2.1. Coordinates, kinematics and dynamics

We consider a slender self-propelled object moving in a plane, which we denote as our $x{-}y$ plane, as illustrated in figure 1. The shape of the slender filamentous object is described by its centreline curve. The length of the filament, $L$ , may change over time due to self-compression/extension. To represent this effect, we introduce a Lagrangian label $s_0\in [0, L_0]$ as the arc length of the filament in the reference state, where $L_0$ is the reference length of the filament (figure 1 a). We then consider the compression/extension at time $t$ in the form

(2.1) \begin{equation} \frac {\partial s}{\partial s_0}=1+\eta p(s_0, t), \end{equation}

where $s=s(s_0, t)$ is the arc length of the current configuration at time $t$ (figure 1 b). Here, we introduce a constant $\eta$ for later use. The body-fixed frame is introduced as a frame attached to the leftmost point of the slender body, as shown in figure 1(b). The position labelled by $s_0$ is deformed to $\tilde {\boldsymbol{x}}(s_0, t)$ in the body-fixed frame, and the tangent angle is given by $\tilde {\theta }(s_0, t)$ . We describe self-deformation of the swimmer as

(2.2) \begin{equation} \tilde {\theta }(s_0, t)=\epsilon q(s_0, t) , \end{equation}

where the constant $\epsilon$ is introduced for later use.

Figure 1.

Schematic of a filament in three frames of reference. (a) A filament in a reference state. The arc length $s_0\in [0, L_0]$ is used for parametrisation of the curve. (b) A filament in the body-fixed frame at time $t$ . The point labelled by $s_0$ is mapped to $\tilde {\boldsymbol{x}}(s_0, t)$ , where the distance along the filament is denoted by $s(s_0, t)$ , and the local tangent angle is represented as $\tilde {\theta }(s_0, t)$ . (c) A filament in the laboratory frame, which is obtained by rigid body transformation, translation with $\boldsymbol{X}(t)$ , and rotation by $\varTheta (t)$ , from the filament in (b). The point $\tilde {\boldsymbol{x}}(s_0, t)$ is mapped to $\boldsymbol{x}(s,t)$ .

The body shape in the laboratory frame is obtained by a rigid body transformation, translation with $\boldsymbol{X}(t)$ , and rotation by $\varTheta (t)$ , as in figure 1(c). The point $\tilde {\boldsymbol{x}}(s_0,t)$ in the body-fixed frame is then mapped to $\boldsymbol{x}(s, t)$ in the laboratory frame by $\boldsymbol{x}=\boldsymbol{X}+\boldsymbol {R}_\varTheta \tilde {\boldsymbol{x}}$ , where $\boldsymbol {R}_{\varTheta }$ is the two-dimensional rotation matrix by an angle $\varTheta$ . In this study, we consider the kinematic problem amounting to solving the displacement $\boldsymbol{X}(t)$ and $\varTheta (t)$ , where the shape gait is provided by the two intrinsic functions, $p(s_0, t)$ and $q(s_0, t)$ , that physically correspond to local extensibility and local bending angle, respectively.

The velocity of the local point on the body in the laboratory frame is obtained as the Lagrangian time derivative defined as ${\textrm D} \boldsymbol{x}/{\textrm D}t=\partial \boldsymbol{x}(s_0,t)/\partial t$ . To compute this, we introduce the linear velocity $\boldsymbol{U}=(U_x, U_y)^{\textrm{T}}$ and the angular velocity $\boldsymbol{\varOmega }=\varOmega \boldsymbol{e}_z$ , with $\varOmega =\textrm{d}\varTheta /\textrm{d}t$ ; the superscript $\textrm{T}$ denotes the transpose. The local velocity is then given by

(2.3) \begin{equation} \frac {{\textrm D} \boldsymbol{x}}{{\textrm D}t} = \boldsymbol{U}+\varOmega \boldsymbol{e}_z\times \tilde {\boldsymbol{x}}+\boldsymbol {R}_{\varTheta }\frac {{\textrm D}{\tilde {\boldsymbol{x}}}}{{\textrm D}t}. \end{equation}

To determine the motion of the object, we calculate the drag forces acting on an infinitesimal segment of the filament via the empirical RFT. With $C_{\parallel }\ (\gt 0)$ and $C_{\bot }\ (\gt 0)$ being the tangential and normal hydrodynamic drag coefficients per unit length, the drag from the environment is obtained as

(2.4) \begin{equation} \boldsymbol{f}(s, t)=-C_{\|}\left (\frac {{\textrm D}\boldsymbol{x}}{{\textrm D}t} \boldsymbol{\cdot }\boldsymbol{e}_{\|}\right )\boldsymbol{e}_{\|}-C_{\perp }\left (\frac {{\textrm D}\boldsymbol{x}}{{\textrm D}t} \boldsymbol{\cdot }\boldsymbol{e}_{\perp }\right )\boldsymbol{e}_{\perp } , \end{equation}

where $\boldsymbol{e}_{\|}(s_0, t)$ and $\boldsymbol{e}_{\perp }(s_0, t)$ are tangential and normal unit vectors at the segment of the filament, respectively. The ratio $\gamma :=C_{\perp}/ C_{\|}$ is called the anisotropy ratio, and the particular case $\gamma =1$ is called isotropic drag. In this study, we take this as an arbitrary positive value. For swimming in low Reynolds number flow, it is known that $\gamma \rightarrow 2$ in the slender filament limit (Gray & Hancock Reference Gray and Hancock1955), and empirically $\gamma =1.5{-}1.8$ is used for flagella and cilia (Lighthill Reference Lighthill1976; Friedrich et al. Reference Friedrich, Riedel-Kruse, Howard and Jülicher2010). When the flagellum possesses an accessory hairy structure known as mastigonemes, however, the effective anisotropy ratio $\gamma$ may be lower than unity (Brennen Reference Brennen1976; Asadzadeh et al. Reference Asadzadeh, Walther, Andersen and Kiørboe2022). In contrast, locomotion of C. elegans on a gel-like structure is well described by a large anisotropy ratio $\gamma \approx 70$ (Keaveny & Brown Reference Keaveny and Brown2017).

To close the system and calculate the locomotion velocity, we employ the force-free $\tilde {\boldsymbol{F}}=(\tilde {F}_x, \tilde {F}_y)^{\textrm{T}}=\boldsymbol{0}$ and torque-free $\tilde {\boldsymbol{M}}=M\boldsymbol{e}_z=\boldsymbol{0}$ conditions in the body-fixed frame:

(2.5) \begin{equation} \int _0^{L(t)}\tilde {\boldsymbol{f}}(s, t)\,\textrm{d}s=\int _0^{L(t)}\tilde {\boldsymbol{x}}\times \tilde {\boldsymbol{f}}(s, t)\,\textrm{d}s=\boldsymbol{0}, \end{equation}

where we employ the torque around the point $\boldsymbol{X}(t)$ . Equations (2.4) and (2.5) yield a closed form for the kinematic problem, allowing us to uniquely determine $\boldsymbol{X}$ and $\varTheta$ from a given deformation $p$ and $q$ .

We denote the translational velocity in the body-fixed frame by $\tilde {\boldsymbol{U}}=(\tilde {U}_x, \tilde {U}_y)^{\textrm{T}}$ , and introduce a $3\times 3$ matrix representation of the Lie group $\mathcal{G}\in \textrm{SE}(2)$ , also known as homogeneous representation, as

(2.6) \begin{equation} \mathcal{G}=\begin{pmatrix} \boldsymbol {R}_{\varTheta } & \boldsymbol{X} \\ (0, 0) & 1 \end{pmatrix}, \end{equation}

and the associated Lie algebra ${\textrm{se}}(2)$ , whose elements $\hat {\mathcal{A}}$ act on $\textrm{SE}(2)$ as $\dot {\mathcal{G}}=\mathcal{G}\hat {\mathcal{A}}$ , with the matrix representation

(2.7) \begin{equation} \hat {\mathcal{A}}=\begin{pmatrix} 0 & \varOmega & \tilde {U}_x \\ -\varOmega & 0 &\tilde {U}_y \\ 0 & 0 & 0 \end{pmatrix}. \end{equation}

The ‘hat’ notation of $\hat {\mathcal{A}}$ is standard in solid mechanics and refers to the correspondence between the matrix representation of the Lie algebra ${\textrm{se}}(2)$ and its twist coordinates stacked in a vector (representation) $\mathcal{A} = (\hat {\mathcal{A}})^{\vee } = ( \tilde {U}_x, \tilde {U}_y, \varOmega )^{\textrm{T}}$ .

To derive the expression in the laboratory frame, we directly solve the differential equation on the rigid body motion, and represent its solution as

(2.8) \begin{equation} \mathcal{G}(T)=\mathcal{G}(0)\,\overline {\textrm{P}}\exp \left [\int _0^T \hat {\mathcal{A}}(t)\,\textrm{d}t \right ], \end{equation}

where the symbol $\overline {\textrm{P}}$ denotes the reverse path ordering operator (Shapere & Wilczek Reference Shapere and Wilczek1989). Using the Magnus expansion, this formal solution (2.8) may be expanded in a Lie brackets series as

(2.9) \begin{align} \mathcal{G}(T)&=\mathcal{G}(0)\exp \left [\int _0^T \hat {\mathcal{A}}(t_1)\,\textrm{d}t_1 +\frac {1}{2}\int _0^T\,\textrm{d}t_1\int _0^{t_1}\,\textrm{d}t_2 \, \left[\hat {\mathcal{A}}(t_1), \hat {\mathcal{A}}(t_2)\right] \right. \nonumber \\[3pt] & \left. +\, \frac {1}{6}\int _0^T\!\!\!\textrm{d}t_1\int _0^{t_1}\!\!\! \textrm{d}t_2\int _0^{t_2}\!\!\! \textrm{d}t_3 \big([\hat {\mathcal{A}}(t_1), [\hat {\mathcal{A}}(t_2), \hat {\mathcal{A}}(t_3)]]+[\hat {\mathcal{A}}(t_3), [\hat {\mathcal{A}}(t_2), \hat {\mathcal{A}}(t_1)]]\big)+\cdots \right ], \end{align}

where the bracket symbol represents the matrix commutator $[\hat {\mathcal{A}}(t_1), \hat {\mathcal{A}}(t_2)]=\hat {\mathcal{A}}(t_1)\,\hat {\mathcal{A}}(t_2)-\hat {\mathcal{A}}(t_2)\,\hat {\mathcal{A}}(t_1)$ .

