# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on March 30, 2018, 12:23:34 PM

Title: Q7-T0801
Post by: Victor Ivrii on March 30, 2018, 12:23:34 PM
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)\\ &d\frac{dy}{dt} = x - y + y(x^2 + y^2) \end{aligned}\right.
Title: Re: Q7-T0801
Post by: Darren Zhang on March 30, 2018, 01:26:18 PM
(a)The critical points are given by the solution set of the equations $$-2x-x-y-x(x^2+y^2) = 0$$ $$x-y+y(x^2+y^2) = 0$$ It is clear that the origin is a critical point. Solving the first equation for C , we find that $$y = \frac{-1 \pm \sqrt{1-8x^2-4x^4}}{2x}$$
Substitution of these relations into the second equation results in two equations of the form $f_{1}(x) = 0$ and $f_{2}(x) = 0$ . Plotting these functions, we note that only$f_{1}(x) = 0$ has real roots. ('It follows that the additional critical points are at (-0.33076,1.0924) and (0.33076,-1.0924)
(b,c) Given that $$F(x,y) = -2x-x-y-x(x^2+y^2)$$ $$G(x,y) = x-y+y(x^2+y^2)$$
the Jacobian matrix of the vector field is $$\begin{pmatrix} -2-3x^2-y^2 & -1-2xy \\ 1+2xy & -1+x^2+3y^2 \end{pmatrix}$$
with complex conjugate eigenvalues $r_{1,2} =((-3 \pm i \sqrt{3})/2)$ . Hence the critical point is a stable spiral, which is asymptotically stable. At the point (-0.33076,1.0924) , the coefficient matrix of the linearized system is $$J(-0.33076,1.0924) = \begin{pmatrix} -3.5216 & -0.27735 \\ 0.27735 & 2.6895 \end{pmatrix}$$
With eigenvalues $r_1 = -3.5092$ and $r_2 = 2.6771$ The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. Identical results hold for the point at (0.33076,-1.0924).

Attached is the part(d)
Title: Re: Q7-T0801
Post by: Victor Ivrii on March 31, 2018, 07:56:09 AM
Please, do not replace $\sqrt{3}$ by approximate decimals. When grading, it becomes very difficult to check (and technically is incorrect) so grader  mark as an error and ignore everything below