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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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8a. Setting the right sides of gives the critical points and possibly
([e 1G2 - e2ai]/[GiG2 - aia2], [e2S1 -eia2]/[SiS2 - aia2]).
(The last point can be obtained from Eq.(36) also). The conditions e2/a2 > ei/si and e2/s2 > ei/ai imply that e2si - eia2 > 0 and eis2 - e2ai < 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' = eix 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 (eisi, 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 ei/Si > e2/a2 and ei/a1 > e2/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)
e 1 e 1 B B
52 a 2 1 g 2
y' = e 2y(1 - —y x) = e 2y(1 - —y - —x). The coexistence
e 2 e 2 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 inidcated by heavy dots.
As B is decreased,
X decreases, Y increases (as indicated above) and the point of intersection moves closer to (0,R). If B/gi < R coexistence is not possible, and the only * *^2 *
critical points are (0,0),(0,R)and (B,0).
It can be shown that (0,0) and (B,0) are unstable and (0,R) is asymptotically stable. Hence we conlcude, when coexistence is no longer possible, that x ^ 0 and y ^ R
and thus the bluegill population will die out.
12a. Setting each equation equal to zero, we obtain x = 0 or
(4 - x - y) = 0 and y = 0 or (2 + 2a - y - ax) = 0. Thus we
have (0,0), (4,0), (0,2 + 2a), and the intersection of
x + y = 4 and ax + y = 2 + 2a. If a n 1, this yields (2,2) as the fourth critical point.
Section 9.5
191
12b. For a = .75 the linear system is
7 Y u
-2 -2 Yu
v-1.5 -2 Xvy
, which
has the characteristic equation r2 + 4r + 1 = 0 so that r = -2 ± ‘\f3 . Thus the critical point is an asymptotically
stable node. For a = 1.25, we have
7 Y u
7 n n V i
?2 -2 Y u
-2.5 -2 Av
so
r2 + 4r -1 = 0 and r = -2 ± yf~5 . Thus (2,2) is an unstable saddle point.
12c. Letting x = u+2 and y = v+2 yields
u' = (u+2)(4 - u-2 - v-2) = -2u - 2v - u2 - uv and v' = (v+2)(2 + 2a - v-2 - au - 2a) = -2au - 2v - v2 - auv. Thus the approximate linear system is u' = -2u - 2v and v' = -2au - 2v.
12d. The eigenvalues are given by -2-r -2
-2a -2-r
= r + 4r + 4
4a = 0, or r = -2 ± 2y a
Thus for 0 < a < 1 there are 2 negative real roots (asymptotially stable node) and for a > 1 the real roots differ in sign, yielding an unstable saddle point. a = 1 is the bifurcation point.
v
v
Section 9.5, Page 509
3b. We have x = 0 or (1 - .5x - .5y) = 0 and y = 0 or
(-.25 + .5x) = 0 and thus we have three critical points: (0,0), (2,0) and (1/2,3/2).
3c. For (0,0) the linear system is dx/dt = x and
7 -,
dy/dt = -.25y and hence A =
which has
1 0 0 -1/4 _
eigenvalues r1 = 1 and r2 = -1/4 and corresponding
V 1 ^ r 0 ?
eigenvectors and
v 0 7 V 1 ,
Thus (0,0) is an unstable
For (2,0), we let x = 2 + u and y = v in the given
du 1
equations and obtain — = -(u+v) - —u(u+v) and
dt
2
dv 3 1
— = —v + —uv. The linear portion of this has matrix
dt 4 2
192
Section 9.5
A =
' -i -i A
0 3/4J
, which has the eigenvalues r1 = -1,
1 ] -4 !
r2 = 3/4 and corresponding eigenvectors and
v 0y v 7 ,
Thus (2,0) is also an unstable saddle point.
For
'1 3^
2 2
V y
we let x = 1/2 + u and y = 3/2 + v in the
du
given equations, which yields ------------
dt
as the linear portion. Thus A
1 1 dv 3
- —u - —v, = —u
4 4 dt 4
1
4
3
— 0
4
, which has
eigenvalues r12 = (-1 ± y!1i)/8. Thus
13
2 2
V y
is an
asymptotically stable spiral point since the eigenvalues are complex with negative real part. Using r1 = (-1 + \/11 i)/8 we find that one eigenvector is
2
and by Section 7.6 the second eigenvector is
1 + y 11 i 2
1 - \fn i
3e.
3f. For (x,y) above the line x + y = 2 we see that x' < 0 and thus x must remain bounded. For (x,y) to the right of x = 1/2, y' > 0 so it appears that y could grow large asymptotic to x = constant. However, this implies a contradiction (x = constant implies x' = 0, but as y gets larger, x' gets increasingly negative) and hence we conclude y must remain bounded and hence (x,y) ^ (1/2,3/2) as t ^ •, again assuming they start in the first quadrant.
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