An alternative path to the computation of $\mathcal{G}$ resorts to gauge field theory (Shapere & Wilczek Reference Shapere and Wilczek1989; Zhao et al. Reference Zhao, Bittner, Clifton, Gravish and Revzen2022). By linearity of the RFT, the matrix $\hat {\mathcal{A}}$ may be decomposed as

(2.10) \begin{equation} \hat {\mathcal{A}}=\sum _\alpha \hat {\mathcal{H}}_\alpha \dot {\sigma }_\alpha , \end{equation}

where $\sigma _\alpha$ ( $\alpha =1,2,\ldots ,N$ ) is the shape variable, and $N$ is the number of degrees of freedom in the shape space, with the overdot symbol denoting the time derivative. Here, $\hat {\mathcal{H}}_\alpha$ is called a gauge field or Stokes connection, as it depends only on the shape variables. A local connection maps a shape variation to a velocity for a given body shape, independent of swimmer position and orientation (Kelly & Murray Reference Kelly and Murray1995). Using the matrix representation of $\textrm{SE}(2)$ , the connection $\hat {\mathcal{H}}_{\alpha }$ is represented by a third-rank tensor of dimensions $3\times 3 \times N$ ; we may also ‘drop the hats’ in (2.10), using the twist coordinates $\mathcal{A}$ , which would, in turn, represent the Stokes connection $\mathcal{H}_{\alpha }$ as a $3 \times N$ matrix $\mathcal{H}=(\boldsymbol{H}_x; \boldsymbol{H}_y; \boldsymbol{H}_\theta )$ , where the semicolon denotes vertical concatenation. The integral representation (2.8) is then rewritten as

(2.11) \begin{equation} \mathcal{G}(T)=\mathcal{G}(0)\,\overline {\textrm{P}}\exp \left [ \sum _\alpha \int _C \hat {\mathcal{H}}_\alpha (\boldsymbol{\sigma })\,\textrm{d}\sigma _\alpha \right ] , \end{equation}

where $C$ is a path in the shape space. We are particularly interested in the net locomotion caused by a periodic shape gait, which is represented by a closed loop in the shape space. In the small-amplitude case, by neglecting terms above third order in the expansions, the net locomotion is well described by a surface integral of the curvature in the shape space as (Hatton & Choset Reference Hatton and Choset2015)

(2.12) \begin{equation} \mathcal{G}(T) \approx \mathcal{G}(0)\exp \left [\sum _{\alpha , \,\beta } \int _{S_C} \hat {\mathcal{F}}_{\alpha \beta }(\boldsymbol{\sigma })\,\textrm{d}\sigma _\alpha \,\textrm{d}\sigma _\beta \right ] , \end{equation}

where $S_C$ is a region in the shape space enclosed by the close loop $C$ , and the associated curvature form, also called field strength or Stokes curvature, is defined as

(2.13) \begin{equation} \hat {\mathcal{F}}_{\alpha \beta }=\frac {\partial \hat {\mathcal{H}}_\alpha }{\partial \sigma _\beta }-\frac {\partial \hat {\mathcal{H}}_\beta }{\partial \sigma _\alpha }+ \big[\hat {\mathcal{H}}_\alpha , \hat {\mathcal{H}}_\beta \big]. \end{equation}

When pulled back to the vector representation of ${\textrm{se}}(2)$ , the curvature field $\mathcal{F}$ has three components, $\mathcal{F}=(\mathcal{F}_x, \mathcal{F}_y, \mathcal{F}_{\theta })^{\textrm{T}}$ , which can be interpreted as the effect of small loop gaits on each spatial direction according to Stokes’ theorem. Finally, according to the Ambrose–Singer theorem (Ambrose & Singer Reference Ambrose and Singer1953), we have (Shapere & Wilczek Reference Shapere and Wilczek1989)

(2.14) \begin{equation} \overline {\textrm{P}}\exp \left [\int _0^T \hat {\mathcal{A}}(t)\,\textrm{d}t \right ] \approx 1+\frac {1}{2}\sum _{\alpha , \,\beta } \int _0^T\hat {\mathcal{F}}_{\alpha \beta }\sigma _\alpha \dot {\sigma }_\beta \,\textrm{d}t , \end{equation}

up to the second order of amplitude.

2.2. Locomotion with isotropic drag

In this subsection, we focus on the particular case of isotropic drag, i.e. $\gamma =1$ . Note that the arguments below are not limited to planar motion, but are valid for general three-dimensional motion without external forces and torques.

Isotropic drag has been known to hinder locomotion. The classical RFT study by Gray & Hancock (Reference Gray and Hancock1955) considered simple sinusoidal transverse deformation, and theoretically showed that drag anisotropy allows net locomotion. Becker, Koehler & Stone (Reference Becker, Koehler and Stone2003) later extended this argument to a general self-propelled free-swimming slender object, and showed that no net motion is possible in isotropic drag with only bending deformation. These studies, however, neglected rotational dynamics; it was then proven that isotropic drag may be compatible with net rotation (Koens & Lauga Reference Koens and Lauga2016). Finally, considering extensible slender swimmers, Pak & Lauga (Reference Pak and Lauga2011) argued that time variation of the total length can generate net locomotion in isotropic drag.

However, ambiguity remains regarding the need for total length variation, and the exact role played by compression in allowing locomotion in isotropic drag. Here, we answer precisely with the proposition below.

Proposition 1. Assume that $\gamma =1$ . Let $\overline {\boldsymbol{X}}$ be the centre of geometry of the swimmer, defined as

(2.15) \begin{align} \overline {\boldsymbol{X}}(t)=\frac {1}{L}\int _0^{L(t)} \boldsymbol{x}(s, t)\,\mathrm{d}s.\end{align}

Then the following statements hold:

1. (i) If the compression is spatially uniform ( $\partial p / \partial s_0 = 0$ for all times), then $\dot {\overline {\boldsymbol{X}}}=0$ , i.e. no net motion is possible.

2. (ii) Otherwise ( $\partial p / \partial s_0 \neq 0$ ), net motion is indeed possible.

Of particular note, case (ii) includes the case $\dot {L}=0$ of constant total length, contrary to the statement by Pak & Lauga (Reference Pak and Lauga2011), which restrictively assumed that the arc length $s$ is a materially conserved quantity. This, however, is not always the case for active compression, as seen in the proof of Proposition1 below.

Proof of Proposition1. Let us compute the time derivative of $\overline {\boldsymbol{X}}$ , recalling that the arc length $s$ is provided as a function of the Lagrangian label $s_0$ as $s=s(s_0, t)$ , which yields the following change of variable in the integral (2.15):

(2.16) \begin{equation} \frac {\textrm{d} \overline {\boldsymbol{X}}}{\textrm{d}t}= \frac {\textrm{d}}{\textrm{d}t}\left [\frac {1}{L}\int _0^{L_0} \boldsymbol{x}(s_0, t)\left (1+\eta\, p(s_0, t)\right )\,\textrm{d}s_0\right ], \end{equation}

where $L_0$ is the body length in the reference frame. The right-hand side of (2.16) is then computed as

(2.17) \begin{eqnarray} -\frac {\dot {L}}{L^2}\int _0^L\boldsymbol{x}(s,t)\,\textrm{d}s+\frac {1}{L}\int _0^{L_0}\frac {\partial \boldsymbol{x}(s_0, t)}{\partial t}\left (1+\eta p\right )\,\textrm{d}s_0+\frac {\eta }{L} \int _0^{L_0}\boldsymbol{x}(s_0, t)\frac {\partial p}{\partial t}\,\textrm{d}s_0. \end{eqnarray}

Assuming isotropic drag $(C_\|=C_\perp =C)$ , the second term vanishes, because one has

(2.18) \begin{equation} \int _0^{L_0}\frac {\partial \boldsymbol{x}(s_0, t)}{\partial t}(1+\eta\, p(s_0, t))\,\textrm{d}s_0 =\int _0^{L}\frac {{\textrm D} \boldsymbol{x}}{{\textrm D} t}\,\textrm{d}s =\frac {1}{C}\int _0^{L}\boldsymbol{f}\,\textrm{d}s=\boldsymbol{0} \end{equation}

from the force balance equations (2.4) and (2.5). By writing $\dot {L}$ as

(2.19) \begin{equation} \frac {\textrm{d}L}{\textrm{d}t}=\frac {\textrm{d}}{\textrm{d}t}\int _0^{L_0}\left [1+\eta\, p(s_0,t)\right ]\,\textrm{d}s_0=\eta \int _0^{L_0}\frac {\partial p(s_0, t)}{\partial t}\,\textrm{d}s_0, \end{equation}

we may summarise the motion of the centre of geometry in the isotropic drag case as

(2.20) \begin{eqnarray} \frac {\textrm{d} \overline {\boldsymbol{X}}}{\textrm{d}t} =\frac {\eta }{L}\left [ \int _0^{L_0}\left (\boldsymbol{x}(s_0, t)\frac {\partial p(s_0, t)}{\partial t}\right )\,\textrm{d}s_0- \overline {\boldsymbol{X}}\int _0^{L_0}\frac {\partial p(s_0, t)}{\partial t}\,\textrm{d}s_0\right ] , \end{eqnarray}

which is clearly non-zero in general, hence point (ii) of the proposition. Note that we can also recover, from (2.20), the zero motion for an inextensible object, by setting $\eta =0$ .

