4.5. Predictor-Corrector Methods#
Implicit methods are useful for so-called stiff equations, to be discussed later. However, another use of implicit methods is to combine an explicit with an implicit formula to form a predictor-corrector method. A commonly used predictor-corrector` method is the combination of the fourth–order Adams methods:
First a predicted value \(y^{(p)}_{j+1}\) of \(y_{j+1}\) is computed by the Adams–Bashforth formula, then \(y^{(p)}_{j+1}\) is used to give \(f_{j+1}\) (i.e can be denoted as \(f^{(p)}_{j+1}\)). The evaluated value of \(f^{(p)}_{j+1}\) is used in the Adams–Moulton formula to correct the approximation given by the Adams–Bashforth formula, and to evaluate an improved value of \(f_{j+1}\). This is the most common procedure for applying predictor-corrector method used and is denoted by PECE; P for computing the Predictor value of \(y_{j+1}\), E for Evaluating the function \(f^{(p)}_{j+1}\), C for applying the Corrector formula and E for a new evaluation of the function.
It is common to take additional corrector steps in order to improve the solution. For example, one possibility is PECECE (or PE(CE)\(^2\)) procedure (i.e. two fixed point iteration per step). Alternatively, corrector iterations can be carried out until the difference between two successive iterations is less than a pre-specified tolerance.
Example 4.2
Consider the initial–value problem
Find approximate solutions using the Adams–Bashforth–Moulton fourth–order predictor-corrector method, with \(\,h=0.1\,\). Find solution in the interval \([0,~~1]\) and apply only one corrector iteration for each step.
Hint
Step 1 - find the starting values
First, we calculate the starting values from the exact solution \(\,y(x) = e^{-x} + x\,\) using \(\,h = 0.1\,\) (we could use any one–step method to find the starting values, see Matlab programs attached).
Step 2 - calculate the predictor solution
Second, we use the \(4^\text{th}\)–order Adams–Bashforth method to predict a value for \(y_{j+1}\):
\[\begin{aligned} y_{j+1} ~&=~ y_j + \frac{h}{24}\bigl[ 55f_j - 59f_{j-1} + 37f_{j-2} - 9f_{j-3} \bigr] \end{aligned}\]Step 3 - calculate the corrector solution and then use the \(4^\text{th}\)–order Adams–Moulton method:
\[\begin{aligned} y_{j+1} ~&=~ y_j + \frac{h}{24}\bigl[ 9f_{j+1} + 19f_j - 5f_{j-1} + f_{j-2} \bigr], \end{aligned}\]to correct the estimated value of \(y_{j+1}\), using one corrector evaluation only.
Solution
Step 1 - using the exact solution, \(\,y(x) = e^{-x} + x\,\), the calculated starting values \(y_0\), \(y_1\), \(y_2\) and \(y_3\), and the corresponding function values \(f(x,y)\) are shown in the table below:
\(j\)
\(x\)
\(y\)
\(f(x,y)\)
0
0.0
1.0000000
0.0000000
1
0.1
1.0048374
0.0951626
2
0.2
1.0187308
0.1812692
3
0.3
1.0408182
0.2591818
Step 2 - using the starting values we can now advance the solution to \(x = {0.4}\), i.e. to calculate \(y_4\) using the following \(4^\text{th}\) Adams–Bashforth formula:
\[\begin{aligned} y_4^p ~&=~ y_3 + \frac{{h}}{24}\bigl[ 55f_3 - 59f_2 + 37f_1 - 9f_0 \bigr] \end{aligned}\]and then use the \(4^\text{th}\)–order Adams–Moulton formula:
\[\begin{aligned} y_4 ~&=~ y_3 + \frac{{h}}{24}\bigl[ 9f_4^p + 19f_3 - 5f_2 + f_1 \bigr], \end{aligned}\]\[\begin{split}\begin{aligned} y_4^p ~&=~ 1.0408182 + \frac{0.1}{24}\bigl[ 55(0.2591818) - 59(0.1812692) + 37(0.0951626) - 9(0) \bigr] \\ ~&=~ 1.0408182 + \frac{0.1}{24}\bigl[ 14.254998 - 10.694886 + 3.521016 + 0 \bigr] \\ ~&=~ {1.0703229} \end{aligned}\end{split}\]To calculate the function value, we substitute \(\,y_4^p = 1.0703229\,\) and \(\,x = 0.4\,\) in \(\,f(x,y) = -y + x + 1\,\), to get \(f_4^p = 0.3296771\). This completes the \(PE\) stage.
Step 3 - We now use \(\,f_4^p = 0.3296771\) in the corrector formula to obtain a corrector value for \(y_4\):
\[\begin{split}\begin{aligned} y_4^c ~&=~ 1.0408182 + \frac{0.1}{24}\bigl[ 9(0.3296771) + 19(0.2591818) - 5(0.1812692) + 0.0951626 \bigr] \\ ~&=~ 1.0408182 + \frac{0.1}{24}\bigl[ 2.9670937 + 4.9244538 - 0.906346 + 0.0951626 \bigr] \\ ~&=~ {1.0703197} \end{aligned}\end{split}\]We now substitute the corrector value of \(y_{4}^c = 1.0703197\) and \(x = 0.4\) in \(f(x,y) ~=~ -y + x + 1\), to get \(f_4^c = 0.3296803\). This completes the \(CE\) stage. Check this value in the following completed table of results for \(x_j = 0.4\).
We then advance the solution to \(x = 0.5\) by repeating the \(PE\) procedure to calculate \(y^{p}_{5}\) and \(f^{p}_5\), then \(y_{5}\) of corrector, and then \(f_5\), and so on, (i.e. repeating steps 2 and 3). Remember, once the corrector value of \(y\) is calculated for each step in \(x\), we take this as new \(y\) value and cross out the predictor estimates.
The computed solution, including the absolute error, up to \(x=1.0\) are displayed in the following table. (Reproduce all the results using step-by-step \(PECE\) calculation to ensure that you follow the method completely).
\(j\)
\(x_j\)
\(y^{AB}_j\)
\(f^{AB}_j\)
\(y^{AM}_j\)
\(f^{AM}_j\)
\(y_{ex}\)
\(|y_{ex} - y^{AM}_j|\)
4
0.4
1.0703229
0.3296771
1.0703197
0.3296803
1.0703200
0.0000003
5
0.5
1.1065330
0.3934670
1.1065301
0.3934699
1.1065307
0.0000006
6
0.6
1.1488135
0.4511865
1.1488109
0.4511891
1.1488116
0.0000007
7
0.7
1.1965868
0.5034132
1.1965844
0.5034156
1.1965853
0.0000009
8
0.8
1.2493301
0.5506699
1.2493279
0.5506721
1.2493290
0.0000010
9
0.9
1.3065705
0.5934295
1.3065685
0.5934315
1.3065697
0.0000011
10
1.0
1.3678800
0.6321200
1.3678783
0.6321217
1.3678794
0.0000012
Note that the method produces an absolute error of the order \(10^{-7}\) to \(10^{-6}\), which is expected for a \(4^\text{th}\)order ABM method, using \(h = 0.1\) – remember, the error should be of the \(O(h^5)\) from the truncation error terms listed for each ABM formulae above.