Author Topic: FE-P5  (Read 589 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1855
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
FE-P5
« on: April 11, 2018, 08:47:26 PM »
For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = x (x -y+1)\, , \\
&y'_t = y (x - 2)\,.
\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.


Tim Mengzhe Geng

  • Full Member
  • ***
  • Posts: 21
  • Karma: 6
    • View Profile
Re: FE-P5
« Reply #1 on: April 12, 2018, 12:34:35 AM »
For part(a)
Let
\begin{equation}
    x(x-y+1)=0
\end{equation}
and
\begin{equation}
    y(x-2)=0
\end{equation}
We will have all the three critical points
\begin{equation}
    (x,y)=(0,0),(2,3) or (-1,0)
\end{equation}
For part(b)
\begin{equation}
   F=x(x-y+1)
\end{equation}
\begin{equation}
   G=y(x-2)
\end{equation}
Therefore, the Matrix $J$
\begin{equation}
   J={
\left[\begin{array}{ccc}
2x-y+1 & -x \\
y & x-2
\end{array}
\right ]},
\end{equation}
At point(0,0), we have
\begin{equation}
   J[0,0]={
\left[\begin{array}{ccc}
1 & 0 \\
0 & -2
\end{array}
\right ]},
\end{equation}
Eigenvalues are
\begin{equation}
   \lambda_1=-2
\end{equation}
\begin{equation}
   \lambda_2=1
\end{equation}
Therefore, (0,0) is a saddle point and thus unstable.
At point(2,3), we have
\begin{equation}
   J[2,3]={
\left[\begin{array}{ccc}
2 & -2 \\
3 & 0
\end{array}
\right ]},
\end{equation}
\begin{equation}
   \lambda_3=1+\sqrt{5}i
\end{equation}
\begin{equation}
   \lambda_4=1-\sqrt{5}i
\end{equation}
Therefore, (2,3) is a spiral point and is unstable. The orientation is counterclockwise.
\begin{equation}
   J[-1,0]={
\left[\begin{array}{ccc}
-1 & 1 \\
0 & -3
\end{array}
\right ]},
\end{equation}
\begin{equation}
   \lambda_5=-1
\end{equation}
\begin{equation}
   \lambda_6=-3
\end{equation}
Therefore, (-1,0) is a node and is asymptotically stable.

Nikola Elez

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 2
    • View Profile
Re: FE-P5
« Reply #2 on: April 12, 2018, 12:38:13 AM »
For part c/d

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1855
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: FE-P5--solution
« Reply #3 on: April 18, 2018, 06:48:47 AM »
a.  Solving $x(x-y+1)=0$, $y(x-2)=0$ we get cases
\begin{align*}
&x=y=0  &&\implies A_1=(0,0),\\
&x=x-2=0  &&\implies \text{impossible}\\
&y=x-y+1=0 &&\implies A_2=(-1,0),\\
&x-y+1=x-2=0 &&\implies A_3=(2,3).
\end{align*}
b. Linearizations at these points have matrices
\begin{align*}
&
\begin{pmatrix}
1 &\ \ 0\\
0 &-2
\end{pmatrix}
&&
\begin{pmatrix}
-1 &\ \ 1\\
0 &-3
\end{pmatrix}
&&
\begin{pmatrix}
2 &-2\\
3 &0
\end{pmatrix}
\\[5pt]
\text{with eigenvalues    }&\{1,-2\} &&\{-1,-3\} && \{1-\sqrt{5}i,1+\sqrt{5}i \}
\end{align*}
and therefore
* $A_1$ is a saddle,
* $A_2$ is a stable node, and
* $A_3$ is unstable focal point and since left bottom number is $3>0$ it is counterclockwise oriented.


c. Axis are:
in $A_1$:  $\mathbf{e}_1=(1,0)^T$ unstable ($\lambda_1=1$), $\mathbf{e}_2=(0,1)^T$ stable ($\lambda_2=-2$).

in $A_2$: $\mathbf{f}_1=(1,0)^T$ ($\lambda_1=-1$), $\mathbf{f}_2=(1,-2)^T$ ($\lambda_1=-3$). Since $\lambda_1 >\lambda_2$, all trajectories have an entry directions $\pm \mathbf{f}_1$ (except two, which have entry directions $\pm \mathbf{f}_2$). Then we draw trajectories near critical points (See attachment  P5-loc.png).

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 (see attachment), impose local pictures on it and finally draw a global portrait
« Last Edit: April 18, 2018, 11:01:01 AM by Victor Ivrii »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1855
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
FE-P5 Comments
« Reply #4 on: April 19, 2018, 01:16:12 PM »
a. Some students missed some stationary points and/or reported wrong points. All further analysis in the wrong points was ignored as irrelevant. For all three correct points found I gave 10pts, with a reduction of 3 pts for each missed points, and 1pts for extra points (only for those who found 3 correct points).

b. Linearization was easy, but some borked it. Finding eigenvalues was supposed to be a breeze but ...
One does not need to solve any equations to find eigenvalues of the diagonal or triangular matrices (some students wrote wrong equations and found wrong eigenvalues). Also eigenvectors of the diagonal matrices are obvious, and of the triangular are easy.

Not everyone found correctly eigenvalues of the matrix at $(2,3)$. T recall, that they are $1\pm i\sqrt{5}$. One does not need to look for eigenvectors, however one should look at the sign of bottom left element of the matrix and conclude what is the direction of rotation (I have not subtracted points for missing justification that it is counter-clockwise)

c. Drawing of the local pictures. At (0,0) mant=y draw incoming and outgoing lines like X instead of +, others indicated the wrong directions. Point $(-1,0)$ was more difficult. And in $(2,3)$ some draw "hairy monsters"

d. Even when all local pictures were drawn correctly, some students draw intersecting lines (trajectories do not intersect!) and not everyone observed that $x=0$ and $y=0$ consist of trajectories (see "skeleton" in my post above)



In some papers with no calculations or with calculations, leading to wrong conclusions) there are "miraculously" correct pictures. Those were discarded because "only solutions (not just answers) are evaluated". 
« Last Edit: April 20, 2018, 05:55:56 AM by Victor Ivrii »