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For constant opacity, the ratio $T^3/\rho =3c\lambda \mathcal{R}/(4\sigma )$ cancels the height dependence in the radiative diffusion coefficient. Since $T_z=-{\Omega }^{2}z/[4(1+\lambda )\mathcal{R}]$, the radiative flux isConsequently $F_z=c\lambda {\Omega }^2/[\kappa (1+\lambda )]$. Equating this to the viscous heating rate givesThus the dynamic viscosity is uniform in height even though pressure and mass density decrease. Its units are mass per length per time, since $c/\kappa$ has those units. A pointwise pressure-proportional alpha prescription would be incompatible with this exact profile, which is why the subsequent alpha relation is vertically averaged.

The first relation is vertical hydrostatic equilibrium. In a thin disc, the central object's gravity near the midplane is $g_z=-{\Omega }^{2}z$; the pressure gradient balances it. The second is an equation of state combining ideal gas pressure with isotropic radiation pressure. Write $\mathcal{R}=k/({\mu }_m{m}_p)$, so $p_g=\mathcal{R}\rho T$ and $p_r=4\sigma T^4/(3c)$; $\sigma$ is the Stefan-Boltzmann constant and ${\mu }_m$ the mean molecular weight in proton-mass units.

The third is local energy balance: the increase of upward radiative flux with height equals viscous dissipation. Here $\mu$ is dynamic viscosity, not mean molecular weight. A Keplerian disc has shear $rd\Omega /dr=-3\Omega /2$, so its heating rate is $\mu (r{\Omega }^{\prime }{)}^2=9\mu {\Omega }^2/4$. The fourth is optically thick radiative diffusion, with opacity per unit mass $\kappa$ and diffusion directed down the temperature gradient. Midplane symmetry gives $F(0)=0$ and oppositely directed fluxes from the two faces. The four relations express vertical force balance, gas–radiation thermodynamics, local heating, and diffusive cooling.

The required pressure integral uses ${\int }_0^1(1-x^2{)}^{4}dx=128/315$. With $\Sigma =(32/35){\rho }_{0}H$ and $p_0/{\rho }_0={\Omega }^2{H}^2/8$,The vertically averaged alpha closure for a mixed-pressure layer therefore givesThe height-integrated dynamic viscosity is $2H\mu =8cH\lambda /[9\kappa (1+\lambda )]$. Equate it to $\bar{\nu }\Sigma$ to obtainThese are integrated equalities, not an imposed pointwise relation $\mu (z)\propto p(z)$.

Now keep $\alpha$, the opacity and molecular constants independent of radius, and use Keplerian rotation $\Omega \propto r^{-3/2}$. In the gas-dominated limit, $\lambda \ll 1$ but positive, the last equality gives $\lambda \propto H\Omega \Sigma$. The column-density relation gives $\lambda \Sigma \propto H^7{\Omega }^6$. Eliminating $\lambda$ yields $H^6\propto {\Sigma }^2{\Omega }^{-5}$, whenceThis is the gas-pressure branch of a mixed-pressure alpha disk. In the radiation-dominated limit, $\lambda /(1+\lambda )\to 1$, so $H\propto (\Omega \Sigma {)}^{-1}$ andFor viscous stability of an accretion disk, linearize the Keplerian viscous diffusion equation at fixed radius: a local mass density perturbation has diffusion coefficient $3{\partial }_{\Sigma }(\bar{\nu }\Sigma )$. The gas branch has $\bar{\nu }\Sigma \propto {\Sigma }^{5/3}$ and a positive derivative, so it smooths perturbations. The radiation branch has $\bar{\nu }\Sigma \propto {\Sigma }^{-1}$ and a negative derivative, giving radiation-pressure viscous instability. The gas-pressure branch is viscously stable; the radiation-pressure branch is viscously unstable in this closure. This concerns radial mass-transport stability on wavelengths where the vertically averaged thin-disc model applies.

