Find the limiting circle of a Hopf Bifurcation point of a non linear oscillator This problem deals with the nonlinear oscillator
$$
y''+(y^2-\lambda)y'+(y+1)(y^2+y-\lambda)=0
$$Find the steady states and determine the values of $\lambda$ at which they are stable. Sketch the bifurcation diagram and classify the three bifurcation points.
Find a first-term approximation of the limit cycle that appears at Hopf bifurcation point.
My attemps:
the steady-states are
$$y=-1$$
$$y=-\frac12+\frac12\sqrt{1+4\lambda}$$
$$y=-\frac12-\frac12\sqrt{1+4\lambda}$$
we can draw out the bifurcation diagram as below

According to the solution key in the textbook i am reading, there are three bifurcation points. I can work out that the point $(0,-1)$ as a transcritical point. the details are:
Let
$$y\sim-1+\delta y_1+\cdots$$
substitute into the above differential equation, we have
$$y_1''+(1-\lambda)y_1'-\lambda y_1=0$$
and therefore, the two roots of the corresponding auxiliary equation is $r=\lambda$ and $r=-1$. And therefore, when $\lambda>0$, it is stable while when $\lambda<0$, it is unstable. We conclude that $(0,-1)$ is a transcritical point. Am i correct?
For the second steady state, i tried let
$$y\sim -\frac12+\frac12\sqrt{1+4\lambda}+\delta y_1$$
Substitute into the above differential equation, we have
$$y_1''+\left(\frac12-\frac12\sqrt{1+4\lambda}\right)y_1'+\left(\frac12\sqrt{1+4\lambda}+\frac12+2\lambda\right)y_1=0$$
From here, i worked out the two corresponding roots as well, it turns out that the real part of the two roots are
$$Re(r_-)=Re(r_+)=-\frac14+\frac14\sqrt{1+4\lambda}$$
Therefore, it is unstable when $\lambda>0$ while stable when $\lambda<0$. Am i correct?
Since when $\lambda=0$, the two roots are purely imaginary. Therefore, $(0,0)$ should be a Hobf bifurcation point.
I don't know why $\left(-\frac14,\frac12\right)$ is a saddle point. And I don't know how to work out the limiting circle as well. Anyone can help me work out?
 A: For the third steady state, let
$$y\sim -\frac12-\frac12\sqrt{1+4\lambda}+\delta y_1$$
Substitute into the above differential equation, we have
$$y_1''+\left(\frac12+\frac12\sqrt{1+4\lambda}\right)y_1'-\left(\frac12\sqrt{1+4\lambda}-\frac12-2\lambda\right)y_1=0$$
The two roots are $$r_\pm=-\frac14-\frac14\sqrt{1+4\lambda}\pm\frac12\sqrt{\frac52\sqrt{1+4\lambda}-\frac32-7\lambda}$$ we see that $r_+\ge0$ if $\lambda\in\left[-\frac14,0\right]$ and $r_+<0$ if $\lambda\in\left(0,\infty\right)$, while $r_-<0$ for $\lambda\in\left[-\frac14,\infty\right)$. Therefore, this steady state is unstable if $\lambda\in\left[-\frac14,0\right]$ and stable if $\lambda\in\left(0,\infty\right)$. From here we conclude that $\left(-\frac14,-\frac12\right)$ is a saddle-node and $\left(0,-1\right)$ is a transcritical point.
For the limiting circle, let $\lambda=\varepsilon>0$, and $y\sim -\frac12+\frac12\sqrt{1+4\lambda}+\delta y_1$ After subtituting into the orginal differential equation we have
$$y_1''+\left(\frac12-\frac12\sqrt{1+4\varepsilon}\right)y_1'+\left(\frac12\sqrt{1+4\varepsilon}+\frac12+2\varepsilon\right)y_1=0$$ By binomial expansion we have the following differential equation,
$$y_1''-\varepsilon y_1'+\left(1+3\varepsilon\right)y_1=0$$This equation is weakly damped for small $\varepsilon$. Based on experience, a multi-scale expansion with time scale $t_1=t$ and $t_2=\varepsilon t$ are used. and by assuming that $y_1\sim Y_1\left(t_1,t_2\right)+\varepsilon Y_2\left(t_1,t_2\right)+\cdots$
The final solution for this after some tedious calculation is as below
$$y_1\sim \frac2{\sqrt{1+ce^{-\lambda t}}} \cos\left( t+\phi_0 \right)$$
$$y_1'\sim -\frac2{\sqrt{1+ce^{-\lambda t}}} \sin\left( t+\phi_0 \right)$$ where $c,\phi_0$ are constants that depend on the initial conditions.
Final note to take so as to make the analysis complete, for the second steady state, the two roots of the auxiliary equations are
$$r_\pm=-\frac14+\frac14\sqrt{1+4\lambda}\pm\frac12\sqrt{-\frac52\sqrt{1+4\lambda}-\frac32-7\lambda}$$ When $\lambda>0$, the two roots are complex conjugates with positive real part, thus unstable focus. When $\lambda=0$, the two roots are conjugate and purely imaginary, this is the Hopf bifurcation point. When $\lambda\in\left(-0.2478,0\right)$, the two roots are conjugate with negative real parts, thus stable focus. Lastly, when $\lambda\in\left(-\frac14,-0.2478\right)$, the two roots are real negative, thus stable node.
