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The component of traction normal to a surface with unit normal $n$. For an incompressible Newtonian fluid, $\sigma =-pI+2\mu e$, so ${\sigma }_{nn}=-p+2\mu n\cdot en$. A free-interface normal-stress jump includes the curvature force from surface tension.

In the thin-gap, laminar approximation, a rotating disc of radius $a$ separated by $d$ from a stationary disc shears a Newtonian fluid. The resisting torque is $\pi \eta \omega a^4/(2d)$, neglecting edge effects and secondary flows. This gives a way of inferring dynamic viscosity from a measured torque.

For a dilute spherical no-slip probe of radius $a$ in an isotropic incompressible linear medium at equilibrium, the bead's displacement per unit step force is $J_b(t)=J(t)/(6\pi a)$, where $J$ is the bulk shear creep compliance. The fluctuation-dissipation theorem gives one-coordinate displacement variance $2k_{B}T{J}_b(t)$. Adding the three equal Cartesian contributions proves the displayed relation. In a Newtonian fluid, $J=t/\eta$, so it becomes $⟨\Delta r^2⟩=6Dt$ with the Stokes–Einstein relation $D=k_{B}T/(6\pi \eta a)$.

The particle must sample a continuum response on its scale and remain in thermal equilibrium. A two-dimensional image records only two of the three displacement components; isotropy then gives $⟨\Delta r^2{⟩}_{3d}=(3/2)⟨\Delta r^2{⟩}_{2d}$. Localization noise, confinement and unresolved fast relaxation can otherwise be mistaken for elastic response.

Eliminating the shear stress from the two linear equations gives the Maxwell shear-wave dispersion equation in physical space,where $\nu$ is the kinematic viscosity. Substitution of a Fourier mode $e^{iky+\sigma t}$ givesThese are temporal exponents: the oscillation angular frequency is the imaginary part of $\sigma$, with the sign determined by the selected time convention. For $4\tau \nu k^2<1$ both exponents are real and nonpositive. At equality the repeated exponent is $-1/(2\tau )$; generic solutions include $t{e}^{-t/(2\tau )}$. Above the threshold the modes are damped propagating shear disturbances:At large $|k|$ their phase velocity tends to $\sqrt{\nu /\tau }=\sqrt{G_{\mathrm{el}}/{\rho }_0}$.

For the limit $\tau \to 0$ at fixed $\mu ,{\rho }_0$ and $k$, expansion of the square root givesAfter a short initial relaxation, the surviving branch obeys the diffusion equation $u_t=\nu u_{yy}$ of a Newtonian fluid. The fast branch disappears on the vanishing viscoelastic relaxation time; at $k=0$ the two exponents are exactly $0$ and $-1/\tau$. This limit is not uniform in $k$: modes with $k$ of order $(\tau \nu {)}^{-1/2}$ still sample the oscillatory transition. If instead $G_{\mathrm{el}}=\mu /\tau$ is held fixed as $\tau \to 0$, then $\mu \to 0$ and the fixed-$k$ slow exponent tends to zero. The ordinary Newtonian fluid limit therefore specifies which material parameter is held fixed.

The $xy$ component of the stationary constitutive equation isInsert $T_{yy}=-\alpha g{T}_{xy}$ and $T_{xx}=(2\tau -\alpha )g{T}_{xy}$. The coefficient of $g^2{T}_{xy}$ becomes $\tau \alpha +\alpha (\tau -\alpha )=\alpha (2\tau -\alpha )$, givingFor $C=\alpha (2\tau -\alpha )>0$, the small-rate shear stress is $T_{xy}=\mu \dot{\gamma }-\mu C{\dot{\gamma }}^3+O({\dot{\gamma }}^5)$. At large positive rate, $T_{xy}\sim \mu /(C\dot{\gamma })$ and the shear viscosity decreases as $\mu /(C{\dot{\gamma }}^2)$. The shear stress is odd in rate, reaches a positive maximum atand decreases thereafter. The original plot displays this homogeneous constitutive curve. Its decreasing branch can permit shear banding; the local model alone supplies neither an interface-selection rule nor a guaranteed stable homogeneous realization.

