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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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Sometimes it is convenient to make use of the special fundamental matrix, denoted by (-), whose columns are the vectors x(1)(t),..., x(n)(t) designated in Theorem
7.4.4. Besides the differential equation (1) these vectors satisfy the initial conditions
x(j)(t0) = e(J j),
(12)
7.7 Fundamental Matrices
395
EXAMPLE
2
where e(j) is the unit vector, defined in Theorem 7.4.4, with a one in the jth position and zeros elsewhere. Thus ^) has the property that
/
(*0) =
V 0 0
0 \ 0
1
(13)
We will always reserve the symbol to denote the fundamental matrix satisfying the initial condition (13) and use ^ when an arbitrary fundamental matrix is intended. In terms of ^) the solution of the initial value problem (1), (7) is even simpler in appearance; since ԗx(t0) = I, it follows from Eq. (10) that
x = ^).
(14)
While the fundamental matrix (t) is often more complicated than t), it is especially
helpful if the same system of differential equations is to be solved repeatedly subject to many different initial conditions. This corresponds to a given physical system that can be started from many different initial states. If the fundamental matrix (^ has been determined, then the solution for each set of initial conditions can be found simply by matrix multiplication, as indicated by Eq. (14). The matrix (^ thus represents a transformation of the initial conditions x0 into the solution x(t) at an arbitrary time t. By comparing Eqs. (10) and (14) it is clear that (0 = ^(t)^x(t0).
I
For the system (3),
x=l4 1,x
in Example 1, find the fundamental matrix such that (0) = I.
The columns of are solutions of Eq. (3) that satisfy the initial conditions
x(1)(0) = Q , x(2)(0) = . (15)
Since the general solution of Eq. (3) is
x=*(2)e3t+c*(2)e',
we can find the solution satisfying the first set of these initial conditions by choosing cx = c2 = 1; similarly, we obtain the solution satisfying the second set of initial conditions by choosing q = 4 and c2 = 4. Hence
11 e3t + 2 et 4 e3t 1 et \
=(2e>>2- 4,3,+4a6)
Note that the elements of ^) are more complicated than those of the fundamental matrix ^(t) given by Eq. (4); however, it is now easy to determine the solution corresponding to any set of initial conditions.
396
Chapter 7. Systems of First Order Linear Equations
The Matrix exp(^^. Recall that the solution of the scalar initial value problem
X = ax, x (0) = x0, (17)
where a is a constant, is
x = x0exp(at). (18)
Now consider the corresponding initial value problem for an n x n system, namely,
x' = Ax, x(0) = x0, (19)
where A is a constant matrix. Applying the results of this section to the problem (19), we can write its solution as
x = (- )x0, (20)
where (0) = I. Comparing the problems (17) and (19), and their solutions, sug-
gests that the matrix (-) might have an exponential character. We now explore this possibility.
The scalar exponential function exp(at) can be represented by the power series
antn
exp(at) = 1 + 2 , (21)
n!
n=1
which converges for all t. Let us now replace the scalar a by the n x n constant matrix A, and consider the corresponding series
^ Antn . A2t2 Antn ,
I + ^2 i- =I + + ҳ + "' + j + . (22)
n! 2! n!
n=1
Each term in the series (22) is an n x n matrix. It is possible to show that each element of this matrix sum converges for all t as n to. Thus the series (22) defines as its sum a new matrix, which we denote by exp (At); that is,
to An -
exp (At) = I + 2 j-, (23)
n!
analogous to the expansion (21) of the scalar function exp(at).
By differentiating the series (23) term by term we obtain
- to a n tn 1
-t[exp(At)] = E -ò = A
n=1 v 7
to \n j-
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