# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:09:07 PM

Title: Q7 TUT 0601
Post by: Victor Ivrii on November 30, 2018, 04:09:07 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} = 1 - y,\\ &\frac{dy}{dt} = x^2 - y^2. \end{aligned}\right.

Bonus: Computer generated picture
Title: Re: Q7 TUT 0601
Post by: Zhiya Lou on November 30, 2018, 04:30:46 PM
Solution
Title: Re: Q7 TUT 0601
Post by: Jerry Qinghui Yu on November 30, 2018, 04:39:09 PM
Solve $1-y = 0\Rightarrow y=1$ and $x^2 - y^2 = (x+y)(x-y) = 0$, substituting $y=1$, we have 2 critical points $(-1, 1), (1,1)$

Computing the Jacobian yields
\begin{align}
J =
\begin{bmatrix}
F_x & F_y\\
G_x & G_y
\end{bmatrix}
=
\begin{bmatrix}
0 & -1\\
2x & -2y
\end{bmatrix}
\end{align}
Plugging in both critical points we have 2 linear systems, first
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
-2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + \sqrt{3}\, \lambda_2 = -1 - \sqrt{3}$, since $\sqrt{3} > 1$, we have $\lambda_1 > 0 > \lambda_2$, so we conclude locally $(-1, 1)$ is a saddle.
Plug in $(1,1)$
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + i\, \lambda_2 = -1 - i$, a complex conjugate with negative real parts, we conclude locally $(1, 1)$ is an asymptotically stable spiral.
Title: Re: Q7 TUT 0601
Post by: Mengfan Zhu on December 01, 2018, 02:54:58 AM
For this question,
there are two critical points: (-1,1) (1,1)
At (-1,1), it's an unstable saddle point.
At (1,1), it's an asymptotically stable spiral point.