This is a Johnson-Segalman fluid without an added Newtonian fluid solvent contribution. To identify conventions explicitly, set $a=1-\alpha /\tau$ and $L=G^T=E+W$, with $W$ the spin tensor. The derivative appearing in the constitutive equation becomesthe Gordon--Schowalter derivative. Its usual slip interval $|a|\le 1$ corresponds to $0\le \alpha \le 2\tau$, while a nonsingular genuinely shear thinning curve requiresAt $\alpha =0$ the model is the Upper-convected Maxwell model; at $\alpha =2\tau$ it is the lower-convected derivative counterpart, and both have constant shear viscosity in this test. At $\alpha =\tau$ the derivative is the Jaumann derivative, giving the corotational Maxwell fluid. Merely requiring $\alpha >0$ is insufficient: for $\alpha =3\tau$ the denominator is $1-3{\tau }^2{\dot{\gamma }}^2$, which vanishes at $|\dot{\gamma }|=1/(\sqrt{3}\tau )$ and then gives a negative steady shear viscosity. There is no solvent-induced large-rate plateau in the stated equation.

Let $p_T=n{k}_{B}T$, $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$, and $M=I+\alpha \sigma$, so the single-bead mobility is $M/\zeta$. Assume an incompressible solvent, constant $n,H,T,\zeta$, negligible bead inertia, and a symmetric positive-definite $M$. The last requirement makes the resistance invertible and Brownian diffusion physically admissible. At a given position and time, $M$ depends on the moment $C$ but not on the individual configuration $\mathbf{R}$.

Use the source's gradient convention $K_{ij}={\partial }_i{u}_j$, and write $G=K^T$ for the component-first gradient. Then $\mathbf{u}({\mathbf{r}}_2)-\mathbf{u}({\mathbf{r}}_1)\simeq G\mathbf{R}$. The spring forces on the two beads are ${\mathbf{F}}_1=H\mathbf{R}$ and ${\mathbf{F}}_2=-H\mathbf{R}$. Their overdamped probability currents areHere the Brownian term written as $-k_{B}T\nabla \psi$ is a force weighted by probability density. Its corresponding conditional osmotic force is $-k_{B}T\nabla \log \psi$; treating the given term as an unweighted deterministic force per bead would not produce the stated Fokker–Planck equation.

With midpoint $\mathbf{x}=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$, the derivatives are ${\nabla }_{r_1}=\frac{1}{2}{\nabla }_x-{\nabla }_R$ and ${\nabla }_{r_2}=\frac{1}{2}{\nabla }_x+{\nabla }_R$. Subtracting the bead currents gives the configuration currentThe midpoint current is $\mathbf{u}\psi$ after neglecting the specified spatial Brownian diffusion. Configuration probability conservation therefore givesSince $G\mathbf{R}$ has components $R_j{\partial }_j{u}_i$, this agrees with the source's $\mathbf{R}\cdot \nabla \mathbf{u}$. The factor two in both spring relaxation and configuration diffusivity comes from relative motion of both beads.

The second moment $C$ is the unnormalized conformation tensor. Now multiply by $R_i{R}_j$ and integrate over configuration space. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+C{G}^T$; spring drift gives $-(2H/\zeta )(MC+CM)$; two integrations by parts of diffusion give $(4k_{B}T/\zeta )M$. HenceBecause $\sigma =nHC-p_{T}I$, $M$ is a polynomial in $C$ and commutes with it. ThusThis is the Hookean dumbbell conformation equation with anisotropic mobility, an exact second-moment closure within the stated model; no Gaussian approximation to $\psi$ was needed.

