Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 334 / 1 / f / Solution

At the sheet, the second-order conditions are ${\psi }_{2x}(0)=0$ and ${\psi }_{2y}(0)=-f_d^{\prime \prime }(0){\sin }^2\theta$. At the wall, ${\psi }_{2x}(d)=0$ and ${\psi }_{2y}(d)={\overline{U}}_2$. The general solution has a polynomial mean mode and a second harmonic with exponentials $e^{\pm 2y}$, as in the unbounded calculation.

Let ${\overline{u}}_2(y)=⟨{\psi }_{2y}⟩$. The averaged Stokes equation, with periodic pressure and no imposed mean pressure gradient, gives ${\overline{u}}_2^{\prime \prime }=0$. The total horizontal hydrodynamic force on the sheet is opposite to the mean shear force on the flat wall. Since the swimmer is force-free, ${\overline{u}}_2^{\prime }(d)=0$, making the mean velocity constant. Matching its values at the sheet and wall therefore givesThe previous part gives $f_d^{\prime \prime }(0)=1-2{\sinh }^{2}d/\Delta =-({\sinh }^{2}d+d^2)/\Delta$, henceThis determines the speed without solving the second-harmonic flow. The force constraint matters: prescribing a translating sheet without imposing zero net force would allow a mean shear and would not determine a unique swimming speed.