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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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174
CHAPTER 9
Section 9.1, Page 468
For Problems 1 through 16, once the eigenvalues have been found, Table 9.1.1 will, for the most part, quickly yield the type of critical point and the stability. In all cases it can be easily verified that A is nonsingular.
1a. The eigenvalues are found from the equation det(A-rI)=0.
3-r -2
Substituting the values for A we have
2 -2-r
r2 - r - 2 = 0 and thus the eigenvalues are r1 = -1 and
f 4 -2 Y?1 H ( 0 í
r2 = 2. For r1 = -1, we have | = and thus
(4 -2 ^ VX11 V 01
v 2 -1 J U2 J v 0 ,
1 ] V1 -2 ¹ Í V 01
and for r2 we have =
V 2 Ó v 2 -4 A?2 J V 0 ó
and thus
2
1b. Since the eigenvalues differ in sign, the critical point is a saddle point and is unstable.
1d.
4a. Again the eigenvalues are given by
1-r -4
4 -7-r
are solutions of
just one eigenvector X
0 and thus r1 = r2 =
(4 -4 Hf ×1 ] Í0
of =
V 4 Jv Ë>2 j V 0 ó
-4 Ä
and hence there is
1
1
4b. Since the eigenvalues are negative, (0,0) is an improper node which is asymptotically stable. If we had found that there were two independent eigenvectors then (0,0) would have been a proper node, as indicated in Case 3a.
1 M
2
Section 9.1
175
4d.
7a.
7b.
7d.
10a.
In this case det(A - rl) = r2 - 2r + 5 and thus the
eigenvalues are r12 = 1 ± 2i. For r1 = 1 + 2i we have
2 -2 1
1
ISJ
H-
4 -2i , vX
1
ISJ
X (1) = 1 ë
1-i°
f 2+2i -2 1 rX
^4 -2+2i í vX
2-2i -2
Xi
X2 -
0
and thus
Similarly for r2 = 1-2i we have and hence X(2)
0
1
l+i
Since the eigenvalues are complex with positive real part, we conclude that the critical point is a spiral point and is unstable.
Again, det(A-rl) = r2 + 9 and thus we have r12 = ±3i.
For r1 = 3i we have
1-3i 2
v -5 -1-3iy
Xi
X2 -
0
and thus
:(1)
-1 + 3i
V Ó
Likewise for r2 = -3i,
1 + 3i 2 V-5 -1+3i
1 ×1ë f î ] so that X(2) = f 2 1
î ISJ v 0 , v-1-3i,
l
0
0 î
0
2
10b. Since the eigenvalues are pure imaginary the critical point is a center, which is stable.
176
Section 9.1
lOd.
13.
17.
0 ' /
If we let x = x + u then x = u and thus the system
which will be in
/ f 1 1 ] x0 + f 1 1 ¯ f 2 1
becomes u = u -
V 1 -1 Ó V 1 -1 ó v 0 ,
the form of Eq.(2) if
1 1ë x f 2 I
Î
=
V 1 -1 ó v 0 .
. Using row
operations, this last set of equations is equivalent to
f 1 1
v 0 -2 /
u =
x0 =
2
-2
and thus x1 = 1 and x2 = 1. Since
1 1
u has (0,0) as the critical point, we
conclude that (1,1) is the critical point of the original system. As in the earlier problems, the eigenvalues are
1-r 1 2 /--------------------------------------
given by = r2 - 2 = 0 and thus r12 = ±y 2 .
1 -1-r
Hence the critical point (1,1) is an unstable saddle point.
The equivalent system is dx/dt = y,dy/dt = -(k/m)x-(c/m)y which is written in the form of Eq.(2) as
d
dt
/ \ x
V yO
0 1 -k/m -c/m
The point (0,0) is clearly a
V yO
critical point, and since A is nonsingular, it is the only one. The characteristic equation is r2+(c/m)r+k/m=0
so r1,r2 = [-c ± (c2
4km) 1/2]/2m. In the underdamped
case c2 - 4km < 0, the characteristic roots are complex with negative real parts and thus the critical point (0,0) is an asymptotically stable spiral point. In the overdamped case c2 - 4km > 0, the characteristic roots are real, unequal, and negative and hence the critical point (0,0) is asymptotically stable node. In the critically damped case c2 - 4km = 0, the characteristic roots are equal and negative. As indicated in the solution to Problem 4, to determine whether this is an improper or proper node we must determine whether there
a -1Î
Section 9.1
177
´ c/2m 1 ^ ×1 ^ V 0 ¯
v -k/m -c/2my v?2 V V 0,
are one or two linearly independent eigenvectors. The eigenvectors satisfy the equation
which has just one solution if
-k/m -c/2my VX2 C ¨ 0J
c2 - 4km = 0. Thus the critical point (0,0) is an asymptotically stable improper node.
18a. If A has one zero eigenvalue then for r = 0 we have
det(A-rl) = detA = 0. Hence A is singular which means Ax = 0 has infinitely many solutions and consequently there are infinitely many critical points.
18b. From Chapter 7, the solution is x(t) = c1?<1) + c2?<2)er2t,
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