g=y_{1}v$. Substitution into the linear ordinary differential equation cancels the terms proportional to $v$ and givesAn integrating factor for this first-order equation for $v^{\prime }$ yieldsThe far-field boundary condition selects the convergent integral from $-\infty$:Any additional multiple of $1+{\zeta }^2$ would grow at $-\infty$. The chosen expression tends to zero: for $r=-\zeta \ge 1$,Here $(1+t^2{)}^{-2}\le (1+r^2{)}^{-2}$ and ${\int }_r^{\infty }{e}^{-t^2/2}dt\le r^{-1}{\int }_r^{\infty }t{e}^{-t^2/2}dt$. Thus the far-field boundary condition is verified without a cancellation of large terms.

At $\zeta =0$, differentiating gives $g^{\prime }(0)=C$, so $C=1$. Using the antiderivative supplied in the paper gives the explicit answerHere $\mathrm{erf}$ is the error function. Finally,recovers the unique correction satisfying $f_1(0)=0$. The decay bound also shows that $f_1$ approaches a finite constant at $-\infty$, as allowed by the normal velocity matching. This is the weak surface-shear perturbation of stagnation flow.

If the printed $+\infty$ condition is instead imposed literally on a positive half-line, reduction of order givesIt satisfies $g_{+}^{\prime }(0)=1$ but grows as $\zeta \to -\infty$, so it cannot describe the fluid domain specified in this paper. This makes the consequence of the printed sign error explicit.

At zero surface shear stress, the solution connected to part (a) isIndeed, it has $f_0^{\prime }=1$, $f_0^{\prime \prime }=f_0^{\prime \prime \prime }=0$ and $f_0(0)=0$. Insert the regular perturbation $f=\zeta +\gamma f_1+O({\gamma }^2)$ into the nonlinear ordinary differential equation. The terms areThe coefficient of $\gamma$ therefore givesExpanding the surface boundary conditions and the far-field matching condition givesThe printed $+\infty$ limit is a sign error. It is present in the native PDF on paper page 3, not introduced by transcription. The fluid occupies $z<0$, the inner variable is $\zeta =z/\delta <0$, and part (b)(iii) correctly matches at $-\infty$. We retain that physical domain throughout. A change to the positive coordinate $\eta =-\zeta$ would also change the differential equation and surface derivative signs; it cannot be made by changing only the far-field sign.

The inequality ${\delta }^2\ll \gamma$ ensures that effects discarded at order ${\delta }^2$ in the boundary-layer scaling are smaller than the retained order-$\gamma$ perturbation. The inequality $\gamma \ll 1$ justifies this regular perturbation. The two conditions control different approximations.

The flat plate moves tangentially and the curved plate is stationary. The no-slip boundary condition and no-penetration boundary condition giveAt the curved plate, impermeability can also be written $w=u{h}_x$; together with zero tangential velocity, this gives both components zero. The native PDF writes $v^{\prime }=0$ in the scaled wall conditions although the normal component is defined as $w^{\prime }$. We use $w^{\prime }$ consistently.

The parabolic lubrication gap changes by order $h_0$ when $x^2/R$ is order $h_0$. Its natural horizontal scale is thereforeThe proposed normal velocity scale $h_{0}U/L$ then makes the two terms of the continuity equation comparable. Balancing the horizontal pressure gradient against the transverse viscous term in the Navier-Stokes equation requires $P/L=\mu U/h_0^2$, soHere pressure is measured from a common reference value, which is chosen at the outer fluid in part (b).

For clarity, define the dimensionless material derivativeSubstitution of all the scalings gives the exact dimensionless equationsFor example, the factor in the last left side is obtained by dividing the vertical acceleration scale $\rho h_0{U}^2/L^2$ by the vertical pressure gradient scale $P/h_0$.

Taking $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$ leaves the lubrication approximationThe scaled gap and boundary conditions areThe Reynolds number itself need not be small: it is ${\delta }^2\mathrm{Re}$ that measures inertia relative to transverse viscous resistance in this thin geometry.

Write $H(x^{\prime })=h^{\prime }(x^{\prime })$ and $p_x^{\prime }=d{p}^{\prime }/d{x}^{\prime }$ below, so the subscript $x$ in this shorthand denotes differentiation with respect to $x^{\prime }$. The vertical pressure gradient is zero, and integrating $u_{z^{\prime }{z}^{\prime }}^{\prime }=p_x^{\prime }$ twice with the wall boundary conditions gives the local Couette-Poiseuille flowThe first two terms are Couette flow due to the moving plate; the last is plane Poiseuille flow due to the local pressure gradient. Its dimensionless volume flux per unit width isUse the Leibniz integral rule and incompressibility to differentiate this integral:Thus the flux is independent of position, andThe normal velocity can then be recovered from incompressibility:It is zero on the lower wall by construction and zero on the upper wall by the constant-volume flux relation. This supplies the remaining component of the leading velocity field.

The endpoint pressure conditions are matching conditions, not an assertion that the lubrication approximation remains valid at arbitrarily large dimensional $x$. On an outer length scale $R$, a viscous pressure variation is of size $\mu U/R$, whose ratio to the inner scale isEven an outer inertial pressure scale $\rho U^2$ has ratioConsequently the gap pressure is much larger than either outer variation. Matching to fluid with a common ambient pressure at both ends, and taking that value as zero gauge pressure, givesThere is an overlap region $1\ll |x^{\prime }|\ll {\delta }^{-1}$ before the wall slope becomes order one. The infinite endpoints represent matching through that region. An externally imposed pressure difference comparable with $P$ would change these conditions and the resulting volume flux; equal outer pressure is the physical assumption being used here.

For $w=e^{\alpha z/2}W$, direct differentiation givesThus the transformed partial differential equation and its impermeable-wall boundary conditions areThe positive exponential factor does not change where the vertical velocity vanishes.