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The biharmonic equation is ${\nabla }^{4}f=0$, where the biharmonic operator is the square of the Laplacian. In planar Stokes flow, taking the curl of the momentum equation and using a Cartesian streamfunction produces this equation. Its decaying spatial Fourier series modes have form $(A+By)e^{-|k|y}{e}^{ikx}$.

For the simple shear flow, the vertical vorticity isFor $\beta \ne 0$, this is not an irrotational flow. A complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, analytic in the flow region, has complex velocity $w^{\prime }=v_x-i{v}_y$. The Cauchy-Riemann equations imply both incompressibility and zero vorticity, so this shear has no such complex potential. It does have a Cartesian streamfunction, $\psi =\beta y^2/2$. In the degenerate case $\beta =0$, the resting fluid has a constant complex potential.

For the linear planar flow with strain and rotation, ${\partial }_{x}u+{\partial }_{y}v=ϵ-ϵ=0$, proving incompressibility. The Cartesian streamfunction convention gives $u={\psi }_y$, $v=-{\psi }_x$, and direct integration givesup to an irrelevant additive constant. Along a streamline, $\mathbf{u}\cdot \nabla \psi ={\psi }_y{\psi }_x-{\psi }_x{\psi }_y=0$. Sincethe stated quadratic expression is constant along each streamline. The three cases below distinguish strain, rotation and their degenerate shear limit.

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.

At the rigid surface, the no-slip boundary condition and no penetration giveAt the horizontal free surface, the kinematic boundary condition and the stress boundary condition areHere surface tension is absent and the inviscid air exerts no tangential traction.

The incompressible Stokes equation isUse the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$ and setThenand the wall and zero-shear conditions becomeThe two momentum equations readEquality of mixed derivatives gives $f^{(4)}=0$, so the boundary conditions implyIntegrating for the pressure and imposing the normal-stress condition at $y=h$ determinesThe complete instantaneous velocity and pressure fields are thereforeIndeed, with the viscous stress tensor $\sigma =-pI+\mu (\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)$, these expressions satisfy both traction conditions. Evaluating the vertical velocity at the material surface gives the pressure-driven thinning of a uniform Stokes layer:Thus

Use the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the incompressible Stokes equation eliminates pressure and gives the biharmonic stream function for planar Stokes flow equationThe material points have no horizontal velocity in the swimming frame and have vertical velocity $-ϵ\cos \theta$. The no-slip boundary condition is thereforeAt infinity the laboratory fluid is at rest, so in this translating frame it moves with $U{\mathbf{e}}_x$:Velocity and its perturbations are periodic in $x$, with period $2\pi$, and the pressure has no imposed mean gradient. The Taylor swimming sheet is force-free; the unbounded problem's bounded far-field velocity enforces the absence of mean shear. An additive constant in the streamfunction has no physical effect.

LetUsing the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$ givesAt fixed $y$,The terms proportional to $\eta f^{\prime }{f}^{\prime \prime }$ cancel, leavingSince $U=-Q/x$, $U{U}^{\prime }=-Q^2/x^3$, while $\nu u_{yy}=Q^2{f}^{\prime \prime \prime }/x^3$. The boundary-layer equation therefore becomesChoosing the wall value of the streamfunction as zero, imposing no slip, and matching to $U=-Q/x$ give

Let a viscous layer occupy $0<y<h(t)$, with no slip at $y=0$, zero tangential traction at $y=h$, and exterior pressure $p_0-{\rho }_a{E}^2{x}^2/2$. The Cartesian streamfunction ansatz $\psi =xf(y)$ reduces the Stokes equations to $f^{(4)}=0$. The resulting surface velocity isso the positive thinning rate is $-\dot{h}={\rho }_a{E}^2{h}^3/(3\mu )$.