Small-slope elastohydrodynamic filament equation ($c_{\perp }{y}_t=-B{y}_{xxxx}$)

For a straight, unloaded inextensible filament with small transverse displacement, resistive-force theory gives viscous force density $-c_{\perp }{y}_t$. Variation of the bending energy gives $-B{y}_{xxxx}$, while the induced filament tension is higher order. Their instantaneous balance yields $c_{\perp }{y}_t=-B{y}_{xxxx}$. It is an overdamped fourth-order diffusion equation: short-wavelength bends relax much faster than long ones. Wiggins and Goldstein's flexive-propulsion paper develops the elastic-wave mechanism for low-Reynolds-number propulsion.

Let positive transverse displacement point along a uniform load per unit length $w$. The filament bending modulus $B$ gives energy ${\int }_0^L[B(h_{xx}{)}^2/2-wh]dx$. The variational derivative is $B{h}_{xxxx}-w$, so resistive-force theory gives ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}+w$. Clamping imposes $h(0)=h_x(0)=0$, while a free tip imposes $h_{xx}(L)=h_{xxx}(L)=0$. The steady solution isIts tip displacement is $w{L}^4/(8B)$. For a submerged cylindrical filament under Newtonian gravity, buoyancy gives $w=({\rho }_f-\rho )\pi a^{2}g$. This linear approximation requires small slope and negligible inertia.

For a uniform-load bending of a clamped filament, total load $F=wL$ and tip displacement $\delta =w{L}^4/(8B)$ obey Hooke's law in the form $F=k_{\mathrm{eff}}\delta$, with $k_{\mathrm{eff}}=8B/L^3$. This stiffness depends on the specified distributed loading. A concentrated tip force instead gives $3B/L^3$, so equal total forces with different distributions are not interchangeable.

The anisotropic-drag contribution to axial force from a small-slope transverse motion is $F=(c_{\perp }-c_{\parallel })\int y_x{y}_{t}dx$. Using the small-slope elastohydrodynamic filament equation and integration by parts givesThus this propulsive contribution is determined by endpoint slope, bending moment, and shear. A separately imposed axial translation adds ordinary longitudinal drag. Exact second-order longitudinal motion required by inextensibility also contributes to instantaneous total drag, but its time-derivative contribution averages to zero over a deformation cycle.

In units of elastohydrodynamic penetration length and inverse forcing frequency, $y_t=-y_{xxxx}$. Prescribing $y(0,t)=y_0\cos t$, zero bending moment $y_{xx}(0,t)=0$, and decay at infinity givesThe two components have phase velocities $-1/S$ and $1/C$. Their unequal spatial attenuation produces a nonreciprocal shape cycle even though the endpoint actuation is reciprocal.