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.

I'm trying to find the equilibrium points for the following system: \begin{align} \frac{dx}{dt} &= x-xy \\ \frac{dy}{dt} &= x+a-y^2 \end{align}

For $a = -1.5,-1,-0.5,0,0.5,$ and $1$.

I know there are two bifurcation points, and I used the phase portraits to determine that they were 0 and 1, but I'm not entirely sure that that's correct. Can anyone help me find the bifurcation points of the system at the points defined for $a$?

Thank you!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The critical points of the system are $\big(0,\sqrt{a}\big)$, $\big(0,-\sqrt{a}\big)$, and $\big(1-a,1\big)$. Defining $$ f(x,y) = \pmatrix{x(1-y) \\ x + a - y^2} $$ you have that $$ f(x,y) = f(x_0,y_0) + \pmatrix{(1-y_0) x -x_0 y \\ x -2y_0 y} + \ldots $$ and the linearization of the problem is $$ \vec{X'} = \pmatrix{ x' \\ y' } = \pmatrix{1-y_0 & -x_0 \\ 1 & -2y_0} \pmatrix{x - x_0\\ y - y_0} = \textbf{A} \, \left(\vec{X} - \vec{X}_0\right) $$ To find out if there are bifurcations, one has to look for the eigenvalue changes of $\textbf{A}$.

If $(x_0,y_0) = (0, \sqrt{a})$, the eigenvalues are $$ \lambda_1 = -2\sqrt{a}, \quad \lambda_2 = 1-\sqrt{a} $$ and $a = 0$ and $a = 1$ are bifurcation points (assuming $a \in \mathbb{R}$).

If $(x_0,y_0) = (0, -\sqrt{a})$, the eigenvalues are $$ \lambda_1 = 2\sqrt{a}, \quad \lambda_2 = 1+\sqrt{a} $$ and $a = 0$ is the only bifurcation point

Finally, if $(x_0,y_0) = (1-a,1)$, then $$ \lambda_1 = -1-\sqrt{a}, \quad \lambda_2 = -1 + \sqrt{a} $$ and the bifurcation points are $a = 0$ and $a = 1$.

Summarizing, $a=0,1$ are the bifurcation points.

share|improve this answer
1  
Why is $a = -1$ a bifurcation point for the last equilibrium? –  Antonio Vargas Dec 6 '12 at 4:41
    
@AntonioVargas Because I'm tired and I need sleep? I was on my way to edit it when your comment came out. Thanks for pointing it out. –  Pragabhava Dec 6 '12 at 4:49

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.