Author Topic: Q7 TUT 0601  (Read 2155 times)

Victor Ivrii

• Administrator
• Elder Member
• Posts: 2599
• Karma: 0
Q7 TUT 0601
« 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

Zhiya Lou

• Jr. Member
• Posts: 12
• Karma: 12
Re: Q7 TUT 0601
« Reply #1 on: November 30, 2018, 04:30:46 PM »
Solution

Jerry Qinghui Yu

• Jr. Member
• Posts: 14
• Karma: 13
Re: Q7 TUT 0601
« Reply #2 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.
« Last Edit: November 30, 2018, 04:52:08 PM by Jerry Qinghui Yu »

Mengfan Zhu

• Jr. Member
• Posts: 10
• Karma: 5
Re: Q7 TUT 0601
« Reply #3 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.