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 $(1+2\lambda )mg/(3CL)$.

Use a constant porosity $\varphi$ and let $v$ denote pore velocity. Darcy law gives the Darcy velocity $q=\varphi v$. If the velocity convention has already absorbed porosity, set $\varphi =1$ in the following formulas. Assume positive permeability of a porous medium throughout the layer, so $k_0>0$ and $k_0+k_1>0$.Here $\mu$ is dynamic viscosity. The flux-weighted residence time in a porous layer is the pore volume divided by throughput. Equivalently, weighting each streamline's transit time $L/v(y)$ by its inlet flux $v(y)dy$ givesThis is also the advective travel scale after cross-layer mixing has homogenized the concentration. For resident-weighted transit time in a layered flow, sampling the initial fluid uniformly by resident volume, while suppressing cross-stream exchange, is a different experiment: its mean iswith the continuous $k_1=0$ limit $\mu \varphi L^2/(\Delta P{k}_0)$. Specifying the sampling convention avoids confusing the reciprocal of the mean speed with the mean reciprocal speed.

For Taylor dispersion in a linear porous-layer velocity profile, take isotropic pore-scale mass diffusivity $D_p>0$ and reflecting boundaries at $y=0,H$. Write $v-U=\gamma (y-H/2)$, where $\gamma =\Delta P{k}_1/(\mu \varphi LH)$. Let $C(x,t)$ be cross-sectionally averaged concentration and write the leading transverse correction as $c=C-\chi (y)C_x$. Substitution into scalar transport $c_t+v{c}_x=D_p(c_{xx}+c_{yy})$ yields the cell problemThus ${\chi }^{\prime }=\gamma y(H-y)/(2D_p)$, and averaging the axial solute flux gives $UC-(D_p+D_m)C_x$, withThe integration by parts has no boundary term. It also proves nonnegativity, even when $k_1<0$. The total effective mass diffusivity is $\mathcal{D}=D_p+D_m$. For anisotropic pore dispersion, the denominator uses the transverse coefficient and the added molecular term uses the axial coefficient.

There are two separate tests for significance. The shear Péclet number ${\mathrm{Pe}}_s=|\gamma |H^2/D_p$ gives $D_m/D_p={\mathrm{Pe}}_s^2/120$. A large value implies substantial enhancement of axial spreading. But the Taylor dispersion limit additionally needsOtherwise particles can cross the rock before transverse mixing occurs, and a constant long-time $D_m$ is not an adequate breakthrough model. The relative front width is of order $\sqrt{\mathcal{D}/(UL)}$ once the long-time model applies.

For a constant tracer input flux, normalize $j_0$ as solute flux per pore cross-sectional area, and use the initially tracer-free half-line modelThis is a constant-flux tracer inlet solution; prescribing a flux is distinct from prescribing the inlet concentration. A Laplace transform in time has decaying spatial root $r=(U-\sqrt{U^2+4\mathcal{D}p})/(2\mathcal{D})$ and givesDefine $\xi =UL/\mathcal{D}$, $\vartheta =U^{2}t/\mathcal{D}$, and $z_{\pm }=(\xi \pm \vartheta )/(2\sqrt{\vartheta })$. Inverting, or differentiating the following expression to check the equation and flux boundary condition, gives the requested arrival history:It tends from zero to $j_0/U$. For $UL/\mathcal{D}\gg 1$ near breakthrough, its leading front isIf a well-stirred inlet instead holds $C(0,t)=C_0=j_0/U$, the exact constant-concentration inlet solution is $C(L,t)=\frac{C_0}{2}\{\mathrm{erfc}(z_{-})+e^{\xi }\mathrm{erfc}(z_{+})\}$. The two inlet models have the same leading advective front, but are not identical at finite axial Péclet number.

