Hey guys so I'm looking at this two dimension nonlinear dynamical system:
$\dot x = ax-xy$
$\dot y = -y + x^2$
the fixed points are at $(0,0)$, $(\sqrt{a},a)$,$(-\sqrt{a},a)$, the $x$ nullclines are $x = 0$ and $y = a$, the $y$ nullcline is $y = x^2$
So I was wondering how you would go about analyzing this two dimensional autonomous system? What kind of bifurcations would this have?
For the origin, you can use Lyapunov's method of linearization or you can use Lyapunov's direct method by utilizing the positive definite and radially unbounded Lyapunov function
$$L(x,y)=x^2+y^2.$$
The time derivative is given by
\begin{equation} \begin{split} \dfrac{d}{dt}L(x,y)=&2x\dot{x}+2y\dot{y}\\ =&2ax^2-2y^2.\\ \end{split} \end{equation}
If $a<0$ then we can show that the origin is globally asymptotically stable because the other equilibria do not exist in this case.
Now, in order to show instability of the origin for $a>0$ we can use Lyapunov's method of linearization, to see that the equilibrium is unstable.
The linearized system for an arbitrary equilibirum point $(x_\text{eq},y_\text{eq})$ is given by $$\Delta \dot{x}=(a-y_{\text{eq}})\Delta x - x_{\text{eq}}\Delta y$$ $$\Delta \dot{y}=2x_\text{eq}\Delta x - \Delta y,$$
in which $\Delta x = x -x_\text{eq}$ and $\Delta y = y -y_\text{eq}$.
First equilibrium: Using $(x_\text{eq}=0,y_\text{eq}=0)$ gives the linearized system $$\Delta \dot{x}=a\Delta x $$ $$\Delta \dot{y}=- \Delta y,$$
which is unstable for $a>0$.
Second equilibrium: Using $(x_\text{eq}=\sqrt{a},y_\text{eq}=a)$ gives the linearized system $$\Delta \dot{x}=- \sqrt{a}\Delta y$$ $$\Delta \dot{y}=2\sqrt{a}\Delta x - \Delta y.$$
The characteristic equation for the eigenvalues is given by
$$\chi(\lambda)=\lambda^2+\lambda+2a.$$
Using the Hurwitz criterion for a second-order polynomial (all coefficients have to be positive in order to have eigenvalues with strictly negative real part) we can conclude that for $a>0$ this equilibrium is asymptotically stable.
Thrid equilibrium: Using $(x_\text{eq}=-\sqrt{a},y_\text{eq}=a)$ gives the linearized system $$\Delta \dot{x}=\sqrt{a}\Delta y$$ $$\Delta \dot{y}=-2\sqrt{a}\Delta x - \Delta y.$$
The characteristic equation for the eigenvalues is given by
$$\chi(\lambda)=\lambda^2+\lambda+2a.$$
Using the Hurwitz criterion for a second-order polynomial (all coefficients have to be positive in order to have eigenvalues with strictly negative real part) we can conclude that for $a>0$ this equilibrium is asymptotically stable.