Find the fixed points of the system, and sketch the trajectories of the system

Question

I am given the following system:

$$x' = [(x-1)^2 + y^2]y$$

$$y' = -[(x-1)^2 + y^2]x \tag{*}$$

where $x = x(t), y = y(t)$.

I am supposed to

Find the fixed points of the system, and sketch the trajectories.

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Definitions in book:

1. Trajectory

Let $(u(t),v(t))$ be a solution of the autonomous system:

$$x' = f(x,y)$$

$$y' = g(x,y)$$

with solution curve $C$ in $(t,x,y)$-space. The projection of $C$ onto the $(x,y)$-plane is called the trajectory of $(u(t),v(t))$.

2. Fixed point

The trajectory of a constant solution is a point called a fixed point (or a singular point or critical point).

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So, to get the fixed points we set $x' = 0 = y'$.

We have 4 systems of equations and 4 corresponding solution sets:

1. $$(0,0)$$

2. $$(1,0)$$

3. $$0 = (x-1)^2 + y^2$$

4. $$\emptyset$$

So the constant solutions to the system are given by

$$(u(t),v(t)) = (0,0)$$

and

$$(u(t),v(t)) = (1,0)$$

while their corresponding trajectories are the fixed points $(x,y) = (0,0)$ and $(x,y) = (1,0)$

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As for the other trajectories, there is this fact in the book that says

the following systems have the same trajectories:

$$x' = M(x,y)f(x,y)$$

$$y' = M(x,y)g(x,y)$$

and

$$x' = f(x,y)$$

$$y' = g(x,y)$$

if $M > 0$ and $M_x, M_y$ are continuous.

So, if we choose $M = [(x-1)^2 + y^2] > 0$ where $2(x-1), 2y$ are continuous, then I guess the system $(*)$ has the same trajectories as

$$x' = y$$

$$y' = -x$$

If so, then

$$x'' = y' = -x$$

$$\to x(t) = c_2\sin(t) + c_1\cos(t)$$

and

$$y(t) = c_2\cos(t) - c_1\sin(t) + c_3$$

Then we have trajectories by setting $t=0$:

$$x(0) = c_1$$

$$y(0) = c_2 + c_3$$

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Any mistakes?

Answer 1

Hint: Note that $r'=0$ and $\theta'=-[(x-1)^2+y^2]$. So, all but three orbits are periodic, right?

Added: Here $r,\theta$ are polar coordinates, and we are rewriting the differential equations using the formulas $$ r'=\left(\sqrt{x^2+y^2}\right)'=\frac{xx'+yy'}{r} $$ and $$ \theta'=\left(\arctan\frac yx\right)'=\frac{y'x-x'y}{r^2} $$ (it isn't really $\arctan(y/x)$ that should be there, due to the second and third quadrants, but the final formula still holds). Other than this, $r'=0$ indicates that all trajectories are contained in circles centered at the origin, while $\theta'\le0$ with $\theta'=0$ if and only if $x=1$ and $y=0$. Now you should make a drawing with circles being traversed in the negative direction, taking into account the two fixed points.