### Author Topic: FE-P5  (Read 6056 times)

#### Victor Ivrii ##### FE-P5
« on: December 14, 2018, 08:03:41 AM »
Typed solutions only. Upload only pictures (at all stationary points on one picture and a general phase portrait  on another; for general one can use computer generated)

For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'  = x(3x +2y -30)\, , \\
&y'  = y(2y-x-6)\,.
\end{aligned}\right.
\end{equation*}

(a) Describe the locations of all critical points.

(b) Classify their types (including whatever relevant: stability, orientation, etc.).

(c)  Sketch the phase portraits near the critical points.

(d)   Sketch the full phase portrait of this system of ODEs.

Hint: avoid redundancy: asymptotically (un)stable node, unstable node, stable center

« Last Edit: December 14, 2018, 08:06:31 AM by Victor Ivrii »

#### Doris Zhuomin Jia

• Jr. Member
•  • Posts: 6
• Karma: 2 ##### Re: FE-P5
« Reply #1 on: December 14, 2018, 10:15:12 AM »
a) 𝑥(3𝑥+2𝑦−30)=0,𝑦(2𝑦−𝑥−6)=0
The critical points are (0,0), (0,3), (10,0) and (6,6)

#### Jerry Qinghui Yu

• Jr. Member
•  • Posts: 14
• Karma: 13 ##### Re: FE-P5
« Reply #2 on: December 14, 2018, 10:37:07 AM »
$$J=\begin{bmatrix} 6x+2y-30 & 2x\\ -y & 4y-x-6 \end{bmatrix}$$

at (0,0):
$$J=\begin{bmatrix} -30 & 0\\ 0 & -6 \end{bmatrix}$$
diagonal matrix with negative eigenvalues => stable node

at (0,3):
$$J=\begin{bmatrix} -24 & 0\\ -3 & 6 \end{bmatrix}$$
triangular matrix with eigenvalues -24 and 6 => saddle

at (10,0):
$$J=\begin{bmatrix} 30 & 20\\ 0 & -16 \end{bmatrix}$$
diagonal matrix with eigenvalues 30 and -16 => saddle

at (6,6):
$$J=\begin{bmatrix} 18 & 12\\ -6 & 12 \end{bmatrix}$$
eigenvalues are $15+3i\sqrt{7}, 15-3i\sqrt{7}$ => unstable spiral

#### Yvette Yu

• Newbie
• • Posts: 2
• Karma: 1 ##### Re: FE-P5
« Reply #3 on: December 14, 2018, 10:39:38 AM »
Here is diagram for (c) and (d)

#### Jingze Wang

• Full Member
•   • Posts: 30
• Karma: 25 ##### Re: FE-P5
« Reply #4 on: December 14, 2018, 10:42:58 AM »
This is the computer generated global phase portrait.

We already know that Wolfram Alpha provides rather crappy pictures here. V.I.
« Last Edit: December 15, 2018, 04:43:41 AM by Victor Ivrii »

#### Jerry Qinghui Yu

• Jr. Member
•  • Posts: 14
• Karma: 13 ##### Re: FE-P5
« Reply #5 on: December 14, 2018, 10:43:52 AM »
a clearer picture for d

#### Doris Zhuomin Jia

• Jr. Member
•  • Posts: 6
• Karma: 2 ##### Re: FE-P5
« Reply #6 on: December 14, 2018, 10:55:45 AM »
Sorry i don't know how to type math symbols, so I do part c.

#### Victor Ivrii ##### Re: FE-P5
« Reply #7 on: December 15, 2018, 04:36:09 AM »
So far nobody explained

1) why rotation in the focal (spiral) point is clockwise
2) why stable node at $(0,0)$ and saddles at $(0,3)$ and $(10,0)$ look this way (actually Yvette did but it should be typed)

(since I am grading this problem I decided to check). I will deal with your submissions at a later time
« Last Edit: December 15, 2018, 11:24:42 AM by Victor Ivrii »

#### Victor Ivrii ##### Re: FE-P5 - OFFICIAL
« Reply #8 on: December 15, 2018, 06:38:50 AM »
(a) Solving $x(3x +2y -30)=0$, $y(2y-x-6)=0$ we get cases
\begin{align*}
&x=0,\ y=0  &&\implies A_1=(0,0),\\
&x=0,\ 2y-x-6=0  &&\implies A_2=(0,3)\\
&y=0,\ 3x+2y-30=0 &&\implies A_3=(10,0),\\
&3x+2y-30=0,\ 2y-x-6=0 &&\implies A_4=(6,6).
\end{align*}
(b)  Linearizations at these points have matrices
\begin{align*}
&A_1&&A_2&&A_3&&A_4\\
&
\begin{pmatrix}
-30 &\ \ 0\\
0 &-6
\end{pmatrix}
&&
\begin{pmatrix}
-24 &\ \ 0\\
-3 &6
\end{pmatrix}
&&
\begin{pmatrix}
30 &20\\
0 &-16
\end{pmatrix}
&&
\begin{pmatrix}
18 &12\\
-6 &12
\end{pmatrix}\\
\text{with eigenvalues}\\
&\{-30,-6\} &&\{-24,6\} && \{30,-16\} && \{15-\sqrt{63}i,15-\sqrt{63}i \}
\end{align*}
correspondingly.

