Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 75 / 4

Work in the body frame and approximate the narrow annulus locally by a planar gap. Let $y=0$ be the cylindrical wall, moving at $-U$ in this frame, and $y=h(x)$ the stationary cell surface. The red blood cell is represented by the given compliant body; the surrounding fluid is modeled as incompressible and Newtonian. Assume $h\ll a$, small local gap slope, and negligible inertia. The lubrication equations are $p_y=0$ and $\mu w_{yy}=p_x$, with no-slip conditions $w(0)=-U$, $w(h)=0$. Integration givesMass conservation makes the body-frame flux constant. With the sign convention of the problem,so the lubrication model of a compliant translating plug givesThe circumferential multiplier is $2\pi a$ to leading order, giving the relative volume flux $-2\pi aQ$.

Let $p_d$ be the pressure in front of the cell. Match the film to the reservoirs by $p(L)=p_d$ and $p(-L)=p_d+\Delta p$. Use $h=a-r$ and symmetric end matching for the parabolic-gap reduction. The elastic law givesFor $r_0(-L)=r_0(L)$ this gives $h(-L)-h(L)=\Delta p/\alpha$. If the outer unstressed shape is not symmetric at the matching ends, the general relation instead includes $h(-L)-h(L)=\Delta p/\alpha -[r_0(-L)-r_0(L)]$. The proposed reduced end equation assumes the symmetric matching used here. The gap must remain positive. To determine the unknown flux and translating speed one also needs zero net axial force on the freely moving body, and a specified absolute downstream pressure relative to $p_0$, not merely a pressure difference. If an external force pulls the body, it must replace the force-free condition. The stated elasticity law fixes $p_0$; treating it as an unknown internal pressure would require a further cell-volume constraint.

The force-free condition is most easily obtained from a control volume containing both body and surrounding film. The pressure force is $\pi a^2\Delta p$, and the resisting wall traction is $2\pi a\mu {\int }_{-L}^L{w}_y(0)dx$. The local velocity solution gives $w_y(0)=4U/h-6Q/h^2$, soThis is the force-free pressure drop of an annular lubrication plug. It includes the film pressure and shear contributions without separately approximating the deformed body's surface normal.

Put $s=2Q/U>0$, $\ell =\sqrt{s/\kappa }$, $h=sH$ and $x=\ell X$. The local parabolic elastic law is $p=p_0+\alpha [r_{00}-a-\kappa x^2/2+h]$, whence $p_x=\alpha \sqrt{\kappa s}(H_X-X)$. Substitution into the local lubrication equation givesThe derivative is with respect to the scaled coordinate $X$; a derivative with respect to dimensional $x$ would be inconsistent with this nondimensional equation.

Let $C=s/a$ and $\tilde{L}=L/\ell$. The elastic end condition gives $\Delta p=\alpha s[H(-\tilde{L})-H(\tilde{L})]$. Scaling the force integral and eliminating $\Delta p$ then givesThis completes the requested formulation and both dimensionless equations. The local parabolic approximation is used in the region contributing most strongly to the lubrication integrals; matching to the outer shape and reservoirs supplies the stated end conditions.

The unstressed shape has equal radii at $x=\pm L$, so its contributions cancel when subtracting the elastic end pressures. The pressure difference is thereforeThe multiplier is a pressure scale: $\alpha$ has dimensions pressure per length and $2Q/U$ is a length.

In the narrowest part of the film, $H=O(1)$, so $h\sim s=2Q/U$. Thereforeis exactly the thin-annulus requirement needed to replace the local circumference by $2\pi a$ and neglect curvature across the film. The axial lubrication scale is $\ell =\sqrt{s/\kappa }$, so the small-slope assumption additionally requires $s/\ell =\sqrt{\kappa s}\ll 1$. The assumed negligible inertia imposes its corresponding small inertial-to-viscous ratio, of order $\rho U{s}^2/(\mu \ell )\ll 1$. Thus small $C$ is necessary, while slope and inertia conditions are independent parts of self-consistency. If the lubrication approximation is required all the way to large scaled endpoints, the leading parabolic gap further requires $C{\tilde{L}}^2\ll 1$ and $\sqrt{\kappa s}\tilde{L}\ll 1$. Alternatively the thin neck can be matched to a separate outer solution; its pressure contribution must then be assessed.

Take the small-$\lambda$ and large-$\tilde{L}$ limits at fixed $C$ first, then expand in small $C$. Seek $H=H_0+\lambda H_1+\cdots$. The leading equation is $H_{0X}=X$, soAt the next order,For large $\tilde{L}$, extend the integrals determining the localized film resistance to the whole line. The even leading profile has equal end values, so the force condition at order $\lambda$ becomesThe left side follows from $H_1(-\tilde{L})-H_1(\tilde{L})=-\int H_{1X}dX$. Using the supplied integrals gives $I_1=2\pi /b$, $I_2=2\pi /b^3$, $I_3=3\pi /b^5$. HenceThis fixes the integration constant in $H_0$. In terms of $z=b^{-2}$ it reads $2z-3z^2=C(8/3-2z)$. The root near $z=2/3$ givesThe correction has a minus sign. The printed $1+C$ is inconsistent with the supplied differential equation and force condition: substituting $z=(2/3)(1+C)$ gives $2z-3z^2=-(4/3)C+O(C^2)$, whereas $C(8/3-2z)=(4/3)C+O(C^2)$. The discrepancy is not a choice of flux sign, because the positive-gap scaling here requires $s=2Q/U>0$. The leading value $b^{-2}=2/3$ is unaffected by this error.

Use the force integral to calculate the leading pressure drop, retaining its first nonzero order in both $C$ and $\lambda$:At $b^2=3/2$, the bracket is $4\pi /(3b)$. With the definitions of $C$ and $\lambda$ this gives the small-speed compliant-plug lubrication asymptoticsThus the requested leading pressure-drop expression remains valid despite the sign error in its subsidiary correction.

Finally, the absolute downstream pressure selects $s$ and hence $Q/U$. At the downstream matching region the leading shape givesbecause $s{X}^2/2$ cancels the unstressed parabolic term. For example, if the downstream excess pressure is fixed andthen $s\simeq 2d/b^2\simeq 4d/3$, andThis is a linear pressure-speed law on the specified small-$\lambda$ branch. If $p_d-p_0$ depends on speed or cell state, its relation must instead be substituted to obtain the corresponding law. If $d\le 0$, the stated leading positive-gap branch does not exist. Without the absolute-pressure relation, the problem gives a family parameterized by $Q/U$, rather than a unique $\Delta p(U)$. Errors from finite $\tilde{L}$, higher powers of $\lambda$ and higher powers of $C$ have been omitted consistently in this leading approximation. Resolving the displayed $O(C)$ correction to $b^{-2}$ requires the omitted finite-domain and higher-$\lambda$ corrections to be smaller than $C$; otherwise only its leading value is justified.