MAT244--2018F > Quiz-7

Q7 TUT 0601

(1/1)

**Victor Ivrii**:

(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**:

Solution

**Jerry Qinghui Yu**:

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.

**Mengfan Zhu**:

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.

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