Multiple Iterations over a system of linear equations
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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
1 Comment
Accepted Answer
Walter Roberson
on 17 Oct 2020
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.
More Answers (1)
Rafael Hernandez-Walls
on 17 Oct 2020
I think the problem is where the plot function, maybe you need to put another command to plot in each iteration, something like this:
...
plot(x1,y)
drawnow
end
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