For the horizontal film, part (a) gives the lubrication theory equationThe length scale is fixed at $L$, so balancing $h/t$ against $K{h}^4/L^2$ gives $h\sim (L^2/(Kt){)}^{1/3}$. The draining viscous film on a finite horizontal plate therefore has the long-time similarity solutionAn additive shift of the time origin can represent the initial transient without changing the long-time law. The positive profile is even, with $H^{\prime }(0)=0$ and $H(\pm 1)=0$. Substitution yieldsTo integrate, put $q=H^3{H}^{\prime }$ and use $q^{\prime }=qdq/dH/H^3$. ThenFor $0<\eta <1$ the decreasing branch has $q<0$, so inversion givesThe negative half of the profile follows by symmetry.

At the edge, $\eta (0)=1$. Set $u=(s/H_0{)}^5$ in the integral to obtainHere $B$ is the beta function. Thus $H_0^3=10/(3C^2)$, and the centre thickness is

For $H\ll H_0$, the implicit integral remaining between $0$ and $H$ givesConsequently the edge region of a draining viscous film hasIts slope satisfies $|h_x|\sim (h_0/L)(1-x/L{)}^{-3/4}$ up to a constant factor, so it becomes order one atThe local height is then also $O(\ell )$, violating the shallow geometry required by lubrication theory. A full local flow is needed to turn the fluid over the edge.

The imposed zero thickness is nevertheless a consistent leading outer boundary condition: $\ell /h_0\sim (h_0/L{)}^{1/3}\ll 1$, so the unresolved edge height is small compared with the film's bulk height. It should not be interpreted as an exact pointwise prediction inside the edge region. Moreover, $H^3{H}^{\prime }\to -\sqrt{2/15}{H}_0^{5/2}$ at the right edge, so the outward volume flux remains finite even as the outer thickness tends to zero. The singular slope is what allows this outer solution to describe drainage.

For an axisymmetric viscous gravity current on a cone, the cylindrical radius at slant distance $x$ is $x\cos \alpha$. A surface annulus therefore has area $2\pi x\cos \alpha dx$. The local downslope volume flux is the same as in part (a), but conservation of mass must account for this growing circumference. ThusThis is a leading thin-film description away from the apex, with $0<\alpha <\pi /2$ and thickness small relative to the local curvature and streamwise length scales.

The original PDF uses the long-time criterion $x_N\gg (V\csc \alpha {)}^{1/3}$; the local TeX incorrectly reads $V\cos \alpha$. To see why the PDF criterion is the relevant one, scale $h\sim V/(x_N^2\cos \alpha )$. In the bulk, the hydrostatic-gradient term relative to direct downslope gravity isHence the leading outer evolution, with $\beta =K\sin \alpha$, isFor a fixed-volume similarity solution, let $x_N\propto t^b$ and $h\propto t^a$. Conservation of mass gives $a+2b=0$, while the evolution equation gives $a-1=3a-b$. Thus $b=1/5$, $a=-2/5$. Writing $h=t^{-2/5}H(\xi )$, $\xi =x/t^{1/5}$, yieldsThe positive branch $H=c\sqrt{\xi }$ satisfies this equation when $c^2=1/(5\beta )$. It givesThe ideal outer current has zero thickness beyond its front. Its mass integral isThereforeThe reduced equation is a conservation law after multiplication by $x$. Its front obeys the Rankine-Hugoniot condition, ${\dot{x}}_N=\beta h_N^2=x_N/(5t)$, consistent with the mass-derived front position. The outer profile has a finite jump at the front; the omitted hydrostatic-gradient term resolves a narrow front region. There is also a small apex region where the outer gradient and conical curvature violate the local approximations. Neither region sets the leading conserved-volume scaling above.

