MAT244-2013S > Quiz 5

Day Section's Quiz - Problem 1

**Sabrina (Man) Luo**:

(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.

**Victor Ivrii**:

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.

**Alexander Jankowski**:

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.

**Victor Ivrii**:

Yes. this is a stable limit cycle. In fact it is stable from both inside and outside.

**Iven Poon**:

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.

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