The upper-convected derivative of the identity is $I^{\nabla }=-(G+G^T)$. Applying it to the polymer stress therefore givesConsequently the coefficients and the closed stress equation arewhere ${\sigma }^{\nabla }=D_t\sigma -\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a polymer relaxation time, unrelated to the inverse-length parameter of question c; $\mu$ is the polymer viscosity scale. The coefficient $\alpha$ has inverse-stress units. In the dimensional mobility parameter in the Giesekus model, the dimensionless factor is $b=\alpha p_T$, not the dimensional $\alpha$ printed in this problem.

The upper-convected rate removes the affine stretching and rotation of material connectors. If $\dot{F}=GF$, then a tensor transported purely as $C=F{C}_0{F}^T$ has $C^{\nabla }=0$. Under a time-dependent rigid rotation $Q$, the rate transforms as $Q{C}^{\nabla }{Q}^T$: this is material frame indifference, not an ordinary componentwise time derivative. The resulting equation is the Giesekus model in these dimensional conventions.

The isotropic subtraction is fixed by zero polymer extra stress at thermal equilibrium. With no flow and positive mobility, the stationary distribution isso $\boxed{{\sigma }_{\mathrm{eq}}=0}$. The constitutive equation correspondingly has a zero-stress quiescent state and a Newtonian low-rate limit $\sigma \simeq \mu (K+K^T)$. Without subtracting exactly $p_{T}I$, the spring contribution would retain an equilibrium isotropic stress and the equation would acquire corresponding reference terms. Because mobility itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate fluid pressure enforcing incompressibility is not included in this polymer extra stress.

Using the given linear momentum balance, multiply by the velocity and integrate over the fixed volume:This is integration by parts component by component, or the divergence theorem. The first term is boundary stress power. With component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the second is the double contraction $\Phi =\sigma :K$.

For a Newtonian fluid, $\sigma =-pI+2\mu e$, where $e=(K+K^T)/2$ is the rate-of-strain tensor and $\mu \ge 0$ is dynamic viscosity. Incompressibility removes the pressure contribution, and the symmetric tensor $e$ is orthogonal to the antisymmetric part of $K$. ThereforeThe displayed energy identity follows for the stated linear balance. A full advective momentum equation in a fixed control volume would also contribute an advective kinetic energy flux. The nonnegativity statement here is for a purely Newtonian viscous stress; a viscoelastic stress can exchange stored free energy with the flow.

The stress tensor of the incompressible Newtonian fluid is $\mathbf{\sigma }=-pI+2\mu e$, where $e=(\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)/2$ is the rate-of-strain tensor. The homogeneous Stokes equations give $\nabla \cdot \mathbf{\sigma }=0$. Apply the divergence theorem to the fluid between the body and a surrounding sphere $S_R$. If ${\mathbf{n}}_B$ points from the body into the fluid, the fluid-domain normal on the inner boundary is $-{\mathbf{n}}_B$. HenceThis is the force flux at infinity in Stokes flow. A nonzero resultant force must be supported by a stress of order $R^{-2}$ on a surface of area of order $R^2$. More precisely, the leading exterior solution is a Stokeslet with force on the fluid $\mathbf{f}=-\mathbf{F}$:Thus $\mathbf{u}=O(R^{-1})$, $p-p_{\infty }=O(R^{-2})$ and $e=O(R^{-2})$. These are leading generic orders for a translating body with nonzero drag; the integral alone does not determine every angular component, and force-free disturbances may decay faster.

Let $x$ measure distance down the incline and $y$ distance perpendicular to it, with the wall at $y=0$ and the free surface at $y=h$. A uniform steady falling film flow has $\mathbf{u}=u(y){\mathbf{e}}_x$. Its convective acceleration vanishes because $\mathbf{u}\cdot \nabla =u{\partial }_x$ and ${\partial }_{x}u=0$. A flat surface exposed to constant atmospheric pressure hasThus ${\partial }_{x}p=0$, and the downslope component of the Navier-Stokes equations reduces to $0=g\sin \alpha +(\eta /\rho )u^{\prime \prime }$. The printed pressure-free equation must be understood as this downslope projection: its full vector version would incorrectly omit the normal hydrostatic pressure gradient that balances $g\cos \alpha$.

