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Take $x$ tailward, so forward swimming has laboratory velocity $-U\mathbf{i}$. For an inextensible travelling waveform, material motion in the swimmer frame is $V\mathbf{i}-c\mathbf{t}$: a material point passes through the wave along its tangent while the wave pattern translates at $V$. Thus the relative velocity of otherwise stationary fluid isPut $q=V-U$ and choose $\mathbf{t}=(X_s,Y_s)$, $\mathbf{n}=(-Y_s,X_s)$, with $X_s^2+Y_s^2=1$. ThenOn the forward-swimming branch $0<U<V<c$ and $X_s>0$, the tangential slip is positive. A law for reversed tangential slip would also need its sign, not just its absolute magnitude.

The quadratic normal law represents inertial cross-flow drag: dynamic pressure scales as fluid density times $w_n^2$, while $|w_n|w_n$ chooses the force direction. Its coefficient contains the cross-sectional size and an empirical drag coefficient. The $|w_t{|}^{3/2}$ exponent is characteristic of an attached laminar boundary layer: its thickness scales as $\delta \sim (\nu s/|w_t|{)}^{1/2}$ and its shear as $\mu |w_t|/\delta$. However, this gives local shear proportional to $|w_t{|}^{3/2}{s}^{-1/2}$, not $s^{1/2}$. The positive power of $s$ is genuinely printed in this paper and repeated in its target formulas. Below it is treated as the prescribed phenomenological weight, rather than misidentified as standard local laminar skin-friction scaling along a slender swimmer.

The normal axial force on an element isThe prescribed tangential force gives $d{F}_{T,x}=K_T(c-q{X}_s{)}^{3/2}{X}_s{s}^{1/2}ds$. Zero mean axial force therefore gives the nonlinear resistive-force balance for a travelling filamentThe absolute values are present in the original PDF. In a periodically beating, constant-mean-speed model, phase-average the integrals where necessary; the leading approximation below is phase independent. Local force laws neglect nonlocal hydrodynamic interactions and assume nearly steady local response.

For the sinusoidal displacement, let $\epsilon =\beta k\ll 1$. Then $Y_s=\epsilon \cos ks$ and $X_s=(1-Y_s^2{)}^{1/2}=1+O({\epsilon }^2)$. Over one wavelength,Replace $X_s$ by one in the leading tangential term and balance the forces. After multiplication by $3/2$,This replacement also requires $c{\epsilon }^2\ll c-V+U$ for a uniformly small relative error inside the fractional power. If the tangential slip is itself $O(c{\epsilon }^2)$, its variation along the body must be retained. Periodicity gives $V/c=⟨X_s⟩=1-{\epsilon }^2/4+O({\epsilon }^4)$, so small geometric slope alone does not guarantee that additional slip condition.

For Lighthill elongated-body theory, put $h(x,t)\simeq \beta \sin [k(x-Vt)]$ and $w_{\perp }=h_t+U{h}_x$. The transverse fluid kinetic energy per unit length is $M_A{w}_{\perp }^2/2$. In the usual tail-flux approximation, with negligible leading-end contribution, the lateral work flux is $M_{A}U⟨h_t{w}_{\perp }⟩$, whereas kinetic energy carried into the wake is $M_{A}U⟨w_{\perp }^2⟩/2$. Their difference is the useful propulsive power $TU$. ConsequentlyBalancing this against the drag of the printed tangential model, $D=(2/3)K_T{L}^{3/2}(c-V+U{)}^{3/2}$, givesFor a genuine local laminar skin-friction estimate, write $dD=C_T{u}_t^{3/2}{s}^{-1/2}ds$, where $C_T$ has the appropriate different dimensions. Its integrated drag is $2C_T{u}_t^{3/2}\sqrt{L}$, and the corresponding elongated-body balance is instead $M_A{\beta }^2{k}^2(V^2-U^2)/8=C_T\sqrt{L}(c-V+U{)}^{3/2}$. The printed $L^{3/2}$ expression is therefore a result of its stated force model, not of that standard boundary-layer law.

The two thrust models describe different mechanisms: cross-flow resistive-force theory uses dissipative local drag, while elongated-body theory uses reactive acceleration of added mass. Both suppress details of the wake, body-body interactions, finite slenderness, skin-friction unsteadiness, separation and possible turbulent transition. The tail-only reactive result also requires a negligible leading-end flux, usually supplied by a nose taper or small leading-end motion; a blunt constant-cross-section body waving equally at both ends cannot automatically omit its leading-end contribution. Speed fluctuations and recoil may matter, and a planar constant-amplitude waveform is an idealization of a real sea snake.

