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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 $-\dot{h}={\rho }_a{E}^2{h}^3/(3\mu )$.