14
Verifying the Big Theorems and an
Introduction to Differential Operators
There are two parts to this chapter. The first (sections 14.1 and 14.2) is a discussion of the proofs
of the main theorems in the last chapter. To be precise, section 14.1 contains a fairly complete
verification of theorem 13.3 (the big theorem on second-order, homogeneous linear differential
equations), along with a partial verification of theorem 13.6 on Wronskians, while section 14.2
contains a rather brief discussion on generalizing the results just verified in section 14.1.
The rest of the chapter is devoted to a fairly elementary development of “linear differential
operators”. This material provides a slightly different perspective on linear differential equations,
and can be enlightening to the careful reader (and confusing to the less careful reader). In
addition, this material will make it easier to prove a few more advanced results later on in this
text.
To be honest, most beginning students of differential equations can probably skip this chapter
and go straight to the next chapter where we actually develop methods for completely solving
a large number of important differential equations. In fact, it may be better to do so, promising
yourself to return to this chapter as the need or interest arises.
14.1
Verifying the Big Theorem on Second-Order,
Homogeneous Equations
Our main goal is to verify theorem 13.3 on page 287. Accordingly, throughout this section
we will assume the basic assumptions of that theorem, namely, that we have an open interval
(α, β) and functions a , b and c that are continuous on (α, β) with a never being zero on this
interval. With these assumptions, we will explore what we can about solutions to the second-order
homogeneous linear differential equation
ay ′′ + by ′ + cy = 0
.
Incidentally, along the way, we will also verify the claim concerning Wronskians in theorem 13.6
for the case where N = 2 .
299
300
Verifying the Big Theorems and an Introduction to Differential Operators
Linear Independence, Initial-Value Problems and Wronskians
Much of our analysis will reduce to being able to solve any initial-value problem
ay ′′ + by ′ + cy = 0
with
y(x0 ) = A
y ′ (x0 ) = B
and
(14.1)
where x0 is some point in (α, β) , and A and B are any two real numbers. Recall that lemma
13.2 on page 283 assures us that any such initial-value problem has one and only one solution.
Solving Initial-Value Problems
Until further notice, we will assume that we have a pair of solutions over (α, β)
{ y1 , y2 }
to our differential equation
ay ′′ + by ′ + cy = 0
.
By the principle of superposition (theorem 13.1 on page 280), we know that any linear combination of these two solutions
y(x) = c1 y1 (x) + c2 y2 (x)
for all x in (α, β)
is also a solution to our differential equation. Now let us consider solving initial-value problem
(14.1) using this linear combination.
As already noted, this linear combination satisfies our differential equation. So all we need
is to do is to find constants c1 and c2 such that
A = y(x0 ) = c1 y1 (x0 ) + c2 y2 (x0 )
and
B = y ′ (x0 ) = c1 y1 ′ (x0 ) + c2 y2 ′ (x0 )
.
That is, we need to solve the linear algebraic system of two equations
c1 y1 (x0 ) + c2 y2 (x0 ) = A
c1 y1 ′ (x0 ) + c2 y2 ′ (x0 ) = B
for c1 and c2 . But this is easy. Start by multiplying each equation by y2 ′ (x0 ) or y2 (x0 ) , as
appropriate:
c1 y1 (x0 ) + c2 y2 (x0 ) = A y2 ′ (x0 )
c1 y1 (x0 )y2 ′ (x0 ) + c2 y2 (x0 )y2 ′ (x0 ) = Ay2 ′ (x0 )
H⇒
c1 y1 ′ (x0 ) + c2 y2 ′ (x0 ) = B y2 (x0 )
c1 y1 ′ (x0 )y2 (x0 ) + c2 y2 ′ (x0 )y2 (x0 ) = By2 (x0 )
Subtracting the second equation from the first (and looking carefully at the results) yields
c1 y1 (x0 )y2 ′ (x0 ) − y1 ′ (x0 )y2 (x0 ) + c2 y2 (x0 )y2 ′ (x0 ) − y2 ′ (x0 )y2 (x0 )
|
{z
}
|
{z
}
W (x 0 )
0
= Ay2 ′ (x0 ) − By2 (x0 ) .
