MAT244-2013S > Quiz 5

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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: