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Write $q=\dot{\gamma }\tau$ and $f=1+\alpha \mathrm{tr}A$. With $L_{ij}={\partial }_j{v}_i$, the steady homogeneous upper-convected derivative is $-LA-A{L}^T$. Its diagonal and shear equations giveOn the physical branch $f>0$, the other off-diagonal components vanish too. Thus $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ andFor $\alpha \ge 0$ the right side has derivative $1+6\alpha a_{12}^2>0$, so the shear stress determines a unique shear rate. The shear viscosity is $G_0\tau /f$ and decreases with shear magnitude when $\alpha >0$. This is the shear part of shear and pipe flow of an affine linear PTT fluid.

For fully developed pipe flow take $\mathbf{v}=w(r)\hat{\mathbf{z}}$ and the signed axial pressure gradient $dp/dz=\Delta p$. Regularity at the axis and the supplied momentum balance determine the shear stress without using the constitutive equation:Locally the constitutive equations have the same shear form, with axial flow replacing the Cartesian flow direction and $\dot{\gamma }=w^{\prime }(r)$ carrying its sign. HenceIntegrating and imposing no-slip boundary conditions at $r=R$ givesA negative pressure gradient produces positive axial flow. At $\alpha =0$ this is the ordinary Hagen-Poiseuille flow; positive $\alpha$ increases the flux at a fixed pressure drop through shear thinning. The nonzero axial normal stress varies radially, but its axial derivative vanishes in fully developed flow, so it does not change the axial momentum balance used here.

With its material state fixed by a prescribed deformation history, the differential pom-pom model is a simple-fluid constitutive law. The stress depends on the local history of a material element through its molecular stretch and orientation, without gradients of that history between neighbouring elements. A differential representation does not prevent a material from being a simple fluid.

To see the history dependence explicitly, use the component-first velocity gradient $K$, so the tensor evolution is $D_{t}A=KA+A{K}^T-(A-I)/{\tau }_1$. Let $F(t,s)$ be the relative deformation gradient, satisfying ${\partial }_{t}F=K(t)F$ and $F(s,s)=I$. Along a material trajectory, an integrating factor givesThus $A$ is a functional of the local deformation history and its prepared initial state. If the initial $A$ is a positive-definite matrix, it retains that property, so $B=A/\mathrm{tr}A$ is well defined. Given this $B(t)$, the scalar stretch solves $\dot{\lambda }=r(t)\lambda +1/{\tau }_2$, where $r=B:K-1/{\tau }_2$. Its variation of constants formula again depends only on that local history:The differential pom-pom model also satisfies material frame indifference. Under an observer rotation, $A$ and $B$ transform as objective tensors, while $\lambda$ is a scalar. The extra skew part of the transformed velocity gradient has zero contraction with symmetric $B$, so the stretch equation is unchanged. The tensor evolution uses an upper-convected derivative, and the final stress transforms objectively. Initial values cannot be independently forgotten when specifying a finite-past experiment: a fully specified prepared state, such as rest, is part of the history specification.

Write $\mathbf{\tau }$ for the Kirchhoff stress tensor and $S$ for the second Piola-Kirchhoff stress tensor. Since $\mathbf{\tau }=FS{F}^T$ and $\dot{F}=LF$ for the velocity gradient,Therefore the upper-convected rate as a stress push-forward is

For the simple shear flow, put $\gamma =f_{,2}$ and $g=\dot{\gamma }$. The velocity gradient and rate-of-strain tensor areTheir upper-convected derivative givesDenote the symmetric extra stress by $E={\sigma }^d$, and the relaxation time by ${\tau }_r$ to distinguish it from $\mathbf{\tau }$. The Oldroyd-B model in the question becomesFor the steady constitutive response, $E_{22}=E_{33}=E_{13}=E_{23}=0$, $E_{12}=\mu g$ and $E_{11}=2(\mu -{\mu }_r){\tau }_r{g}^2$. HenceA constant imposed shear rate can still have startup transients; these values describe the steady response.

For circular Couette flow $v(r)e_{\theta }$, the velocity gradient in the orthonormal polar basis $(e_r,e_{\theta })$ isFor completeness, in fixed Cartesian coordinates let $\omega (r)=v(r)/r$ and $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Then $v=\omega Jx$ and $L=\omega J+({\omega }^{\prime }/r)(Jx)\otimes x$. Along a particle, the polar basis rotates at angular speed $\omega$. Its contribution to the material derivative cancels the rigid-rotation part of $L$ in the upper-convected derivative. The rotating-frame reduction of circular viscoelastic shear therefore leavesThe flow direction is now $e_{\theta }$ and the gradient direction is $e_r$. With $g=v^{\prime }-v/r$, the same simple-shear constitutive calculation gives

Angular force balance implies $r^2\mu g$ is constant. Integrating $v^{\prime }-v/r\propto r^{-2}$ gives $v=Ar+B/r$. With $v(a)=V_a$ and $v(b)=0$,The radial force balance is $-p^{\prime }-2(\mu -{\mu }_r){\tau }_r{g}^2/r=0$. Thus the inertialess Oldroyd-B circular Couette pressure isThe constant $p_0$ is fixed by a pressure or normal-traction boundary condition.

An objective time derivative transforms as a physical tensor under a time-dependent rigid change of observer. If $x^{\ast }=Q(t)x+c(t)$, with $Q$ orthogonal and $A^{\ast }=QA{Q}^T$, objectivity requiresThis means the constitutive equation makes the same physical prediction in translating and rotating frames, rather than requiring individual tensor components to be invariant.

