Boyce/DiPrima 9 ed, Ch 7.7: Fundamental Matrices
Transcript of Boyce/DiPrima 9 ed, Ch 7.7: Fundamental Matrices
Boyce/DiPrima 9th ed, Ch 7.7: Fundamental
MatricesElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
Suppose that x(1)(t),…, x(n)(t) form a fundamental set of
solutions for x' = P(t)x on α < t < β.
The matrix
,
)()(
)(
)(
1
)1(
1
=
txtx
t
n
⋮⋱⋮
⋯
Ψ
whose columns are x(1)(t),…, x(n)(t), is a fundamental matrix
for the system x' = P(t)x. This matrix is nonsingular since its
columns are linearly independent, and hence detΨΨΨΨ ≠ 0.
Note also that since x(1)(t),…, x(n)(t) are solutions of x' = P(t)x,
ΨΨΨΨ satisfies the matrix differential equation ΨΨΨΨ' = P(t)ΨΨΨΨ.
,
)()(
)()()1(
=
txtx
tn
nn ⋯
⋮⋱⋮Ψ
Example 1:
Consider the homogeneous equation x' = Ax below.
In Chapter 7.5, we found the following fundamental
xx
=′
14
11
In Chapter 7.5, we found the following fundamental
solutions for this system:
Thus a fundamental matrix for this system is
ttetet
−
−=
=
2
1)(,
2
1)( )2(3)1( xx
−=
−
−
tt
tt
ee
eet
22)(
3
3
Ψ
Fundamental Matrices and General Solution
The general solution of x' = P(t)x
can be expressed x = ΨΨΨΨ(t)c, where c is a constant vector with
components c1,…, cn:
)()1(
1 )( n
nctc xxx ++= ⋯
components c1,…, cn:
==
n
n
nn
n
c
c
txtx
txtx
t ⋮
⋯
⋮⋱⋮
⋯ 1
)()1(
)(
1
)1(
1
)()(
)()(
)( cΨx
Fundamental Matrix & Initial Value Problem
Consider an initial value problem
x' = P(t)x, x(t0) = x0
where α < t0 < β and x0 is a given initial vector.
Now the solution has the form x = ΨΨΨΨ(t)c, hence we choose c
so as to satisfy x(t ) = x0. so as to satisfy x(t0) = x0.
Recalling ΨΨΨΨ(t0) is nonsingular, it follows that
Thus our solution x = ΨΨΨΨ(t)c can be expressed as
0
0
10
0 )()( xΨcxcΨ tt−=⇒=
0
0
1 )()( xΨΨx tt−=
Recall: Theorem 7.4.4
Let
=
=
=
0
0
0
,,0
1
0
,0
0
1
)()2()1(⋮…
⋮⋮
neee
Let x(1),…, x(n) be solutions of x' = P(t)x on I: α < t < β that
satisfy the initial conditions
Then x(1),…, x(n) are fundamental solutions of x' = P(t)x.
100
βα <<== 0
)(
0
)()1(
0
)1( ,)(,,)( tttnn exex …
Fundamental Matrix & Theorem 7.4.4
Suppose x(1)(t),…, x(n)(t) form the fundamental solutions given
by Thm 7.4.4. Denote the corresponding fundamental matrix
by ΦΦΦΦ(t). Then columns of ΦΦΦΦ(t) are x(1)(t),…, x(n)(t), and hence
010
001
⋯
⋯
Thus ΦΦΦΦ-1(t0) = I, and the hence general solution to the
corresponding initial value problem is
It follows that for any fundamental matrix ΨΨΨΨ(t),
I=
=Φ
100
010)( 0
⋯
⋮⋱⋮⋮
⋯t
00
0
1 )()()( xΦxΦΦx ttt == −
)()()()()()( 0
100
0
1tttttt
−− =⇒== ΨΨΦxΦxΨΨx
The Fundamental Matrix ΦΦΦΦand Varying Initial Conditions
Thus when using the fundamental matrix ΦΦΦΦ(t), the general
solution to an IVP is
This representation is useful if same system is to be solved for
many different initial conditions, such as a physical system
00
0
1 )()()( xΦxΦΦx ttt == −
many different initial conditions, such as a physical system
that can be started from many different initial states.
