= Solution
Use $u=\partial_y\psi$ and $v=-\partial_x\psi$ in the frame translating with the sheet. At each instant, the incompressible <Stokes equations> are
$$
-\nabla p+\mu\nabla^2\mathbf u=0,\qquad\nabla\cdot\mathbf u=0.
$$
Taking their curl gives the <biharmonic stream function for planar Stokes flow>:
$$
\boxed{\nabla^4\psi=0}
$$
in the fluid above the actual sheet. The <no-slip boundary condition> is imposed on material points, whose label is $x$ and whose actual horizontal position is $x_s$. With $q=x-t$,
$$
\psi_y(x_s,y_s,t)=-3\epsilon a\cos3q,\qquad
\psi_x(x_s,y_s,t)=\epsilon b\cos q.
$$
The second sign follows from $v=\partial_t y_s=-\epsilon b\cos q$. Far above the sheet,
$$
\psi_y\to U,\qquad\psi_x\to0,
$$
with bounded velocity and no imposed mean pressure gradient. The positive far-field velocity arises because the sheet moves with laboratory velocity $-U\mathbf e_x$. All fields are $2\pi$-periodic in $x$; an additive function of time in $\psi$ is a harmless gauge. These conditions formulate the <Taylor swimming sheet> problem with both types of travelling deformation.
Expand $U=\epsilon U_1+\epsilon^2U_2+\cdots$ and $\psi=\epsilon\psi_1+\epsilon^2\psi_2+\cdots$. At first order the boundary can be evaluated at $y=0$:
$$
\nabla^4\psi_1=0,\qquad
\psi_{1y}(x,0,t)=-3a\cos3q,\qquad
\psi_{1x}(x,0,t)=b\cos q.
$$
The far-field conditions are $\psi_{1y}\to U_1$ and $\psi_{1x}\to0$. For a nonzero <Fourier mode> of wavenumber $k$, bounded velocity selects $(E+Gy)e^{-ky}e^{ikq}$; the exponentially growing modes are excluded. Matching the modes $k=1$ and $k=3$ gives
$$
\boxed{\psi_1=b(1+y)e^{-y}\sin q-3ay e^{-3y}\cos3q}.
$$
An additive gauge has been set to zero. The first term is a <transverse mode of a Taylor swimming sheet>, and the second is a <longitudinal mode of a Taylor swimming sheet>. Their velocities decay at infinity. The mean first-order boundary velocity is zero, and the bounded zero <Fourier mode> is a constant plus a multiple of $y$, so \b[$U_1=0$].
For the second-order <Taylor-expanded no-slip boundary condition>, the first-order displacement is
$$
\boldsymbol\xi=(a\sin3q,b\sin q).
$$
There is no explicit second-order material velocity in the prescribed motion. Hence, with all first-order derivatives evaluated at $(x,0,t)$,
$$
\nabla^4\psi_2=0,\qquad
u_2=-\boldsymbol\xi\cdot\nabla u_1,\qquad
v_2=-\boldsymbol\xi\cdot\nabla v_1,
$$
where $u_j=\psi_{jy}$ and $v_j=-\psi_{jx}$. Differentiation of the first-order solution gives
$$
u_{1x}=9a\sin3q,\quad u_{1y}=18a\cos3q-b\sin q,\quad
v_{1x}=b\sin q,\quad v_{1y}=-9a\sin3q
$$
on the flat reference boundary. Therefore
$$
\boxed{u_2(x,0,t)=-9a^2\sin^23q-18ab\sin q\cos3q+b^2\sin^2q,\qquad
v_2(x,0,t)=8ab\sin q\sin3q}.
$$
The boundary conditions for the <streamfunction> are $\psi_{2y}=u_2$ and $\psi_{2x}=-v_2$. Their far-field counterparts are $\psi_{2y}\to U_2$ and $\psi_{2x}\to0$.
Products of the two distinct first-order <Fourier modes> generate only the nonzero wavenumbers $2,4,6$, as well as a mean. Thus the bounded general solution compatible with these data has the form
$$
\boxed{\psi_2=C_0(t)+U_2y+\operatorname{Re}\left\{\sum_{k\in\{2,4,6\}}(E_k+G_ky)e^{-ky}e^{ikq}\right\}}.
$$
The discarded zero-mode terms $y^2,y^3$ would give unbounded velocity, and the discarded nonzero modes grow exponentially. The <zero-mean normal velocity in periodic half-space Stokes flow> is consistent with a bounded periodic flow and no far-field vertical flux. For example, integrating $\psi_{2x}=-v_2$ along the boundary yields
$$
\psi_2(x,0,t)=-2ab\sin2q+ab\sin4q+C_0(t).
$$
This fixes the $E_k$; the nonconstant part of $u_2$ fixes $G_k-kE_k$. All coefficients are thereby determined, but none is needed to find the mean swimming speed.
Indeed, <mean boundary velocity determines Taylor-sheet swimming speed>. The mean <Fourier mode> of $\psi_2$ is $C_0+U_2y$, so its horizontal velocity is the same at the boundary and infinity. <Fourier orthogonality of sheet swimming modes> removes the mixed product, giving
$$
\boxed{U_2=\langle u_2(x,0,t)\rangle_x=\frac{b^2-9a^2}{2},\qquad
U=\frac{\epsilon^2}{2}(b^2-9a^2)+O(\epsilon^3)}.
$$
Here $\langle\cdot\rangle_x$ is the average over one spatial period; a temporal average gives the same answer for these travelling waves. The laboratory swimming velocity is $-U\mathbf e_x$. The transverse wave drives motion opposite to its direction of propagation, whereas the longitudinal wave drives it in the propagation direction. Their wavenumbers enter quadratically. Thus \b[$b/a=3$ cancels the swimming speed at order $\epsilon^2$]. This <different-wavenumber cancellation in sheet swimming> is not a claim that all higher-order swimming terms vanish.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-334-sheet-cancellation.png]
{title=The leading sheet swimming coefficient scaled by a squared changes sign at b over a equal to three, where longitudinal and transverse contributions cancel}
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