Math 128A Spring 2003 Week 11 Solutions Burden & Faires 5.6: 1b, 3b, 7, 9, 12 Burden & Faires 5.7: 1b, 3b, 5 Burden & Faires 5.

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1 Math 128A Spring 2003 Week 11 Solutions Burden & Faires 5.6: 1b, 3b, 7, 9, 12 Burden & Faires 5.7: 1b, 3b, 5 Burden & Faires 5.8: 1b, 3b, 4 Burden & Faires 5.6. Multistep Methods 1. Use all the Adams-Bashforth methods to approximate the solutions to the following initial-value problem. In each case use exact starting values, and compare the results to the actual values. y = 1 + (t y) 2, 2 t 3, y(2) = 1, with h = 0.2; actual solution y(t) = t t. The exact starting values are given by: y(t 0 ) = y(t 1 ) = y(t 2 ) = y(t 3 ) = y(t 4 ) = For the Adams-Bashforth two-step method, we have: w i+1 = w i + h 2 [3f(t i, w i ) f(t i 1, w i 1 )] for i 1. Thus we get: w 2 = w 3 = w 4 = w 5 = w 6 = w 7 = w 8 = w 9 = w 10 = For the Adams-Bashforth three-step method, we have: w i+1 = w i + h 12 [23f(t i, w i ) 16f(t i 1, w i 1 ) + 5f(t i 2, w i 2 )] 1

2 for i 2. Thus we get: w 3 = w 4 = w 5 = w 6 = w 7 = w 8 = w 9 = w 10 = For the Adams-Bashforth four-step method, we have: w 3 = w i+1 = w i + h 24 [55f(t i, w i ) 59f(t i 1, w i 1 ) + 37f(t i 2, w i 2 ) 9f(t i 3, w i 3 )] for i 3. Thus we get: w 3 = w 4 = w 5 = w 6 = w 7 = w 8 = w 9 = w 10 = For the Adams-Bashforth five-step method, we have: w 3 = w 4 = w i+1 = w i + h 720 [1901f(t i, w i ) 2774f(t i 1, w i 1 ) f(t i 2, w i 2 ) 1274f(t i 3, w i 3 ) + 251f(t i 4, w i 4 )] 2

3 for i 4. Thus we get: w 3 = w 4 = w 5 = w 6 = w 7 = w 8 = w 9 = w 10 = The actual results are as follows: y(t 0 ) = y(t 1 ) = y(t 2 ) = y(t 3 ) = y(t 4 ) = y(t 5 ) = y(t 6 ) = y(t 7 ) = y(t 8 ) = y(t 9 ) = y(t 10 ) = Use each of the Adams-Bashforth methods to approximate the solution to the following initial-value problem. In each case use starting values obtained from the Runge-Kutta method of order four. Compare the results to the actual values. y = 1 + y/t + (y/t) 2, 1 t 3, y(1) = 0, with h = 0.2; actual solution y(t) = t tan(ln t). The Runge-Kutta method starting values are given by: y(t 0 ) = y(t 1 ) = y(t 2 ) = y(t 3 ) = y(t 4 ) = Thus applying the methods, we get the results: 3

4 t i 2-step 3-step 4-step 5-step exact The initial-value problem has solution y = e y, 0 t 0.20, y(0) = 1 y(t) = 1 ln(1 et). Applying the three-step Adams-Moulton method to this problem is equivalent to finding the fixed point w i+1 of g(w) = w i + h 24 [9ew + 19e wi 5e wi 1 + e wi 2 ]. a. With h = 0.01, obtain w i+1 by functional iteration for i = 2,..., 19 using exact starting values w 0, w 1, and w 2. At each step use w i to initially approximate w i+1. Will Newton s method speed the convergence over functional iteration? Proof. a. Using functional iteration we get: Using Newton s method, we get: t i w i iterations y(t i )

5 t i w i iterations y(t i ) So, with Newton s method, one less iteration is necessary at each step. 9. a. Derive Eq. (5.32) by using the Lagrange form of the interpolating polynomial. Derive Eq. (5.34) by using Newton s backward-difference form of the interpolating polynomial. a. Letting P (t) be the Lagrange form of the interpolating polynomial, we calculate: ti+1 y(t i+1 ) y(t i ) + t i ti+h P (t) dt [ (f(ti, y(t i )) f(t i 1, y(t i 1 )))t = y(t i ) + t i h + f(t ] i, y(t i ))h + f(t i 1, y(t i 1 ))t i f(t i, y(t i ))t i dt h = y(t i ) + 1 2h (f(t i, y(t i )) f(t i 1, y(t i 1 ))) ( (t i + h) 2 t 2 ) i + f(ti, y(t i ))h +f(t i 1, y(t i 1 ))t i f(t i, y(t i ))t i = y(t i ) + h 2 (3f(t i, y(t i )) f(t i 1, y(t i 1 ))) Letting P (t) be Newton s backward-difference form of the interpolating polynomial, we cal- 5

6 culate: ti+1 y(t i+1 ) y(t i ) + t i 3 = y(t i ) + k=0 3 ( ) ( 1) k s k f(t k i, y(t i ))h ds k=0 k f(t i, y(t i ))h( 1) k 1 0 ( ) s ds k [ = y(t i ) + h f(t i, y(t i )) f(t i, y(t i )) + 5 = y(t i ) + h { f(t i, y(t i )) [f(t i, y(t i )) f(t i 1, y(t i 1 ))] 12 2 f(t i, y(t i )) + 3 ] 8 3 f(t i, y(t i )) [f(t i, y(t i )) 2f(t i 1, y(t i 1 )) + f(t i 2, y(t i 2 ))] + 3 } 8 [f(t i, y(t i )) 3f(t i 1, y(t i 1 )) + 3f(t i 2, y(t i 2 )) f(t i 3, y(t i 3 ))] = y(t i ) + h 24 [55f(t i, y(t i )) 59f(t i 1, y(t i 1 )) + 37f(t i 2, y(t i 2 )) 9f(t i 3, y(t i 3 ))] 12. Derive Simpson s method by applying Simpson s rule to the integral y(t i+1 ) y(t i 1 ) = ti+1 t i 1 f(t, y(t)) dt. y(t i+1 ) y(t i 1 ) = ti+1 t i 1 f(t, y(t)) dt h 3 [f(t i 1, y(t i 1 )) + 4f(t i, y(t i )) + f(t i+1, y(t i+1 ))] Burden & Faires 5.7. Variable Step-Size Multistep Methods 1. Use the Adams Variable Step-Size Predictor-Corrector Algorithm with tolerance TOL = 10 4, hmax = 0.25, and hmin = to approximate the solution to the given initial-value problem. Compare the results to the actual values. y = 1 + (t y) 2, 2 t 3, y(2) = 1; actual solution y(t) = t + 1/(1 t). Applying the Adams Variable Step-Size Predictor-Corrector Algorithm, we get the following results: 6

7 t i w i h i y(t i ) Use the Adams Variable Step-Size Predictor-Corrector Algorithm with TOL = 10 6 to approximate the solution to the following initial-value problem: y = 1 + y/t + (y/t) 2, 1 t 3, y(1) = 0; actual solution y(t) = t tan(ln t). Applying the Adams Variable Step-Size Predictor-Corrector Algorithm, we get the following results: t w h y(t)

8 t w h y(t) An electrical circuit consists of a capacitor of constant capacitance C = 1.1 farads in series with a resistor of constant resistance R 0 = 2.1 ohms. A voltage E(t) = 110 sin t is applied at time t = 0. When the resistor heats up, the resistance becomes a function of the current i, and the differential equation for i(t) becomes ( 1 + 2k ) di i R 0 R(t) = R 0 + ki, where k = 0.9, dt + 1 R 0 C i = 1 R 0 C 8 de dt.

9 Find i(2), assuming that i(0) = 0. Running the Adams Variable Step-Size Predictor-Corrector Algorithm with tolerance 10 3 with minimum and maximum mesh size given by and 0.5 respectively, we find that i(2) Burden & Faires 5.8. Extrapolation Methods 1. Use the Extrapolation Algorithm with tolerance TOL = 10 4, hmax = 0.25, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values. y = 1 + (t y) 2, 2 t 3, y(2) = 1; actual solution y(t) = t + 1/(1 t). Applying the Extrapolation Algorithm, we get the following results: t i w i h i k i y(t i ) Use the Extrapolation Algorithm with TOL = 10 6, hmax = 0.5, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values. y = 1 + y/t + (y/t) 2, 1 t 3, y(1) = 0; actual solution y(t) = t tan ln t. Applying the Extrapolation Algorithm, we get the following results: t i w i h i k i y(t i ) Let P (t) be the number of individuals in a population at time t, measured in years. If the average birth rate b is constant and the average death rate d is proportional to the size of the population (due to overcrowding), then the growth rate of the population is given by the logistic equation dp (t) dt = bp (t) k[p (t)] 2, where d = kp (t). Suppose P (0) = 50, 976, b = , and k = Find the population after 5 years. The population after 5 years is approximately

9.6 Predictor-Corrector Methods

SEC. 9.6 PREDICTOR-CORRECTOR METHODS 505 Adams-Bashforth-Moulton Method 9.6 Predictor-Corrector Methods The methods of Euler, Heun, Taylor, and Runge-Kutta are called single-step methods because they use