Also, once ΦΦΦΦ(t) has been determined, the solution to each set
of initial conditions can be found by matrix multiplication, as
indicated by the equation above.
Thus ΦΦΦΦ(t) represents a linear transformation of the initial
conditions x0 into the solution x(t) at time t.
Example 2: Find ΦΦΦΦ(t) for 2 x 2 System (1 of 5)
Find ΦΦΦΦ(t) such that ΦΦΦΦ(0) = I for the system below.
Solution: First, we must obtain x(1)(t) and x(2)(t) such that
xx
=′
14
11
We know from previous results that the general solution is
Every solution can be expressed in terms of the general
solution, and we use this fact to find x(1)(t) and x(2)(t).
ttecec
−
−+
=
2
1
2
12
3
1x
=
=
1
0)0(,
0
1)0( )2()1( xx
Example 2: Use General Solution (2 of 5)
Thus, to find x(1)(t), express it terms of the general solution
and then find the coefficients c1 and c2.
ttecect
−
−+
=
2
1
2
1)( 2
3
1
)1(x
1 2
To do so, use the initial conditions to obtain
or equivalently,
=
−+
=
0
1
2
1
2
1)0( 21
)1(ccx
=
− 0
1
22
11
2
1
c
c
Example 2: Solve for x(1)(t) (3 of 5)
To find x(1)(t), we therefore solve
by row reducing the augmented matrix:
=
− 0
1
22
11
2
1
c
c
Thus
2/1
2/1
2/110
2/101
2/110
111
240
111
022
111
2
1
=
=→
→
→
−−→
−
c
c
−
+=
−+
=
−
−−
tt
tttt
ee
eeeet
3
33)1(
2
1
2
1
2
1
2
1
2
1
2
1)(x
Example 2: Solve for x(2)(t) (4 of 5)
To find x(2)(t), we similarly solve
by row reducing the augmented matrix:
=
− 1
0
22
11
2
1
c
c
Thus
4/1
4/1
4/110
4/101
4/110
011
140
011
122
011
2
1
−=
=→
−→
−→
−→
−
c
c
+
−=
−−
=
−
−
−
tt
tt
tt
ee
eeeet
2
1
2
14
1
4
1
2
1
4
1
2
1
4
1)(
3
3
3)2(x
Example 2: Obtain ΦΦΦΦ(t) (5 of 5)
The columns of ΦΦΦΦ(t) are given by x(1)(t) and x(2)(t), and thus
from the previous slide we have
+−
−+=
−−
−−
tttt
tttteeee
t114
1
4
1
2
1
2
1
)(33
33
Φ
Note ΦΦΦΦ(t) is more complicated than ΨΨΨΨ(t) found in Ex 1.
However, now that we have ΦΦΦΦ(t), it is much easier to
determine the solution to any set of initial conditions.
+− −− tttteeee
2
1
2
1 33
−=
−
−
tt
tt
ee
eet
22)(
3
3
Ψ
Matrix Exponential Functions
Consider the following two cases:
The solution to x' = ax, x(0) = x0, is x = x0eat, where e0 = 1.
The solution to x' = Ax, x(0) = x0, is x = ΦΦΦΦ(t)x0, where ΦΦΦΦ(0) = I.
Comparing the form and solution for both of these cases, we
might expect ΦΦΦΦ(t) to have an exponential character. might expect ΦΦΦΦ(t) to have an exponential character.
Indeed, it can be shown that ΦΦΦΦ(t) = eAt, where
is a well defined matrix function that has all the usual
properties of an exponential function. See text for details.
Thus the solution to x' = Ax, x(0) = x0, is x = eAtx0.
∑∑∞
=
∞
=
+==10 !! n
nn
n
nnt
n
t
n
te
AI
AA
Coupled Systems of Equations
Recall that our constant coefficient homogeneous system
,2211
12121111
nnnnnn
nn
xaxaxax
xaxaxax
+++=′
+++=′
…
⋮
…
written as x' = Ax with
is a system of coupled equations that must be solved
simultaneously to find all the unknown variables.
