= Solution
In the travelling coordinate $X=x-ct$, <mass conservation> becomes $[\alpha(u-c)]'=0$. Thus the conserved relative flux is $Q$ and
$$
\boxed{u=c+Q/\alpha.}
$$
For a periodic <peristaltic pumping> solution, the neglected-inertia momentum equation is
$$
\widetilde P'(\alpha)\alpha'+\epsilon kP_e\cos kX=-R(\alpha)(c+Q/\alpha).
$$
Take $\alpha_0$ to be the spatial mean area, so $\langle\alpha_1\rangle=\langle\alpha_2\rangle=0$. This normalization, and zero net <pressure> drop over a period, are implicit in the specified periodic forcing. The constant leading state has no <pressure gradient>, giving $Q_0=-\alpha_0c$. Put $B=\widetilde P'_0$ and $d=R_0c/\alpha_0$. At first order,
$$
B\alpha_1'+d\alpha_1=-kP_e\cos kX-R_0Q_1/\alpha_0.
$$
Period averaging gives $Q_1=0$, and the zero-mean solution is
$$
\alpha_1=-\frac{P_e[(d/k)\cos kX+B\sin kX]}{B^2+(d/k)^2},\qquad
\langle\alpha_1^2\rangle=\frac{P_e^2}{2[B^2+(d/k)^2]}.
$$
For the second-order mean balance, expand the <velocity> and friction:
$$
u_1=c\alpha_1/\alpha_0,\qquad
u_2=c\alpha_2/\alpha_0-c\alpha_1^2/\alpha_0^2+Q_2/\alpha_0,\qquad
R_1=-nR_0\alpha_1/\alpha_0.
$$
Averaging the exact <pressure> derivative gives zero, so $\langle R_0u_2+R_1u_1\rangle=0$. Hence the <weak peristaltic pumping of a collapsible tube> result is
$$
\boxed{Q_0=-\alpha_0c,\qquad Q_1=0,\qquad Q_2=\frac{c(n+1)}{2\alpha_0}\frac{P_e^2}{\widetilde P_0'^2+(R_0c/(k\alpha_0))^2}.}
$$
The mean laboratory <volumetric flow rate> is $c\langle\alpha\rangle+Q=\epsilon^2Q_2+O(\epsilon^3)$, forward with the wave when $c>0$. The negative leading relative flux is therefore consistent with positive mean pumping.
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