# 2D bifurcation problem

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 $$\beta \not = 0$$

Hint.

Analyzing the intersections of

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

we can foresight the equilibrium points qualification.

For $$\lambda \approx -1$$ For $$\lambda\approx -5$$ 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 $$\lambda\approx -35$$ For $$\lambda \approx 1$$ NOTE

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

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