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For an incompressible flow, write the Newtonian fluid stress tensor as $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation gives ${\partial }_j{\sigma }_{ij}=0$. Symmetry of the Cauchy stress tensor therefore givesIntegrating and applying the divergence theorem expresses the viscous dissipation as boundary power:At a moving rigid body, the surface velocity is $U+\Omega \times (x-x_c)$. Its boundary-power contribution is $U\cdot F+\Omega \cdot G$, with the force and torque evaluated using the fluid's outward normal vector. Both resultants vanish for a force-free, torque-free inclusion. Thus the new viscous dissipation comes entirely from the unchanged outer boundary velocity. Subtracting the particle-free boundary power givesApply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the fluid outside the inclusion. On the outer boundary $u^{\prime }=0$, soWriting $n_p=-n$ for the particle's outward normal vector gives the required extra dissipation due to a rigid inclusion:This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local viscous dissipation.

For the sphere, the ambient rate-of-strain tensor $E$ is symmetric and trace free. The sphere in a uniform straining Stokes flow has zero translational and rotational velocity: inversion symmetry eliminates its force, and symmetry of $E$ eliminates its torque. To derive the disturbance, put $r=|x|$, $Q=x\cdot Ex$ and seekThe incompressible flow condition and Stokes equation reduce toA decaying family satisfying these equations is $f=A{r}^{-5}$, $g=B{r}^{-5}-5A{r}^{-7}/2$, $j=2B{r}^{-5}$. The no-slip boundary condition requires $f(a)=-1$ and $g(a)=0$, so $A=-a^5$ and $B=-5a^3/2$. ConsequentlyThe total velocity is zero at $r=a$, and the disturbance decays as $r^{-2}$. The pressure constant has been set to zero. These boundary and far-field conditions, together with Uniqueness of Stokes flow, establish the solution. There is no rotational background in a pure strain flow.

On the sphere $u^{\prime }=-u_0=-aE{n}_p$. Contracting the supplied Newtonian fluid stress tensor with $n_p$ makes its two terms proportional to $(n_p\cdot E{n}_p)n_p$ cancel, leaving $\sigma n_p=5\mu E{n}_p$. Hence the extra dissipation due to a rigid inclusion isHere ${\int }_{S^2}{n}_j{n}_{k}d\Omega =4\pi {\delta }_{jk}/3$. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.

If $N_v$ denotes the number of spheres per unit volume, $N_v(4\pi a^3/3)=\varphi$. ThusAdd this to the particle-free viscous dissipation density $2\mu E:E$. Comparing with $2{\mu }_{\mathrm{eff}}E:E$ gives the Einstein viscosity formula for a dilute suspension:to first order in $\varphi$. The hydrodynamic interactions neglected here contribute beyond that dilute order.

For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.

For the requested sedimentation drift of a weighted two-rod body sketch, a full parametric trajectory follows by also integrating $U_v/\dot{\theta }$. Put $\psi =\theta -\pi /4$ and set the initial height to zero:It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda <1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda >1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda )mg/(3CL)$.

Take $\Delta \rho >0$ and $\tau >0$, with $x$ pointing along the imposed shear stress, and put $G=\Delta \rho g$. Relative to the upper fluid's hydrostatic reference, the hydrostatic pressure in the current is $p=G(h-z)$. Thus the horizontal pressure gradient is $G\nabla h$. The lubrication theory momentum balance, lower no-slip boundary condition, and imposed upper shear stress giveIts depth-integrated volume flux isDefine $D=G/(3\mu )$ and $A=\tau /(2\mu )$. The continuity equation gives the shear-driven viscous gravity current equationThe two terms have opposite roles: imposed shear stress carries fluid downstream, whereas the hydrostatic pressure gradient spreads it from thick regions towards thin ones.

For the approximation, a representative horizontal velocity is $U\sim \tau H/\mu +G{H}^3/(\mu L)$, with vertical velocity $UH/L$. Small slopes require $H/L\ll 1$. Horizontal fluid inertia relative to vertical viscous resistance is ${\rho }_{c}U{H}^2/(\mu L)$, where ${\rho }_c=\rho +\Delta \rho$. Hence sufficient small parameters, expressed without an unknown velocity, areFor the hydrostatic pressure and normal-stress approximation also require $\mu U/(GHL)\ll 1$. Its gravity-driven part is already $(H/L{)}^2$; its shear-driven part addsThis condition controls viscous normal stress, vertical viscous corrections and the normal projection of shear at a slightly tilted interface relative to $GH$. With it and the previous inertia bounds, vertical inertia is small as well. Time variations here are on the transport timescale $L/U$; separately imposed rapid forcing would need its own unsteady inertia bound. A small ordinary Reynolds number is a stronger sufficient restriction, but the reduced inertia ratios above are what the thin-layer momentum balance directly requires.

