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For radius $b$, arc length $L$, tangent angle $\varphi$ to the horizontal plane, and local drag coefficient $C$, resistive-force theory givesThe sign of $B$ selects the handedness convention. With the force exerted on the fluid taken positive, $AD-B^2=C^2{L}^2{b}^2/2>0$. Translation-rotation coupling supplies propulsion for a rotating helical microswimmer with a spherical head.

Local resistive-force theory for a helix gives $AD-B^2={\ell }^2{a}^2{\xi }_{\parallel }{\xi }_{\perp }$. Positive local drag therefore guarantees a positive-definite axial resistance matrix and positive viscous power loss. The identity also keeps the denominator in counterrotating two-helix balances positive.

With drag ratio $\rho =c_{\parallel }/c_{\perp }$, squared-tangent average $\beta ={\Lambda }^{-1}\int x_s^{2}ds$, and axial wave speed $V$, resistive-force theory gives the displayed axial swimming ratio. Pure axial free swimming additionally requires vanishing cross-tangent average, as in reflection-symmetric waveforms; otherwise a transverse force or translation is present.

In resistive-force theory with $0<\gamma <1$, head drag $\delta \ge 0$ and the ideal constant-inclination condition ${\alpha }^2=\beta$, put $r=1-\gamma$, $d=1+\delta$, $A=r({\delta }^2-\gamma )$ and $B=\gamma d^2$. At prescribed swimming speed $U$, eliminating wave speed using linear resistive-force propulsion of a travelling filament givesIts derivative has numerator $A{\beta }^2+2B\beta -B$. This numerator is strictly increasing on $(0,1)$, is negative at zero, and equals $(\delta +\gamma {)}^2$ at one. Thus its unique zero is the displayed global minimum. When $A=0$ the formula gives ${\beta }_{\ast }=1/2$; when $\delta =0$ it gives $1/(1+\sqrt{\gamma })$. A planar sawtooth realizes constant inclination in the ideal model; smooth rounding and elastic costs change that idealization.

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.

A rigid rod with moving centre and changing orientation has material velocity $\dot{\mathbf{X}}+s\dot{\theta }\mathbf{n}$. Its centred rotational velocity is odd in arclength and has no resultant force, while its orientation-dependent resistance tensor couples translation to transverse force. Resistive-force theory uses per-length coefficients $c_{\parallel },c_{\perp }$.

The local resistive-force theory coefficients ${\xi }_{\parallel }$ and ${\xi }_{\perp }$ describe drag along and across a filament tangent. For very slender bodies their leading logarithmic values have ratio ${\xi }_{\perp }/{\xi }_{\parallel }\simeq 2$. Geometry, ends and nonlocal fluid interactions affect the corrections. Anisotropy enables propulsion by rotating helices.

Choose an explicit orientation convention, since the supplied PDF does not display the referenced diagram. Let ${\mathbf{e}}_z$ point upward and ${\mathbf{e}}_2$ complete a right-handed frame with the horizontal flow direction ${\mathbf{e}}_1$. Usewhere $\mathbf{i}$ points from the head towards the tail. A tail point has position $\mathbf{r}=(a+s)\mathbf{i}$. Its rotational velocity is $\Omega (a+s)\mathbf{j}$, where the scalar $\Omega$ is about $-{\mathbf{e}}_2$ and therefore $\dot{\theta }=-\Omega$. Keeping that sign is essential for stability.

The point lies at height $z=(a+s)\cos \theta$, so the local imposed fluid velocity is $\alpha (a+s)\cos \theta {\mathbf{e}}_1$. Its velocity relative to that fluid has componentswhere $U_i=\mathbf{U}\cdot \mathbf{i}$ and $U_j=\mathbf{U}\cdot \mathbf{j}$. These follow by projection; rigid rotation has no tangential component.