Now we assume uniform compression, i.e. $\partial p/\partial s_0=0$ , so $\partial p(s_0, t)/\partial t=\dot {p}$ . Equation (2.20) then becomes

(2.21) \begin{equation} \frac {\textrm{d}\overline {\boldsymbol{X}}}{\textrm{d}t}=\frac {\eta \dot {p}}{L}\left [ \int _0^{L_0}\boldsymbol{x}(s_0, t)\,\textrm{d}s_0-L_0\overline {\boldsymbol{X}}\right ] =\frac {\eta \dot {p}}{L}\left [ \frac {L_0}{L}\int _0^{L_0}\boldsymbol{x}(s_0, t)(1+\eta p)\,\textrm{d}s_0-L_0\overline {\boldsymbol{X}}\right ] =\boldsymbol{0}, \end{equation}

which proves point (i). $\def\lumina\hspace*{23pc}\square$

The first term on the right-hand side of (2.20) clearly highlights how the coupling between the local position and the compression rate contributes to the net motion, and this means in particular that a change of total length is not essential for the generation of motion. In the following sections, we quantitatively investigate the role of this coupling.

3.1. Model

We consider an idealised planar swimmer made of two rigid slender rods $S_1$ , $S_2$ of lengths $\ell _1$ , $\ell _2$ , connected by a junction, and assume that the swimmer is situated in an $x{-}y$ plane. The position of the swimmer with respect to the fixed frame $(\boldsymbol{e}_x, \boldsymbol{e}_y, \boldsymbol{e}_z)$ is represented by the position of the junction between the two rods, and is denoted by $\boldsymbol{x}_0 = (x_0,y_0)$ . Additionally, we consider two moving frames $(\boldsymbol{e}_{1,\parallel },\boldsymbol{e}_{1,\bot })$ and $(\boldsymbol{e}_{2,\parallel },\boldsymbol{e}_{2,\bot })$ attached to each rod. We denote by $\theta _0$ the angle between $\boldsymbol{e}_{x}$ and $\boldsymbol{e}_{1,\parallel }$ , and by $\alpha$ the angle between $\boldsymbol{e}_{1,\parallel }$ and $\boldsymbol{e}_{2,\parallel }$ . These angles are taken positive for an anticlockwise rotation from $\boldsymbol{e}_x$ and $\boldsymbol{e}_{1,\parallel }$ , respectively.

Here, we assume that the angle $\alpha$ and the lengths $\ell _1$ , $\ell _2$ evolve in time according to some prescribed gait, and aim to derive the effect of such a gait on the swimmer configuration in space: position $\boldsymbol{x}_0$ and orientation $\theta$ . To track the current length of segment $S_i$ with respect to its reference length $\ell _i^0$ , we introduce $\beta _i$ such that $\ell _i = \beta _i \ell _i^0$ .

In particular, we focus on two subcases of interest for the compression. The first case is uniform compression. Unicellular swimming microorganisms such as Stentor and Lacrymaria exhibit morphological changes of their helically coiled structure during the compression and extensions, and it is reasonable to assume uniform compression/extension as a simple model of their deformation. In the two-link model, this means $\ell _1 =\ell _2$ at all times, or

(3.1) \begin{equation} \beta _2 = \frac {\ell _1^0}{\ell _2^0} \beta _1 , \end{equation}

which yields $\beta _1 = \beta _2$ at all times if $\ell _1^0 = \ell _2^0$ .

The second subcase is the constant total length, which fits case (ii) of Proposition1 and may model distributed muscular contraction. In the two-link model, it translates as $\ell _1 + \ell _2 = \ell _1^0 + \ell _2^0$ at all times, or

(3.2) \begin{equation} \beta _2 = \left ( 1 + \frac {\ell _1^0}{\ell _2^0} \right ) - \frac {\ell _1^0}{\ell _2^0} \beta _1 , \end{equation}

which means $\beta _1 + \beta _2 = 2$ at all times if $\ell _1^0 = \ell _2^0$ .

For $s_0 \in [0,\ell _i^0]$ , let $\boldsymbol{x}_{i}(s_0)$ be the coordinates of the point of Lagrangian label $s_0$ on segment $i$ , with the origin of coordinates being taken at the $S_1{-}S_2$ junction. Then it holds that

(3.3) \begin{align} \boldsymbol{x}_{1}(s_0) & = \boldsymbol{x}_0 - s_0 \beta _1 \boldsymbol{e}_{1,\parallel }, \end{align}

(3.4) \begin{align} \boldsymbol{x}_{2}(s_0) & = \boldsymbol{x}_0 + s_0 \beta _2 \boldsymbol{e}_{2,\parallel }, \end{align}

and it follows, for the local velocity at $\boldsymbol{x}_{i}(s_0)$ , that

(3.5) \begin{align} \dot {\boldsymbol{x}}_1 (s_0) & = \dot {\boldsymbol{x}}_0 - s_0 \dot {\beta }_1 \boldsymbol{e}_{1,\parallel } - s_0 \beta _1 \dot {\theta } \boldsymbol{e}_{1,\bot }, \end{align}

(3.6) \begin{align} \dot {\boldsymbol{x}}_2 (s_0) & = \dot {\boldsymbol{x}}_0 + s_0 \dot {\beta }_2 \boldsymbol{e}_{2,\parallel } + s_0 \beta _2 (\dot {\theta } + \dot {\alpha }) \boldsymbol{e}_{2,\bot }. \end{align}

Using (2.4), we can express the local force density exerted by the fluid on the swimmer at point $\boldsymbol{x}_{i}(s_0)$ :

(3.7) \begin{equation} \boldsymbol{f}_i (s_0) = - C_{\parallel } ( \dot {\boldsymbol{x}}_i \boldsymbol{\cdot }\boldsymbol{e}_{\parallel } ) \boldsymbol{e}_{\parallel } - C_{\bot } ( \dot {\boldsymbol{x}}_i \boldsymbol{\cdot }\boldsymbol{e}_{\bot } ) \boldsymbol{e}_{\bot }. \end{equation}

Then one can compute the total hydrodynamic force on each segment, scaled by extension $\beta _i$ :

(3.8) \begin{equation} \boldsymbol{F}_i = \int _{S_i} \boldsymbol{f}_i \textrm{d} S_i = \int _0^{\ell ^0_i} \beta _i\, \boldsymbol{f}_i(s_0)\, \textrm{d} s_0, \end{equation}

which yields

(3.9) \begin{align} \boldsymbol{F}_1 &= - \dot {x} \beta _1 \ell _1^0 (C_{\parallel } \cos \theta _0\, \boldsymbol{e}_{1,\parallel } - C_{\bot } \sin \theta _0\, \boldsymbol{e}_{1,\bot } ) \nonumber \\[2pt] &\quad -\, \dot {y} \beta _1 \ell _1^0 (C_{\parallel } \sin \theta _0\, \boldsymbol{e}_{1,\parallel } + C_{\bot } \cos \theta _0\, \boldsymbol{e}_{1,\bot } ) \nonumber \\[2pt] &\quad +\, \dot {\theta }_0 C_{\bot } \beta _1^2 \frac {({\ell _1^0})^2}{2} \boldsymbol{e}_{1,\bot } + C_{\parallel } \beta _1 \dot {\beta }_1 \frac {({\ell _1^0})^2}{2} \boldsymbol{e}_{1,\parallel }, \end{align}

(3.10) \begin{align} \boldsymbol{F}_2 & = - \dot {x}_0 \beta _2 \ell _2^0 \, (C_{\parallel } \cos (\theta _0 + \alpha )\, \boldsymbol{e}_{2,\parallel } - C_{\bot } \sin (\theta _0 + \alpha )\, \boldsymbol{e}_{2,\bot } ) \nonumber \\[2pt] &\quad -\, \dot {y}_0 \, \beta _2 \ell _2^0 \, (C_{\parallel } \sin (\theta _0 + \alpha )\, \boldsymbol{e}_{2,\parallel } + C_{\bot } \cos (\theta _0 + \alpha )\, \boldsymbol{e}_{2,\bot } ) \nonumber \\[2pt] &\quad +\, ( \dot {\theta }_0 + \dot {\alpha } ) C_{\bot } \beta _2^2 \frac {({\ell _2^0})^2}{2} \boldsymbol{e}_{2,\bot } + C_{\parallel } \beta _2 \dot {\beta }_2 \frac {({\ell _2^0})^2}{2} \boldsymbol{e}_{2,\parallel }. \end{align}

Similarly, the torques $T_1$ and $T_2$ on each segment with respect to the junction point, and projected on $\boldsymbol{e}_z$ , are given by

(3.11) \begin{align} T_1 & = \beta _1^2 C_{\perp } \frac {({\ell _1^0})^2}{2} \left ( - \dot {x}_0 \cos \theta _0 + \dot {y}_0 \cos \theta _0 \right ) + \dot {\theta }_0 \beta _1^3 \frac {({\ell _1^0})^3}{3}, \end{align}

(3.12) \begin{align} T_2 & = \beta _2^2 C_{\perp } \frac {({\ell _2^0})^2}{2} \left ( \dot {x}_0 \cos ( \theta _0 + \alpha ) + \dot {y}_0 \cos ( \theta _0 + \alpha ) \right ) + (\dot {\theta }_0 + \dot {\alpha }) \beta _2^3 \frac {({\ell _2^0})^3}{3}. \end{align}

Since inertia is negligible, we can write balance of force and torques at all times:

(3.13) \begin{equation} \left \{ \begin{array}{l} \boldsymbol{F}_1 + \boldsymbol{F}_2 = 0, \\ T_1 + T_2 = 0. \end{array} \right . \end{equation}

Using (3.9)–(3.13), and plugging the expression for $\beta _2$ with respect to $\beta _1$ , one can derive the Stokes connection that links shape velocities $(\dot {\alpha },\dot {\beta }_1)$ to rigid motion velocities in the reference frame $(\dot {x}_0,\dot {y}_0,\dot {\theta }_0)$ :

