0
$\begingroup$

I need to write these equations in steady state form (time derivatives set to 0), and then substitute in values for the parameters (e,d,q,f). However, I don't understand how to do this as if I set the derivatives to 0, when solving the equations, the parameters no longer appear?

e x'[t] == (1 - x[t])*x[t] + q y[t] - 
   x[t] y[t]
d y'[t] == -q y[t] - x[t] y[t] + 
   2 f z[t] 
z'[t] == x[t] - z[t]
$\endgroup$
1
$\begingroup$

Try rule /. _'[t]->0

eqn = {e x'[t] == (1 - x[t])*x[t] + q y[t] - x[t] y[t],d y'[t] == -q y[t] - x[t] y[t] + 2 f z[t],z'[t] == x[t] - z[t]} /. _'[t]->0
(*{0 == (1 - x[t]) x[t] + q y[t] - x[t] y[t], 0 == -q y[t] - x[t] y[t] + 2 f z[t], 0 == x[t] - z[t]}*)

These equations might be solved for {x[t],y[t],z[t]}

The two parameters e,d are irrelevant for the solution!

$\endgroup$
1
$\begingroup$

The equations reduce to algebraic :

{0 == (1 - x)*x + q y - x y,
0 == -q - x y + 2 f z,
0 == x - z}

You can use Solve or NSolve.

$\endgroup$
3
  • $\begingroup$ I'd divide both sides by d and e first, to not lose them when you set x'[t] and y'[t] to zero. $\endgroup$
    – Chris K
    Mar 16 '20 at 13:00
  • 2
    $\begingroup$ @ChrisK How does that make any difference? Once you start solving the equations, you multiply by d and e and they're gone again. $\endgroup$ Mar 16 '20 at 13:34
  • 1
    $\begingroup$ @SjoerdSmit Umm, good point :) Maybe I was thinking ahead to a stability analysis, where d and e would matter. $\endgroup$
    – Chris K
    Mar 16 '20 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.