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Very common assumption (not important, but exclusively used for simplicity hereafter): constant time step \( t_{n+1}-t_n\equiv\Delta t \)
$$ \begin{align} u^{n+1} &= (1 - a\Delta t )u^n \quad (\hbox{FE}) \tag{15}\\ u^{n+1} &= \frac{1}{1+ a\Delta t} u^n \quad (\hbox{BE}) \tag{16}\\ u^{n+1} &= \frac{1-\half a\Delta t}{1 + \half a\Delta t} u^n \quad (\hbox{CN}) \tag{17}\\ u^{n+1} &= \frac{1 - (1-\theta) a\Delta t}{1 + \theta a\Delta t}u^n \quad (\theta-\hbox{rule}) \tag{18} \end{align} $$
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