# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:09:43 PM

Title: Q7 TUT 0701
Post by: Victor Ivrii 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
Title: Re: Q7 TUT 0701
Post by: Tzu-Ching Yen 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.
Title: Re: Q7 TUT 0701
Post by: Tianyu Guo on November 30, 2018, 04:56:58 PM
The phase portrait is shown below.