# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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6d. To sketch the required trajectories, we must find the

eigenvectors for each of the linearized systems and then analyze the behavior of the linear solution near the critical point. Using this approach we find that the

solution near (0,0) has the form

/ \ x

¨ ÓÓ

1

et +

c2

0

1

5t/2 and thus the origin is approached only for

large negative values of t. In this case ec dominates e5t/2 and hence in the neighborhood of the origin all trajectories are tangent to the x-axis except for one pair (ci = 0) that lies along the y-axis.

For (2,2) we find the eigenvector corresponding to r = (-5 + \pi )/2 = -1.63 is given by (1^^3)X1/2 + X2 = 0

and thus

V 1 > ( 1 1

V (\J~3 -1)/2 r V .37 ,

is one eigenvector. For

r = (-5 -^/"3)/2 = -3.37 we have (1 + ^3)^/2 + X2 = 0 and

thus

-(VT +1)/2

1 1 ¦1.37,

is the second eigenvector.

v ³ V J _r-L' / ^ /

Hence the linearized solution is

f - Ë

/ \ u

= c

1

1

.37

e-1.63t + c2

1

1.37

,-3.37t

For large

positive values of t the first term is the dominant one and thus we conclude that all trajectories but two approach (2,2) tangent to the straight line with slope .37. If c1 = 0, we see that there are exactly two (c2 > 0 and c2 < 0) trajectories that lie on the straight line with slope -1.37.

In similar fashion, we find the linearized solutions near (1,0) and (0,5/3) to be, respectively,

0

1

v

Section 9.4

189

> V 1 ] V 1 1

u = c1 e-t + 02 e11t/4

vJ V 0 , v15/2,

and

¥ \ = ci V 0 ] e-5t/2 + c2 r 1 ^

u

V vJ V 11 V5/52 1

ait/e

which, along with the above analysis, yields the sketch shown.

6e.From the above sketch, it appears that (x,y) ^ (2,2) as 6f.t ^ • as long as (x,y) starts in the first quadrant.

To ascertain this, we need to prove that x and y cannot become unbounded as t ^ • . From the given system, we can observe that, since x > 0 and y > 0, that dx/dt and dy/dt have the same sign as the quantities 1 - x + y/2 and 5/2 - 3y/2 + x/4 respectively. If we set these quantities equal to zero we get the straight lines y = 2x - 2 and y = x/6 + 5/3, which divide the first quadrant into the four sectors shown. The signs of x' and ó are indicated, from which it can be concluded that x and y must remain bounded [and in fact approach (2,2)] as t ^ • . The discussion leading up to Fig.9.4.4 is also useful here.

8a. Setting the right sides of gives the critical points and possibly

([e 1G2 - e2ai]/[SiS2 - aia2], [e2S1 -eia2]/[SiS2 - aia2]).

(The last point can be obtained from Eq.(36) also). The conditions e2/a2 > e1/s1 and e2/s2 > e1/a1 imply that e 2s1 - e 1a 2 > 0 and e 1s2 - e 2a 1 < 0. Thus either the x coordinate or the y coordinate of the last critical point is negative so a mixed state is not possible. The linearized system for (0,0) is x' = e1x and y' = e2y and thus (0,0) is an unstable equilibrium point. Similarly, it can be shown [by linearizing the given system or by using Eq.(35)] that (0, e2/s2) is an asymptotically stable critical point and that (e1s1, 0) is an unstable critical point. Thus the fish represented by y(redear) survive.

the equations equal to zero (0,0), (0, e 2/S2), (e i/CTi,0),

190

Section 9.4

8b. The conditions e 1/s1 > e 2/a 2 and e 1/a 1 > e 2/s2 imply that e 2s1 - e 1a 2 < 0 and e 1s2 - e 2a 1 > 0 so again one of the coordinates of the fourth point in 8a. is negative and hence a mixed state is not possible. An analysis similar to that in part(a) shows that (0,0) and (0,e2/s2) are unstable while (e1/s1,0) is stable. Hence the bluegill (represented by x) survive in this case.

51 a 1 1 g 1

9a. x' = e 1x(1 - —x - ---y) = e 1x(1 - —x - —y)

e1 e1 B B

52 a 2 1 g 2

y' = e 2y(1 - —y x) = e 2y(1 - —y - —x). The coexistence

e2 e2 R R

1 g 1 g 2 1

equilibrium point is given by —x + —y = 1 and —x + —y =

B B R R

1. Solving these (using determinants) yields

X = (B - g 1R)/(1 - g 1g 2) and Y = (R - g 2B)/(1-g 1g 2).

9b. If B is reduced, it is clear from the answer to part(a)

that X is reduced and Y is increased. To determine whether the bluegill will die out, we give an intuitive argument which can be confirmed by doing the analysis.

Note that B/g 1 = e 1/a 1 > e 2/s2 = R and

R/g 2 = e 2/a 2 > e 1/s1 = B so that the graph of the lines

1 - x/B - g1y/B = 0 and 1 - y/R - g2x/R = 0 must appear as indicated in the figure, where critical points are ú/óõ

inidcated by heavy dots.

As Â is decreased,

X decreases, Y increases (as indicated above) and the point of intersection moves closer to (0,R). If B/y2. < R coexistence is not possible, and the only * *^2 *

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