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Integrating a Stokeslet distribution along a straight slender filament explains the leading drag anisotropy. Axial motion samples the tensor component $2/(8\pi \mu |z|)$, while transverse motion samples $1/(8\pi \mu |z|)$. Integrating on both sides from radius $a$ to cutoff $\ell$ gives ${\xi }_{\parallel }\sim 2\pi \mu /\ln (\ell /a)$ and ${\xi }_{\perp }\sim 4\pi \mu /\ln (\ell /a)$. End effects and nonlocal hydrodynamic interactions supply the corrections retained by slender-body theory.

In the small-amplitude slender-body theory approximation, an undulating swimmer accelerates an added mass of fluid per unit length $m(x)$. With $\mathcal{D}={\partial }_t+U{\partial }_x$ and transverse displacement $h$, the lateral force on the body is $-\mathcal{D}(m\mathcal{D}h)$. Its reaction on the fluid has the opposite sign. For a periodic stroke and the usual negligible leading-edge added-mass contribution, the mean trailing-edge thrust is $m⟨h_t^2-U^2{h}_x^2⟩/2$. This is a reactive inertial model, distinct from viscous resistive-force theory.

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.

Let $\eta (x,t)$ be an arbitrary smooth variation periodic in both variables. The first variation of the constrained functional isUsing integration by parts in $x$ and $t$, every boundary term cancels by periodic boundary conditions. ThereforeThe Euler-Lagrange equation isFor a nontrivial stroke with nonzero thrust, $\Gamma \ne 0$. Set $c=C/(\Gamma {\xi }_{\perp })$ and $v=y_t$. Then $v_t+c{v}_x=0$, whose method of characteristics solution is $v=V(x-ct)$. Integrating in time gives the travelling-wave stationarity of planar filament propulsion:Here $g$ is an arbitrary static periodic component. It contributes neither power nor mean propulsion because the time average of $y_t$ vanishes. It can be set to zero when describing the active stroke. The wave must be compatible with both specified periods: $f(x-cT)=f(x)$.

For a pure travelling wave, its leading force and power areThus positive $x$ propulsion uses a wave travelling in the negative $x$ direction. Any smooth interior maximizer satisfying only the stated power constraint must have this travelling active deformation. If the drag were isotropic, $C=0$ and there would be no leading propulsion.

There is a qualification to the printed maximization claim: the variation proves stationarity, not the existence of an unrestricted global maximum. The bandwidth limitation in fixed-power filament optimization can be seen directly. Set $k=2\pi /\lambda$, $\Omega =2\pi /T$, and consider $y_m=a\sin (mkx+\Omega t)$. The leading expressions giveAt fixed power the formal quadratic objective increases with $m$. Eventually that sequence violates the small-slope approximation, but the approximation supplies no numerical slope or wavelength cutoff with which to define the global optimization. Even within a strict small-slope neighbourhood, a sufficiently small admixture of a higher spatial harmonic can improve a candidate interior maximum while keeping power fixed. For example, $y=a[\sqrt{1-{\eta }^2}\sin (kx+\Omega t)+\eta \sin (mkx+\Omega t)]$ retains the fundamental periods and exactly the same leading power, while its force is multiplied by $1+(m-1){\eta }^2$. For fixed $m>1$, taking $a$ and $\eta$ small keeps every slope small.

With an explicit admissible Fourier bandwidth or an additional geometric constraint, the intended conclusion can be made precise. For a finite set of nonstatic Fourier series modes proportional to $e^{i(k_{j}x-{\Omega }_{j}t)}$, with wavenumbers $k_j$ and frequencies ${\Omega }_j$,This is a weighted average of $-C{k}_j/({\xi }_{\perp }{\Omega }_j)$. A largest allowed ratio is attained by a single travelling mode, or by modes sharing the same phase speed; their superposition is still a travelling wave. Thus the Euler-Lagrange calculation yields the requested travelling-wave form, while a global maximum requires a specified admissible shape class.

Let $e=(\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)/2$ be the rate-of-strain tensor and $\mathbf{\sigma }=-pI+2\mu e$ the Newtonian fluid stress tensor. The viscous dissipation is $D[\mathbf{u}]=2\mu {\int }_{V}e:edV$. The Minimum-dissipation theorem for Stokes flow compares the Stokes flow with every sufficiently regular incompressible flow in the same domain having the same prescribed boundary velocity, with no body force or with a conservative body force absorbed into the fluid pressure. A trial field need not satisfy the Stokes flow equations. Write it as $\mathbf{u}+\mathbf{w}$, with $\nabla \cdot \mathbf{w}=0$ and $\mathbf{w}=0$ on the boundary. Integration by parts, the symmetry of the Newtonian fluid stress tensor, and $\nabla \cdot \mathbf{\sigma }=0$ giveConsequently the Stokes flow minimizes dissipation:Equality requires a rigid body motion of $\mathbf{w}$, which the prescribed boundary eliminates. Extending the flow inside the inserted particle as its rigid body velocity produces an admissible comparison field with zero internal rate-of-strain tensor; hence the extra dissipation due to a rigid inclusion is nonnegative.

