I'm a bit stuck in solving the following exercise: consider the dynamical system
$$\dot{x} = 2 + 3\mu x – x^3$$
I have to find the equilibrium points and the stability type, and the the bifurcations and their nature.
Attempts
I don't know how to calculate the equilibrium points, for I should solve $\dot{x} = 0$ but I am not able to solve $x^3 + 3\mu x – 2 = 0$, so I'm already stuck here…
That $2$ really hit me hard.
My notes then say that in order to classify an equilibrium point I shall find the topology near each one of them, hence by calculating the eigenvalues of the Jacobian matrix (evaluated at the critical points found).
For what concerns the bifurcations, I might have done some work, but I'm not sure about.
If $3 \mu > 0$ hence $\mu > 0$ then I have a tangent bifurcation (two critical points). For the max:
$$\frac{d}{dx}(3\mu x – x^3) = 3\mu – x^2 \rightarrow x = \pm\sqrt{3\mu}$$
Hence $x_{max} = \sqrt{3\mu}$ and we have
$$3\mu x_{max} – x_{max}^3 = 3\mu\sqrt{3\mu} – 9\mu^2\sqrt{3\mu} = 3\mu\sqrt{3\mu}(1 – 3\mu)$$
Let's call this $H(\mu)$
Tangent bifurcation for $2 = \pm H(\mu)$. If we call $\mu\sqrt{3\mu} = t$ then
$$2 = t – t^2$$
Yet this has no solution in $\mathbb{R}$.
Can you please help me in solving this problem?
Best Answer
Let $p(x)=2+3\mu x-x^3$. Note that $p(0)=2$ and $p(x)\to-\infty$ if $x\to\infty$, hence there must be at least one equilibrium $\hat{x}_1$ for positive $x$.
Next, $p'(x)=3\mu-3x^2$ and whence $p'(x)=0$ if and only if $x=\pm\sqrt{\mu}$. This also shows that $p$ has only one root $\hat{x}_1$, which must be asymptotically stable, if $\mu<0$ (make a sketch!).
Moreover, $p(\sqrt{\mu})=2(1+\mu\sqrt{\mu})>0$ and grows with growing $\mu$, hence it is maximum of $p$, and $p(-\sqrt{\mu})=2(1-\mu\sqrt{\mu})$ and decreases with increase of $\mu$, hence it is minimum of $p$ (make another sketch). The graph of $p$ will touch the $x$-axis when $p(-\sqrt{\mu})=0$, i.e., when $\mu=1$ (make another sketch).
To summarize, your system has a unique asymptotically stable equilibrium $\hat{x}_1$ if $\mu<1$, has three equlibria $\hat{x}_2<\hat{x}_3<0<\hat{x}_1$ if $\mu>1$, where $\hat{x}_1$ and $\hat{x}_2$ are attractive, and $\hat{x}_3$ is unstable, and hence $\mu=1$ is the point of tangent bifurcation in your system.