Linear stability

 

Figure 7.1: Solution of linearized equation for $\beta=0.08$, Ca=21.0 and selected values of ${\rm De}$. The solid line is $\vert\phi/\phi_{0}\vert$, broken lines are $\vert\tau_{11}/\tau_{11}^{0}\vert$,and $\vert\tau_{33}/\tau_{33}^{0}\vert$ with the latter the larger of the two.  

The equations linearized about the ideal uniaxial solution are given in Appendix 2. We consider symmetric flows where W is an odd function of $\xi$ while $\phi, \tau_{11}$ and $\tau_{33}$ are even in $\xi$. First let $\xi =(z+1/2)$ so that the domain becomes $-1/2 \leq z \leq 1/2$ and then seek normal mode solutions of the form

$$W=x_1(t)\sin 2n\pi z,\quad (\phi,\tau_{11},\tau_{33})=(x_{2}(t),x_{3}(t), x_{4}(t)) \cos 2n\pi z$$

Inserting the above in (7.12)-(7.15) we obtain

$$x_{1} =(-a(t)x_{2}+x_{3}-x_{4})/[6n\pi (1-\beta)]$$ (229)

and for ${\rm De} \neq 0$  

$$\frac{d {\bf X}}{d t}={\bf A}{\bf X}$$ (230)

where ${\bf X} =(x_{2}, x_{3}, x_{4})^{T}$ and the matrix ${\bf A}$ is defined as follows

$${\bf A}= \left ( \begin{array} {ccc} a(t)e & -e&e \\ 2a(... ...m De}&2-\frac{1}{{\rm De}} -4c(t)e/{\rm De}\end{array} \right )$$

Here we have $b(t)=\beta +{\rm De}\tau_{11}^{0}$, $c(t)=\beta +{\rm De}\tau_{33}^{0}$ and $e=1/[6(1-\beta)]$. For the Newtonian case, ${\rm De} =0$, the above system reduces to the single equation  

$$\frac{\partial x_{2}}{\partial t}=(1+{\mbox{Ca}^{-1}}e^{t/2})x_{2}.$$ (231)

Since the coefficient of x₂ in (7.6) is positive it follows that the ideal uniaxial elongation flow is unstable for a Newtonian fluid. Because equations (7.5) have variable coefficients a closed form analytical solution is not possible hence we solve them numerically. Results for selected values of the parameters are shown in figure 7.1. From figure 7.1, we see that the flow is stable for De sufficiently large and unstable for De small.