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For a no-slip channel with half-width $h$, a harmonic pressure gradient $\Delta p{e}^{i\omega t}$ and complex viscosity $\hat{\mu }=G_0\tau /(1+i\omega \tau )$, put $z^2=i\omega \rho h^2/\hat{\mu }$. Solving $\hat{\mu }{U}^{\prime \prime }-i\omega \rho U=\Delta p$ gives $U=-\Delta p[1-\cosh (zy/h)/\cosh z]/(i\omega \rho )$; integration gives the displayed flux per unit width. At low frequency it approaches $-2\Delta p{h}^3/(3G_0\tau )$. At fixed positive $h,\tau ,G_0,\rho$, the high-frequency leading flux is $-2\Delta ph/(i\omega \rho )$, although shear waves can persist in the local profile. If instead $\omega \tau \gg 1$ while $\rho {\omega }^2{h}^2/G_0\ll 1$, the negligible-inertia elastic limit is $Q\simeq -2i\omega \Delta p{h}^3/(3G_0)$. The last two limits impose different scale orderings.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.

Write the background mass density as $\bar{\rho }(z)={\rho }_0+{\rho }_0^{\prime }z$ and its hydrostatic pressure as $\bar{p}(z)$, whereApply an Eulerian perturbation of a fluid variable to the background fields:Thus $\pi$ is the Eulerian pressure perturbation, in pascals; it is not the background hydrostatic pressure.

For incompressible flow, mass conservation becomes $D{\rho }_{\mathrm{tot}}/Dt=0$. Linearizing about rest givesThe Euler equations are ${\rho }_{\mathrm{tot}}D\mathbf{u}/Dt=-\nabla p_{\mathrm{tot}}-{\rho }_{\mathrm{tot}}g\hat{\mathbf{z}}$. The background pressure gradient and weight cancel. Dropping products of perturbations givesUnder $|{\rho }_0^{\prime }z|\ll {\rho }_0$, replace $\bar{\rho }$ by ${\rho }_0$ in inertia but retain the background mass density gradient in the displacement-induced buoyancy. This is the Boussinesq approximation. Its linear Boussinesq equations here areThe absolute value makes explicit that the small-density-variation condition concerns magnitude, including the usual negative ${\rho }_0^{\prime }$ of stable density stratification.

Write $P(x)$ for the outer pressure extrapolated to the wall. The normal balance in the thin boundary layer is $p_y={\overline{F}}_y$, givingThe spatially varying magnetic pressure deficit drives a tangential pressure gradient. For fixed forcing as $\omega \to \infty$, layer inertia is small compared with viscous dissipation; explicitly the ratio is $U{\delta }^2/(\nu b)$. The leading tangential balance isThe outer gradient $P^{\prime }$ contributes only at higher inner order. Matching the leading outer shear gives $u_y\to 0$, and the physical no-slip boundary condition gives $u(x,0)=0$. Two integrations yield the magnetic skin-layer streaming slip:For the present field amplitude,Using incompressibility and $v(x,0)=0$ also givesHence $v/u=O(\delta /b)$ in the layer, except at symmetry zeros where a componentwise ratio is inappropriate. In the matching region $\delta \ll y\ll b$, $u\sim U$ and $v\sim -y{U}^{\prime }$, up to the smaller displacement term $\delta U^{\prime }/2$.

The effective outer boundary conditions are tangential slip $u_{\mathrm{bulk}}(x,0)=U(x)$ and zero leading normal velocity $v_{\mathrm{bulk}}(x,0)=0$. These describe the bulk flow extrapolated through the unresolved magnetic skin layer; the actual wall remains at rest. For $x>0$ the slip is negative, and for $x<0$ it is positive. Fluid converges along the wall toward the wire, turns upward near $x=0$, and returns outward farther above the wall.

The following original streamline sketch uses the small-Reynolds number bulk Stokes flow for this slip. It illustrates that circulation without claiming that the unspecified bulk Reynolds number fixes a unique complete flow. With $X=x/b$, $Y=y/b$, $Z=X+iY$, a dimensionless stream function isIts factor in brackets is holomorphic in $Y>0$, so $\Psi$ solves the biharmonic equation; its wall derivative is ${\Psi }_Y(X,0)=-X/(1+X^2{)}^3$. Thus it satisfies both effective wall conditions and gives an exact creeping-flow illustration of the derived slip.

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.

