# Toronto Math Forum

## MAT244-2013S => MAT244 Math--Tests => Quiz 5 => Topic started by: Sabrina (Man) Luo on April 03, 2013, 09:11:21 AM

Title: Day Section's Quiz - Problem 1
Post by: Sabrina (Man) Luo on April 03, 2013, 09:11:21 AM
(1) For the system
\begin{equation*}
\left\{\begin{aligned}
&dx/dt=y+x(1-x^2-y^2),\\
&dy/dt=-x+y(1-x^2-y^2)
\end{aligned}
\right.\end{equation*}
determine all critical points, linearize around each critical point, and determine what conclusion can be made about the nonlinear system at each critical point based on the linearization. Draw a phase portrait for the nonlinear system.

Title: Re: Day Section's Quiz - Problem 1
Post by: Victor Ivrii on April 03, 2013, 09:29:47 AM
Sabrina, thanks for posting problems (I "LaTeXed" them). Also you should post different problems in different topics.

Pictures have too low resolution to be read. However in Problem 1 at least one can see a correctly sketched equilibrium point.

PS: After you sent to me actual jpg files (each more than 1MB I removed colour but result is not as good as if the original scan was B/W). It should be < 100 kb per page

Bonus question to this problem: are there any closed trajectories? Hint: use polar coordinates.
Actually this bonus question deals with a notion introduced in more advanced section of Chapter 9.
Title: Re: Day Section's Quiz - Problem 1
Post by: Alexander Jankowski on April 03, 2013, 10:37:32 AM
The critical points can be determined by solving the equations
\begin{equation*}
0 = y + x(1 - x^2 - y^2), \qquad 0 = -x + y(1 - x^2 - y^2).
\end{equation*}
It turns out that the only critical point is $(0,0)$. Now, let
\begin{equation*}
F(x,y) = y + x(1 - x^2 - y^2), \qquad G(x,y) = -x + y(1 - x^2 - y^2).
\end{equation*}
The Jacobian matrix of these functions is
\begin{equation*}
J = \left( \begin{array}{cc} 1 - 3x^2 - y^2 & 1 - 2xy \\ -1 - 2xy & 1 - x^2 - 3y^2 \end{array} \right).
\end{equation*}
Evaluating it at $(0,0)$, we find that
\begin{equation*}
J = \left( \begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array} \right).
\end{equation*}
The linear system at $(0,0)$ is then
\begin{equation*}
\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array} \right)
\left( \begin{array}{c} x \\ y \end{array} \right).
\end{equation*}
The eigenvalues of the coefficient matrix are
\begin{equation*}
\text{det} \left( \begin{array}{cc} 1 - \lambda & 1 \\ -1 & 1 - \lambda \end{array} \right) = \lambda^2 - 2\lambda + 2 = 0
\Longleftrightarrow \lambda = 1 \pm i.
\end{equation*}
The eigenvalues are complex with positive real part. Therefore, $(0,0)$ is an unstable spiral point. We can conclude that the non-linear system behaves similarly. To determine the orientation of the spiral, we use the vector $(x,y)^T = (0,1)$ in the equation of the system to find that
\begin{equation*}
\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \end{array} \right).
\end{equation*}
This tells us that the spiral is oriented clockwise. A stream plot of the system is attached. Note that there is a closed trajectory $x^2 + y^2 = 1$ that exhibits orbital stability. It is the limit cycle of the system and is approached by all trajectories as $t\rightarrow \infty$. This can be shown rigorously by applying the same procedure that is given in Example 1 of Section 9.7--in fact, the polar system of equations that must be solved is the exact same.
Title: Re: Day Section's Quiz - Problem 1
Post by: Victor Ivrii on April 03, 2013, 11:04:42 AM
Yes. this is a stable limit cycle. In fact it is stable from both inside and outside.

Title: Re: Day Section's Quiz - Problem 1
Post by: Iven Poon on April 03, 2013, 02:16:21 PM
I already written my solution down on Monday, and I planned to post on Thursday when Professor Ivrii uploaded the questions...

Anyway, I would still like to share how I did this question.

Title: Re: Day Section's Quiz - Problem 1
Post by: Victor Ivrii on April 03, 2013, 04:30:52 PM
I already written my solution down on Monday, and I planned to post on Thursday when Professor Ivrii uploaded the questions...

Anyway, I would still like to share how I did this question.

Not meâ€”Sabrina. I just improved her post and split it into 2