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An active swimming beam has a total internal bending moment $G$ containing passive elasticity and muscular actuation. Neglecting cross-sectional rotary inertia, slice angular momentum balance identifies the internal transverse force with $G_x$. Slice linear momentum balance then gives $G_{xx}=m_b{h}_{tt}-F_{\mathrm{body}}$. In Lighthill elongated-body theory, $F_{\mathrm{body}}=-\mathcal{D}(m\mathcal{D}h)$, where $\mathcal{D}={\partial }_t+U{\partial }_x$. The body and its hydrodynamic added mass therefore contribute with the same sign to the acceleration coefficient. The positive quantity $\mathcal{D}(m\mathcal{D}h)$ is the reaction on the fluid.

At leading order in slope, the hydrodynamic torque on the distal filament is $-\mu K_N{\int }_0^{L}s(Y_t-V)ds$. The displayed opposite torque is required at its base. It is the total internal bending moment; identifying it with the active moment assumes no separate passive elastic moment. If a passive moment is specified, subtract that contribution to determine the required active moment.

For $G_{xx}=R$, zero bending moment and transverse force at the leading end give $G(x)={\int }_0^x(x-s)R(s)ds$. The two trailing-end free-end bending boundary conditions hold exactly when the displayed two integrals vanish. They express total transverse force and torque balance. Adding unknown rigid translation and yaw to a prescribed deformation supplies the two degrees of freedom of the swimming recoil correction. Its principal inertial matrix is the weighted moment matrix of $1$ and $x-x_c$ with weight $m_b+m$, hence is positive definite for a nondegenerate positive mass distribution.

A freely swimming body has neither an imposed end couple nor an imposed end transverse force. Its end bending moment is therefore zero, and the transverse force $S=G_x$ is also zero at both ends. These are the four free-end bending boundary conditions; they do not require the end displacement or slope to vanish.

Put $R=m_b{h}_{tt}+\mathcal{D}(m\mathcal{D}h)$, using the physically consistent sign. Integrating $G_{xx}=R$ with the two leading-end conditions givesThe two trailing-end conditions consequently requireThese free-end compatibility for a swimming beam conditions are total transverse force and torque balances. They explain why an arbitrary plausible prescribed $h$ does not suffice: two integration constants can satisfy only two end conditions unless the forcing already satisfies both compatibility integrals.

The missing freedom is the swimming recoil correction. An observed or prescribed deformation $h_d$ relative to the body must be supplemented by a lateral translation $Y(t)$ and a small yaw $\Theta (t)$:To identify $Y$ with the actual centre of mass displacement, first remove the body-mass-weighted mean from $h_d$. One may also remove its body-mass-weighted first moment about $x_c$ to define the yaw reference uniquely. This changes the decomposition of the observed motion, not the physical displacement.

For $\xi =x-x_c$, the recoil contribution to the forcing isInsert $R=R_d+R_{\mathrm{rec}}$ into the two compatibility integrals, using $1$ and $\xi$ as weights. This gives two coupled ordinary differential equations for the two recoil motions. Their principal acceleration matrix isFor a nondegenerate positive mass distribution it is a positive-definite matrix: its quadratic form at $(a,b)$ is ${\int }_0^L(m_b+m)(a+b\xi {)}^{2}dx>0$ unless $a=b=0$. Thus initial translation and yaw determine the recoil through these equations, and the corrected $h$ supplies the two compatibility conditions. The four free-end conditions determine the bending moment together with two rigid recoil motions; they do not overdetermine a physically complete swimming motion.

Choose the internal transverse force $S(x,t)$ so that the net internal transverse force on a slice is $[S(x+dx,t)-S(x,t)]$. Its linear momentum balance is thenThe coordinate $x$ labels body slices, so their lateral material acceleration is $h_{tt}$; the fluid transport operator $\mathcal{D}$ must not be substituted for this body acceleration.

Let $G$ be the total internal bending moment, including both passive elastic and active muscular contributions. With the compatible moment orientation, the net couple on the slice is $(G_x-S)dx$. The cross-sectional rotary inertia is smaller than the translational inertia by the square of the transverse-to-longitudinal length ratio, and is neglected in this slender-beam approximation. Slice angular momentum balance therefore gives $S=G_x$. Eliminating $S$ from the slice linear momentum balance provesThis equation is the balance law for an active bending beam for a swimming fish; a constitutive relation for the passive bending moment or the muscle actuation is not needed to derive it. With the force-on-body convention established above it becomesAs a check, constant $m$ and rigid lateral acceleration give the effective inertia $m_b+m$. Combining the printed positive force in part (i) with the minus sign in the beam equation would instead give $m_b-m$. That is the same sign defect, rather than a negative hydrodynamic added mass. If the positive quantity in part (i) is called $F_y$ as a force on the fluid, the beam equation must contain $+F_y$.

The overdamped filament bending equation is linear, so sinusoidal forcing has a sinusoidal steady response at the same frequency. Write $h=h_0\mathrm{Re}[e^{i\omega t}F(\eta )]$ with $\eta =x/\ell$. Substitution givesUsing the elastohydrodynamic penetration length from the previous part reduces this ordinary differential equation toThe boundary conditions become $F(0)=1$, $F^{\prime \prime }(0)=0$, and decay of $F$ and its derivatives at infinity. The derivative condition expresses zero bending moment at the driven end, rather than a clamped slope. This separation of variables constructs the periodic response; an unspecified initial shape can add a transient solution of the homogeneous evolution problem.

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.