Choose the helix handedness so its unit tangent is $\mathbf{t}=\cos \alpha \mathbf{i}+\sin \alpha {\mathbf{e}}_{\varphi }$. Let $q={\Omega }_0-\Omega$; the body rotates at $+\Omega \mathbf{i}$, while the flagellum rotates at $-q\mathbf{i}$ in the laboratory, since its relative motor rotation is $-{\Omega }_0\mathbf{i}$. Both translate at $-U\mathbf{i}$. Thus the relative fluid velocity on a flagellar element is $\mathbf{w}=U\mathbf{i}+bq{\mathbf{e}}_{\varphi }$.

Low-Reynolds-number resistive-force theory gives force per unit length on the flagellumIts axial and azimuthal components areDefine the integrated axial resistance matrix of a slender helix coefficientsThe sphere has translational resistance $A_b=6\pi \mu a$ by the Stokes drag law and rotational resistance $H_b=8\pi \mu a^3$. Its hydrodynamic force is $+A_{b}U\mathbf{i}$ and its torque $-H_b\Omega \mathbf{i}$. Integration of $f_x$ and $b{f}_{\varphi }$ gives the zero total force and torque conditionsDividing the first by $L$ and expanding the coefficients gives exactly the two requested balances. The internal motor torque cancels in the total torque balance; it is not an external applied torque.

Eliminate $U$ and use $q+\Omega ={\Omega }_0$. The effective flagellar rotational resistance is $H_e=H-C^2/(A+A_b)$, soThe determinant of helical resistance in resistive-force theory is $AH-C^2=b^2{L}^2{K}_N{K}_T>0$, hence the denominator is positive. Opposite handedness changes the sign of the translation-rotation coupling and reverses the swimming direction.

For $K_T=K$, $K_N=2K$, $\alpha =\pi /4$ and $a=b$, the coefficients are $A=3KL/2$, $C=KaL/2$, $H=3K{a}^{2}L/2$. ThereforeEquivalently, with $h=KL/(\pi \mu a)$, $U=a{\Omega }_{0}4h/(2h^2+21h+48)$. These are positive speed magnitudes for the selected handedness; the translational velocity is $-U\mathbf{i}$.

This helical microswimmer with a spherical head approximation neglects head-tail hydrodynamic interactions as permitted, treats the flagellum as slender and rigid, and averages away transverse forces associated with finite helical ends. The flow is quasistatic Stokes flow, so neither body inertia nor fluid inertia enters the balance.

Let $\hat{u}(\hat{r})=-W_0(1-{\hat{r}}^2/R^2)+\hat{w}(\hat{r})$ and $\mathbf{p}=\sin \theta {\mathbf{e}}_r+\cos \theta \mathbf{k}$, with $\cos \theta >0$ for the upward-stable orientation. Cell conservation in a swimming suspension gives radial fluxRegularity at the axis and impermeability at the wall imply ${\hat{j}}_r=0$, hence $D{\hat{n}}^{\prime }=V_C\hat{n}\sin \theta$. Axial advection has no divergence in this fully developed, axially uniform state.

The fluid is incompressible. Use dynamic stresslet strength $S_d$ and dynamic viscosity ${\mu }_d$ first, so the swimming contribution to its stress is $S_d\hat{n}(\mathbf{pp}-I/3)$. The isotropic part can be absorbed into the effective pressure. With the ordinary fluid hydrostatic pressure removed, the axial momentum equation isThe final term is the downward excess weight of the cells. Sedimentation is neglected in their swimming flux, not in this buoyancy force. Fluid convective inertia vanishes for the specified fully developed velocity even before a low-Reynolds-number approximation is invoked.

There is a dimensional convention to make explicit. The printed definition of $\gamma$ contains the extra fluid density $\rho$. All the printed coefficients are consistent if this question uses density-normalized pressure, kinematic viscosity and stresslet strength:Then ${\hat{p}}^{\prime }=4\mu W_0/R^2$ generates the stated background Poiseuille flow. If $\mu$ and $S$ instead mean dynamic viscosity and a physical force-dipole strength, the correct buoyancy parameter is $g\Delta \rho \nu N_0{R}^2/({\mu }_d{W}_0)$, without a further density denominator. The following derivation retains the consistent density-normalized interpretation of the printed parameters.

