## 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 nk   = a 0 D n + a 1 D n1 ++ a n1 D+ a n                     (8.9)

where D is the derivative operator and ak 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 ( nk )

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 nk                                     (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 pA and pB, 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     k0 .  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 pA(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 pA(r) are real, then the roots rk must either be real or else occur in conjugate pairs: α±iβ .  In the latter case, we have

( rαiβ )( rα+iβ )= r 2 2αr+ α 2 + β 2

Thus, pA(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   =( r2 )( r3 )  and

A=( D2 )( D3 )

### 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   = ( r1 ) 2 ( r 2 +1 )

and

A= ( D1 ) 2 ( D 2 +1 )