= Solution
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 gives
$$
w(y)=-U\left(1-\frac yh\right)+\frac{p_x}{2\mu}y(y-h).
$$
<Mass conservation> makes the body-frame flux constant. With the sign convention of the problem,
$$
\int_0^h w\,dy=-\frac{Uh}{2}-\frac{p_xh^3}{12\mu}=-Q,
$$
so the <lubrication model of a compliant translating plug> gives
$$
\boxed{p_x=-\frac{6\mu U}{h^2}+\frac{12\mu Q}{h^3}}.
$$
The 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 gives
$$
\begin{aligned}
h(L)&=a-r_0(L)+(p_d-p_0)/\alpha,\\
h(-L)&=a-r_0(-L)+(p_d+\Delta p-p_0)/\alpha,
\end{aligned}
$$
For $r_0(-L)=r_0(L)$ 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 $2\pi a\mu\int_{-L}^Lw_y(0)\,dx$. The local <velocity> solution gives $w_y(0)=4U/h-6Q/h^2$, so
$$
\boxed{\Delta p=\frac{2\mu}{a}\int_{-L}^L\left(\frac{4U}{h}-\frac{6Q}{h^2}\right)dx}.
$$
This 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 gives
$$
\boxed{H_X+\lambda(H^{-2}-H^{-3})=X},\qquad\lambda=\frac{6\mu U}{\alpha\sqrt\kappa\,s^{5/2}}.
$$
The 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 $\widetilde L=L/\ell$. The elastic end condition gives $\Delta p=\alpha s[H(-\widetilde L)-H(\widetilde L)]$. Scaling the <force> integral and eliminating $\Delta p$ then gives
$$
\boxed{H(-\widetilde L)-H(\widetilde L)=C\lambda\int_{-\widetilde L}^{\widetilde L}\left(\frac4{3H}-\frac1{H^2}\right)dX}.
$$
This 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.
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