= Solution
When deformation is stopped, the <rate-of-strain tensor> and <velocity gradient> vanish, leaving $\mathsf T+\tau\partial_t\mathsf T=0$. Hence
$$
\boxed{\mathsf T(t)=\mathsf T(0)e^{-t/\tau}.}
$$
The positive constant $\tau$ is the <viscoelastic relaxation time>. For slow weak deformation, all the transport and quadratic terms are smaller than $\mathsf T$, so $\mathsf T\simeq2\mu\mathsf E$. Thus $\mu>0$ is the zero-rate <dynamic viscosity>, measured in <stress> times time. The corresponding short-time <shear modulus> of the <Maxwell fluid> response is $G_{\rm el}=\mu/\tau$. The source's $\alpha$ has units of time, since $\alpha\mathsf T\mathsf E$ must have units of <stress>.
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