Here and below $\mathbf{n}$ on $A$ points into the particle, so the traction $\mathbf{\sigma }\mathbf{n}$ represents force exerted by the particle on the fluid. Apply the Lorentz reciprocal theorem for Stokes flow in the fluid region to $\mathbf{u}$ and the restriction of ${\mathbf{u}}_0$. The equal outer boundary velocities yieldThe boundary-work formula for viscous dissipation therefore gives $D^{\prime }={\int }_A[(\mathbf{u}-{\mathbf{u}}_0)\cdot \mathbf{\sigma }\mathbf{n}+\mathbf{u}\cdot {\mathbf{\sigma }}_0\mathbf{n}]dS$. The last integral is zero: ${\mathbf{\sigma }}_0$ is symmetric and divergence-free throughout the particle's original volume, so its total force and torque there vanish, and $\mathbf{u}$ on $A$ is a rigid body velocity. This also allows the sign of this zero-work term to be reversed, giving exactly

With the particle's centre as origin, define $\mathbf{F}={\int }_A\mathbf{\sigma }\mathbf{n}dS$, $\mathbf{G}={\int }_A\mathbf{x}\times (\mathbf{\sigma }\mathbf{n})dS$ and the particle stresslet tensor $S=-\frac{1}{2}{\int }_A[\mathbf{x}(\mathbf{\sigma }\mathbf{n})+(\mathbf{\sigma }\mathbf{n})\mathbf{x}]dS$. Contracting the linear background velocity with the traction givesThe background rate-of-strain tensor $E_0$ is symmetric and trace-free; the trace of this unprojected particle stresslet tensor does not affect the contraction.

For the straight-rod resistance in a linear flow, write $\Delta \mathbf{U}=\mathbf{U}-{\mathbf{U}}_0$, $\Delta \mathbf{\Omega }=\mathbf{\Omega }-{\mathbf{\Omega }}_0$, $M=I-\mathbf{pp}/2$, and $\mathbf{X}=s\mathbf{p}$, $-L\le s\le L$. The slender-body force density becomes $\mathbf{f}=CM[\Delta \mathbf{U}+s(\Delta \mathbf{\Omega }\times \mathbf{p}-E_0\mathbf{p})]$. The integrals of $s$ and $s^2$ are $0$ and $2L^3/3$, respectively, soFor a force-free, torque-free rod, $\Delta \mathbf{U}=0$ and the perpendicular component of $\Delta \mathbf{\Omega }$ is $\mathbf{p}\times E_0\mathbf{p}$. Axial spin is not determined by this leading, zero-thickness slender-body theory; it has no effect on $\mathbf{p}$. Put $a_p=\mathbf{p}\cdot E_0\mathbf{p}$. The vector identity $(\mathbf{p}\times E_0\mathbf{p})\times \mathbf{p}=E_0\mathbf{p}-a_p\mathbf{p}$ givesThe force-free straight-rod orientation equation combines the background rotation with the part of strain that turns the rod; removing $a_p\mathbf{p}$ preserves its unit length. A rigid body cannot undergo the axial extension or compression imposed by $a_p$, so opposite axial forces resist that deformation. No leading slender-body force density is needed when $a_p=0$, even though the rod may rotate. Its particle stresslet tensor and viscous dissipation are

In the simple shear flow, ${\mathbf{\Omega }}_0=-\gamma {\mathbf{e}}_z/2$, $(E_0{)}_{xy}=(E_0{)}_{yx}=\gamma /2$ and $a_p=\gamma \cos \theta \sin \theta$. Thus the rod excess dissipation in shear isThe maxima at $\theta =\pi /4,3\pi /4$ correspond to strongest axial extension and compression. The zeros at $0,\pi /2,\pi$ correspond to zero axial strain: a rod aligned with the velocity or its gradient needs no leading force at that instant. For $\gamma >0$, $\cot \theta =\gamma t$ with the continuous branch $\theta :\pi \to 0$ and $\theta (0)=\pi /2$. Reversing the shear flow reverses the traversal. Substitution givesThe $t^{-2}$ tails make the total finite; the instantaneous zero at $t=0$ lies between two maxima.

For a dilute rod suspension, the added bulk stress is the number density times the orientation average of the particle stresslet tensor. In this ideal infinitely slender, non-interacting model, rods approach alignment with the shear flow and their excess viscous dissipation decays. A real finite-aspect-ratio rod continues to tumble; rotational diffusion and interactions also maintain a spread of orientations. One therefore expects increased effective viscosity, with alignment offering a mechanism for reduced excess viscosity at strong shear flow. This is a qualitative expectation, not a claim that a finite rod suspension has zero steady excess stress.

In resistive-force theory (RFT), the hydrodynamic force per unit arclength exerted by the fluid on a slender filament isHere $\mathbf{t}$ is the unit tangent and $\mathbf{u}$ the local velocity relative to the background fluid. The parallel and perpendicular drag coefficients of a slender filament are positive and generally satisfy ${\xi }_{\perp }>{\xi }_{\parallel }$; their leading logarithmic ratio is about two. This local approximation assumes a small filament radius compared with length and curvature scales, negligible fluid inertia, and a Newtonian fluid. It represents drag by the local tangent direction and neglects nonlocal hydrodynamic interactions between separated filament segments. Boundaries, close approaches and end corrections may require slender-body theory or a more complete flow calculation. In this problem the background fluid is at rest, and the drag coefficients are taken uniform along the filament.

The rigid-body velocity in a deforming swimmer frame is the sum of translation, rotation and material deformation. With $\mathbf{r}=x{\mathbf{e}}_x+ϵg{\mathbf{e}}_y$,so, in instantaneous swimmer-frame components,The reference-frame conditions $g(0,t)=g_x(0,t)=0$ attach the origin and orientation to the filament's base and base tangent. They prevent arbitrary shape translations or tilts from being absorbed into the definition of $g$.

The exact tangent and arclength element areSince $U,V,\Omega =O(ϵ)$, the local velocity is $O(ϵ)$. Changing the drag tensor by its $O(ϵ)$ tangent correction therefore changes $\mathbf{f}$ only at $O({ϵ}^2)$. The arclength correction is smaller still at this order. Also $-ϵ\Omega g=O({ϵ}^2)$. Thus $\mathbf{t}={\mathbf{e}}_x$ is sufficient at order $ϵ$. This assumes the small-slope expansion is uniform, $ϵ|g_x|\ll 1$.

Put $G_0(t)=⟨g_t{⟩}_x$ and $G_1(t)=⟨x{g}_t{⟩}_x$, with $⟨h{⟩}_x=L^{-1}{\int }_0^{L}hdx$. The first-order local force isIts total hydrodynamic forces areTaking the moment about the swimmer-frame origin, $M_z=\int (x{f}_y-ϵg{f}_x)ds$, the second term is beyond first order. HenceFor force-free and torque-free motion, the three leading coefficients vanish. The longitudinal force immediately gives $U_1=0$. The two transverse equations areTheir determinant is $L^2/12>0$. Solving yields the first-order free swimming of a planar filament:The resistive-force theory coefficient cancels because both equations use the same transverse drag. Another interpretation is the least-squares projection of filament deformation velocity: $-V_1-{\Omega }_{1}x$ is the best affine approximation to $g_t$ on $[0,L]$. Zero transverse force and torque mean that the residual is orthogonal to $1$ and $x$.

If $g$ is sufficiently differentiable and periodic with period $T$, each $g_t(x,t)$ has zero temporal mean. More explicitly,Both bracketed quantities return to their initial values over a period. Thus the first-order periodic transverse swimming velocity satisfiesThese results concern first order and swimmer-frame components. Changes of orientation can affect laboratory displacements at second order; higher-order net swimming is not ruled out.

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.

In resistive-force theory, the force per unit length is local and linear in the filament velocity relative to the fluid:The distinct parallel and perpendicular drag coefficients encode the anisotropic resistance of a slender filament. Neglecting nonlocal hydrodynamic interactions makes this a leading logarithmic approximation to slender-body theory.

For a straight rod $\mathbf{X}=s\mathbf{p}$, $-L<s<L$, leading slender-body theory uses $\mathbf{f}=C(I-\mathbf{pp}/2)[\Delta \mathbf{U}+s(\Delta \mathbf{\Omega }\times \mathbf{p}-E\mathbf{p})]$. Integration gives $\mathbf{F}=2CL(I-\mathbf{pp}/2)\Delta \mathbf{U}$ and $\mathbf{G}=(2C{L}^3/3)[(I-\mathbf{pp})\Delta \mathbf{\Omega }-\mathbf{p}\times E\mathbf{p}]$. Axial spin is outside this centreline approximation.