1550078481-Ordinary_Differential_Equations__Roberts_

PORTRAIT User's Guide 535

Next, we generated solutions to system (4) which corresponded to the ini-

tial conditions (iv) x(O) = 1.5, y(O) = 0, (v) x(O) = 2, y(O) = 0, and

(vi) x(O) = 5, y(O) = 0. The trajectories corresponding to these solutions

begin on the x -axis- the lower boundary of R. The trajectory which begins
at (1.5, 0) heads toward the point (0 .5, 0) on the boundary of R. We had to
rerun the IVP (iv) and limit the interval of integration to [O, 0.53] so that its
graph would remain in the rectangle. The trajectories corresponding to the
initial conditions (v) and (vi) proceed toward the critical point (2, 3). A graph
of the trajectories corresponding to the initial conditions (i) through (vi) are
displayed in Figure B.9.

'" PORTRAIT - - - l!ili!Jra

Enter the 19stem defining functions F1. F2

11 = F1 = l3'y2·2"yl'J'1-1

Ente1 the number of initial value problems lo solve

1'2 = FZ = ja•yJ·y2)'2-7

For each initial value problem i. eiiter lhe interval of integralion [ai. bi]. the initial value ci.

and ttle initial conditions dl i and d2i.

a i"' bia d1i•

1.1

2.5

d2i.

2.5

Horizontal Range

Hmin ·I,_ ___ ,

Hmax•j

, Vertical --~-­Range

Vmin = 0

Vmax = 5

Figure B.9 Phase-Plane Graph of System (4) Using

Initial Conditions (i) Through (vi)

We completed the phase-plane portrait by generating solutions to the IVPs

consisting of system (4) and the initial conditions (vii) x(O) = 1, y(O) = 5,

(viii) x(O) = 3, y(O) = 5, (ix) x(O) = 5, y(O) = 4.5, and (x) x(O) = 5,

y(O) = 3. These traj ectories all begin on the boundary of the rectangle and

proceed toward the critical point (2, 3). The resulting phase-plane portrait
appears in Figure B.10.