Toronto Math Forum
MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:10:42 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.
$$\left\{\begin{aligned}
&\frac{dx}{dt} = y +x(1-x^2 - y^2),\\
&\frac{dy}{dt} = -x + y(1-x^2 - y^2)
\end{aligned}\right.$$
Bonus: Computer generated picture
-
(a)
\begin{equation}
\left\{
\begin{array}{**lr**}
y+x-x^{3}-xy^{2}=0 & \\
-x+y-x^{2}y-y^{3}=0\\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
x^{2}+y^{2}=0
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
x=0 & \\
y=0\\
\end{array}
\right.
\end{equation}
Therefore, the only critical point is (0,0)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
1-3x^{2}-y^{2} & 1-2xy \\
-1-2xy & 1-x^{2}-3y^{2}
\end{bmatrix}\\
~\\
J(0,0) &= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
\end{align*}
(c)
\begin{align*}
For (0,0), let A&= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
-1 & 1-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda-1)^{2}+1=0\\
~\\
\lambda &= 1 \pm i \\
~\\
Then \ the \ system \ has \ a \ clockwise \ spiral \ outwards \ at \ (0,0) \\
\end{align*}
(d) In the attachment.
-
There is my solution
-
Yulin, your picture sucks.
Doris, please do not post several minor pdfs
On the picture attached you can see, indeed, unstable focal point, clockwise and a stable limit cycle