The no-slip boundary condition gives $u(0)=0$. Neglecting air shear stress makes the surface shear-free, so $u^{\prime }(h)=0$. Two integrations give the uniform gravity-driven Newtonian film profileThe velocity increases up to the free surface, whereThis assumes a uniform laminar Newtonian fluid film, a stationary wall and negligible surface shear; a varying thickness or externally imposed downslope pressure gradient would alter the balance.

For a homogeneous Newtonian fluid with constant kinematic viscosity $\nu >0$, assume the same steady purely azimuthal flow. In cylindrical coordinates, the azimuthal Navier-Stokes equation has no pressure forcing or convective term in this state, and becomesEquivalently, ${\Omega }^{\prime \prime }+3{\Omega }^{\prime }/r=0$, or $(r^3{\Omega }^{\prime }{)}^{\prime }=0$. Two integrations give the steady circular Couette profile,This is the basic Taylor-Couette flow between rotating cylinders. The constants depend on boundary angular velocities; for radii $R_{<},R_{>}$ and prescribed angular velocities ${\Omega }_{<},{\Omega }_{>}$,If the domain includes $r=0$, regularity requires $B=0$, leaving solid-body rotation. On an annulus a nonzero $B$ is allowed and wall torques maintain the shear.

The restriction concerns a steady profile. An arbitrary initial angular velocity does not cease to exist when viscosity is introduced; within the purely azimuthal ansatz it evolves according to ${\partial }_t\Omega =\nu ({\Omega }^{\prime \prime }+3{\Omega }^{\prime }/r)$ until compatible boundary conditions and angular-momentum constraints determine its long-time state.

Given an admissible velocity field, recover the pressure gradient from the Stokes equation, $\nabla p=\mu {\nabla }^2\mathbf{v}$, and integrate it, fixing the additive constant by matching to the bath pressure near the open ends. The biharmonic equation ensures the compatibility of this integration. For a freely moving hinge, the laboratory velocity field also contains the unknown rigid translation and rotation. Solve the interior and exterior Stokes flows with these wall velocities, then impose zero total external force and torque on the complete apparatus if its motor supplies only internal actuation. The pinned solution alone cannot determine that free motion.

At a fixed angle, the actuation velocity scale is $U\sim L|\dot{\alpha }|$. The Stokes equation gives $\Delta p\sim \mu U/L\sim \mu |\dot{\alpha }|$. Integrating normal traction over a plate of length $L$ gives a force per unit width of scalewith a dimensionless signed coefficient depending on geometry and exterior matching. Normal viscous stress has the same dimensional scale and is included in $C$. The force reverses when the actuation reverses. A narrow gap can make $|C|$ large.

More specifically, the corrected local streamfunction gives, up to a reference pressure,Indeed, inserting $\mathbf{v}=(-\dot{\alpha }r{G}^{\prime }/2,\dot{\alpha }rG)$ into the vector Laplacian gives $p_r=-\mu \dot{\alpha }(G^{\prime \prime \prime }+4G^{\prime })/(2r)=\mu \dot{\alpha }K/r$ and $p_{\theta }=0$. For $\alpha <\pi /2$, opening produces reduced pressure toward the hinge when matched at $r_{\ast }\sim L$. The logarithmic singularity is integrable: ${\int }_0^L\log (L/r)dr=L$, consistent with the stated force scale. A real hinge regularizes the singularity; the local formula is not a full end-matched solution.

The scallop theorem applies to a single periodically actuated hinge in a stationary Newtonian fluid with negligible fluid and body inertia. At each shape, Linearity of Stokes flow makes its rigid velocity proportional to $\dot{\alpha }$. Closing retraces the sequence of shapes and reverses the corresponding motion, irrespective of different opening and closing rates. The hinge may move during each half-cycle, butA motor can generate instantaneous forces and consume energy through viscous dissipation without producing sustained swimming. Additional independently controlled shape changes, appreciable inertia, or a fluid with memory would require a different analysis.

