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This is solution to one of problems in Numerical Analysis. This is matlab code. Its helpful to students of Computer Science, Electrical and Mechanical Engineering. This code also help to understand algorithm and logic behind the problem. This code includes: Adams, Fourth, Order, Predictor, Corrector, Algorithm, Equally, Space, Points, Interval, Approximation, Bashforth, Function
Typology: Exercises
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% 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','y', '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 an integer > 3 for the number of subintervals\n'); N = input(' '); if N <= 3 fprintf(1,'Number must be at least 4.\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: % STEP 3 AND 4 % compute starting values using Runge-Kutta method T(I+1) = T(I)+H; K1 = HF(T(I), W(I)); K2 = HF(T(I)+0.5H, W(I)+0.5K1); K3 = HF(T(I)+0.5H, W(I)+0.5K2); K4 = HF(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+IH; % predict W(I) Part1 = 55.0F(T(4),W(4))-59.0F(T(3),W(3))+37.0F(T(2),W(2)); Part2 = -9.0F(T(1),W(1)); W0 = W(4)+H(Part1+Part2)/24.0; % correct W(I) Part1 = 9.0F(T0,W0)+19.0F(T(4),W(4))-5.0F(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: 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);