Unformatted text preview: gin, we expand f (tn 1 +
cih Yi) using (3.1.6) to obtain
f (tn 1 + cih Yi) = f + ft cih + fy (Yi ; y(tn 1)) + 1 ftt (cih)2 + 2fty (cih)(Yi ; y(tn 1))+
; ; ; ; fyy (Yi ; y(tn 1))2] + O(h3):
; All arguments of f and its derivatives are at (tn 1 y(tn 1)). They have been suppressed
Substituting the exact ODE solution and the above expression into (3.2.3a) yields
; y(tn) = y(tn 1) + h
i=1 ; bi f + ft cih + fy (Yi ; y(tn 1)) + O(h2)]:
; The expansion of Yi ; y(tn 1 ) will, fortunately, only require the leading term thus, using
Yi ; y(tn 1) = h aij f + O(h2):
; ; j =1 Hence, we have y(tn) = y(tn 1) + h
i=1 bi f + ft cih + hffy s
j =1 aij + O(h2)]: Equating terms of this series with the Taylor's series (3.1.8) of the exact solution yields
(3.2.5c) with l = 1, (3.2.5c) with l = 2, and (3.2.7) with k = 2. We have demonstrated
the equivalence of these conditions when (3.2.5b) is satis ed.
Remark 1. The results of Theorem 3.2.1 and conditions (3.2.5) and (3.2.7) apply to
both explicit and implicit methods.
Let us conclude thi...
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- Spring '14
- Numerical Analysis, yn, Tn, Numerical ordinary differential equations