Adams bashford predictor corrector, Assignments of Mathematical Methods for Numerical Analysis and Optimization

This is the MATLAB code for the Adams Bashfor predictor corrector method

Typology: Assignments

2019/2020

Uploaded on 04/17/2020

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Question 7
Table of Contents
i) ...................................................................................................................................... 1
ii) ..................................................................................................................................... 1
iii) .................................................................................................................................... 1
iv) .................................................................................................................................... 1
function ............................................................................................................................. 2
i)
f = @(t,y) (t*exp(3*t)-2*y);
a = 0;
b = 1;
n = 5;
alpha = 0;
abm4(f,a,b,alpha,n);
ii)
f = @(t,y) (1+(t-y).^2);
a = 2;
b = 3;
n = 5;
alpha = 1;
abm4(f,a,b,alpha,n);
iii)
f = @(t,y) (1+(y/t));
a = 1;
b = 2;
n = 5;
alpha = 2;
abm4(f,a,b,alpha,n);
iv)
f = @(t,y) (cos(2*t)+sin(3*t));
a = 0;
b = 1;
n = 5;
alpha = 1;
abm4(f,a,b,alpha,n);
1
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Question 7

Table of Contents

i) ...................................................................................................................................... 1 ii) ..................................................................................................................................... 1 iii) .................................................................................................................................... 1 iv) .................................................................................................................................... 1 function ............................................................................................................................. 2

i)

f = @(t,y) (texp(3t)-2*y); a = 0; b = 1; n = 5; alpha = 0; abm4(f,a,b,alpha,n);

ii)

f = @(t,y) (1+(t-y).^2); a = 2; b = 3; n = 5; alpha = 1; abm4(f,a,b,alpha,n);

iii)

f = @(t,y) (1+(y/t)); a = 1; b = 2; n = 5; alpha = 2; abm4(f,a,b,alpha,n);

iv)

f = @(t,y) (cos(2t)+sin(3t)); a = 0; b = 1; n = 5; alpha = 1; abm4(f,a,b,alpha,n);

Question 7

function

function [y t] = abm4(f,a,b,alpha,n)

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

for i = 4:n t0 = a+ih; 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; 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; fprintf('%5.4f %11.8f\n', t0, w0); for j = 1: t(j) = t(j+1); w(j) = w(j+1); end t(4) = t0; w(4) = w0; end end

Published with MATLAB® R2019b