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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?

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  • $\begingroup$ what is the stability of the critical point? $\endgroup$ – user147258 May 25 '14 at 3:18
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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 $(u-1)$.

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  • $\begingroup$ So the bifurcation is b=u=1? $\endgroup$ – user147258 May 12 '14 at 0:57

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