Q7 TUT 0501

[Victor Ivrii]

(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.
$$\left\{\begin{aligned}
&\frac{dx}{dt} = -2x - y -x(x^2 + y^2),\\
&\frac{dy}{dt} = x - y + y(x^2 + y^2).
\end{aligned}\right.$$

Bonus: Computer generated picture

[Jingze Wang]

To find critical points, let $x'=0, y'=0$
Thus, $-2x-y-x(x^2+y^2)=0, x-y+y(x^2+y^2)=0$
$ (0,0)$ is one crtical point
Also, $(0.330757,1.09242), (0.330757,-1.09242)$ are also critical points

Let $f(x)=-2x-y-x(x^2+y^2), g(x)=x-y+y(x^2+y^2)$
$f_x=-2-3x^2-y^2, f_y=-1-2xy$
Also, $ g_x=1+2xy, g_y=-1+x^2+3y^2$

$J=\begin{bmatrix}
-2-3x^2-y^2&-1-2xy\\
1+2xy&-1+x^2+3y^2\\
\end{bmatrix}$

Plug in the critical points to find eigenvalues of each linear system
For $(0,0)$,

We get $\begin{bmatrix}
-2&--1\\
1&-1\\
\end{bmatrix}$,
$r=\frac{-3\pm\sqrt{-3}i}{2}$
Therefore, it is stable at $(0,0)$

Similarly, for others critical points,
Plug in and eigenvalues are $-3.5092, 2.6771$
Thus, it is a saddle point therefore unstable

[Guanyao Liang]

This is my answer

[Victor Ivrii]

I am not sure why in the textbook was the problem with difficult to calculate stationary points. The easiest way:
\begin{align*}
-&2x-y-x(x^2+y^2)=0,\\
&x-y+y(x^2+y^2)=0.
\end{align*}
Multiplying these equations by $y$ and $x$ correspondingly and adding we get $-y^2-3xy +x^2=0$ and we can find $x = (3\pm \sqrt{13})y/2$. Then plugging to one of the equations in addition to $(0,0)$ we get two other points.

I will instruct TAs to consider only $(0,0)$.

Guanyao:  calculation of eigenvalues  missing; picture wrong

Jingze: again, wrong range for variables in the picture