Let $z$ measure distance normal to the plane, with the wall at $z=0$ and the free surface at $z=h$. Take density $\rho$, dynamic viscosity $\mu$, and gravitational acceleration $g$. In lubrication theory, normal momentum balance gives hydrostatic pressure while inertia and streamwise viscous derivatives are smaller. With negligible surface tension and negligible air shear,Apply the no-slip boundary condition $u=v=0$ at $z=0$, and zero leading tangential stress $u_z=v_z=0$ at $z=h$. The velocity profiles areDepth integration gives the volume flux, with $K=\rho g/(3\mu )$:Combining incompressibility with the wall and surface kinematic conditions gives conservation of mass, $h_t+{\partial }_x{q}_x+{\partial }_y{q}_y=0$. Therefore the two-dimensional version of the gravity-driven thin film on an incline satisfiesDirect downslope gravity supplies the advective term, while gradients of the hydrostatic pressure spread the film along the surface. At $\alpha =0$ this reduces to nonlinear gravitational spreading in a horizontal film.

Take $\mathbf{F},\mathbf{G}$ to be the force and couple exerted by the body on the fluid, equivalently the external force and couple needed to maintain its motion. This fixes the sign for a positive hydrodynamic resistance matrix. If one uses the fluid's force on the body, both resultants have the opposite sign.

For two Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\tilde{\mathbf{u}},\tilde{\mathbf{\sigma }})$ in the same fluid domain $\mathcal{D}$, with zero body force, the Lorentz reciprocal theorem for Stokes flow isHere $\mathbf{n}$ points out of the fluid. To see the identity, take the divergence of the difference of the two cross-work fluxes. The stress divergences vanish, while incompressibility and symmetry of the Newtonian fluid stress tensor reduce the remaining terms to $2\mu (\mathbf{e}:\tilde{\mathbf{e}}-\tilde{\mathbf{e}}:\mathbf{e})=0$. The divergence theorem proves the result.

On the body, the no-slip boundary condition is $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{r}$, and the far-field contribution vanishes for decaying exterior flows. Reciprocity becomesWith $q=(\mathbf{U},\mathbf{\Omega }{)}^T$ and $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}q$, this is $q^T\mathsf{R}\tilde{q}={\tilde{q}}^T\mathsf{R}q$ for all pairs, so ${\mathsf{R}}^T=\mathsf{R}$. The power identity givesEquality would force zero strain throughout the connected exterior domain, hence a rigid fluid motion. Decay at infinity makes that motion zero, and the no-slip boundary condition then gives $q=0$. This proves the symmetry and positivity of a rigid-body resistance matrix:

For the helix, assume $0<\varphi <\pi /2$ and use cylindrical unit vectors. Its arclength element and tangent areIntegrating to $\theta =L\cos \varphi /b$ gives $\boxed{\text{total arc length}=L}$. In a combined axial translation and rotation, $\mathbf{V}=U{\mathbf{e}}_z+b\Omega {\mathbf{e}}_{\theta }$. The slender-body force density, or local resistive-force theory, givesWriting $s_{\varphi }=\sin \varphi$, $c_{\varphi }=\cos \varphi$, its axial and azimuthal components areThe axial couple density is $b{f}_{\theta }$. Integration over arclength gives the axial resistance matrix of a slender helix:Thus pure translation gives $(F_z,G_z)=(AU,BU)$, and pure rotation gives $(F_z,G_z)=(B\Omega ,D\Omega )$. The mixed coefficients agree, as reciprocity requires, and $AD-B^2=C^2{L}^2{b}^2/2>0$. The sign of $B$ follows the handedness in the parametrization; reversing handedness reverses propulsion.

For the helical microswimmer with a spherical head, the diagram's rotation relation is ${\Omega }_0=\Omega -\omega$. Neutral buoyancy and the absence of external forcing make the whole swimmer force-free and torque-free. Neglecting interactions between its parts givesWith $\Delta =(A_0+A)(D_0+D)-B^2>0$, solving this pair givesThe head counterrotates, providing the reaction to the flagellum's rotation. The figure fixes this relative rotation convention; it is duplicated and corrupted in the local TeX.

For a very large head, the coefficient regime is $A_0\gg A$ and $D_0\gg D$. ConsequentlyThe large translational resistance $A_0$ makes the microswimmer slow even though the flagellum rotates almost at the motor rate.

For a very small head, $A_0\ll A$ and $D_0\ll D$. The appropriate denominator retains the translation-rotation coupling:Both tend to zero with $D_0\propto a^3$, while ${\Omega }_0\simeq -\omega$. The motor mainly rotates the low-resistance head, and produces little flagellar motion relative to the fluid.

