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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 secondorder di erence equation with constant (independent of n) coe cients.
These di erence equations have many similarities to secondorder, constantcoe cient
ODEs. In this case, the di erence equation may be solved by assuming a solution of the
form
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 nontrivial 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,
n
n
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,
1 Thus,
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 secondorder di erence equation. If > 0, the
parasitic solution
(;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 restating 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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 Spring '14
 JosephE.Flaherty
 The Land

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