Write $q=\dot{\gamma }\tau$ and $f=1+\alpha \mathrm{tr}A$. With $L_{ij}={\partial }_j{v}_i$, the steady homogeneous upper-convected derivative is $-LA-A{L}^T$. Its diagonal and shear equations giveOn the physical branch $f>0$, the other off-diagonal components vanish too. Thus $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ andFor $\alpha \ge 0$ the right side has derivative $1+6\alpha a_{12}^2>0$, so the shear stress determines a unique shear rate. The shear viscosity is $G_0\tau /f$ and decreases with shear magnitude when $\alpha >0$. This is the shear part of shear and pipe flow of an affine linear PTT fluid.

For fully developed pipe flow take $\mathbf{v}=w(r)\hat{\mathbf{z}}$ and the signed axial pressure gradient $dp/dz=\Delta p$. Regularity at the axis and the supplied momentum balance determine the shear stress without using the constitutive equation:Locally the constitutive equations have the same shear form, with axial flow replacing the Cartesian flow direction and $\dot{\gamma }=w^{\prime }(r)$ carrying its sign. HenceIntegrating and imposing no-slip boundary conditions at $r=R$ givesA negative pressure gradient produces positive axial flow. At $\alpha =0$ this is the ordinary Hagen-Poiseuille flow; positive $\alpha$ increases the flux at a fixed pressure drop through shear thinning. The nonzero axial normal stress varies radially, but its axial derivative vanishes in fully developed flow, so it does not change the axial momentum balance used here.

The Darcy law here is $\mathbf{u}=-k\nabla p/\mu$, with hydrostatic contributions absorbed into $p$ if present. Since $k$ is uniform and the fluid is incompressible, $\nabla \cdot \mathbf{u}=0$ gives $\Delta p=0$ in the porous matrix.

Pressure is uniform across the thin crack at leading order. No-slip lubrication theory gives its flux per unit depth $q=-h^3{p}_x/(12\mu )$. Conservation in a short crack segment says $q_x=-[u_y{]}_{-}^{+}$, while Darcy flow gives $[u_y{]}_{-}^{+}=-(k/\mu )[p_y{]}_{-}^{+}$. ThereforeThe other interfacial condition is pressure continuity, $[p{]}_{-}^{+}=0$. The far pressure gradient is $-\mu U{\mathbf{e}}_x/k$; outside the slit both pressure and normal Darcy flux are continuous. The leading tapered-crack flux vanishes at its endpoints. The assumptions $h\ll a$ and $h\gg \sqrt{k}$ justify transverse pressure uniformity and neglect of porous-wall slip on scale $\sqrt{k}$. Close to the sharp tips the reduced model is an outer approximation, not a uniformly resolved tip solution. This is lubrication boundary condition for a crack in a Darcy medium.

These are planar elliptic coordinates. The curve $\xi =0$ is the slit $[-a,a]$, traced on its upper and lower sides. The curve $\xi =1$ is the ellipse with semiaxes $a\cosh 1$ and $a\sinh 1$. For fixed noncardinal $\eta$, the coordinate curve is the appropriate branch of $x^2/(a^2{\cos }^2\eta )-y^2/(a^2{\sin }^2\eta )=1$. The cardinal angles give the two exterior horizontal rays and the upper/lower vertical rays. The requested original sketch is:

On the upper face $0<\eta <\pi$, the supplied derivatives reduce to ${\partial }_x=-(a\sin \eta {)}^{-1}{\partial }_{\eta }$ and ${\partial }_y=(a\sin \eta {)}^{-1}{\partial }_{\xi }$. The same point on the lower face has angle $2\pi -\eta$, so its ${\partial }_y$ has the opposite sign. ThusInsert this and the tangential derivative into the crack condition and multiply by $12a^2\sin \eta$ to obtainPressure continuity becomes $p(0,\eta )=p(0,2\pi -\eta )$. The uniform-flow background is $p_0=-\mu Ux/k=-(\mu Ua/k)\cosh \xi \cos \eta$. Its derivatives immediately giveThe harmonic equation is $p_{\xi \xi }+p_{\eta \eta }=0$ because the metric multiplier is nonzero outside the slit.