In the steady far-downstream regime, cross-stream derivatives dominate gravity-driven spreading. Dropping the smaller downstream gravity flux leavesWrite $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim (D/A)h_c^{2}x$, while conservation of the source volume flux gives $Q\sim A{h}_c^{2}Y$. Therefore the downstream similarity of a shear-driven gravity current hasThe omitted downstream gravity flux relative to the imposed shear flux is $D{h}_c^2/(Ax)\sim (Y/x{)}^2\ll 1$, so this approximation is self-consistent in the stated regime.

For the full profile put $w=h^2$. Then $h^3{h}_y=(w^2{)}_y/4$ andThis is a porous medium equation with downstream distance as its evolution coordinate. Conservation of $\int wdy=Q/A$ and a similarity solution $w=x^{-1/3}f(\eta )$, $\eta =y/x^{1/3}$, givewhere symmetry gives zero integration constant at $\eta =0$. Inside the positive region, $f^{\prime }=-2A\eta /(3D)$, so $f=(A/(3D))({\eta }_N^2-{\eta }^2)$. Requiring the total downstream volume flux to be $Q$ fixes the coefficient:Consequentlyfor $|y|<y_N$, with $h=0$ outside. The squared height is a parabolic Barenblatt solution; the height cross-section is a semicircular profile after rescaling its axes. The cross-stream volume flux vanishes at the edges even though the height slope becomes singular there. The profile describes the outer lubrication theory region, not a resolved microscopic front.

Near the point source the two horizontal dimensions are comparable, say $\ell$. Upstream spreading arrests where outward gravity-driven volume flux balances downstream shear-driven volume flux:Eliminating $h_{\ast }$ givesThis is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_{\ast }\sim (\mu Q/(\Delta \rho g){)}^{1/4}$. The same horizontal scale is obtained by setting $y_N\sim x$ in the downstream similarity solution, confirming where that solution fails.

For the line source the steady volume flux per unit width is $q=A{h}^2-D{h}^3{h}_x$. On the downstream constant-height branch, $q=Q_{2d}$ gives $h_d^2=Q_{2d}/A$. Upstream there is no net flux through the finite nose, so $q=0$. Thus $h{h}_x=A/D$, and continuity of height at the source givesThe line-source upstream reach under imposed shear is thereforeIn the upstream part, shear-driven and gravity-driven volume fluxes cancel. The volume flux per unit width jumps by exactly $Q_{2d}$ at the source. As in the point-source profile, the square-root nose is a formal outer solution with an unresolved steep front; its finite upstream reach follows from flux balance, not from assuming a small slope all the way to the nose.

Take a steady parallel velocity $(u(y),0)$ and interpret the specified gradient as $p_x=G>0$. The horizontal Navier-Stokes equation reduces to $\mu u^{\prime \prime }=G$: the pressure force opposes positive $x$ motion. The no-slip boundary condition imposes $u(0)=0$, $u(h)=U$. Integrating twice givesThe velocity gradient is $u^{\prime }=G(y-h/2)/\mu +U/h$, an increasing function of $y$. It is negative somewhere in the channel exactly when its lower-wall limit is negative:Under this condition the negative-gradient interval is $0<y<h/2-\mu U/(Gh)$.

The volume flux per unit span isThus the zero-flux Couette-Poiseuille flow requiresThe profile starts at zero, reaches $u=-U/3$ at $y=h/3$, crosses zero at $y=2h/3$, and reaches $U$ at the moving wall. Negative and positive flux contributions cancel, even though the local velocity is nonzero.

Keep the prescribed outer slip velocity fixed when perturbing the unsteady Prandtl equation. Cancelling the base equation and retaining terms linear in the disturbance gives the linearized unsteady Prandtl equationThe no-slip boundary condition supplies $\tilde{u}=\tilde{v}=0$ at the wall; the fixed outer velocity supplies $\tilde{u}\to 0$ at infinity. There is no prescribed zero normal-velocity disturbance at infinity: a finite displacement-related value is permitted.

With the frozen parallel profile $U=U(y)$, $V=0$, the coefficients are invariant under translations in $x$ and $t$. Normal modes therefore separate those variables. For the mode convention used here the equations areEliminate $u=-v^{\prime }/(i\alpha )$ to obtain the Prandtl normal-mode equationwith a bounded far-field $v$. Requiring $v(\infty )=0$ would incorrectly eliminate the proposed outer profile. For real $U$ and real $\alpha$, complex conjugation of the equation and its boundary conditions replaces $(\alpha ,c,v)$ by $(-\alpha ,c^{\ast },v^{\ast })$. This proves the stated eigenvalue symmetry. The paired modes have the same temporal growth rate $\alpha \mathrm{Im}c$, so choosing positive wavenumber loses no real physical disturbance.

