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In uniaxial extensional flow, the conformation equations give $A_{xx}=1/(1-2\alpha \tau \dot{ϵ})$ and $A_{yy}=A_{zz}=1/(1+\alpha \tau \dot{ϵ})$. Their stress difference gives the displayed extensional viscosity. A stable positive steady conformation requires both denominators positive; a first pole marks loss of a bounded steady response. At zero extension rate the Trouton ratio is three relative to the zero-shear viscosity.

For uniaxial extensional flow use $\mathbf{v}=\dot{\gamma }(x,-y/2,-z/2)$, so $L=E=\dot{\gamma }\mathrm{diag}(1,-1/2,-1/2)$. Again put $q=\dot{\gamma }\tau$ and $f=1+\alpha \mathrm{tr}A$. A steady diagonal tensor hasSet $T=f-2q$. ThenSince $f-1=\alpha \mathrm{tr}A$ and $f=T+2q$, multiplication givesThis is the cubic for uniaxial extension of an affine linear PTT fluid. For $\alpha >0$, choose its physical root $T>\max (0,1-2q)$, continuous from $T=1$ at rest. The product on the left grows strictly from zero to infinity in that interval, so this root is unique. The other algebraic roots need not describe an admissible steady material state.

The extensional viscosity is the axial-minus-transverse normal stress divided by the principal extension rate. Isotropic pressure cancels, leavingAt small $q$, $f=1+6\alpha q^2+O(q^3)$ and $T=1-2q+O(q^2)$. Thusthe Trouton ratio of three relative to the zero-shear viscosity. At large $q$ with fixed $\alpha >0$, the positive cubic root remains bounded and tends to $\alpha$: dividing the cubic by $6q^2$ gives $T[1+O(q^{-1})]=\alpha$. ThereforeThe nonlinear relaxation prevents the divergent high-rate response of the affine Maxwell model. If $\alpha =0$, the physically stable steady branch instead has $f=1$ and ends at $q=1/2$; no finite positive high-rate limit follows. The finite plateau consequently presumes the usual positive PTT relaxation parameter, as well as $G_0,\tau >0$.

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 ${\mu }_0\dot{\gamma }$.

Under the plug-like extensional viscosity approximation, incompressibility gives $w_z=-u_x$. Symmetry about $z=0$ fixes $w=-z{u}_x$. Remove the reference hydrostatic stress of the middle fluid, and write $S$ for its leading vertical normal stress. Vertical traction balance at the upper surface givesThe middle fluid's horizontal normal stress exceeds $S$ by $4\mu u_x$: the horizontal strain is $u_x$ and the vertical strain is $-u_x$. HenceAt a slightly inclined upper interface, tangential traction continuity gives, to the order retained in the small-outer-viscosity extensional approximation,The slope term is needed even though the interface is nearly horizontal: dropping it would miss the derivative of $h$ in the extensional resultant. Terms involving the outer deviatoric normal-stress difference times the slope are smaller in this approximation. Integrating horizontal force balance over the symmetric layer, or over its upper half, now givesSubstitution yields the extensional-flow equationThe full layer thickness is $2h$, so its conservation of mass is $(2h{)}_t+(2hu{)}_x=0$, orThis explains the factor $4\mu$ in the extensional stress rather than the shear viscosity coefficient alone.

For a Fourier mode of wave number $k>0$, set $h=h_0+\eta e^{ikx-st}$ and $u=\hat{u}{e}^{ikx-st}$. Neglect ${\sigma }_{zz}^{+}$ as stipulated. The Fourier traction map for a viscous half-space, with outer dynamic viscosity $\lambda \mu$, gives ${\sigma }_{xz}^{+}=-2\lambda \mu k\hat{u}{e}^{ikx-st}$. The two linear equations areEliminating $\hat{u}$ proves the positive decay rate for relaxation of a symmetric viscous layer:When $\lambda \ll \kappa$, inner longitudinal extension dominates the viscous dissipation, and $s\sim \Delta \rho g{h}_0/(4\mu )$. When $\kappa \ll \lambda$ while the plug approximation remains valid, tangential motion of the semi-infinite outer fluids supplies the main resistance, and $s\sim \Delta \rho g{h}_0\kappa /(2\lambda \mu )$. The PDF's second inequality is $\kappa \ll \lambda$; the TeX converts it incorrectly to $\kappa \ll 1$. At much larger outer viscosity the plug assumption fails, as the next calculation shows.

Setting $\alpha =0$ gives the Upper-convected Maxwell modelIn steady incompressible uniaxial extension of rate $\dot{ϵ}$, its tensile and transverse stresses giveThe extensional viscosity rises above the Trouton ratio value $3\eta$ and diverges at $\lambda \dot{ϵ}=1/2$, the ideal model's extensional catastrophe. An extensional rheometer can locate that rapid growth and estimate $\lambda \simeq 1/(2{\dot{ϵ}}_c)$. More robustly, impose a small deformation, stop the flow, and fit the exponential stress decay $\tau \propto e^{-t/\lambda }$.

For the uniaxial extensional flowthe spin tensor vanishes andThe flow is steady and homogeneous, so with $\xi =1$ the Jaumann derivative of $\mathbf{\alpha }$ vanishes. The structure equation yieldsWriting $k=c_2/c_1$ and using
$\dot{\mathbf{\gamma }}:\dot{\mathbf{\gamma }}=6{\dot{ϵ}}^2$ givesConsequentlyThe extensional viscosity is the tensile stress difference divided by $\dot{ϵ}$:For $b_2=0$ this is the constantThe factor three is the Trouton ratio associated with the effective zero-rate shear viscosity.

For uniaxial extension,The diagonal steady conformation tensor iswhere physical solutions require $F>2a$. The extensional viscosity isThe implicit closure is

A layer of half-thickness $h_0$, dynamic viscosity $\mu$ and density $\rho$, between semi-infinite fluids of viscosity $\lambda \mu$ and densities $\rho \mp \Delta \rho$, relaxes a symmetric thickness Fourier mode. In the plug-like extensional viscosity regime, mass conservation and $4\mu (h{u}_x{)}_x=h\Delta \rho g{h}_x-{\sigma }_{xz}^{+}$ together with the Fourier traction map for a viscous half-space give the displayed rate, where $\kappa =|k|h_0$. Inner longitudinal extension dominates for $\lambda \ll \kappa$; outer tangential resistance dominates for $\kappa \ll \lambda$ while the plug approximation remains valid.

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.

For uniaxial extensional flow put $q=\tau \dot{\gamma }>0$, $f=1+\alpha \mathrm{tr}A$ and $T=f-2q$. The diagonal components are $a_{11}=2q/T$, $a_{22}=a_{33}=-q/(T+3q)$. The identity $f-1=\alpha \mathrm{tr}A$ gives the cubic equation in the title. For $\alpha >0$ its physical root has $T>\max (0,1-2q)$ and connects to $T=1$ at rest. Its extensional viscosity is ${\eta }_E=G_0\tau [2/T+1/(T+3q)]$, tending to $3G_0\tau$ at small $q$ and $2G_0\tau /\alpha$ at large $q$. At $\alpha =0$ the stable steady Maxwell extension instead ends at $q=1/2$; the finite high-rate limit must not be extended to that degenerate case.