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The leapfrog scheme has been very widely used, but it has some serious disadvantages, as will be discussed later. 2 Non-iterative schemes. 49 m = 1, l = 1 αn – 1 Here there is no gain in accuracy. The highest accuracy (second order) is obtained for = 0 (the leapfrog scheme). m = 1 , l > 1 (Nystrom schemes) We can increase the order of accuracy by choosing appropriate α's if l > 1 . Schemes with m > 1 are not of much interest and will not be discussed here. 2 Implicit schemes ( β ≠ 0 ) Here we should be able to achieve accuracy at least as high as l + 2 .

This is a clue that -------- is an important quantity. ∆x ∆x Suppose that the line through the point ( j∆x, n∆t ) , x – ct = x 0 , where x 0 is a constant, does not lie in the domain of dependence. This is the situation shown in the figure. In general, there is no hope of obtaining smaller discretization error, no matter how c∆t small ∆x and ∆t become, so long as -------- is unchanged, because the true solution ∆x u ( j∆x, n∆t ) depends only on the initial value of u at the single point ( x 0 , 0) which n cannot influence u j .

A famous example of an iterative scheme is the fourth-order Runge-Kutta scheme. This is an excellent scheme when f has a simple form, but it is not economically practical when f is complicated. The scheme is given by: q n+1 n 1 = q + ∆t --- ( k 1 + 2k 2 + 2k 3 + k 4 ) , 6 n k1 ∆ t  -, n + 1--- ∆ t , k 2 = f q + ---------2  2 n k 1 = f ( q , n∆t ) , n k 2 ∆t  -, n + 1--- ∆t , k 3 = f q + ---------2  2 n k 4 = f ( q + k 3 ∆ t, ( n + 1 )∆ t ) . 18) Each of the k ’s can be interpreted as an approximation to f .

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