That is,
c1 W (x0 ) = Ay2 ′ (x0 ) − By2 (x0 )
.
(14.2a)
Verifying the Big Theorem
301
where W is the Wronskian of function pair {y1 , y2 } , which (as you may recall from section
13.3), is the function on (α, β) given by
W = W [y1 , y2 ] = y1 y2 ′ − y1 ′ y2
.
Similar computations yield
c2 W (x0 ) = By1 (x0 ) − Ay1 ′ (x0 ) .
(14.2b)
Observe that, if W (x0 ) 6= 0 , then there is exactly one possible value for c1 and one possible
value for c2 , namely,
c1 =
Ay2 ′ (x 0 ) − By2 (x 0 )
W (x 0 )
and
By1 (x 0 ) − Ay1 ′ (x 0 )
W (x 0 )
c2 =
.
On the other hand, if W (x0 ) = 0 , then system (14.2) reduces to
0 = Ay2 ′ (x0 ) − By2 (x0 )
and
0 = By1 (x0 ) − Ay1 ′ (x0 )
which cannot be solved for c1 and c2 . If the right sides of these last two equations just happen
to both be 0 , then the values of c1 and c2 are irrelevant — any values work. And if either
right-hand side is nonzero, then no values for c1 and c2 will work.
The fact that the solvability of the above initial-value problem depends entirely on whether
W (x0 ) is zero or not is an important fact. Let us enshrine this fact in a lemma for later reference.
Lemma 14.1 (Wronskians and initial-value problems)
Let x0 , A and B be any three fixed real values with x0 in (α, β) , and assume {y1 , y2 } is a
pair of functions differentiable at x0 . Also, let W be the Wronskian of {y1 , y2 } ,
W = y1 y2 ′ − y1 ′ y2
.
Then the problem of finding constants c1 and c2 such that
y = c1 y1 + c2 y2
satisfies the system
y(x0 ) = A
and
y ′ (x0 ) = B
has exactly one solution (i.e., exactly choice for c1 and exactly one choice for c2 ) if and only if
W (x0 ) 6= 0 .
Wronskians and Linear Independence
It turns out that the Wronskian of {y1 , y2 } provides an alternative test for determining whether
this pair of solutions is linearly dependent or linearly independent. To start, suppose {y1 , y2 } is
linearly dependent. Then either at least one of these functions, say, y1 , is a constant multiple of
the other function; that is, for some constant κ ,
y1 (x) = κ y2 (x)
for all x in (α, β) .
This, of course, means that
y1 ′ (x) = κ y2 ′ (x)
for all x in (α, β) .
302
Verifying the Big Theorems and an Introduction to Differential Operators
Thus,
W = y1 y2 ′ − y1 ′ y2 = κ y2 y2 ′ − κ y2 ′ y2 = 0
.
So,
{y1 , y2 } is linearly dependent over (α, β)
H⇒
(14.3a)
W = 0 everywhere on (α, β) .
With a little thought, you’ll realize that this is equivalent to
W 6= 0 somewhere on (α, β)
H⇒
{y1 , y2 } is linearly independent over (α, β) .
(14.3b)
Showing that the above implications still hold with the arrows reversed requires a bit more
work. To simplify that work, let’s first prove the following two corollaries of lemma 13.2 on
page 283:
Corollary 14.2
Let x0 be a point in an interval (α, β) , and assume a , b and c are continuous functions on
(α, β) with a never being zero in (α, β) . Then the only solution to
ay ′′ + by ′ + cy = 0
with
y(x0 ) = 0
and
y ′ (x0 ) = 0
.
is the trivial solution,
for all x in (α, β) .
y(x) = 0
PROOF: The trivial solution is certainly a solution to the given initial-value problem, and
lemma 13.2 assures us that this solution is the only solution.
Corollary 14.3
Let x0 be a point in an interval (α, β) , and assume a , b and c are continuous functions on
(α, β) with a never being zero in (α, β) . Assume that {y1 , y2 } is a pair of solutions on (α, β)
to
ay ′′ + by ′ + cy = 0 .