The bottom-heavy spherical-cell orientation dynamics isIts scalar tilt equation is $\dot{\theta }=-\sin \theta /B-{\hat{u}}^{\prime }/2$. Steady gravitational-viscous torque balance therefore gives $\sin \theta =-B{\hat{u}}^{\prime }/2$. This equilibrium is stable on the $\cos \theta >0$ branch and exists only when $|B{\hat{u}}^{\prime }/2|<1$.

Put $r=\hat{r}/R$, $w=\hat{w}/W_0$, $n=\hat{n}/N_0$. Subtract the background pressure-driven balance and divide by $\mu W_0/R^2$. The governing system iswhereThe last parameter contains $B$, as in the PDF, not the diffusivity substituted by the converted TeX. Boundary and normalization conditions areThe wall cell-flux condition is already incorporated in $n^{\prime }=\chi n\sin \theta$; there is no prescribed wall concentration or additional fixed-volume-flux condition. The radial divergence of the stresslet stress can be balanced by a radial pressure depending only on $r$. It neither produces radial flow in this ansatz nor changes the imposed axial pressure gradient, so it is eliminated from the axial problem, not assumed to vanish.

For $\gamma \ll 1$, $\sigma =\gamma {\sigma }_1$, let $w=\gamma w_1+O({\gamma }^2)$, $n=n_0+\gamma n_1+O({\gamma }^2)$. The leading tilt hasThe supplied $B{W}_0/R<1$ ensures this leading orientation remains on the stable branch. Put $a=\lambda \chi$. Integrating the concentration equation and imposing the average concentration gives the gyrotactic focusing in a downward pipe flow profileAt $a=0$ the continuous limit is $n_0=1$.

At first order the axial equation isIntegrate from the regular axis, where both boundary terms vanish:Since $\sin 2{\theta }_0=-4\lambda r\sqrt{1-4{\lambda }^2{r}^2}$, the stresslet correction to gyrotactic pipe flow iswith its regular limiting value at $r=0$, andThe orientation and concentration corrections, if needed to complete the expansion, follow without another differential boundary problem:These enforce $2{\int }_0^1{n}_{1}rdr=0$.

For $\lambda >0$, write the two positive integrand coefficients as $F(s)=A(1-e^{-a{s}^2})/(2as)$ and $H(s)=4\lambda s{n}_0(s)\sqrt{1-4{\lambda }^2{s}^2}$. ThenThus $w_1<0$ throughout the interior for ${\sigma }_1\ge 0$, and at any given radius a positive correction is possible only ifA simple sufficient condition for negativity is ${\sigma }_1\ge -1/(8\lambda )$: indeedWhen $\lambda =0$, the stresslet contribution vanishes and the buoyancy correction is negative regardless of ${\sigma }_1$.

Physically, downward shear tilts upward swimmers toward the axis, concentrating excess weight there and enhancing the downward flow. Positive $S$ in the specified $+Sn\mathbf{pp}$ stress convention reinforces this effect; sufficiently negative $S$ opposes it. A positive $w$ means reduced downward velocity, not necessarily reversal of the total pipe flow. The deterministic orientation closure fails at excessive shear, and scalar diffusivity neglects directional and shear-dependent dispersion.

Let $f$ be probability density per unit solid angle, with ${\int }_{S^2}fd\Omega =1$ and isotropic initial condition $f(\mathbf{p},0)=1/(4\pi )$. In still fluid, bottom-heavy spherical-cell orientation dynamics gives $\dot{\mathbf{p}}=B^{-1}[\mathbf{k}-(\mathbf{k}\cdot \mathbf{p})\mathbf{p}]$. Adding isotropic rotational diffusion gives the gyrotactic orientation Fokker-Planck equationThe derivatives are tangential to the unit sphere. Rotational symmetry about $\mathbf{k}$ is preserved by both the equation and the initial condition, so $f$ depends only on $\theta ,t$. Since $\dot{\theta }=-\sin \theta /B$, the polar probability flux is $j_{\theta }=-\sin \theta f/B-D_r{f}_{\theta }$. Expanding ${\partial }_{t}f+{\sin }^{-1}\theta {\partial }_{\theta }(\sin \theta j_{\theta })=0$ yieldsRegular solutions have no probability flux through the poles. The mean swimming velocity is ${\mathbf{V}}_c=V_s⟨\mathbf{p}⟩$, with only a vertical component by symmetry.

