Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 77 / 2

The Laplacian and the added drag term in the Brinkman equation must have the same dimensions. ThereforeThe length ${\alpha }^{-1}$ is a hydrodynamic screening length. When the viscous coefficients are identical in the two terms, the corresponding permeability is ${\alpha }^{-2}$.

Scale time by ${\omega }^{-1}$, length by $k^{-1}$, velocity by $\omega /k$, pressure by $\mu \omega$, and streamfunction by $\omega /k^2$. Write $A=\alpha /k$, $\xi =x-t$, and drop stars from dimensionless coordinates. The surface is $y=ϵ\sin \xi$ and its material velocity in the sheet frame is $(0,-ϵ\cos \xi )$. The no-slip boundary condition is thereforeThe stationary matrix identifies a preferred frame. In the sheet frame it translates with $\mathcal{U}{\mathbf{e}}_x$, so matrix-relative Brinkman velocity gives the consistent equationsEquivalently one may keep the printed right-hand side $A^2\mathbf{u}$ after defining $\tilde{p}=p-A^2\mathcal{U}x$; then the far-field pressure has the gradient needed to balance that term. Using the printed equation with a uniform far-field flow and no such adjustment is inconsistent. The distinction first matters at second order, since the first-order swimming speed is zero. We solve the upper half-space; the lower half-space is its reflected counterpart.

Taking the curl of the Brinkman equation eliminates pressure and the constant matrix velocity. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$ and vorticity $-{\nabla }^2\psi$, the Brinkman swimming sheet equation isThe surface conditions are conditions on partial derivatives evaluated on the moving boundary:At infinity, ${\psi }_y\to \mathcal{U}$ and ${\psi }_x\to 0$. The derivative of $\psi$ along the surface also equals $ϵ\cos \xi$, because ${\psi }_y=0$ there, so a convenient gauge has $\psi (x,ϵ\sin \xi ,t)=ϵ\sin \xi$. Decaying perturbations and the force-free condition determine the swimming solution.

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.

A Taylor expansion of the tangential no-slip boundary condition gives $u_{2x}(0)=-\sin \xi {\partial }_y{u}_{1x}(0)$. Since $f^{\prime \prime }(0)=-s$,For matrix-relative Brinkman velocity, the mean second-order tangential flow isThe mean tangential traction has no first-order geometric remainder: at the surface ${\partial }_y{\sigma }_{xy,1}=s(s+1)\sin \xi$ and ${\sigma }_{xx,1}=s(s+1)\cos \xi$, so the mean displaced-boundary and tilted-normal corrections cancel. Thus the force-free condition gives ${\overline{u}}_{2x}^{\prime }(0)=0$, hence $C=0$. In the $A=0$ limit boundedness likewise excludes mean shear. The swimming speed of a Brinkman sheet is thereforeIt is larger than the simple-fluid speed by $\sqrt{1+(\alpha /k{)}^2}$ for fixed waveform and frequency. The matrix offers a reaction structure for the transverse stroke and more strongly confines the induced motion; it acts as a footing against which the wave pushes. This is a prescribed-stroke comparison, not a guarantee at fixed motor power. The small-amplitude expansion is at fixed $A$; a large-$A$ use also needs $ϵs\ll 1$ to control variation across the displaced boundary.

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.