For the specified profile, $h^3=H_0^3\sin \eta$ on the upper face. Its coefficient in the tangential operator is constant, so the imposed cosine mode separates. With $P_0=\mu Ua/k$, write $p=P_{0}F(\xi )\cos \eta$. Harmonicity gives $F^{\prime \prime }=F$, the interfacial condition gives $F^{\prime }(0)=\alpha F(0)$, and the far field fixes the growing coefficient to $-1/2$. ThereforeReflection symmetry makes the two face derivatives equal. Other homogeneous cosine modes decaying as $e^{-m\xi }$ would require $-m=\alpha m^2$ at the slit, impossible for $m>0$ and $\alpha \ge 0$. The constant pressure gauge is immaterial. This justifies the separated crack pressure in elliptic coordinates, not only a candidate satisfying selected conditions.

At the midpoint $F(0)=-1/(1+\alpha )$ and $p_x=-\mu U/[k(1+\alpha )]$. HenceFor $\alpha \ll 1$, $Q\sim H_0^{3}U/(12k)$ and the crack barely perturbs the imposed Darcy pressure gradient. For $\alpha \gg 1$, $Q\sim 2aU$ and the slit is nearly equipotential: its high axial conductance collects flow from the surrounding matrix rather than sustaining a substantial pressure gradient. The parameter compares crack conductance $H_0^3/\mu$ with matrix conductance $ka/\mu$. In fact $q(x)=Q\sqrt{1-x^2/a^2}$, so the tip flux is zero in this outer model.

Subtracting the exact uniform background gives $\delta p=P_0\alpha e^{-\xi }\cos \eta /(1+\alpha )$. Since $r\sim a{e}^{\xi }/2$ and $\eta \sim \theta$ at infinity,The printed far-field coefficient is too large by a factor of two. The factor $1/2$ is forced by the stated coordinate map. An independent check is the large-$\alpha$ solution $p=-(\mu U/k)\mathrm{Re}\sqrt{z^2-a^2}$, whose expansion has precisely $\delta p\sim \mu U{a}^2\cos \theta /(2kr)$. This is the elliptic crack pressure dipole.

The phrase “suitably chosen curve” matters for the pressure-only dissipation integral. Put $C_d=\alpha \mu U{a}^2/[2(1+\alpha )k]$ and use a rectangle with end faces $x=\pm X$ and transverse extent $|y|\le H$, taking $a\ll X\ll H$. The horizontal faces have $n_x=0$. The two end-face integrals of $\delta p{n}_x$ totalThis is a fixed-throughflow normalization: the integrated velocity perturbation on an infinite end face is zero, since it is proportional to ${\partial }_x\int \delta pdy={\partial }_x(\pi C_d\mathrm{sgn}x)=0$. Work on the transverse faces vanishes as $H/X\to \infty$. Thus the requested line functional givesA bare circular cutoff would give $-\pi U{C}_d$ instead; it does not enforce this same throughflow normalization, so its pressure-only term cannot be substituted into the fixed-throughflow dissipation comparison. Keeping the control surface explicit resolves that ambiguity. This is fixed-throughflow dissipation reduction of a conductive crack.

Let $s=\pi \varphi a^2\alpha /(1+\alpha )\ll 1$. The noninteracting average dissipation per area is $\mu U^2/k+\varphi \Delta \mathcal{D}=(\mu U^2/k)(1-s)$. Equating this with $\mu U^2/k_{xx}^{\ast }$ givesOnly the first-order density term is controlled by the noninteraction assumption; the rational form is not an all-orders result. It increases permeability along the aligned cracks and saturates as their conductance becomes large. This is dilute permeability enhancement by aligned cracks.

Potential-vorticity inversion is the diagnostic recovery of the flow from its advected vortical scalar, using the balance relation and specified boundary conditions. Its usefulness is a separation of tasks: transport the scalar by the current velocity, invert to find the new velocity, and repeat. The inverse is spatially nonlocal even though the transport law is local. The four models differ chiefly in the operator being inverted and the information needed at boundaries.

In two-dimensional vortex dynamics, planar incompressibility permits $\mathbf{u}=(-{\psi }_y,{\psi }_x)$, and vertical vorticity is $\zeta ={\nabla }^2\psi$. The curl of inviscid momentum givesThere is no vortex stretching in strictly planar motion, so vorticity is materially conserved. The two-dimensional potential-vorticity inversion is the Poisson equation ${\nabla }^2\psi =\zeta$. On the plane, with suitable integrability and far-field behavior, its Green's function is $(2\pi {)}^{-1}\log |\mathbf{x}-{\mathbf{x}}^{\prime }|$, givingDifferentiation gives the planar Biot-Savart kernel. Here ${\psi }_h$ is a harmonic function fixed by boundary, circulation and far-field data. A rigid impermeable wall is a streamline, so $\psi$ is constant on it; constants or circulations around distinct holes require specification. The vorticity alone cannot distinguish two flows differing by a permitted harmonic contribution, such as a uniform current on an unbounded plane. A constant added to $\psi$ has no physical effect.

