Homework Assignment 7

1a. I wrote a (C) program which uses the Runge-Kutta algorithm to solve the given initial value problem. It outputs the solution for the requested values of t. The output is as follows:

t        y
0.000000 0.500000
0.200000 0.830627
0.400000 1.219705
0.600000 1.662463
0.800000 2.153075
1.000000 2.684424
1.200000 3.247815
1.400000 3.832632
1.600000 4.425906
1.800000 5.011798
2.000000 5.570960

cumlative error for h= 0.005000: 0.638096 255.876437
cumlative error for h= 0.010000: 0.719357 144.590754
cumlative error for h= 0.020000: 0.762486 77.011071
cumlative error for h= 0.040000: 0.789045 40.241273
cumlative error for h= 0.200000: 0.857770 9.435465

1b. The comments in the program explain how it works. The values stored in the array 'N' are chosen to provide the requested time steps for the interval from t=0 to 2. The cumlative error is the sum of the absolute values of the differences between the value of the current solution and the value of the reference solution (h=0.001) at the appropiate time point.

To measure the convergence order, the output for this part contains an extra column, which is the cumlative error for that value of h multiplied by the corresponding value of N. Note that this is proportional to the number that one would obtain by dividing by h, so for our purposes the effect is equivalent. We can see that the values for the cumlative error (which is a global error) are almost constant originally, so that the global error is almost zeroth order in h! (This means that the global error decays as h^zero=1 as h decays, i.e. error doesn't decrease with h.) Since the second column is (proportional to) the global error divided by h, and it grows rapidly with decreasing h, we know that the convergence order is much less than first order. I in fact performed a detailed calculation to estimate the convergence order more accurately, and got something around 0.1. (I did this by investigating how the differences in successive cumlative errors above increases with N.)

Since the Runge-Kutta method is claimed to have fourth order global error, it would appear that something is wrong. To try to determine exactly what is wrong, I tried a number of things.

Compare using cumlative error versus error just at y(2). Using error in y(2) I get slightly greater than order h convergence. This is reasonable, as this is a local error, which should be about a factor of h smaller than the global cumlative error.
Perform same analysis with function exp(t). Here I got global error decay of order h^3. This would correspond to about a fourth order local error decay.
Calculate an anyltical solution to the given differential equation, to see if comparing with the h=0.001 solution rather than the true solution made much difference. The analytical solution is
y(t) = -(1/2)e^t + t^2 + 2t + 1. Comparing each solution to this analytical solution gave very similar results to comparing each solution to the h=0.001 case. This was also true for the function y=e^t.
Considering that this analytical function isn't much different from the simple e^t I don't see why the Runge-Kutta method should yield such a different convergence order. I used identical code in each case.

Many of the extra files in this directory were used to perform these investigations.

2a. I wrote a program which uses the Adams Fourth-Order Predictor Corrector algorithm to solve the same problem. (It only contains comments original to this version.) The output is as follows:

t        y        y ref.   difference
0.000000 0.500000 0.500000 0.000000
0.200000 0.830627 0.829299 0.001328
0.400000 1.219705 1.214088 0.005617
0.600000 1.662463 1.648941 0.013523
0.800000 2.143972 2.127230 0.016742
1.000000 2.661189 2.640859 0.020330
1.200000 3.204792 3.179942 0.024850
1.400000 3.762755 3.732400 0.030355
1.600000 4.320556 4.283484 0.037072
1.800000 4.860453 4.815176 0.045277
2.000000 5.360770 5.305472 0.055298

cumlative error for h= 0.005000: 0.000235  37.622188
cumlative error for h= 0.010000: 0.000937  37.489285
cumlative error for h= 0.020000: 0.003673  36.733170
cumlative error for h= 0.040000: 0.014083  35.207219
cumlative error for h= 0.200000: 0.250392  25.039242

2b. Note that here the extra error column is multiplied by N^2, so that this method is a little worse that order h^2. (It may converge to order h^2 in the limit as h --> 0. Since this method uses the Runge-Kutta method for just a portion, it is reasonable to expect that it will perform better. One would also expect the process of predicting and correcting to lead to faster convergence to the true solution. My results confirm this; the Adams method appears to be two orders faster than the Runge-Kutta. Unfortunately the same problem as in question 1 seems to be present, as I am not getting the claimed fourth order convergence.