,,
)(
)(
)(
1
1111
=
=
nnn
n
n aa
aa
tx
tx
t
⋯
⋮⋱⋮
⋯
⋮ Ax
Uncoupled Systems & Diagonal Matrices
In contrast, if each equation had only one variable, solved for
independently of other equations, then task would be easier.
In this case our system would have the form
00
00 21111 n
xxdxx
xxxdx
+++=′
+++=′
…
…
or x' = Dx, where D is a diagonal matrix:
,00
00
21
21112
nnnn
n
xdxxx
xxdxx
+++=′
+++=′
…
⋮
…
=
=
nn
nd
d
d
tx
tx
t
⋯
⋮⋱⋮⋮
⋯
⋯
⋮
00
00
00
,
)(
)(
)(22
11
1
Dx
Uncoupling: Transform Matrix T
In order to explore transforming our given system x' = Ax of
coupled equations into an uncoupled system x' = Dx, where D
is a diagonal matrix, we will use the eigenvectors of A.
Suppose A is n x n with n linearly independent eigenvectors
ξξξξ(1),…, ξξξξ(n), and corresponding eigenvalues λ ,…, λ .ξξξξ(1),…, ξξξξ(n), and corresponding eigenvalues λ1,…, λn.
Define n x n matrices T and D using the eigenvalues &
eigenvectors of A:
Note that T is nonsingular, and hence T-1 exists.
=
=
n
n
nn
n
λ
λ
λ
ξξ
ξξ
⋯
⋮⋱⋮⋮
⋯
⋯
⋯
⋮⋱⋮
⋯
00
00
00
,2
1
)()1(
)(
1
)1(
1
DT
Uncoupling: T-1AT = D
Recall here the definitions of A, T and D:
=
=
=n
nn
n
nnn
n
aa
aa
λ
λ
λ
ξξ
ξξ
⋯
⋮⋱⋮⋮
⋯
⋯
⋯
⋮⋱⋮
⋯
⋯
⋮⋱⋮
⋯
00
00
00
,,2
1
)()1(
)(
1
)1(
1
1
111
DTA
Then the columns of AT are Aξξξξ(1),…, Aξξξξ(n), and hence
It follows that T-1AT = D.
n
nnnnn aaλ
ξξ⋯
⋯⋯00
1
TDAT =
=)()1(
1
)(
1
)1(
11
n
nnn
n
n
ξλξλ
ξλξλ
⋯
⋮⋱⋮
⋯
Similarity Transformations
Thus, if the eigenvalues and eigenvectors of A are known,
then A can be transformed into a diagonal matrix D, with
T-1AT = D
This process is known as a similarity transformation, and A
is said to be similar to D. Alternatively, we could say that Ais said to be similar to D. Alternatively, we could say that A
is diagonalizable.
=
=
=
n
n
nn
n
nnn
n
aa
aa
λ
λ
λ
ξξ
ξξ
⋯
⋮⋱⋮⋮
⋯
⋯
⋯
⋮⋱⋮
⋯
⋯
⋮⋱⋮
⋯
00
00
00
,,2
1
)()1(
)(
1
)1(
1
1
111
DTA
Similarity Transformations: Hermitian Case
Recall: Our similarity transformation of A has the form
T-1AT = D
where D is diagonal and columns of T are eigenvectors of A.
If A is Hermitian, then A has n linearly independent
orthogonal eigenvectors ξξξξ(1),…, ξξξξ(n), normalized so that orthogonal eigenvectors ξξξξ(1),…, ξξξξ(n), normalized so that
(ξξξξ(i), ξξξξ(i)) =1 for i = 1,…, n, and (ξξξξ(i), ξξξξ(k)) = 0 for i ≠ k.
With this selection of eigenvectors, it can be shown that
T-1 = T*. In this case we can write our similarity transform as
T*AT = D
Nondiagonalizable A
Finally, if A is n x n with 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 A is not
diagonlizable.