The required boundary model is a fracture that drains after the storm, with $h(0,t)=0$, and a current of finite extent whose flux vanishes at its advancing nose. Introduce mass density $\rho$, gravity $g$, constant porosity $\varphi$ and dynamic viscosity $\mu$. Hydrostatic Darcy law gives the depth-integrated discharge $q=-k\rho gh{h}_x/\mu$. Fluid-volume conservation then gives the Boussinesq equation for an unconfined aquiferThe fluid volume per fracture length is $V=\varphi {\int }_0^{\infty }hdx$; omitting the constant $\varphi$ does not change its fractional decay rate.

The conserved first moment of a draining porous current follows directly:The finite outlet discharge does not contribute to $[xh{h}_x]$ at $x=0$. This establishes $D=\int xhdx$ constant. An unspecified nonzero fracture head would instead give $\dot{D}=\kappa h(0,t{)}^2/2$; the drained boundary is essential.

For the dipole similarity solution of a draining porous current, set $h=a{t}^{\alpha }f(\eta )$ and $\eta =x/(d{t}^{\beta })$. The conserved first moment requires $\alpha +2\beta =0$, while matching the time powers in the evolution equation gives $\alpha -1=2\alpha -2\beta$. HenceChoose the nose at $\eta =1$ and the normalization $\kappa a/d^2=1$. The profile equation becomesFor $f=b{\eta }^2+c{\eta }^{1/2}$, the two sides have coefficients $-b,-5c/8$ and $6b^2,15bc/4$, respectively. The nonzero positive profile therefore has $b=-1/6$; its zero at $\eta =1$ gives $c=1/6$. Its normalization is determined byThus a complete choice of the constants isSet $h=0$ beyond the nose. The square-root behavior at the outlet allows finite drainage: $f{f}^{\prime }\to 1/72$ as $\eta \downarrow 0$, while $f{f}^{\prime }\to 0$ at the nose.

Since ${\int }_0^{1}fd\eta =1/18$, the volume isThe drainage volume decay in a dipole porous current shows that the printed positive rate cannot describe drainage and its factor is also inconsistent with the conserved-moment scaling. This solution supplies a direct counterexample to that rate and the corrected one. If the initial volume $V_0$ is prescribed at the start of a similarity phase, replace $t$ by $t+t_0$ with $t_0=(\varphi ad/(18V_0){)}^4$. A general post-storm initial shape need not be exactly self-similar; this is the dipole similarity profile and its long-time scaling, not an asserted exact profile for every initial condition. The initial volume alone does not specify $D$.

Take $A>0$ for the repulsive molecular interaction, $p_0$ as the common reference pressure, and let $T_i=T_m-\delta T$ be the ice-film temperature. All ratios involving $T_m$ use its absolute value, approximately $273.15\mathrm{K}$, rather than the numerical Celsius value zero. The premelting film permits water to reach and freeze at the bottom of the ice while the upper surface is displaced upward without freezing there.

For equal phase mass densities, equality of the solid and liquid chemical potentials gives the Clapeyron pressure relation for equal-density phases. Indeed, $d\mu =-sdT+dp/\rho$ and $s_l-s_s=L/T_m$, so expansion around coexistence at $(T_m,p_0)$ yieldsThis is a pressure difference between phases, not a same-pressure Clausius-Clapeyron slope. The molecular disjoining pressure supports this difference:Treat the normal solid load as its effective pressure in this planar model. The gravitational load corresponding to hydrostatic pressure is $p_s=p_0+\rho gh$, so $p_l=p_0+\rho gh-p_T$. Refer the bath pressure to the sheet's upper gravitational datum, or neglect the sheet-scale hydrostatic head. The Darcy flow law then gives the upward supply and the ice growth rateHere $\Pi$ is permeability of a porous medium, $\mu$ is dynamic viscosity, and equal densities identify supplied liquid volume with added ice volume to leading order in $a/h$.

