Find $\lim\limits_{t\to\infty}x(t)$ if $x'= (x-y)(1-x^2-y^2)$, $y' = (x+y)(1-x^2-y^2)$ Suppose $x_0,y_0$ are reals such that $x_0^2+y_0^2>0.$ Suppose $x(t)$ and $y(t)$ satisfy the following:


*

*$\frac{dx}{dt} = (x-y)(1-x^2-y^2),$

*$\frac{dy}{dt} = (x+y)(1-x^2-y^2),$

*$x(0) = x_0,$

*$y(0) = y_0.$


I am asked to find $\lim_{t\to\infty}x(t).$
Dividing the two differential equations, we have $$\frac{dy}{dx} = \frac{x+y}{x-y} = \frac{1 + \frac{y}{x}}{1-\frac{y}{x}}.$$
Let $v(x)=\frac{y}{x},$ so that $y = xv(x),$ and $$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$ This implies that $$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v}.$$ Rearranging and integrating, we get $$\log(x) = \arctan(v)-\frac{1}{2}\log(1+v^2)+c,$$ so $$\log(x) = \arctan\left(\frac{y}{x}\right)-\frac{1}{2}\log\left(1+\left(\frac{y}{x}\right)^2\right)+c.$$ By writing $1+\left(\frac{y}{x}\right)^2$ as $\frac{x^2+y^2}{x^2},$ we can simplify the above equation to $$\frac{1}{2}\log(x^2+y^2) = \arctan\left(\frac{y}{x}\right) + c.$$
However, I'm not sure how to proceed from this implicit equation. Any help would be much appreciated!
 A: We first pass to polar coordinates, viz:
$x = r\cos \theta; \; y = r \sin \theta; \tag 1$
$\dot x = \dot r \cos \theta - r\dot \theta \sin \theta; \tag 2$
$\dot y = \dot r \sin \theta + r \dot \theta \cos \theta; \tag 3$
$\begin{pmatrix} \dot x \\ \dot y \end{pmatrix} = \begin{bmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r\cos \theta \end{bmatrix} \begin{pmatrix} \dot r \\ \dot \theta \end{pmatrix}; \tag 4$
$\begin{bmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r\cos \theta \end{bmatrix}^{-1} = \dfrac{1}{r} \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} ; \tag 5$
$\begin{pmatrix} \dot r \\ \dot \theta \end{pmatrix} = \dfrac{1}{r} \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{pmatrix} \dot x \\ \dot y \end{pmatrix}; \tag 6$
$\dot x = (x - y)(1 - x^2 - y^2) = r(\cos \theta - \sin \theta)(1 - r^2); \tag 7$
$\dot y = (x + y)(1 - x^2 - y^2) = r(\cos \theta + \sin \theta)(1 - r^2); \tag 8$
$\begin{pmatrix} \dot r \\ \dot \theta \end{pmatrix} = \dfrac{1}{r} \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{pmatrix} r(\cos \theta - \sin \theta)(1 - r^2) \\   r(\cos \theta + \sin \theta)(1 - r^2)\end{pmatrix}$
$= (1 - r^2) \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{pmatrix} \cos \theta - \sin \theta \\ \cos \theta + \sin \theta \end{pmatrix} = (1 - r^2) \begin{pmatrix} r \\ 1 \end{pmatrix}; \tag{10}$
$\dot r = r(1 - r^2); \tag{11}$
$\dot \theta = 1 - r^2. \tag{12}$
Since
$x_0^2 + y_0^2 > 0, \tag{13}$
we have
$r_0^2 = x_0^2 + y_0^2 > 0 \Longrightarrow r_0 > 0; \tag{14}$
by (11), 
$0 < r < 1 \Longrightarrow \dot r > 0; \tag{15}$
$1 < r \Longrightarrow \dot r < 0; \tag{16}$
