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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.
Download (direct link): elementarydifferentialequations2001.pdf
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t 2 xt
S'/s/s///
--^ S s' S / /
^ /
FIGURE 7.6.1 A direction field for the system (11).
To find a fundamental set of solutions we assume that
x = gert
and obtain the set of linear algebraic equations
2 Ã
1
— — — r 2 1
É=(0
(12)
(13)
1
x
2
2
386
Chapter 7. Systems of First Order Linear Equations
for the eigenvalues and eigenvectors of A. The characteristic equation is
— 2 — r 1
1
— - — r 2
(14)
therefore the eigenvalues are r1 = — 2 + i and r2 = — j — i. From Eq. (13) a straightforward calculation shows that the corresponding eigenvectors are
?(1) =
?(2) =
i
Hence a fundamental set of solutions of the system (11) is
x(1)(t) =
,(—1/2+i )t
x(2)(t) =
1
i
(—1/2—i)t.
(15)
(16)
To obtain a set of real-valued solutions, we must find the real and imaginary parts of either x(1) or x(2). In fact,
x(1) (t) = ( j ) e v2(cos t + i sin t) =
1 \ —t/2(
e—t/2coS -e~t/2 sin t
I J t/2 sir + ^ e—t/2cos t
Hence
u(t) = e
- e—1/2
cos t sint
v(t) = e
- e—t/2
sint cos t
(17)
(18)
is a set of real-valued solutions. To verify that u(t) and v(t) are linearly independent, we compute their Wronskian:
W(u, v)(t) =
e t/2 cos t
- e~t/2 oin t
e—/2 sin t
e—1/2 coc t
= e t (cos2 t + sin2 t) = e t.
Since the Wronskian is never zero, it follows that u (t) and v( t) constitute a fundamental set of (real-valued) solutions ofthe system (11).
The graphs of the solutions u(t) and v(t) are shown in Figure 7.6.2a. Since
u(0) =
v(0) =
the graphs of u(t) and v(t) pass through the points (1, 0) and (0, 1), respectively. Other solutions of the system (11) are linear combinations of u(t) and v(t), and graphs of a few of these solutions are also shown in Figure 7.6.2a. In all cases the solution approaches the origin along a spiral path as t ^ to; this is due to the fact that the solutions (18) are products of decaying exponential and sine or cosine factors. Some typical graphs of ^ versus t are shown in Figure 7.6.2b; each one represents a decaying oscillation in time.
Figure 7.6.2a is typical of all second order systems x' = Ax whose eigenvalues are complex with negative real part. The origin is called a spiral point and is asymptotically stable because all trajectories approach it as t increases. For a system whose eigenvalues have a positive real part the trajectories are similar to those in Figure 7.6.2a, but the direction of motion is away from the origin and the trajectories become unbounded. In this case, the origin is unstable. If the real part of the eigenvalues is zero, then the trajectories neither approach the origin nor become unbounded but instead traverse
1
1
i
1
i
7.6 Complex Eigenvalues
FIGURE 7.6.2 (a) Trajectories of the system (11); the origin is a spiral point. (b) Plots of x1
versus t for the system (11).
repeatedly a closed curve about the origin. In this case the origin is called a center and is said to be stable, but not asymptotically stable. In all three cases the direction of motion may be either clockwise, as in this example, or counterclockwise, depending on the elements of the coefficient matrix A.
EXAMPLE
2
The electric circuit shown in Figure 7.6.3 is described by the system of differential equations
d ( I\ (-1 -A ( A
dt \V)~\ 2 -1; \VJ’ (19)
where I is the current through the inductor and V is the voltage drop across the capacitor. These equations were derived in Problem 19 of Section 7.1. Suppose that at time t = 0 the current is 2 amperes and the voltage drop is 2 volts. Find I(t) and V(t) at any time.
R = 1 ohm
FIGURE 7.6.3 The circuit in Example 2.
388
Chapter 7. Systems of First Order Linear Equations
Assuming that
V
we obtain the algebraic equations
A = gert, (20)
- Vr --- r) (I)=(0)- (21)
The eigenvalues are determined from the condition
= r2 + 2r + 3 = 0; (22)
-1 - r -1
2 -1 - r
thus rl = -1 + ë/21 and r2 = — 1 — ë/2³. The corresponding eigenvectors are then found from Eq. (21), namely,
g'1' =(-/2,) Š gO' (23)
The complex-valued solution corresponding to rl and g(1) is
g(1)er1t e(-1+/21)1
= ^ /2 e-t (cosV2 t + ³ sin ë/21)
-t/ co^V2 f\,ie-f/ sinV2t \ (24)
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