= Solution
The constitutive convolution is the response of a <linear viscoelastic fluid> close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the <stress> is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic <incompressible flow>, the <deviatoric stress> is proportional to the history of the <rate-of-strain tensor>; an arbitrary isotropic <pressure> is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega\tau\ll1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear <normal-stress differences> are outside this formula's scope.
If a small step of strain $\varepsilon_0$ is imposed at time zero and then held, $E(t)=\varepsilon_0\delta(t)$ to linear order. The convolution consequently gives $\sigma'(t)=2G(t)\varepsilon_0$ for $t>0$. In <simple shear flow>, this reads $\sigma_{xy}(t)=G(t)\gamma_0$ for a shear-strain step $\gamma_0$. Thus \b[$G(t)$ is the remaining shear stress per unit imposed step of shear strain], which explains the name <relaxation modulus> and directly describes <stress relaxation>.
Use the convention $E(t)=\operatorname{Re}[\widehat E e^{i\omega t}]$. Substitution into the convolution defines the <complex viscosity>:
$$
\widehat\sigma'=2\widehat\mu(\omega)\widehat E,\qquad
\boxed{\widehat\mu(\omega)=\int_0^\infty G(s)e^{-i\omega s}\,ds.}
$$
Write $\widehat\mu=\mu'-i\mu''$. Its real part $\mu'$ multiplies the component in phase with the <rate-of-strain tensor>, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^*=i\omega\widehat\mu$, so the <storage modulus> and <loss modulus> are
$$
G'=\omega\mu''=-\omega\operatorname{Im}\widehat\mu,\qquad G''=\omega\mu'=\omega\operatorname{Re}\widehat\mu.
$$
For shear-rate amplitude $\widehat{\dot\gamma}$ the mean dissipated power per unit volume is $\tfrac12\mu'|\widehat{\dot\gamma}|^2$. For the exponential <Maxwell fluid> memory kernel, direct integration gives
$$
\boxed{\widehat\mu=\frac{G_0\tau}{1+i\omega\tau},\quad
\mu'=\frac{G_0\tau}{1+\omega^2\tau^2},\quad
\mu''=\frac{G_0\omega\tau^2}{1+\omega^2\tau^2}.}
$$
The <zero-shear viscosity> is $\mu_0=G_0\tau$, and the high-frequency <storage modulus> tends to $G_0$.
For the channel, let $v_x(y,t)=\operatorname{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of $\partial_xp$, not the positive pressure drop. The <Cauchy momentum equation> and <no-slip boundary conditions> give
$$
i\omega\rho U=-\Delta p+\widehat\mu U'',\qquad U(\pm h)=0.
$$
Define $k^2=i\omega\rho/\widehat\mu$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution is
$$
U(y)=-\frac{\Delta p}{i\omega\rho}\left[1-\frac{\cosh(ky)}{\cosh(kh)}\right].
$$
Its integral across the channel gives the <oscillatory channel flux of a linear Maxwell fluid>:
$$
\boxed{Q=\int_{-h}^h U(y)\,dy=-\frac{2\Delta p h}{i\omega\rho}\left(1-\frac{\tanh z}{z}\right),\qquad
z^2=\frac{i\omega\rho h^2(1+i\omega\tau)}{G_0\tau}.}
$$
This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.
As $\omega\to0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. Hence
$$
Q=-\frac{2\Delta p h^3}{3\widehat\mu}\left[1-\frac25z^2+O(z^4)\right]
=-\frac{2\Delta p h^3}{3\mu_0}\left[1+i\omega\left(\tau-\frac{2\rho h^2}{5\mu_0}\right)+O(\omega^2)\right].
$$
In particular \b[$Q\to-2\Delta p h^3/(3G_0\tau)$], the usual <plane Poiseuille flow> flux with viscosity $\mu_0$. The fluid has enough time to relax and behaves as a <Newtonian fluid> at leading order.
For fixed positive $h,\tau,\rho,G_0$, let $c=\sqrt{G_0/\rho}$. At large frequency the square root with positive real part satisfies
$$
z=\frac{h}{2c\tau}+i\frac{\omega h}{c}+O(\omega^{-1}).
$$
Its real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O(\omega^{-1})$ and
$$
\boxed{Q=-\frac{2\Delta p h}{i\omega\rho}[1+O(\omega^{-1})]\qquad(\omega\to\infty).}
$$
The leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the <Maxwell fluid> is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.
Finally take the different ordering $\omega\tau\gg1$ but $\rho\omega^2h^2/G_0\ll1$. Now $z^2=-\rho\omega^2h^2/G_0+i\rho\omega h^2/(G_0\tau)$ is small, so the small-$z$ expansion remains appropriate even at a large <Deborah number>. The result is
$$
Q\simeq-\frac{2\Delta p h^3(1+i\omega\tau)}{3G_0\tau},\qquad
\boxed{Q\sim-\frac{2i\omega\Delta p h^3}{3G_0}.}
$$
Here inertia is negligible throughout the narrow gap. Dividing the <velocity> by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho\omega^2h^2/G_0\ll1$ eventually fails.
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