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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 givesThe 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 $y=y_0\sin (kx-\omega t)$,At infinity, ${\Psi }_y\to U$ and ${\Psi }_x\to 0$, 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 $U^{\ast }=kU/\omega$. The two dimensionless parameters are the small-slope approximation parameter $ϵ=k{y}_0$ and dimensionless slip length $\delta =k\gamma$. With $\theta =X-\tau$, the dimensionless biharmonic stream function for planar Stokes flow satisfiesExpand $\psi =ϵ{\psi }_1+{ϵ}^2{\psi }_2+\cdots$ and $U^{\ast }=ϵU_1+{ϵ}^2{U}_2+\cdots$. At first order,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 givesThus $(1+2\delta )(B-1)=0$ and $B=1$ for every $\delta \ge 0$. The spatially averaged first-order streamfunction is affine in $Y$; its Navier slip boundary condition forces $U_1=0$. ThereforeThis 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 yieldsDirect differentiation givesHenceAverage 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 $⟨Q_2{⟩}_X$ is zero. The mean boundary velocity determines Taylor-sheet swimming speed principle then givesThe even remainder follows because changing the sign of the amplitude is equivalent to a half-period phase shift. In dimensional form,For the Navier-slip Taylor swimming sheet, the ratio of speed to the no-slip value is $1+2\delta >1$ whenever $\delta >0$: slip increases the swimming speed. The expansion is for fixed $\delta$ as $ϵ\to 0$; it is not a uniform prediction of arbitrarily large speed if the slip length is allowed to diverge with the inverse amplitude.

Use $u={\partial }_y\psi$ and $v=-{\partial }_x\psi$ in the frame translating with the sheet. At each instant, the incompressible Stokes equations areTaking their curl gives the biharmonic stream function for planar Stokes flow: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 $q=x-t$,The second sign follows from $v={\partial }_t{y}_s=-ϵb\cos q$. Far above the sheet,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 $U=ϵU_1+{ϵ}^2{U}_2+\cdots$ and $\psi =ϵ{\psi }_1+{ϵ}^2{\psi }_2+\cdots$. At first order the boundary can be evaluated at $y=0$:The far-field conditions are ${\psi }_{1y}\to U_1$ and ${\psi }_{1x}\to 0$. 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 $k=3$ givesAn 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 $U_1=0$.

For the second-order Taylor-expanded no-slip boundary condition, the first-order displacement isThere is no explicit second-order material velocity in the prescribed motion. Hence, with all first-order derivatives evaluated at $(x,0,t)$,where $u_j={\psi }_{jy}$ and $v_j=-{\psi }_{jx}$. Differentiation of the first-order solution giveson the flat reference boundary. ThereforeThe 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 ${\psi }_{2x}\to 0$.

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 formThe 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 yieldsThis fixes the $E_k$; the nonconstant part of $u_2$ fixes $G_k-k{E}_k$. 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 $C_0+U_{2}y$, so its horizontal velocity is the same at the boundary and infinity. Fourier orthogonality of sheet swimming modes removes the mixed product, givingHere $⟨\cdot {⟩}_x$ 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 $b/a=3$ cancels the swimming speed at order ${ϵ}^2$. This different-wavenumber cancellation in sheet swimming is not a claim that all higher-order swimming terms vanish.

At second order the Taylor-expanded no-slip boundary condition iswith ${\nabla }^4{\psi }_2=0$, ${\psi }_{2y}\to U_2$, and ${\psi }_{2x}\to 0$ at infinity. Set the irrelevant boundary constant to zero. The general relevant solution contains a zero Fourier mode and a second harmonic:Bounded velocity eliminates $C,D$ and growing harmonics. Averaging the tangential boundary condition gives $B=⟨{\sin }^2\theta ⟩=1/2$. This is the mean boundary velocity determines Taylor-sheet swimming speed principle: the remaining mean velocity is uniform and equals the far-field velocity. ThereforeThe 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 ${\psi }_2=y/2-(y/2)e^{-2y}\cos 2\theta$.

For a transverse Taylor swimming sheet below a flat no-slip boundary condition at height $d$, the first-order amplitude $f(y)$ satisfies $({\partial }_y^2-1{)}^{2}f=0$, $f(0)=1$, $f^{\prime }(0)=f(d)=f^{\prime }(d)=0$. Writing $\Delta ={\sinh }^{2}d-d^2$ givesThe mean boundary velocity determines Taylor-sheet swimming speed, yieldingThis is greater than the unbounded value $1/2$ for every $d>0$. In a narrow gap it scales as $3/d^2$, with the small-amplitude calculation requiring $ϵ\ll d$ as well as $ϵ\ll 1$. Taylor's swimming sheet near a soft boundary recovers this rigid-wall limit while studying how compliance changes propulsion.