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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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To solve the system observe that the coefficient matrix A is the same as the matrix in Example 1. Thus we know that r = 2 is a double eigenvalue and that it has only a single corresponding eigenvector, which we may take as ?T = (1, -1). Thus one solution of the system (8) is
x(1)(t) = (_1) e2t, (9)
but there is no second solution of the form x = ?ert.
Based on the procedure used for second order linear equations in Section 3.5, it may be natural to attempt to find a second solution of the system (8) of the form
x = ?te2t, (10)
where ? is a constant vector to be determined. Substituting for x in Eq. (8) gives
2?te2t + ?e2t - A?te2t = 0. (11)
For Eq. (11) to be satisfied for all t it is necessary for the coefficients of te2t and e2t each to be zero. From the term in e2t we find that
? = 0. (12)
Hence there is no nonzero solution of the system (8) of the form (10).
Since Eq. (11) contains terms in both te2t and e2t, it appears that in addition to ?te2t the second solution must contain a term of the form ^e2t; in other words, we need to assume that
x = ?te2t + ^e2t, (13)
where ? and ^ are constant vectors. Upon substituting this expression for x in Eq. (8) we obtain
2?te2t + (? + 2-n)e2t = A(?te2t + ^). (14)
Equating coefficients of te2t and e2t on each side ofEq. (14) gives the conditions
(A - 2I)? = 0 (15)
Chapter 7. Systems of First Order Linear Equations
(A — 2I)-n = g (16)
for the determination of g and ^. Equation (15) is satisfied if g is an eigenvector of A corresponding to the eigenvalue r = 2, that is, gT = (1, —1). Since det(A — 2I) is zero, we might expect that Eq. (16) cannot be solved. However, this is not necessarily true, since for some vectors g it is possible to solve Eq. (16). In fact, the augmented matrix for Eq. (16) is
— 1 —1 I 1)
1 1 I —V'
The second row of this matrix is proportional to the first, so the system is solvable. We have
— Ο1 — Ο2 = 1
so if n1 = k, where k is arbitrary, then n2 = ~k — 1. If we write
—1 — k —1 —1
then by substituting for g and ^ in Eq. (13) we obtain
A +„2t
x =(v— 1) teZt +(_ij e + k{ — 1) eZt. (18)
The last term in Eq. (18) is merely a multiple of the first solution x(1)(t), and may be ignored, but the first two terms constitute a new solution:
Λ2)(Λ_² Λ te2t,( 0\
x2(t) =(^— te2t +^—e2t. (19)
An elementary calculation shows that W[x(1), x(2)](t) = — e4t, and therefore x(1) and x(2) form a fundamental set of solutions of the system (8). The general solution is
x = C1x(1)(t) + C2x(2)(t)
1 \ „ / ο\ λ "
_1) e2t + c2
-0 te2t A-?! e2t
The graph of the solution (20) is a little more difficult to analyze than in some of the previous examples. It is clear that x becomes unbounded as t —to and that x — 0 as t ——to. It is possible to show that as t ——to, all solutions approach the origin tangent to the line x2 = —x1 determined by the eigenvector. Similarly, as t — to, each trajectory is asymptotic to a line of slope —1. The trajectories of the system (8) are shown in Figure 7.8.2a and some typical plots of x1 versus t are shown in Figure 7.8.2b. The pattern of trajectories in this figure is typical of second order systems x' = Ax with equal eigenvalues and only one independent eigenvector. The origin is called an improper node in this case. If the eigenvalues are negative, then the trajectories are similar, but are traversed in the inward direction. An improper node is asymptotically stable or unstable, depending on whether the eigenvalues are negative or positive.
7.8 Repeated Eigenvalues
xz(t) x4' , .
-2 -1 1 2 x1
FIGURE 7.8.2 (a) Trajectories of the system (8); the origin is an improper node. (b) Plots of
x1 versus t for the system (8).
One difference between a system of two first order equations and a single second order equation is evident from the preceding example. Recall that for a second order linear equation with a repeated root r1 of the characteristic equation, a term cerj t in the second solution is not required since it is a multiple of the first solution. On the other hand, for a system of two first order equations, the term ^erit of Eq. (13) with r1 = 2 is not a multiple of the first solution gerjt, so the term ^erjt must be retained.
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