Coding Euler's Method!!

Question

Sharif 2014-11-8

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/161808-coding-euler-s-method

回答： Mohammad Abouali 2014-11-9

Hello everyone so I am trying to code Euler's method to solve this matrix A and I can't figure it out. Here is my code so far:

k_AB = 0.4;
k_BC = 0.1;
k_CA = 0.4;
k_ba = 0.2;
k_cb = 0.45;
k_ac = 0.16;
A = [ (-k_AB - k_ac), k_ba, k_CA;
      k_AB, (-k_BC - k_ba), k_cb;
      k_ac, k_BC, (-k_CA - k_cb); ];
x0 = [1;0;0];
h = 0.1;
for t = 0:0.1:1
     xNEW = x0 + h*(A*x0);
      x0 = xNEW
end
x0 = [2;0;0];

It seems that my first step has the 3 correct values, but every step after that is incorrect. If anyone who is familiar with Euler's Explicit Method can help I'd be extremely grateful! I've been trying this for hours on my Friday night :(

So step size = 0.1, initial conditions = [1;0;0] from time interval 0:0.1:1

Any help works!

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Sharif 2014-11-8

I tried solving it by hand and trust my hand-written answers more than my MATLAB code. And can you elaborate about what you said about the derivative of matrix A?

Sharif 2014-11-9

Anybody think they can help? :\

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Answer 1

Mohammad Abouali 2014-11-9

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/161808-coding-euler-s-method#answer_158245

在 MATLAB Online 中打开

Your code is correct. What is x0 = [2;0;0]; at the end though.

Here is a modified version of your code. It is modified in such a way that you plot the changes in time at the end for each element of x

clear;
clc;
close all;
k_AB = 0.4;
k_BC = 0.1;
k_CA = 0.4;
k_ba = 0.2;
k_cb = 0.45;
k_ac = 0.16;
A = [ (-k_AB - k_ac), k_ba, k_CA;
      k_AB, (-k_BC - k_ba), k_cb;
      k_ac, k_BC, (-k_CA - k_cb); ];
x(:,1) = [1;0;0];
h = 0.1;
t=0:h:1;
for i=2:numel(t)
     x(:,i) = x(:,i-1) + h*(A*x(:,i-1));
end
plot(t,x(1,:),'b')
hold on
plot(t,x(2,:),'r')
plot(t,x(3,:),'k')
xlabel('Time')