Take $x$ along a horizontal gap and $s$ perpendicular to its plates, with $0<s<b$. Assume a Newtonian fluid, stationary plates, a steady flow, a laminar flow and a fully developed flow away from the entrances and side edges. The no-slip boundary condition holds on both plates. Any transverse gravity is absorbed in the hydrostatic pressure; there is no imposed transverse through-flow. Translational symmetry and the no-penetration boundary condition then give $\mathbf{u}=(u(s),0,0)$. The Navier-Stokes equation reduces toTwice integrating gives the plane Poiseuille flow:The volumetric flow rate through one gap isThere are $n-1$ internal gaps. Their total cross-sectional area is $(n-1)wb$, because the plate thickness is negligible, so the Darcy velocity isThus the permeability of a porous medium has the dimensions of squared length. If finite plate thickness were included, the Darcy velocity averaged over the whole manifold would also contain its porosity.

Both the local acceleration and the convective acceleration vanish: ${\partial }_t\mathbf{u}=0$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=u_x{\partial }_x\mathbf{u}=0$. The dominant balance is therefore between the pressure gradient and viscous stress, with no inertia contribution in this fully developed solution. This exact cancellation does not guarantee that arbitrary large-Reynolds number realizations remain laminar flows; entrances, disturbances and turbulence can invalidate the assumptions.

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

A simple fluid reduces to a linear viscoelastic fluid when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The relaxation modulus $G(t)$ is the shear-stress response to a unit small step of shear strain.

Choose the harmonic convention $e^{i\omega t}$. If the imposed shear rate is $\mathrm{Re}(\hat{\dot{\gamma }}{e}^{i\omega t})$, the corresponding shear stress is $\mathrm{Re}({\eta }^{\ast }(\omega )\hat{\dot{\gamma }}{e}^{i\omega t})$. Substitution into the constitutive convolution gives the complex viscosity:For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, so $G^{\prime }$ is the storage modulus, $G^{\prime \prime }$ the loss modulus, and ${\eta }^{\ast }=G^{\prime \prime }/\omega -i{G}^{\prime }/\omega$ in this convention.

Creep is the progressive deformation produced by a held stress. For a small step $S_{0}H(t)$ of shear stress, define the creep compliance $J$ by $\gamma (t)=S_{0}J(t)$ from a relaxed state. Writing hats for Laplace transforms, the same constitutive law gives $\hat{S}=p\hat{G}\hat{\gamma }$. Since $\hat{S}=S_0/p$, this yieldsHere $H$ is the Heaviside step function. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep convolution includes that jump through ${\int }_{0^{-}}^{t}G(t-s)dJ(s)=1$ for $t>0$.

For the Maxwell fluid, $\hat{G}(p)=G_0/(p+1/\tau )$. Thus the two material responses areThe creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency shear viscosity is $\eta =G_0\tau$.

For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of stress by $S$. The incompressibility constraint is satisfied, and the advective part of the material derivative vanishes because no field depends on $x$. With no imposed longitudinal pressure gradient, momentum conservation and the linear Maxwell law giveThe initial conditions are $u=S=0$ in $y>0$; the no-slip boundary condition is $u(0,t)=UH(t)$, with decay as $y\to \infty$. Transforming these equations givesChoose the analytic square-root branch in $\mathrm{Re}p>0$ with $\mathrm{Re}q>0$, so that the spatially increasing solution is excluded. The Maxwell start-up shear layer consequently hasThe Bromwich inversion formula now givesThe vertical Bromwich contour is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.

The Newtonian limit holds $\eta =G_0\tau$ fixed while $\tau \to 0$, so $G_0\to \infty$. Then $q(p)\to \sqrt{\rho p/\eta }$. With $\nu =\eta /\rho$, the inverse Laplace transform gives the complementary error function profileThis is diffusive momentum penetration in a Newtonian fluid, with thickness of order $\sqrt{\nu t}$.

