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Choose the extending direction as $x$ and write $s=\dot{\gamma }>0$. The uniaxial extensional flow, satisfying incompressibility, hasStarting from rest, symmetry preserves $A=\mathrm{diag}(a,b,b)$. The differential pom-pom model tensor equation becomesPut $x=s{\tau }_1$. For $x\ne 1/2$, the solutions areIf $x<1/2$, both transients decay, givingIf $x>1/2$, the axial component is instead $a=(2x{e}^{(2x-1)t/{\tau }_1}-1)/(2x-1)$ and grows exponentially without bound, while $b\to 1/(1+x)$. At the boundary $x=1/2$, which lies between the two stated regimes, $a=1+t/{\tau }_1$ grows linearly and $b\to 2/3$.

Normalization makes the molecular orientation finite even when the unnormalized conformation tensor diverges. For $0\le x<1/2$, direct division by $a+2b$ givesFor $x>1/2$, the divergent axial component dominates the matrix trace, soThe same normalized limit holds at $x=1/2$, agreeing continuously with the limit of the subcritical expression. The limiting orientations have nonnegative eigenvalues and unit matrix trace, as required for the normalized molecular orientation.

Now set $S(t)=B(t):K=s(a-b)/(a+2b)$. Its limiting value isThe scalar stretch equation is $\dot{\lambda }=[S(t)-1/{\tau }_2]\lambda +1/{\tau }_2$. If $S_{\infty }<1/{\tau }_2$, its coefficient is eventually bounded above by a negative constant. The variation of constants formula gives a bounded solution, converging toIf $S_{\infty }>1/{\tau }_2$, the coefficient is eventually positive, and the positive stretch grows without bound. At equality a bounded stretch would imply $\dot{\lambda }\to 1/{\tau }_2>0$, which contradicts boundedness. Thus equality is not an additional bounded steady regime.

The function $S_{\infty }(s)$ is continuous and strictly increasing from zero to infinity. Consequently there is a unique critical extension rate defined byand the molecular stretch remains bounded precisely for $\dot{\gamma }<{\dot{\gamma }}_c$. This is distinct from the tensor threshold $\dot{\gamma }{\tau }_1=1/2$. If ${\tau }_2<2{\tau }_1$, for example, the normalized orientation can remain well defined and the stretch remain bounded for some rates above the tensor threshold; divergence of $A$ alone does not prove divergence of the physical stress.

In the requested doubly subcritical regime, write $D=1-s{\tau }_1-s^2{\tau }_1{\tau }_2>0$. The steady uniaxial extension of the differential pom-pom model hasThe isotropic pressure cancels from the tensile normal-stress difference. Hence the extensional viscosity isorIts zero-rate limit is $G{\tau }_1=3{\mu }_0$, recovering the Trouton ratio of three. The restrictions exclude both loss of a finite tensor steady state and loss of a finite stretch steady state; neither denominator may be continued through its singularity as a physical steady constitutive response.

Let two observers be related by ${\mathbf{x}}^{\ast }=\mathbf{c}(t)+Q(t)\mathbf{x}$, with Q a proper rotation matrix. A spatial second-rank tensor has material frame indifference if its values at corresponding events satisfy $\boxed{J^{\ast }=QJ{Q}^T}$. This defines an objective second-rank tensor, and allows the observer's rotation to depend on time.

For Cauchy stress tensor, consider a material surface element with unit normal n and contact traction $\mathbf{t}=\sigma \mathbf{n}$. A rigid observer change preserves its area and rotates both its normal and actual contact force: ${\mathbf{n}}^{\ast }=Q\mathbf{n}$, ${\mathbf{t}}^{\ast }=Q\mathbf{t}$. Thus ${\sigma }^{\ast }Q\mathbf{n}=Q\sigma \mathbf{n}$ for every n, proving ${\sigma }^{\ast }=Q\sigma Q^T$. Fictitious body forces of an accelerated observer do not alter this transformation of physical contact traction.

To keep the vorticity convention explicit, write $L_{ij}={\partial }_j{v}_i$, $E=(L+L^T)/2$ and $\Omega =(L^T-L)/2$, the negative of the usual component-first spin tensor. Equivalently, with the derivative-index-first $\nabla v$ used in the paper, $\Omega =[\nabla v-(\nabla v{)}^T]/2$. Put $R=\dot{Q}{Q}^T$, which is antisymmetric. The transformed velocity is ${\mathbf{v}}^{\ast }=\dot{\mathbf{c}}+\dot{Q}\mathbf{x}+Q\mathbf{v}$, and direct differentiation givesThe additional observer spin cancels from the symmetric part, proving that the rate-of-strain tensor is objective.

Along the same material particle, differentiating $J^{\ast }=QJ{Q}^T$ givesThe two extra terms cancel those from ${\Omega }^{\ast }{J}^{\ast }-J^{\ast }{\Omega }^{\ast }$. Moreover $E^{\ast }{J}^{\ast }+J^{\ast }{E}^{\ast }=Q(EJ+JE)Q^T$. Hence the complete objective time derivative satisfiesThis proves covariance for every constant scalar $\alpha$; changing the spin convention without changing the commutator sign would not. It is a Gordon--Schowalter derivative with the paper's slip-parameter convention.