=
=
=
n
n
nn
n
nnn
n
aa
aa
λ
λ
λ
ξξ
ξξ
⋯
⋮⋱⋮⋮
⋯
⋯
⋯
⋮⋱⋮
⋯
⋯
⋮⋱⋮
⋯
00
00
00
,,2
1
)()1(
)(
1
)1(
1
1
111
DTA
Example 3:
Find Transformation Matrix T (1 of 2)
For the matrix A below, find the similarity transformation
matrix T and show that A can be diagonalized.
=
14
11A
We already know that the eigenvalues are λ1 = 3, λ2 = -1
with corresponding eigenvectors
Thus
−=
=
2
1)(,
2
1)( )2()1(
tt ξξ
−=
−=
10
03,
22
11DT
Example 3: Similarity Transformation (2 of 2)
To find T-1, augment the identity to T and row reduce:
=→
→
−→
−−→
−
− 4/12/14/12/101
4/12/110
0111
1240
0111
1022
0111
1T
Then
Thus A is similar to D, and hence A is diagonalizable.
−
=→
−
→4/12/14/12/110
1T
D
ATT
=
−=
−
−=
−
−=−
10
03
26
13
4/12/1
4/12/1
22
11
14
11
4/12/1
4/12/11
Fundamental Matrices for Similar Systems (1 of 3)
Recall our original system of differential equations x' = Ax.
If A is n x n with n linearly independent eigenvectors, then A
is diagonalizable. The eigenvectors form the columns of the
nonsingular transform matrix T, and the eigenvalues are the
corresponding nonzero entries in the diagonal matrix D.corresponding nonzero entries in the diagonal matrix D.
Suppose x satisfies x' = Ax, let y be the n x 1 vector such that
x = Ty. That is, let y be defined by y = T-1x.
Since x' = Ax and T is a constant matrix, we have Ty' = ATy,
and hence y' = T-1ATy = Dy.
Therefore y satisfies y' = Dy, the system similar to x' = Ax.
Both of these systems have fundamental matrices, which we
examine next.
Fundamental Matrix for Diagonal System (2 of 3)
A fundamental matrix for y' = Dy is given by Q(t) = eDt.
Recalling the definition of eDt, we have
( )
∑
∞
=∞∞ n
n
n
n
nn n
t
tt
λλ 00
!000
1
1D
( )
=
=
==
∑
∑
∑∑∞
=
=∞
=
∞
=
t
t
n
n
n
n
n
n
n
n
n
nn
ne
e
n
t
n
n
t
n
tt
λ
λ
λλ
λ
00
00
00
!00
00
!
!00
00
00
!)(
1
0
0
0
1
0
⋱
⋱⋱D
Q
Fundamental Matrix for Original System (3 of 3)
To obtain a fundamental matrix ΨΨΨΨ(t) for x' = Ax, recall that the
columns of ΨΨΨΨ(t) consist of fundamental solutions x satisfying
x' = Ax. We also know x = Ty, and hence it follows that
tnttn neee
λλλξξξξ )(
1
)1(
1
)(
1
)1(
111 00 ⋯⋯
The columns of ΨΨΨΨ(t) given the expected fundamental solutions
of x' = Ax.
=
==tn
n
t
n
tn
nn
nn eeeλλλ
ξξξξ )()1(
11
)()1(
11
100
00
⋯
⋮⋱⋮
⋯
⋱
⋯
⋮⋱⋮
⋯
TQΨ
Example 4:
Fundamental Matrices for Similar Systems
We now use the analysis and results of the last few slides.
Applying the transformation x = Ty to x' = Ax below, this
system becomes y' = T-1ATy = Dy:
yyxx
−=′⇒
=′10
03
14
11
A fundamental matrix for y' = Dy is given by Q(t) = eDt:
Thus a fundamental matrix ΨΨΨΨ(t) for x' = Ax is
yyxx
−
=′⇒
=′1014
=
−t
t
e
et
0
0)(
3
Q
−=
−==
−
−
− tt
tt
t
t
ee
ee
e
et
220
0
22
11)(
3
33
TQΨ