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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]
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2 Answers 2

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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!

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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.

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  • $\begingroup$ 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
    Commented Mar 16, 2020 at 13:00
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    $\begingroup$ @ChrisK How does that make any difference? Once you start solving the equations, you multiply by d and e and they're gone again. $\endgroup$
    – Sjoerd Smit
    Commented Mar 16, 2020 at 13:34
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    $\begingroup$ @SjoerdSmit Umm, good point :) Maybe I was thinking ahead to a stability analysis, where d and e would matter. $\endgroup$
    – Chris K
    Commented Mar 16, 2020 at 14:01

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