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The Computable Differential Equation Lecture ... - Bruce E. Shapiro

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80 CHAPTER 4. IMPROVING ON EULER’S METHOD<br />

Figure 4.5: Result of the forward Euler method to solve y ′ = −100(y−sin t), y(0) = 1<br />

with h = 0.001 (top), h = 0.019 (middle), and h = 0.02 (third). <strong>The</strong> bottom figure<br />

shows the same equation solved with the backward Euler method for step sizes of<br />

h = 0.001, 0.02, 0.1, 0.3, left to right curves, respectively<br />

.<br />

method:<br />

For the scalar test equation this gives<br />

y n = y n−1 + h n f(t n , y n ) (4.98)<br />

y n = y n−1 + hλy n (4.99)<br />

Solving for y n ,<br />

1<br />

y n =<br />

1 − hλ y n−1 (4.100)<br />

<strong>The</strong> absolute stability requirement |y n /y n−1 | < 1 gives<br />

or (substituting z = hλ = x + iy,<br />

|1 − hλ| ≥ 1 (4.101)<br />

(1 − x) 2 + y 2 ≥ 1 (4.102)<br />

Math 582B, Spring 2007<br />

California State University Northridge<br />

c○2007, B.E.<strong>Shapiro</strong><br />

Last revised: May 23, 2007

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