Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetics Modeling

  1. Jul 13, 2006 #1
    In a paper, I encountered a system modeled by six coupled first-order differential equations like so:


    [tex]\frac{dp_i}{dt}=-\beta(p_i-m_i)[/tex] , where i=1,2,3 and j=3,1,2.

    According to the paper, the system has a unique steady state which becomes unstable when [tex]\frac{(\beta+1)^2}{\beta}<\frac{3X^2}{4+2X}[/tex], where X is defined [tex]X=-\frac{\alpha n p^(n-1)}{(1+p^n)^2}[/tex]and p is the solution to [tex]p=\frac{\alpha}{1+p^n}+\alpha_0[/tex].

    Lacking a textbook, I have had very little success in seeing how the steady state was derived. I intend to model a similar system. Can someone point me in the right way to understand these equations or show the derivation outright?

    Thank you in advance.
  2. jcsd
  3. Jul 14, 2006 #2


    User Avatar
    Science Advisor

    Finding the steady state solution (also called an "equilibrium solution") is very simple. The "steady state" solution is one that doesn't change which means that the derivatives are 0. You must have
    [tex]\frac{dm_i}{dt}=-m_i+\frac{\alpha}{1+p^n_j}+\alpha_0= 0[/tex]
    [tex]\frac{dp_i}{dt}=-\beta(p_i-m_i)= 0[/tex]
    From the second equation, we obviously have [itex]p_i= m_i[/itex]
    Putting that into the second equation, we have
    [tex]-p_i}+\frac{\alpha}{1+p^n_j}+\alpha_0= 0[/tex]
    If we let p be the function that satisfies
    [tex]-p}+\frac{\alpha}{1+p^n}+\alpha_0= 0[/tex]
    (just the equation above with pn replaced by p)
    then the steady state solution is pi(t)= mi(t)= p for all i and t.

    The "hard" part is determining when the steady state solution is stable or unstable. It is stable if, when the solution is slightly off the steady state solution, it tends to go toward it, and unstable if it tends to move away. In particular, taking p to be the solution to the equation above,
    if pi>p, [itex]\frac{dp_i}{dt}[/itex] must be negative so the solution will tend downward toward p. Conversely, if pi< p, then the derivative must be postive. Of course, as long as the solution is not the steady state solution, we cannot assume that all the pi(t) are the same not that pi= mi. You can perhaps see from the form of the condition- a complicated equation involving X and then a complicated function defining X itself- that determining when the steady state solution is stable or unstable is not a simple problem!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook