Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 355 / 3 / a

Let the unit normal $\mathbf{n}$ point from the fluid toward the free surface, let $\mathbf{t}$ lie along the surface, and write the swimmer direction asReflecting the swimmer position across the plane and reflecting its orientation toconstructs the free-surface image of a force dipole. At the surface the two dipoles have equal tangential velocity and opposite normal velocity. Their sum therefore satisfies the no-penetration boundary condition; tangential velocity is even across the plane and normal velocity is odd, so the tangential traction vanishes and the stress-free boundary condition is also satisfied.

The vector from the image to the swimmer is $\mathbf{r}=-2h\mathbf{n}$, for which $({\mathbf{p}}^{\ast }\cdot \mathbf{r}{)}^2=4h^2{\sin }^2\theta$. Evaluating the image force-dipole flow at the swimmer givesThere is no tangential image velocity at the swimmer in this point-dipole approximation. With the convention that $\mathcal{P}>0$ is an extensile pusher microswimmer, a nearly parallel pusher is attracted toward the surface, whereas a nearly parallel contractile puller microswimmer is repelled. For $|\sin \theta |>1/\sqrt{3}$ the normal drift reverses because the image samples the axial rather than equatorial part of the dipolar flow.