Differential Equations Solutions 111

Differential Equations Solutions 111 - 121 The last two...

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Unformatted text preview: 121 The last two terms can be combined and represent an effective local error. Therefore, the global error is magnified if |(1 − ha/2)/(1 + ha/2)| > 1. Conversely, the method is stable when 1 − ha/2 < 1, 1 + ha/2 which holds for all h > 0. CHALLENGE 20.7. Recall that Euler’s method is yn+1 = yn + hf (tn , yn ), and backward Euler is yn+1 = yn + hf (tn+1 , yn+1 ). P : y = 1 + .1(12 ) = 1.1, E : f = (1.1)2 − .5 = 0.71, C : y = 1 + .1 ∗ .71 = 1.071, E : f = (1.071)2 − 0.5. The predicted value is quite close to the corrected value; this is an indication that the stepsize is small enough to obtain some accuracy in the computed solution. CHALLENGE 20.8. f (t, y ) = 10y 2 − 20. P: y P = y (0) + .1f (0, y (0)) = 1 + .1(−10) = 0. E: f P = f (.1, y P ) = 10 ∗ 0 − 20 = −20. C: y C = y (0) + .1f P = 1 − 2 = −1. E: f C = f (.1, y C ) = 10 − 20 = −10. Note that the predicted and corrected values are quite different, so neither can be trusted; we should reduce the stepsize and recompute. The true value is y (.1) ≈ −0.69. ˜ CHALLENGE 20.9. Suppose y is the result of the predictor and y is the result ˜ of the corrector. Assuming y is much more accurate than y, ˜ y − ytrue ≈ y − y ≡ δ If δ > τ , reduce h and retake the step: ...
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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