Consider a model (very similar to Lotka-volterra prey-predator -model, exception $h$).

$$ \frac{dx}{dt} = h+x(\alpha -\beta y)$$ $$ \frac{dy}{dt} = -y(\gamma - \rho x). $$

Let's write this in matrix form:

$$ \left(\begin{array}{c}\dot{x} \\ \dot{y} \end{array} \right) = \left( \begin{array}{cc} \alpha & -\beta x \\ \rho y & -\gamma \end{array} \right) \left(\begin{array}{c} x \\ y \end{array} \right) + \left(\begin{array}{c} h \\ 0 \end{array} \right) = \left(\begin{array}{c} 0 \\ 0 \end{array} \right) $$

where the final right hand-side is due to the definition of the equilibrium. So we have a system of the form $\bar{0} = A \bar{v} + \bar{h}$. The equation $\dot{f} = A \bar{v} + \bar{h}$ is non-linear ODE, more here, so I need to investigate the $J_{f} := \left[ \frac{\partial f_{i}}{\partial y_{j}} \right]$ where $f(\bar{x}, \bar{y}) = \frac{d \bar{v_{i}}}{d t}$ (read this link, clarifies things a lot!) So the Jacobian is needed for the stability more here. I am trying to solve the problem 5 here.

Clarifications needed, Some Helper Questions

  1. Fix-points? Some material mention things called "fixpoints". $F(\bar{p}) =0$ and then checking some condition, thinking -- really, no need to check it like the history?

  2. linearization? Confused (see the history).

  • 7
    $\begingroup$ Incomprehensible. What do $z$ and $p$ have to do with the rest of the statement? What do you mean by "a situation with log(negative)"? Whatever you mean by that, how does it relate to $x\gt0,y\gt0,h\gg0$? How do you intgrate with respect to time? What is a complex equilibrium? "eigenvalues" only come up when you have a matrix, and you haven't presented one. "Negative real part" - of what? Please think for a few minutes about what you actually want to know, and how to communicate it so that others might understand you. $\endgroup$ Commented Oct 23, 2011 at 5:44
  • 4
    $\begingroup$ @hhh: You seem to completely have misunderstood the question. Equilibrium means that $dx/dt=dy/dt=0$, so you should find $(x,y)$ such that the right-hand sides become zero. Then you investigate the stability of this equilibrium by linearization (and that's where the "negative real part" stuff enters), but I'm not going to explain all that here. It should be in your textbook. $\endgroup$ Commented Oct 23, 2011 at 9:25
  • 1
    $\begingroup$ "Let's write this in matrix form." The matrix form you give yields $dx/dt=h-\mu x+\epsilon y^2$, which is not $dx/dt=h+x(\alpha-\beta y)$, not even close, so best rethink your matrix form. "Linearization" just means dropping the nonlinear terms; in this case, the terms with $xy$. But, really, you're going to have to read this on your own somewhere - there are dozens of textbooks that do systems of first-order ordinary differential equations - or else hire a tutor. $\endgroup$ Commented Oct 23, 2011 at 11:09
  • 3
    $\begingroup$ @hhh: I didn't find much on Wikipedia, except a very short description here, but there are lots of lecture notes and other descriptions on the internet. Search for "linearization ode" or "linear stability analysis ode", for example. $\endgroup$ Commented Oct 23, 2011 at 12:17
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    $\begingroup$ @hhh If you are going to dive into these topics more, then you have to, no you MUST!!! read Strogatz' book. It is a wonderful book and shows all these steps in a progressive Strogatz way. $\endgroup$
    – user13838
    Commented Oct 23, 2011 at 21:35

1 Answer 1


First you have to find the equilibrium points of this system. One possibility is $x=\frac{-h}{\alpha}, y=0$ and another one is $x=\frac{\gamma}{\rho}, y = \frac{\alpha}{\beta}+\frac{\rho h}{\beta\gamma}$ (plug these values in to see it for yourself).

What the idea of linearization simply stating is: Take the 1st-order Taylor series approximation of the solutions. Around some(!) neighborhood of any point in the state space, the nonlinear system $\dot x = f(x)$ behaves like a linear system $\dot x = Ax$. For this you have to use the Jacobian (first term of Taylor approx.) evaluated at the point of linearization. If this point moreover, is an eq. point then the constant term vanishes and you are left with a linear system, otherwise a drift term stays. Thus the overall system describes $$ \frac{d\Delta x}{dt} = A (x_0+\Delta x) $$ i.e. the small perturbation dynamics around the eq. point $x_0$. If $A$ is stable your system goes back to the eq. point or it goes to another eq. point or blows up.

Take the first eq. point as an example: $$ \left.J\right|_{eq1} = \left.\pmatrix{\alpha-\beta y &-\beta x\\ \rho y &\rho x-\gamma}\right|_{x=\frac{-h}{\alpha}, y=0}=\pmatrix{\alpha & \frac{\beta h}{\alpha}\\ 0 &-\rho\frac{h}{\alpha}-\gamma } $$ Note that you have changed the sign of the variables so I can't directly tell you if this is a stable linearization, for example $\alpha = -\mu$. Check if it really is (question 3 asks this). Now you can plug in the second equilibrium point and you will get: $$ \left.J\right|_{eq2} = \left.\pmatrix{\alpha-\beta y &-\beta x\\ \rho y &\rho x-\gamma}\right|_{x=\frac{\gamma}{\rho}, y = \frac{\alpha}{\beta}+\frac{\rho h}{\beta\gamma}}=\pmatrix{\frac{-\rho h}{\gamma} &-\frac{\beta\gamma}{\rho}\\ \frac{\rho\alpha}{\beta}+\frac{\rho^2 h}{\beta\gamma} &0 } $$ The rest is simple since the eigenvalues are the solutions of $$ \lambda^2 + \frac{\rho h}{\gamma}\lambda + (\gamma\alpha +\rho h) = 0 $$ From this you can compute the eigenvalues and see when these variables make the discriminant negative.

  • $\begingroup$ hey but is it really $\rho x - \gamma$ with the Jacobian 2x2 point? Should it be $-\gamma$? Sorry I may have miscorrected your text, have to re-read... $\endgroup$
    – hhh
    Commented Oct 23, 2011 at 19:10
  • 1
    $\begingroup$ @hhh I have fixed it now. Note that they are partial derivatives, hence $-\gamma y +\rho x y$ leads to that. $\endgroup$
    – user13838
    Commented Oct 23, 2011 at 19:13
  • $\begingroup$ ...is it actually possible to prove the steady state by plugging the values into the formula and notice that $\dot{x}=0$ and $\dot{y} = 0$ or do I miss something if I do it that way? $\endgroup$
    – hhh
    Commented Oct 23, 2011 at 21:07
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    $\begingroup$ That would only prove that it is an eq. point. Think of a pendulum. The upright position is also an eq. point but an unstable one. The smallest disturbance cause the solution to converge to the downward solution. $\endgroup$
    – user13838
    Commented Oct 23, 2011 at 21:12
  • $\begingroup$ Wow this became a CW ?! I think you can shorten your question a lot if you get what needs to be done. ;) $\endgroup$
    – user13838
    Commented Oct 23, 2011 at 21:14

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