(a) Determine all critical points of the given system of equations.
(b) Find the corresponding linear system near each critical point.
(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?
(d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
$\left\{\begin{array}{l}{\dfrac{dx}{dt}=-2x-y-x(x^2+y^2)} \\ {\dfrac{dy}{dt}=x-y+y(x^2+y^2)}\end{array}\right.$
To find critical points, let
$\dfrac{dx}{dt}=0\;,\; \dfrac{dy}{dt}=0$
Therefore,
$\left\{\begin{array}{l}{-2x-y-x(x^2+y^2)=0} \\ {x-y+y(x^2+y^2)=0}\end{array}\right.$
Then we have critical points
$(0,0),\quad \bigg(-\sqrt{\dfrac{4}{\sqrt{13}}-1}\;,\ (\dfrac{1}{2}+\dfrac{3}{2\sqrt{13}})\sqrt{4\sqrt{13}-13}\bigg),\quad \bigg(\sqrt{\dfrac{4}{\sqrt{13}}-1} \;,\ (-\dfrac{1}{2}-\dfrac{3}{2\sqrt{13}})\sqrt{4\sqrt{13}-13}\bigg)$
Which is
$(0,0),\quad (-0.330757,1.09242),\quad (0.330757,-1.09242)$
let
$\left\{\begin{array}{l}{f(t)=-2x-y-x(x^2+y^2)} \\ {g(t)=x-y+y(x^2+y^2)}\end{array}\right.$
$J[f(t),g(t)]=$$
\left [
\begin{matrix}
f_x & f_y \\
g_x & g_y
\end{matrix}
\right ]
$$
=$$
\left [
\begin{matrix}
-2-3x^2-y^2 & -1-2xy \\
1+2xy & -1+x^2+3y^2
\end{matrix}
\right ]
$
Plug in the critical points to find eigenvalues of each linear system.
For $(0,0)$,
$J(0,0)=$$
\left [
\begin{matrix}
-2 & -1 \\
1 & -1
\end{matrix}
\right ]
$
$det(A-\lambda I)=$$det
\left [
\begin{matrix}
-2-\lambda & -1 \\
1 & -1-\lambda
\end{matrix}
\right ]=\lambda^2+3\lambda+3
$
Solve for
$\lambda^2+3\lambda+3=0$
$\lambda_1=\dfrac{-3+\sqrt{-3}i}{2}\;,\; \lambda_2=\dfrac{-3-\sqrt{-3}i}{2}$
Therefore, the system is a spiral and stable at $(0,0)$.
for $(-0.330757,1.09242)$,
$J(-0.330757,1.09242)=$$
\left [
\begin{matrix}
-3.521576422 & -0.2773500983 \\
0.2773500983 & 2.689526127
\end{matrix}
\right ]
$
$det(A-\lambda I)=$$det
\left [
\begin{matrix}
-3.521576422-\lambda & -0.2773500983 \\
0.2773500983 & 2.689526127-\lambda
\end{matrix}
\right ]=\lambda^2+0.832050295\lambda-9.394444
$
Solve for
$\lambda^2+0.832050295\lambda-9.394444=0$
$$
\begin{align}
\lambda_1&=\dfrac{-0.832050295+\sqrt{38.27010}}{2}=2.677116459 \notag \\
\notag \\
\lambda_2&=\dfrac{-0.832050295-\sqrt{38.27010}}{2}=-3.509166754 \notag \\
\notag \\
\end{align}
$$
Therefore, the system is a saddle point and unstable at $(-0.330757,1.09242)$.
Identical results hold for the point at $(0.330757,-1.09242)$.