(3.14) \begin{equation} \boldsymbol{A}(\theta _0,\alpha ,\beta _1)\begin{pmatrix} \dot {x}_0 \\ \dot {y}_0 \\ \dot {\theta }_0 \end{pmatrix} = \boldsymbol{B} (\theta _0,\alpha ,\beta _1) \begin{pmatrix} \dot {\alpha } \\ \dot {\beta _1} \end{pmatrix}. \end{equation}

To highlight the existence of a connection, we rewrite the resistance matrices in a moving frame $(\boldsymbol{e}_{a,\parallel },\boldsymbol{e}_{a,\bot })$ representing the average orientation of the swimmer, namely at $\theta _0 + \alpha /2$ . As noted by Hatton & Choset (Reference Hatton and Choset2015) and Bass, Ramasamy & Hatton (Reference Bass, Ramasamy and Hatton2022), this frame has the important advantage of minimising the error accumulated by higher-order terms when integrating the Stokes curvature (2.10) over a loop trajectory in the shape space. Hence we define $\boldsymbol{A}_a = \boldsymbol{R}_{\theta _0+\alpha /2}^{-1} \boldsymbol{A} \boldsymbol{R}_{\theta _0+\alpha /2}$ and $\boldsymbol{B}_a = \boldsymbol{R}_{\theta _0+\alpha /2}^{-1} \boldsymbol{B}$ , and write (3.14) in the form,

(3.15) \begin{equation} \boldsymbol{R} ^{-1}_{\theta _0+\alpha /2} \begin{pmatrix} \dot {x}_0 \\ \dot {y}_0 \\ \dot {\theta }_0 \end{pmatrix} = \mathcal{H}\begin{pmatrix} \dot {\alpha } \\ \dot {\beta _1} \end{pmatrix}, \end{equation}

where the connection $\mathcal{H}$ is given by $\mathcal{H}=\boldsymbol{A}_a^{-1} \boldsymbol{B}_a$ .

Now let us give detailed expressions of these matrices in both cases of interest, (3.1) and (3.2). For simplicity, we assume $\ell ^0_1 = \ell ^0_2 = \ell$ . In the case of uniform compression (3.1), one obtains

(3.16) \begin{equation} \mathcal{H} = \begin{pmatrix} 0 & 0\\ -\cfrac {\beta _1 \gamma \ell \cos \left (\dfrac {\alpha }{2}\right )}{4{\left [1+(\gamma -1) \cos ^2 \left (\dfrac {\alpha }{2}\right )\right ]}} & \cfrac {\ell \sin \left (\dfrac {\alpha }{2}\right )}{2{\left [\gamma +(1-\gamma ) \sin ^2\left ( \dfrac {\alpha }{2}\right )\right ]}}\\ -1/2 & 0 \end{pmatrix}. \end{equation}

Then one can compute the velocity of the geometric centre from its definition as $ \boldsymbol{x}_m = ({1}/{4}) ( 2 \boldsymbol{x}_0 + \boldsymbol{x}_1(\ell ) + \boldsymbol{x}_2(\ell ) )$ , which, expressed in the rotating basis $(\boldsymbol{e}_{a,\parallel },\boldsymbol{e}_{a,\bot })$ , yields the motion only in the $y_m$ direction, with its velocity given by

(3.17) \begin{eqnarray} \dot {y}_m= \displaystyle \frac {\ell \cos \left (\dfrac {\alpha }{2}\right )\,{\left (\gamma -1\right )}{\left [{{\beta }_1}\sin \alpha + \dfrac {\mathrm{\alpha }\beta _1 }{2}(\cos \alpha -1) \right ]}}{2\gamma -2\cos \alpha +2\gamma \cos \alpha +2}, \end{eqnarray}

from which we deduce that any translational motion occurs along the $y_m$ direction. Moreover, it is clear that isotropic drag ( $\gamma = 1$ ) forbids any translation, in agreement with case (i) of Proposition1.

In the case of constant total length (3.2), one has

(3.18) {{{ \begin{equation} \boldsymbol{A}_a = \left (\!\!\begin{array}{ccc} \ell {\left (\gamma +\cos \alpha -\gamma \cos \alpha +1\right )} & \ell \sin \alpha {\left (\beta _1 -1\right )}{\left (\gamma -1\right )} & -\gamma {\ell }^2 \sin \left (\alpha /2\right ){\left ({\beta _1 }^2 -2\beta _1 +2\right )} \\[1.5pt] \ell \sin \alpha {\left (\beta _1 -1\right )}{\left (\gamma -1\right )} & \ell {\left (\gamma -\cos \alpha +\gamma \cos \alpha +1\right )} & -2\gamma {\ell }^2 \cos \left (\alpha /2\right ){\left (\beta _1 -1\right )} \\[1.5pt] -\gamma {\ell }^2 \sin \left (\alpha /2\right ){\left ({\beta _1 }^2 -2\beta _1 +2\right )} & -2\gamma {\ell }^2 \cos \left (\alpha /2\right ){\left (\beta _1 -1\right )} & -(2/3)\gamma {\ell }^3 {\left (3{\beta _1 }^2 -6\beta _1 +4\right )} \end{array}\!\! \right ), \end{equation}}}}

and

(3.19) \begin{equation} \boldsymbol{B}_a = \begin{pmatrix} -(\gamma /2) {\ell }^2 \sin \left (\alpha /2\right ){{\left (\beta _1 -2\right )}}^2 & {\ell }^2 \cos \left (\alpha /2\right )\\[1.5pt] (\gamma /2) {\ell }^2 \cos \left (\alpha /2\right ){{\left (\beta _1 -2\right )}}^2 & -{\ell }^2 \sin \left (\alpha /2\right ){\left (\beta _1 -1\right )}\\[1.5pt] (\gamma /3) \,{\ell }^3 \,{{\left (\beta _1 -2\right )}}^3 & 0 \end{pmatrix}, \end{equation}

leading to the connection $\mathcal{H} = \boldsymbol{A}_a^{-1} \boldsymbol{B}_a$ in the form

(3.20) \begin{equation} \mathcal{H} = \frac {1}{Q(\cos \alpha )} \begin{pmatrix} 2 \gamma \ell \beta _1^2 (\beta _1-2)^2 (\beta _1-1) \sin (\alpha /2)\, P_{11} (\alpha ) &\!\! \ell \cos (\alpha /2)\, P_{12} (\alpha ) \\[1.5pt] \gamma \ell \beta _1^2 (\beta _1-2)^2 \cos (\alpha /2)\, P_{21} (\alpha ) &\!\! \gamma \ell (\beta _1-1) \sin (\alpha /2)\, P_{22}(\alpha ) \\[1.5pt] (\beta _1-2)^2\, P_{31}(\alpha ) &\!\! \beta _1 (\beta _1-2)\, P_{32}(\alpha ) \end{pmatrix}, \end{equation}

where $P_{ij}$ are polynomials in $\cos \alpha$ , $\sin \alpha$ , $\beta _1$ (whose expressions are too lengthy to write fully) and $\gamma$ , and $Q$ is defined as

(3.21) \begin{align} Q(x) &= \ell ^4 \big[ 3 \beta _1^4 \big(7 \gamma ^2 {x}^2 -7 \gamma ^2 -11 \gamma {x}^2 -2 \gamma x+13 \gamma +4 {x}^2 -4\big)\nonumber \\[1.5pt] &\quad + 12 \beta _1^3 \big(- 7 \gamma ^2 {x}^2 +7 \gamma ^2 +11 \gamma {x}^2 +2 \gamma x-13 \gamma -4 {x}^2 +4\big) \nonumber \\[1.5pt] &\quad + 16 \beta _1 ^2 \big( 7 \gamma ^2 {x}^2 - 7 \gamma ^2 - 11 \gamma {x}^2 - \frac {3}{2} \gamma x+ \frac {37}{2} \gamma +4 {x}^2 -4\big) \nonumber \\[1.5pt] &\quad + 8 \beta _1 \big( - 7 \gamma ^2 {x}^2 + 7 \gamma ^2 +11 \gamma {x}^2 - 35 \gamma -4 {x}^2 +4\big) +112\gamma\big]. \end{align}

The expression for the motion of the geometric centre can be calculated with symbolic calculus software but is too tedious to be reproduced here, even in the particular case of isotropic drag. We can, however, evaluate it at specific shapes in the case $\gamma = 1$ to ensure that motion is indeed possible. For isotropic drag ( $\gamma = 1$ ) and flat shape ( $\alpha = 0,\ \theta _0 = 0$ ), one obtains

(3.22) \begin{equation} \dot {x}_m=\frac {\ell }{2} \dot {\beta _1},\quad \dot {y}_m=-\frac {7\,\beta _1 \ell {{\left (\beta _1 -1\right )}}^2 {\left (\beta _1 -2\right )}}{48\beta _1 ^2 -96\beta _1 +56} \dot {\alpha }, \end{equation}

and for links of equal length ( $\beta _1 =1$ ), one obtains

(3.23) \begin{equation} \dot {x}_m=\frac {\ell \cos (\alpha + \theta _0)}{2} \dot {\beta _1},\quad \dot {y}_m=\frac {\ell \sin (\alpha + \theta _0)}{2} \dot {\beta _1}. \end{equation}

In both cases, it is clear that the geometric centre does not stay fixed, despite the total length of the swimmer remaining constant. This simple example therefore provides a constructive proof for point (ii) of Proposition1.

3.2. Gait prediction

From the expression of the connection, one may obtain the curvature of the gauge field $\mathcal{F}$ following (2.13). Since the squeeze-me-bend-you swimmer possesses only two degrees of freedom, each scalar component of $\mathcal{F}$ can be easily visualised on a two-dimensional plane $(\alpha ,\beta )$ . Moreover, any deformation gait can be represented in the same plane, and (2.10) indicates that the net locomotion produced by the gait will be, at first order, proportional to $\mathcal{F}$ integrated over the (algebraic) area enclosed by the gait cycle in the shape space. In the following, we use this principle to design simple gaits achieving elementary motion in each available direction, demonstrating the swimming capabilities of the squeeze-me-bend-you swimmer model.

