Non-stationary nonlinear transport model

In a nonlinear model, Eq. (2.8) is modified to

$$\mbox{\boldmath$K$} ( \mbox{\boldmath$r$} ) \mbox{\boldmath$r$} + \mbox{\boldmath$C$} ( \mbox{\boldmath$r$} )\frac{{\rm d}\mbox{\boldmath$r$}}{{\rm d} t} = \mbox{\boldmath$F$}(\mbox{\boldmath$r$}),$$ (2.16)

Time discretization is the same as in but the assumption in Eq. (2.15) is not true anymore. Let us assume that Eq. (2.16) should be satisfied at time $\tau\in\langle t,t+\Delta t \rangle$. By substituting of into Eq. (2.16) leads to the following equation

$$\left[(1-\alpha)\mbox{\boldmath$r$}_t + \alpha\mbox{\boldmath$r$}... ...box{\boldmath$r$}_\tau) = \mbox{\boldmath$F$}_{\tau}(\mbox{\boldmath$r$}_\tau).$$ (2.17)

Eq. (2.17) is non-linear and the Newton method is used to obtain the solution. First, the Eq. (2.17) is transformed into a residual form with the residuum vector $R$$_{\tau}$, which should converge to the zero vector

$$\mbox{\boldmath$R$} _{\tau} = \left[(1-\alpha)\mbox{\boldmath$r$}_t + \alpha\mbox{\bo... ...- \mbox{\boldmath$F$}_{\tau}(\mbox{\boldmath$r$}_\tau) \to \mbox{\boldmath$0$}.$$ (2.18)

A new residual vector at the next iteration, $R$$_\tau^{i+1}$, can determined from the previous residual vector, $R$$_{\tau}^i$, and its derivative simply by linearization. Since the aim is getting an increment of solution vector, $\Delta$$r$$_{\tau}^i$, the new residual vector $R$$_{\tau}^{i+1}$ is set to zero

$$\mbox{\boldmath$R$} _\tau^{i+1} \approx \mbox{\boldmath$R$} _{\tau}^i+\frac{\partial{\mbox{\boldmath$R$}_{\tau}^i}}{\partial\mbox{\boldmath$r$}_t} \Delta\mbox{\boldmath$r$}_{\tau}^i = \mbox{\boldmath$0$},$$ (2.19)

$$\Delta \mbox{\boldmath$r$} _{\tau}^i = - \left[\frac{\partial{\mbox{\boldmath$R$}_{\tau}^i}}{\partial\mbox{\boldmath$r$}_t}\right]^{-1} \mbox{\boldmath$R$}_{\tau}^i.$$ (2.20)

Deriving Eq. (2.18) and inserting to Eq. (2.20) leads to

$$\mbox{\boldmath$\tilde K$} _\tau^i = \frac{\partial{\mbox{\boldmath$R$}_{\tau}^i}}{\partial\mbox{\bold... ...playstyle\frac{1}{\Delta t}$}\mbox{\boldmath$C$}_{\tau}^i(\mbox{\boldmath$r$}),$$ (2.21)

$$\Delta \mbox{\boldmath$r$} _{\tau}^i = - \left[\mbox{\boldmath$\tilde K$}_\tau^i\right]^{-1} \mbox{\boldmath$R$}_{\tau}^i,$$ (2.22)

which gives the resulting increment of the solution vector $\Delta$   $r$$_{\tau}^i$

$$\begin{split}\Delta \mbox{\boldmath$r$}_{\tau}^i = - \left[\m... ...oldmath$F$}_{\tau}(\mbox{\boldmath$r$}_\tau) \Big\},\end{split}$$ (2.23)

and the new total solution vector at time $t + \Delta t$ is obtained in each iteration

$$\mbox{\boldmath$r$} _{t+\Delta t}^{i+1}= \mbox{\boldmath$r$} _{t+\Delta t}^{i} + \Delta \mbox{\boldmath$r$} _{\tau}^i.$$ (2.24)

There are two options how to initialize the solution vector at time $t + \Delta t$. While the first case applies linearization with a known derivative, the second case simply starts from the previous solution vector. The second method in Eq. (2.26) is implemented in OOFEM.

$$\mbox{\boldmath$r$} _{t+\Delta t}^{0} = \mbox{\boldmath$r$} _t + \Delta t\frac{\partial{\mbox{\boldmath$r$}_t}}{\partial t},$$ (2.25)

$$\mbox{\boldmath$r$} _{t+\Delta t}^{0} = \mbox{\boldmath$r$} _t.$$ (2.26)

Note that the matrices $K$$($$r$$_\tau),$   $C$$($$r$$_\tau)$ and the vector $F$$($$r$$_\tau)$ depend on the solution vector $r$$_\tau$. For this reason, the matrices are updated in each iteration step (Newton method) or only after several steps (modified Newton method). The residuum $R$$_\tau^{i}$ and the vector $F$$_\tau($$r$$_\tau)$ are updated in each iteration, using the most recent capacity and conductivity matrices.