2D bifurcation problem

Solution 1:

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 $

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

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

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