We first examine the uniform compression case, in which only the $y$ component of $\mathcal{F}$ does not vanish, and takes a remarkably simple form depending only on $\alpha$ (and notably not on $\beta$ ):

(3.24) \begin{equation} \mathcal{F}_y = \dfrac {\cos \left (\dfrac {\alpha }{2}\right )-3\,{\cos \left (\dfrac {\alpha }{2}\right )}^3 }{{{\left (2\,{\cos \left (\dfrac {\alpha }{2}\right )}^2 +2\right )}}^2 }. \end{equation}

The function $\mathcal{F}_y$ is even and $2 \unicode{x03C0}$ -periodic. On the interval $[-\unicode{x03C0} ,\unicode{x03C0} ]$ , it vanishes at $\pm \unicode{x03C0}$ and $\pm \alpha _m$ , with $\alpha _m= \arccos (- 1/3 )$ (which, serendipitously, happens to be the ‘tetrahedral angle’ measured between a diagonal of a cube and its adjacent edge). It is therefore of constant sign on the interval $[-\alpha _m,\alpha _m]$ , suggesting that to ensure maximal displacement, the swimmer must fold between these two values, with compression $\beta$ alternating between any two set values with the same periodicity in the meantime. Such a gait is represented in figure 3(a).

Figure 3.

Curvature fields for the squeeze-me-bend-you swimmer, as defined in (2.13). Dotted black lines indicate zero-curvature level set. Suggested gaits are indicated by a continuous black line, and the corresponding deformation sequence of the swimmer is represented on the right of each curvature plot: (a) $y$ -displacement for uniform compression, (b) $x$ -displacement, (c) $y$ -displacement, and (d) $\theta$ -displacement, with time progress shown by the arrow and colours.

In the constant length case, the symbolic expression of $\mathcal{F}$ is not nearly nice enough to be analysed in the same way, or even reproduced here. We may nonetheless evaluate it numerically, and its three components are plotted in figure 3(b–d). Then it is possible to predict the qualitative behaviour of the swimmer after some gait by leveraging the symmetries of $\mathcal{F}$ (Shammas, Choset & Rizzi Reference Shammas, Choset and Rizzi2007), as is classically done in analogous studies on minimal swimmer models such as the Purcell swimmer (Avron & Raz Reference Avron and Raz2008). In figure 3(b), one may first observe that $\mathcal{F}_x$ is odd in $\alpha$ . For this reason, a simple circular or elliptical gait centred at $(0,1)$ would fail to produce any net locomotion in the direction $x$ . A good candidate for a successful gait would be a ‘butterfly’ gait as shown in figure 3(b). Note that in the meantime, no motion in $y$ or $\theta$ would result from this butterfly gait, due to the even parity in $\alpha$ of $\mathcal{F}_y$ and $\mathcal{F}_{\theta }$ visible in figure 3(c,d). On the other hand, to generate net motion purely in the $y$ direction, an elliptical gait centred at $\alpha = 0$ may be considered, but needs to be shifted away from $\beta = 1$ to avoid producing net rotation at the same time, as shown in figure 3(b).

As one can infer from $\mathcal{F}_{\theta }$ in figure 3(d), elliptical gaits centred at $(0,1)$ will rotate the swimmer in the reference plane. We note that the net rotation after one cycle is outstandingly high when compared to the capabilities of its cousin in the family of minimal swimmer models, the Purcell swimmer, which features only bending and no compression. Indeed, under the reasonable assumption that $\alpha \in [-\unicode{x03C0} ,\unicode{x03C0} ]$ , a simple optimisation routine on elliptical gaits shows that the Purcell swimmer can reach a net rotation of around $\unicode{x03C0} /2$ rad in one cycle, at most, while the squeeze-me-bend-you swimmer can achieve a rotation of nearly $3\unicode{x03C0} /2$ . This high orientational manoeuvrability suggests that compression, even in a small amount, plays a prominent role in the orientational dynamics of slender swimmers. We further quantitatively evaluate this enhancement of manoeuverability in § 7.

4.1. Leading-order calculations

We first consider $\tilde {F}_x=0$ by substituting (2.1)–(2.2) into (2.5), to obtain, up to the second order of expansions,

(4.9) \begin{align} \int _0^L\left (\tilde {U}_x+\frac {{\textrm D}\tilde {x}}{{\textrm D}t}\right )\,\textrm{d}s &= \epsilon \varOmega \int _0^L q\tilde {y}\,\textrm{d}s +\epsilon (\gamma -1) \int _0^L q\left (\tilde {U}_y+\varOmega \tilde {x}+\frac {{\textrm D}\tilde {y}}{{\textrm D}t}\right ){\textrm d}s \nonumber\\ &\quad +\, O(\epsilon ^3, \epsilon ^2\eta ) . \end{align}

Here, we find that the right-hand side of (4.9) is of the second order of small parameters, while the left-hand side is of the first order of magnitudes. Hence at the leading order of the expansions, we have

(4.10) \begin{eqnarray} \tilde {U}^{(1)}_x =-\frac {\eta }{L}\int _0^{L_0}\left [\int _0^{s_0}\frac {\partial p}{\partial t}(s^{\prime}_0, t)\,\textrm{d}s^{\prime}_0\right ] \textrm{d}s_0. \end{eqnarray}

We then consider the force balance in the normal direction by imposing $\tilde {F}_y=0$ , which is written as

(4.11) \begin{eqnarray} \epsilon (\gamma -1) \int _0^L q\left (\tilde {U}_x+\frac {{\textrm D}\tilde {x}}{{\textrm D}t}\right )\textrm{d}s=\gamma \int _0^L \left ( \tilde {U}_y+\varOmega \tilde {x}+\frac {{\textrm D}\tilde {y}}{{\textrm D}t}\right )\textrm{d}s+O(\epsilon ^3, \epsilon ^2\eta ) . \end{eqnarray}

Similarly, the torque balance $\tilde {M}=0$ reads

(4.12) \begin{align} &\epsilon (\gamma -1)\int _0^L q\tilde {x}\left (\tilde {U}_x+\frac {{\textrm D}\tilde {x}}{{\textrm D}t}\right )\textrm{d}s+ \int _0^L\tilde {y} \left (\tilde {U}_x+\frac {{\textrm D}\tilde {x}}{{\textrm D}t}\right )\textrm{d}s \nonumber \\ &\quad = \gamma \int _0^L \tilde {x}\left ( \tilde {U}_y+\varOmega \tilde {x}+\frac {{\textrm D}\tilde {y}}{{\textrm D}t}\right )\textrm{d}s+O(\epsilon ^3, \epsilon ^2\eta ) . \end{align}

By similar order estimates for (4.11)–(4.12), we find that in the leading order, only the right-most integral contributes in each equation. Hence we obtain

(4.13) \begin{align} L\tilde {U}_y^{(1)}+\frac {L^2}{2}\varOmega ^{(1)}+\epsilon \int _0^{L_0}\frac {\partial q}{\partial t}(s_0, t)\,\textrm{d}s_0 &= 0, \end{align}

(4.14) \begin{align} \frac {L^2}{2}\tilde {U}_y^{(1)}+\frac {L^3}{3}\varOmega ^{(1)}+\epsilon \int _0^{L_0}s_0\frac {\partial q}{\partial t}(s_0, t)\,\textrm{d}s_0 &= 0, \end{align}

yielding the first-order velocities

(4.15) \begin{align} \tilde {U}_y^{(1)} &= -\frac {6\epsilon }{L^2}\int _0^{L_0}\left (\frac {2L}{3}-s_0\right )\frac {\partial q}{\partial t}\,\textrm{d}s_0 , \end{align}

(4.16) \begin{align} \varOmega ^{(1)} &= \frac {12\epsilon }{L^3}\int _0^{L_0}\left (\frac {L}{2}-s_0\right )\frac {\partial q}{\partial t}\,\textrm{d}s_0. \end{align}

4.2. Second-order calculations

We then proceed to the second-order velocities, by substituting the leading-order results into the force and torque balance equations (4.9)–(4.12).

We start with the tangential force balance. The second-order contributions are calculated as

(4.17) \begin{equation} L\tilde {U}_x^{(2)}+\int _0^L\big(\tilde {W}_x-\varOmega ^{(1)}\tilde {y}\big)\,\textrm{d}s+\epsilon (1-\gamma )\int _0^Lq\left (\tilde {U}_y^{(1)}+\varOmega ^{(1)}\tilde {x}+\frac {{\textrm D}\tilde {y}}{{\textrm D}t} \right )\textrm{d}s=0 , \end{equation}

where we introduced

(4.18) \begin{equation} \tilde {W}_x(s_0, t):=\tilde {U}^{(1)}_x+\frac {{\textrm D} \tilde {x}}{{\textrm D}t}. \end{equation}

From (4.17), we may readily obtain the second-order tangential velocity $\tilde {U}_x^{(2)}$ . Similarly, the normal force balance and torque balance equations are expanded up to second order as

(4.19) \begin{eqnarray} \epsilon (\gamma -1) \!\!\int _0^L q\tilde {W}_x\,\textrm{d}s=\gamma \!\!\int _0^L \! \left( \tilde {U}^{(1)}_y+\varOmega ^{(1)}\tilde {x}+\frac {{\textrm D}\tilde {y}}{{\textrm D}t}\right )\textrm{d}s +\gamma \!\left(\! L\tilde {U}^{(2)}_y +\frac {L^2}{2}\varOmega ^{(2)}\right ) \end{eqnarray}

and

(4.20) \begin{align} \epsilon (\gamma -1)\int _0^L q\tilde {x}\tilde {W}_x\,\textrm{d}s &= -\int _0^L\tilde {y} \tilde {W}_x\,\textrm{d}s +\gamma \int _0^L \tilde {x}\left ( \tilde {U}^{(1)}_y+\varOmega ^{(1)}\tilde {x} +\frac {{\textrm D}\tilde {y}}{{\textrm D}t}\right )\textrm{d}s \nonumber \\ &\quad +\gamma \left ( \frac {L^2}{2}\tilde {U}^{(2)}_y +\frac {L^3}{3}\varOmega ^{(2)}\right ) , \end{align}

respectively. Hence $\tilde {U}^{(2)}_y$ and $\varOmega ^{(2)}$ may be calculated by evaluating the integrals in (4.19)–(4.20) up to second order.

