Author Topic: Q7 TUT 0701  (Read 1999 times)

Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
Q7 TUT 0701
« on: November 30, 2018, 04:09:43 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 - y^2, \\ &\frac{dy}{dt} = y - x^2. \end{aligned}\right.

Bonus: Computer generated picture

Tzu-Ching Yen

• Full Member
• Posts: 31
• Karma: 22
Re: Q7 TUT 0701
« Reply #1 on: November 30, 2018, 04:28:06 PM »
$0 = y - x^2, 0 = x - y^2$
$x = x^4, (x,y) = (1, 1), (0, 0)$

$F = x - y^2, F_x = 1, F_y = -2y$
$G = y - x^2, G_x = -2x, G_y = 1$
at $(x, y) = (1, 1)$
$\left[ {\begin{array}{cc} 1 & -2 \\ -2 & 1 \\ \end{array} } \right]$
gives characteristic equation
$r^2 - 2r - 3 = 0, r = 3, -1$
Solutions are real but opposite sign so near (1, 1) it's a saddle point

at $(x, y) = (0, 0)$
$\left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$

$r = 1$
Solutions are real and positive. Near (0, 0) is a unstable node.

Tianyu Guo

• Newbie
• Posts: 2
• Karma: 2
Re: Q7 TUT 0701
« Reply #2 on: November 30, 2018, 04:56:58 PM »
The phase portrait is shown below.