Stability of the trivial solution of a system of differential equations I am trying to determine the stability properties of the equilbrium solution $(x,y) = (0,0)$ of the following system of ODEs:
$$
\dot x = x - y + kx(x^2+y^2), \\
\dot y = x - y + ky(x^2+y^2),
$$
where $k \neq 0$ is a constant. Transforming to polar coordinates $(x,y) = (r\cos\theta, r\sin\theta)$ produces the following system:
$$
\dot r = r\cos 2\theta + kr^3, \\
\dot\theta = 1 - \sin 2\theta. 
$$
By numerical investigation of the direction field and some orbits, I suspect that all solutions go to zero or infinity if $k < 0$ or $k >0$ respectively. Also, from the second equation it follows that $\theta$ increases monotonically to one of its equilibrium points. I also know that the second equation is seperable and solvable, and that the first equation can be rewritten using an extra transformation $u = r^{-2}$ to get a first order linear ODE, which in principle is also solvable. However, the expressions get too nasty for me to say anything concrete about them. 
If my suspicions about these stability properties are correct, what could be a good way to show it?
 A: Hint.
The central manifold can be determined by making $y = h(x) = \sum_{k=1}^n a_k x^k$ and equating
$$
h_x(x)(x-h(x)+k x(x^2+h^2(x))-(x-h(x)+k h(x)(x^2+h^2(x)))=0
$$
Solving for the $a_k$'s with $n = 3$ we obtain
$$
\cases{h_1(x) = x + \frac{4k x^3}{3}+O(x^4)\\
h_2(x) = x}
$$
Follows a plot showing the stream plot for $k = \frac 14$

The flows are respectively
$$
\cases{\dot x_1 \approx \frac{2kx_1^3}{3}+\frac{8 k^2x_1^5}{3}+\frac{16k^3x_1^7}{36}\\
\dot x_2 = \frac{2k x_2^3}{9}
}
$$
both unstable flows for $k = \frac 14$.
A: Let's take a look at the equation for $\theta$ first. It is clear that $\dot{\theta} \geq 0.$ So, $\theta$ will always increase until it hits a value $\theta=\frac{\pi}{4}+n\pi$ at which point it will stop changing since $1-\sin(2(\frac{\pi}{4}+n\pi))=0.$ Now let's look at the equation for $r$. Note $r\geq 0$. For large enough values for $r$, $|kr^3| \gg |r\cos(2\theta)|$, so we can say $\dot{r}\approx kr^3.$ If $k>0$, then as long as $r \neq 0$ it is fairly obvious that $r$ will get arbitrarily large. If $k<0$, then $r$ will tend towards the critical point $r=0$. 
