Night Sections

[Victor Ivrii]

\begin{equation*}
\left\{\begin{aligned}
&\frac{dx}{dt} = x(1.5-0.5x-y),\\
&\frac{dy}{dt} = y(2-y-1.125x).
\end{aligned}\right.
\end{equation*}


This problem can be interpreted as describing the interaction of two species with population densities $x$ and $y$.


(b)  Find the critical points.

(c)  For each critical point find the corresponding linear system.  Find the eigenvalues and eigenvectors of the linear system;   classify each critical point as to type, and determine whether  it is asymptotically stable, stable, or unstable.

(d) Sketch the trajectories in the neighborhood of each critical point.

(e) [Bonus]  Find, if possible, solution in the form $H(x,y)=C$ and sketch the phase portrait.

[Frank (Yi) Gao]

(b) Finding the critical points means solving:
 \begin{equation*}
  0 = x(1.5 - 0.5x - y),
  0 = y(2 - x - 1.125y)
 \end{equation*}
There are four possibilities then,
 \begin{equation*}
 (x,y) = (0,0),(0,2),(3,0),(\frac{4}{5},\frac{11}{10})
 \end{equation*}
(c) The Jacobian for this question is then,
 \begin{equation*}
 J = \left( \begin{array}{cc} 1.5-1.5x-y & -x \\ -1.125y & 2-2y-1.125x \end{array} \right).
 \end{equation*}
For the point (0,0):
 \begin{equation*}
 J= \left( \begin{array}{cc} 1.5 & 0 \\ 0 & 2 \end{array} \right).
 \end{equation*}
So,
 \begin{equation*}
 \lambda_{1} = 1.5, \lambda_{2} = 2 \\
 \xi_{1} = \left(\begin{array}{cc} 1 \\ 0 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 0 \\ 1 \end{array} \right) 
 \end{equation*}
and thus, (0,0) represents a unstable node. For (0,2),
 \begin{equation*}
  J= \left( \begin{array}{cc} -0.5 & 0 \\ -2.25 & -2 \end{array} \right).
 \end{equation*}
So,
  \begin{equation*}
  \lambda_{1} = -0.5, \lambda_{2} = -2 \\
  \xi_{1} = \left(\begin{array}{cc} 3 \\ 1 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 0 \\ 1 \end{array} \right)
 \end{equation*}
and thus (0,2) its a node which is asymptotically stable. For (3,0),
 \begin{equation*}
  J= \left( \begin{array}{cc} -1.5 & -3 \\  0 & -1.375 \end{array} \right).
 \end{equation*}
So,
  \begin{equation*}
  \lambda_{1} = -1.5, \lambda_{2} = -1.375 \\
  \xi_{1} = \left(\begin{array}{cc} 1 \\ 0 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 24 \\ -1 \end{array} \right)
  \end{equation*}
and thus (3,0) is also an asymptotically stable node. Finally, for (4/5,11/10),
\begin{equation*}
  J= \left( \begin{array}{cc} -\frac{2}{5} & -\frac{4}{5} \\ -\frac{99}{80} & -\frac{11}{10} \end{array} \right).
 \end{equation*}
So,
  \begin{equation*}
  \lambda^{2} + \frac{3}{2}\lambda - \frac{11}{20} = 0 \\   
  \lambda_{1} = -1.8, \lambda_{2} = 0.3 \\
  \xi_{1} = \left(\begin{array}{cc} 1 \\ -1.4 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 1.57 \\ -1 \end{array} \right)
  \end{equation*}
and thus (4/5,11/10) is a saddle point (unstable).

[Benny Ho]

solution

[Devangi Vaghela]

Part One of Answer

[Devangi Vaghela]

Part Two of Answer

[Victor Ivrii]

Benny, nice try, but ...

Devangi – good, Frank—even better. But phase portrait is still missing (and bonus question too)

[Sabrina (Man) Luo]

Benny, nice try, but ...

Devangi – good, Frank—even better. But phase portrait is still missing (and bonus question too)

[Sabrina (Man) Luo]

Benny, nice try, but ...

Devangi – good, Frank—even better. But phase portrait is still missing (and bonus question too)

[Alexander Jankowski]

To complete this thread, here is a stream plot. Overall, the analysis tells us that in this case, one of the competing species must eventually die out--unless, by chance, the initial populations lie on the separatrix that passes through $(4/5,11/10)$.

[Victor Ivrii]

To complete this thread, here is a stream plot. Overall, the analysis tells us that in this case, one of the competing species must eventually die out--unless, by chance, the initial populations lie on the separatrix that passes through $(4/5,11/10)$.

But it exactly "one of" depending on initial conditions.