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Unformatted text preview: frog scheme is O(h2), so it is certainly consistent. Since the test
equation is linear, it su ces to forgo perturbations and examine solution of
1.5 1 0.5 0 −0.5 −1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Figure 5.6.1: Solution of Example 5.6.1 with = ;2 and h = 1=32. yn = yn;2 + 2h yn;1 y0 = : This is a second-order di erence equation with constant (independent of n) coe cients.
These di erence equations have many similarities to second-order, constant-coe cient
ODEs. In this case, the di erence equation may be solved by assuming a solution of the
yn = c n:
29 Substituting this assumed solution into the leap frog di erence equation yields c n=c n;2 + 2h c n;1 : Seeking non-trivial solutions, it is permissible to divide by c n;2 to obtain
2 ; 2h ; 1 = 0: The roots of this quadratic equation are
12 =h p 1 + (h )2: The two roots satisfy the di erence equation and the general solution is a linear combination of both of them thus,
yn = c1 1 + c2 2 :
The values of c1 and c2 are determined from the initial condition y0 = and the starting
value y1 however, for our purposes, it won't be necessary to explicitly calculate c1 and
c2. If, however, y0 and y1 are O( ) then the c1 and c2 will also be O( ).
Let's obtain more explicit information by assuming that jh j 1. Then, expanding
the square root,
or = 1 + h + O((h )2) 2 = ; 1 ; h + O((h )2)]: yn = c1 1 + h + O((h )2)]n + (;1)nc2 1 ; h + O((h )2)]n
yn = c1e hn + (;1)n c2e; hn + O(n(h )2): The rst term on the right approximates the exact ODE solution which, of course, is y(tn) = e tn = e hn:
The second solution is called \parasitic" or \extraneous" and arises because the rstorder ODE has been approximated by a second-order di erence equation. If > 0, the
(;1)nc2 e; hn
30 decays as n increases and causes little harm. However, if < 0, the parasitic solution
grows with increasing n and eventually dominates the approximation of the exact solution. The parasitic solution oscillates from time step to time step as indicated by the
presence of the (;1)n factor (Figure 5.6.1).
Let's continue our investigation of stability by re-stating the basic concept (Section
2.1) as it applies to a LMM.
De nition 5.6.6. A LMM is stable if there exists an h for each f (t y) such that a change in th...
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