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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??

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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
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2 Answers 2

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.

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Thank's a lot!! –  Nico Mar 22 '11 at 20:08
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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.

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Thank's a lot, really useful! –  Nico Mar 22 '11 at 20:10
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