4.3. Third-order calculations

To obtain the third-order progressive velocity, one needs to expand the position $\tilde {x}$ and $\tilde {y}$ , and the vectors $\boldsymbol{n}$ and $\boldsymbol{t}$ , up to the third order of small parameters $\epsilon$ and $\eta$ . The force balance equations in the $\tilde {x}$ direction (4.9) are therefore evaluated as

(4.21) \begin{align} \tilde {U}^{(3)}_xL &- \tilde {\varOmega }^{(2)}\int _0^L\tilde {y}\,\textrm{d}s+\int _0^L\left [\tilde {W}_x+\tilde {U}^{(2)}_y-\tilde {\varOmega }^{(1)}\tilde {y}\right ] \left(1-\epsilon ^2q^2\right)\,\textrm{d}s +\epsilon ^2\gamma \int _0^Lq^2\tilde {W}_x \,\textrm{d}s \nonumber \\[2pt] & \quad +\, \epsilon (1-\gamma )\int _0^L\left [ \tilde {U}^{(1)}_y+\tilde {U}^{(2)}_y+\bigl (\tilde {\varOmega }^{(1)}+\tilde {\varOmega }^{(2)}\bigr )\tilde {x}+\frac {{\textrm D}\tilde {y}}{{\textrm D}t}\right ]\textrm{d}s =0, \end{align}

by neglecting the fourth- and higher-order terms. Since the first- and second-order velocities are all calculated, by substituting these expressions into (4.21) and evaluating the integrals, we may obtain the expression for the third-order tangential velocity $\tilde {U}^{(3)}_x$ .

5.1. Uniform compression

As an illustrative example, we first consider locomotion with bending wave and uniform compression, given by

(5.1) \begin{eqnarray} p(s_0, t)=\sin (\omega t)\quad\textrm{and}\quad q(s_0, t)=\sin (ks_0+\omega t+\phi ) , \end{eqnarray}

where $\phi \in [0, 2\unicode{x03C0} )$ denotes the phase difference between the two modes. For simplicity, we assume that the filament contains an integer number of waves, i.e. $kL_0=\pm 2\unicode{x03C0} , \pm 4\unicode{x03C0} , \pm 6\unicode{x03C0} , \ldots$ .

The first-order calculations provide

(5.2) \begin{align} \tilde {U}_x^{(1)} &= -\frac {\eta \omega L_{0}}{2}\cos \omega t, \end{align}

(5.3) \begin{align} \tilde {U}_y^{(1)} &= \frac {\epsilon \omega }{k}\sin (\omega t+\phi ) -\frac {6\epsilon \omega }{k^2L_0}\cos (\omega t+\phi ), \end{align}

(5.4) \begin{align} \varOmega ^{(1)} &= \frac {12\epsilon \omega }{k^2L_0^2}\cos (\omega t+\phi ), \end{align}

which vanish after averaging over a time period.

We then proceed to calculating the second-order contributions. The tangential velocity at the second order may be calculated from (4.17) as

(5.5) \begin{equation} \tilde {U}^{(2)}_x=\frac {\epsilon ^2\omega }{2k}\left [ (\gamma -1)+\frac {1}{2}\cos (2\omega t+2\phi )\right ] -\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -2)\left [1+\cos (2\omega t+2\phi )\right ]. \end{equation}

By calculating the integrals and solving the linear problem at the second-order equations (4.19)–(4.20), we obtain

(5.6) \begin{align} \tilde {U}_y^{(2)} &= -\frac {\epsilon \eta \omega }{2 k}\left [ \left (\frac {\gamma -1}{\gamma } \right )\cos \phi +\left (\frac {3\gamma -1}{\gamma }\right )\cos (2\omega t+\phi )\right ]\nonumber\\ &\quad -\frac {3\epsilon \eta \omega }{k^2L_0^3}\left [ \left (\frac {2\gamma -3}{\gamma }\right )\sin \phi +\left (\frac {4\gamma -3}{\gamma }\right )\sin (2\omega t+\phi )\right ]\end{align}

(5.7) \begin{align} \varOmega ^{(2)} &= \frac {18\epsilon \eta \omega }{k^2L_0^2}\left (\frac {\gamma -1}{\gamma }\right )\left [\sin \phi +\sin (2\omega t+\phi ) \right ]. \end{align}

By taking the time average, we have

(5.8) \begin{align} \langle \tilde {U}_x\rangle &= \frac {\epsilon ^2\omega }{2k}(\gamma -1)-\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -2), \end{align}

(5.9) \begin{align} \langle \tilde {U}_y\rangle &= -\frac {\epsilon \eta \omega }{2k}\left (\frac {\gamma -1}{\gamma }\right )\cos \phi -\frac {3\epsilon \eta \omega }{k^2L_0}\left (\frac {2\gamma -3}{\gamma } \right )\sin \phi , \end{align}

and for the angular velocity,

(5.10) \begin{equation} \langle \varOmega \rangle = \frac {18\epsilon \eta \omega }{k^2L_0^2}\left ( \frac {\gamma -1}{\gamma }\right )\sin \phi , \end{equation}

where we introduced the time average symbol, $\langle \bullet \rangle =(1/T)\int _0^T \bullet \,\textrm{d}t$ , and deformation time period $T=2\unicode{x03C0} /\omega$ . To obtain the expressions in the laboratory frame to second order, one needs to take the non-commutative effects into account as in (2.9). Instead of using the gauge-theoretical method, we directly compute the Magnus expansion (2.9), as the shape variable is not obvious in the continuous filament model. The final results are then obtained as

(5.11) \begin{align} \langle U_x\rangle &= \frac {\epsilon ^2\omega }{2k}(\gamma -1)-\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -1), \end{align}

(5.12) \begin{align} \langle U_y\rangle &= -\frac {\epsilon \eta \omega }{2k}\left (\frac {\gamma -1}{\gamma }\right )\cos \phi +\frac {9\epsilon \eta \omega }{k^2L_0}\left (\frac {\gamma -1}{\gamma } \right )\sin \phi , \end{align}

while the rotational velocity is unchanged from (5.10). We clearly find no net locomotion with the isotropic drag ( $\gamma =1$ ) for the uniform compression as in the general theory in § 2.2.

The corrections by the Lie bracket appear in the second terms in (5.11)–(5.12), which represent the finite-size effect of the body, and vanish in the limit $kL\rightarrow \infty$ . Many mathematical models with bending motions including the Taylor sheet and a slender filament model usually assume one-dimensional motion, while this simplification is only valid for an infinitely long body, because the instantaneous lateral and rotational motion, in general, does not vanish. This result highlights the importance of the motion non-commutativity for a finite-sized swimmer with lateral oscillation, the so-called yawing motion.

Although the compression itself does not generate progressive velocity, the bending–compression coupling in the $O(\epsilon \eta )$ terms generates both normal translation and rotation. This notably underlines the enhancement of manoeuvrability through body compression. In fact, by manipulating the phase shift $\phi$ , the swimmer is able to make a turn in both directions.

When $\sin \phi =0$ , as found from (5.10), no net rotation is generated. To analyse this no-net-rotation property in the large-amplitude case, we numerically examine the swimmer dynamics, by discretising the slender swimmer by $N$ small segments with length $\ell _i$ ( $i=1,\ldots, N$ ), known as the $N$ -link model (Moreau, Giraldi & Gadêlha Reference Moreau, Giraldi and Gadêlha2018). In the case of the current kinematic problem, we prescribe the segment length and the angle between the $i$ th and $(i+1)$ th segments, which we denote by $\alpha _i$ ( $i=1,\ldots, N-1$ ). The force balance and torque balances (2.5) are then computed by summing the contributions from each segment. These balance relations then form linear equations for the translational and rotational velocities. In the numerical calculations below, we use $N=40$ , which guarantees satisfying numerical accuracy for our purpose. Although the non-local hydrodynamic interactions are not negligible outside the small-amplitude regime, we use the resistive drag relation as empirical modelling, with a focus on the effects of the bending–compression coupling and the mechanical roles of drag anisotropy. Sample trajectories of the swimmer are shown with its shape gait in figure 4. The no-net-motion property still holds even outside the small-amplitude regime, as seen in figure 4(a).

Figure 4.

Top: sample trajectories of a swimmer under uniform compression with different phase shifts, (a) $\phi =0$ and (b) $\phi =\unicode{x03C0} /2$ . The parameters are $\epsilon =\eta =0.4$ , $\gamma =2$ , $L_0=1,\ kL_0=2\unicode{x03C0}$ and $\omega =1$ . With the initial position $(0,0)$ (marked by a red dot) and the initial angle $\theta =0$ , we drew the orbits of the leftmost end of the filament from $t=0$ to $t=10$ . The configuration at $t=10$ is also shown. Bottom: time sequence of swimmer shape from $t=0$ to $t=1$ with the local extension visualised by the colours.

Figure 5.

Schematics for the symmetry arguments of the no-net-rotation property: (a) the original shape in the body fixed frame with velocities $\tilde {\boldsymbol{U}}$ and $\varOmega$ ; (b) the shape after the time reversal with a shift $c=kL_0/\omega -T/2$ ; (c) the head-to-tail inversion; (d) the $\unicode{x03C0}$ -rotation of the system.

