Chapter 6.3, Problem 8P

Problems 8 through 10 deal with the competition system $\begin{array}{l}\frac{dx}{dt}=60x-3x^2-4xy\\ \frac{dy}{dt}=42y-3y^2-2xy\end{array}$ (3) in which $c_1{c}_2=8<9=b_1{b}_2$, so the effect of inhibition should exceed that of competition. The linearization of the system in (3) at (0, 0) is the same as that of (2). This observation and Problems 8 through 10 imply that the four critical points (0, 0), (0, 14), (20, 0), and (12, 6) of (3) resemble those shown in Fig. 6.3. 13—a nodal source at the origin, a saddle point on each coordinate axis, and a nodal sink interior to the first quadratic. In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Finally, construct a first-quadrant phase plane portrait for the nonlinear system in (3). Do your local and global portraits look consistent?

Show that the linearization of (3) at (0, 14) is $u\text{'}=4u,v\text{'}=-28u-42v$. Then show that the coefficient matrix of this linear system has the positive elgenvalue ${\lambda }_1=4$ and the negative eigenvalue ${\lambda }_2=42$. Hence (0, 14) is a saddle point for the system in (3).