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The elastohydrodynamic penetration length balances the transverse viscous force $c_{\perp }\omega y$ against the bending force $By/{\ell }_{\omega }^4$ in a periodically driven filament. The sperm number $L/{\ell }_{\omega }$ distinguishes nearly rigid, weakly deformed filaments from long filaments whose response is localized near the actuator.

A periodic displacement with angular frequency $\omega$ has $y_t\sim \omega y$. Balancing this viscous forcing with bending over length ${\ell }_{\omega }$ gives $c_{\perp }\omega y\sim By/{\ell }_{\omega }^4$. Thus the elastohydrodynamic penetration length and sperm number areSince $[B/(c_{\perp }\omega )]={\mathsf{L}}^4$, the fourth root has dimensions length, as required.

With $X=x/{\ell }_{\omega }$, $\tau =\omega t$, and $Y=y/{\ell }_{\omega }$, the bending equation becomesThe defining balance cancels both coefficients, giving $\boxed{Y_{\tau }=-Y_{XXXX}}$. A large sperm number means that the periodically forced response decays before reaching the distant end.

On the upper half-filament, write $\mathbf{r}(s)=(h(s),s)$ with $0\le s\le L$. To first order in slope, resistive-force theory supplies the uniform transverse load $f_x={\zeta }_{\perp }U$, and Euler--Bernoulli balance givesMidpoint symmetry and clamping impose $h(0)=h^{\prime }(0)=0$; the free-end force and moment conditions are $h^{\prime \prime \prime }(L)=h^{\prime \prime }(L)=0$. With $H=h/L$ and $\xi =s/L$,Four integrations giveIn particular, $H(1)=\mathrm{Sp}/8$ and $H^{\prime }(1)=\mathrm{Sp}/6$. The small-slope approximation first fails when the end slope is order one, giving the estimate $\mathrm{Sp}\sim 6$; parametrically, breakdown occurs at Sperm number of order one.