Differential pom-pom model ($\sigma =G{\lambda }^{2}A/\mathrm{tr}A-pI$)

This differential model separates normalized molecular orientation $B=A/\mathrm{tr}A$ from scalar stretch $\lambda$. With component-first velocity gradient $K$, it evolves by $D_{t}A=KA+A{K}^T-(A-I)/{\tau }_1$ and $D_t\lambda =\lambda B:K-(\lambda -1)/{\tau }_2$. Its relaxed state is $A=I$, $B=I/3$, $\lambda =1$. The displayed constant-relaxation stretch law is an idealization; variants can add saturation or nonlinear relaxation.

For extension rate $s>0$ and $x=s{\tau }_1<1/2$, the steady conformation tensor is $\mathrm{diag}((1-2x{)}^{-1},(1+x{)}^{-1},(1+x{)}^{-1})$. Its normalized orientation has axial component $(1+x)/[3(1-x)]$ and transverse components $(1-2x)/[3(1-x)]$. Stretch is finite only if $s^2{\tau }_1{\tau }_2<1-x$, when ${\lambda }_{\infty }=(1-x)/(1-x-s^2{\tau }_1{\tau }_2)$ and the displayed extensional viscosity follows. Above $x=1/2$, the unnormalized axial tensor grows exponentially but $B\to \mathrm{diag}(1,0,0)$; then stretch remains bounded only if $s{\tau }_2<1$. Orientation normalization and stretch relaxation therefore impose different thresholds.

Linearizing the differential pom-pom model about its relaxed state leaves no first-order stretch perturbation in incompressible flow. Its anisotropic orientation obeys $\dot{a}+a/{\tau }_1=A_1$, producing the displayed relaxation modulus. In a slow-flow expansion, $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Trace normalization and the first stretch correction affect only isotropic stress at this order, giving the displayed second-order fluid coefficients. Linear small-amplitude response allows arbitrary frequency, whereas this local second-order expansion also requires slow time dependence.