$r = 1 \Longrightarrow \dot r = 0; \tag{17}$
it is relatively easy to see that (14)-(17) in concert imply that
$\displaystyle \lim_{t \to \infty} r = 1; \tag{18}$
also, (11) and (12) together yield
$r \ne 0 \Longrightarrow \dot{(\ln r)} = \dfrac{\dot r}{r} = 1 - r^2 = \dot \theta, \tag{19}$
whence, integrating over $t$,
$\ln r - \ln r_0 = \theta - \theta_0, \tag{20}$
or
$\theta = \ln r - \ln r_0 + \theta_0, \tag{21}$
and so via (18)
$\displaystyle \lim_{t \to \infty} \theta = \lim_{t \to \infty} \ln r - \ln r_0 + \theta_0 = \theta_0 - \ln r_0; \tag{22}$
finally,
$\displaystyle \lim_{t \to \infty} x(t) = \lim_{t \to \infty} ( r \cos \theta) = (\lim_{t \to \infty} r)( \lim_{t \to \infty} \cos \theta) = \cos (\theta_0 - \ln r_0). \tag{23}$
A: Let $r^2 = x^2+y^2$, $x = rcos\theta$, and $y = rsin\theta$.
Then, you get the following equations:
$$\begin{align}\frac{dr}{dt} &= r(1-r^2) \\
\frac{d\theta}{dt} &= (1-r^2)
\end{align}$$
The solution to the first equation is: $r^2=\frac{ke^{2t}}{1+ke^{2t}}$, where $k=\frac{r^2_0}{1-r_0^2}$.
Then, the solution to the second equation is: $\theta-\theta_0=log\ \frac{r}{r_0}$,
where $r_0^2=x_0^2+y_0^2$ and $tan\ \theta_0= \frac{y_0}{x_0}$.
A: If you only want to compute the limit, you don't even need to solve the equation explicitly (at least for $r$).
Following @player100's suggestion, you can transform the system into
$$
\begin{cases}
\dot r = r(1-r^2)\\
\dot \theta = 1-r^2
\end{cases},
$$
with $r^2 = x^2+y^2$ and $\theta = \arctan\frac{y}{x}.$
Then we can observe that our system rests at the origin as well as on the unit circle. Since $\dot r > 0$ when $0 < r < 1,$ and $\dot r < 0$ when $r>1$, we may conclude that any solution starting in $(x_0,y_0)$ with $x_0^2+y_0^2 > 0$ will be attracted to the unit circle, so that
$$\lim_{t \to \infty} r(t) = 1.$$
Hence our limit becomes
$$ \lim_{t\to \infty} x(t) = \lim_{t\to \infty} r(t)\cos\theta(t) = \lim_{t\to \infty} \cos\theta(t) = \cos \left(\lim_{t\to \infty}\theta(t)\right). $$
Now
\begin{align}
\theta(t) - \theta(t_0) &= \int_{t_0}^t \dot\theta(s) \, ds = \int_{t_0}^t 1-r(s)^2 \, ds,
\end{align}
and changing the integration variable to $u = r(s),$ so that
$$du = \dot r(s) \, ds = r(s)(1-r(s)^2) \, ds = u(1-u^2) \, ds,$$
we obtain
\begin{align}
\int_{t_0}^t 1-(r(s))^2 \, ds &= \int_{r(t_0)}^{r(t)} \frac{1}{u} \, du
= \log(r(t))-\log(r(t_0)).
\end{align}
This gives
\begin{align}
\lim_{t\to \infty}\theta(t)
&= \lim_{t\to \infty}\theta(t) - \theta(t_0) + \theta(t_0)
= \lim_{t\to \infty} \log(r(t))-\log(r(t_0)) + \theta(t_0)\\
&= \log(1)-\log(r(t_0)) + \theta(t_0)
= \arctan{\frac{y_0}{x_0}} - \log\sqrt{x_0^2+y_0^2},
\end{align}
so that
$$\lim_{t\to \infty} x(t) = \cos \left( \arctan{\frac{y_0}{x_0}} - \log\sqrt{x_0^2+y_0^2} \right). $$
