1a. Use the Runge-Kutta (Order Four) algorithm to solve the initial value problem:
y' = y - (t** 2) + 1
0 <= t ,= 2
y(0) = 0.5
Solutions should be generated for the time points t= 0.0, 0.2, 0.4...2.0
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1b. Provide empirical evidence that the Runge-Kutta (order four) algorithm exhibits fourth-order convergence. Use the following strategy. Solve the above initial value problem using the 0.001, 0.005, 0.010, 0.020 and 0.040 time steps. Assume that the results for the 0.001 time step are correct. Calculate the cumulative errors for each of the 0.005, 0.010, 0.020 and 0.040 time step series relative to the 0.001 time step. Finally,compare the 0.005, 0.010, 0.020 and 0.040 time step cumulative errors to determine the convergence order.
####################################################### # # # Using Runge -Kutta (Order Four) algorithm to solve # # y' = y - (t**2) + 1 # # 0 <= t <=2 , y(0) = 0.5 # # time steps h = 0.001, 0.005, 0.010, 0.020, 0.040 # # # # Assume that the result for 0.001 time step are # # correct and calculate the cummulative errors for # # each of the time steps series relative to the 0.001 # # # ####################################################### ************ when h=0.001, N=2000 ************ This is to be used as a standard one the cumulative errors relative to itself h=0.001 is 0.0 ************** when h=0.005, N=400 ************** the cumulative errors relative to h = 0.001 is 7.0634681437198310E-009 ************** when h=0.01, N=200 ************** the cumulative errors relative to h = 0.001 is 5.6670162784655531E-008 ************** when h=0.02, N=100 ************** the cumulative errors relative to h = 0.001 is 4.5491147537823906E-007 ************** when h=0.04, N=50 ************** the cumulative errors relative to h = 0.001 is 3.6631778929097081E-006"cerror" = the cumulative errors relative to h = 0.001 If we make every cerror/(h**i), i= 1, 2, 3, 4, 5 , and then we will get the following results :
------- the result of that cerror/(h**1) --------- c2[1] (h=0.005) ====> 1.4126936287439662E-06 c3[1] (h=0.01) ====> 5.6670162784655531E-06 c4[1] (h=0.02) ====> 2.2745573768911953E-05 c5[1] (h=0.04) ====> 9.1579447322742702E-05 ------- the result of cerror/(h**2) --------- c2[2] (h=0.005) ====> 2.8253872574879324E-04 c3[2] (h=0.01) ====> 5.6670162784655531E-04 c4[2] (h=0.02) ====> 1.1372786884455977E-03 c5[2] (h=0.04) ====> 2.2894861830685675E-03 ------- the result of cerror/(h**3) --------- c2[3] (h=0.005) ====> 5.6507745149758641E-02 c3[3] (h=0.01) ====> 5.6670162784655524E-02 c4[3] (h=0.02) ====> 5.6863934422279876E-02 c5[3] (h=0.04) ====> 5.7237154576714182E-02 ------- the result of cerror/(h**4) --------- c2[4] (h=0.005) ====> 1.1301549029951728E+01 c3[4] (h=0.01) ====> 5.6670162784655522E+00 c4[4] (h=0.02) ====> 2.8431967211139937E+00 c5[4] (h=0.04) ====> 1.4309288644178546E+00 ------- the result of cerror/(h**5) --------- c2[5] (h=0.005) ====> 2.2603098059903455E+03 c3[5] (h=0.01) ====> 5.6670162784655520E+02 c4[5] (h=0.02) ====> 1.4215983605569969E+02 c5[5] (h=0.04) ====> 3.5773221610446363E+01
By the observation of the results above, when i=3, the value of cerror/(h**3) in all the time steps seems to converge to a same constant, that is, the cumulative errors are very likely to converge when order=3. Thus, we can guess the convergence order equal to 3. However, it may depends on other factors, because we just use the h=0.001 as our standard value of comparision, and it does not mean that this value is the exact correct value.
