Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For the following system find the fixed points \begin{cases} x'(t) &= x^2 - y,\\ y'(t) &= x-y. \end{cases}

I got $y=x^2$ and $y=x$.

These are non linear systems and so we need to compute the fixed points at its Jacobian matrix.

However, I am not sure on how to do this since I don't know the stability at the fixed points. Hence, I will not be able to draw a phase portrait for it.

share|improve this question
2  
Can you solve the equation $x^2-x=0$? –  Artem Dec 1 '12 at 19:07
    
i beleive i understand what you are trying to say. Are the fixed points(0,0) and (1,1) since i got x=0,1 after solving the equation –  Saba Di Dec 1 '12 at 19:22
1  
Right. Do you know what the Jacobi matrix is? –  Artem Dec 1 '12 at 19:23
    
yes i have solved the jacobian matrix and have obtained at (0,0) it is stable, spiral at(1,1) it is a saddle point thank you for the help –  Saba Di Dec 1 '12 at 19:29
    
This is correct. You can post the details of your calculations as an answer and accept it in some time. –  Artem Dec 1 '12 at 19:36
add comment

1 Answer

The Jacobian is $$\begin{pmatrix} 2x & -1 \\ 1 & -1\end{pmatrix}$$ which has trace $2x-1$ and determinant $1-2x$.

  • At $(0,0)$, with trace $-1$ and determinant $1$, there is a stable spiral.
  • At $(1,1)$, with trace $1$ and determinant $-1$, there is a saddle point.
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.