Math 128A Spring 2003 Week 10 Solutions

Transcription

1 Math 128A Spring 2003 Week 10 Solutions Burden & Faires 5.4: 1a, 3a, 11a, 15, 17 Burden & Faires 5.5: 1a the equation should be y = te 3t 2y), 5 Burden & Faires 5.4. Runge-Kutta Methods 1. Use the Modified Euler method to approximate the solution to the following intial-value problem, and compare the result to the actual value. y = te 3t 2y, 0 t 1, y0) = 0, with h = 0.5; actual solution yt) = 1 5 te3t 1 25 e3t e 2t. Using the modified Euler method, we have: t 0 = 0.0 w 0 = yt 0 ) = )e30.0) 1 25 e30.0) e 20.0) = t 1 = 0.5 w 1 = w 0 + h 2 [ft 0, w 0 ) + ft 1, w 0 + hft 0, w 0 ))] = [f0.0, ) + f0.5, f0.0, ))] = 0.25[0.5e 1.5 ] = yt 1 ) = )e30.5) 1 25 e30.5) e 20.5) = t 2 = 1.0 w 2 = [f0.5, ) + f1.0, f0.5, ))] = [0.5e ) + 1.0e e )))] = yt 2 ) = )e31.0) 1 25 e31.0) e 21.0) = Repeat Exercise 1 using the Midpoint method. 1

2 Using the Midpoint method, we have: t 0 = 0.0 w 0 = yt 0 ) = t 1 = 0.5 w 1 = w 0 + hf t 0 + h 2, w 0 + h ) 2 ft 0, w 0 ) = f0.25, f0.0, )) = 0.5[0.25)e )0)] = yt 1 ) = t 2 = 1.0 w 2 = w 1 + hf t 1 + h 2, w 1 + h ) 2 ft 1, w 1 ) = f0.75, f0.5, )) = [ 0.75)e [ 0.5)e ) ])] = yt 2 ) = Use the Runge-Kutta method of order four to approximate the solution to the following initial-value problem, and compare the result to the actual value. y = y/t y/t) 2, 1 t 2, y1) = 1, with h = 0.1; actual solution yt) = t/1 + ln t). Applying the Runge-Kutta method of order four we obtain the following results. t i w i yt i ) The irreversible chemical reaction in which two molecules of solid potassium dichromate K 2 Cr 2 O 7 ), two molecules of water H 2 O), and three atoms os solid sulfur S) combine to yield three molecules of gas sulfur dioxide SO 2 ), four molecules of solid potassium hydroxide KOH), and two molecules of solid chromic oxide Cr 2 O 3 ) can be represented symbolically by the stoichiometric equation: 2K 2 Cr 2 O 7 + 2H 2 O + 3S rkoh + 2Cr 2 O 3 + 3SO 2. 2

3 If n 1 molecules of K 2 Cr 2 O 7, n 2 molecules of H 2 O, and n 3 molecules of S are originally available, the following differential equation describes the amount xt) of KOH after time t: dx n dt = k 1 x ) 2 n 2 x ) 2 n 3 3x ) 3, where k is the velocity constant of the reaction. If k = , n 1 = n 2 = , and n 3 = , how many units of potassium hydroxide will have been formed after 0.2 s? Running the Runge-Kutta method with 100 subintervals gives and with 1000 subintervals gives , so we see that there will be 2079 units of potassium hydroxide. 17. The Runga-Kutta method of order four can be written in the form w 0 = α, w i+1 = w i + h 6 ft i, w i ) + h 3 ft i + α 1 h, w i + δ 1 hft i, w i )) Find the values of the constants + h 3 ft i + α 2 h, w i + δ 2 hft i + γ 2 h, w i + γ 3 hft i, w i ))) + h 6 ft i + α 3 h, w i + δ 3 hft i + γ 4 h, w i + γ 5 hft i + γ 6 h, w i + γ 7 hft i, w i )))). α 1, α 2, α 3, δ 1, δ 2, δ 3, γ 2, γ 3, γ 4, γ 5, γ 6, and γ 7. w 0 = α k 1 = hft i, w i ) k 2 = hf t i + h 2, w i + 1 ) 2 k 1 = hf t i + h 2, w i + 1 ) 2 hft i, w i ) k 3 = hf t i + h 2, w i + 1 ) 2 k 2 = hf t i + h2, w i + 12 hf t i + h 2, w i + 1 )) 2 hft i, w i ) k 4 = hft i+1, w i + k 3 ) = hf t i+1, w i + hf t i + h2, w i + 12 hf t i + h 2, w i + 1 ))) 2 hft i, w i ) w i+1 = w i + 1 hft i, w i ) + 2hf t i + h 6 2, w i + 1 ) 2 hft i, w i ) +2hf t i + h2, w i + 12 hf t i + h 2, w i + 1 )) 2 hft i, w i ) +hf t i+1, w i + hf t i + h2, w i + 12 hf t i + h 2, w i + 1 )))) 2 hft i, w i ) = w i + h 6 ft i, w i ) + h t 3 f i + h 2, w i + 1 ) 2 hft i, w i ) + h 3 f t i + h 2, w i + 1 t 2 hf i + h 2, w i + 1 )) 2 hft i, w i ) + h t 6 f i+1, w i + hf t i + h2, w i + 12 hf t i + h 2, w i + 1 ))) 2 hft i, w i ) 3

