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A persistent barrier to whole-mantle circulation adds an internal thermal bottleneck and permits the deep mantle to retain a larger temperature excess. In a single well-mixed layer there is one interior temperature and one surface heat-loss law. The coupled thermal histories of layered mantle convection arise because separately mixed layers can have different temperatures, exchanging heat through an interfacial boundary layer as well as losing heat at the surface. Their thermal histories then have coupled relaxation modes rather than the single mode in (c).

For example, let ${\theta }_l,{\theta }_u$ be lower- and upper-layer temperature excesses, $H_l,H_u$ their heat capacities per area, andWith no internal heating the budgets are $H_l{\dot{\theta }}_l=F-J$ and $H_u{\dot{\theta }}_u=J-E$. At steady state,Thus a fixed basal flux still leaves the surface eventually with $E=F$, but the lower layer is hotter than it would be in the single-temperature model. For a given deep temperature instead, the extra interfacial drop reduces outward heat transfer. This favours delayed loss of deep primordial heat and distinct upper- and lower-mantle cooling histories. A numerical change in cooling time cannot be obtained from layer count alone: interface transport, layer thicknesses, temperature-dependent viscosity, internal heating, and whether material can cross the interface matter. Merely replacing $d$ by $d/2$ in (c) misses the added interface and is not a model of two-layer cooling.

Let $\delta =R{e}^{-1/2}$, stretch $Y=y/\delta$ and write the original transverse velocity as $v=\delta V$. Continuity then gives $u_x+V_Y=0$. In the streamwise momentum equation $u_{yy}/Re=u_{YY}$ is order one, while $u_{xx}/Re$ is negligible. The transverse equation gives $p_Y=0$ to leading order, so pressure is supplied by the outer Euler flow. Its streamwise balance is $p_x=-U{U}^{\prime }$. ConsequentlyFor a stationary impermeable plate, $\boxed{u(x,0)=V(x,0)=0}$, with matching $\boxed{u(x,Y)\to U(x)}$ as $Y\to \infty$. The transverse velocity must match the outer displacement flow; it is generally wrong to impose $V\to 0$ independently at infinity.

At the leading edge the wall no-slip data meet the external stream abruptly, and the boundary-layer approximation is nonuniform. For a constant external speed, the thickness is $O(\sqrt{x/Re})$, whose ratio to $x$ ceases to be small when the local Reynolds number is order one. A leading-edge region retaining neglected derivatives is needed there. An adverse pressure gradient can also cause separation farther downstream; a large global Reynolds number alone does not prevent it.

The Von Mises boundary-layer transformation uses $\Psi$ as transverse coordinate. With $u=U(x,\Psi )$ and ${\Psi }_x=-V$, ${\Psi }_Y=u=U$, the chain rule givesThe streamwise advection terms reduce to $u{u}_x+V{u}_Y=U{U}_x$, so, for $Y>0$ where $U>0$, division by $U$ givesContinuity was already enforced by the stream function. Substitute $U=g(\eta )$ with $\eta =\Psi /\sqrt{x}$. Then $U_x=-\eta g^{\prime }/(2x)$ and ${\partial }_{\Psi }(U{U}_{\Psi })=(g{g}^{\prime }{)}^{\prime }/x$, yieldingThe wall coordinate is zero because $\Psi$ was set to zero there, and the far condition matches unit external speed. This is the transformed Blasius similarity problem. One must not additionally prescribe a finite $g^{\prime }(0)$: the coordinate change degenerates at the no-slip wall, and the regular physical shear corresponds to $g\propto \sqrt{\eta }$ near zero. This similarity reduction is obtained for the attached monotone boundary layer assumed in the question.

Existence of a positive normalized profile can also be seen directly. Put $q=g{g}^{\prime }$ and choose $q(0)=a>0$. In $g$ coordinates the regular initial-value system is $dq/dg=-\eta /2$, $d\eta /dg=g/q$, with $\eta (0)=0$. It starts as $g\sim \sqrt{2a\eta }$. For finite positive $\eta$, $q^{\prime }/q=-\eta /(2g)$ keeps $q$ positive and $q\le a$, while $(g^2{)}^{\prime }=2q$ gives $g\le \sqrt{2a\eta }$. Consequently $q^{\prime }/q\le -\sqrt{\eta }/(2\sqrt{2a})$, so $q$ is integrable at infinity. It follows that $g$ increases to a finite positive limit $M$. The rescaled profile $\tilde{g}(\eta )=M^{-1}g(\sqrt{M}\eta )$ satisfies the same equation and has limit one. This supplies the requested similarity solution with both boundary conditions, rather than only its formal differential equation.

