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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.

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  • $\begingroup$ 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
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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 ?

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  • $\begingroup$ 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
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    $\begingroup$ About the clockwise/anti-clockwise stuff, check the $θ$ ;) $\endgroup$ – Rebellos Nov 21 '17 at 20:39
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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.

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

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    $\begingroup$ 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
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    $\begingroup$ @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
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If

$\dot x = y \tag 1$

and

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

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

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

or

$r(t) = r_0 \tag 5$

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

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

whence

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

or

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

since

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

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

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

and

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

then

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

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.

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