Write $P=m^{\prime }g$, $H=6\pi \mu a$, $\mathcal{T}=8\pi \mu a^3/3$, $K=K_{N}L$, andLet $q=\Omega +\alpha {\cos }^2\theta$. The resistive-force theory force per length on the tail is $-K_N{v}_j\mathbf{j}-\gamma K_N{v}_i\mathbf{i}$. Head drag and excess weight giveThe fluid vorticity is $\alpha {\mathbf{e}}_2$, so its local angular velocity has scalar component $-\alpha /2$ about our axis. Sphere rotation resistance therefore contributes $-\mathcal{T}(\Omega +\alpha /2)$. Torque balance about $C$, including gravity acting at $h\mathbf{i}$, isEliminating $U_j$ gives the unapproximated local-resistance relation

Now retain head and tail translation resistance at the same order, as intended by $\sigma =O(L/a)$, but use $a/L\ll 1$ in the lever-arm integrals. Thus $I_1\simeq L^2/2$, $I_2\simeq L^3/3$. Sphere rotation resistance is smaller than tail rotation resistance by order $(a/L{)}^2$, so omit $\mathcal{T}$ at fixed scaled shear. Also neglect $h$ relative to the tail's hydrodynamic force-centre distance $K_N{I}_1/(H+K)=O(L)$. Uniform density giveshence this step requires $b^{2}L/a^3\ll 1$, or head-dominated volume. It is consistent with the distinguished logarithmic ordering, but $b\ll a\ll L$ by itself would not ensure it. The model further neglects hydrodynamic interaction between head and tail, and uses low-Reynolds-number local drag.

Under these approximations,soBecause $\dot{\theta }=-\Omega$, its equilibria satisfy $\alpha (1-{\sin }^2\theta )=2\beta \sin \theta$. Definewith $s_{\ast }=0$ at zero shear. The two orientations modulo $2\pi$ are ${\theta }_s=\arcsin s_{\ast }$ and ${\theta }_u=\pi -{\theta }_s$. The derivative of $\dot{\theta }=\alpha {\cos }^2\theta -2\beta \sin \theta$ at an equilibrium isThe second factor is positive, so the branch $\cos {\theta }_s>0$ is stable and the other is unstable. On the stable branch the tail tip is above $C$: the head is lower than the far tail end.

The corresponding leading translation velocities areThe second formula uses $q=2\beta \sin \theta$ in the normal force balance. In the equilibrium orientation $\alpha {\cos }^2\theta =2\beta \sin \theta$, and $U_z=U_i\cos \theta +U_j\sin \theta$ simplifies toThis is shear-dependent settling of a sphere-and-tail body. Since $0<\gamma <1$ for a slender filament, $s_{\ast }^2$ increases with the magnitude of shear and the signed upward-coordinate velocity $U_z$ increases towards zero. Equivalently, increasing shear decreases the downward sedimentation speed. The formula remains negative throughout this leading model. If $\gamma =1$ there is no shear dependence; reversing the drag anisotropy would reverse the trend. At very large shear, the omitted sphere rotation term proportional to $\mathcal{T}\alpha$ can become important, so the approximation is not uniform in arbitrarily large $\alpha$.

The Lighthill elongated-body theory approximation treats each cross-section of a slender swimmer as a locally two-dimensional potential flow problem. The transverse dimension is small compared with the length and the scale over which cross-sections vary. The displacement and slope are small enough to linearize the impermeability condition about a straight body. The Reynolds number is sufficiently large for lateral loading to be dominated by fluid inertia; viscosity, axial friction and higher-order end corrections are neglected at this order. This reactive approximation must be distinguished from the viscous resistive-force theory used for a flagellum.

In coordinates translating with the swimmer, the undisturbed fluid passes in the positive $x$ direction at speed $U$. Applying the impermeability condition to the moving centre plane gives the transverse fluid velocityThe local added mass $m(x)$ is defined so that the transverse fluid hydrodynamic impulse per unit axial length is $P_y=m(x)v_y$. A fixed body-coordinate interval receives an impulse flux $U{P}_y$ at its leading end and loses the corresponding flux at its trailing end. Its rate of change of impulse plus the net outward impulse flux is thereforeThis is the transverse force exerted by the body on the fluid. By Newton's third law, the transverse force exerted by the fluid on the body has the opposite sign:The distinction is essential here: the printed positive expression is labelled as the force on the fish, so that label and expression are inconsistent. For example, take a uniform cross-section and a rigid lateral displacement $h=H(t)$. The swimmer must accelerate fluid of added mass $m$, which reacts on it with force $-m{H}^{\prime \prime }$, not $+m{H}^{\prime \prime }$. Thus the positive expression is recovered exactly with the force-on-fluid convention; the physically consistent force on the fish is the negative expression. In the following parts, $F_y$ denotes $F_{\mathrm{body}}$ so that the beam balance uses a genuine external force on the fish.

