= Solution
At rest the tensor equation relaxes $A$ toward $I$ and the stretch equation relaxes $\lambda$ toward one. Thus
$$
\boxed{A_0=I,\qquad B_0=\frac I3,\qquad\lambda_0=1.}
$$
The equilibrium nonpressure <stress> $GI/3$ is isotropic and can be absorbed into the arbitrary <pressure> in <incompressible flow>.
For the linear response, put $A=I+a+O(\varepsilon^2)$, $\lambda=1+\ell+O(\varepsilon^2)$ and take the <velocity gradient> to be $O(\varepsilon)$. The first-order tensor equation is
$$
D_ta+\frac a{\tau_1}=K+K^T=A_1.
$$
Its <matrix trace> has no forcing, because <incompressibility> gives $\operatorname{tr}K=0$. Starting relaxed, $\operatorname{tr}a=0$. The first-order stretch equation is $D_t\ell+\ell/\tau_2=(I/3):K=0$, hence $\ell=0$. Consequently $B=I/3+a/3+O(\varepsilon^2)$, and its anisotropic <stress> is $Ga/3$. From the relaxed remote past,
$$
\sigma^{\rm lin}=-p_*I+\frac G3\int_{-\infty}^t e^{-(t-s)/\tau_1}A_1(s)ds
=-p_*I+2\int_{-\infty}^t\frac G3e^{-(t-s)/\tau_1}E(s)ds.
$$
The <linear and second-order response of the differential pom-pom model> therefore has
$$
\boxed{G_{\rm rel}(t)=\frac G3e^{-t/\tau_1}\quad(t\ge0),\qquad\mu_0=\int_0^\infty G_{\rm rel}(t)dt=\frac{G\tau_1}{3}.}
$$
The stretch time $\tau_2$ does not enter the linear shear <relaxation modulus>.
For the <second-order fluid> constants, additionally make the slow-flow expansion: count $K$ as first order and $D_t$ as raising the order by one. Iterating
$$
A-I=\tau_1(KA+AK^T-D_tA)
$$
gives
$$
A=I+\tau_1A_1+\tau_1^2(KA_1+A_1K^T-D_tA_1)+O(3).
$$
Using the <Rivlin-Ericksen tensor> convention $A_2=D_tA_1+A_1K+K^TA_1$, the identity
$$
KA_1+A_1K^T+A_1K+K^TA_1=2A_1^2
$$
reduces this to $A=I+\tau_1A_1+\tau_1^2(2A_1^2-A_2)+O(3)$. Since $\operatorname{tr}A_2=\operatorname{tr}A_1^2$, normalization gives
$$
B=\frac I3+\frac{\tau_1}{3}A_1+\frac{\tau_1^2}{3}(2A_1^2-A_2)
-\frac{\tau_1^2}{9}\operatorname{tr}(A_1^2)I+O(3).
$$
Stretch first changes at second order. Its equation gives $\lambda-1=\tau_1\tau_2\operatorname{tr}(A_1^2)/6+O(3)$. Multiplication by $\lambda^2$ therefore adds only an isotropic <stress> at second order; multiplying its second-order increment by the anisotropic first-order term would be third order. Absorb these isotropic corrections into $p_*$ to obtain
$$
\sigma=-p_*I+\frac{G\tau_1}{3}A_1+\frac{G\tau_1^2}{3}(2A_1^2-A_2)+O(3).
$$
Comparison with the <viscometric functions> parametrization yields
$$
\boxed{\mu_0=\frac{G\tau_1}{3},\qquad\psi_1=\frac{2G\tau_1^2}{3},\qquad\psi_2=0.}
$$
Equivalently, in $\sigma=-pI+\mu_0A_1+\alpha_1A_2+\alpha_2A_1^2$, the constants are $\alpha_1=-G\tau_1^2/3$ and $\alpha_2=2G\tau_1^2/3$. The general small-amplitude linear response above permits fast oscillation; the local <second-order fluid> expansion also requires the slow-flow ordering just used.
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