Assume a steady, incompressible, inviscid flow with a nearly uniform velocity $u(x)$ across each water section. The channel varies slowly enough for vertical accelerations to be negligible, pressure to be hydrostatic and the water surface to be nearly horizontal across the width. Atmospheric pressure is uniform; neglect surface tension, wall friction and energy losses. The hydraulic equations apply away from jumps and short nonhydrostatic regions.

The Euler equation for horizontal momentum, with hydrostatic pressure, givesIntegrating gives a constant Bernoulli equation head $\mathcal{B}=H+h+u^2/(2g)$. Conservation of mass gives the constant volume discharge $Q=bhu$. Thus the shallow-water specific energy, measured above the local bed, isDefine the Froude number by $F=u/\sqrt{gh}$ for the positive flow direction. ThenAt fixed discharge and width, energy tends to infinity at both zero and infinite depth. It has a unique minimum atIndeed $E_{hh}=3Q^2/(g{b}^2{h}^4)>0$. Above the minimum there are two depth branches: the shallower one is supercritical flow and the deeper one is subcritical flow.

The local long-wave propagation speeds from the shallow water equations are $u\pm \sqrt{gh}$. In subcritical positive flow, one wave family travels upstream and one downstream. In supercritical flow both travel downstream. At critical flow the upstream-directed speed is zero, so the flow can provide hydraulic control separating upstream adjustment from downstream conditions. This statement concerns long waves in the slowly varying hydraulic approximation, rather than all short-wave disturbances.

Let $S>0$ denote volume loss per unit downstream length and take $Q_0>0$. Mass balance gives $Q_x=-S$, soAssume fluid is removed carrying its local axial velocity, with no additional axial force on the retained flow. Under this inviscid seepage approximation, the horizontal momentum equation and its Bernoulli head are unchanged. On the flat bed, hydraulic control with prescribed seepage therefore usesThe calculation applies on a reach with positive remaining discharge. Constant loss cannot be extended indefinitely past $x=Q_0/S$ while preserving this positive-flow model.

At a regular control, $F=1$ and the same differentiated energy condition requires $q_x=0$, where $q=(Q_0-Sx)/b(x)$. Differentiating givesWrite $D=\sqrt{Q_0^2+S^2{L}^2}$. The two algebraic roots are $(Q_0\pm D)/S$. At the plus root $Q=-D<0$, so it is outside the stipulated positive-flow branch. The admissible control candidate isIt lies upstream of the geometric throat. At this point $Q(x_c)=D$ andIt is a maximum of positive $q$: differentiating the numerator of $q_x$ at $x_c$ gives $2S{x}_c-2Q_0=-2D<0$. ConsequentlyA control exists only if the reach includes this point and boundary conditions supply the critical conserved head $\mathcal{B}=3h_c/2$ and a suitable transcritical connection. Merely prescribing $Q_0$ and $S$ does not force that head. At $S=0$ the limiting location is zero and the usual throat depth is recovered.

In the same idealized momentum model, write $S=-|S|$. The discharge now increases downstream. The admissible maximum of $Q/b$ moves to $x_c>0$, beyond the geometric throat, because added discharge initially offsets the widening channel. The formula $x_c=-S{L}^2/(Q_0+D)$ remains valid, and the critical depth depends on $S^2$ through $D$. The positive-flow reach is now restricted upstream by $Q_0+|S|x>0$, rather than downstream by exhaustion of the discharge. The same critical-head and boundary-connection conditions are still needed.

Actual injection introduces a further physical issue: newly added fluid need not have the stream's axial velocity. Let its axial velocity be $u_a$ and let $A=bh$. The momentum correction for lateral mass addition isRemoval at the local velocity, or ideal co-moving addition, gives the conserved-head analysis above. Addition at rest instead accelerates new fluid and decreases the head. It cannot be modeled solely by changing the sign of $S$ in a constant-head calculation.

For example, flat-bed addition at rest gives $(1-F^2)h_x=F^{2}h[b_x/b-2Q_x/Q]$. Its regular-control compatibility is $b_x/b=2Q_x/Q$, rather than the co-moving condition $b_x/b=Q_x/Q$; in the parabolic channel it gives the different candidate $x_c=-S{L}^2/Q_0$. Thus the sign reversal moves the ideal control downstream, but its quantitative prediction depends on the momentum carried by the injected water. The source specifies the volume rate only, so an injection-velocity assumption must accompany any definite constant-energy answer.