The elastic limit instead holds $G_0$ fixed while $\tau \to \infty$. Write $c=\sqrt{G_0/\rho }$. On the chosen branch, $q(p)\to p/c$, so the Laplace transform time-shift rule yieldsAn elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear strain is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.

The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$. The material derivative simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear strain. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho /G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll 1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary simple fluid.

The cell conservation equation equates the change of cell number density $n$ to minus the divergence of its total flux. The flux $n\mathbf{u}$ is advection by the water, $n{V}_s⟨\mathbf{p}⟩$ is the mean swimming flux, and $-\mathbf{D}\cdot \nabla n$ is the dispersal flux. The tensor $\mathbf{D}$ describes effective translational diffusion, whereas $D_r$ in part (i) describes rotational diffusion; they have different dimensions. This conservation law neglects birth, death and sources at the walls.

The equation $\nabla \cdot \mathbf{u}=0$ expresses incompressible flow. In the momentum equation, $\rho$ is the reference suspension density, $\rho [{\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}]$ is material acceleration times density, $-\nabla p_e$ is the pressure force after subtracting the pure-water hydrostatic pressure, and $\mu {\nabla }^2\mathbf{u}$ is the viscous force for constant dynamic viscosity $\mu$ in a Newtonian fluid. Since $n$ counts cells per volume, $g^{\prime }$ in $-g^{\prime }n{\mathbf{e}}_z$ is excess downward weight per cell; it is not an acceleration. The small cell-fluid density difference permits the Boussinesq approximation: retain that difference in buoyancy while using constant $\rho$ elsewhere. Diluteness permits neglect of cell-cell interactions, changes of viscosity, and active cell stresses. Any separate settling speed is assumed negligible or already incorporated in the specified mean drift. The light strength and swimming speed are uniform.

Consider the intended long-channel bulk approximation. At the sidewalls impose zero normal cell flux and the no-slip boundary condition. The closed ends require zero total fluid flux through each horizontal section, so ${\int }_{-a}^{a}wdx=0$; localized turning regions are omitted from the bulk equations. The mean concentration is ${\int }_{-a}^{a}ndx=2a{n}_0$. The final paragraph below distinguishes these bulk conditions from exact closed-end cell conservation.

For $\mathbf{u}=w(x){\mathbf{e}}_z$, the vorticity is $-w^{\prime }(x){\mathbf{e}}_y$. Hence $\epsilon =-w^{\prime }/(2D_r)$, and the leading horizontal mean swimming direction is $K_0{\mathbf{e}}_x$. With $\mathbf{D}=DI$ and $n=n(x)$, steady cell conservation equation says that $J_x=V_s{K}_{0}n-D{n}^{\prime }$ is constant. Its wall value is zero, so $n^{\prime }=\beta n$ with $\beta =V_s{K}_0/D$. The phototactic concentration in a vertical channel is thereforeThis result neglects the $O({\epsilon }^2)$ correction to horizontal swimming. For positive phototaxis, $\beta >0$ and cells accumulate towards the illuminated wall $x=a$.

The steady inertial terms vanish for $w=w(x)$, since $\mathbf{u}\cdot \nabla =w{\partial }_z$. The horizontal momentum equation gives ${\partial }_x{p}_e=0$. The vertical momentum equation then requires a constant excess pressure gradient $G={\partial }_z{p}_e$ and gives $\mu w^{\prime \prime }=G+g^{\prime }{n}_1{e}^{\beta x}$. Twice integrating producesSet $b=a\beta$. The two no-slip boundary conditions give $A=-C\sinh b/a$ and $B=-C\cosh b-G{a}^2/(2\mu )$. The zero fluid-flux condition follows by integrating this expression:Consequently the remaining constants areThe resulting velocity can also be written without separate polynomial constants asHere $G$ is the zero-flux pressure gradient in a cell-driven channel. It is negative for $b>0$: pressure decreases upwards, creating the upward pressure force that balances the excess cell weight and drives the return flow. Its magnitude is determined by zero net fluid transport, not by prescribing pressure independently. In the zero-light limit, the continuous result is $n=n_0$, $G=-g^{\prime }{n}_0$ and $w=0$.