To recover the printed law, use the usual hydraulic control of frost heave idealization: the sheet's top is at $T_m$, the liquid film and ice have a common thermal conductivity $K$, and the flow is slow enough that latent-heat production is small compared with the conductive heat passing through the layer. In addition to a quasi-steady ice temperature field, this needsThe leading heat flux is consequently continuous through the film and ice. Their thermal resistances giveLet $B_T=\rho L\Delta T/T_m$, a pressure scale. The thermomolecular pressure and the molecular law then giveSubstitute this pressure into the Darcy flow law:The first term draws water toward the undercooled ice; the increasing gravitational load eventually cancels it.

Quasi-steady temperature alone is not sufficient to fix this exact prefactor. With film conductivity $K_l$, ice conductivity $K_s$, and nonnegligible latent heat, the appropriate additional balance isTogether with the Darcy equation, this is the implicit growth model. Even in the slow-flow limit, a conductivity ratio $K_s/K_l=r$ changes the leading driving pressure by $r^{3/4}$. Taking $r=2$ is a counterexample to obtaining the printed prefactor from quasi-steadiness alone. Thus the boxed equation is the intended equal-conductivity, thin-film, hydraulically limited model, rather than a consequence of only the stated quasi-steady assumption. This distinction is the heat-balance correction to a premelted-film growth model.

For that reduced equation defineThe scale $h_{\ast }$ balances $F{h}_{\ast }^{-3/4}$ against $\rho g{h}_{\ast }$; $F$ has units of pressure times length to the power $3/4$, so $h_{\ast }$ is a length and $t_{\ast }$ is a time. They yield

Set $u={\eta }^{7/4}$. The equation becomes linear:For initial thickness $\eta (0)={\eta }_0\ge 0$, the explicit solution of gravity-limited frost heave isThe usual zero-initial-thickness sketch usesIts early and late behaviour areFor $0<\eta <1$, the derivative is positive and ${\eta }^{\prime \prime }=(-\frac{3}{4}{\eta }^{-7/4}-1){\eta }^{\prime }<0$, so the curve is increasing and concave downward. It approaches the stable thickness $h_{\ast }$ exponentially. Initial thickness above $h_{\ast }$ instead relaxes downward; at $h_{\ast }$ it is stationary.

The singular slope at zero belongs to the formal reduced solution. Since $a/h\propto h^{-3/4}$, the thin-film assumption eventually fails as $h\to 0$. The early power law therefore describes an intermediate continuum regime after any microscopic startup, not arbitrarily early physical times. A finite positive ${\eta }_0$ supplies a regular initial condition when that regime begins. Gravity limits the ultimate thickness; molecular suction and the Darcy resistance set the growth toward it.

Let $K$ be intrinsic permeability of a porous medium, $\varphi$ the mobile pore fraction, and let $h$ measure thickness normal to the sloping cap. Define cold and reservoir-temperature properties byThe imposed inequality makes ${\delta }_c>0$, so the cold fluid is still buoyant. Assume $K>0$, $\sin \theta >0$, positive densities and viscosities, and the usual positive thermal-expansion and viscosity coefficients. Away from localized fronts, neglect the along-current thickness gradient. Hydrostatic water pressure and Darcy's law then give the upslope Darcy velocitiesTypically $u_h>u_c$: warming increases buoyancy and decreases dynamic viscosity.

Current depths and the moving moving thermal front in a porous current. For this slope-driven porous gravity current, the injected mass flux per unit well length fixes the cold plateau thickness:The moving thermal front in a porous current moves at $V_T=\Gamma u_c$. Since $u_c$ is explicitly a Darcy velocity, the fluid mass stored per unit horizontal area is $\varphi \rho h$. Apply mass conservation across the moving moving thermal front in a porous current:Writing $\gamma =\varphi \Gamma$ and $r=u_c/u_h$, this gives the distal warm depthFor a retarded moving thermal front in a porous current take $0\le \gamma <1$; otherwise this two-plateau geometry needs reconsideration. The factor $(1-\gamma )/(1-\gamma r)$ is important: the distal mass flux is not generally $Q$, because mass is being stored as the cold region replaces the warm region. Setting the two fluxes equal would silently assume a stationary moving thermal front in a porous current.

