# Using 222 and 223 for example we obtain the local

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Unformatted text preview: d implicitly by the backward Euler method (2.2.3), we must generally determine it by solving a nonlinear algebraic equation. This will typically require iteration. Perhaps, the simplest procedure is functional iteration as described by the algorithm of Figure 2.2.4. We will subsequently show that functional iteration converges when d Oh y j y ( )j 1 hf ty < for all that occur during the iteration. Bounding j y j by a Lipschitz constant yields y f L jj1 hL : Unfortunately, is large for sti systems, so the step size would have to be restricted to be small in order to achieve convergence of the iteration. This defeats the purpose of using the backward Euler method. L h 24 Newton's method provides an alternative to functional iteration that usually produces better results. Thus, let ( )= F yn yn ; yn;1 ; ( n n) hf t y : and calculate successive iterates by ( +1) yn or ( +1) yn = () yn ; = () yn () yn () ; ( n( )) ( ) Fy yn ; yn;1 1; with =0 1 Fy ; (n ( () hfy tn yn (0) yn () hf t = ) yn ) ::: =0 1 ::: yn : Remark 1. In practice, fy need not be reevaluated after each step of the iteration, and may approximated by nite di erences as () ( 1) f (tn yn ) ; f (tn yn ): () fy (tn yn ) () ( 1) ; ; yn yn ; (0) Remark 2. Newton iteration converges provided that the initial iterate yn is su ciently close to yn and h is su ciently small. However, typically h can be much larger than that required for functional iteration. Remark 3. It is usually not wise to perform too many Newton iterations per time step. Slow convergence of Newton's method often indicates that the time step is too large thus, it is better to halt the iteration, reduce the time step, and calculate a new solution. The issues raised above will be discussed in more quantitative terms in subsequent chapters. For the moment, let us reconsider the linear problem of Example 2.1.4 that we failed to solve by Euler's method because of the stability restriction (2.1.16) Example 2.2.1. Consider the solution of the test equation y 0 = y y (0) = 1 with = ;100 and = 0 05. Using the backward Euler method (2.2.3), we have h : yn = yn;1 25 + h yn tn n 0 1 2 3 4 5 6 yn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 en 1.00000 0.16667 0.02778 0.00463 0.00077 0.00013 0.00002 0.00000 -0.15993 -0.02773 -0.00463 -0.00077 -0.00013 -0.00002 Table 2.2.1: The solution and error of = , (0) = 1, obtained by the backward Euler method with = ;100 and = 0 05. y h 0 y y : or yn;1 = 1; h t = yn;1 =1 2 0=1 6 Solutions are shown for 2 0 0 3] in Table 2.2.1. Although solutions aren't particularly accurate, they do not display any of the violent oscillations that were observed with the forward Euler method. yn n ::: y : : Problems 1. When ( ) 0, the solution of the backward Euler equation is a much closer approximation to the solution of (2.1.15) than the forward Euler method. For simplicity, suppose is real and negative. Then, the ratio of solutions of the ODE at two successive times satis es ( n) = (tn tn;1 ) = h ( n 1) For 0, h 1. Use (2.2.3) and a Taylor's...
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