## 8.7. The Algebra Of Constant-Coefficient Operators

A **constant-
coefficient operator** *A* is a
linear operator of the form

A=
∑
k=0
n
a
k
D
n−k
=
a
0
D
n
+
a
1
D
n−1
+⋯+
a
n−1
D+
a
n
(8.9)

where *D*
is the derivative operator and *a*_{k}
are real constants. If
a
0
≠0
,
then *A* is of order *n*.
We can apply *A* to a function *y* defined on some interval *J* to obtain another function on *J* given by

A(
y
)=
∑
k=0
n
a
k
y
(
n−k
)

In this section, we shall consider only
functions the derivative of which exists for every order on the interval
(
−∞,∞
)
. The set of all such functions is denoted by
C
∞
and is called the class of **infinitely differentiable functions**. If
y∈
C
∞
,
so is
A(
y
)
.

It is easy to show that a constant coefficient
operator is also a linear transformation, as defined in §4.1. Furthermore,
any 2 such operators *A* and *B* commute, i.e.,
AB=BA
. Given *A*
in (8.9), its **associated polynomial**
is defined as

p
A
(
r
)=
∑
k=0
n
a
k
r
n−k
(8.9a)

Conversely, given any polynomial of real
coefficients, there is a corresponding operator *A* with the same constant coefficients.

### Theorem 8.6.

Let *A*
and *B* be constant coefficient
operators with associated polynomials *p*_{A}
and *p*_{B}, respectively. Let *l* be any real number, then

(a)
A=B
iff
p
A
=
p
B

(b)
p
A+B
=
p
A
+
p
B

(c)
p
AB
=
p
A
p
B

(d)
p
λA
=λ
p
A

In other words, the association between
operators and polynomials is 1-1 onto and they satisfy the same algebraic
relations.

#### Proof

Parts (b)-(d) follow immediately from
defintion (8.9a). Hence, we shall prove only (a).

Let
p
A
=
p
B
,
then they must have the same degree as well as coefficients. This means *A* and *B* have the same
order and coefficients. Hence,
A=B
. This proves the ‘if’ part of (a).

Next, given
A=B
,
we have
A(
y
)=B(
y
)
∀ y∈
C
∞
. Let
y=
e
rx
,
where *r* is a constant. Then
y
(
k
)
=
r
k
e
rx
=
r
k
y
∀ k≥0
. Hence,

A(
y
)=
p
A
(
r
)
e
rx
=B(
y
)=
p
B
(
r
)
e
rx

Since
e
rx
≠0
,
we have
p
A
=
p
B
,
which proves the ‘only if’ part of (a).

### Comment

According to Theorem 8.6, constant
coefficient operators can be manipulated like polynomials. For example, if *p*_{A}(*r*) can be
factorized as

p
A
(
r
)=
a
0
∏
k=1
n
(
r−
r
k
)
(8.10)

so can *A*:

A=
a
0
∏
k=1
n
(
D−
r
k
)
(8.10a)

[Note that the factors on the right sides
commute.] Now, the fundamental theorem
of algebra guarantees the factorization (8.10).
Furthermore,
if the coefficients of *p*_{A}(*r*) are real, then the roots *r*_{k} must either be real or
else occur in conjugate pairs:
α±iβ
. In the latter case, we have

(
r−α−iβ
)(
r−α+iβ
)=
r
2
−2αr+
α
2
+
β
2

Thus, *p*_{A}(*r*) can always be factorized into a
product of linear and quadratic factors of *real*
coefficients. Ditto *A*.

### Example 1.

Let
A=
D
2
−5D+6
. Then
p
A
=
r
2
−5r+6
=(
r−2
)(
r−3
)
and

A=(
D−2
)(
D−3
)

### Example 2.

Let
A=
D
4
−2
D
3
+2
D
2
−2D+1
. Then

p
A
(
r
)=
r
4
−2
r
3
+2
r
2
−2r+1
=
(
r−1
)
2
(
r
2
+1
)

and

A=
(
D−1
)
2
(
D
2
+1
)