% ADAMS-FOURTH ORDER PREDICTOR-CORRECTOR ALGORITHM 5.4
%
% To approximate the solution of the initial value problem
% y' = f(t,y), a <= t <= b, y(a) = alpha,
% at N+1 equally spaced points in the interval [a,b].
%
% INPUT: endpoints a,b; initial condition alpha; integer N.
%
% OUTPUT: approximation w to y at the (N+1) values of t.
syms('F', 'OK', 'A', 'B', 'ALPHA', 'N', 'FLAG', 'NAME', 'OUP');
syms('H', 'T', 'W', 'I', 'K1', 'K2', 'K3', 'K4', 'T0', 'W0', 'J');
syms('t', 's','Part1','Part2');
TRUE = 1;
FALSE = 0;
T = zeros(1,4);
W = zeros(1,4);
fprintf(1,'This is Adams-Bashforth Predictor Corrector Method\n');
fprintf(1,'Input the function F(t,y) in terms of t and y\n');
fprintf(1,'For example: y-t^2+1 \n');
s = input(' ','s');
F = inline(s,'t','y');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input left and right endpoints on separate lines.\n');
A = input(' ');
B = input(' ');
if A >= B
fprintf(1,'Left endpoint must be less than right endpoint\n');
else
OK = TRUE;
end;
end;
fprintf(1,'Input the initial condition\n');
ALPHA = input(' ');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input a positive integer for the number of subintervals\n');
N = input(' ');
if N <= 0
fprintf(1,'Number must be a positive integer\n');
else
OK = TRUE;
end;
end;
if OK == TRUE
fprintf(1,'Choice of output method:\n');
fprintf(1,'1. Output to screen\n');
fprintf(1,'2. Output to text file\n');
fprintf(1,'Please enter 1 or 2\n');
FLAG = input(' ');
if FLAG == 2
fprintf(1,'Input the file name in the form - drive:\\name.ext\n');
fprintf(1,'For example A:\\OUTPUT.DTA\n');
NAME = input(' ','s');
OUP = fopen(NAME,'wt');
else
OUP = 1;
end;
fprintf(OUP, 'ADAMS-BASHFORTH FOURTH ORDER PREDICTOR CORRECTOR METHOD\n\n');
fprintf(OUP, ' t w\n');
% STEP 1
H = (B-A)/N;
T(1) = A;
W(1) = ALPHA;
fprintf(OUP, '%5.3f %11.7f\n', T(1), W(1));
% STEP 2
for I = 1:3
% STEP 3 AND 4
% compute starting values using Runge-Kutta method
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;
% STEP 5
fprintf(OUP, '%5.3f %11.7f\n', T(I+1), W(I+1));
end;
% STEP 6
for I = 4:N
% STEP 7
% T0, W0 will be used in place of t, w resp.
T0 = A+I*H;
% predict W(I)
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;
% correct W(I)
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;
% STEP 8
fprintf(OUP, '%5.3f %11.7f\n', T0, W0);
% STEP 9
% prepare for next iteration
for J = 1:3
T(J) = T(J+1);
W(J) = W(J+1);
end;
% STEP 10
T(4) = T0;
W(4) = W0;
end;
end;
% STEP 11
if OUP ~= 1
fclose(OUP);
fprintf(1,'Output file %s created successfully \n',NAME);
end;
