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The Giesekus model adds the nonlinear anisotropic-drag term $\alpha \lambda {\mathbf{\tau }}^2/\eta$ to the Upper-convected Maxwell model. Positive dimensionless mobility factor $\alpha$ produces shear thinning and a nonzero second normal-stress difference.

The $xy$ component of the stationary constitutive equation isInsert $T_{yy}=-\alpha g{T}_{xy}$ and $T_{xx}=(2\tau -\alpha )g{T}_{xy}$. The coefficient of $g^2{T}_{xy}$ becomes $\tau \alpha +\alpha (\tau -\alpha )=\alpha (2\tau -\alpha )$, givingFor $C=\alpha (2\tau -\alpha )>0$, the small-rate shear stress is $T_{xy}=\mu \dot{\gamma }-\mu C{\dot{\gamma }}^3+O({\dot{\gamma }}^5)$. At large positive rate, $T_{xy}\sim \mu /(C\dot{\gamma })$ and the shear viscosity decreases as $\mu /(C{\dot{\gamma }}^2)$. The shear stress is odd in rate, reaches a positive maximum atand decreases thereafter. The original plot displays this homogeneous constitutive curve. Its decreasing branch can permit shear banding; the local model alone supplies neither an interface-selection rule nor a guaranteed stable homogeneous realization.

This is a Johnson-Segalman fluid without an added Newtonian fluid solvent contribution. To identify conventions explicitly, set $a=1-\alpha /\tau$ and $L=G^T=E+W$, with $W$ the spin tensor. The derivative appearing in the constitutive equation becomesthe Gordon--Schowalter derivative. Its usual slip interval $|a|\le 1$ corresponds to $0\le \alpha \le 2\tau$, while a nonsingular genuinely shear thinning curve requiresAt $\alpha =0$ the model is the Upper-convected Maxwell model; at $\alpha =2\tau$ it is the lower-convected derivative counterpart, and both have constant shear viscosity in this test. At $\alpha =\tau$ the derivative is the Jaumann derivative, giving the corotational Maxwell fluid. Merely requiring $\alpha >0$ is insufficient: for $\alpha =3\tau$ the denominator is $1-3{\tau }^2{\dot{\gamma }}^2$, which vanishes at $|\dot{\gamma }|=1/(\sqrt{3}\tau )$ and then gives a negative steady shear viscosity. There is no solvent-induced large-rate plateau in the stated equation.

For steady uniaxial extension of the Giesekus model, the component-first velocity gradient is diagonal, $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$. For a steady spatially uniform diagonal stress, the material derivative vanishes and the convected terms are $-2\gamma {\sigma }_{11}$ and $+\gamma {\sigma }_{22}$. The off-diagonal equations vanish identically, and symmetry permits equal transverse components. The remaining equations areSet $W=\lambda \gamma \ge 0$, $b=\alpha p_T$ and $s_i={\sigma }_{ii}/p_T$, with $\mu =\lambda p_T$. For $b>0$, the branches continuously connected to zero stress at zero rate areA sufficient nondegenerate range for physical solutions at every nonnegative rate is $0<b<1$. Indeed $M=(1-b)I+bA$ with $A=HC/(k_{B}T)$, so positive conformation implies positive mobility. The longitudinal quadratic equation is nonpositive at $s_1=0$ and has a nonnegative root of a polynomial. The transverse quadratic equation has values $b-1<0$ at $s_2=-1$ and $W\ge 0$ at $s_2=0$, so its larger root of a polynomial lies in $(-1,0]$. Thus the displayed branches satisfy $s_i>-1$, giving $A_i=1+s_i>0$. The other algebraic roots do not provide a physical branch connected to the equilibrium distribution. For constant positive diagonal $A$, the associated stationary configuration distribution is a Gaussian with covariance $C=(k_{B}T/H)A$.

For positive $\alpha$ on the physical branch, the stretching stress remains finite through $W=1/2$. At that rate,At large extension rate,Thus the stress continues to grow, but the uniaxial extensional viscosity ${\eta }_E=({\sigma }_{11}-{\sigma }_{22})/\gamma$ approaches the finite plateau $2\lambda /\alpha$ rather than developing a finite-rate pole. The transverse stress approaches $-p_T$ as its conformation tensor variance becomes small. At small rate the two stresses are $2\mu \gamma$ and $-\mu \gamma$, so ${\eta }_E\to 3\mu$.

When $\alpha =0$, the physical steady branch instead isThe stretching stress diverges as $W\to 1/2$ from below. Above that threshold its algebraic continuation has $1+s_1=1/(1-2W)<0$ and cannot be a conformation covariance. There is no physical finite steady stretching state. The transient equationgives linear growth at $W=1/2$ and exponential growth at larger rates, the Hookean coil-stretch failure of the upper-convected Maxwell limit.

Parameter qualification. Positivity of $\alpha$ alone, without a restriction on $b=\alpha n{k}_{B}T$, does not ensure real physical steady solutions for the entire tensor. For example $b=2$, $W=1$ makes the transverse equation $2s_2^2+2s_2+1=0$, which has no real root. Its diffusion/resistance interpretation is already inadmissible for some conformations. The physically positive-mobility regime stated above avoids that counterexample; the source's positive-$\alpha$ case must not be taken as a guarantee for arbitrary parameters.

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 uniaxial extensional flow with $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$, write $W=\lambda \gamma$, $b=\alpha p_T$, and the stress as $p_T\mathrm{diag}(s_1,s_2,s_2)$. The displayed longitudinal equation is supplemented by $b{s}_2^2+(1+W)s_2+W=0$. Choose the roots tending to zero at $W=0$. For $0<b<1$ these roots have $1+s_i>0$ at every nonnegative rate. The stretching stress is finite at $W=1/2$ and the extensional viscosity $({\sigma }_{11}-{\sigma }_{22})/\gamma$ tends to $2\lambda /\alpha$ as $\gamma \to \infty$. In the zero-$\alpha$ Upper-convected Maxwell model, the physical steady longitudinal branch terminates at $W=1/2$. Arbitrary positive $b$ is insufficient: $b=2,W=1$ gives the impossible real transverse equation $2s_2^2+2s_2+1=0$.

The Upper-convected Maxwell model evolves viscoelastic stress according toIt has constant shear viscosity and predicts a divergent extensional viscosity at a finite extension rate.