Predictor Corrector Method using MATLAB

MATLAB Programs:

% Adams-Bashforth Predictor Corrector Method

 % Approximate the solution to the initial-value problem

 % dy/dt=y-t^2+1 ; 0<=t<=2 ; y(0)=0.5;

 f = @(t,y) (y-t^2+1);

 a = input('Enter left end ponit, a:  ');

 b = input('Enter right end point, b:  ');

 n = input('Enter no. of subintervals, n: ');               

 alpha = input('Enter the initial condition, alpha:  ');

 h = (b-a)/n;

 t(1) = a;

 w(1) = alpha;

 fprintf('    t         w\n');

 fprintf('%5.4f  %11.8f\n', t(1), w(1));

 for i = 1:3

    t(i+1) = t(i)+h;

    k1 = h*f(t(i), w(i));

    k2 = h*f(t(i)+0.5*h, w(i)+0.5*k1);

    k3 = h*f(t(i)+0.5*h, w(i)+0.5*k2);

    k4 = h*f(t(i+1), w(i)+k3);

    w(i+1) = w(i)+(k1+2.0*(k2+k3)+k4)/6.0;

    fprintf('%5.4f  %11.8f\n', t(i+1), w(i+1));

 end

 for i = 4:n

    t0 = a+i*h;

    part1 = 55.0*f(t(4),w(4))-59.0*f(t(3),w(3))+37.0*f(t(2),w(2));

    part2 = -9.0*f(t(1),w(1));

    w0 = w(4)+h*(part1+part2)/24.0;

    part1 = 9.0*f(t0,w0)+19.0*f(t(4),w(4))-5.0*f(t(3),w(3))+f(t(2),w(2));

    w0 = w(4)+h*(part1)/24.0;

    fprintf('%5.4f  %11.8f\n', t0, w0);

    for j = 1:3

       t(j) = t(j+1);

       w(j) = w(j+1);

    end

    t(4) = t0;

    w(4) = w0;

 end

OUTPUTs:

>> Predictor_corrector_c
Enter left end ponit, a:  0
Enter right end point, b:  2
Enter no. of subintervals, n: 10
Enter the initial condition, alpha:  0.5
    t         w
0.0000   0.50000000
0.2000   0.82929333
0.4000   1.21407621
0.6000   1.64892202
0.8000   2.12720563
1.0000   2.64082860
1.2000   3.17990264
1.4000   3.73235048
1.6000   4.28342082
1.8000   4.81509636
2.0000   5.30537067
>>