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At rest the tensor equation relaxes $A$ toward $I$ and the stretch equation relaxes $\lambda$ toward one. ThusThe equilibrium nonpressure stress $GI/3$ is isotropic and can be absorbed into the arbitrary pressure in incompressible flow.

For the linear response, put $A=I+a+O({\epsilon }^2)$, $\lambda =1+\ell +O({\epsilon }^2)$ and take the velocity gradient to be $O(\epsilon )$. The first-order tensor equation isIts matrix trace has no forcing, because incompressibility gives $\mathrm{tr}K=0$. Starting relaxed, $\mathrm{tr}a=0$. The first-order stretch equation is $D_t\ell +\ell /{\tau }_2=(I/3):K=0$, hence $\ell =0$. Consequently $B=I/3+a/3+O({\epsilon }^2)$, and its anisotropic stress is $Ga/3$. From the relaxed remote past,The linear and second-order response of the differential pom-pom model therefore hasThe 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. IteratinggivesUsing the Rivlin-Ericksen tensor convention $A_2=D_t{A}_1+A_{1}K+K^T{A}_1$, the identityreduces this to $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Since $\mathrm{tr}{A}_2=\mathrm{tr}{A}_1^2$, normalization givesStretch first changes at second order. Its equation gives $\lambda -1={\tau }_1{\tau }_2\mathrm{tr}(A_1^2)/6+O(3)$. 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 $p_{\ast }$ to obtainComparison with the viscometric functions parametrization yieldsEquivalently, in $\sigma =-pI+{\mu }_0{A}_1+{\alpha }_1{A}_2+{\alpha }_2{A}_1^2$, 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.