Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following system describing a free-market economy:

$$\frac{dp}{dt}=\alpha r,\qquad \frac{dr}{dt}=-\beta r$$

Where $\alpha$ and $\beta$ are positive constants, $p$ is the price and $r$ the difference between demand and supply. Describe the nature of the fixed points and determine whether the price tends to these fixed points.

Is the best way to approach to consider solving for $p(t)$ and noting that $\alpha$ is $>0$ and so if the difference $r$ is positive as $t\to\infty$ then our $p\to\infty$ and vice versa for all the other possibilities. Many thanks in advance.

share|improve this question
At a fixed point all derivatives are zero. –  Ross Millikan May 2 '12 at 15:41
@RossMillikan so were looking for our steady states? –  user24930 May 2 '12 at 15:47
That is how I understand a fixed point. –  Ross Millikan May 2 '12 at 15:50

1 Answer 1

up vote 1 down vote accepted

Your system has only 1 fixed point, $r=0$. This value makes all derivatives equal to 0. Since your system is a 2nd order LTI dynamical system, you can write it as $\mathbf{\dot{x}}=\mathbf{A}\cdot \mathbf{x}$:

$${\dot{p} \brack \dot{r}}=\left[ \begin{array}{rcl}\begin{array}{cc} 0 & \alpha \\ 0 & -\beta \\ \end{array} \end{array} \right] {p \brack r}$$

The eigenvalues of matrix are $\lambda_1=0$ and $\lambda_2=-\beta$. Since $\beta>0$, you got non-isolated fixed points since every point of the form $(p,0)$ with $(p>0)$ will be a fixed point (line of fixed points). As far as stability is concerned, your fixed points are globally stable since $-\beta<0$.

share|improve this answer

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.