If there is a point x0 in (α, β) and a constant κ such that
y1 (x0 ) = κ y2 (x0 )
and
y1 ′ (x0 ) = κ y2 ′ (x0 ) ,
then
y1 (x) = κ y2 (x)
for all x in (α, β)
and, hence, the pair {y1 , y2 } is linearly dependent on (α, β) .
PROOF:
where
Observe that both y = y1 and y = κ y2 satisfy the same initial-value problem,
ay ′′ + by ′ + cy = 0
with
A = y1 (x0 )
and
y(x0 ) = A
y ′ (x0 ) = B
and
B = y1 ′ (x0 )
.
Verifying the Big Theorem
303
Again, lemma 13.2 tells us that this initial-value problem only has one solution. Hence y = y1
and y = κ y2 must be the same, and the claims in the corollary follow immediately.
Now suppose the Wronskian of our pair {y1 , y2 } of solutions to our differential equation is
zero at some point x0 in (α, β) . That is,
W (x0 ) = y1 (x0 )y2 ′ (x0 ) − y2 (x0 )y1 ′ (x0 ) = 0
for some x0 in (α, β) .
This, of course, means that
y1 (x0 )y2 ′ (x0 ) = y2 (x0 )y1 ′ (x0 )
.
(14.4)
Let us consider all the possibilities. For simplicity, we will start by assuming y2 (x0 ) 6= 0 :
1.
If y2 (x0 ) 6= 0 and y1 ′ (x0 ) 6= 0 , then equation (14.4) yields
y1 (x0 )y2 ′ (x0 ) = y2 (x0 )y1 ′ (x0 ) 6= 0
,
implying that y1 (x0 ) 6= 0 and y2 ′ (x0 ) 6= 0 . Consequently, we can divide both sides of
this equation by these two values and set
κ =
y1 ′ (x 0 )
y1 (x 0 )
=
y2 (x 0 )
y2 ′ (x 0 )
.
Thus, κ is a constant such that
y1 (x0 ) = κ y2 (x0 )
y1 ′ (x0 ) = κ y2 ′ (x0 ) ,
and
and corollary 14.3 tells us that {y1 , y2 } is linearly dependent.
2.
If y2 (x0 ) 6= 0 and y1 ′ (x0 ) = 0 , then equation (14.4) yields
y1 (x0 )y2 ′ (x0 ) = y2 (x0 )y1 ′ (x0 ) = 0
telling us that y1 (x0 ) = 0 or y2 ′ (x0 ) = 0 .
(a) But if y1 (x0 ) = 0 , then y1 is the solution to
ay ′′ + by ′ + cy = 0
with
y(x0 ) = 0
and
y ′ (x0 ) = 0
,
which, as noted in corollary 14.2 means that y1 is the trivial solution, automatically
making {y1 , y2 } linearly dependent.
(b)
On the other hand, if y2 ′ (x0 ) = 0 , then we can set
κ =
y1 (x 0 )
y2 (x 0 )
.
This, along with the fact that y1 ′ (x0 ) = 0 and y2 ′ (x0 ) = 0 gives us
y1 (x0 ) = κ y2 (x0 )
and
y1 ′ (x0 ) = κ y2 ′ (x0 ) ,
and corollary 14.3 tells us that {y1 , y2 } is linearly dependent.
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Verifying the Big Theorems and an Introduction to Differential Operators
From the above it follows that {y1 , y2 } must be linearly dependent if we have both W (x0 ) = 0
and y2 (x0 ) 6= 0 . Now let’s assume W (x0 ) = 0 and y2 (x0 ) = 0 . Equation (14.4) still holds,
but, this time, yields
y1 (x0 )y2 ′ (x0 ) = y2 (x0 )y1 ′ (x0 ) = 0 · y1 ′ (x0 ) = 0
,
which means that y1 (x0 ) = 0 or y2 ′ (x0 ) = 0 .
1.
If y2 (x0 ) = 0 and y2 ′ (x0 ) = 0 , then y2 satisfies
ay ′′ + by ′ + cy = 0
with
y(x0 ) = 0
y ′ (x0 ) = 0
and
.
