At second order the <Taylor-expanded no-slip boundary condition> is
$$
\psi_{2y}(0)=-\sin\theta\,\psi_{1yy}(0)=\sin^2\theta=\frac12(1-\cos2\theta),\qquad \psi_{2x}(0)=-\sin\theta\,\psi_{1xy}(0)=0,
$$
with $\nabla^4\psi_2=0$, $\psi_{2y}\to U_2$, and $\psi_{2x}\to0$ at infinity. Set the irrelevant boundary constant to zero. The general relevant solution contains a zero <Fourier mode> and a second harmonic:
$$
\psi_2=A+By+Cy^2+Dy^3+\operatorname{Re}\{[E e^{-2y}+F e^{2y}+y(G e^{-2y}+J e^{2y})]e^{2i\theta}\}.
$$
Bounded velocity eliminates $C,D$ and growing harmonics. Averaging the tangential boundary condition gives $B=\langle\sin^2\theta\rangle=1/2$. This is the <mean boundary velocity determines Taylor-sheet swimming speed> principle: the remaining mean velocity is uniform and equals the far-field velocity. Therefore
$$
\boxed{U_2=\frac12,\qquad \mathbf V_{\rm swim}=-\frac{\epsilon^2}{2}\mathbf e_x+O(\epsilon^4).}
$$
The absence of odd powers follows because changing the amplitude sign is just a half-wavelength translation. Although unnecessary for the speed, the oscillatory solution can also be written explicitly as $\psi_2=y/2-(y/2)e^{-2y}\cos2\theta$.

Solution

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

{c}

For a displaced wall $y=\epsilon h_1(x,t)$, write fluid velocity as $\mathbf u=\epsilon\mathbf u_1+\epsilon^2\mathbf u_2+\cdots$ and prescribed wall velocity as $\epsilon\mathbf v_1+\epsilon^2\mathbf v_2+\cdots$. A <Taylor expansion> about $y=0$ gives $\mathbf u_1(0)=\mathbf v_1$ and $\mathbf u_2(0)=\mathbf v_2-h_1\partial_y\mathbf u_1(0)$. The second-order correction converts first-order velocity gradients into a mean boundary flow.

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