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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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T =
(1)
(29)
ev
Since the columns of T are linearly independent vectors, detT = 0; hence T is nonsingular and T -1 exists. A straightforward calculation shows that the columns of the matrix AT are just the vectors A?(1),..., A?(n). Since A?(k) = kk?(k), it follows that
AT
where
D
fk1^(1) kn ^
kn ev
k1 0 0 \
0 k2 ... 0
0 0 ... kJ
= TD,
(30)
(31)
is a diagonal matrix whose diagonal elements are the eigenvalues of A. From Eq. (30) it follows that
T-1AT = D.
(32)
Thus, if the eigenvalues and eigenvectors of A are known, A can be transformed into a diagonal matrix by the process shown in Eq. (32). This process is known as a similarity
398
Chapter 7. Systems of First Order Linear Equations
EXAMPLE
3
transformation, and Eq. (32) is summed up in words by saying that A is similar to the diagonal matrix D. Alternatively, we may say that A is diagonalizable. Observe that a similarity transformation leaves the eigenvalues of A unchanged and transforms its eigenvectors into the coordinate vectors e(1),..., e(n).
If A is Hermitian, then the determination of T -1 is very simple. We choose the eigenvectors ?(1),..., ^(n) of A so that they are normalized by (%(l), ?,(l)) = 1 for each i, as well as orthogonal. Then it is easy to verify that T -1 = T*; in other words, the inverse of T is the same as its adjoint (the transpose of its complex conjugate).
Finally, we note that if A has fewer than n linearly independent eigenvectors, then there is no matrix T such that T-1AT = D. In this case, A is not similar to a diagonal matrix, and is not diagonalizable.
Consider the matrix
A =
1 1 4 1
(33)
Find the similarity transformation matrix T and show that A can be diagonalized.
In Example 1 of Section 7.5 we found that the eigenvalues and eigenvectors of A are
r1 = 3,
i(1) =
=-1. r2) = ( _2
(34)
Thus the transformation matrix T and its inverse T are
T
22
T-1_____
Consequently, you can check that
(35)
T-1AT =
0 -1
D.
(36)
1
2
Now let us turn again to the system
x = Ax, (37)
where A is a constant matrix. In Sections 7.5 and 7.6 we have described how to solve such a system by starting from the assumption that x = ?ert. Now we provide another viewpoint, one based on diagonalizing the coefficient matrix A.
According to the results stated just above, it is possible to diagonalize A whenever A has a full set of n linearly independent eigenvectors. Let ?(1), ? ? ?, ?(n) be eigenvectors of A corresponding to the eigenvalues r1, ? ? ?, rn and form the transformation matrix T
whose columns are ?(1), ? ? ?, ?(n). Then, defining a new dependent variable y by the
relation
x = Ty, (38)
we have from Eq. (37) that
Ty7 = ATy.
(39)
7.7 Fundamental Matrices
399
Multiplying by T we then obtain
y' = (T-1AT )y,
or, using Eq. (32),
y = Dy.
(40)
(41)
Recall that D is the diagonal matrix with the eigenvalues rv , rn of A along the
diagonal. A fundamental matrix for the system (41) is the diagonal matrix (see Problem 16)
Q(t ) = exp(Dt) =
/ er1t 0 . .0
0 ef2l . 0
0 0 ? ? ernl)
(42)
A fundamental matrix ^ for the system (37) is then found from Q by the transformation (38),
^ = TQ;
that is,
W(t) =
(43)
(44)
Equation (44) is the same result that was obtained in Section 7 . 5. This diagonalization procedure does not afford any computational advantage over the method of Section 7.5, since in either case it is necessary to calculate the eigenvalues and eigenvectors of the coefficient matrix in the system of differential equations. Nevertheless, it is noteworthy that the problem of solving a system of differential equations and the problem of diagonalizing a matrix are mathematically the same.
Consider again the system of differential equations
x' Ax,
(45)
where A is given by Eq. (33). Using the transformation x = Ty, where T is given by Eq. (35), you can reduce the system (45) to the diagonal system
y'= 3 -?) y=Dy-
(46)
Obtain a fundamental matrix for the system (46) and then transform it to obtain a fundamental matrix for the original system (45).
By multiplying D repeatedly with itself, we find that
D2
9 0
0 1
D3
27 0
01
(47)
400
Chapter 7. Systems of First Order Linear Equations
Therefore it follows from Eq. (23) that exp(Dt) is a diagonal matrix with the entries e3t and e-t on the diagonal, that is,
_Dt (e3t 0'
eDt = (, 0 e-V (48)
Finally, we obtain the required fundamental matrix 0( t) by multiplying T and exp(Dt):
*(t)=(2 -2) (e3d e-) = G? -?) (49)
Observe that this fundamental matrix is the same as the one found in Example 1.
PROBLEMS In each of Problems 1 through 10 find a fundamental matrix for the given system of equations. s In each case also find the fundamental matrix O(f) satisfying 0(0) = I.
1. x' =
3. x' = 5. x' = 7. x' =
9. x' =
13.
14.
15.
-2
-1
-2
-5
-2
-1
1
3
2. x' = 4. x' = 6. x' =
4 2
2 8 -4
0 1
4 -2
-1 -4
1 -1
1 -1 53
1 1' 1 -1
>-8 -5 -3,
=.2
11. Solve the initial value problem
x'
-2
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