In resistive-force theory, the flagellum is slender and moves at a small Reynolds number, so local viscous resistance is linear in its velocity and fluid and body inertia are neglected. Normal and tangential resistance coefficients are treated as positive constants, usually ${\kappa }_N>{\kappa }_T$. The approximation neglects distant hydrodynamic interactions along the filament, head-tail interactions, nearby boundaries and self-contact. The centreline is an inextensible filament with prescribed shape. Its local resistance tensor isSince the filament velocity relative to distant fluid is $-\mathbf{w}$, the fluid force on a filament element is $K\mathbf{w}ds$. The equal and opposite force on the fluid has total axial component $T$. Projecting onto $\mathbf{i}$ givesThis establishes the thrust expression with its force-on-fluid convention.

For a uniform travelling wave, write the moving centreline as $\mathbf{r}(s,t)={\mathbf{r}}_{\mathrm{wave}}(s-ct)+(V-U)t\mathbf{i}$, where $s$ is material arc length and $|{\mathbf{r}}_{\mathrm{wave}}^{\prime }|=1$. Differentiating at fixed material label gives ${\mathbf{r}}_t=(V-U)\mathbf{i}-c\mathbf{t}$. HenceLet $q=\mathbf{t}\cdot \mathbf{i}=X_s$, $\alpha =⟨q⟩$ and $\beta =⟨q^2⟩$, where these brackets are contour-length averages. The axial wavelength divided by contour wavelength is $\alpha =V/c$. For a filament comprising whole wave periods, or after the usual period averaging, $c{\int }_0^{L}qds=VL$. This averaging is needed to avoid end-phase corrections for an arbitrary fraction of a wavelength. Substitution of $\mathbf{w}$ into the thrust expression now givesFor $\gamma ={\kappa }_T/{\kappa }_N$ and $r=1-\gamma$, this is $T/({\kappa }_{N}L)=Vr(1-\beta )-U(1-r\beta )$. The swimming body is force-free, so the reaction thrust on the head-tail assembly balances its head drag. Thus $T=\delta {\kappa }_{N}LU$, and linear resistive-force propulsion of a travelling filament givesWe henceforth assume $0<\gamma <1$, $\delta \ge 0$ and $0<\beta <1$. Isotropic resistance, $\gamma =1$, gives no propulsion in this model.

The flagellar contribution to viscous dissipation is the positive quadratic form $P_f=\int \mathbf{w}\cdot K\mathbf{w}ds$. Expanding it, and using $c⟨q⟩=V$, givesThere is also head dissipation $\delta {\kappa }_{N}L{U}^2$. Equivalently, the work of the flagellar deformation relative to the translating head is $P_f+UT$; the force balance makes $UT$ precisely the head dissipation. Consequently, with $\alpha =V/c$,This derives the rate of working, rather than merely inferring it from the thrust.

The Cauchy-Schwarz inequality gives ${\alpha }^2=⟨q{⟩}^2\le ⟨q^2⟩=\beta$. Equality requires $q$ to be constant almost everywhere. For a planar periodic filament this is an ideal sawtooth with alternating transverse tangent and fixed axial tangent, or a limit with negligibly rounded corners. It is not an identity for a general sinusoidal wave, and the ideal calculation omits the elastic cost of sharp corners.

For the stipulated equality ${\alpha }^2=\beta$, put $d=1+\delta$ and eliminate $V$ at fixed $U$ using $V/U=(d-r\beta )/[r(1-\beta )]$. Direct simplification yieldsThe derivative isWriting $A=r({\delta }^2-\gamma )$ and $B=\gamma d^2$, the numerator is $A{\beta }^2+2B\beta -B$. It is negative at zero and equals $(\delta +\gamma {)}^2>0$ at one. Its derivative is positive throughout $(0,1)$: this is immediate if $A\ge 0$; if $A<0$, its minimum derivative is $2(A+B)=2(\delta +\gamma {)}^2>0$. Since $F$ diverges at both endpoints, the unique stationary point is the global fixed-speed energy optimum for a planar flagellum:This form includes the case $A=0$, when ${\beta }_{\ast }=1/2$, without a removable division by zero. With no head drag it reduces to ${\beta }_{\ast }=1/(1+\sqrt{\gamma })$.

The printed final optimality formula does not follow from the preceding thrust and power equations. For a concrete check, $\gamma =1/2$, $\delta =0$ gives $F=(2-\beta )/[\beta (1-\beta )]$. The derived minimum is at $\beta =2-\sqrt{2}$. The printed formula instead gives ${\beta }^2=\sqrt{6}-2$, hence $\beta \approx 0.67044$, where $F^{\prime }\ne 0$ and the expenditure is greater. Thus the thrust and working formulas can both be established as above, but the last printed conclusion must be replaced by the derived optimum.

Take $x$ tailward, so forward swimming has laboratory velocity $-U\mathbf{i}$. For an inextensible travelling waveform, material motion in the swimmer frame is $V\mathbf{i}-c\mathbf{t}$: a material point passes through the wave along its tangent while the wave pattern translates at $V$. Thus the relative velocity of otherwise stationary fluid isPut $q=V-U$ and choose $\mathbf{t}=(X_s,Y_s)$, $\mathbf{n}=(-Y_s,X_s)$, with $X_s^2+Y_s^2=1$. ThenOn the forward-swimming branch $0<U<V<c$ and $X_s>0$, the tangential slip is positive. A law for reversed tangential slip would also need its sign, not just its absolute magnitude.

The quadratic normal law represents inertial cross-flow drag: dynamic pressure scales as fluid density times $w_n^2$, while $|w_n|w_n$ chooses the force direction. Its coefficient contains the cross-sectional size and an empirical drag coefficient. The $|w_t{|}^{3/2}$ exponent is characteristic of an attached laminar boundary layer: its thickness scales as $\delta \sim (\nu s/|w_t|{)}^{1/2}$ and its shear as $\mu |w_t|/\delta$. However, this gives local shear proportional to $|w_t{|}^{3/2}{s}^{-1/2}$, not $s^{1/2}$. The positive power of $s$ is genuinely printed in this paper and repeated in its target formulas. Below it is treated as the prescribed phenomenological weight, rather than misidentified as standard local laminar skin-friction scaling along a slender swimmer.

The normal axial force on an element isThe prescribed tangential force gives $d{F}_{T,x}=K_T(c-q{X}_s{)}^{3/2}{X}_s{s}^{1/2}ds$. Zero mean axial force therefore gives the nonlinear resistive-force balance for a travelling filamentThe absolute values are present in the original PDF. In a periodically beating, constant-mean-speed model, phase-average the integrals where necessary; the leading approximation below is phase independent. Local force laws neglect nonlocal hydrodynamic interactions and assume nearly steady local response.

For the sinusoidal displacement, let $\epsilon =\beta k\ll 1$. Then $Y_s=\epsilon \cos ks$ and $X_s=(1-Y_s^2{)}^{1/2}=1+O({\epsilon }^2)$. Over one wavelength,Replace $X_s$ by one in the leading tangential term and balance the forces. After multiplication by $3/2$,This replacement also requires $c{\epsilon }^2\ll c-V+U$ for a uniformly small relative error inside the fractional power. If the tangential slip is itself $O(c{\epsilon }^2)$, its variation along the body must be retained. Periodicity gives $V/c=⟨X_s⟩=1-{\epsilon }^2/4+O({\epsilon }^4)$, so small geometric slope alone does not guarantee that additional slip condition.

For Lighthill elongated-body theory, put $h(x,t)\simeq \beta \sin [k(x-Vt)]$ and $w_{\perp }=h_t+U{h}_x$. The transverse fluid kinetic energy per unit length is $M_A{w}_{\perp }^2/2$. In the usual tail-flux approximation, with negligible leading-end contribution, the lateral work flux is $M_{A}U⟨h_t{w}_{\perp }⟩$, whereas kinetic energy carried into the wake is $M_{A}U⟨w_{\perp }^2⟩/2$. Their difference is the useful propulsive power $TU$. ConsequentlyBalancing this against the drag of the printed tangential model, $D=(2/3)K_T{L}^{3/2}(c-V+U{)}^{3/2}$, givesFor a genuine local laminar skin-friction estimate, write $dD=C_T{u}_t^{3/2}{s}^{-1/2}ds$, where $C_T$ has the appropriate different dimensions. Its integrated drag is $2C_T{u}_t^{3/2}\sqrt{L}$, and the corresponding elongated-body balance is instead $M_A{\beta }^2{k}^2(V^2-U^2)/8=C_T\sqrt{L}(c-V+U{)}^{3/2}$. The printed $L^{3/2}$ expression is therefore a result of its stated force model, not of that standard boundary-layer law.