In fact, this can be predicted by considerations on the symmetry of the shape gait. Let us consider two linear changes of variables: time reversal with a phase shift $t\longmapsto t^{\prime}=-t+T/2+kL_0/\omega$ , followed by head-to-tail inversion $s_0\longmapsto s_0^{\prime}=L_0-s_0$ (figures 5 a–c). Then the bending and compression functions are unchanged: $q(s_0^{\prime}, t^{\prime})=q(s_0, t)$ and $p(s_0^{\prime}, t^{\prime})=p(s_0, t)$ , from the assumption on $kL_0$ . Hence the positions in the body-fixed frame also remain the same, after taking a $\unicode{x03C0}$ -rotation of the system as illustrated in figure 5(d). With this transformation of $s_0$ and $t$ , the time derivatives of $p$ and $q$ have an additional minus sign, $\partial q^{\prime}/\partial t^{\prime} = -\partial q/\partial t$ and $\partial p^{\prime}/\partial t^{\prime} =-\partial p/\partial t$ , yielding again the same shape velocity after the additional $\unicode{x03C0}$ -rotation of the system. Therefore, we recover the same shape and shape deformation velocity, while the rotation direction is reversed after the sequences of transformations (figure 5 d). To compensate for the uniqueness of the motion with given boundary conditions, the net rotation must vanish, while the instantaneous rotation is still allowed by the phase shift in the transformation of $t$ .

With $\sin \phi \neq 0$ , the above symmetry arguments are no longer available. Hence, as in the case $\phi =\unicode{x03C0} /2$ of figure 4(b), net rotational motions are generated by the bending–compression coupling, allowing the swimmer to turn.

5.2. Symmetric uniform compression

We then consider a symmetrical situation, where the functions $p(s_0, t)$ and $q(s_0, t)$ are given by

(5.13) \begin{eqnarray} p(s_0, t)=\sin (2\omega t)\quad\textrm{and}\quad q(s_0, t)=\sin (ks_0+\omega t+\phi ) , \end{eqnarray}

to focus on progressive velocity. The assumption of an integer number of waves is again employed here.

By expanding the velocities up to second order, direct calculations then lead to the progressive velocity in the body-fixed coordinates

(5.14) \begin{equation} \tilde {U}^{(2)}_x=\frac {\epsilon ^2\omega }{2k}\left [ (\gamma -1)+\frac {1}{2}\cos (2\omega t+2\phi )\right ] -\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -2)\left [1+\cos (2\omega t+2\phi )\right ], \end{equation}

as well as normal and rotational velocities

(5.15) \begin{align} \tilde {U}_y^{(2)} &= -\frac {\epsilon \eta \omega }{2 k}\left [ \left (\frac {3\gamma -2}{\gamma } \right )\cos (\phi -\omega t)+\left (\frac {5\gamma -2}{\gamma }\right )\cos (3\omega t+\phi )\right ]\nonumber\\ &\quad -\frac {3\epsilon \eta \omega }{k^2L_0}\left [ \left (\frac {5\gamma -6}{\gamma }\right )\sin (\phi -\omega t)+\left (\frac {7\gamma -6}{\gamma }\right )\sin (3\omega t+\phi )\right ]\end{align}

(5.16) \begin{align} \varOmega ^{(2)} &= \frac {36\epsilon \eta \omega }{k^2L_0^2}\left (\frac {\gamma -1}{\gamma }\right )\left [\sin (\phi -\omega t)+\sin (3\omega t+\phi ) \right ]. \\[6pt] \nonumber \end{align}

By taking the time average of the second-order velocities, we obtain

(5.17) \begin{eqnarray} \langle \tilde {U}^{(2)}_x\rangle =\frac {\epsilon ^2\omega }{2k}(\gamma -1)-\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -2), \end{eqnarray}

and $\langle \tilde {U}_y\rangle =\langle \varOmega \rangle =0$ as expected by symmetry. After including the non-commutative effects between the normal translation and the rotation, we recover the same expression as in the asymmetrical case,

(5.18) \begin{eqnarray} \langle U_x^{(2)}\rangle &=&\frac {\epsilon ^2\omega }{2k}(\gamma -1)-\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -1). \end{eqnarray}

Now let us examine the third-order contribution for the tangential velocity. We need to expand $\tilde {x}(s_0, t)$ and $\tilde {y}(s_0, t)$ as well as the vectors $\boldsymbol{e}_{\|}(s_0, t)$ and $\boldsymbol{e}_{\perp }(s_0, t)$ up to third orders of $\epsilon$ and $\eta$ . The third-order tangential velocity in the body-fixed frame is therefore obtained by substituting the first- and second-order velocities into (4.21), and evaluating the integrals. Since the expression is lengthy, we present only the time-averaged velocity $\langle \tilde {U}^{(3)}_x\rangle$ , given by

(5.19) \begin{equation} \langle \tilde {U}^{(3)}_x\rangle =\frac {\epsilon ^2\eta \omega }{4k}(\gamma -1)\sin 2\phi -\frac {3\epsilon ^2\eta \omega }{k^3 L_0^2}\left ( \frac{5\gamma-6}{\gamma} \right) (\gamma-2)\sin 2\phi . \end{equation}

We then evaluate the Lie brackets in (2.9) from the matrix $\hat {\mathcal{A}}$ by employing the first- and second-order velocities as well as $\tilde {U}^{(3)}_x$ . The averaged tangential velocity in the laboratory frame may be calculated up to third order in the form $\langle U_x\rangle =\langle U_x^{(2)}\rangle +\langle U_x^{(3)}\rangle$ , with

(5.20) \begin{equation} \langle U_x^{(3)}\rangle =\frac {\epsilon ^2\eta \omega }{4k}(\gamma -1)\sin 2\phi -\frac {3\epsilon ^2\eta \omega }{k^3 L_0^2}\left ( \frac {5\gamma -8}{\gamma }\right )(\gamma -1)\sin 2\phi . \end{equation}

Hence, thanks to the corrections coming from the non-commutativity of the rigid motion, we recover case (i) of Proposition1.

The effect of uniform compression may be found at third order with an increase or decrease of velocity depending on the phase difference $\phi$ . We also numerically examine the effect of the uniform compression outside of the small-amplitude region, and plot the ratio of the averaged velocities with and without compression, $\langle U_{x}\rangle /\langle U_{x}|_{\eta = 0)}\rangle$ , in figure 6. For small amplitude, we obtain that this ratio is expressed, up to third order, by a linear function of $\eta$ as

(5.21) \begin{eqnarray} \frac {\langle U_{x}\rangle }{\langle U_{x}|_{\eta = 0}\rangle }=1+\frac {\langle U^{(3)}_x\rangle }{\langle U^{(2)}_x\rangle } =\frac {\eta }{2}\left (1-\frac {15-24\gamma ^{-1}}{k^2 L^2_0}\right )\left (1-\frac {6}{k^2L_0^2}\right )^{-1}. \end{eqnarray}

As seen in the figure, the impact of the compression is linearly proportional to the compression size $\eta$ , which agrees well with what the small-amplitude theory predicts. The quantitative agreements are more precise as the wavenumber $k$ increases (see figure 6 b).

Figure 6.

Impacts of uniform compression on the swimmer velocity. The averaged velocity is relative to a non-compressive bending swimmer. We used symmetrical uniform compression with $\omega =2\unicode{x03C0}$ and $\gamma =2$ for different values of $\eta$ and $\phi \in \{0, \pm \unicode{x03C0} /4\}$ . We used wavenumbers (a) $k=2\unicode{x03C0}$ and (b) $k=6\unicode{x03C0}$ .

Remarkably, the plots in figure 6 are only weakly affected by the size of $\epsilon$ , although the small-amplitude theory cannot apply for large $\epsilon$ .

6.1. The $\phi =0$ case

When $\phi =0$ , the compression and bending, $p$ and $q$ , are given by

(6.2) \begin{eqnarray} p(s_0, t)=q(s_0, t)=\sin (ks_0+\omega t) . \end{eqnarray}

An example shape of the swimmer is shown in figures 7(a) and 8(a), and the local extension $p(s_0, t)$ is visualised by the different colours from blue (compressed) to red (extended).

Figure 8.

Top: sample trajectories of a swimmer with a bending–compression wave with (a) $\phi =0$ and $\phi =\unicode{x03C0} /2$ . The parameters are the same as in figure 7. With the initial position $(0,0)$ (marked by a red dot) and the initial angle $\theta =0$ , we drew the orbits of the leftmost end of the filament from $t=0$ to $t=10$ . The configuration at $t=10$ is also shown. Bottom: time sequence of swimmer shape from $t=1$ to $t=10$ . The colour of the swimmer represents the local extension with the same colour as in figure 7.

Using the results derived in the previous section, (4.10), (4.15) and (4.16), we readily obtain the leading-order velocities by direct calculations as

(6.3) \begin{align} \tilde {U}^{(1)}_x &= \frac {\eta \omega }{k}\sin \omega t , \end{align}

(6.4) \begin{align} \tilde {U}^{(1)}_y &= \frac {\epsilon \omega }{k}\sin \omega t-\frac {6\epsilon \omega }{k^2 L_0}\cos \omega t , \end{align}

(6.5) \begin{align} \varOmega ^{(1)} &= \frac {12\epsilon \omega }{k^2 L_0^2}\cos \omega t . \end{align}

The time average of these velocities all vanish at this order, $\langle \tilde {U}^{(1)}_x\rangle =\langle \tilde {U}^{(1)}_y\rangle =\langle \varOmega ^{(1)}\rangle =0$ .

