# Systems of differential equations and equilibrium points...

• May 28th 2011, 04:29 AM
MaverickUK82
Systems of differential equations and equilibrium points...
Hi again guys, hope you can help with this simple question...

I am up to systems of differential equations and their resulting equilibrium points.

Now, when faced with finding their equilibrium points, is the a simple way of knowing how many points we will find or are looking for:

I am currently only up to working on two differential equations. The ones am working on are also linear.

I am guessing that because they both only linear, I will have 2x equilibrium points?
• May 28th 2011, 07:15 AM
TheEmptySet
Quote:

Originally Posted by MaverickUK82
Hi again guys, hope you can help with this simple question...

I am up to systems of differential equations and their resulting equilibrium points.

Now, when faced with finding their equilibrium points, is the a simple way of knowing how many points we will find or are looking for:

I am currently only up to working on two differential equations. The ones am working on are also linear.

I am guessing that because they both only linear, I will have 2x equilibrium points?

In general if you have the the system ODE's

$\frac{dx}{dt}=P(x,y)$

$\frac{dy}{dt}=Q(x,y)$

This equilibrium points are the solutions to the system of equation

$\begin{cases}P(x,y)=0 \\ Q(x,y)=0 \end{cases}$

If you have a plane autonomous system the linear system in P and Q will have exactly one solution, and hence one equilibrium point.
• May 29th 2011, 09:49 AM
HallsofIvy
A system of linear equation has either a unique solution, an infinite number of solutions, or no solution.

That is, a system of any number of linear differential equations has either a single equilibrium point, an infinite number of equilibrium points (forming a subspace), or none.
• May 29th 2011, 01:38 PM
MaverickUK82
That's bad because i've found two equilibrium points for the system I'm working on. Have I explained it correctly.
• May 29th 2011, 05:37 PM
TheEmptySet
Quote:

Originally Posted by MaverickUK82
That's bad because i've found two equilibrium points for the system I'm working on. Have I explained it correctly.

What is the system of ODE's
• May 30th 2011, 02:06 PM
MaverickUK82
x'=-axy+bx
y'=cxy-dy

the x' and y' are supposed to be liebniz notation, i.e., an x with a single dot above it but I can't do it in latex...
• May 30th 2011, 02:15 PM
TheEmptySet
Quote:

Originally Posted by MaverickUK82
x'=-axy+bx
y'=cxy-dy

the x' and y' are supposed to be liebniz notation, i.e., an x with a single dot above it but I can't do it in latex...

First off the equations are not linear! You have products of x and y.

you can you \dot for time derivatives

$\begin{cases}\dot{x}=-axy+bx \\ \dot{y}=cxy-dy \end{cases}$

$P(x,y)=axy+bx=x(ay+b)$ and $Q(x,y)=cxy-dy=y(cx-d)$

and yes there are two critical points