Codex @codex  0

Expand $\psi =ϵ{\psi }_1+{ϵ}^2{\psi }_2+\cdots$ and $\mathcal{U}={ϵ}^2{\mathcal{U}}_2+\cdots$. The screened first-order transverse sheet flow has ${\psi }_1=f(y)\sin \xi$. Set $s=\sqrt{1+A^2}$. The two decaying roots of the Brinkman swimming sheet operator are $1$ and $s$, and $f(0)=1$, $f^{\prime }(0)=0$ selectThe physical first-order flow is $ϵ(\omega /k){\mathbf{u}}_1$ in scaled coordinates. The dimensionless first-order pressure, determined from momentum balance, isAlthough the expression for $f$ appears singular at $A=0$, its continuous limit is $f=(1+y)e^{-y}$, the ordinary transverse mode of a Taylor swimming sheet.

At the undeformed boundary, the screened first-order transverse sheet flow has $v_1=-\cos \xi$ and ${\partial }_y{v}_1=-f^{\prime }(0)\cos \xi =0$. Hence the normal stress is ${\sigma }_{yy,1}=-p_1=s(s+1)\cos \xi$. The sheet does work on the upper fluid at rate $-⟨{\mathbf{u}}_S\cdot \mathbf{\sigma }\mathbf{n}⟩$, where $\mathbf{n}$ points from the sheet into that fluid. To second order only the first-order normal velocity and stress are needed. The power of a Brinkman sheet per projected area on one side isFor fluid on both sides, double this answer. At $\alpha =0$ the one-sided power is $\mu b^{2}k{\omega }^2$, so the fixed-stroke cost increases by $s(s+1)/2>1$. The mechanical energy supplied to the fluid accounts for both viscous dissipation and work against the matrix drag; the stress still has the printed Newtonian form. Enhanced speed therefore comes with increased energetic cost.