rakhi said:
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
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.