A calculus problem by Rocco Dalto

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

Converting to polar coordinates, let $x = r cos\theta, y = r sin\theta$ $(1): \: x^2 + y^2 = r^2, (2): \: \theta = \arctan(\dfrac{y}{x})$ $$

Taking the derivatives with respect to $t$ of $(1)$ and $(2)$ we obtain: $$

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

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

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

Adding the above equations we obtain: $$

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

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

Adding the above equations we obtain: $$

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

The system of differentials above has a single critical point at $r = 0$ . Assuming $r > 0$ , 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 region $R = A - B$ .

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

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

$\therefore$ 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 $R$ . $$

Note: I would have solved the indefinite integral $\int \dfrac{dr}{r(3 - e^{r^2})} \: dr$ if it could, and found a solution $x = x(t), y= y(t)$ . $$

A yes/no problem is not very interesting.