Multiple Iterations over a system of linear equations

Question

JD 2020-10-17

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/617048-multiple-iterations-over-a-system-of-linear-equations

评论： JD 2020-10-17

Hello all. I am trying to solve the system of linear equations define by CX = K over multiple iterations (300*delta_t). But my plot only plots the first time step. Can someone help me with understanding how I can fix this in my code below?

Thanks so much!

clear all 
close all 
clc
N=5; 
u(1:1:N) = 0;
u(N+1) = 1;
delta_t = 20;

Below is the for loop im having trouble with

for t = 1:delta_t:300*delta_t
    G(t) = t/20;
    A(t) = G(t)/2;
    B(t) = 1 + G(t);
    
  for j = 2:1:N
        k(j) = ((1-G(t))*u(j))+((G(t)/2)*(u(j+1)+u(j-1)));
  end  
    C = [A(t) B(t) 0 0;
     A(t) B(t) A(t) 0;
     0 A(t) B(t) A(t);
     0 0 A(t) B(t)];
 
    k = [k(2); k(3); k(4); k(5)-A(t)];
    X = C\k;
    x1 = [0; X; 1];
    y = 0:1/5:1;
    
    plot(x1,y)  
end

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Ameer Hamza 2020-10-17

Can you write down your problem in mathematical form?

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Answer 1

Walter Roberson 2020-10-17

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https://ww2.mathworks.cn/matlabcentral/answers/617048-multiple-iterations-over-a-system-of-linear-equations#answer_516653

在 MATLAB Online 中打开

u(1:1:N) = 0;
u(N+1) = 1;

So all of your u values are 0 except for the last one

k(j) = ((1-G(t))*u(j))+((G(t)/2)*(u(j+1)+u(j-1)));

Since all of your u are 0 except for the last one, u(j) is going to be 0 throughout that loop, and u(j+1)+u(j-1) is going to be 0 except when j = N at which point you are indexing u(N+1)+u(N-1) which would be 1-0 which would be 1. 0 times anything is 0, so k(j) is 0 except for when j = N.

For the last case, j = N, we can see that k(N) = G(t)/2 * (1-0) = G(t)/2

    A(t) = G(t)/2;
    %...
    k = [k(2); k(3); k(4); k(5)-A(t)];

We know that k(2), k(3), k(4) are all 0, and that k(5) = G(t)/2 and A(t) = G(t)/2 . G(2)/2 - G(t)/2 = 0. Therefore you are replacing k with a vector of 4 zeros.

The solution for C\k when k is all zero is going to be a vector of 0.

Therefore your solutions are all the same for every iteration, so you are going to end up plotting the same line many times.