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2a .Use the Adasm Fourth-Order Predictor-Corrector algorithm to solve the initial value problem:
y' = y - (t** 2) + 1
0 <= t <= 0.5
Solutions should be generated for the time points t = 0.0, 0.2, 0.4 ...2.0
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2b. Provide empirical evidence that the Adams Fourth-Order Predictor-Corrector algorithm exhibits fourth-order convergence. Use the following strategy. Solve the above initial value problem using the 0.001, 0.005, 0.010, 0.020 and 0.040 time step series relative to the 0.001 time step. Finally, compare the 0.005, 0.010, 0.020 and 0.040 time step cumulative errors to determine the convergence order.
####################################################### # # # Using Adams Fourth-Order Predictor-Corrector # # algorithm to solve # # y' = y - (t**2) + 1 # # 0 <= t <=2 , y(0) = 0.5 # # time steps h = 0.001, 0.005, 0.010, 0.020, 0.040 # # # # Assume that the result for 0.001 time step are # # correct and calculate the cummulative errors for # # each of the time steps series relative to the 0.001 # # # ####################################################### ************ when h=0.001, N=2000 ************ This is to be used as a standard one the cumulative errors relative to itself h=0.001 is 0.0 ************** when h=0.005, N=400 ************** the cumulative errors relative to h = 0.001 is 1.3499270368200200E-008 ************** when h=0.01, N=200 ************** the cumulative errors relative to h = 0.001 is 1.0568627406559727E-007 ************** when h=0.02, N=100 ************** the cumulative errors relative to h = 0.001 is 8.0781187472567240E-007 ************** when h=0.04, N=50 ************** the cumulative errors relative to h = 0.001 is 5.9083192336384371E-006"cerror" = the cumulative errors relative to h = 0.001 If we make every cerror/(h**i), i= 1, 2, 3, 4, 5 , and then we will get the following results :
------- the result of cerror/(h**1) --------- c2[1] (h=0.005) ====> 2.6998540736400400E-06 c3[1] (h=0.01) ====> 1.0568627406559727E-05 c4[1] (h=0.02) ====> 4.0390593736283620E-05 c5[1] (h=0.04) ====> 1.4770798084096093E-04 ------- the result of cerror/(h**2) --------- c2[2] (h=0.005) ====> 5.3997081472800801E-04 c3[2] (h=0.01) ====> 1.0568627406559727E-03 c4[2] (h=0.02) ====> 2.0195296868141810E-03 c5[2] (h=0.04) ====> 3.6926995210240232E-03 ------- the result of cerror/(h**3) --------- c2[3] (h=0.005) ====> 1.0799416294560160E-01 c3[3] (h=0.01) ====> 1.0568627406559727E-01 c4[3] (h=0.02) ====> 1.0097648434070905E-01 c5[3] (h=0.04) ====> 9.2317488025600580E-02 ------- the result of cerror/(h**4) --------- c2[4] (h=0.005) ====> 2.1598832589120317E+01 c3[4] (h=0.01) ====> 1.0568627406559727E+01 c4[4] (h=0.02) ====> 5.0488242170354525E+00 c5[4] (h=0.04) ====> 2.3079372006400143E+00 ------- the result of cerror/(h**5) --------- c2[5] (h=0.005) ====> 4.3197665178240641E+03 c3[5] (h=0.01) ====> 1.0568627406559726E+03 c4[5] (h=0.02) ====> 2.5244121085177261E+02 c5[5] (h=0.04) ====> 5.7698430016000358E+01By the observation of the results above, when i=3, the value of cerror/(h**3) in all the time steps seems to converge to a same constant, that is, the cumulative errors are very likely to converge when order=3. Thus, we can guess the convergence order equal to 3. However, as the discussion of 1b., it may depends on other factors, because we just use the h=0.001 as our standard value of comparision, and it does not mean that this value is the exact correct value.
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Ho-Yen Chang