Updating the parameter of ODE45

2 views (last 30 days)
alpedhuez
alpedhuez on 26 May 2020
Commented: Star Strider on 26 May 2020
I have an ODE system such as
b=1;
f = @(t,x) -(b/N)*x(1)*x(2);
ODE45
I want to write a loop that changes the value of b. But it does not seem to me that just changing the value of b will not update the equation f thus the solution.
b=2;
f = @(t,x) -(b/N)*x(1)*x(2)
ODE45
What should be done here?
  2 Comments
Star Strider
Star Strider on 26 May 2020
That is the best solution. Include both constants in the argument list:
f = @(t,x,b,N) -(b/N)*x(1)*x(2);
then proceed as in the documentation example.

Sign in to comment.

Accepted Answer

Star Strider
Star Strider on 26 May 2020
Actually, ‘f’ needs to be a (2x1) matrix, since there are two values of ‘x’, and they both need to be defined, so something like this will work:
f = @(t,x,b,N) [x(1); -(b/N)*x(1)*x(2)];
tspan = [0 5];
ic = rand(2,1);
b = rand;
N = 42;
[t,x] = ode45(@(t,x)f(t,x,b,N), tspan, ic);
I am not certain what you are doing, so this may need to be changed. However to integrate two dependent variables, the ODE function has to have that structure, even if you only use the ‘x(:,2)’ result. I would need to know the original differential equation to know if this is correct for it.
  4 Comments
alpedhuez
alpedhuez on 26 May 2020
Edited: alpedhuez on 26 May 2020
This seems to work:
f = @(t,x,b,d,r,N) [-(b/N)*x(1)*x(2);(b/N)*x(1)*x(2)-(r+d)*x(2);r*x(2);d*x(2)]
Star Strider
Star Strider on 26 May 2020
That definitely makes sense!
I am still not certain what you are modeling, however that is definitely the correct way to write the ODE function.

Sign in to comment.

More Answers (0)

Categories

Find more on Programming in Help Center and File Exchange

Tags

Products


Release

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!