Interpret ${\rho }_2$ as the ambient density: the PDF's phrase assigning it a “depth” is a typographical error. Take $0<{\rho }_1-{\rho }_2\ll {\rho }_2$, and use a common reference density in inertia while retaining the density difference in buoyancy. This is the Boussinesq approximation, with reduced gravity $g^{\prime }=g({\rho }_1-{\rho }_2)/{\rho }_{\mathrm{ref}}>0$. For the shallow water approximation, the lower-layer depth must be small compared with longitudinal variation scales, vertical accelerations must be negligible and pressure hydrostatic. The deep ambient is approximated as a hydrostatic pressure reservoir away from the nose; assume a uniform velocity over each lower-layer cross-section and neglect drag and entrainment.

Write the prismatic triangular area as $A(h)=k{h}^2$ with fixed $k>0$, so the width at height $z$ is $2kz$. The excess hydrostatic pressure force divided by reference density isThe prismatic-channel shallow water equations are thereforeFor positive depth, divide the first equation by $2kh$ and use it in the second to obtainThe coefficient matrix in variables $(h,u)$ is $\left(\begin{array}{cc}u& h/2\\ g^{\prime }& u\end{array}\right)$. Its two distinct real characteristic speeds areThus the system is strictly hyperbolic for $h>0$ and $g^{\prime }>0$, though it degenerates at zero depth. Since $dc/dh=c/(2h)$, direct substitution in the two equations givesHence $R_{\pm }=u\pm 4c$ are the Riemann invariants conserved along their respective characteristic curves. The factor four, rather than the rectangular-channel factor two, comes from the triangular relation $A\propto h^2$.

The layer equations describe long-wave adjustment behind the nose, but do not resolve the short region where the dense fluid pushes aside the ambient. A finite-height front cannot be determined by simply adjoining a vacuum solution of the one-layer momentum equation. The ambient pressure and turning flow provide additional nose physics.

In the nose frame, the material condition makes the adjacent current velocity equal to the front speed in the laboratory frame. The two interior characteristic speeds relative to that front are then $\pm c_f$: one characteristic brings information from the body and one brings a boundary condition from the nose. This is why a gravity-current front condition is needed. With a deep ambient, fixed channel shape and negligible viscosity, the local buoyancy depth supplies the velocity scale $\sqrt{g^{\prime }{h}_f}$. A dimensionless closure therefore has the formwhere a constant $F$ is an idealization for a fixed geometry and chosen nose model. Dimensional analysis identifies the form but does not determine the coefficient.

To determine an appropriate coefficient theoretically, take a control volume moving with the nose and bounded by sections far enough ahead and behind for hydrostatic pressure and approximately uniform section velocities to apply. Use the actual triangular wetted areas and integrated pressures. Conservation of fluid volume fixes the ambient return-flow velocities; integrated horizontal momentum balances the incoming and outgoing momentum fluxes against the pressure forces. A Bernoulli relation on an ambient streamline, with an explicit nose-loss assumption if needed, supplies the pressure or energy closure. The density-anomaly pressure contribution of the lower layer contains the integral $g^{\prime }k{h}_f^3/3$ derived above. Solving those balances gives $u_f^2/(g^{\prime }{h}_f)$ and hence $F^2$.

Alternatively, measured front speeds and consistently defined adjacent heights in a geometrically similar experiment can calibrate the closure. A rectangular-channel numerical coefficient must not be transferred to a triangular channel without checking those balances. The height definition and dissipative nose structure also matter; a primary study of gravity-current front conditions explains why different height conventions can lead to different inferred coefficients. No numerical value of $F$ is required for the following solution.

Let $c_0=\sqrt{g^{\prime }{h}_0/2}$, as printed in the PDF. The converted TeX misrenders this expression. The undisturbed reservoir has $u=0$, $c=c_0$. The expansion fan moving into it is a simple wave with $R_{+}=u+4c=4c_0$ constant. On the other characteristic family, a self-similar fan satisfiesSolving these two equations givesThe fan begins at $\xi =-c_0$, where it matches the initially resting reservoir.

