Toronto Math Forum
MAT2442013S => MAT244 MathTests => Quiz 5 => Topic started by: Victor Ivrii on April 03, 2013, 08:07:49 PM

\begin{equation*}
\left\{\begin{aligned}
&\frac{dx}{dt} = x(1.50.5xy),\\
&\frac{dy}{dt} = y(2y1.125x).
\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.

(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} 1.51.5xy & x \\ 1.125y & 22y1.125x \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).

solution

Part One of Answer

Part Two of Answer

Benny, nice try, but ...
Devangi â€“ good, Frankâ€”even better. But phase portrait is still missing (and bonus question too)

Benny, nice try, but ...
Devangi â€“ good, Frankâ€”even better. But phase portrait is still missing (and bonus question too)

Benny, nice try, but ...
Devangi â€“ good, Frankâ€”even better. But phase portrait is still missing (and bonus question too)

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 outunless, by chance, the initial populations lie on the separatrix that passes through $(4/5,11/10)$.

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 outunless, 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.