For the Johnson-Segalman fluid, first use the standard uniaxial extensional flow $\mathbf{v}=(\dot{ϵ}x,-\dot{ϵ}y/2,-\dot{ϵ}z/2)$, with $\dot{ϵ}>0$. Here $\Omega =0$ and E is diagonal. The steady conformation equation giveswith vanishing off-diagonal components on the stable steady branch. The tensile normal-stress difference is $\alpha G_0(A_{xx}-A_{yy})+3{\mu }_0\dot{ϵ}$. Dividing by the extension rate gives the Johnson-Segalman extensional response:The conformation tensor evolution in each principal direction is ${\dot{A}}_{ii}+(1/\tau -2\alpha E_{ii})A_{ii}=1/\tau$. Therefore a positive, attracting steady conformation requires both displayed denominators positive. For $\alpha >0$ the first pole is at $\dot{ϵ}=1/(2\alpha \tau )$; for $\alpha <0$ it is at $\dot{ϵ}=1/(|\alpha |\tau )$ in the transverse components. At a pole the corresponding component grows linearly in time, and beyond it generically grows exponentially. The model has a finite-rate extensional catastrophe; the algebraic viscosity cannot be continued as a physical steady response past the first pole. At zero rate, ${\eta }_E=3({\mu }_0+{\alpha }^2{G}_0\tau )$, giving Trouton ratio three. For $\alpha =0$ the polymer contribution to stress disappears and the response is simply Newtonian.

The question does not specify extension geometry. If a planar convention $\mathbf{v}=(\dot{ϵ}x,-\dot{ϵ}y,0)$ is intended instead, the same diagonal calculation gives $({\sigma }_{xx}-{\sigma }_{yy})/\dot{ϵ}=4{\mu }_0+4{\alpha }^2{G}_0\tau /[1-4{\alpha }^2{\tau }^2{\dot{ϵ}}^2]$, valid on the stable branch $|2\alpha \tau \dot{ϵ}|<1$. The preceding boxed formula uses the standard uniaxial definition of extensional viscosity.

For simple shear flow $v_x=\dot{\gamma }y$, write $A_{xx}=X$, $A_{yy}=Y$, $A_{xy}=A_{yx}=Z$, and $A_{zz}=1$. The xx, yy and xy components of the steady constitutive equation giveSubstituting the first two relations into the third yields $Z=\alpha \tau \dot{\gamma }/D$, with $D=1+(1-{\alpha }^2){\tau }^2{\dot{\gamma }}^2$. Define the viscometric functions by ${\sigma }_{xy}=\mu \dot{\gamma }$, $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, using x for flow, y for the shear gradient and z for vorticity. The Johnson-Segalman steady viscometric functions are thereforeThe solvent creates no normal-stress differences in this flow. For nonzero $\alpha$, the fluid is shear thinning with positive first and negative second normal-stress differences.

To classify the stress curves, set $x=\tau \sqrt{1-{\alpha }^2}\dot{\gamma }$ and $b={\mu }_0/({\alpha }^2{G}_0\tau )$. The scaled shear stress is $S=x[b+(1+x^2{)}^{-1}]$, andThe second term has minimum $-1/8$ at $x^2=3$. Thus the shear stress is monotone for $b\ge 1/8$ and has a decreasing segment for $\boxed{0<b<1/8}$. At equality it has a horizontal tangent but no decreasing interval. Below the threshold, its maximum and minimum occur atwith the smaller root giving the maximum. Negative-rate curves are obtained by odd reflection.

For small b, polymer stress dominates the low-rate response, while the solvent restores a positive high-rate slope. The Johnson-Segalman negative-slope shear threshold signals a constitutive instability of homogeneous shear: different local shear rates can carry the same shear stress, permitting shear banding. A stress plateau or a selected interface position cannot be inferred from the local curve alone; spatial regularization and boundary/history information matter. This limitation is illustrated in a primary diffusive Johnson-Segalman study. If $\alpha =0$, the scaling by b is inapplicable and the stress is just ${\mu }_0\dot{\gamma }$.

A plain material derivative of the conformation tensor is not an objective time derivative: an observer undergoing a time-dependent rigid rotation would infer a different constitutive response, and even rigid-body rotation could appear to change polymer deformation. The upper-convected derivative subtracts deformation and rotation carried by the velocity gradient and is frame indifferent.

The Oldroyd-B model is often inadequate because its Hookean dumbbells are infinitely extensible. It predicts constant shear viscosity rather than shear thinning, zero second normal-stress difference, and an unbounded extensional viscosity at a finite extension rate. Real polymer chains have finite extensibility and commonly exhibit shear thinning, bounded extensional stress, multiple relaxation times, and nonlinear solvent or concentration effects.

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.