Author Topic: FE5  (Read 533 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
FE5
« on: December 08, 2014, 04:16:29 PM »
For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = x (5-2x-3y)\, , \\
&y'_t = y (5-3x-2y)
\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 phase portrait of this system of ODEs.



Solution

(a) Solving $x (5-2x-3y=0$, $y (5-3x-2y )$ we have 4 cases $x=y=0$, $x=5-3x-2y=0$, $y=5-2x-3y=0$ and $5-3x-2y =5-2x-3=0$ giving us 4 points $(0,0)$, $(0,\frac{5}{2})$, $(\frac{5}{2},0)$ and $(1,1)$.

(b) Let  $f= x (5-2x-3y)=5x-2x^2-3xy$, $g=y (5-3x-2y)=5y-2y^2-3xy$. Then $f_x=5-4x-3y$, $f_y=-3x$, $g_x=-3y$, $g_y=5-4y-3x$.

  • $(0,0)$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals $\begin{pmatrix} 5 & 0 \\ 0 &5\end{pmatrix}$ with eigenvalues $r_1=r_2=5$;  and eigenvectors $(1, 0)^T$ and $(0,1)^T$; unstable node;
  • $(0,\frac{5}{2})$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals  $\begin{pmatrix} -\frac{5}{2} & 0 \\ -\frac{15}{2} &-5\end{pmatrix}$ with eigenvalues $r_1=-\frac{5}{2},\ r_2=-5$ and eigenvectors $(1,-3)^T$ and $(1,0)^T$ respectively;  stable node;
  • $(\frac{5}{2},0)$; the same as in (2);
  • $(1,1)$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals  $\begin{pmatrix} -2 & -3 \\ -3 &-2\end{pmatrix}$ with eigenvalues $r_1=-5$ and $r_2=1$ and eigenvectors $(1,1)^T$ and $(1,-1)^T$ respectively; saddle.


(c-d) Plotting


Remark This is ``two competing species'' system.
« Last Edit: December 14, 2014, 04:09:03 PM by Victor Ivrii »