Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 49 / 3

For the specified affine linear PTT fluid, perturb about the relaxed state $A=0$, $\mathbf{v}=0$. Products of $A$ with the velocity gradient and the term $\alpha (\mathrm{tr}A)A$ are second order in the perturbation. The material-advection term is also second order. The linearized equation is thereforeWith vanishing stress in the remote past, an integrating factor givesSince ${\sigma }^{\prime }=G_{0}A$, comparison with the linear viscoelastic fluid convolution yields $G(s)=G_0{e}^{-s/\tau }$ and zero-shear viscosity $G_0\tau$. The nonlinear relaxation parameter does not appear at linear order; the response is that of a Maxwell fluid.

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.

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 $G_0,\tau >0$.