### Author Topic: Q7-T0201  (Read 1727 times)

#### Victor Ivrii

• Elder Member
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##### Q7-T0201
« 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.

« Last Edit: March 30, 2018, 12:21:34 PM by Victor Ivrii »

#### Jared Jubas-Malz

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##### Re: Q7-T0201
« Reply #1 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.
« Last Edit: March 30, 2018, 01:42:37 PM by Jared Jubas-Malz »