Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am an economist and I am trying to figure out the stability of this system,

$\frac{\partial x}{\partial t} = \frac{a}{y} + x b$

$\frac{\partial y}{\partial t} = C + y c + \frac{d}{x} $

I have really no clues about how to deal with the nonlinearities. Any help? And how to find the equilibrium point??

share|cite|improve this question
you can linearize them near a point $(x_0,y_0)$ where $1/x$ is approx. $x_0-(x-x_0)/x_0^2$ and similarly for $1/y$ – yoyo Mar 22 '11 at 18:29
up vote 4 down vote accepted

There is a single equilibrium point $ x={\frac {ac-db}{Cb}}$, $ y=-{\frac {aC}{ac-db}}$ (assuming the denominators are nonzero). The Jacobian matrix at that point is $\left[ \begin {array}{cc} b&-{\frac { \left( ac-db \right) ^{2}}{a{C}^{2}}}\\ -{\frac {d{C}^{2}{b}^{2}}{ \left( ac-db \right) ^{2}}}&c\end {array} \right]$. This has determinant $D = {\frac {b \left( ac-db \right) }{a}}$ and trace $T = b+c$. If $D > 0$ and $T < 0$ the equilibrium is asymptotically stable. If $D > 0$ and $T = 0$ the linearization is a centre (stable but not asymptotically stable), and the nonlinear system might be either stable or unstable. If $T > 0$ or $D < 0$, the equilibrium is unstable.

share|cite|improve this answer
Thank's a lot!! – Nico Mar 22 '11 at 20:08

To solve explicitly this system, one could use the change of variables $z=xy$ to get the equivalent system $z'=(b+c)z+a+d+Cx$ and $x'=x(b+a/z)$, from which one can deduce a differential equation in $z$ only.

But to study the stability of a differential system $x'=f(x,y)$, $y'=g(x,y)$, one does not need to solve it. Instead, a phase diagram is often sufficient. The idea is to consider the plane $(x,y)$ and to point an arrow at each point $(x,y)$ in the direction of $(f(x,y),g(x,y))$. The path $t\mapsto (x(t),y(t))$ followed by a solution is then visible and its asymptotics (involving the fixed points of the system) may become obvious.

In your case, much will depend on the signs of the parameters $a$, $b$, $c$, $d$ and $C$, and maybe of the initial point $(x(0),y(0))$. My guess is that you are considering positive solutions hence useful conditions might be that $a$, $d$ and $C$ are positive.

For an example and for some more detailed explanations, see here.

share|cite|improve this answer
Thank's a lot, really useful! – Nico Mar 22 '11 at 20:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.