Take the tangential length and velocity scales as order one. In the boundary layer, continuity reads $u_x+({\delta }_2/{\delta }_1)V_Y=0$, so a nontrivial balance requires ${\delta }_2\sim {\delta }_1$. Tangential inertia is order one, and the normal viscous term has size ${\mathrm{Re}}^{-1}{\delta }_1^{-2}$. Balancing them givesup to constants absorbable in the scaled variables. Streamwise diffusion is then smaller by ${\mathrm{Re}}^{-1}$. The normal momentum equation has a pressure term of size ${\delta }_1^{-1}{p}_Y$ and inertial/viscous terms of size ${\delta }_1$, hence $p_Y=0$ at leading order. The remaining Prandtl boundary-layer equations areThe boundary conditions are no slip and no penetration at the plate, and matching to the outer tangential velocity:The outer inviscid tangential momentum equation supplies $p_x=-U_s{U}_s^{\prime }$; pressure is shared between the boundary layer and outer flow because it has no leading normal variation.

An attached similarity profile must approach $f^{\prime }\to A$ as $\eta \to \infty$. Linearizing about $f\sim A\eta +\text{constant}$ gives, for the small velocity correction $g=f^{\prime }-A$,When $\alpha A>0$, one independent far-field correction decays with a Gaussian factor; when $\alpha A<0$, the two nontrivial asymptotic modes grow, one algebraically and one exponentially, so a nonconstant wall profile cannot decay into the prescribed outer flow. Thus the attached exponential similarity construction requiresEquivalently the outer speed increases along the direction of motion, giving a favorable pressure gradient. In the intended flow away from the leading edge along positive $x$, this means $A>0$ and $\alpha >0$. The sign-reversed choice $A<0,\alpha <0$ describes locally favorable flow in the opposite direction; it is not a layer developing downstream from the stated leading edge.

The asymptotic approximation also requires a geometrically thin layer, away from the leading-edge region. Comparing its thickness with the streamwise variation length $1/|\alpha |$ gives the local condition $\mathrm{Re}|A|e^{\alpha x}/|\alpha |\gg 1$. An exponentially decaying outer speed eventually violates this condition on an infinite downstream interval. These scale and orientation requirements distinguish formal similarity reduction from a uniformly valid attached boundary layer.

For a constant-mass density incompressible flow with no imposed body force, the two-dimensional Navier-Stokes equation isA conservative body force, if retained, may be incorporated into a modified pressure. Using scales $L,U_0,L/U_0$ and $\rho U_0^2$ givesHere $\nu =\mu /\rho$ is the molecular kinematic viscosity. At high Reynolds number, inertia and the pressure gradient dominate on the outer length scale, while molecular viscous diffusion is small there. Thin boundary layers, narrow shear flows and fine turbulent scales can still have sufficiently large velocity gradients for viscous effects to matter. High Reynolds number permits instability and turbulence, but does not alone prove that every flow is turbulent. Molecular viscous dissipation remains essential at the small end of the turbulent cascade.

Let $\delta \ll L$ be a representative mixing-layer thickness, ${ϵ}_m=\delta /L$, and let $U_s$ be a representative streamwise velocity. Use the scalesThe transverse velocity scale follows from the mean continuity equation. For turbulent transverse transport to balance streamwise inertia,With this ordering, the streamwise eddy-stress divergence isThus transverse diffusion of streamwise linear momentum is retained, while streamwise diffusion and the $V_x$ contribution are smaller by ${ϵ}_m^2$. Molecular kinematic viscosity is neglected where ${\nu }_T\gg \nu$.

In the transverse momentum equation, inertia and the retained eddy stresses have scale ${ϵ}_m{U}_s^2/L$, whereas an order-one transverse pressure variation would give $P_y/\rho \sim U_s^2/({ϵ}_{m}L)$. Hence the leading dimensionless transverse balance is $P_y=0$. The resulting boundary layer equations for the planar turbulent mixing layer areHere ${\nu }_T={\mu }_T/\rho$. The effective pressure convention explained in part (a)(iii) is understood. Since the outer streams have uniform horizontal velocities, their inviscid momentum equations give $P_x=0$; choosing the far-field gauge pressure as zero gives $P=0$ throughout this approximation.

Assume $U_c=(U_a+U_b)/2>0$ and set $\Delta U=U_a-U_b>0$. With ${\nu }_T=C\Delta U\delta (x)$ independent of $y$, balancing streamwise transport with transverse mixing gives a layer width proportional to $x$. Write its constant slope asHere $S>0$ is a spreading constant associated with the chosen definition of thickness; it is unrelated to the capillary $S$ in Question 1. The origin $x=0$ selects zero virtual-origin offset.

Define $a=\sqrt{S/C}$, $\xi =ay/\delta$, and introduce the stream functionThis automatically satisfies incompressibility and yieldsFor the momentum equation, use ${\xi }_x=-\xi {\delta }^{\prime }/\delta$ and ${\xi }_y=a/\delta$:Because $U_c^2{\delta }^{\prime }=S\Delta U{U}_c$, the common coefficient cancels, giving the turbulent mixing-layer similarity equationThe boundary layer approximation also requires $S\Delta U/U_c\ll 1$, consistently with the assumed slow spreading.

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.

At the interface, the tangential velocity is $u(x,0)=x{f}^{\prime }(0)$ and the normal velocity remains zero. The preceding error function expression givesHence, within the leading boundary layer model,Allowing the formally discarded finite-Reynolds number terms gives an additional order-${\delta }^2$ error, smaller than the order-$\gamma$ correction under the assumed ordering. The coefficient is $\sqrt{\pi /8}\approx 0.626657$.

Thus positive surface shear increases the outward surface speed linearly: for $x>0$ it increases $u$, and for $x<0$ it makes $u$ more negative. The stagnation point remains at $x=0$. A negative coefficient would result from using the printed $+\infty$ condition literally, and would describe the different half-line problem identified in part (ii).

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.

In a stationary, horizontally homogeneous wall boundary layer, Hamiltonian ray-tracing equations conserve $\omega$ and $k_x$. If $U$ increases away from the wall, downstream-oriented $k_x>0$ waves have decreasing positive intrinsic frequency and decreasing $k_y^2$. At $U=\omega /k_x-c_0$, the transverse wavenumber vanishes and the acoustic ray turns. For $U(0)=0$, a downstream launch angle ${\beta }_0$ above the wall can reach a uniform stream of Mach number $M_{\infty }$ only if $\cos {\beta }_0\le (1+M_{\infty }{)}^{-1}$. This uses the positive acoustic branch, constant sound speed and no intervening critical layer; a turning point itself requires a uniform local approximation rather than the ordinary WKB method.

For a positive small parameter $ϵ$, an order-one mismatch between the reduced solution $y_0=f(x)$ and a prescribed right endpoint value $y(1)=b$ creates a right boundary layer. Balancing $ϵy^{\prime \prime }$ against the order-one reaction term gives width $\sqrt{ϵ}$. In the inner variable $X=(1-x)/\sqrt{ϵ}$, the leading equation is $Y_0^{\prime \prime }=Y_0-f(1)$. Its boundary and matching conditions are $Y_0(0)=b$ and $Y_0\to f(1)$ as $X\to \infty$, soThe leading additive composite expansion is $f(x)+(b-f(1))e^{-(1-x)/\sqrt{ϵ}}$, obtained by subtracting the common overlap value. For the nonlinear perturbation $ϵy^{\prime \prime }=y-x^2+ϵ(2+y^2)$ with zero endpoint values, the first two outer expansion terms are $x^2-ϵx^4$, and the leading right-layer term is $1-e^{-X}$.

A shear instability is growth of disturbances that draw kinetic energy from spatial variation of a base-flow velocity. The energy-transfer term involves the mean velocity gradient and perturbation velocity correlations. A velocity gradient alone does not guarantee instability: viscosity, stratification, boundaries and the profile shape matter. Kelvin-Helmholtz instability is a central interfacial example. For a smooth inviscid parallel shear flow, Rayleigh's inflection-point theorem makes an interior inflection point necessary for an exponentially growing two-dimensional normal mode; it is not sufficient. Applying this local parallel-flow criterion to a slowly varying boundary layer is a stability heuristic rather than a complete viscous stability calculation.

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

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

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

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

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.

For fixed $k>0$ and endpoint values $y(0)=0$, $y(1)=1$, the outer solution is $y\sim {\epsilon }^2{k}^2$. The left boundary layer has width $\epsilon$. At the right, a layer of width $\sqrt{\epsilon }$ carries the order-one rise, and a translated layer of width $\epsilon$ connects its inner edge to the small plateau. The dominant balance depends on both amplitude and derivative scale; using the same scale at both ends loses a necessary match.

For smooth radial variation on scale $r$, $c_s\sim \Omega H$, $\nu \sim \alpha \Omega H^2$, $u_r\sim \nu /r$ and $u_z\sim H{u}_r/r$. The exact point-mass vertical gravity differs from ${\Omega }_K^{2}z$ by relative order $(H/r{)}^2$. Radial pressure support changes the rotation and dominant shear heating by the same order. Meridional viscous forces, radial radiative-flux divergence, thermal advection and compressional work are at most order $(H/r{)}^2$ relative to their retained vertical terms for $\alpha \lesssim 1$. These estimates require secular disk evolution and exclude sharp radial boundary layers or independently rapid vertical motion.

// Target: gravitational-instability-of-an-astrophysical-disk.bigb

A boundary layer in which turbulent velocity fluctuations transport momentum and scalars. A near-wall constant-stress region supports the law of the wall; buoyancy modifies it through Monin-Obukhov similarity theory. A viscous or roughness sublayer and an outer-layer scale bound the range of either approximation.

An approximation is uniform on a specified domain when its error bound holds throughout that domain as the small parameter tends to zero. In a matched asymptotic expansion, adding the inner and outer approximations and subtracting their common overlap produces a composite approximation. Checking only a fixed point away from a boundary layer does not establish uniformity.

A viscous boundary layer is a boundary layer next to a surface in which viscosity remains important even when the outer flow has a large Reynolds number. It reconciles the outer inviscid velocity with the no-slip boundary condition.