= Solution
Use the <Cartesian streamfunction> convention $u_x=\partial_y\Psi$, $u_y=-\partial_x\Psi$. In the creeping-flow regime implicit in the question's <biharmonic equation> hint, the <Stokes equations> are $\mu\nabla^2\mathbf u=\nabla p$, $\nabla\cdot\mathbf u=0$. Taking their curl eliminates pressure and gives
$$
\boxed{\nabla^4\Psi=0.}
$$
The fluid occupies the region above the actual sheet; the perturbation calculation expands its boundary about $y=0$. In the swimming frame the material velocity is $(0,-\omega y_0\cos(kx-\omega t))$. The vertical kinematic condition and the supplied <Navier slip boundary condition> therefore give, at $y=y_0\sin(kx-\omega t)$,
$$
\Psi_x=\omega y_0\cos(kx-\omega t),\qquad
\Psi_y=\gamma(\Psi_{yy}-\Psi_{xx}).
$$
At infinity, $\Psi_y\to U$ and $\Psi_x\to0$, with bounded velocity and no imposed shear; all fields are periodic in $x$ with period $2\pi/k$. The sign is important: a sheet translating at $-U\mathbf e_x$ sees the remote fluid translating at $+U\mathbf e_x$.
Set $X=kx$, $Y=ky$, $\tau=\omega t$, $\psi=k^2\Psi/\omega$, and $U^*=kU/\omega$. The two dimensionless parameters are the small-slope approximation parameter $\epsilon=ky_0$ and dimensionless <slip length> $\delta=k\gamma$. With $\theta=X-\tau$, the dimensionless <biharmonic stream function for planar Stokes flow> satisfies
$$
\begin{aligned}
(\partial_X^2+\partial_Y^2)^2\psi&=0,\\
\psi_X&=\epsilon\cos\theta,\qquad
\psi_Y=\delta(\psi_{YY}-\psi_{XX})
&&\text{at }Y=\epsilon\sin\theta,\\
\psi_Y&\longrightarrow U^*,\qquad\psi_X\longrightarrow0
&&\text{as }Y\longrightarrow\infty.
\end{aligned}
$$
Expand $\psi=\epsilon\psi_1+\epsilon^2\psi_2+\cdots$ and $U^*=\epsilon U_1+\epsilon^2U_2+\cdots$. At first order,
$$
(\psi_1)_X(X,0,\tau)=\cos\theta,\qquad
(\psi_1)_Y(X,0,\tau)
=\delta[(\psi_1)_{YY}-(\psi_1)_{XX}](X,0,\tau).
$$
The decaying first harmonic of the <biharmonic equation> has the form $(A+BY)e^{-Y}\sin\theta$. The vertical condition fixes $A=1$ and excludes a cosine component. The horizontal condition gives
$$
B-1=\delta[(1-2B)-(-1)]
=-2\delta(B-1).
$$
Thus $(1+2\delta)(B-1)=0$ and $B=1$ for every $\delta\geq0$. The spatially averaged first-order <streamfunction> is affine in $Y$; its <Navier slip boundary condition> forces $U_1=0$. Therefore
$$
\boxed{\psi_1=(1+Y)e^{-Y}\sin(X-\tau),\qquad U_1=0.}
$$
This <first-order slip independence of a transverse sheet> is independent of <slip length>. Its surface tangential velocity and surface shear are both zero: $(\psi_1)_Y(0)=0$ and $(\psi_1)_{YY}(0)-(\psi_1)_{XX}(0)=0$. This explains why the first-order <streamfunction> agrees with the <no-slip boundary condition>.
To find the <slip-enhanced swimming speed of a transverse sheet>, expand only the <Navier slip boundary condition> through second order. Put $Q_j=(\psi_j)_{YY}-(\psi_j)_{XX}$. <Taylor expansion> at the displaced boundary yields
$$
(\psi_2)_Y-\delta Q_2
=-\sin\theta\,(\psi_1)_{YY}
+\delta\sin\theta\,\partial_YQ_1
\quad\text{at }Y=0.
$$
Direct differentiation gives
$$
(\psi_1)_{YY}(0)=-\sin\theta,\qquad
Q_1=2Ye^{-Y}\sin\theta,\qquad
\partial_YQ_1(0)=2\sin\theta.
$$
Hence
$$
(\psi_2)_Y(X,0,\tau)-\delta Q_2(X,0,\tau)
=(1+2\delta)\sin^2\theta.
$$
Average over $X$. The zero <Fourier series> mode of a <biharmonic streamfunction> is a cubic polynomial in $Y$; bounded velocity at infinity removes the quadratic and cubic terms. Thus its velocity is the constant $U_2$, and its averaged shear $\langle Q_2\rangle_X$ is zero. The <mean boundary velocity determines Taylor-sheet swimming speed> principle then gives
$$
\boxed{U_2=\frac{1+2\delta}{2},\qquad
U^*=\frac{\epsilon^2}{2}(1+2\delta)+O(\epsilon^4).}
$$
The even remainder follows because changing the sign of the amplitude is equivalent to a half-period phase shift. In dimensional form,
$$
\boxed{\mathbf V_{\rm sheet}
=-\frac{\omega k y_0^2}{2}(1+2k\gamma)\,\mathbf e_x
+O\!\left(\frac{\omega}{k}\epsilon^4\right).}
$$
For the <Navier-slip Taylor swimming sheet>, the ratio of speed to the no-slip value is $1+2\delta>1$ whenever $\delta>0$: \b[slip increases the swimming speed]. The expansion is for fixed $\delta$ as $\epsilon\to0$; it is not a uniform prediction of arbitrarily large speed if the <slip length> is allowed to diverge with the inverse amplitude.
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