1550078481-Ordinary_Differential_Equations__Roberts_

PORTRAIT User's Guide 531

The solution values of the component yl (the prey population) decrease from

the initial value of 5 at time 0 to a minimum value of 2.8 56719 at time 0.8055.

Then the prey population increases to a value of 17.23472 at time 3.8745

and then decrease again on the interval [3.8745, 4.5]. The solution values of

the component y2 (the predator population) decreases from an initial value

of 5 at time 0 to a value of 0.4183726 at time 2.6055. Then the predator

population increases on the interval [2.6055, 4 .5]. These values were obtained

by clicking the OPTIONS button at the upper right corner of the graph and

then selecting option three, PRINT solution on the monitor, setting i = 1,

pressing the Enter key, and scrolling down and up the table of solution values

to find component minima and maxima and associated times.

We still want to produce a phase-plane graph of y(t) versus x(t)-that is,

y2(x) versus y1(x)- for both IVP (3)(i) and (3)(ii), so we click the OPTIONS

button. This causes the set of options to reappear. Since we have not yet

solved the IVP (3)(ii), we select the fifth option, INPUT CONDITIONS for

next IVP to be solved, and click OK Since for (ii) the interval of integration

is [O, 5] and the initial conditions are y 1 (0) = 10 and Y2 (0) = 5, for i = 2

in the box under bi = we enter 5, in the box under dli = we enter 10, and

in the box under d2i = we enter 5. After checking to see that the values for

the interval of integration and the initial conditions are entered correctly, we

click the VERIFY button and then click the INTEGRATE button. When the

option screen appears on the monitor, we select option two, PHASE-PLANE

graph of any subset of IVPs, and click the OK button. Check boxes for the

IVPs to graph appear. Since we wish to produce the phase-plane graph of

the solution of both IVPs, we click the box to the left of 1, we click the box

to the left of 2, and then we click the OK button. An Information for Graph-

ing box and values for the horizontal range [Hmin, Hmax] and vertical range

[Vmin, Vmax] are displayed. The values which appeared in this instance were

Hmin = 2.856719E + 00 , Hmax = l.723472E + 01, Vmin = 4.183726E - 01 ,

and Vmax = 5.493573E + 00. We click the OK button in the Information

for Graphing box. Since the problem instructed us to display the phase-

plane graph on the rect angle R with Hmin = 0, Hmax = 20 , Vmin = 0, and

Vmax = 6, we enter these values in the appropriate boxes and click the
VERIFY HMIN, HMAX, VMIN, VMAX button. This causes the graph
shown in Figure B.6 to appear on the monitor. The outside curve in Fig-
ure B.6 is the phase-plane graph of the first IVP (3)(i) and the inside curve is
the phase-plane graph of the IVP (3)(ii). The initial point (5, 5) of the outside
curve is the left endpoint of the curve and corresponds to the initial conditions

yl(O) = 5 and y2(0) = 5 of the IVP (3)(i). As time increases the phase-plane

solution of the IVP (3)(i) proceeds around the outside curve in the counter-

clockwise direction. Likewise, as time increases the phase-plane solution of the

IVP (3) (ii) proceeds around the inside curve in the counterclockwise direc-

tion.