= Solution
\b[With its material state fixed by a prescribed deformation history, the differential pom-pom model is a simple-fluid constitutive law.] The <stress> depends on the local history of a material element through its molecular stretch and orientation, without gradients of that history between neighbouring elements. A differential representation does not prevent a material from being a <simple fluid>.
To see the history dependence explicitly, use the component-first <velocity gradient> $K$, so the tensor evolution is $D_tA=KA+AK^T-(A-I)/\tau_1$. Let $F(t,s)$ be the relative <deformation gradient>, satisfying $\partial_tF=K(t)F$ and $F(s,s)=I$. Along a material trajectory, an <integrating factor> gives
$$
A(t)=e^{-(t-t_0)/\tau_1}F(t,t_0)A(t_0)F(t,t_0)^T
+\frac1{\tau_1}\int_{t_0}^t e^{-(t-s)/\tau_1}F(t,s)F(t,s)^Tds.
$$
Thus $A$ is a functional of the local deformation history and its prepared initial state. If the initial $A$ is a <positive-definite matrix>, it retains that property, so $B=A/\operatorname{tr}A$ is well defined. Given this $B(t)$, the <scalar> stretch solves $\dot\lambda=r(t)\lambda+1/\tau_2$, where $r=B:K-1/\tau_2$. Its <variation of constants> formula again depends only on that local history:
$$
\lambda(t)=\lambda(t_0)e^{\int_{t_0}^t r(q)dq}
+\frac1{\tau_2}\int_{t_0}^t e^{\int_s^t r(q)dq}ds.
$$
The <differential pom-pom model> also satisfies <material frame indifference>. Under an observer rotation, $A$ and $B$ transform as objective tensors, while $\lambda$ is a scalar. The extra skew part of the transformed <velocity gradient> has zero contraction with symmetric $B$, so the stretch equation is unchanged. The tensor evolution uses an <upper-convected derivative>, and the final <stress> transforms objectively. Initial values cannot be independently forgotten when specifying a finite-past experiment: a fully specified prepared state, such as rest, is part of the history specification.
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