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SSG6.5-6%20%20Stability%20and%20Bifurcations

SSG6.5-6%20%20Stability%20and%20Bifurcations - 6.5 PHASE...

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6.5. PHASE LINES AND CLASSIFYING EQUILIBRIA 805 6.5 Phase Lines and Classifying Equilibria In this section and the next section, we focus on autonomous (independent of time) differential equations dy dt = f ( y ) In the previous section we noted that the slope field for an autonomous differential equation is time-independent. Since each vertical line in the slope field contains all the information about the slopes, the slope fields contains an infinite amount of redundancy. In this section we trim off this redundancy using phase lines and discuss classifying equilibria , the y -values for which f ( y ) = 0. Phase lines In the last section we sketched slope fields by determining where the slope is zero (nullcline), and where it is positive and where it is negative. In this section, we consider a phase-line diagram that collapses the two-dimensional slope field to the y -axis without losing any information regarding the qualitative behavior of solutions to the differential equation dy dt = f ( y ) (e.g. see Figure 6.26). The following procedure creates a phase line. Phase Lines To draw a phase line for dy dt = f ( y ), Step 1. Draw a vertical line corresponding to the y -axis. Step 2. Draw solid circles on this line corresponding to the equilibria of dy dt = f ( y ) . That is, y -values where f ( y ) = 0. Step 3. Draw an upward arrow on intervals where f ( y ) > 0. On these intervals, solutions of the differential equation are increasing. Step 4. Draw a downward arrow on intervals where f ( y ) < 0 . On these intervals, the solutions of the differential equation are decreasing. Example 1. Phase lines for clonal genotypes Consider two clonally reproducing lines of the same species ( i.e. individual replicate themselves rather than reproducing sexually) exhibiting two genotypes a and A and whose per-capita growth rates are r a and r A , respectively. Suppose these two clonal lines are growing together in the same population and let y denote the proportion of genotype © 2010 Schreiber, Smith & Getz
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806 6.5. PHASE LINES AND CLASSIFYING EQUILIBRIA Figure 6.26: An illustration of how the three qualitatively different solution zones y < 0, 0 < y < 1, and y > 1, separated by the two equilibrium solutions y = 0 and y = 1 associated with the logistic equation dy dt = y (1 y ), can be collapsed on the y axis by removing (or projecting down) the time axis t . a in this population. It it left as an exercise(see Problem 39) to show that the variable y satisfies the equation dy dt = ( r a r A ) y (1 y ) . a. Draw the phase line for this equation when r a > r A . b. Draw the phase line for this equation when r a < r A . c. Discuss why this makes sense. Solution. a. Begin by drawing the y -axis. The equilibria are determined by the solutions of 0 = ( r a r A ) y (1 y ) Since the equilibria are y = 0 and y = 1, we draw solid circles on the y axis at these y -values. Since r a > r A , we have dy dt > 0 for 0 < y < 1 and we draw an upward arrow on this interval. Since dy dt < 0 for y > 1 and y < 0, we draw downward arrows on these intervals. This results in the phase line illustrated in Figure 6.27a.
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