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Quiz 5 / Re: Night Sections

« on: April 03, 2013, 08:49:26 PM »

(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).