quote:
Goed gezien. Wacht, lees dit:
6). Thus there are infinitely many fundamental matrices for a given system.
However, there is one fundamental matrix that we already know, namely,
Φ(n) =
n−1
i=n0
A
, with Φ(n0) = I
(Exercises 3.2, Problem 5). In the autonomous case when A is a constant
matrix, Φ(n) = An−n0, and if n0 = 0, then Φ(n) = An. Consequently, it
would be much more suitable to use the Putzer algorithm to compute the
fundamental matrix for an autonomous system.
Theorem 3.7. There is a unique solution Ψ(n) of the matrix (3.2.4) with
Ψ(n0) = I.
Proof. One may think of the matrix difference equation (3.2.4) as a
system of k2 first-order difference equations. Thus, to complete the point,
we may apply the “existence and uniqueness” Theorem 3.4 to obtain a k2-
vector solution ν such that ν(n0) = (1, 0, . . . , 1, 0, . . .)T , where 1’s appear at
the first, (k+2)th, (2k+3)th, . . . slots and 0’s everywhere else. The vector
ν is then converted to the k × k matrix Ψ(n) by grouping the components
into sets of k elements in which each set will be a column. Clearly, Ψ(n0) =
I.
We may add here that starting with any fundamental matrix Φ(n), the
fundamental matrix Φ(n)Φ−1(n0) is such a matrix. This special fundamental
matrix is denoted by Φ(n, n0) and is referred to as the state transition
matrix.
One may, in general, write Φ(n,m) = Φ(n)Φ−1(m) for any two positive
integers n,m with n ≥ m. The fundamental matrix Φ(n,m) has some
agreeable properties that we ought to list here. Observe first that Φ(n,m)
is a solution of the matrix difference equation Φ(n + 1,m) = A(n)Φ(n,m)
(Exercises 3.2, Problem 2). The reader is asked to prove the following
statements:
Φ−1(n,m) = Φ(m, n) (Exercises 3.2, Problem 3).
(ii) Φ(n,m) = Φ(n, r)Φ(r,m) (Exercises 3.2, Problem 3).
(iii) Φ(n,m) =
n−1
i=m A
(Exercises 3.2, Problem 3).
Corollary 3.8. The unique solution of x(n, n0, x0) of (3.2.1) with
x(n, n0, x0) = x0 is given by
x(n, n0, x0) = Φ(n, n0)x0. (3.2.5)
Checking the linear independence of a fundamental matrix Φ(n) for n ≥
n0 is a formidable task. We will instead show that it suffices to establish
linear independence at n0.