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+ g"(X )x /2 and c(x) = c(0) + c'(X )x, where 1 2
0 < X , X < x and g(0) = 0. 12
-k- = (-g(0) - g' (0)x)
and thus the system can be written as
[g"(X1)x2/2 - c'(X2)xy]
d / \ x S' 1—1 o / \ x r 0 'Y
dt IYJ v -g' (0) -c(0) j I YJ v-g"(^1)x2/2 - c'(^2)xy ,
from which the results follow.
Section 9.4, Page 501
3b. x(1.5 - .5x - y) = 0 and y(2 - y - 1.125x) = 0 yield (0,0),
(0,2) and (3,0) very easily. The fourth critical point is the intersection of .5x + y = 1.5 and 1.125x + y = 2, which is (.8,1.1).
3c. From Eq (5)
we get — dt
1.5-x0-y0 -x0 Yu
-1.125y0 2-2y0-1.125x0^ v
(0,0) we get u' = 1.5u and v' = 2v, so r = 3/2 and r = 2, and thus (0,0) is an unstable node. For (0,2) we have u' = -.5u and v' = -2.25u-2v, so r = -.5, -2 and thus (0,2) is an asymptotically stable node. For (3,0) we get u' = -1.5u-3v and v' = -1.375v, so r = -1.5, -1.375 and hence (3,0) is an symptotically stable node. For (.8,1.1) we have u' = -.4u -.8v and v' = -1.2375u - 1.1v which give r = -1.80475, .30475 and thus (.8,1.1) is an unstable saddle point.
5b. The critical points are found by setting dx/dt = 0 and
dy/dt = 0 and thus we need to solve x(1 - x - y) = 0 and
y(1.5 - y - x) = 0. The first yields x = 0 or y = 1 - x
and the second yields y = 0 or y = 1.5 - x. Thus (0,0), (0,3/2) and (1,0) are the only critical points since the two straight lines do not intersect in the first quadrant (or anywhere in this case). This is an example of one of
the cases shown in Figure 9.4.5 a or b.
6b. The critical points are found by setting dx/dt = 0 and
dy/dt = 0 and thus we need to solve x(1-x + y/2) = 0 and
y(5/2 - 3y/2 + x/4) = 0. The first yields x = 0 or
y = 2x - 2 and the second yields y = 0 or y = x/6 + 5/3. Thus we find the critical points (0,0), (1,0), (0,5/3)
and (2,2). The last point is the intersection of the two
straight lines, which will be used again in part d.
6c. For (0,0) the linearized system is x' = x and y' = 5y/2,
which has the eigenvalues r1 = 1 and r2 = 5/2. Thus the
origin is an unstable node. For (2,2) we let x = u + 2 and y = v + 2 in the given system to find
(since x' = u' and y' = v') that
du/dt = (u+2)[1 - (u+2) + (v+2)/2]
dv/dt = (v+2)[5/2 - 3(v+2)/2 + (u+2)/4]
Hence the linearized equations are
f Y u
= (u+2)(-u+v/2) and
= (v+2)(u/4 - 3v/2).
r -2 1 '
which has the eigenvalues r1,2 = (-5 ± <\/~3)/2. Since these are both negative we conclude that (2,2) is an asymptotically stable node. In a similar fashion for
( Y u
(1,0) we let x = u + 1 and y = v to obtain the linearized ^-1 1/2 0 11/4,
^ = -1 and r2 = 11/4 as eigenvalues and thus (1,0) is an unstable saddle point. Likewise, for (0,5/3) we let
x = u, y = v + 5/3 to find
( Y u
corresponding linear system. Thus ^ = 11/6 and
r2 = -5/2 and thus (0,5/3) is an unstable saddle point.
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
? y )
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 + \/~3 )/2 = -1.63 is given by (1^^3)X1/2 + X2 = 0
f 1 ^ ? 1 1
v (\J~3 -1)/2 7 V .37 ,
is one eigenvector. For
r = (-5 - )/2 = -3.37 we have (1 + j^3)Xi/2 + X2 = 0 and
1 1 ?1.37,
is the second eigenvector.
v 1 v J _r-L' / ^ /
Hence the linearized solution is
/ \ u
e-1.63t + c2
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,
\ f i ] f i 1
u = ci e-t + c2 eiit/4
vJ V 0 , vi5/2,
f \ u = ci f 0 ] e-5t/2 + c2 ? i ^
V vJ f i 1 f5/52J
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 y 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.