Q7-T0401

[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.
$$
\left\{\begin{aligned}
&\frac{dx}{dt} = 1 - xy\\
&\frac{dy}{dt} = x - y^3
\end{aligned}\right.$$

[Darren Zhang]

(a)The critical points are given by the solution set of the equations.$$1-xy = 0$$ $$x-y^3=0$$
After multiplying the second equation by y , it follows that $y=1/-1$ . Hence the critical points of the system are at (-1,1) and (-1,-1).

(b,c) Note that $F(x,y) = 1-xy$ and $G(x,y) = x-y^3$ . The Jacobian matrix of the vector field is $$ J = \begin{pmatrix} -y & -x \\ 1 & -3y^2 \end{pmatrix} $$
At the critical point (1,1), the coefficient matrix of the linearized system is$$ J(1,1) = \begin{pmatrix} -1 & -1 \\ 1 & -3 \end{pmatrix}$$

with eigenvalues  $r_1 = r_2 = -2$ . The eigenvalues are real and equal. It is easy to show that there is only one linearly independent eigenvector. Hence the critical point is a stable improper node. 

 At the point (-1,-1), the coefficient matrix of the linearized system is $$ J(-1,-1) = \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix}$$

with eigenvalues $r_1 = -1+\sqrt{5}$, $r_2 = -1- \sqrt{5}$. The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable.

Attached is the part(d).

[Nikola Elez]

Small error, you stated that (-1,1) is a critical point, when it is (1,1)

[Syed Hasnain]

I think that for the critical point (1,1) it is
an asymptotically stable spiral

[Victor Ivrii]

Indeed, there was a misprint in one place, but later considered was a correct point. I inserted bold into formatting to show where each point is analyzed