How can I solve this question: Let $X=\left ( X_{1},X_{2} \right )$ field in $\mathbb{R}^{2}$, where
$\begin{matrix} X_{1}=x_{2}+x_{1}\left ( 1-x_{1}^{2}-x_{2}^{2} \right )\\ \, \, \, \, \, X_{2}=-x_{1}+x_{2}\left ( 1-x_{1}^{2}-x_{2}^{2} \right ) \end{matrix}$
Show that this field has a unique periodic orbit.
Thanks four your help
We are given the vector field
$\vec X = \begin{pmatrix} X_1 \\ X_2 \end{pmatrix} = \begin{pmatrix} x_{2}+x_{1}(1-x_{1}^{2}-x_{2}^{2}) \\ -x_{1}+x_{2}( 1-x_{1}^{2}-x_{2}^{2}) \end{pmatrix}; \tag 1$
we define the radial vector field
$\vec r = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}; \tag 2$
let
$r = \Vert \vec r \Vert = \sqrt{x_1^2 + x_2^2} \tag 3$
be he radial co-ordinate; then
$r^2 = \vec r \cdot \vec r = x_1^2 + x_2^2; \tag 4$
now let $\vec r(t)$ satisfy the differential equation
$\vec r^\prime (t) = \vec X; \tag 5$
we have
$2 r r^\prime = (r^2)^\prime = (\vec r \cdot \vec r)^\prime = 2 \vec r \cdot \vec r^\prime; \tag 6$
since $\vec r(t)$ satisfies the differetial equation
$\vec r^\prime = \vec X, \tag 7$
(6) yields
$2 r r^\prime = 2 \vec r \cdot \vec r^\prime = 2 \vec r \cdot \vec X = 2(x_1 X_1 + x_2 X_2), \tag 8$
$r r^\prime = x_1 X_1 + x_2 X_2 = x_1 (x_{2}+x_{1}(1-x_{1}^{2}-x_{2}^{2})) + x_2 ( -x_{1}+x_{2}( 1-x_{1}^{2}-x_{2}^{2}))$ $= x_1 x_2 + x_1^2 (1-x_{1}^{2}-x_{2}^{2}) - x_2 x_1 + x_2^2 (1-x_{1}^{2}-x_{2}^{2})$ $= (x_1^2 + x_2^2)(1-x_{1}^{2}-x_{2}^{2}) = r^2(1 - r^2), \tag 9$
whence, as long as $r > 0$,
$r^\prime = r(1 - r^2). \tag{10}$
It follows from (10) that
$r^\prime > 0, \; 0 < r < 1; \tag{11}$
$r^\prime < 0, \; r > 1; \tag{12}$
$r^\prime = 0, \; r = 1; \tag{13}$
we thus see that $r$ is increasing for $0 < r < 1$, decreasing if $r > 1$, and constant if $r = 1$; therefore no trajectory with $r \ne 1$ can be periodic, since $r$ will never take the same value more than once, but with $r = 1$ at any time the orbit will remain in the circle $\Vert \vec r \Vert = 1$ forever. To show that such an integral curve is actually periodic, we must examine $\theta^\prime$; this is given by the component of $\vec X$ normal to $\vec r$. From (2), we may take
$\vec r^\bot = \begin{pmatrix} -x_2 \\ x_1 \end{pmatrix}; \tag{14}$
then
$\vec r \cdot \vec r^\bot = 0 \tag{15}$
the unit vector in the direction of $\vec r^\bot$ is
$\dfrac{ \vec r^\bot}{r}, \tag{16}$
since $\Vert \vec r \Vert = \Vert \vec r^\bot \Vert = r$; we obtain the angular vecocity $\theta^\prime$ by taking the inner product of $\vec X$ with (16); we have
$\vec r^\bot \cdot \vec X = -x_2 X_1 + x_1 X_2$ $= -x_2(x_{2}+x_{1}(1-x_{1}^{2}-x_{2}^{2})) + x_1 ( -x_{1}+x_{2}(1-x_{1}^{2}-x_{2}^{2})) = -(x_1^2 + x_2^2) = - r^2; \tag{17}$
thus
$\theta^\prime = \dfrac{ \vec r^\bot}{r} \cdot \vec X = -r; \tag{18}$
when $r = 1$ we see that
$\theta^\prime = -1; \tag{19}$
it follows that a system point with $r = 1$ is rotating about the origin with constant angular velocity $-1$, so it traverses the unit circle in time $2\pi$, which is evidently it's period.
Thus the circle $r = 1$ is the unique periodic orbit of the given system, and its period is $2\pi$.
The only point in the $x_1-x_2$ plane we haven't addressed is the origin $(0, 0)$. But since that is an equilibrium, the system point sticks there like glue. No periodic trajectories here, folks, move on . . .