- #1
pondzo
- 169
- 0
Homework Statement
$$\dot{x_1}=x_2-x_2^3,~~~~~~\dot{x_2}=-x_1-3x_2^2+x_1^2x_2+x_2$$
I need help in determining the type and stability of the fixed points in this system.
Homework Equations
The Jordan Normal Form[/B]
Let A be a 2x2 matrix, then there exists a real and non singular matrix M such that ##J=M^{-1}AM##, J is said to be in the jordan normal form.
Linearization Theorem
In the neighbourhood of a fixed point which has a simple linearization, the phase portraits of the non linear system and its linearization are qualitatively the same. This applies only if the Jacobian matrix is non-singular and have non-zero real part at the fixed point.
3. The Attempt at a Solution
The following points are fixed: ##(0,0),(-1,1) \text{ and } (2,1)##
I will call the Jacobian matrix A:
$$A(x_1,x_2)=\begin{bmatrix}0&&1-3x_2^2\\2x_1x_2-1&&x_1^2+1-6x_2\end{bmatrix}$$
I will only be talking about the fixed point ##(0,0)##
$$A(0,0)=\begin{bmatrix}0&&1\\-1&&1\end{bmatrix}$$
The transformation matrix M consists of the eigenvectors of A: $$M=\begin{bmatrix}1&&1\\\frac{1+\sqrt{3}i}{2}&&\frac{1-\sqrt{3}i}{2}\end{bmatrix}\text{ thus, }J=\begin{bmatrix}\frac{1+\sqrt{3}i}{2}&&0\\0&&\frac{1-\sqrt{3}i}{2}\end{bmatrix}$$
My first problem is that the I have not seen this jordan normal form before. The closest form to this that we have discussed is ##J=\begin{bmatrix}\alpha&&-\beta\\\beta&&\alpha\end{bmatrix}##, where ##\alpha## and ##\beta## are real, which is a stable focus. My second problem is that the transformation matrix M is not real, and according to the first theorem M should be real. And even if the theorem I have is incorrect, and M can be complex, I am not sure how I can transform the phase portrait back to the original coordinates using M, since it is a complex transformation. And help would be greatly appreciated! Cheers.
Last edited: