I want to solve the following exercise:
Determine the critical points of the system
\begin{align}
\dot{x} &= x^2- y^3\\
\dot{y} &= 2x(x^2 – y)
\end{align}
Are there attractors in this system? Determine a first integral. Do periodic solutions exist?
What I've tried so far:
I think that I've found the critical points of this system. If I inspect
\begin{align}
x^2 – y^3 = 0\\
2x(x^2 – y) = 0
\end{align}
I find that $x = 0$ or $x = \pm \sqrt{y}$.
Case $x = 0$:
$-y^3 = 0$ so that $y = 0$. Hence the first critical point that I find is $(0,0)$.
Case $x = \pm\sqrt{y}$:
$y – y^3 = 0$ so $y=1$ or $y = -1$. If $y = -1$ then $x = \pm i$. If $y = 1$ then $x = \pm1$. Hence I find the critical points $(1, 1), (i, -1), (-1,1),$ and $(-i,-1)$. I think that I've found a first integral by looking at the equation
$$
\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{dy}{dx} = \dfrac{2x^3 – 2xy}{x^2 – y^3}
$$
Integrating this final equation gives
$$
\int \dfrac{2x^3 – 2xy}{x^2 – y^3}dx = y(y^2 – 1)\log(y^3 – x^2) + x^2 + C
$$
Hence a first integral of this system is
$$
F(x,y) = y(y^2 – 1)\log(y^3 – x^2) + x^2
$$
I tried plotting this function to find out if the system of equations has periodic solutions but I don't get any wiser looking at the plots.
I have the following definitions of an attractor:
A critical point $x = a$ of the equation $\dot{x} = f(x)$ in $\mathbb{R}^n$ is called a positive attractor if there exists a neighborhood $\Omega_a\subset \mathbb{R}^n$ of $x = a$ such that $x(t_0)\in\Omega_a$ implies $\lim_{t\to\infty} x(t) = a$. If a critical point $x = a$ has this property for $t \to – \infty$ then $x = a$ is called a negative attractor.
I think that in order to determine whether either of the critical points is an attractor I need to know what the solution $x(t)$ looks like. I'm not sure how to solve this system though and if I enter the equations in wolfram then I don't get a solution either (I might be missing something). So therefore I tried to linearize the equations. I have:
$$
f(x, y) = \begin{pmatrix}x^2 – y^3\\2x(x^2 – y)\end{pmatrix}, \dfrac{\partial f(x,y)}{\partial(x,y)} = \begin{pmatrix}2x & -3y^2\\6x^2& -2x\end{pmatrix}, \dfrac{\partial^2f(x,y)}{\partial (x,y)^2} = \begin{pmatrix}2 & -6y\\12x & 0\end{pmatrix}, \\\dfrac{\partial^3 f(x,y)}{\partial (x,y)^3} = \begin{pmatrix}0 & -6\\12 & 0\end{pmatrix}
$$
Hence, using the Taylor expansion around $(x,y) = (0,0)$ I get
\begin{align}
\dot{x} &= x^2 – y^3 + \ldots\\
\dot{y} &= 2x^3 + \ldots
\end{align}
but I'm not really sure how I can use this to learn anything new..
Question: How can I learn more about this system so that I will be able to say more about the existence of attractors and periodic solutions? Is my reasoning thus far correct?
Best Answer
$$\begin{align} \dot{x} &= x^2- y^3\\ \dot{y} &= 2x(x^2 - y) \end{align}$$ $$\frac{dy}{dx}=\frac{2x(x^2 - y)}{x^2- y^3}$$ Let $X=x^2$ $$\frac{dy}{dX}=\frac{X - y}{X- y^3}$$ $$\frac{dX}{dy}=\frac{X- y^3}{X-y}$$ Let $X=y+u$ $$\frac{dX}{dy}=1+\frac{du}{dy}=\frac{y+u- y^3}{u}$$ $$\frac{du}{dy}=\frac{y- y^3}{u}$$ $$2udu=2(y-y^3)dy$$ $$u^2=y^2-\frac12 y^4+c$$ $$(X-y)^2=(x^2-y)^2=y^2-\frac12 y^4+c$$ $$x^4-2x^2y=-\frac12 y^4+c$$ The equation of the trajectory is : $$\boxed{y^4-4x^2y+2x^4=2c}$$ The analytical study of this equation shows that $c\geq -\frac12$ and that the trajectory is a closed curve.