Consider the equations:

$ \frac{dx}{dt} = y$ and $ \frac{dy}{dt} = -x $

By transforming variables, obtain $ \frac{dr^2}{dt}=0 $ and $ \frac{d \theta}{dt} = -1$

I know that if I say let $r^2 =x^2 +y^2 \cdot \frac{dr^2}{dt} = tx \frac{2x}{dt} + 2y \frac{dy}{dt} =2xy =2xy-2xy = 0$, which would mean $r^2$ is neutrally stable, but I don't understand where this gets me, much less what a change of variables is.

Moreover, what can one conclude about whether these are circular solutions, their direction, and their stability. I have not seen this mentioned before.

  • $\begingroup$ What do $r$ and $\theta$ stand for? Polar coordinates? If yes, then better mention it somewhere. $\endgroup$ – user499203 Nov 21 '17 at 17:07
  • $\begingroup$ @ThePirateBay It seems that it does refer to polar coordinates, at least the good looking answers seem to think so. Sorry, I should've specified, as that was half of my question, as I've heard of polar coordinates, but haven't learned how to use them before. I understand the general concept though, an angle and a vector. $\endgroup$ – Bad at algebra and proofs Nov 21 '17 at 17:31
  • $\begingroup$ @HRM4321 The change of variables you have refers to polar coordinates, as I mentioned below! $\endgroup$ – Rebellos Nov 21 '17 at 17:33
  • $\begingroup$ Why is there a factor of $t$ in front of $x$ in your formula for $\frac{dr^2}{dt}$? Indeed, what does $\frac{2x}{dt}$ mean? $\endgroup$ – Robert Lewis Nov 21 '17 at 18:06

Assuming you're using polar coordinates, a change of variables to the polar coordinates system corresponds to $x=r\cosθ,y=r\sin θ,r=x^2+y^2$ :

It is :

$$r^2 = x^2 + y^2 \Rightarrow 2rr' = 2xx' + 2yy' \Rightarrow rr' = xx' + yy' $$

Substituting $x',y'$ from your given system :

$$rr'= xy - yx = 0$$

To find the angle $θ$, we take :

$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$

Using the quotient rule, we get :

$$\theta' = \dfrac{x y' - y x'}{r^2}$$

Substituting $x',y'$ as before, from your given system :

$$θ = \frac{-x^2 -y^2}{r^2} =\frac{-(x^2 + y^2)}{r^2}=\frac{-r^2}{r^2}=-1$$

This is how you get the expressions.

Going over stability, it's easy to see that $O(0,0)$ is a stationary point of your given system and also the only one.

Since you have $rr' = 0$, you can conclude that $O(0,0)$ is clarified as a center for your system.

You can easily double check that by doing the common stability way :

$$J(x,y) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$

Which means that :

$$J(0,0) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$

and that :

$$\det(J(0,0) - λI) = 0 \Leftrightarrow \cdots \Leftrightarrow λ^2 + 1=0 \Leftrightarrow λ = \pm i$$

Since you have purely imaginary eigenvalues, $O(0,0)$ is a center for your system, thus we have double checked our finding.

Can you now make a conclusion about the direction and the circularity of the solutions ?

  • $\begingroup$ Great answer, thanks! This book has been minimal with the higher level calc problems, being mostly algebra related, so I was unsure if maybe I had missed something significant in the last section. $\endgroup$ – Bad at algebra and proofs Nov 21 '17 at 17:45
  • $\begingroup$ This is a common exercise regarding differential equation systems and dynamical systems ... Pretty weird that you're doing it on a wider broad lesson ! $\endgroup$ – Rebellos Nov 21 '17 at 17:52
  • $\begingroup$ If I could just pick your brain a little more, I believe the direction is clockwise, as the Eigens are opposite, but if the question asks "conclude that there are circular solutions," doesn't the fact that it is a center simply mean that the solutions follow a circular pattern? That is, just as one would draw a normal phase diagram? $\endgroup$ – Bad at algebra and proofs Nov 21 '17 at 17:53
  • $\begingroup$ Sorry for replying late. Yes, the direction field is consisted of circles around the center (which is $O(0,0)$), you're correct ! $\endgroup$ – Rebellos Nov 21 '17 at 20:34
  • 1
    $\begingroup$ About the clockwise/anti-clockwise stuff, check the $θ$ ;) $\endgroup$ – Rebellos Nov 21 '17 at 20:39

