# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on March 30, 2018, 12:17:11 PM

Title: Q7-T0201
Post by: Victor Ivrii on March 30, 2018, 12:17:11 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} = x + x^2 + y^2\\ &\frac{dy}{dt} = y - xy \end{aligned}\right.

Title: Re: Q7-T0201
Post by: Jared Jubas-Malz on March 30, 2018, 01:28:49 PM
The critical points would be when $x'=0$ and $y'=0$:
\begin{align}0=x+x^2+y^2\end{align}
\begin{align}0=y-xy\end{align}
From $(1)$ and $(2)$ the critical points would be $(0,0)$ and $(-1,0)$.
The Jacobian matrix would be:
\begin{align}J=\begin{pmatrix}F_{x}&F_{y}\\G_{x}&G_{y}\end{pmatrix}=\begin{pmatrix}2x+1&2y\\-y&1-x\end{pmatrix}\end{align}
At $(0,0)$:
\begin{align}J=\begin{pmatrix}1&0\\0&1\end{pmatrix}\end{align}
Since it is a triangular matrix, the eigenvalues would just be the diagonal entries so $r_{1}=r_{2}=1$. This means that $r_{1}=r_{2}>0$ so the nonlinear system would be an unstable proper node or spiral point.
At $(-1,0)$:
\begin{align}J=\begin{pmatrix}-1&0\\0&2\end{pmatrix}\end{align}
Since it is a triangular matrix, the eigenvalues would just be the diagonal entries so $r_{1}=-1$ and $r_{2}=2$. Since $r_{1}<0<r_{2}$ the nonlinear system would be an unstable saddle point.