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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
f (tn yn ) ; f (tn yn
fy (tn yn )
; ; 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
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
6 yn 0.00
0.30 en 1.00000
-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
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)
1. Use (2.2.3) and a Taylor's...
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- Spring '14