Fall 2003 Math 308/501–502
9 Matrix Methods for Linear Systems
9.4 Linear Systems in Normal Form
Wed, 12/Nov
c
±
2003, Art Belmonte
Summary
Deﬁnition of a linear system; matrix notation
A
linear system
is one that may be written in the
normal form
x
0
(
t
)
=
Ax
(
t
)
+
f
(
t
)
or more brieﬂy,
x
0
=
Ax
+
f
.Here
x
is an
n
×
1
column
vector
function,
A
is an
n
×
n
matrix function, and
f
is an
n
×
1
column
vector function (the
forcing term
). If
f
=
0
,the
n
×
1 zero vector
function, then the system is
homogeneous
; otherwise it is said to
be
nonhomogeneous
.
The
j
th row in this vector differential equation is
x
0
j
=
f
j
(
t
)
+
n
±
k
=
1
a
jk
(
t
)
x
k
(
t
).
The sum on the righthand side consists of
j
th element of the
forcing term and the dot product of the
j
th row of
A
with the
column vector
x
. Note that the
x
k
appear solely to the ﬁrst power
(hence the phrase “linear”). Moreover, while the
a
and
f
k
functions may depend on the independent variable
t
(or be
constants), they do
not
depend on the dependent variables (the
x
k
).
YALO (“Yet Another Linear Operator”)
Rewrite
x
0
=
Ax
+
f
as
x
0

Ax
=
f
and let
L
[
x
]
=
x
0

Ax
.
Then the system becomes
L
[
x
]
=
f
. (NOTE: In this form, it’s
actually easier to see why the linear system is called homogeneous
if
f
=
0
and nonhomogeneous if
f
6=
0
.) That
L
is a
linear
operator follows immediately from the linearity of differentiation
and matrix muliplication.
Initial value problem for a linear system of ODEs
This consists of the differential equation
x
0
=
Ax
+
f
together
with the initial condition
x
(
t
0
)
=
x
0
.
Existence and Uniqueness Theorem
With
A
(
t
)
and
f
(
t
)
deﬁned as above and continuous on an interval
I
,let
t
0
∈
I
and
x
0
∈
R
n
. Then the initial value problem
x
0
(
t
)
=
A
(
t
)
x
(
t
)
+
f
(
t
),
x
(
t
0
)
=
x
0
has a unique solution deﬁned for
all t
∈
I
.
Converting higherorder linear ODEs to systems
Given the
n
th order linear equation in standard form
y
(
n
)
(
t
)
+
n

1
±
j
=
0
p
j
(
t
)
y
(
j
)
(
t
)
=
g
(
t
),
let
x
1
=
y
,
x
2
=
y
0
,
x
3
=
y
00
,
...
,
x
n
=
y
(
n

1
)
. We thus have
x
0
k
=
y
(
k
)
=
x
k
+
1
,
k
=
1
,
2
,...
n

1. Moreover,
x
0
n
=
y
(
n
)
=
g
(
t
)

n

1
±
j
=
0
p
j
(
t
)
y
(
j
)
(
t
).
Therefore, we have the linear system
x
0
1
=
x
2
x
0
2
=
x
3
···
x
0
n

1
=
x
n
x
0
n
=
g
(
t
)

n
±
j
=
1
p
j
(
t
)
y
(
j
)
(
t
).
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 Spring '08
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 Linear Algebra, Derivative, Linear Systems, Vector Space

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