# 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!

• Use polar coordinates... – player100 Aug 24 '18 at 4:02
• @player100 well using polar, I can get $r = Ae^\theta,$ but I'm not sure how that helps me find $\lim_{t \to \infty}x(t)..?$ – dhk628 Aug 24 '18 at 4:45
• Note that $(r^2)'=(x^2+y^2)'=2(xx'+yy')$. And you can substitute your equations for $x'$ and $y'$ into that expression, and simplify. – player100 Aug 24 '18 at 4:58
• Eliminating $t$ leads you nowhere as you are asked the limit for $t\to\infty$. – Yves Daoust Aug 24 '18 at 6:34
• The title was wrong (you are not asked to solve the system, only to find the limit of $x(t)$) and uninformative, please see modified version. – Did Aug 24 '18 at 6:44

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 $$\lim_{t\to \infty} x(t) = \cos \left( \arctan{\frac{y_0}{x_0}} - \log\sqrt{x_0^2+y_0^2} \right).$$

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,

$\displaystyle \lim_{t \to \infty} x(t) = \lim_{t \to \infty} ( r \cos \theta) = (\lim_{t \to \infty} r)( \lim_{t \to \infty} \cos \theta) = \cos (\theta_0 - \ln r_0). \tag{23}$

• +1 really nice and complete answer @Robert – Isham Aug 24 '18 at 13:40
• @Isham: thank you sir! – Robert Lewis Aug 24 '18 at 14:26

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 $tan\ \theta_0= \frac{y_0}{x_0}$.