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186     4 Initial value problems



























                   Figure 4.10 Dynamic concentration profiles for the activated reaction mechanism with k 2 = 100,
                                               ∗
                                                      ∗
                   obtained using the QSSA (A + M → A + M, A → B).
                   This includes as special cases the explicit (forward) Euler method for θ = 0, the implicit
                   (backward) Euler method for θ = 1, and the Crank–Nicholson method for θ = 1/2.


                   Numerical accuracy and the order of an integration method

                   We begin first with a discussion of accuracy. Let us form Taylor series expansions around
                   x(t k ) and x(t k+1 ) in the forward and reverse time directions respectively, with t k+1 = t k +
                    t,
                                                   ( t) 2     ( t) 3
                                               [k]
                        x(t k+1 ) = x(t k ) + ( t) f x  +  ¨ x(t k ) +  ¨ x˙(t k ) +· · ·  (4.151)
                                                    2           6
                                                   ( t) 2       ( t) 3
                                             [k+1]
                      x(t k ) = x(t k+1 ) − ( t) f x  +  ¨ x(t k+1 ) −  ¨ x˙(t k+1 ) +· · ·  (4.152)
                                                    2            6
                   Multiplying (4.151) by (1 − θ), (4.152) by (−θ), and adding the resulting two equations,
                   we obtain

                                                        [k]       [k+1]
                           x(t k+1 ) − x(t k ) = ( t) (1 − θ) f x  + θ f x  + LE     (4.153)
                   where the local error introduced by truncating the two expansions is
                             ( t) 2                     ( t) 3
                       LE =      [(1 − θ)¨x(t k ) − θ ¨x(t k+1 )] +  [(1 − θ)¨x˙(t k ) + θ ¨x˙(t k+1 )] +· · ·
                               2                          6
                                                                                     (4.154)
                                                                  2
                   As the leading term in the local error is proportional to ( t) , we write
                                               [k]           [k+1]       2
                      x(t k+1 ) = x(t k ) + ( t) θ f x  + (1 − θ) f x  + O[( t) ]    (4.155)
                   By this analysis, we see that at every time step, there is a new local error, proportional
                                                                     [2]
                        2
                                                                 [1]
                   to ( t) , introduced into our numerical state trajectory {x , x ,...}. Over the course of
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