In the intermediate coefficient regime $A_0\ll A$ and $D_0\gg D$, the head supplies a strong rotational reaction with little translational penalty. ThenThe head radius has dropped out. To find the optimal pitch of a helical microswimmer, put $z=\tan \varphi$. The dimensionless speed becomes $z/(z^2+2)$ and has derivative $(2-z^2)/(z^2+2{)}^2$. Hence, for $\omega >0$ with the chosen handedness,The angle is measured from the horizontal plane. An intermediate pitch combines axial and transverse tangent directions, allowing anisotropic drag to convert rotation into translation; either extreme removes that coupling.

The geometric head-size regimes quoted in the paper hold with the logarithmic slenderness factor treated as fixed. More precisely, $A_0\ll A$ requires $a\ll L(1+{\cos }^2\varphi )/(3|\log ϵ|)$, while $D_0\gg D$ requires $a^3\gg L{b}^2(1+{\sin }^2\varphi )/(4|\log ϵ|)$. These coefficient inequalities state the validity of the intermediate approximation when the logarithm is quantitatively important. The many-turn and local-drag assumptions of slender-body theory also remain in force.

Let $T_q\phi (x)={\sum }_{k=0}^q{\phi }^{(k)}(0)x^k/k!$ denote the Taylor polynomial at zero. On $0<|x|<1$, the integrand defining the Hadamard finite-part reciprocal-power distribution isThe first term is odd, so it cancels under symmetric truncation. The Taylor remainder in the second term has absolute value at most $|x{|}^m\Vert {\phi }^{(m)}{\Vert }_{\infty }/m!$. The integral near zero therefore has a finite limit. For $|x|\ge 1$, the original terms are integrable: $x^{-m}\phi$ has compact support, and each subtracted power has exponent $k-m\le -2$.

This proves existence and linearity. It also gives the global estimateThusIn particular the order of a distribution is bounded by one integer on all compact sets, not merely separately on each of them.

For the recurrence, write the truncated Hadamard finite-part integral aswhere parity evaluates all subtraction integrals explicitly:Set $C_1=0$ and ${\Lambda }_1=\mathrm{pv}(1/x)$. On the two truncated intervals, integration by parts givesThere are no boundary contributions at infinity. The coefficients of the subtraction terms satisfyThis is exactly the negative of the boundary term's Taylor polynomial through degree $m$; the degree-$m$ term itself vanishes by parity. HenceTaking the limit proves the full distributional identityThe same calculation includes $m=1$. Starting with the distributional derivative of the logarithmic modulus from part (i), induction yieldsThis identity holds on all of $\mathbb{R}$ as a distributional identity. It is stronger than matching the ordinary derivatives away from zero, which would leave possible differentiated Dirac delta distributions at zero undetermined.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when all functions eventually have compact support in one compact set $K$ and every derivative converges uniformly:A distribution is a continuous linear functional on this space of test functions. Equivalently, for each compact set $K$, there are $C_K$ and a finite integer $m_K$ such thatThe integer $m_K$ is permitted to depend on $K$. The usual weak convergence of distributions isThus the convergence convention on the distribution space is its weak dual topology.

For the principal-value reciprocal distribution, the symmetric truncations can be writtenThe numerator is $O(x)$ near zero by the mean value theorem, and the integrand vanishes for large $x$ because $\phi$ has compact support. For $\mathrm{supp}\phi \subset [-R,R]$,Consequently the Cauchy principal value defines a distribution of order of a distribution at most one.

The function $\log |x|$ has local integrability, since ${\int }_0^1|\log x|dx<\infty$, and therefore defines a distribution. For its distributional derivative, remove $(-\epsilon ,\epsilon )$ and use integration by parts on both remaining intervals:The boundary term is $O(\epsilon |\log \epsilon |)$ and tends to zero. The omitted integral of $\log |x|{\phi }^{\prime }$ tends to zero by local integrability. Hence the distributional derivative of the logarithmic modulus satisfiesSymmetric truncation is essential to this normalization of the principal-value reciprocal distribution.