Time Discretization

By applying finite elements to (1) we have derived the semi-discrete problem $\mathchoice{\hbox{\boldmath $\displaystyle M$}} {\hbox{\boldmath $\textstyle M... ...} {\hbox{\boldmath $\scriptstyle 0$}} {\hbox{\boldmath $\scriptscriptstyle 0$}}$, which we now must solve. To do so we introduce a partition $0 = t_0 < t_1 < t_2 < \ldots t_M = T$ of the time interval $0 \leq t \leq T$ into $M$ subintervals of equal length $k$ and replace the time derivative $\mathchoice{\hbox{\boldmath $\displaystyle \dot \xi$}} {\hbox{\boldmath $\text... ...h $\scriptstyle \dot \xi$}} {\hbox{\boldmath $\scriptscriptstyle \dot \xi$}}(t)$ by a difference quotient viz.,

$${\mathchoice{\hbox{\boldmath$\displaystyle M$}} {\hbox{\boldmath$... ...} {\hbox{\boldmath$\scriptstyle 0$}} {\hbox{\boldmath$\scriptscriptstyle 0$}}},$$ (13)

where $\mathchoice{\hbox{\boldmath $\displaystyle \xi$}} {\hbox{\boldmath $\textstyle... ...\boldmath $\scriptstyle \xi$}} {\hbox{\boldmath $\scriptscriptstyle \xi$}}(t_n)$. Rearranging terms we then have the iteration scheme

$$(\mathchoice{\hbox{\boldmath$\displaystyle M$}} {\hbox{\boldmath$... ...oldmath$\scriptstyle \xi$}} {\hbox{\boldmath$\scriptscriptstyle \xi$}}^{(n-1)},$$ (14)

which is also known as the Backward Euler Method. There are different choises of initial data $\mathchoice{\hbox{\boldmath $\displaystyle \xi$}} {\hbox{\boldmath $\textstyle... ...{\boldmath $\scriptstyle \xi$}} {\hbox{\boldmath $\scriptscriptstyle \xi$}}^{0}$ but the simplest is to let $\xi_j^0 = \pi_h u_0(x)$ that is to use the linear interpolant of $u_0(x)$. Thus one starts from the initial data and successivly compute solution approximations $\mathchoice{\hbox{\boldmath $\displaystyle \xi$}} {\hbox{\boldmath $\textstyle... ...boldmath $\scriptstyle \xi$}} {\hbox{\boldmath $\scriptscriptstyle \xi$}}^{(1)}$, $\mathchoice{\hbox{\boldmath $\displaystyle \xi$}} {\hbox{\boldmath $\textstyle... ...boldmath $\scriptstyle \xi$}} {\hbox{\boldmath $\scriptscriptstyle \xi$}}^{(2)}$, $\ldots$ at times $t_1$, $t_2$, $\ldots$ using the algorithm (14).