To examine the net locomotion, we proceed to the second-order calculations. The tangential velocity $\tilde {U}^{(2)}_x$ may be obtained by substituting the leading-order expressions into (4.17). After some manipulations, we have

(6.6) \begin{equation} \tilde {U}^{(2)}_x=\frac {\omega }{2k}\left [ \epsilon ^2(\gamma -1) - \eta ^2 + \frac {\epsilon ^2}{2} \cos (2 \omega t)\right ]-\frac {6\omega \epsilon ^2}{k^3L^3_0} (\gamma -2)(1+\cos (2\omega t)). \end{equation}

From the linear equations for $\tilde {U}^{(2)}_y$ and $\varOmega ^{(2)}$ in  (4.19)–(4.20), we may derive the second-order contributions:

(6.7) \begin{align} \tilde {U}^{(2)}_y &= -\frac {\epsilon \eta \omega }{2k}\left [\frac {1+\gamma \cos (2\omega t)}{\gamma }\right ]-\frac {3\epsilon \eta \omega }{2k^2L_0}\left (\frac {\gamma +1}{\gamma }\right )\sin (2\omega t), \end{align}

(6.8) \begin{align} \varOmega ^{(2)} &= \frac {3\epsilon \eta \omega }{k^2L_0^2}\left (\frac {\gamma +1}{\gamma }\right )\sin (2\omega t). \end{align}

To obtain the locomotion after a full deformation cycle from the instantaneous velocities, we consider the extra contribution from the non-commutative Lie algebra (see (2.9)) in addition to the time average of the velocities in the body-fixed frame, which gives

(6.9) \begin{align} \langle U^{(2)}_x\rangle &= \frac {\omega }{2k}\big[(\gamma -1)\epsilon ^2-\eta ^2\big]-(\gamma -1)\frac {6\omega }{k^3L_0^2}\epsilon ^2, \end{align}

(6.10) \begin{align} \langle U^{(2)}_y\rangle &= -\frac {\epsilon \eta \omega }{2\gamma k}+\frac {6\epsilon \eta \omega }{k^3L_0^2}, \end{align}

and the average angular velocity is simply the time average in the body-fixed frame, $\langle \varOmega \rangle =0$ .

The tangential velocity $\langle U^{(2)}_x\rangle$ contains an additional compression term of $O(\eta ^2)$ , which remains for a purely compression/extension wave without bending. Note that the direction of swimming from the compression wave is opposite to the locomotion from the bending wave.

The normal velocity $\langle U^{(2)}_y\rangle$ does not vanish. This notable effect of the drift velocity is generated by the bending–compression coupling, seen as the $O(\epsilon \eta )$ term. The rotational motion, however, vanishes after taking the time average. To examine the dynamics outside the small-amplitude theory, we numerically solve the system and plot an example trajectory in figure 8(a). As seen in the figure, the no-net-rotation property still holds for a large amplitude, and this property is derived by the symmetry arguments as in the uniform compression case, where we consider the $\unicode{x03C0}$ -rotation of the system after the time reversal with a phase shift and the head-to-tail inversion $t\longmapsto t^{\prime}=-t+T/2-kL_0/\omega$ and $s_0\longmapsto s_0^{\prime}=L_0-s_0$ (see also figure 5). In this case, the symmetry arguments follow for any wavenumber $k$ .

6.2. The $\phi =\unicode{x03C0} /2$ case

We then consider the situation with $\phi =\unicode{x03C0} /2$ , where the contraction of one side of the body travels down, and an example shape is shown in figures 7(b) and 8(b), with different colours illustrating the compressed (blue) and extended (red) regions.

We start with the small-amplitude regime, where we follow similar calculations as in the previous sections. By executing the calculations up to second order, and calculating the Lie brackets in (2.9), we may derive the expressions of the velocities in the laboratory frame as

(6.11) \begin{align} \langle U^{(2)}_x\rangle &= \frac {\omega }{2k}\big[(\gamma -1)\epsilon ^2-\eta ^2\big]-(\gamma -1)\frac {6\omega }{k^3L_0^2}\epsilon ^2, \end{align}

(6.12) \begin{align} \langle U^{(2)}_y\rangle &= -\frac {3\epsilon \eta \omega }{k^2L_0}\left (\frac {\gamma +1}{\gamma }\right ), \end{align}

(6.13) \begin{align} \langle \varOmega \rangle &= \frac {6\epsilon \eta \omega }{k^2L_0^2}\left (\frac {\gamma +1}{\gamma }\right ). \end{align}

The velocity $\langle U_x\rangle$ has the same expression as that of the $\phi =0$ case, while the normal and rotational velocities are qualitatively different. In particular, net rotational motion is generated. An example trajectory with its shape gait is shown in figure 7(b).

6.3. Symmetric bending–compression wave

Muscular contraction is typically generated symmetrically to avoid drifting and turning motions. In this subsection, therefore, we consider a symmetric bending–compression wave, given by

(6.14) \begin{eqnarray} p(s_0, t)=\sin (ks_0+2\omega t)\quad\textrm{and}\quad q(s_0, t)=\sin (ks_0+\omega t+\phi ) . \end{eqnarray}

For brevity, we again assume that the filament contains an integer number of waves, i.e. $kL=\pm 2\unicode{x03C0} , \pm 4\unicode{x03C0} , \pm 6\unicode{x03C0} , \ldots\,$ .

We first consider the case with $\phi =0$ . The first-order calculations provide

(6.15) \begin{align} \tilde {U}_x^{(1)} &= \frac {2\eta \omega }{k}\sin (2\omega t), \end{align}

(6.16) \begin{align} \tilde {U}_y^{(1)} &= \frac {\epsilon \omega }{k}\sin (\omega t) -\frac {6\epsilon \omega }{k^2L_0}\cos (\omega t), \end{align}

(6.17) \begin{align} \varOmega ^{(1)} &= \frac {12\epsilon \omega }{k^2L_0^2}\cos (\omega t), \end{align}

which yields no net motion or rotation. Up to second order, we may obtain explicit forms by estimating the integrals in (4.17) as

(6.18) \begin{align} \tilde {U}^{(2)}_x &= -\frac {\eta ^2\omega }{k}+\frac {\epsilon ^2\omega }{4k}\left [ 3-2\gamma -2\cos ^2\omega t \right ]-\frac {12\epsilon ^2\omega }{k^3L_0^2}(\gamma -2)\cos ^2\omega t, \end{align}

(6.19) \begin{align} \tilde {U}^{(2)}_y &= \frac {\epsilon \eta \omega }{4 k}\left [ 2\left (\frac {\gamma -2}{\gamma }\right )\cos \omega t-3\cos (3\omega t)\right ]\nonumber \\ &\quad +\, \frac {\epsilon \eta \omega }{4 k^2 L_0}\left [ 12\left (\frac {\gamma +1}{\gamma }\right )\sin \omega t-3\left (\frac {\gamma +4}{\gamma }\right )\sin (3\omega t)\right ], \end{align}

(6.20) \begin{align} \varOmega ^{(2)} &= \frac {\epsilon \eta \omega }{2}\sin \omega t+\frac {\epsilon \eta \omega }{2 k^2 L_0^2}\left [6\left (\frac {\gamma +3}{\gamma }\right )\cos \omega t -3\left (\frac {3\gamma +4}{\gamma }\right )\right ]\sin \omega t . \end{align}

By taking the time average, we derive

(6.21) \begin{align} \langle \tilde {U}^{(2)}_x\rangle =\frac {\omega }{2k}\left [ (\gamma -1)\epsilon ^2-2\eta ^2\right ]-\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -2), \end{align}

and zero normal and angular velocities

(6.22) \begin{equation} \langle \tilde {U}^{(2)}_y\rangle =0,\quad \langle \varOmega ^{(2)}\rangle =0, \end{equation}

as expected from the symmetry.

After the non-commutative algebra, the averaged velocity for the tangential velocity in the reference frame is calculated as

(6.23) \begin{align} \langle U_x^{(2)}\rangle =\frac {\omega }{2k}\left [ (\gamma -1)\epsilon ^2-2\eta ^2\right ]-\frac {6\epsilon ^2\omega }{k^3L_0^2}(\gamma -1). \end{align}

We then proceed to the third-order contributions for the progressive velocity, using (4.21) and the second-order expressions (6.18)–(6.20). Direct calculations lead to the time-averaged velocity

(6.24) \begin{equation} \langle \tilde {U}^{(3)}_x\rangle =\frac {3\epsilon ^2\eta \omega }{2k^2L^2_0}, \end{equation}

which is, however, found to be cancelled out in the laboratory frame after calculating the Lie brackets such that

(6.25) \begin{equation} \langle U^{(3)}_x\rangle = 0. \end{equation}

The zero velocity contributions at third order are similar to those of the classical Taylor sheet (Taylor Reference Taylor1951), but are different from the uniform compression case.

Figure 9.

Numerical simulation of averaged swimming velocity $\langle U_x \rangle$ (in the reference frame) of the swimmer driven by a symmetrical compression wave with $L_0=1$ , $kL=2\unicode{x03C0}$ , $\phi =\unicode{x03C0} /2$ and $\omega =2\unicode{x03C0}$ for different values of $\epsilon$ and $\eta$ . The dashed line is a null curve of the velocity predicted by the small-amplitude theory up to third order, which is in excellent agreement when $\epsilon \lesssim 0.4$ .

The case $\phi =\unicode{x03C0} /2$ can also be calculated in a similar manner. By executing the second-order calculations, we found that the expression of the averaged progressive velocity $\langle U_x\rangle$ is identical to that obtained in the $\phi =0$ case (6.23), together with zero normal and rotational velocities due to the symmetry. At third order, the progressive velocity has a form similar to (6.24), i.e.

(6.26) \begin{equation} \langle \tilde {U}^{(3)}_x\rangle =-\frac {3\epsilon ^2\eta \omega }{2k^2L^2_0}, \end{equation}

and after including the non-commutative effects from the Lie bracket, this contribution vanishes again: $\langle U^{(3)}_x\rangle = 0$ .

Hence both when $\phi =0$ and $\phi =\unicode{x03C0} /2$ , according to the analysis above (see  (6.23)), net locomotion disappears when

(6.27) \begin{equation} \left |\frac {\eta }{\epsilon }\right |= \sqrt {\left (1-\frac {12}{k^2L_0^2}\right )\frac {\gamma -1}{2}} . \end{equation}

To test these theoretical results, we numerically compute the averaged swimming velocity for a finite amplitude with different $\epsilon$ and $\eta$ , with the same numerical scheme as in § 5. The results are shown in figure 9, together with the theoretical prediction of the null curve by the small-amplitude theory (6.27). As seen in the figure, the theoretical prediction is in excellent agreement when $\epsilon \lesssim 0.4$ .