For the requested large-$b$ sketch, cells occupy a boundary layer of thickness ${\beta }^{-1}$ near $x=a$. Put $\xi =x/a$ and $W=g^{\prime }{n}_0{a}^2/(\mu b)$. The exact dimensionless profile above isAway from the illuminated-wall layer it tends to $(1+\xi )(1-3\xi )/2$. There is broad upward return flow for $-1<\xi <1/3$ and downward flow on the illuminated side. The outer maximum is $2W/3$ at $\xi =-1/3$. Near $\xi =1$, the exponential term is approximately $2e^{-b(1-\xi )}$ and restores the no-slip boundary condition from the negative outer velocity to zero at the wall. The finite-$b$ profile shown below retains that layer. The small-vorticity assumption must still be enforced, for example by sufficiently weak excess weight or a sufficiently dilute suspension; $b\gg 1$ alone does not ensure $|w^{\prime }|/(2D_r)\ll 1$.

There is a genuine global qualification to the bulk construction. For an exactly steady channel with impermeable ends as well as impermeable sidewalls, the closed-channel cell-flux constraint also requires $\int J_{z}dx=0$ at every height. In this bulk ansatz the first-order orientation closure givessince integration by parts, $n^{\prime }=\beta n$ and $w(\pm a)=0$ imply $\int n{w}^{\prime }=-\beta \int nw$. Moreover, multiply $\mu w^{\prime \prime }=G+g^{\prime }n$ by $w$ and integrate. The wall values and zero fluid flux givefor a nontrivial profile. Thus zero axial cell flux generally fails unless the extra factor $1+V_s\beta K_1/(2D_r)$ vanishes, or the flow is trivial. A complete steady closed-channel solution generically needs axial concentration variation. The explicit profile solves the stated bulk equations, sidewall conditions, concentration normalization and zero fluid-flux condition; it is not, for arbitrary parameters, an exact globally closed steady cell distribution. Ignoring end regions gives the intended bulk approximation, but does not by itself prove the stronger global existence claim.

A generalized Newtonian fluid can approximate the shear response of a fluid with viscoelasticity when the flow varies slowly relative to its stress-memory time scales and elastic stresses are negligible, or in locally steady unidirectional flow where a fitted steady shear viscosity captures the relevant stress and elastic normal stresses do not significantly influence the motion. It is not a general replacement in rapidly changing flows or in flows controlled by elastic stresses. Two specific failures are stress relaxation after deformation stops, since this law has no memory, and nonzero normal-stress differences in steady simple shear flow, since the only nonzero nonpressure components are off diagonal. A rate-dependent viscosity alone cannot repair either limitation.

Take the usual physical power-law fluid range $k>0$, $n>0$, and $G>0$. The steady unidirectional Couette-Poiseuille flow satisfies $0=G+{\partial }_y{\sigma }_{xy}$. The stated upper-wall stress thus givesPut $Y=y/h$. Inverting this monotone stress law gives $u_y=(Gh/k{)}^{1/n}\mathrm{sgn}(\alpha -Y)|\alpha -Y{|}^{1/n}$. Integration from the lower no-slip boundary condition $u(-h)=-U$ gives the power-law Couette-Poiseuille flow profileThe upper no-slip boundary condition requiresThe bracket is a strictly increasing function of $\alpha$ with value zero at $\alpha =0$, so the equation determines a unique positive $\alpha$. The absolute values are essential: for sufficiently strong pressure driving, the shear can reverse inside the channel. For a Newtonian fluid, $n=1$, $\mu =k$, and the bracket reduces to $4\alpha$, yieldingFor zero pressure gradient the limit is pure Couette flow; the parameterization of wall stress by $Gh\alpha$ then becomes singular rather than indicating zero Couette stress.