Q7-T0301

[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.
$$
\left\{\begin{aligned}
&\frac{dx}{dt} = y +x(1-x^2 - y^2)\\
&\frac{dy}{dt} = -x + y(1-x^2 - y^2)
\end{aligned}\right.$$

[Darren Zhang]

(a)The critical points are solns of these questions.
$$y+x(1-x^2-y^2) = 0$$ $$-x+y(1-x^2-y^2)=0$$
Multiply the first equation by y and the second equation by x . The difference of the two equations gives $x^2+y^2=0$. Hence the only critical point is at the origin.
(b,c) With $F(x,y) = y+x(1-x^2-y^2)$ and $G(x,y) = -x+y(1-x^2-y^2)$, the Jacobian matrix of the vector field is
$$ J = \begin{pmatrix} 1-3x^2-y^2 & 1-2xy \\ -1-2xy & 1-x^2-3y^2 \end{pmatrix} $$
At the origin, the coefficient matrix of the linearized system is
$$ J(0,0) = \begin{pmatrix} 1 & 1  \\ -1 & 1\end{pmatrix} $$
with complex conjugate eigenvalues $r_{1} = 1+i$, $r_{2} = 1-i$,  . Hence the origin is an unstable spiral.

Attached is the part (d)

[Syed Hasnain]

I have attached my solution...
thanks

[Victor Ivrii]

Junjie is right. There is one stationary point (unstable focal with clockwise rotation) but also one can see a stable limit cycle $x^2+y^2=1$.