The printed plus commutator uses the derivative-index-first velocity gradient, $G_{ij}={\partial }_i{u}_j$. This is the transpose of the component-index-first convention $L_{ij}={\partial }_j{u}_i$ often used for the velocity gradient. In this notation $\omega =G-G^T$, and the rate is the Jaumann derivativeTo check the covariance of the Jaumann derivative, let $R=\dot{Q}{Q}^T$. Direct transformation givesThe observer-rotation terms cancel in the displayed corotational rate. Equivalently, in the $L$ convention the physical spin tensor is $W=(L-L^T)/2=-\omega /2$, and the same rate is $D_{t}A-WA+AW$. One must transpose the gradient and reverse the commutator together; using $L$ in the printed plus-sign expression would not be objective.

Another example is the upper-convected derivative, written consistently in the $L$ convention asThe extra terms in $L^{\ast }=QL{Q}^T+R$ cancel those in $D_t^{\ast }{A}^{\ast }$, so this also transforms covariantly. In contrast, the uncorrected material derivative $D_{t}A$ of a second-rank tensor is not objective: its transformation contains $R{A}^{\ast }-A^{\ast }R$. For instance a constant anisotropic tensor in a stationary fluid has zero material derivative in its original frame, but its components have a nonzero commutator derivative in a rotating observer's frame. The ordinary partial time derivative is likewise not an objective tensor rate.

In the Giesekus model, $\lambda$ is the viscoelastic relaxation time: after deformation stops, polymeric stress relaxes on that timescale. The parameter $\eta$ has dimensions of dynamic viscosity and is the model's zero-rate viscosity scale. Since $\lambda {\mathbf{\tau }}^2/\eta$ already has dimensions of stress, $\alpha$ is dimensionless.

The symbol $\stackrel{▽}{\tau }$ is the upper-convected derivativeThe convective and velocity-gradient terms account for translation, rotation, and affine stretching of material elements. They make the constitutive law objective under time-dependent rigid changes of observer.

An objective time derivative transforms as a tensor under every superposed time-dependent rigid translation and rotation of the observer. A constitutive prediction using it is therefore frame-indifferent. The upper-convected derivative, lower-convected derivative, and Jaumann derivative are objective; the componentwise partial time derivative, and generally the uncorrected material derivative of a tensor, are not objective under rotating observers.

The Johnson--Segalman--Oldroyd model uses the Gordon--Schowalter derivativeHere $\eta$ is the zero-shear viscosity, ${\lambda }_1$ is the stress-relaxation time, and ${\lambda }_2$ is the retardation time associated with the more rapidly responding viscous contribution. Passive Oldroyd-type materials normally have $0\le {\lambda }_2\le {\lambda }_1$.

A low-rate steady-shear experiment measures $\eta$. A step strain followed by stress relaxation determines ${\lambda }_1$ from the exponential decay. Creep and recovery, or a frequency sweep in small-amplitude oscillatory shear, then separates ${\lambda }_2$ from ${\lambda }_1$; fitting both storage modulus and loss modulus is especially direct.

Write the velocity gradient aswhere $\mathbf{\Omega }$ is the spin tensor. Expanding the upper-convected derivative in the structure equation givesFor $\xi =0$, the first four terms reproduce the upper-convected derivative. Settingtherefore givesWith polymeric stress ${\mathbf{\tau }}_p=b_1\mathbf{\alpha }$, this is the Oldroyd-B model. For $c_1>0$, its relaxation time and polymer viscosity areIndeed the total stress obeys

For $\xi =2$, the coefficient of
$\dot{\mathbf{\gamma }}\mathbf{\alpha }+\mathbf{\alpha }\dot{\mathbf{\gamma }}$
is $+1/2$, so the objective derivative becomes the lower-convected derivative. Hencerecovers the Oldroyd-A model, again with $c_1>0$ for a finite positive relaxation time. Parameter choices such as $b_1{c}_2=0$ give the degenerate Newtonian limit.

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.

The affine linear PTT fluid considered here uses a symmetric dimensionless tensor $A$ with $\stackrel{\nabla }{A}+(1+\alpha \mathrm{tr}A)A/\tau =2E$ and polymer stress $G_{0}A$. Here $\stackrel{\nabla }{A}$ is the upper-convected derivative, $G_0,\tau >0$ and the nonlinear relaxation parameter $\alpha$ is usually positive. Linearizing about rest gives the Maxwell fluid relaxation modulus $G(s)=G_0{e}^{-s/\tau }$. The PTT family also admits other relaxation functions and nonaffine derivatives; their nonlinear predictions need not equal those of this specified version.

Using the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$, the Rivlin-Ericksen tensors are defined by the displayed recurrence. In particular $A_1$ is twice the rate-of-strain tensor. These tensors transform objectively under time-dependent rigid observer changes. With this convention the upper-convected derivative of $A_1$ is $A_2-2A_1^2$, an identity useful in the slow-flow expansion of a second-order fluid.

For azimuthal velocity $v(r)e_{\theta }$, the velocity gradient in the rotating polar basis is $\left(\begin{array}{cc}0& -v/r\\ v^{\prime }& 0\end{array}\right)$. The basis rotates along a particle at angular speed $v/r$. Accounting for that rotation in the material derivative replaces $L$ by $L-(v/r)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ in the upper-convected derivative. Its remaining shear rate is $v^{\prime }-v/r$, so the local viscoelastic normal-stress difference follows the same constitutive algebra as simple shear flow.

Differentiate the Kirchhoff stress tensor relation $\mathbf{\tau }=FS{F}^T$, using $\dot{F}=LF$ for the velocity gradient. Subtracting the two deformation terms gives the displayed upper-convected derivative. This removes the transport generated solely by deformation of a fixed material stress tensor.