Solution Method

By using a first order finite difference approximation wit step size $ dt>0$ 6.41 get the form

$\displaystyle \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)$ (141)

and

$\displaystyle D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'-\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'}$ (142)

where $ \sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With

$\displaystyle \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' + \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{-'}\right)^2}$ (143)

we have

$\displaystyle \tau = \eta\hackscore{eff} \cdot \dot{\gamma}$ (144)

where

$\displaystyle \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1} , \eta\hackscore{max})$    with $\displaystyle \eta\hackscore{max} = \left\{ \begin{array}{rcl} \frac{\tau\hacks...
...ot{\gamma}>0 \\ &\mbox{ if } \\ \infty & & \mbox{otherwise} \end{array} \right.$ (145)

The upper bound $ \eta\hackscore{max}$ makes sure that yield condtion 6.47 holds. With this setting the eqaution 6.53 takes the form

$\displaystyle \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' + \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)$ (146)

After inserting 6.57 into 6.48 we get

$\displaystyle -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) \...
...c{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij}^{'-} \right)\hackscore{,j}$ (147)

Combining this with the incomressibilty condition 6.40 we need to solve a Stokes problem as discussed in section 6.1.1 in each time step.

If we set

$\displaystyle \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}$ (148)

we need to solve the nonlinear problem

$\displaystyle \eta\hackscore{eff} - min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff}) , \eta\hackscore{max}) =0$ (149)

We use the Newton-Raphson Scheme to solve this problem

$\displaystyle \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max}, \eta\hack...
...^{(n)} \cdot \eta'( \tau^{(n)} ) } {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )$ (150)

where $ \eta'$ denotes the derivative of $ \eta$ with respect of $ \tau$ and $ \tau^{(n)} = \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.

Looking at the evaluation of $ \eta$ in 6.59 it makes sense formulate the iteration 6.61 using $ \Theta=\eta^{-1}$. In fact we have

$\displaystyle \eta' = - \frac{\Theta'}{\Theta^2}$    with $\displaystyle \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'$ (151)

As

$\displaystyle \left(\frac{1}{\eta^{q}} \right)' = \frac{n^{q}-1}{\eta^{q}\hacks...
...{(\tau\hackscore{t}^q)^{n^{q}-1}} = \frac{n^{q}-1}{\eta^{q}}\cdot\frac{1}{\tau}$ (152)

we have

$\displaystyle \Theta' = \frac{1}{\tau} \omega$    with $\displaystyle \omega = \sum\hackscore{q}\frac{n^{q}-1}{\eta^{q}}$ (153)

which leads to

$\displaystyle \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max}, \eta\hack...
...(n)} + \omega^{(n)} } {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })$ (154)

esys@esscc.uq.edu.au