In nonrotating layerwise-two-dimensional vortex dynamics, strong stable density stratification constrains motion to almost horizontal layers. In the ideal independent-layer limit, vertical velocity vanishes at leading order, $(u,v)=(-{\psi }_y,{\psi }_x)$ andhold separately at each fixed $z$. Height is a parameter: the horizontal inverse Laplacian contains no vertical derivative. With background buoyancy $N^{2}z$, the leading Ertel potential vorticity is $Q=\mathbf{\omega }\cdot \nabla b\simeq N^2\zeta$, so the materially advected normalized scalar is $Q/N^2\simeq \zeta$. The same horizontal Poisson equation is inverted layer by layer, using each layer's boundary data. This is an asymptotic constrained model, not arbitrary three-dimensional Euler motion with vertical velocity set to zero: its validity requires small Froude number and controlled vertical scales. At finite buoyancy frequency, the weak vertical motion and tilted isopycnals restore coupling. The inversion may be independent by layer while the reconstructed fields vary with height. The horizontal Euler equations for an inviscid fluid determine the pressure gradient, while hydrostatic balance relates its vertical derivative to buoyancy.

In shallow-water quasi-geostrophic inversion, begin with the exact shallow-water potential vorticity $(f+\zeta )/h$. Linearizing its anomaly about depth $H$ gives, after multiplying by $H$, $\zeta -f_0\eta /H$. For small Rossby number, geostrophic balance gives $\psi =g\eta /f_0$, while $\zeta ={\nabla }_h^2\psi$. Keeping the beta plane background gradient givesThus $q-\beta y$ is inverted with the modified Helmholtz equation $({\nabla }_h^2-L_D^{-2})\psi =q-\beta y$, under the specified boundary and far-field conditions. The velocity follows from $\psi$ and surface elevation from $\eta =f_0\psi /g$. A horizontal Fourier mode of the anomaly satisfiesThe Rossby deformation radius therefore limits the reach of the balanced response. Layer stretching, represented by $-\psi /L_D^2$, is absent from ordinary planar vorticity inversion. Unlike the pure Poisson equation, the finite-deformation operator also responds to the constant part of $\psi$, which specifies mean elevation rather than an arbitrary streamfunction offset. Specified bottom topography adds a known source to $q$; it is subtracted along with the background before inversion.

In stratified quasi-geostrophic inversion, continuously varying vertical displacement replaces the shallow layer's single stretching term. For positive $N^2(z)$ and constant nonzero $f_0$, the three-dimensional quasi-geostrophic potential vorticity isGeostrophic balance and hydrostatic balance give velocity $(-{\psi }_y,{\psi }_x)$ and buoyancy perturbation $b=f_0{\psi }_z$. Invert the anisotropic elliptic boundary value problem using interior $q$, prescribed top and bottom buoyancy as Neumann boundary conditions, and the lateral, far-field and global-flow data. Boundary buoyancy evolves by its own material conservation when the boundary is impermeable. It cannot be discarded: the Eady edge wave in the preceding solution has zero interior potential vorticity anomaly but nonzero boundary buoyancy and a nonzero induced flow.

For constant $N,f_0$, the stretched coordinates for quasi-geostrophic inversion $Z=Nz/|f_0|$ turn the inversion operator into the three-dimensional Laplacian. Its infinite-space Green's function is $-1/(4\pi |\mathbf{X}-{\mathbf{X}}^{\prime }|)$ in $(x,y,Z)$ coordinates. Hence a horizontal disturbance of scale $L$ communicates vertically over scale $|f_0|L/N$, explicitly coupling layers. In a bounded domain with homogeneous difference data, multiplying the difference of two inversions by its streamfunction and applying integration by parts givesThis proves uniqueness up to a constant when the relevant boundary terms vanish; a mean or pressure reference fixes the remaining constant. Compatible boundary fluxes are needed for a purely Neumann boundary-value problem.