Therefore $A_1$ is a stable node, $A_2$ and $A_3$ are saddles, $A_4$ is an unstable focal point and since left bottom number is $-6<0$ it is clockwise oriented.

Directions  are:

$A_1$:  $\mathbf{f}_1=(1,0)^T$, $\mathbf{f}_2=(0,1)^T$ ; Since $-30=\lambda_1 <\lambda_2=-6$, all trajectories have an entry directions $\pm \mathbf{f}_2$ (except two, which have entry directions $\pm \mathbf{f}_1$).

$A_2$: $\mathbf{f}_1=(10,1)^T$ ($\lambda_1=-24$)--stable, $\mathbf{f}_2=(0,1)^T$ ($\lambda_2=6$)--unstable.

$A_3$: $\mathbf{f}_1=(1,0)^T$ ($\lambda_1=30$)--unstable, $\mathbf{f}_2=(1,-2.3)^T$ ($\lambda_2=-16$)--stable.

(c)--(d) One should observe that either $x=0$ in every point of the trajectory, or in no point; and that $y=0$ in every point of the trajectory, or in no point. It allows us to make a skeleton'' of the phase portrait (thick black lines) on the figure; red lines are very approximate

Remark: $A_4$ may not look as a focal point on the computer-generated plot but it is. The reason for this disparity is that for one rotation (so angle increases by $2\pi$) the radius increases in $\exp (2\pi 15/\sqrt{63})\approx 130\,000$ times and this is a really big number.
« Last Edit: December 15, 2018, 11:27:12 AM by Victor Ivrii »

#### Victor Ivrii « Reply #9 on: December 16, 2018, 04:04:01 AM »
I graded 55% (it goes slowly since there are 4 parts and 4 stationary points. Bad handwriting and usage of inappropriate writing tools like dull pencils slows down. I consider this as a disrespect and rudeness not only towards graders but to all class: slowing down grading pushes back marks release and submission of the final grades). In extreme cases unreadable parts were evaluated by 0. Remember, there is no "the benefit of the doubt".

Part (a)
Many students either failed to find all four stationary points or found wrong points. Each correctly found points added .5 to the score. Wrong points were 0 worth (but it could be $-.5$). In the next parts such points were ignored, except (d) which is worth 0 unless all nuts and bolts are in place in (a)--(c) and nothing is extra.

Parts (b) and (c)
1) The most stupid and frustrating: having a triangular or even diagonal matrix instead of simply writing down its diagonal elements as eigenvalues, many write down determinant, open parenthesis, get quadratic equation, solve it, sometimes incorrectly.  Three out of four points are in this category.

2) For point $(6,6)$ one needs to calculate eigenvalues, simple declarations are evaluated by $0$ which propagates to (c) and negates (d).

Most of the students failed to explain why rotation is clockwise. I decided not to punish for this.

Each correctly labelled point adds .75 to the score. Minor computational errors without serious consequences are not punished.

3) Point $(0,0)$ became tricky in (c): one needs to draw all entering lines (except to) as tangent to $y$-axis. Failure to do so reduces contribution of this point from .5 to .25 -- unless corrected in (d)

4) At $(0,3)$ and $(10,0)$ one needs to calculate eigenvectors and draw picture correspondingly. Failure to do so reduces contribution of each such point from .5 to .25 -- unless corrected in (d)

Part (d)
One needs to make a connected picture, not just a "skeleton". Picture should not be too crude. Crucial errors in (a)--(c) remove (d) from consideration.

General.
It looks like some students managed to find internet sites with advices "how to solve", mainly either useless, or harmful. There is no point to "simplify" triangular matrix before calculating its eigenvalues: they are written on the diagonal. What can be simpler than this? Surely, if you have a matrix with integer elements, all divisible by $p$, one can divide by $p$, calculate the eigenvalues of the divided matrix and then multiply them by $p$. You should not forget the last step. If $p>0$ while the picture does not change, it is still an error (even if I do not punish for it). If $p<0$ it shows that you do not completely understand what you are doing and no partial marks are given. Sorry "I made an error but my thinking process was correct"  does not cut here.