## Oiq

Figure 24.5 Phase; diagram tor example 24.14

Figure 24.5 Phase; diagram tor example 24.14

for which the solutions are/-1 = — I /2 and r2 = 1/2. Since the roots are of opposite sign, the steady-state solution is a saddle-point equilibrium.

Construct the phase diagram:

Step I Determine the motion of yBegin by graphing the yi isocline: setting y, = 0 to find the isocline gives the horizontal line y? = 2. Next, we note that y i < 0 below this isocline (when v2 < 2) and v, > 0 above the isocline (when >'2 >2). The appropriate horizontal arrows of motion are shown in figure 24.5. Step 2 Determine the motion ofy2. Begin by graphing the y2 isocline: setting y2 = 0 gives the vertical line yi = 2. To the right of this line (yi > 2). y2 > 0 and to the left of it ( Vj < 2). y2 < 0. The appropriate vertical arrows of motion are shown in the phase diagram.

The arrows of motion in figure 24.5 indicate that trajectories in the southwest and northeast sectors ofthe phase plane definitely move away from the steady state. But the arrows of motion in the northwest and southeast sectors show that trajectories move toward the steady state. What do the trajectories actually look like?

Figure 24.6 shows some representative trajectories. Consideran arbitrary starting point in figure 24.6 such as point o. At this point, the arrows of motion indicate that y i is decreasing and y2 is increasing. The motion is northwesterly, therefore. Follow tins trajectory along its path. As it gets c/ose to the yi isocline where v, = 0, the motion of yi slows down but vj continues to increase. As a result the

Figure 24.6 Phase diagram for example 24.14 showing representative trajectories; A saddle point trajectory bends upward, As it crosses the vi isocline, y( is stationary tor an instant even though y2 keeps increasing. As a result the trajectory must be vertical at the crossing. From there it proceeds into a new isosector in which both Vi and y2 are increasing. Thus the trajectory bends back and goes in a northeasterly direction. It stays in that isosector, traveling ever farther and farther away from the steady state.

Consider a trajectory starting at point b. Like the trajectory that started at a, the motion is northwesterly; however, this time the trajectory gets close to the Vi isocline where y2 = 0. so the motion of y2 slows down while yi continues to decrease. As a result the trajectory bends to the left. As it crosses the y2 isocline, it is horizontal because y2 is stationary at that point, even though yi keeps decreasing. In the new isosector, the trajectory turns southwesterly and continues in that direction, traveling away from the steady state.

Trajectories starting from points c and d are also shown. These have the minor-image properties of the trajectories starting from points a and b. These four arbitrarily chosen trajectories verify that most trajectories end up diverging from a steady state which is a saddle-point equilibrium. These trajectories also demonstrate the very important property that trajectories must obey the arrows of motion and must be horizontal when they cross the y2 isocline and vertical when they cross the yi isocline.

Since the steady state is a saddle-point equilibrium, we know that some trajectories do converge to the steady state, provided they start from initial conditions satisfying theorem 24.5. By theorem 24.5. the saddle path is given by

Figure 24.6 Phase diagram for example 24.14 showing representative trajectories; A saddle point

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