If the model absorbs porosity into storage and uses pore velocity as its transport speed, the same formulas use $\gamma =\Gamma$. The explicit Darcy convention in the PDF instead gives $\gamma =\varphi \Gamma$; porosity must be specified or absorbed consistently. In the stationary-front limit $\gamma =0$, the two steady depths reduce to $Q/({\rho }_c{u}_c)$ and $Q/({\rho }_h{u}_h)$. In the Boussinesq approximation, use a common reference density in the mass factors while retaining ${\delta }_c,{\delta }_h$ in the driving force.

Leakage thresholds. A normal thickness $h_i$ generates cap overpressure ${\Pi }_i={\delta }_{i}g{h}_i\cos \theta$. Before leakage modifies the current, the two plateau values areFor the caprock leakage threshold $\Delta p>0$, the candidate critical mass-injection rates are thereforeCold-only leakage occurs for $Q_c^{\ast }\le Q<Q_h^{\ast }$, provided $Q_c^{\ast }<Q_h^{\ast }$. Both plateaus can leak once $Q\ge \max (Q_c^{\ast },Q_h^{\ast })$. In the usual common-density approximation, ${\mu }_c>{\mu }_h$ and $r<1$ ensure $Q_c^{\ast }<Q_h^{\ast }$, giving the expected sequence: no leakage, cold-only leakage, then leakage from both cold and warm regions. Without that approximation the ordering must be checked; the given inequality $\Delta \rho >\alpha \Delta T$ alone does not establish it. These are onset criteria computed on the nonleaking current, not a post-leakage mass budget.

For small slopes $\cos \theta \simeq 1$ and $\tan \theta \simeq \sin \theta \simeq \theta$. If another thickness convention is used, its hydrostatic column and projected flux must be changed consistently; one should not mix a normal thickness with a vertical-pressure formula lacking the cosine.

Cross-current heat conduction. Heat transfer from the warm formation makes temperature vary across the carbon dioxide depth and introduces a warming time controlled by thickness, thermal diffusivity and the surrounding rock's heat capacity. It smooths the sharp thermal adjustment and causes gradual changes in buoyancy, viscosity and the velocity profile. The cold fluid generally warms, becomes more mobile and requires less depth to carry a prescribed mass flux; the enhanced cold-region overpressure and its distinct leakage zone tend to shrink. Warming may occur before a parcel reaches the idealized advective moving thermal front in a porous current, especially for a thin current. A single supplied $\Gamma$ no longer describes all heat transport. Quantitative depths and thresholds then require a coupled temperature equation and thermal boundary data; their exact changes cannot be inferred from the linear property laws alone.

Introduce water dynamic viscosity $\mu$ and the net driving liquid-pressure difference $\Delta P=p_0-p_f>0$. If $p_c>0$ denotes capillary suction relative to an ambient gas pressure $p_a$, then $p_f=p_a-p_c$ and $\Delta P=p_0-p_a+p_c$. If $p_c$ is instead the prescribed liquid front pressure, use $\Delta P=p_0-p_c$. This makes the capillary sign convention explicit.

Hemispherical front and injection flux. A hemispherical surface at radius $r$ has area $2\pi r^2$. Quasi-steady mass conservation makes the total volume flux independent of $r$ in the saturated region, so Darcy's law givesThe changing pore volume is $2\pi \varphi (R^3-R_s^3)/3$, giving $Q=2\pi \varphi R^2\dot{R}$. HenceFor the initially dry idealization $R(0)=R_s$, integration gives the implicit front lawIts right-hand side increases for $R>R_s$, defining one advancing radius and thus the flux implicitly in time. Equivalently, with $r=R/R_s$ and $t_{\ast }=\mu \varphi R_s^2/(k\Delta P)$, $t/t_{\ast }=r^3/3-r^2/2+1/6$. The infinite initial flux is an idealization of zero wetted resistance; inertia, finite inlet geometry and pore-scale effects regularize its earliest stage.

