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Problem 8 the equations transform to dr/dt = r2 - 2,
d0/dt = -1. This system has a periodic solution r = \p2 ,
0 = -t + t0. If r < \p2 , then dr/dt < 0, and the direction of motion along a trajectory is inward. If r > , then dr/dt > 0, and the direction of motion is
outward. Thus the periodic solution r = \p2 , 0 = -t + t0 is unstable.
11. If F(x,y) = x+y+x3-y2, G(x,y) = -x+2y+x2y+y3/3, then
Fx(x,y) + Gy(x,y) = 1+3x2+2+x2+y2 = 3+4x2+y2. Since the
conditions of Theorem 9.7.2 are satisfied for all x and y, and since Fx + Gy > 0 for all x and y, it follows that the system has no periodic nonconstant solution.
13. Since x = f(t), y = y(t) is a solution of Eqs.(15), we have df/dt = F[f (t),y (t)], dy/dt = G[f (t),y (t)]. Hence on the curve C,
F(x,y)dy - G(x,y)dx = f'(t)y'(t)dt -y7 (t) f'(t)dt = 0. It follows that the line integral around C is zero.
However, if Fx + Gy has the same sign throughout D, then the double integral cannot be zero. This gives a contradiction. Thus either the solution of Eqs.(15) is not periodic or if it is, it cannot lie entirely in D.
16a. Setting x' = 0 and solving for y yields y = x3/3 - x + k. Substituting this into y'= 0 then gives
x + .8(x3/3 - x + k) - .7 = 0 Using an equation solver we obtain x = 1.1994, y = -.62426 for k = 0 and x = .80485, y = -.13106 for k = .5. To determine the type of critical points these are, we use Eq.(13) of Section 9.3 to find the
linear coefficient matrix to be A -
is the critical point. For xc = 1.1994 we obtain complex conjugate eigenvalues with a negative real part, and therefore k = 0 yields an asymptotically stable spiral point. For xc = .80485 the eigenvalues are also complex conjugates, but with positive real parts, so k = .5 yields an unstable spiral point.
16b. Letting k = .1, .2, .3, .4 in the cubic equation of part (a)
and finding the corresponding eigenvalues from the matrix in part (a), we find that the real part of the eigenvalues change sign between k = .3 and k = .4. Continuing to iterate in this fashion we find that for k = .3464 that the real part of the eigenvalue is -.0002 while for k = .3465 the real part is .00005, which indicates k = .3465 is the
critical point for which the system changes from stable to unstable.
16d. You must plot the solution for values of k slightly less than k0, found in part (c), to determine whether a limit cycle exists.
Section 9.8, Page 538
1a. From Eq (6), 1 = -8/3 is clearly one eigenvalue and the other two may be found from 12 + 111 - 10(r-1) = 0 using the quadratic formula.
Ë×1 ^ Ã 0 ¯
X2 = 0
,v^3 7 V 0,
, which requires Xi = X2 = 0
1b. For 1 = 11 we have -10+8/3 -10 0
r -1+8/3 0
v 0 0 0
and X3 arbitrary and thus X(1) = (0,0,1)T.
For 1 = 13 = (-11 + a)/2, where a = ‘sj 81+40r , we have ' 10+(11-a)/2 10 0 Y X1
r -1+(11-a)/2 0 X2
0 0 -8/3+(11-a)/2 j[x3,
The last line implies x3 = 0 and multiplying the first line by
(-9+a)/2 we obtain
4i ^ V 01
v?2^ V 0 ,
Substituting a = 81+40r we have
- 10r -10(9-a)/2
fz Ë ¥
V si' II 0
VX2 J V 0y
9 - yj 81+40r . Thus X(3) = -2r
which is proportional to the answer given in the text. Similar calculations give S(2) .
1c. Simply substitute r = 28 into the answers in parts (a) and (b).
2a. The calculations are somewhat simplified if you let x = b + u, y = b + v, and z = (r-1)+w, where b = V8(r-1)/3 . An alternate approach is to extend Eq.(13) of Section 9.3, which is:
f \ u v
÷ ^ vwy
In this example F = -10x + 10y, G = rx H = -8z/3 + xy and thus
\ / ´ Fx Fy Fz "
vJ = x Gy z
w ) I Hx Hy Hz )
f \ / î 10 0 > /
u \---1 u
v = r -1 1 vv
V wV v y0 x0 3 v w
x0 = y0 = V8(r-1)/3 .
2b. Eq.(9) is found by evaluating
- y - xz and
q.(8) for P2 since
-1-1 -b = 0.
2c. If r = 28, then Eq.(9) is 313 + 4112 + 3041 + 4320 = 0, which has the roots -13.8546 and .093956 ± 10.1945i.
3b. If r1,r2,r3 are the three roots of a cubic polynomial, then the polynomial can be factored as
(x-ri)(x-r2)(x-r3). Expanding this and equating to the given polynomial we have A = -(r!+r2+r3),
B = r1r2 + r1r3 + r2r3 and C = -r^2r3. We are interested in the case when the real part of the complex conjugate roots changes sign. Thus let r2 = a+ib and r3 = a-ib, which yields