First-order free swimming of a planar filament ($V_1=-4G_0+6G_1/L,{\Omega }_1=6G_0/L-12G_1/L^2$)

For $y=ϵg(x,t)$ on $0\le x\le L$, uniform local drag and force/torque balance give $U_1=0$, $V_1=-4⟨g_t{⟩}_x+6⟨x{g}_t{⟩}_x/L$, and ${\Omega }_1=6⟨g_t{⟩}_x/L-12⟨x{g}_t{⟩}_x/L^2$. The tangent can be replaced by the undeformed direction at this order because the velocity is already first order.

The first-order lateral translation and rotation of a prescribed planar filament are time derivatives of fixed linear spatial averages of its shape. A periodic shape therefore gives zero temporal mean of both quantities. Net laboratory displacement can still arise at second order.

At first order, the transverse drag depends on $g_t+V_1+{\Omega }_{1}x$. Its zero force and torque conditions make this residual orthogonal to $1$ and $x$. Thus $-V_1-{\Omega }_{1}x$ is the linear least-squares projection of the prescribed transverse deformation velocity onto the span of $1$ and $x$. This recasts a force-balance calculation as a projection.