Solution Method

By using a first order finite difference approximation wit step size $dt>0$ 7.50 get the form

$$\dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)$$ (148)

and

$$D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'+\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'}$$ (149)

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

$$\dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' - \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}$$ (150)

we have

$$\tau = \eta\hackscore{eff} \cdot \dot{\gamma}$$ (151)

where

$$\eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1} , \eta\hackscore{max}) \eta\hackscore{max} = \left\{ \begin{array}{rcl} \frac{\tau\hacks... ...ot{\gamma}>0 \\ &\mbox{ if } \\ \infty & & \mbox{otherwise} \end{array} \right.$$ (152)

The upper bound $\eta\hackscore{max}$ makes sure that yield condtion 7.56 holds. With this setting the eqaution 7.62 takes the form

$$\sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' - \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)$$ (153)

After inserting 7.66 into 7.57 we get

$$-\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) \... ...F\hackscore{i}- \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}$$ (154)

Combining this with the incomressibilty condition 7.49 we need to solve a Stokes problem as discussed in section 7.1.1 in each time step.

If we set

$$\frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}$$ (155)

we need to solve the nonlinear problem

$$\eta\hackscore{eff} - min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff}) , \eta\hackscore{max}) =0$$ (156)

We use the Newton-Raphson Scheme to solve this problem

$$\eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max}, \eta\hack... ...^{(n)} \cdot \eta'( \tau^{(n)} ) } {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )$$ (157)

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 7.68 it makes sense formulate the iteration 7.70 using $\Theta=\eta^{-1}$. In fact we have

$$\eta' = - \frac{\Theta'}{\Theta^2} \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'$$ (158)

As

$$\left(\frac{1}{\eta^{q}} \right)' = \frac{1-\frac{1}{n^{q}}}{\eta... ...e{t}^q)^{1-\frac{1}{n^{q}}}} = \frac{1-\frac{1}{n^{q}}}{\eta^{q}}\frac{1}{\tau}$$ (159)

we have

$$\Theta' = \frac{1}{\tau} \omega \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}}$$ (160)

which leads to

$$\eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max}, \eta\hack... ...(n)} + \omega^{(n)} } {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })$$ (161)