Across all four models, inversion provides the induced flow, while scalar transport supplies time evolution. The independent horizontal inverse in the first two models becomes a finite-deformation inverse in shallow water and a vertically coupled elliptic inverse in continuous stratification. The principle applies within the stated balance or constrained model. In the unrestricted Boussinesq equations or shallow water equations, gravity waves and divergent motion can share the same potential vorticity, so the vortical scalar alone does not reconstruct every possible flow.

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 first equation is vertical hydrostatic equilibrium: the upward pressure gradient balances the central mass's vertical gravitational force. For $|z|\ll r$, the point-mass acceleration is $-GMz/(r^2+z^2{)}^{3/2}\simeq -{\Omega }_K^{2}z$, where Keplerian rotation gives ${\Omega }_K^2=GM/r^3$. The disc's self-gravity and vertical inertia are neglected.

The second equation is local thermal balance. The divergence of the vertical radiative flux equals viscous dissipation. In the convention of this alpha disk, $\nu =\alpha p/(\rho \Omega )$, and the dominant Keplerian shear has $rd\Omega /dr=-3\Omega /2$. ThereforeThe radiative flux increases from zero at the symmetric midplane to a positive outgoing value at the upper surface. Radial heat transport, thermal advection and rapid storage of thermal energy are excluded from this local balance.

The third equation is radiative diffusion, equivalentlyIt applies to optically thick material, with $\kappa$ the opacity per unit mass and $\sigma$ the Stefan-Boltzmann constant. Here constant electron-scattering opacity represents Thomson scattering; the formula uses a diffusive, nearly thermal radiation field. Positive upward flux requires temperature to decrease upwards.

The final equation is the ideal gas pressure law, with mean molecular weight ${\mu }_m$ in proton-mass units. Writing $\mathcal{R}=k/({\mu }_m{m}_p)$ gives $p=\mathcal{R}\rho T$. The pressure is gas pressure because radiation pressure is assumed negligible. The four equations express vertical support, local heating balance, diffusive cooling and the gas equation of state.

The first equation is mass conservation for an incompressible flow whose area may change: local area storage balances the divergence of the volume flux $\alpha u$. The two left-hand momentum terms are local acceleration and convective acceleration, while $-p_x$ is the driving pressure gradient and $-R(\alpha )u$ models viscous drag per unit mass. The pressure-area law of a collapsible tube expresses transverse wall balance using internal minus prescribed external pressure.

For example, choose a reference area ${\mathcal{A}}_{\ast }$, length $L$ and velocity $U_{\ast }$. If dimensional variables are $\mathcal{A},s,t_d,v,P$, takeThe dimensionless external pressure uses the same scale. A dimensional friction rate $r_f(\mathcal{A})$ becomes $R(\alpha )=L{r}_f({\mathcal{A}}_{\ast }\alpha )/U_{\ast }$, and the dimensional wall pressure law is divided by $\rho U_{\ast }^2$. These choices make both inertia and the pressure-gradient coefficient unity; other equivalent reference scales are possible. On the mechanically stable wall branch, increasing area requires increased transmural pressure, whereas an enlarged tube reduces resistance. Thus $\boxed{{\tilde{P}}^{\prime }(\alpha )>0,R^{\prime }(\alpha )<0}$.

In the travelling coordinate $X=x-ct$, mass conservation becomes $[\alpha (u-c){]}^{\prime }=0$. Thus the conserved relative flux is $Q$ andFor a periodic peristaltic pumping solution, the neglected-inertia momentum equation isTake ${\alpha }_0$ to be the spatial mean area, so $⟨{\alpha }_1⟩=⟨{\alpha }_2⟩=0$. This normalization, and zero net pressure drop over a period, are implicit in the specified periodic forcing. The constant leading state has no pressure gradient, giving $Q_0=-{\alpha }_{0}c$. Put $B={\tilde{P}}_0^{\prime }$ and $d=R_{0}c/{\alpha }_0$. At first order,Period averaging gives $Q_1=0$, and the zero-mean solution isFor the second-order mean balance, expand the velocity and friction:Averaging the exact pressure derivative gives zero, so $⟨R_0{u}_2+R_1{u}_1⟩=0$. Hence the weak peristaltic pumping of a collapsible tube result isThe mean laboratory volumetric flow rate is $c⟨\alpha ⟩+Q={ϵ}^2{Q}_2+O({ϵ}^3)$, forward with the wave when $c>0$. The negative leading relative flux is therefore consistent with positive mean pumping.