Here the prescribed parallel profile is a local frozen-coefficient model. A nontrivial arbitrary $U(y)$ with $V=0$ and constant outer slip is not generally an exact steady solution of the unforced unsteady Prandtl equation; its base equation would require $U^{\prime \prime }=0$. The following stability calculation uses the simplifying local model stipulated for the mode analysis.

Set $c_0=U_c=U(y_c)$. Away from the critical point the dominant equation is $(U-c_0)v_0^{\prime }-U^{\prime }{v}_0=0$, whose solutions are multiples of $U-c_0$. Choosing the coefficient to be zero below $y_c$ and one above it satisfies the wall and tangential far-field conditions. The choice $c_0=U_c$ makes $v_0$ continuous across the joining point; $U^{\prime }(y_c)=0$ also makes its first derivative continuous. Its second derivative jumps. Thus it is a valid leading outer solution away from $y_c$, but viscosity must smooth the join in a critical layer in a shear flow.

Near the join, $U-c_0\sim (y-y_c{)}^2/2$. If the layer width is $\ell$, matching gives $v=O({\ell }^2)$. The advection side of the mode equation scales as $\alpha {\ell }^3$ and $v^{\prime \prime \prime }$ scales as ${\ell }^{-1}$. Balance gives ${\ell }^4\sim {\alpha }^{-1}=\epsilon /k$. Equivalently, the eigenvalue correction ${\epsilon }^{1/2}{c}_1$ balances $(y-y_c{)}^2$, so the quarter-power critical layer hasSubstitution, retaining the leading terms and cancelling their common factor, gives

At order ${\epsilon }^{1/2}$ outside the layer,A solution consistent with the chosen lower branch and the wall conditions is $v_1=0$ below the join. Above it a particular solution is $v_1=-c_1$; a multiple of $U-c_0$ represents an arbitrary amplitude renormalization and may be set to zero. Matching consequently requiresThe corresponding derivatives match as $w_Y\sim Y$, $w_{YY}\to 1$ on the positive side and as zero on the negative side. The matching statements require that growing homogeneous corrections be absent.

To symmetrize these different end conditions and remove $k$ and $i$, chooseThe polynomial $A(z)=z^2/2-C$ is itself a solution of the transformed homogeneous equation. Direct substitution, using $a^4=ik$, givesThe derivative is third order, as required by the original mode equation. Initially the two ends are along the rotated rays $z=aY$ with real $Y$. Continuing them to the real $z$ axis is compatible with the asymptotic end conditions: the decreasing homogeneous correction behaves exponentially as $\exp [-z^2/(2\sqrt{2})]$ and still decreases throughout the rotation from angle $\pi /8$ to zero. Thus the supplied real-axis spectral normalization uses the same recessive conditions; a complex coordinate change should not be mistaken for a real stretching alone.

For each supplied real eigenvalue $C_n=(4n+7)/\sqrt{2}$, the phase-speed correction isThe mode factor has temporal magnitude $e^{\alpha \mathrm{Im}ct}$. ThereforeAmong the listed indices, $n=-2$ gives positive growth $\gamma \sim \sqrt{\alpha }/2$; $n=-1,1,2,\dots$ give decay. Hence the frozen non-monotone flow has high-wavenumber instability of a non-monotone Prandtl layer, with arbitrarily rapid linear growth as the positive wavenumber increases. The normal-mode calculation demonstrates linear instability; it is not by itself a claim about the final nonlinear state.

Scale time by ${\omega }^{-1}$, length by $k^{-1}$, velocity by $\omega /k$, pressure by $\mu \omega$, and streamfunction by $\omega /k^2$. Write $A=\alpha /k$, $\xi =x-t$, and drop stars from dimensionless coordinates. The surface is $y=ϵ\sin \xi$ and its material velocity in the sheet frame is $(0,-ϵ\cos \xi )$. The no-slip boundary condition is thereforeThe stationary matrix identifies a preferred frame. In the sheet frame it translates with $\mathcal{U}{\mathbf{e}}_x$, so matrix-relative Brinkman velocity gives the consistent equationsEquivalently one may keep the printed right-hand side $A^2\mathbf{u}$ after defining $\tilde{p}=p-A^2\mathcal{U}x$; then the far-field pressure has the gradient needed to balance that term. Using the printed equation with a uniform far-field flow and no such adjustment is inconsistent. The distinction first matters at second order, since the first-order swimming speed is zero. We solve the upper half-space; the lower half-space is its reflected counterpart.

