MAT244--2018F > Quiz-7

Q7 TUT 0101

(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} = x + x^2 + y^2, \\

&\frac{dy}{dt} = y - xy.

\end{aligned}\right.$$

Bonus: Computer generated picture

**Chonghan Ma**:

(a)

Set x’ = 0 and y’=0

Then we have critical points (0,0), (-1,0)

(b)

J = \begin{bmatrix}2x+1 & 2y \\-y & 1-x \end{bmatrix}

Linear systems are shown with each critical point:

J(0,0) = \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}

J(-1,0) = \begin{bmatrix}-1 & 0 \\0 & 2 \end{bmatrix}

(c)

Eigenvalues are computed by det(A - tI)= 0

So that

At (0,0): t=1

Critical point is an unstable proper node or spiral point

At (-1,0): t=-1 and 2

Critical point is an unstable saddle point

**Jingze Wang**:

This is computer generated picture :)

**Mengfan Zhu**:

Hello, I solve this question step by step.

Please see the picture below.

If there are any mistakes, reply to me anytime.

Thank you very much.

**Victor Ivrii**:

Jingze,

You took too large range by $y$ and ...

When we have non-zero repeated eigenvalues, there is always a node (stable or unstable). Spiral may appear in this situation only when the right-hand is not smooth, which is not the case. On the picture attached you can see...

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