The two thrust models describe different mechanisms: cross-flow resistive-force theory uses dissipative local drag, while elongated-body theory uses reactive acceleration of added mass. Both suppress details of the wake, body-body interactions, finite slenderness, skin-friction unsteadiness, separation and possible turbulent transition. The tail-only reactive result also requires a negligible leading-end flux, usually supplied by a nose taper or small leading-end motion; a blunt constant-cross-section body waving equally at both ends cannot automatically omit its leading-end contribution. Speed fluctuations and recoil may matter, and a planar constant-amplitude waveform is an idealization of a real sea snake.

Choose the helix handedness so its unit tangent is $\mathbf{t}=\cos \alpha \mathbf{i}+\sin \alpha {\mathbf{e}}_{\varphi }$. Let $q={\Omega }_0-\Omega$; the body rotates at $+\Omega \mathbf{i}$, while the flagellum rotates at $-q\mathbf{i}$ in the laboratory, since its relative motor rotation is $-{\Omega }_0\mathbf{i}$. Both translate at $-U\mathbf{i}$. Thus the relative fluid velocity on a flagellar element is $\mathbf{w}=U\mathbf{i}+bq{\mathbf{e}}_{\varphi }$.

Low-Reynolds-number resistive-force theory gives force per unit length on the flagellumIts axial and azimuthal components areDefine the integrated axial resistance matrix of a slender helix coefficientsThe sphere has translational resistance $A_b=6\pi \mu a$ by the Stokes drag law and rotational resistance $H_b=8\pi \mu a^3$. Its hydrodynamic force is $+A_{b}U\mathbf{i}$ and its torque $-H_b\Omega \mathbf{i}$. Integration of $f_x$ and $b{f}_{\varphi }$ gives the zero total force and torque conditionsDividing the first by $L$ and expanding the coefficients gives exactly the two requested balances. The internal motor torque cancels in the total torque balance; it is not an external applied torque.

Eliminate $U$ and use $q+\Omega ={\Omega }_0$. The effective flagellar rotational resistance is $H_e=H-C^2/(A+A_b)$, soThe determinant of helical resistance in resistive-force theory is $AH-C^2=b^2{L}^2{K}_N{K}_T>0$, hence the denominator is positive. Opposite handedness changes the sign of the translation-rotation coupling and reverses the swimming direction.

For $K_T=K$, $K_N=2K$, $\alpha =\pi /4$ and $a=b$, the coefficients are $A=3KL/2$, $C=KaL/2$, $H=3K{a}^{2}L/2$. ThereforeEquivalently, with $h=KL/(\pi \mu a)$, $U=a{\Omega }_{0}4h/(2h^2+21h+48)$. These are positive speed magnitudes for the selected handedness; the translational velocity is $-U\mathbf{i}$.

This helical microswimmer with a spherical head approximation neglects head-tail hydrodynamic interactions as permitted, treats the flagellum as slender and rigid, and averages away transverse forces associated with finite helical ends. The flow is quasistatic Stokes flow, so neither body inertia nor fluid inertia enters the balance.

Write $\omega =kc$, $q=k(1+i)$, and $E=e^{qL}$. The local resistive-force theory force per unit length on the filament iswhere $\mathbf{t}=(X_s,Y_s)$ is the unit tangent and $\mathbf{v}=(X_t-U,Y_t-V)$ the material velocity relative to the fluid. Inextensibility gives $X_s=(1-Y_s^2{)}^{1/2}=1-Y_s^2/2+\cdots$. With $X(0,t)=0$, $X_t=-{\int }_0^s{Y}_s{Y}_{st}d{s}^{\prime }+\cdots$ is second order and has zero time average for a periodic prescribed stroke. The actual small-slope condition is $k|H|e^{kL}\ll 1$ for fixed $kL$; $kH\ll 1$ alone is not uniform when $kL$ grows.

