Synchronized states of hydrodynamically coupled filaments and their stability

Cilia and flagella are organelles that play central roles in unicellular locomotion, embryonic development, and fluid transport around tissues. In these examples, multiple cilia are often found in close proximity and exhibit coordinated motion. Inspired by the flagellar motion of biflagellate cells, we examine the synchrony exhibited by a filament pair surrounded by a viscous fluid and tethered to a rigid planar surface. A geometrically switching base moment drives filament motion, and we characterize how the stability of synchonized states depends on the base torque magnitude. In particular, we study the emergence of bistability that occurs when the antiphase, breast-stroke branch becomes unstable. Using a bisection algorithm, we find the unstable edge state that exists between the two basins of attraction when the system exhibits bistability. We establish a bifurcation diagram, study the nature of the bifurcation points, and find that the observed dynamical system can be captured by a modified version of Adler's equation. The bifurcation diagram and presence of bistability reveal a simple mechanism by which the antiphase breast stroke can be modulated, or switched entirely to in-phase undulations through the variation of a single bifurcation parameter.

Figure 1

Schematic diagram for the filament system indicating the arclength $s\in [0,l]$, diameter $2a$, base torques ${\mathit{N}}_i$, base angles ${\theta }_i$, and separation distance $B$.

Figure 2

Filament dynamics over one beat period for $M_b=7$ with (a) antiphase oscillations (branch I); (b) stable, phase-shifted oscillations (branch II); (c) unstable, phase-shifted oscillations (branch III), and (d) in-phase oscillations (branch IV). The top panels in each figure show the different base angles, $\mathrm{\Delta }\theta ={\theta }_2-{\theta }_1$, over one period. The symbols correspond to the time points that the filaments are shown.

Figure 3

The synchrony parameter $Q\in [-1,1]$ of the asymptotic state in the $M_b\text{−}b$ plane: (a) $\mathrm{\Delta }\varphi =0.49$; (b) $\mathrm{\Delta }\varphi =0.01$. $Q=1$ corresponds to antiphase beating, and $Q=-1$ corresponds to in-phase beating. In (c), the contours show $\mathrm{\Delta }Q=(Q_{0.49}+Q_{0.01})/2\in [-1,1]$, where the subscript indicates the value of $\mathrm{\Delta }\varphi$ for which $Q$ was computed. $\mathrm{\Delta }Q\approx 0$ in (c) indicates bistability.

Figure 4

Bifurcation of synchronized states represented by $Q$ with $M_b\in [2.5,10]$ and $b=0.5$. Here, the solid and dashed lines indicate stable and unstable states, respectively, and the markers indicate the simulation data.

Figure 5

Phase portrait in ${\theta }_1\text{−}{\theta }_2$ space for the synchronized states for $M_b=7$: (a) branch I (antiphase oscillations); (b) branch II (stable, phase-shifted oscillations); (c) branch III (unstable phase-shifted oscillations corresponding to the edge state); and (d) branch IV (in-phase oscillation). Filament dynamics corresponding to these cases are shown in Fig. 2.

Figure 6

Stability of branch I (antiphase oscillations): (a) $\mathrm{\Gamma }\left(t\right)$ for cases $M_b=2.5,6$, and 10. Phase portraits for (b) $M_b=2.5$, (c) $M_b=6$, and (d) $M_b=10$. The different colored lines show trajectories over one period beginning at period $T_i$ as indicated in the legend. In each case $T_{\infty }$ shows the limit cycle once the oscillations are steady.

Figure 7

$Q\left(t\right)$ for the edge state (solid black line) corresponding to branch III for $M_b=4$. The dash-dotted blue lines indicate the trajectories that approach branch II when performing the bisection, while the dash-dotted red lines show the trajectories that move toward branch IV.

Figure 8

Bifurcation diagram of the modified Adler's model for the pitchfork bifurcation base torque $M_{b,PF}=2.75$ and saddle-node bifurcation value $M_{b,SN}=8.2$. $\mathrm{\Delta }=0$ and $\mathrm{\Delta }=\pi$ correspond to in-phase and antiphase synchronization, respectively. Here, $Q=-1+2\mathrm{\Delta }/\pi$ (see main text).