The best I can interpret the situation is as follows:

$dx/dt = y$ and $dy/dt = -x$ when solved (accompanied with initial and/or boundary conditions, which are absent from your problem statement), will give a solution of the form $x = x(t)$ and $y = y(t)$. When you plot this on x-y plane, you will get a curve, with $t$ being a parameter.

Every curve on x-y plane can be expressed by a polar equation as well, with $x = r \cos \theta$ and $y = r \sin \theta$. Now, in your case, $r$ and $\theta$ will also be functions of $t$. Use these transformations along with chain and product rules for differentiation to get $dr^2/dt = 0$ and $d \theta / dt = -1$.

The parameter $t$ in fact controls how fast a particle moves on that curve. You can have the exact same curve traced by two different particles at different speeds. A simple ODE in $x$ and $y$ will give you only the curve without any information about the speed of the particle.


with $$x'=y$$ and $$y'=-x$$ we get $$x''=-x$$ and you can solve this equation

  • 2
    $\begingroup$ Dr. Sonnhard, If you invest just a little bit more effort in your answers, you'll get much more rep per answer. I saw your last 10 answers and I have to notice that you're not trying to make a problem clearer to OP or to go deep in the concept described in a problem itself. I suggest you to increase the quality and reduce the quantity of your answers. $\endgroup$ – user499203 Nov 21 '17 at 17:37
  • 1
    $\begingroup$ @ThePirateBay With rep being 48.2K I don't think the Doctor needs help on how to gain rep. I guess the answer is a good hint to make OP work it out by her/him self. Not an attempt to get as much rep as possible. $\endgroup$ – 4386427 Nov 21 '17 at 23:08


$\dot x = y \tag 1$


$\dot y = -x, \tag 2$

then the polar $r$-coordinate satisfies

$2r\dot r = \dfrac{d(r^2)}{dt} = \dfrac{d(x^2 + y^2)}{dt} = 2x \dot x + 2y \dot y = 2xy - 2yx = 0; \tag 3$

so if $r \ne 0$ we have

$\dot r = \dfrac{dr}{dt} = 0 \tag 4$


$r(t) = r_0 \tag 5$

is a constant. (4) and (5) imply that any solution of (1)-(2) with $r \ne 0$ must lie in a circle of radius some $r_0 > 0$. As for $\theta$, if $\theta \in [0, 2\pi) \setminus \{-\pi/ 2, \pi / 2 \}$ we have

$\tan \theta = \dfrac{y}{x}, \tag 6$


$\dot \theta \sec^2 \theta = \dfrac{d \tan \theta}{dt} = \dfrac{\dot y x - \dot x y}{x^2} = \dfrac{-x^2 - y^2}{x^2} = -\dfrac{r^2}{x^2}, \tag 7$


$\dot \theta = -\dfrac{r^2}{x^2 \sec^2 \theta}; \tag 8$


$r^2 \cos^2 \theta = x^2, \tag 9$

we obtain

$r^2 = x^2 \sec^2 \theta, \tag{10}$

so (8) becomes

$\dot \theta = - \dfrac{r^2}{r^2} = -1. \tag{11}$

If, on the other hand, $\theta \in [0, 2\pi) \setminus \{ 0, \pi \}$, we can choose

$\cot \theta = \dfrac{x}{y}, \tag{12}$

and in a manner analogous to the above we then have

$-\dot \theta \csc^2 \theta = \dfrac{\dot x y - x \dot y}{y^2} = \dfrac{y^2 + x^2}{y^2} = \dfrac{r^2}{y^2}, \tag{13}$

and once again we find

$\dot \theta = -\dfrac{r^2}{y^2 \csc^2 \theta} = -\dfrac{r^2}{r^2} = -1; \tag{14}$

