Linear Stability

To determine the stability of the base flow we define small perturbations  

$$q=\bar{q}+\hat{q}$$ (192)

where $\bar{q}$ is the base flow given by (6.1)-(6.2), and $\hat{q}$ is the perturbation. Substituting (6.3) in (6.21)-(6.32) and linearizing we obtain the equations in Appendix 2 with the hats dropped.

Introduce the stream function $\chi$

$$u=\frac{\partial \chi}{\partial \psi},\hspace{.1in} v=\rho\frac{\partial \chi}{\partial \rho},$$

and seek separated solutions of the form

$$(\chi,w)=\rho^{ik}e^{\sigma t}(q_{1},q_{2}),$$

$$(\Sigma,\zeta,\gamma,\Gamma,\Pi,\Delta)=\rho^{ik}e^{\sigma t} (q_{3},q_{4},q_{5},q_{6},q_{7},q_{8})$$

where $q_{i}=q_{i}(\psi)$, k is positive and $\sigma$ may be complex. First solve equations (6.37)-(6.42) for $q_{3}\cdots q_{8}$, substitute these in (6.34)-(6.36) and eliminate p to obtain the following equations

   $$\begin{eqnarray} \lefteqn{(\Omega+1)^{2}(\beta \Omega-\Omega-1)(D^{2}-k^{2})^{2}... ...2}q_{1} -2\beta{\rm De}(\Omega+1)(\Omega+2)D^{2}q_{2},\nonumber \end{eqnarray}$$

and  

$$(D^{2}-k^{2})[(\Omega+1)(\beta \Omega -\Omega-1)q_{2} +ik\beta\mbox{We}q_{1}]=0,$$ (193)

with boundary conditions  

$$q_{1}=q_{1}^{'}=0,\hspace{.1in} q_{2}=0 \hspace{.1in} \mbox{for} \hspace{.1in}\psi=0,1$$ (194)

where $D \equiv \frac{d}{d\psi}$ and $\Omega=\sigma {\rm De}$. From (6.5) and (6.6), we have  

$$(\Omega+1)(\beta \Omega -\Omega -1)q_{2}+ik\beta\mbox{We}q_{1}=0.$$ (195)

Note that $\Omega=-1$ is an eigenvalue. The corresponding eigensolution is stable. Using (6.7) in (6.4) and assuming that $\Omega \neq -1$, we obtain for q₁ the following eigenvalue problem  

$$(D^{2}-k^{2})^{2}q_{1}+\lambda D^{2}q_{1}=0.$$ (196)

and the boundary conditions  

$$q_{1}=q_{1}^{'}=0 \hspace{.1in} \mbox{for}\hspace{.1in} \psi=0,1$$ (197)

where  

$$\lambda=\frac{2ik\beta {\cal E}[(1-\beta)(\Omega^{2}+4\Omega)+3-2\beta]} {(\Omega+1)^{2}(\beta\Omega-\Omega-1)^{2}}.$$ (198)

We have introduced a new elasticity parameter ${\cal E}=\mbox{DeWe}.$ One can show that any eigenvalue $\lambda$ must be positive. It then follows that $\Omega=0$ cannot be an eigenvalue so that no real eigenvalue can pass through the origin.

The eigenvalue problem (6.8)-(6.9) can be solved explicitly. First introduce a change of independent variables $x=\psi -1/2$, then we have to solve (6.8) on the interval $-1/2 \leq x \leq 1/2$ with boundary conditions q₁=q₁^'=0 for $x=\pm1/2$. In that case the eigenfunctions separate into even and odd functions. The even eigenfunctions are given by

$$\varphi_{e}(x)=\cos(\nu_{2}/2)\cos(\nu_{1}x)-\cos(\nu_{1}/2)\cos(\nu_{2}x),$$ (199)

where

$$\nu_{1}=\frac{1}{2}(\mu+\sqrt{\mu^{2}-4k^{2}}), \hspace{.1in} \nu_{2}=\frac{1}{2}(\mu-\sqrt{\mu^{2}-4k^{2}}),$$

are positive and $\mu=\sqrt{\lambda}$ satisfies the transcendental equation  

$$\nu_{1}\tan(\nu_{1}/2)=\nu_{2}\tan(\nu_{2}/2).$$ (200)

The odd eigenfunctions are

$$\varphi_{o}(x)=\sin(\nu_{2}/2)\sin(\nu_{1}x)-\sin(\nu_{1}/2)\sin(\nu_{2}x),$$ (201)

where  

$$\nu_{2}\tan(\nu_{1}/2)=\nu_{1}\tan(\nu_{2}/2).$$ (202)

Equations (6.12) and (6.14) possess an increasing sequence of positive roots which we denote by $\{\lambda_{n}\}$. From (6.10) we then solve for $\Omega$. If $\Re(\Omega)$, the real part of $\Omega$, is negative then the base flow is linearly stable. On the other hand if $\Re(\Omega)\gt$ then it is unstable while it is neutrally stable if $\Re(\Omega)=0$.

In order to determine the critical condition for the onset of instability we solve (6.10) for $\Omega$. For simplicity we consider only the case $\beta=1$. In that case $\Omega$ satisfies the quadratic equation  

$$\lambda\Omega^{2}+2\lambda \Omega-(2ik{\cal E}-\lambda)=0.$$ (203)

For neutral stability set $\Omega=i\Omega_{0}$ where $\Omega_{0}$ is real. In that case (6.15) gives

$$\Omega_{0}=1,$$ (204)

and

$${\cal E}=\frac{\lambda}{k}.$$ (205)

The last equation gives the neutral stability curve in the ${\cal E}$-k plane.

It can be shown [3] that for $\beta= 1/2$, the results are the same as those obtained for $\beta=1$. The neutral stability curves for selected values of $\beta$ are given in figure 6.2.

For a given wave number k, we obtain a critical elasticity number ${\cal E}$ (corresponding to $\lambda=\lambda_{1}$), above which the base flow loses stability. As ${\cal E}$ passes through this critical number a pair of complex conjugate eigenvalues $\Omega$ crosses the imaginary axis into the right half-plane. This is the scenario for a Hopf bifurcation.

 

Figure 6.2: Neutral stability curves in ${\cal E}$-k plane for selected values of the retardation parameter $\beta$. 

For each value of $\beta$ there is a minimum value of ${\cal E}$ above which the base flow loses stability. This critical elasticity number depends only on the viscosity ratio $\beta$ and is given by

$${\cal E}_{c}=\kappa \frac{(2-\beta)\Omega_{0}[(\beta -1)\Omega_{0}^{2}+1]} {\beta [(\beta-1)\Omega_{0}^{2}+3-2\beta)]},$$ (206)

where $\Omega_{0}$ is the imaginary part of the critical eigenvalue and $\kappa =\min \frac{\lambda}{k}=21.1791853188$. This minimum is attained at the critical wave number k_c=3.0905.