# Classify the equlibrium points of the system-Pls help me, its !

1. Oct 29, 2012

### rakhi

Classify the equlibrium points of the system-Pls help me, its urgent!

Hi ,

I have my maths exam tomorow and I am not able to understand the concept to classify the equilibrium points of a system.. I will be grateful if anyone could help me with this problem

Classify the equilibrium points of the system :

x'= )y^3)-4x
y'= (y^3)-y-3x

2. Oct 30, 2012

### haruspex

Re: Classify the equlibrium points of the system-Pls help me, its urgent!

The equilibrium points are where the derivatives are zero, right? So what happens if you put the derivatives as zero in those equations?

3. Oct 30, 2012

### Vargo

Re: Classify the equlibrium points of the system-Pls help me, its urgent!

Hello,

He's asking how to classify them, not how to find them.

4. Oct 30, 2012

### HallsofIvy

Staff Emeritus
Re: Classify the equlibrium points of the system-Pls help me, its urgent!

The first thing to do is "linearize" the system close to each equilibrium point. I presume you have found that the equilibrium points are (0, 0), (2, 2), and (-2, -2).

If x and y are close to 0, then, since powers get smaller and smaller close to 0, the linearization is x'= -4x, y'= -y- 3x. You could actually solve the equations to get $x= Ce^{-4t}$, $y= De^{-t}- Ce^{-4t}$ and see from that that both x and y go to zero as t goes to infinity. Another, perhaps simpler, method is to think of it as a matrix equation:
$$\frac{d}{dt}\begin{pmatrix}x(t) \\ y(t)\end{pmatrix}= \begin{pmatrix}-4 & 0 \\ -3 & -1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$$
and observe that its eigenvalues, which are -4 and -1, are both negative.

In fact the simplest thing to do is to form the "Jacobian" of the left side,
$$\begin{pmatrix}\frac{\partial y^3- 4x}{\partial x} & \frac{\partial y^3- 4x}{\partial y} \\ \frac{\partial y^3- y- 3x}{\partial x} & \frac{\partial y^3- y- 3x}{\partial y}\end{pmatrix}= \begin{pmatrix}-4 & 3y^2 \\ -3 & 3y^2- 1\end{pmatrix}$$
and then set x= 0,y= 0, x= 2, y= 2, x= -2, y= -2, to find the eigenvalues.

For example, setting x= 0, y= 0 gives
$$\begin{pmatrix} -4 & 0 \\ -3 & -1\end{pmatrix}$$
which has eigenvalues -4 and -1 as before.

Setting x= 2, y= 2 gives
$$\begin{pmatrix}-4 & 12 \\ - 3 & 11\end{pmatrix}$$
which has eigenvalues 8 and -1. Since one eigenvalue is positive and the other negative, this is a saddle point. Further, (1, 1) is an eigenvector for eigenvalue 8 and eigenvector (2, 1) associated with eigenvalue -1. So if you draw lines through (2, 2) in those directions, you could get a rough graph of the solutions in that region.

Setting x= -2, y= -2 gives the same matrix (because there is no x and y is squared) so the same eigenvalues and eigenvectors. (-2, 2) is also a saddle point.

Last edited: Oct 30, 2012