For a thin wetted shell $\ell =R-R_s\ll R_s$, the law reduces toFor $R\gg R_s$, the hemispherical capillary imbibition model instead predictswith $Q/Q_{\infty }=1+R_s/R+O((R_s/R{)}^2)$. Radial spreading makes the hydraulic resistance approach a finite value, while the increasing front area reduces its advance speed.

Comparison with one-dimensional imbibition. Under the same constant-driving-pressure, negligible-gravity assumptions, a planar front at depth $z$ has Darcy flux $q=k\Delta P/(\mu z)$ and storage law $\varphi \dot{z}=q$. ThereforeA fixed-area one-dimensional flux decays as $t^{-1/2}$, whereas the hemispherical total flux tends to a constant and its radius grows as $t^{1/3}$. If purely vertical imbibition includes gravity, upward penetration eventually approaches a capillary-rise height and downward penetration has a different gravity-driven limit. The hemispherical assumption itself neglects that directional gravity effect, so the power-law comparison is within its capillary-dominated regime.

Evaporation-limited steady radius. Take $F_e$ as volumetric water loss per unit front area per unit time. At equilibrium the incoming flux equals $2\pi R^2{F}_e$, givingIf the stated loss is a mass flux, replace $F_e$ by $F_e/{\rho }_w$. Porosity cancels from this equilibrium but affects the time to approach it. The incoming flux decreases with $R$ while the evaporating area increases, so the equilibrium is stable in this model. As $F_e\to 0$, $R_{\ast }\sim [k\Delta P{R}_s/(\mu F_e){]}^{1/2}$; sufficiently large radii can invalidate the negligible-gravity or semi-infinite homogeneous-medium assumptions.

Let $s$ be arclength, $\mathbf{t}={\partial }_s\mathbf{r}$ the unit tangent, and $\mathbf{v}$ the local filament velocity relative to the fluid. The resistive-force theory force per unit arclength, exerted by the fluid on the filament, isThe parallel and perpendicular drag coefficients of a slender filament are proportional to dynamic viscosity. For radius $a$ and a relevant axial cutoff length $\ell \gg a$, the anisotropic slender-filament drag from Stokeslet integration gives the leading logarithmic formsThe cutoff may be set by filament length or deformation wavelength. End shape and the choice of cutoff change order-one constants inside the logarithms. Slender-body theory also includes nonlocal interactions that the local resistive-force theory approximation omits. Typically ${\xi }_{\perp }>{\xi }_{\parallel }$, with ratio tending to two in the slender limit; this drag anisotropy enables propulsion.

Write $C={\xi }_{\perp }-{\xi }_{\parallel }>0$. For a small-amplitude planar deformation, $\mathbf{t}=(1,y_x)+O(y_x^2)$ and the leading transverse velocity is $y_t$. Exact inextensibility requires the quadratic longitudinal displacement of an inextensible planar filament: if $X(s,t)=s+X_2(s,t)$ and $Y(s,t)=y(s,t)$, then $X_{2,s}=-y_s^2/2$ to this order. Its velocity is quadratic and has zero time average for a periodic material cycle without mean axial drift. Thus the fixed material abscissa in the question is leading-order notation; this correction does not change the averaged propulsive term below. An imposed mean translation would contribute its separate ordinary axial drag.

To leading nonzero order,The time-averaged deformation-induced propulsive force of a periodic planar filament and power of a periodic planar filament are consequentlyThese are the leading quadratic functionals. Their signs refer to force on the filament. For example, $y=a\sin (kx-\Omega t)$ gives negative $F_x$, opposing the direction of wave propagation.