1
$\begingroup$

I am studying the 1961 Fitzugh-Nagumo model paper (download it here), and I am lost in the stability study (p. 450).

Specifically, I do not understand the Taylor series development.

How does he reach the equations (6) p. 450?

From the nullclines (p. 449):

$x = -x + x^3/3 - z$

$y = (a-x)/b$

Fitzhugh reaches by Taylor series (p. 450) ($ ξ = x - x_{1}$ and $ η = y - y_{1} $):

$ dξ/dt = c [η + (1 - x_{1}^2)ξ + x_{1}ξ^2 + ξ^3/3]$

$ dη/dt = -(ξ + bη) / c $

Can someone help me understand?

  • 0
    This looks like a very nice question. It would be even better if you put the equations you reference directly into the question. I know it's a lot of work, but it will make people more likely to help you. Welcome to the site, btw! – 2017-01-31
  • 0
    Hi! Thanks for the feedback. The equations I'm referencing are in the questions. Or maybe I am understanding wrong what you're saying? And thanks!! – 2017-01-31
  • 0
    If everything you reference is in there, then you're good. I just got the sense from the wording that they weren't. – 2017-01-31
  • 0
    Well, $ e $ and $ n $ are actually greek letters in the paper, but I cannot find how to easily do these characters with Markdown... – 2017-01-31
  • 0
    Oh, that's ok. No worries. – 2017-01-31
  • 0
    Cool, thank you! – 2017-01-31
  • 0
    You can write \xi and \eta to get $\xi$ and $\eta$. – 2017-01-31

1 Answers 1

0

You don't get equation (6) from equations (4) and (5); those are only for determining what $x_1$ and $y_1$ are.

What you do is substitute $x(t)=x_1+\xi(t)$ and $y(t)=y_1+\eta(t)$ into the actual system of ODEs, which consists of equations (1) and (2) on p. 447. Then $\dot x(t)$ is replaced by $\dot \xi(t)$ (since $x_1$ is just a constant), $y+x-x^3/3$ is replaced by $(y_1+\eta)+(x_1+\xi)-(x_1+\xi)^3/3$, and so on.

After you've done that, and expanded everything on the right-hand side of the equations, you can simplify them by using that $x_1$ and $y_1$ are solutions of (4) and (5). This is important, since it will make all the constant terms cancel out, leaving only terms which contain powers of $\xi$ and/or $\eta$. And then, to get a linear system, one omits all terms of degree greater than one.

All this is a standard procedure called “linearization at an equilibrium point (or fixed point, or singular point)”, which is explained in any textbook on dynamical systems, so if you're unfamiliar with how to do it or what it's useful for, it's perhaps a good idea to find a book where you can read more about it.

  • 0
    Thank you so much! It definitely helps and I understand the idea, but I still cannot do it myself though. Am I not supposed to do a Taylor series at some point (it is mentioned p.450)? And, in your answer, (how) can I get expressions (only depending on a, b, c, z) for $ x_1 $ and $ y_1 $ thanks to equations (4) and (5)? The cubic is bugging me. What would be the easiest way to proceed? – 2017-02-02
  • 0
    Since the right-hand sides of the ODEs are just polynomials, what you get when you substitute and expand *is* the Taylor series. (It only has finitely many terms in this case.) And you don't have to solve for $x_1$ and $y_1$ explicitly (since this would involve solving a cubic and would be very messy). Just use that you know from (4) and (5) that (some combinations of $x_1$ and $y_1$)$=0$. Exactly these combinations should appear as the constant terms in the Taylor expansions. – 2017-02-02
  • 0
    Okay, now I got it! Thank you so much. I just have some weird coefficients lying around. Some $ 2/3 $ in the $ dξ/dt $ equation and some $ 2 $ in the $ dη/dt $, but anyway, not a big deal. Thanks for your help! – 2017-02-02
  • 0
    My bad, some stupid mistakes. Now it's perfect. Thanks! – 2017-02-02