(5), (11) and (14) together imply that the trajectories are circles centered at the origin, and the angular velocity is a constant $-1 \; \text{rad}/\text{sec}$, assuming the second as the unit of time. We in addition see from (11), (14) that $\theta$ satisfies

$\theta(t) = -t + \theta(t_0). \tag{15}$

One thus concludes that the non-constant solutions are circular orbits centered at $(0, 0)$ in the $xy$-plane, moving about the origin in a clockwise direction with constant angular speed, which is the same for every non-trivial trajectory. $(0, 0)$ is of course a fixed point of (1)-(2); any trajectory initialized such that $(x(t_0), y(t_0)) = (0, 0)$ will satisfy $(x(t), y(t)) = (0, 0)$ for all times $t$.

As far as the stability of the solutions of (1)-(2) is concerned, the above discussion shows that orbits maintain a constant distance from the origin for all time. Therefore, a solution starting within a distance $\epsilon$ from $(0, 0)$ will ever remain within $\epsilon$ of the origin; it cannot diverge to $\infty$; indeed, in cannot leave an open disk about $(0, 0)$ in which it starts; in this sense, the origin is a stable fixed point of (1)-(2). As for the stability of the circular orbits themselves, it is evident that the trajctories

$r(t) = r_1, r(t) = r_2, \theta = -t + \theta(t_0) \tag{16}$

maintain their initial separation both in $r$ and $\theta$ for all $t$; certainly that is manifest for $r_1(t)$ and $r_2(t)$, since they are each constant for all $t$; if

$\theta_1(t) = -t + \theta_1(t_0) \tag{17}$


$\theta_2(t) = -t + \theta_2(t_0), \tag{18}$


$\theta_2(t) - \theta_1(t) = (-t + \theta_2(t_0)) - (-t + \theta_1(t_0)) = \theta_2(t_0) - \theta_1(t_0); \tag{19}$

we thus see that any initial phase difference 'twixt two solutions is maintained for all time. Therefore, if two trajectories $(r_i(t), \theta_i(t))$ differ at some $t_1$ by radial amount $\Delta r$ and phase angle $\Delta \theta$, these differences in phase coordinates will be maintained by the system points $(r_1(t), \theta_1(t)$ and $(r_2(t), \theta_2(t)$ for all time. In this sense, the orbits themselves are stable; any perturbation of the initial conditions gives rise to an orbit whose phase point remains close to the original one; it does not peel off and diverge from the original solution with the passage of time.

This system is not, however, stable against perturbations of the original equations (1)-(2) themselves. For example, if we replace (1)-(2) by modified version,

$\dot x = \mu x + y, \tag{20}$

$\dot y = -x + \mu y, \tag{21}$

then in lieu of (3) we find

$2r\dot r = \dfrac{d(r^2)}{dt} = \dfrac{d(x^2 + y^2)}{dt} = 2x \dot x + 2y \dot y$ $= 2x(\mu x + y) + 2y(-x + \mu y) = 2\mu(x^2 + y^2) = 2\mu r^2, \tag{22}$

whence for $r \ne 0$,

$\dot r = \mu r, \tag{23}$

the solutions of which take the form

$r(t) = r(t_0) e^{\mu t}; \tag{24}$

also, the form of (7) remains

$\dot \theta \sec^2 \theta = \dfrac{d \tan \theta}{dt} = \dfrac{\dot y x - \dot x y}{x^2} = \dfrac{x(-x + \mu y) - y(y + \mu x)}{x^2} = -\dfrac{r^2}{x^2}, \tag{25}$

as does that of (13), so we still have (11)/(14); though the angular speed is left unchanged, the radial coordinate of a solution either diverges to $\infty$ or converges to $0$ according to the sign of $\mu$; we see that the original system (1)-(2) is not stable, in terms of global behavior, against perturbations of the form (20)-(21); circular orbits become spirals.

As a final note, I think it is worth observing that the modified system (20)-(21) is stable against sufficiently small changes in $\mu$, in the sense that as long as $\mu \ne 0$ the diverging or converging spiral orbits will remain so.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.