# locate any bifurcation in the $2D$ system?

bifurcation for the following $2D$ system: $$\left\{\begin{matrix} x′=ux-y+x^3\\ y'=bx-y \end{matrix}\right.$$

I have got $ux-y+x^3=0,\ y=bx$, then $x=0\ \ \text{and}\ x = \pm \sqrt{b-u}$.

But I don't how to continue to find the bifurcation?

• what is the stability of the critical point? – user147258 May 25 '14 at 3:18

Obviously, for $b<u$ there is only one stationary point at $(x,y)=(0,0)$. The Jacobian at this point is $$\begin{bmatrix} u&-1\\b&-1 \end{bmatrix}$$ with eigenvalues given by $$0=z^2-(u-1)z+(b-u)=(z-\tfrac12(u-1))^2-(\tfrac14(u-1)^2-(b-u)).$$ When crossing the point $b=u$, the sign structure of the eigenvalues changes. But much also depends on the sign of $(u-1)$.