True? False?

$\dfrac{dx}{dt} = 3x - y - x(e^{ x^2+ y^2})$

$\dfrac{dy}{dt} = x + 3y - y(e^{ x^2 + y^2})$

Converting to polar coordinates, let $x = r \cos\theta, \:\ y = r \sin\theta,\:$ $x^2 + y^2 = r^2$ and $\theta = \arctan(\dfrac{y}{x})$ $$

Taking the derivatives with respect $t$ $$

$x^2 + y^2 = r^2 \implies r \dfrac{dr}{dt} = x \dfrac{dx}{dt} + y \dfrac{dy}{dt}$

and

$\theta = \arctan(\dfrac{y}{x}) \implies r^2 \dfrac{d\theta}{dt} = x \dfrac{dy}{dt} - y \dfrac{dx}{dt}$ $$

$(1): x * (\dfrac{dx}{dt} = 3x - y - x(e^{ x^2+ y^2}))$

$(2): y * (\dfrac{dy}{dt} = x + 3y - y(e^{ x^2 + y^2}))$

and adding $(1)$ and $(2)$ we obtain: $\boxed{\dfrac{dr}{dt} = r(3 - e^{r^2})}$

$(3): -y * (\dfrac{dx}{dt} = 3x - y - x(e^{ x^2+ y^2}))$

$(4): x * (\dfrac{dy}{dt} = x + 3y - y(e^{ x^2 + y^2}))$

and adding $3$ and $4$ we obtain: $\boxed{\dfrac{d\theta}{dt} = 1}$

The system of differentials above has a single critical point at $r = 0(x = 0, y = 0)$ .

For $r > 0$ , the system above reduces to:

$\dfrac{dr}{dt} = r(3 - e^{r^2})$

$\dfrac{d\theta}{dt} = 1$

Let $A =$ { $(x,y)| x^2 + y^2 <= 2$ } and $B =$ { $(x,y)| x^2 + y^2 <= 1$ }.

Let the annulus region $R = A - B$

For $r > 0, \dfrac{dr}{dt} > 0$ on the inner circle $r = 1$ , and $\dfrac{dr}{dt} < 0$ on the outer circle $r = \sqrt{2}.$ $$

Note: The radial vector points in the direction of $R$ at every boundary point.

$\therefore$ any path through a boundary point will enter $R$ and remain in $R$ as $t \rightarrow \infty$ , and by the Poincare-Bendixson theorem the initial system has a closed path in $R$ .

If we define $r_0= \sqrt{\ln 3}$ , then $e^{r_0^2}=3$ and so $\binom{x}{y} \; = \; \binom{r_0\cos t}{r_0\sin t}$ is a closed solution. The circle, centre $0$ , radius $r_0$ , serves as the region $R$ and also defines the closed path solution. We do not need the Poincare-Bendixson Theorem here.