### Author Topic: Q7-T0501  (Read 2103 times)

#### Victor Ivrii

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##### Q7-T0501
« on: March 30, 2018, 12:22:31 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.

#### Darren Zhang

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##### Re: Q7-T0501
« Reply #1 on: March 30, 2018, 01:13:14 PM »
(a) The critical points are solutions of the equations,$1-y = 0$, $(x-y)(x+y) = 0$
The first equation requires that y= 1 . Based on the second equation, x =1/-1. Hence the critical points are (-1,1) and (1,1).
(b,c) $F(x,y) = 1-y$ and $G(x,y) = x^2-y^2$. The Jacobian matrix of the vector field is $$J = \begin{pmatrix} 0 & -1 \\ 2x & 2y \end{pmatrix}$$
At the critical point $(-1,1)$, the coefficient matrix of the linearized system is $$J(-1,1) = \begin{pmatrix} 0 & -1 \\ -2 & -2 \end{pmatrix}$$
with eigenvalues $r_1 = -1-\sqrt{3}$ and $r_2 = -1+\sqrt{3}$ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (1,1), the coefficient matrix of the linearized system is  $$J = \begin{pmatrix} 0 & -1 \\ 2 & -2 \end{pmatrix}$$ with complex conjugate eigenvalues $r_1 = -1+i$, $r_2 = -1-i$. The critical point is stable spiral, which is asymptotically stable.
Attached is the part(d)