At first order, the transverse force is $f_y=-\mu K_N(Y_t-V)$. Adding the spherical-head drag force $\mu K_{N}L\delta V$ and imposing zero total force givesEquivalently $V(t)=\mathrm{Re}[-i\omega H(E-1)e^{-i\omega t}/(qL(1+\delta ))]$. This form is useful for the subsequent time averages.

At second order, the axial resistive-force theory force isAdding the head drag force gives the head-drag correction to planar flagellar propulsion:since $\overline{X_t}=0$. Directly, $Y_s=kH{e}^{ks}(\cos \vartheta -\sin \vartheta )$ and $Y_t=kcH{e}^{ks}\sin \vartheta$, with $\vartheta =k(s-ct)$. HenceAlso ${\int }_0^L{Y}_{s}ds=Y(L,t)-Y(0,t)$, whose complex amplitude is $H(E-1)$. Using $\overline{\mathrm{Re}(a{e}^{-i\omega t})\mathrm{Re}(b{e}^{-i\omega t})}=\mathrm{Re}(a\overline{b})/2$, together with $\mathrm{Re}(-i\omega /q)=-c/2$, givesAs $|E-1{|}^2=e^{2kL}-2e^{kL}\cos kL+1$, the requested mean speed follows:Anisotropic resistance is essential: $\gamma =1$ removes propulsion at this order. The mean motion is opposite to the direction of wave propagation when the bracket is positive, since the head's axial velocity is $-U$.

For the basal bending moment of a planar flagellum, take positive moment to be the positive out-of-plane torque exerted on the distal filament at its base. The leading hydrodynamic torque on that filament is $-\mu K_N{\int }_0^{L}s(Y_t-V)ds$, so balance requiresAn explicit unsimplified time-dependent answer isThe first term follows from ${\int }_0^{L}s{e}^{qs}ds=[E(qL-1)+1]/q^2$. This is the total internal bending moment required by the hydrodynamic load. It equals the contractile moment when passive elasticity is neglected. If a specified passive moment $M_{\mathrm{pass}}$ also acts, the required active moment is $M_0-M_{\mathrm{pass}}$; no elastic modulus is supplied here.

The prescribed base coordinate needs a geometrical qualification. Literally, a coordinate measured from the attachment point must have $Y(0,t)=0$, whereas the printed waveform has $Y(0,t)=H\cos \omega t$. The preceding calculation reproduces the printed speed by treating that waveform as the prescribed coordinate, including its moving transverse offset. If the intention is instead a fixed attachment at the origin, replace it by $\tilde{Y}=H[e^{ks}\cos k(s-ct)-\cos \omega t]$. ThenThese follow by replacing $Y_t$ with $Y_t-Y_t(0,t)$ in the same force balance. The basal moment likewise uses ${\tilde{Y}}_t$ and $\tilde{V}$. Thus the printed kinematics and its claim that the origin is the actual attachment cannot both be imposed without this correction.

Choose positive tilt towards $+x$, so the rod tangent is $\mathbf{t}=(\sin \theta ,\cos \theta )$ and its normal is $\mathbf{n}=(\cos \theta ,-\sin \theta )$. The opposite angular convention reverses the signed phase optimum later. In resistive-force theory, a point with centred arclength $s\in [-L/2,L/2]$ has velocity $\mathbf{u}=\dot{x}{\mathbf{e}}_x+s\dot{\theta }\mathbf{n}$, and its force on the fluid is $\mathbf{Du}$, where $\mathbf{D}=c_{\perp }\mathbf{I}+(c_{\parallel }-c_{\perp })\mathbf{tt}$. The force on the rod has the opposite sign.

The pure centred rotational contribution is odd in $s$ and therefore gives zero net force instantaneously in either direction. In the combined oscillating rigid rod in resistive-force theory, however, orientation modulates the translational force. After half a period, $\dot{x}\mapsto -\dot{x}$ and $\theta \mapsto -\theta$. Reflection in the $y$ axis takes one configuration and its forcing into the other. A force component in $x$ reverses under this reflection, whereas a component in $y$ does not. Explicitly,The first is half-period antisymmetric and the second is half-period symmetric. Hence $⟨F_x⟩=0$, while symmetry permits $⟨F_y⟩\ne 0$. The possible mean $y$ force comes from the coupled translation and rotation, not from pure centred rotation alone.

