A Maxwell material is a model material obtained by connecting a perfectly elastic solid and a Newtonian liquid in series.
Let $\sigma$ be the stress, $\gamma$ the shear strain, $G$ the shear modulus and $\eta$ the viscosity.
For an elastic solid, we have
$$\sigma (t) = \gamma (t) \ G$$
and for a Newtonian liquid,
$$\sigma (t) = \dot \gamma (t) \ \eta$$
In a Maxwell material, the total shear strain is the sum of the strains of the single elements ($e$=elastic, $N$=Newtonian):
$$\gamma (t)= \gamma_e (t)+ \gamma_N (t) \tag{1}\label{1}$$
and each elements bears the same stress:
$$\sigma (t) = G_M \ \gamma_E (t) = \eta_M \ \dot \gamma_N (t) \tag{2}\label{2}$$
Let us apply a step strain,
$$\gamma(t) = \begin{cases} 0 & t<0\\ \gamma & t \geq 0\\ \end{cases} $$
From \ref{1} and \ref{2} we get
$$\tau \ \dot \gamma_N(t) = \gamma - \gamma_N(t)$$
where we have defined the relaxation time $\tau=\eta_M/G_M$. Using the initial condition $\gamma_N(0)=0$, we get, for $t \geq 0$,
$$\gamma - \gamma_N(t) = \gamma_E(t) = \gamma \exp(-t/\tau)$$
What I don't understand is: why is it reasonable to assume that $\gamma_N(0)=0$? How is this initial condition justified physically?
Addendum
I am starting to think that the problem is actually ill-defined at $t=0$ if we use the above equations, and that the initial condition must actually be intended as
$$\gamma_N (\epsilon) = 0 \ \ \forall \epsilon >0 \tag{3}\label{3}$$
Indeed, using \ref{1} and \ref{2} and introducing Heaviside's function $\theta(t)$ so that we can write $\gamma(t) = \gamma \ \theta(t)$, we have
$$\sigma(t) = \eta_M \frac{d}{dt}[\gamma \ \theta(t) - \gamma_E(t)] = \eta_M[ \gamma \ \delta(t) - \dot \gamma_E (t)]$$
having introduced the Dirac delta $\delta(t)$. So it looks like if we want \ref{1} and \ref{2} to be valid for any $t$, a Dirac delta, and thus an infinite stress, appears at $t=0$, which is unphysical.
On the other hand, if we take \ref{3} as initial condition or interpret \ref{1} and \ref{2} as valid only at $t>0$, then we may still be able to save the reasoning. Of course then one may argue that in reality there will never be a perfect step strain...