4 Thus we have α 1 = 1 2 α 2 = 1 2 α 3 = 1 δ 1 = 1 2 δ 2 = 1 2 δ 3 = 1 γ 2 = 1 2 γ 3 = 1 2 γ 4 = 1 2 γ 5 = 1 2 γ 6 = 1 2 γ 7 = 1 2 Burden & Faires 5.5. Error Control and the Runge-Kutta Method 1. Use the Runge-Kutta-Fehlberg method with tolerance TOL = 10 4, hmax = 0.25, and hmin = 0.05 to approximate the solution to the following initial-value problem. Compare the results to the actual values. y = te 3t 2y, 0 t 1, y0) = 0; actual solution yt) = 1 5 te3t 1 25 e3t e 2t. Running the Runge-Kutta-Fehlberg method we get the following results. t i w i h i yt i ) In the theory of the spread of contagious disease, a relatively elementary differential equation can be used to predict the number of infective individuals in the population at any time, provided appropriate simplification assumptions are made. In particular, let us assume that all individuals in a fixed poplulation have an equally likely chance of being infected and once infected remain in that state. Suppose xt) denotes the number of susceptible individuals at time t and yt) denotes the number of infectives. It is reasonable to assume that the rate at which the number of infectives changes is proportional to the product of xt) and yt) since the rate depends on both the number of infectives and the number of susceptibles present at that time. If the population is large enough to assume that xt) and yt) are continuous variables, the problem can be expressed y t) = kxt)yt), where k is a constant and xt) + yt) = m, the total population. This equation can be rewritten involving only yt) as y t) = km yt))yt). Assuming that m = 100, 000, y0) = 1000, k = , and that time is measured in days, find an approximation to the number of infective individuals at the end of 30 days. b. The differential equation in part a) is called a Bernoulli equation and it can be transformed into a linear differential equation in ut) = yt)) 1. Use this technique to find the exact solution to the equation, under the same assumptions as in part a), and compare the true value of yt) to the approximation given there. What is lim t yt)? Does this agree with your intuition? 4

5 Running Runge-Kutta-Fehlberg with a tolerance of 10 4, minimum mesh size of 0.1 and maximum mesh size of 1.0, we get y30) = b. Applying the substitution, we get the differential equation: dut) 1 ) dt u t) [ut)] 2 = k = k m ut) 1) ut) 1 m 1 ) 1 ut) ut) u t) = kmut) 1) = kmut) + k ut) = ce kmt + 1 m u0) = c + 1 m c = u0) 1 m ut) = u0) 1 ) e kmt + 1 m m. When m = , u0) = , k = , then the solution is given by ut) = ) e 0.2t Thus, we have and yt) = = 99e 0.2t + 1 ) e 0.2t + 1 y30.0) = lim yt) = t 105 = m. It would be expected that eventually everyone would be infected. 5