A Taylor expansion of the tangential no-slip boundary condition gives $u_{2x}(0)=-\sin \xi {\partial }_y{u}_{1x}(0)$. Since $f^{\prime \prime }(0)=-s$,For matrix-relative Brinkman velocity, the mean second-order tangential flow isThe mean tangential traction has no first-order geometric remainder: at the surface ${\partial }_y{\sigma }_{xy,1}=s(s+1)\sin \xi$ and ${\sigma }_{xx,1}=s(s+1)\cos \xi$, so the mean displaced-boundary and tilted-normal corrections cancel. Thus the force-free condition gives ${\overline{u}}_{2x}^{\prime }(0)=0$, hence $C=0$. In the $A=0$ limit boundedness likewise excludes mean shear. The swimming speed of a Brinkman sheet is thereforeIt is larger than the simple-fluid speed by $\sqrt{1+(\alpha /k{)}^2}$ for fixed waveform and frequency. The matrix offers a reaction structure for the transverse stroke and more strongly confines the induced motion; it acts as a footing against which the wave pushes. This is a prescribed-stroke comparison, not a guarantee at fixed motor power. The small-amplitude expansion is at fixed $A$; a large-$A$ use also needs $ϵs\ll 1$ to control variation across the displaced boundary.

Let $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ be the rate-of-strain tensor of an incompressible flow, and let $s(t)$ be the supremum over the conductor of its largest eigenvalue. State Backus' necessary condition for dynamo action with its magnetic boundary conditions: an isolated bounded conductor of uniform positive magnetic diffusivity $\eta$, surrounded by an electrical insulator with a decaying potential exterior field, and no imposed magnetic field or boundary energy input. For definiteness take a no-slip boundary condition on the fluid, which eliminates the stretching surface term. If the conductor lies within a sphere of radius $R$ and $S={\sup }_{t}s(t)$, a necessary condition for a nondecaying dynamo isThe constant is the free-decay spectral bound for an insulating exterior, not a universal constant for every magnetic boundary condition. The condition is necessary, not sufficient, and involves maximum stretching rather than an rms velocity.

To see both the condition and the growth-rate bound, include exterior magnetic energy:This follows from the resistive induction equation and integration by parts, with the stated boundary assumptions. The magnetic free-decay spectral bound is ${\int }_D|\nabla \times \mathbf{B}{|}^{2}dV\ge ({\pi }^2/R^2){\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^{2}dV$. For a sphere its lowest mode is the dipolar poloidal free-decay mode; enclosing a smaller conductor gives the same valid lower bound. Since $B_i{e}_{ij}{B}_j\le s(t)|\mathbf{B}{|}^2$, we obtainIntegrating this differential inequality gives decay whenever $S<\eta {\pi }^2/R^2$. More generally the exponential rate of the field norm, rather than of its squared energy, satisfiesSimply discarding the nonnegative resistive dissipation already proves the requested maximum-strain bound. The energy exponent is twice the field-amplitude exponent.

For the alpha-Omega dynamo model, write $a=|{\alpha }_0|$, $w=|\Omega |$, $k\ge 0$ and $d=\eta k^2$. Direct differentiation givesFor $a,kw>0$, use the weighted energy estimate for two coupled modes and form the positive weighted norm $N=kw|A{|}^2+a|B{|}^2$. The inequality $N\ge 2\sqrt{akw}|A||B|$ givesFor each fixed $k$ and model parameters, this norm is equivalent to the amplitude norm; its square-root exponential rate is therefore bounded by $h(k)=\sqrt{awk}-\eta k^2$, uniformly over all admissible $f(t)$. Maximizing over $k$ givesThe exponent is also achievable in order of magnitude. Choose the admissible constant $f=1$. The growing eigenvalue of the two-component system has real part $-\eta k^2+\sqrt{awk/2}$. Its maximum occurs at $k=(aw/(32{\eta }^2){)}^{1/3}$ and equals $32^{-10/3}(a^2{w}^2/\eta {)}^{1/3}$. Thus the bounded-modulation alpha-Omega growth estimate has the scalingThis means the maximum over allowed modulations and wavenumbers, not that every bounded modulation grows; $f=0$ supplies no regenerating alpha coupling.

The Omega effect rapidly makes toroidal field from poloidal field, but exponential dynamo action also requires the slower alpha effect to regenerate the poloidal component. The coupled amplification rate is of order $\sqrt{a\Omega k}$ rather than $\Omega$; shortening the wavelength to increase it also increases magnetic diffusion as $\eta k^2$. Their optimal balance gives $k\sim {\Omega }^{1/3}$ and growth $\sim {\Omega }^{2/3}$. The Backus' necessary condition for dynamo action estimate controls stretching alone and does not incorporate this regeneration bottleneck. A shear without regeneration can give transient amplification but not this sustained exponential feedback.

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.