= Solution
The weakly nonlinear, unsteady and dispersive core terms must enter at the same order. Axial nonlinearity first contributes at $O(\epsilon^2)$, while unsteadiness contributes $O(\epsilon\lambda St)$ and transverse-inertia <pressure> variation contributes $O(\epsilon\lambda^{-2})$. Thus take the distinguished limit
$$
\boxed{\lambda St=\beta\epsilon,\qquad\lambda^{-2}=D\epsilon,\qquad\beta,D=O(1)>0}.
$$
Equivalently, $\lambda=O(\epsilon^{-1/2})$, $St=O(\epsilon^{3/2})$ and the wall-layer condition becomes $Re\gg\epsilon^{-7/2}$. These orderings imply the initially stipulated long wavelength, low <Strouhal number> and large <Reynolds number>. In the core $\lambda/Re=o(\epsilon^3)$, so perturbation viscous terms are beyond the retained order.
Subtract the undisturbed fully developed state and scale <pressure> by $\rho\hat U^2$. The inviscid core equations are
$$
u_x+\frac1r(rv)_y=0,\qquad\lambda St\,u_t+uu_x+vu_y=-p_x,\qquad p_y=-\lambda^{-2}(\lambda St\,v_t+uv_x+vv_y).
$$
Put $u=u_0+\epsilon u_1+\epsilon^2u_2+\cdots$ and $v=\epsilon v_1+\epsilon^2v_2+\cdots$. At first order $p_{1y}=0$ and
$$
u_0u_{1x}+v_1u_0'=-p_{1x},\qquad u_{1x}+\frac1r(rv_1)_y=0.
$$
The wall's normal movement first enters at second order in this scaling, so $v_1=0$ at both walls. Taking the regular wall limits, where $u_0=0$ and $u_0'\ne0$, gives $p_{1x}=0$. The arbitrary constant <pressure> perturbation can be set to zero.
Define the first-order axisymmetric <stream function> by $u_1=\psi_{1y}/r$ and $v_1=-\psi_{1x}/r$. The axial equation becomes $u_0\psi_{1yx}-u_0'\psi_{1x}=0$, or $(\psi_{1x}/u_0)_y=0$. Consequently $\psi_1=A(x,t)u_0(y)$ after absorbing an $x$-independent profile into the background. This proves the <annular inviscid-core displacement>
$$
\boxed{u_1=\frac{A}{R+y}u_0'(y),\qquad v_1=-\frac{A_x}{R+y}u_0(y)}.
$$
It gives exactly the stated core expansion without requiring an explicit calculation of $u_0(y)$. The factors $1/(R+y)$ arise from cylindrical <mass conservation>.
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