How to run a loop for multiple variations of a variable.

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Westin Messer
Westin Messer 2018-10-22
编辑： Matt J 2018-10-22
Hello, I'm new to MATLAB and coding and I'm trying to vary the parameter 'p' from 0 to 3 in incriminates of 1 inside the loop which should give me three different sets of data for the v_d equation but I can't get it to work. Can anyone see what I am doing wrong? Thanks!
e=0.01; k=4; a=0.1; p = 0:1:3;
i = 0.001;
u=zeros(100000,1);
v=zeros(100000,1);
v_d=zeros(100000,1);
t=zeros(100000,1);
% Initial conditions:
u(1)=0.6;
v(1)=0.0;
v_d(1)=0.0;
t(1)=0;
dt=0.001;
for i=1:1:50000
t(i+1)=t(i)+dt;
u(i+1) = u(i)+ dt*((1/e)*((k*u(i)*(u(i)-a)*(1-u(i)))-v(i)));
v(i+1) = v(i)+ dt*(u(i)-v(i));
v_d(i+1) = v_d(i)+ dt*(u(i)-p*v_d(i));
end
  1 个评论
Matt J
Matt J 2018-10-22
which should give me three different sets of data for the v_d equation
Don't you mean 4 different sets: 0,1,2,3

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采纳的回答

Matt J
Matt J 2018-10-22
v_d=zeros(50000,4); %<--change
t(1)=0;
dt=0.001;
for i=1:1:50000
t(i+1)=t(i)+dt;
u(i+1) = u(i)+ dt*((1/e)*((k*u(i)*(u(i)-a)*(1-u(i)))-v(i)));
v(i+1) = v(i)+ dt*(u(i)-v(i));
v_d(i+1,:) = v_d(i,:)+ dt*(u(i)-p.*v_d(i,:)); %<--change
end
  2 个评论
Westin Messer
Westin Messer 2018-10-22
Thanks! In another part of my code I have:
p = 0:1:3;
upts=(-2:.05:2);
vnullpts_d=upts/p;
And I get the following error:
Matrix dimensions must agree.
Error in BME_721_Midterm (line 50)
vnullpts_d=upts/p;
Do you know how I'd fix this to get four different values of vnullpts?
Matt J
Matt J 2018-10-22
编辑：Matt J 2018-10-22
4 different values per value of upts, you mean? So, you want the result as a 4x81 matrix? If so, one option is
vnullpts_d=upts./p(:);
Although, be mindful that you have a divide-by-zero condition, because of p(1)=0.

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