Quiz 4--Problem (night sections)

[Victor Ivrii]

9.2 p 517, # 10(a,c)

Consider system
\begin{equation*}
\left\{\begin{aligned}
&\frac{dx}{dt}=(2+x)(y-x),\\
&\frac{dy}{dt}=y(2+x-x^2);
\end{aligned}\right.
\end{equation*}
(a - 1 points) Find all critical points (equilibrium solutions) and write  the linearization of the system at each critical point;

(b - 3 points) For the linearized systems at each critical point draw the phase portrait and identify its type (including stability, if applicable,  the orientation, etc.);

(Bonus - 1 point) Describe (by drawing) the basin of attraction for each  asymptotically stable point (if there is any) and determine for each
critical point whether the phase portrait will not change in the nonlinear  system.

[Rudolf-Harri Oberg]

a)For critical points, we just set $x'=0$ and $y'=0$ which will yield four critical points: $P1=(0,0), P2=(-1,-1), P3=(2,2), P4=(-2,0)$. For linearization, we have to compute the Jacobian i.e the matrix of first derivatives of the functions. If $F=(2+x)(y-x), G=y(2+x-x^2)$, then $F_x=y-2x-2, F_y=2+x, G_x=y(1-2x), G_y=2+x-x^2$. Now we have got everything to find corresponding linearized systems, just use equation 13 from page 522 (book).

b) To determine the nature of solutions at each critical point, we just have to find the eigenvalues of the matrices we get from evaluating the Jacobian at the respective critical points.

For the point $P1$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-2$, so the the critical point is a saddle (look picture from handout), this is unstable.

For the point $P2$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+r+3=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P3$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+4r+24=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P4$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-4$, so the the critical point is a saddle, this is unstable.

For visualization of solutions, go to the course homepage and look at the section of learning resources. Pick the link to math.rice, select PPLANE.2005.10, type in the equations and enjoy the picture! You can also see the answer to the bonus question from the picture.

[Matthew Cristoferi-Paolucci]

With linearization matrices at critical points

[Matthew Cristoferi-Paolucci]

Some classifications for part b

[Victor Ivrii]

Let us start from simplest arguments:
1) $x'=(x+2)(x-y)$ and we see that $x=-2$ is a solution. Then $y'=2y$ and therefore movement from point $(-2,0)$.

2) $y'=y(2+x-x^2)$ and we see that $y=0$ is a solution. Then $x'=x(x+2)$ and then movement is to point $(-2,0)$ and from point $(0,0)$.

So we get what I attached on Q5Nschema.png in blue.

Finding equilibrium points is easy $\bigl[x+2=0 \text{ OR } x-y=0\bigr]\text{ AND } \bigl[y=0\text{ OR } x=-1\text{ OR }x=2 \bigr]$ which is equivalent to cases 1,2) $y=0$ and $x=0,-2$ and 3,4) $y=x=2,-1$ which shows 4 points found both by Rudolf-Harri and Matthew (red on the same picture).

Both found eigenvalues but one also needs to find
a) in the case of real eigenvalue also eigenvectors showing directions of stable and unstable separatrices. Actually at $(-2,0)$ both directions are obvious and in $(0,0)$ one is obvious but one needs to find another one and indicate which are stable and which are not (easy).

b) In the case of stable focal points $(2,2)$ and $(-1,-1)$ one needs to say clockwise or counter-clockwise. Since Matthew wrote matrices it would be very easy for him.

c) And I expect from you to do an actual computer simulation and post picture and define basins of two stable focal points

[Sabrina (Man) Luo]

Some classifications for part b

For the Thrid Matrix A: Det(A) should be r^2+r+3=0, and r=-1/2±i sqrt 11/2

[Victor Ivrii]

Some classifications for part b

For the Thrid Matrix A: Det(A) should be $r^2+r+3=0$, and $r=-1/2\pm i \sqrt{ 11}/2$

Which point you are talking about?

[Sabrina (Man) Luo]

I am talking about the second picture (classification of part b), the written solution of the third equilibrium part. There is a calculating error. I just corrected it.

[Sabrina (Man) Luo]

I am talking about the second picture (classification of part b), the written solution of the third equilibrium part. There is a calculating error. I just corrected it.

[Victor Ivrii]

I converted your attachment in .doc format (which is not a graphics format and I allow to post it "just in case") to png. Still the picture is correct near equilibrium points but not completely correct globally. We already know that $x=-2$ consists of solutions and thus cannot be crossed by other solutions.

I am talking about the second picture (classification of part b), the written solution of the third equilibrium part. There is a calculating error. I just corrected it.

We have 4 points $(-2,0)$, $(0,0)$, $(-1,-1)$ and $(2,2)$. Which of them? Why one needs to look at not properly oriented crappy snapshot to learn what point you are referring to?

Still I am waiting for someone to post computer simulation picture. I am also waiting for answer to the bonus problem. I will post by myself during weekend and topic will be closed.

[Victor Lam]

We got saddle points (unstable) for both (0,0) and (-2,0). Also (-1,-1) and (2,2) are both asymptotically stable spirals of clockwise orientation. Solutions can be seen in the visual.
Note as t approaches infinity, all points satisfying x > -2 ultimately approach the particular critical points.

[Victor Lam]

Also, the line y = 0 is not included in the basin of attraction, as we look at the original system for dy/dt, there clearly is a y term there, which will just stay on the line y = 0. Hence, the basin of attraction for the two critical points can be represented by the 2 rectangles (one above y = 0, the other below y = 0).