= Solution
A <simple fluid> reduces to a <linear viscoelastic fluid> when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The <relaxation modulus> $G(t)$ is the shear-stress response to a unit small step of shear <strain>.
Choose the harmonic convention $e^{i\omega t}$. If the imposed <shear rate> is $\operatorname{Re}(\widehat{\dot\gamma}e^{i\omega t})$, the corresponding <shear stress> is $\operatorname{Re}(\eta^*(\omega)\widehat{\dot\gamma}e^{i\omega t})$. Substitution into the constitutive <convolution> gives the <complex viscosity>:
$$
\boxed{\eta^*(\omega)=\int_0^\infty G(s)e^{-i\omega s}ds.}
$$
For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^*=i\omega\eta^*=G'+iG''$, so $G'$ is the <storage modulus>, $G''$ the <loss modulus>, and $\eta^*=G''/\omega-iG'/\omega$ in this convention.
Creep is the progressive deformation produced by a held <stress>. For a small step $S_0H(t)$ of shear stress, define the <creep compliance> $J$ by $\gamma(t)=S_0J(t)$ from a relaxed state. Writing hats for <Laplace transforms>, the same constitutive law gives $\widehat S=p\widehat G\widehat\gamma$. Since $\widehat S=S_0/p$, this yields
$$
\boxed{\widehat J(p)=\frac{1}{p^2\widehat G(p)}.}
$$
Here $H$ is the <Heaviside step function>. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep <convolution> includes that jump through $\int_{0^-}^tG(t-s)dJ(s)=1$ for $t>0$.
For the <Maxwell fluid>, $\widehat G(p)=G_0/(p+1/\tau)$. Thus the two material responses are
$$
\eta^*(\omega)=\frac{G_0\tau}{1+i\omega\tau},\qquad
J(t)=\frac1{G_0}+\frac{t}{G_0\tau}\quad(t\ge0).
$$
The creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency <shear viscosity> is $\eta=G_0\tau$.
For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of <stress> by $S$. The <incompressibility> constraint is satisfied, and the advective part of the <material derivative> vanishes because no field depends on $x$. With no imposed longitudinal <pressure gradient>, <momentum conservation> and the linear Maxwell law give
$$
\rho u_t=S_y,\qquad S_t+\frac S\tau=G_0u_y.
$$
The initial conditions are $u=S=0$ in $y>0$; the <no-slip boundary condition> is $u(0,t)=UH(t)$, with decay as $y\to\infty$. Transforming these equations gives
$$
\widehat S=\frac{G_0}{p+1/\tau}\widehat u_y,\qquad
\widehat u_{yy}=q(p)^2\widehat u,\qquad
q(p)=\sqrt{\frac{\rho p(1+p\tau)}{G_0\tau}}.
$$
Choose the analytic square-root branch in $\operatorname{Re}p>0$ with $\operatorname{Re}q>0$, so that the spatially increasing solution is excluded. The <Maxwell start-up shear layer> consequently has
$$
\widehat u(y,p)=\frac Up e^{-q(p)y}.
$$
The <Bromwich inversion formula> now gives
$$
\boxed{u(y,t)=\frac{U}{2\pi i}\int_{b-i\infty}^{b+i\infty}\frac1p\exp\left[pt-y\sqrt{\frac{\rho p(1+p\tau)}{G_0\tau}}\right]dp,\qquad b>0.}
$$
The vertical <Bromwich contour> is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.
The Newtonian limit holds $\eta=G_0\tau$ fixed while $\tau\to0$, so $G_0\to\infty$. Then $q(p)\to\sqrt{\rho p/\eta}$. With $\nu=\eta/\rho$, the inverse <Laplace transform> gives the <complementary error function> profile
$$
\boxed{u(y,t)=U\operatorname{erfc}\left(\frac{y}{2\sqrt{\nu t}}\right).}
$$
This is diffusive momentum penetration in a <Newtonian fluid>, with thickness of order $\sqrt{\nu t}$.
The elastic limit instead holds $G_0$ fixed while $\tau\to\infty$. Write $c=\sqrt{G_0/\rho}$. On the chosen branch, $q(p)\to p/c$, so the <Laplace transform time-shift rule> yields
$$
\boxed{u(y,t)=U H(t-y/c).}
$$
An elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear <strain> is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.
\b[The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$.] The <material derivative> simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear <strain>. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho/G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary <simple fluid>.
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