= Solution
Eliminating the <shear stress> from the two linear equations gives the <Maxwell shear-wave dispersion> equation in physical space,
$$
\tau\partial_t^2u+\partial_tu=\nu\partial_y^2u,
\qquad \nu=\frac{\mu}{\rho_0},
$$
where $\nu$ is the <kinematic viscosity>. Substitution of a <Fourier mode> $e^{iky+\sigma t}$ gives
$$
\tau\sigma^2+\sigma+\nu k^2=0,
\qquad
\boxed{\sigma_\pm=\frac{-1\pm\sqrt{1-4\tau\nu k^2}}{2\tau}.}
$$
These are temporal exponents: the oscillation <angular frequency> is the imaginary part of $\sigma$, with the sign determined by the selected time convention. For $4\tau\nu k^2<1$ both exponents are real and nonpositive. At equality the repeated exponent is $-1/(2\tau)$; generic solutions include $t e^{-t/(2\tau)}$. Above the threshold the modes are damped propagating shear disturbances:
$$
\sigma_\pm=-\frac1{2\tau}\pm i\sqrt{\frac{\nu k^2}{\tau}-\frac1{4\tau^2}}.
$$
At large $|k|$ their <phase velocity> tends to $\sqrt{\nu/\tau}=\sqrt{G_{\rm el}/\rho_0}$.
For the limit $\tau\to0$ at fixed $\mu,\rho_0$ and $k$, expansion of the <square root> gives
$$
\boxed{\sigma_{\rm slow}=-\nu k^2+O(\tau),\qquad \sigma_{\rm fast}=-\frac1\tau+\nu k^2+O(\tau).}
$$
After a short initial relaxation, the surviving branch obeys the <diffusion equation> $u_t=\nu u_{yy}$ of a <Newtonian fluid>. The fast branch disappears on the vanishing <viscoelastic relaxation time>; at $k=0$ the two exponents are exactly $0$ and $-1/\tau$. This limit is not uniform in $k$: modes with $k$ of order $(\tau\nu)^{-1/2}$ still sample the oscillatory transition. If instead $G_{\rm el}=\mu/\tau$ is held fixed as $\tau\to0$, then $\mu\to0$ and the fixed-$k$ slow exponent tends to zero. The ordinary <Newtonian fluid> limit therefore specifies which material parameter is held fixed.
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