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d0 ( dt ) .
--- ) . Thus
(1/2)mL2(d0/dt)2 = mgL(cos0 - cosa). Thus
(d0/dt)2 = (2g/L)(cos0 - cosa). Since we are releasing
the pendulum with zero velocity from a positive angle a,
the angle 9 will initially be decreasing so d9/dt < 0.
If we restrict attention to the range of 9 from 9 = a to
9 = 0, we can assert d9/dt = -ä/2g/L ä/cos9 - cosa .
Solving for dt gives dt = - VL/2g d9/\/cos9 - cosa .
27b.Since there is no damping, the pendulum will swing from
its initial angle a through 0 to -a, then back through 0 again to the angle a in one period. It follows that 9(T/4) = 0. Integrating the last equation and noting that as t goes from 0 to T/4, 9 goes from a to 0 yields
T/4 = -ä/ L/2g 10 (1/V cos9 - cosa )d9.
28a. If — = y, then dt
= — = -g(x) dt
28b.Under the given assumptions we have g(x) = g(0) + g'(0)x 2
+ g"(X )x /2 and c(x) = c(0) + c'(X )x, where 1 2
0 < x , x < x and g(0) = 0.
= (-g(0) - g' (0)x)
and thus the system can be written as
[g"(Xi)x2/2 - c'(X2)xy]
d / \ 1 ¥ \ Ã 0 ¯
x 0 x
dt I YJ 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 2-2y0-1.125x0ylv 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
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).
´ -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
Ã 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
Ã Y u
corresponding linear system. Thus ^ = 11/6 and
r2 = -5/2 and thus (0,5/3) is an unstable saddle point.