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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!

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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.

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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

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