I come across this problem which is about bifurcation. I am trying to take all the cases. I am expecting Hopf bifurcation to occur here but the last case I could not find the fixed point. Could you please help me?

Consider the vector field on plane \begin{align} \dot{x} &= x - xy +1\\ \label{sys1} \dot{y} &= \alpha y + \beta x^2,\nonumber \end{align}

where $\alpha$ and $\beta$ are parameters. Study the bifurcation of the dynamical system in as much details as possible. You have to study the situation in the neighbourhood of the Bogdonov-Takens bifurcation points.

[Case 1:] When $\alpha=0$ and $\beta =0$ we have not equilibrium point. [Case 2:] Assume $\beta=0$.

\begin{align} \dot{z} &= z - zy +1\\ \label{sys2} \dot{y} &= \alpha y\nonumber \end{align}

When $\alpha=0$ we go back to case 1, however when $\alpha \not = 0$ we have a equilibrium point $(-1,0)$. Shifting the fixed point to the origin by $z= x-1$ we have the system

\begin{equation} \begin{aligned} \dot{x} &= x+y - x y \\ \dot{y} &= \alpha y \end{aligned} \end{equation}

which has the fixed point $(0,0)$. Finding the Jacobian for (\ref{sys3})

$$A_{(0,0)} = \begin{pmatrix} 1&&1\\ 0&& \alpha \end{pmatrix}, $$

we can see that the eigenvalues are $\lambda_1=1$, and $\lambda_2=\alpha$ with the eigenvectors

$$v_1 = \begin{pmatrix} 1\\ 0 \end{pmatrix}, \text{ and } v_2 = \begin{pmatrix} \frac{1}{\alpha -1} \\1 \end{pmatrix}$$ When $\alpha \geq 0$ the equilibrium point (0,0) unstable (source) with two dimension unstable manifold, and when $\alpha <0$ the equilibrium point $(0,0)$ is a hyperbolic unstable (saddle) equilibrium point; and the stable and unstable eigenspaces are;

$$\mathcal{E}^s = \{ (x,y) \in \mathbb{R}^2 | y= \frac{1}{\alpha - 1} x \}, \text{ and } \mathcal{E}^u = \{ (x,y) \in \mathbb{R}^2 | y= 0 \}$$

with one dimension stable manifold and one dimension unstable manifold.

Case 3: Assume $\alpha=0$. When $\beta =0$ we go to case 1. When $\beta \not =0$ we have the system has no fixed points.

Case 4: When $\alpha \not = 0 $, and $\beta \not = 0$


1 Answer 1



Analyzing the intersections of

$$ \cases{ x-x y+1 =0\\ \lambda y +x^2=0 } $$

we can foresight the equilibrium points qualification.

For $\lambda \approx -1 $

enter image description here

For $\lambda\approx -5$

enter image description here

Here the tangency point is solved easily as follows

$$ \lambda x+x^3+\lambda = (x-x_1)^2(x-x_2) $$

so equating to zero the $x$'s powers coefficients

$$ \cases{ \lambda+x_1^2x_2 = 0\\ \lambda -x_1^2-2x_1x_2 = 0\\ 2x_1+x_2=0 } $$

we get $x_1 = -\frac 32, x_2=3, \lambda = -\frac{27}{4}$

For $\lambda\approx -35$

enter image description here

For $\lambda \approx 1$

enter image description here


For $\lambda > 0$ we have one equilibrium point.

For $-\frac{27}{4}<\lambda< 0$ we have one equilibrium point.

For $\lambda < -\frac{27}{4}$ we have three equilibrium points.

For a given equilibrium point $(x_0,y_0)$ the Jacobian is

$$ J=\left( \begin{array}{cc} 1-y_0 & -x_0 \\ 2 x_0 & \lambda \\ \end{array} \right) $$

with eigenvalues

$$ \frac 12\left(\lambda+1-y_0\pm\sqrt{(y_0-1+\lambda)^2-8x_0^2}\right) $$


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