Use the printed Maxwell-filtered local drag law at fixed material arclength, after transients have decayed to a periodic response. Averaging the force equation over a period makes the time-derivative term vanish. Therefore the mean force equals the mean instantaneous resistive-force theory driving force, even though its oscillating part is filtered. In particular,with the same sign convention as before.

At first order the tangent is ${\mathbf{e}}_y$ and both velocity components are single harmonics of frequency $\omega$. The linear Maxwell fluid response multiplies their complex force amplitudes by $(1+i\omega \lambda {)}^{-1}$. Only its in-phase real part contributes to the mean work. ConsequentlyThe leading power is strictly smaller when $\lambda >0$ and $\omega >0$, and equal in the Newtonian limit. The zero-frequency component generating the mean force is unattenuated, whereas elastic storage and phase lag reduce the dissipative harmonic response. This conclusion concerns the specified linear local force law and prescribed kinematics; it is not a claim for every nonlinear viscoelastic constitutive model.

Let $\mathbf{t}=(x_s,y_s)$ be the unit tangent of the inextensible filament. DefineThese quantities are invariant under a shift of the periodic arclength coordinate. Since $x_s^2+y_s^2=1$, the average tangent tensor is $M=\left(\begin{array}{cc}\beta & \delta \\ \delta & 1-\beta \end{array}\right)$. Also $\int \mathbf{t}ds=\lambda {\mathbf{e}}_x$.

For finite-amplitude propulsion of an inextensible periodic filament, distinguish axial phase speed from tangential material speed. The $V$ in the displayed formula is the axial speed of the travelling pattern relative to the mean filament motion. In the wave frame an inextensible material line slides backwards tangentially at the constant speed $Q=V\Lambda /\lambda$: its mean axial sliding is $Q\lambda /\Lambda =V$. If the filament translates at $-U{\mathbf{e}}_x$ in the laboratory, its local fluid-relative velocity isThis relation is the axial and arclength travelling-wave speeds conversion. It ensures the arclength-averaged material velocity is $-U{\mathbf{e}}_x$, rather than confusing wave propagation with material transport.

Integrating the resistive-force theory hydrodynamic force over a period and imposing zero axial force givesSubstitution of $Q\lambda =V\Lambda$ yieldsFor $0<\rho <1$ and $\beta <1$ it is positive: anisotropic drag drives swimming opposite to the wave. A straight waveform has $\beta =1$ and gives no propulsion; isotropic drag $\rho =1$ also gives zero. The denominator $(1-\beta )+\rho \beta$ is positive. If instead $V$ denotes arclength propagation speed, the left-hand ratio acquires the extra factor $\lambda /\Lambda$; the two speeds are not interchangeable.

There is a geometric qualification to the asserted direction. For free swimming strictly along $x$, zero transverse force also requires $\delta =0$, which holds for the usual reflection-symmetric waveforms. The periodicity assumptions alone do not imply that symmetry. A general waveform has cross-resistance of an asymmetric planar waveform and can require a transverse translation. If the mean laboratory velocity is $(-U_x,U_y)$, the full force-free condition isWriting $A_0=1+(\rho -1)\beta$, $B_0=1+(\rho -1)(1-\beta )$ and $\mathcal{D}=A_0{B}_0-(\rho -1{)}^2{\delta }^2$, one obtainsThe printed expression is recovered when $\delta =0$, or as the axial force-balance result if transverse motion is externally constrained. For an explicit permitted asymmetric shape, concatenate tangents $(\sqrt{3}/2,1/2)$ and $(1/2,-\sqrt{3}/2)$ with arclength fractions $p=\sqrt{3}/(1+\sqrt{3})$ and $1-p$. Their mean vertical tangent is zero, but $\delta =\sqrt{3}(2p-1)/4\ne 0$. Repeating the segments gives a periodic graph; the corners can be smoothed without removing the nonzero cross integral. Thus pure opposite-to-wave free swimming also needs a zero cross-resistance assumption.