At the front, write $k_F=\sqrt{2}F$. The nose closure is $u_f=k_F{c}_f$ and the invariant gives $u_f+4c_f=4c_0$. ThereforeA uniform region of this state connects the right edge of the fan, ${\xi }_n=u_f-c_f$, to the advancing front $x_f=u_{f}t$. Its width is $c_{f}t>0$. The complete lower-layer solution for $t>0$ isandAhead of the front there is no lower-layer current, so its velocity is not a physical variable there. Both transitions inside the current are continuous; the finite-height nose is closed by the ambient-front condition, not by an unsupported vacuum jump. These formulas give the front-regularized triangular-channel dam break for any specified positive constant $F$.

Initially the same expansion fan and constant-speed front apply. Its left edge reaches the closed rear wall at $t_w=x_0/c_0$. After that, the infinite-reservoir solution would give a nonzero velocity at the wall and violates the no-through-flow condition. A reflected disturbance adjusts the interior while the finite initial volume is progressively spread over a longer, shallower current.

The nose does not respond instantaneously when the fan reaches the wall. Its first return signal follows a $+$ characteristic from $(x,t)=(-x_0,t_w)$ through the undisturbed fan. There $u+c=(8c_0+3x/t)/5$, so integrating $dx/dt=u+c$ givesIt reaches the uniform nose-adjacent region at $t_n=t_w(c_0/c_f{)}^{5/2}$, where $x=(u_f-c_f)t_n$. Crossing that region at speed $u_f+c_f$, it catches the front at $2t_n$. Thus the initial nose remains uninformed about the rear wall until this return time. Subsequently its speed must adjust rather than remain constant indefinitely.

For late times use a prismatic triangular-channel gravity-current box model with uniform representative depth $\bar{h}$ over length $\ell =x_f+x_0$. The initial volume is $V=k{h}_0^2{x}_0$. With no entrainment or loss,Apply the same front closure using that representative depth:Starting from a matching late-time state $(t_b,{\ell }_b)$, integration gives finite-volume inertial spreading in a triangular channel:Hence $\ell \propto t^{4/5}$, $\bar{h}\propto t^{-2/5}$ and the nose speed decays like $t^{-1/5}$. A simple initialized box approximation can instead set ${\ell }_b=x_0$ and $t_b=0$, but it then deliberately replaces the exact early rarefaction rather than reproducing it.

The box's volume equation is compatible with a velocity increasing linearly from zero at the rear wall to the nose speed: $u(x,t)=\dot{\ell }(x+x_0)/\ell$. A flat-depth box with this velocity is not an exact solution of the full momentum equation during deceleration. It is an integral approximation that leaves the interior shape and nose dynamics unresolved while predicting late-time bulk scaling.

Write $\nu =\mu /\rho$ and average the horizontal Navier-Stokes equation, with zero-mean fluctuations and incompressible flow. The mean flow is stationary, has no vertical velocity and has no horizontal pressure gradient. The mean advection of $U(z)$ is therefore zero. Incompressibility rewrites the averaged nonlinear fluctuation term as a divergence:These correlations transport mean momentum and produce Reynolds stress. Horizontal statistical homogeneity makes the horizontal derivatives of the averaged correlations vanish. It does not require the instantaneous turbulent derivatives themselves to vanish. Only the vertical divergence remains, givingMultiplying by density and integrating once gives constant-stress Reynolds-averaged wall flow:Here ${\tau }_d$ is physical shear stress. Positive mean shear commonly has $\overline{u^{\prime }{w}^{\prime }}<0$, representing downward transport of positive horizontal momentum. The Reynolds term then contributes positively to the total stress.

The printed target formula has an extra vertical derivative in the Reynolds contribution. That term has dimensions of stress per length, while $\mu U_z$ has dimensions of stress, so the formula as printed is inconsistent. The derivative belongs in the stress-divergence equation above; the boxed integrated stress is the corrected result. Part b also uses a kinematic rather than physical stress, so the density factor must be tracked explicitly.

Define the friction velocity $u_{\ast }=\sqrt{{\tau }_d/\rho }$ for positive wall shear. In the near-wall region turbulence is suppressed and molecular viscosity carries the stress. With no slip, $U\simeq u_{\ast }^{2}z/\nu$. The viscous wall coordinate $z^{+}=z{u}_{\ast }/\nu$ becomes order one at the characteristic viscous sublayer lengthThe numerical extent of viscosity dominance depends on the crossover convention and wall condition; the scaling is the required viscous length.