Take steady simple shear $u_x=\dot\gamma y$ and define $W_i=\lambda_i\dot\gamma$. Substitution of the symmetric stress components into the full constitutive equation gives, after solving the three coupled algebraic equations,
$$
\tau_{xy}=\eta\dot\gamma
\frac{1+a(2-a)W_1W_2}{1+a(2-a)W_1^2}.
$$
Thus the <Steady shear viscosity of a Johnson--Segalman--Oldroyd fluid> is
$$
\boxed{\eta_{\rm sh}(\dot\gamma)
=\frac{\tau_{xy}}{\dot\gamma}
=\eta\frac{1+a(2-a)\lambda_1\lambda_2\dot\gamma^2}
{1+a(2-a)\lambda_1^2\dot\gamma^2}.}
$$
For the stated small positive $a$, so that $0<a<2$, this decreases from $\eta$ at zero rate to $\eta\lambda_2/\lambda_1$ at high rate exactly when
$$
\boxed{\lambda_1>\lambda_2.}
$$
Equality gives constant viscosity, while $\lambda_2>\lambda_1$ gives shear thickening.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 352 / 2 / c / Solution

{c}
{title2=$\eta_{\rm sh}$}

For Gordon--Schowalter parameter $0<a<2$, the steady shear viscosity is
$$
\eta_{\rm sh}=\eta
\frac{1+a(2-a)\lambda_1\lambda_2\dot\gamma^2}
{1+a(2-a)\lambda_1^2\dot\gamma^2}.
$$
It is shear thinning when $\lambda_1>\lambda_2$.

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