I am studying a biological system (HIV) and arrived at this simplified dynamical system:

\begin{align} x_1' &= a_1 + a_2x_2 - a_1x_2 - a_4x_1 - a_5\frac{1+a_6x_3}{1+a_7x_3}x_1\\ x_2' &= a_5\frac{1+a_6x_3}{1+a_7x_3}x_1 - a_2x_2\\ x_3' &= a_8x_1 - a_9x_3 \end{align} Where as the context and notation dictate: all coefficients $a_i$ are strictly positive. I also showed all $x_i$ to be strictly positive and bounded.

I have obtained locally-asymptotically stability for the unique positive fixed point. I also tried various lyapunov functions, but without much luck. I would truly appreciate any idea to approach this problem, perhaps a form of Lyapunov function, a direct method, etc that I should try.

Additionally, I have done extensive numerical simulations, which suggests the positive steady state is globally asymptotically stable.

Edit: for those interested in the original system - especially to back up the various claims that I made in my question. Please note that the parameters do not match to the system in the question. It's just a convenient way to write the parameters.

\begin{align} x_1' &= a_1x_2 -a_2x_1\\ x_2' &= -(f(y_2) + a_1) x_2 + a_3x_3 + a_2x_1\\ x_3' &= f(y_2)x_2 - a_3x_3\\ y_1' &= b_1x_2 - (b_2+b_3)y_1\\ y_2' &= b_2y_1 - b_4y_2\\ f(y_2) &= \frac{c_1}{c_2} \frac{1 + c_2(y_2/c_3)^n}{1+(y_2/c_3)^n}. \end{align} The only conditions on the parameters are that they are positive and $c_2>1$. Note that the first 3 equations are conservative, hence it can be reduced. And at any time, $x_1 + x_2 + x_3 = 1$. This system is part of a larger system, but is decoupled from the rest.

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    $\begingroup$ Are you sure that there is no misprint in the first line? $a_2x_2-a_3x_2$ (with the same $x_2$) in the first line looks suspicious unless it is your way to say that the coefficient at $x_2$ can have arbitrary sign, which would make suspicious your claim about the invariance of the .positive cone. $\endgroup$
    – fedja
    Apr 28, 2019 at 19:57
  • $\begingroup$ @fedja Thank you so much. There is indeed a misprint. $-a_3x_2$ should be $-a_1 x_2$, where $x_1$ and $x_2$ are both bounded above by $1$. $\endgroup$
    – Paichu
    Apr 28, 2019 at 20:05
  • $\begingroup$ Ah-oh. The various claims you make about the system do not hold for general positive coefficients $a_j$, so, unless you tell us a bit more about your coefficients, it is hard to give you a good advice. Why don't you post the problem exactly as you have it? $\endgroup$
    – fedja
    Apr 28, 2019 at 23:28
  • $\begingroup$ @fedja I added the original system to back up those claims. $\endgroup$
    – Paichu
    Apr 29, 2019 at 1:32
  • $\begingroup$ Last stupid question: it looks like the product $b_2y_1b_4y_2$ in the equation for $y_2'$ should actually be the difference. Am I right? $\endgroup$
    – fedja
    Apr 29, 2019 at 19:11

1 Answer 1


This is more a long comment than an answer, but I know that a similar problem, also originated from (macromolecular) biology, was studied and solved by Gaetano Fichera, Maria Adelaide Sneider and Jeffreys Wyman. The system of ODEs they studied was the following one $$ \left\{ \begin{split} x_1' &= Lc_1^2 - (\rho_1 + 2Lc_1 + Qc_2)x_1 + (P - 2Lc_1)x_2 — Lc_1x_3 + Lx_1^2 \\ &\quad+ (2L + Q)x_1 x_2 + (L + Q)x_1x_3 + Lx_2^2 + Lx_2x_3,\\ \\ x_2' &=Qc_2x_1- \rho_2 x_2 + Nc_1x_3- Qx_1x_2- (N+Q)x_1x_3- Qx_2x_3,\\ \\ x_3' &=Mc_1c_2- Mc_2x_1+\big[\rho_2 - P- M(c_1+c_2)\big]x_2- \big[\rho_3+Nc_1+M(c_1+c_2)\big]x_3\\ &\quad+ Mx_1x_2+ (M+N)x_1x_3+Mx^2_2+(2M+N)x_2x_3 + Mx_2^2. \end{split} \right. $$ where $\rho_1, \rho_2, \rho_3, c_1, c_2 L, M, P, Q$ are positive constants satisfying only the condition $\rho_2 > P$.
This system of ODEs is related to the behavior of the pigment rhodopsin in the presence of light and the problem studied in [1] (of which [2] is a summary of results) is equivalent to prove the existence of a unique critical point $\xi=(\xi_1,\xi_2,\xi_3)$ and its asymptotic stability.
I read about this problem from the obituary of Maria Adelaide Sneider and from the "Last lesson" of Gaetano Fichera: the initial condition $x(0)$ was required to belong to a given polyhedron $K\in\Bbb R^3$, and in [1] the existence and uniqueness of a critical point was proven without restriction on the coefficients, while the asymptotic stability was proved only locally (and globally only under restrictions on the coefficients). The problem was fully solved only ten years after by Maria Adelaide Sneider in the paper [3], which have been recently digitalized by the Italian Mathematical Union. According to Fichera, she was able to find a continuous Lyapunov function having only piecewise continuous first derivatives: then, by decomposing $\Bbb R^3$ in 16 "pyramidal" regions, she was able to prove that, in the intersection of each of them with the polyhedron $K$, conditions for the global asymptotic stability are attained.

[1] Gaetano Fichera, Maria Adelaide Sneider, Jeffreys Wyman (1977), "On the existence of a steady state in a biological system", Atti della Accademia Nazionale dei Lincei. Memorie. Classe di Scienze Fisiche, Matematiche e Naturali, Serie VII, Sezione III, XIV (1): 1–26, Zbl 0414.92004.

[2] Gaetano Fichera, Maria Adelaide Sneider, Jeffreys Wyman (1977a), "On the existence of a steady state in a biological system", PNAS, 74 (10): 4182–4184.

[3] Sneider, Maria Adelaide (1988), "Steady state in a biological system: global asymptotic stability", Atti della Accademia Nazionale dei Lincei. Rendiconti. Classe di Scienze Fisiche, Matematiche e Naturali, Serie VIII (in Italian), LXXXII (2): 345–352, MR 1152650, Zbl 0964.92502.

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    $\begingroup$ Thank you. While they share some similarity, then non-linearity is different. Additionally, the method established in the reference is rather complicated. I believe for the system that I am studying, certain property on its structure can be used to simplify the analysis. $\endgroup$
    – Paichu
    May 1, 2019 at 19:52
  • $\begingroup$ @Paichu, you're welcome. My "answer" is intended only as a report on a problem similar to the one you are studying, nothing more: and I am sure you will find a method better suited for your problem. $\endgroup$ May 1, 2019 at 20:29

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