Again, corollary 14.2 tells us that y1 is the trivial solution, automatically making {y1 , y2 }
linearly dependent.
2.
On the other hand, if y2 (x0 ) = 0 , y2 ′ (x0 ) 6= 0 and y1 (x0 ) = 0 , then we can set
y1 ′ (x 0 )
y2 ′ (x 0 )
κ =
and observe that, from this and the fact that y1 (x0 ) = 0 = y2 (x0 ) , we have
and
y1 (x0 ) = κ y2 (x0 )
y1 ′ (x0 ) = κ y2 ′ (x0 ) .
which, according to corollary 14.3, means that {y1 , y2 } is linearly dependent.
Thus, we have that {y1 , y2 } is linearly dependent whenever W (x0 ) = 0 whether or not y2 (x0 ) =
0 . That is,
W (x0 ) = 0 for some x0 in (α, β)
{y1 , y2 } is linearly dependent over (α, β) .
H⇒
(14.5a)
Equivalently,
{y1 , y2 } is linearly independent over (α, β)
H⇒
W (x) 6= 0 for every x in (α, β) .
(14.5b)
Linearly Independent Pairs and General Solutions
Now assume that our pair of solutions {y1 , y2 } is linearly independent over (α, β) , and let b
y
be any single solution to our differential equation. Pick some x0 in (α, β) , and consider the
initial-value problem
where
ay ′′ + by ′ + cy = 0
with
A = b
y(x0 )
and
y(x0 ) = A
and
B = b
y ′ (x0 ) .
y ′ (x0 ) = B
Clearly, y = b
y is a solution to the initial-value problem.
On the other hand, implication (14.5b) tells us that, since {y1 , y2 } is linearly independent,
the corresponding Wronskian is nonzero at x0 , and lemma 14.1 and the principle of superposition
then assure us that there are constants c1 and c2 such that
y = c1 y1 + c2 y2
Verifying the Big Theorem
305
is also a solution to this same initial-value problem. So we have “two” solutions to our initialvalue problem,
b
y
and
y = c1 y1 + c2 y2
But (by lemma 13.2 on page 283) we know this initial-value problem only has one solution. This
means our two solutions must be the same,
b
y(x) = c1 y1 (x) + c2 y2 (x)
for every x in (α, β) .
Hence, any solution to our differential equation can be written as a linear combination of y1 and
y1 . This shows that
{y1 , y2 } is linearly independent over (α, β)
H⇒
y = c1 y1 + c2 y2 is a general solution to our differential equation ,
(14.6)
which, incidentally, means that the linearly independent pair {y1 , y2 } is a fundament set of
solutions for our differential equation.
Summarizing the Results So Far
Combining the results given in lemma 14.1 and implications (14.3), (14.5) and (14.6) gives us
the following lemma.
Lemma 14.4
Assume a , b and c are continuous functions on an interval (α, β) with a never being zero in
(α, β) , and let {y1 , y2 } be a pair of solutions on (α, β) to
ay ′′ + by ′ + cy = 0
.
Then all of the following statements are equivalent (i.e., if one holds, they all hold):
1.
The pair {y1 , y2 } is linearly independent on (α, β) .
2.
The Wronskian of {y1 , y2 } ,
W = y1 y2 ′ − y2 y1 ′
,
is nonzero at one point in (α, β) .
3.
The Wronskian of {y1 , y2 } ,
W = y1 y2 ′ − y2 y1 ′
,
is nonzero at every point in (α, β) .
4.
Given any point x0 in (α, β) and any two fixed values A and B , there is exactly one
ordered pair of constants {c1 , c2 } such that
y(x) = c1 y1 (x) + c2 y2 (x)
also satisfies the initial conditions
y(x0 ) = A
and
y ′ (x0 ) = B
.
306
5.
Verifying the Big Theorems and an Introduction to Differential Operators
The arbitrary linear combination of y1 and y2 ,
y = c1 y1 + c2 y2
,
is a general solution for the above differential equation.
6.
The pair {y1 , y2 } is a fundamental set of solutions for the above differential equation.
If you check, you will see that most of our “big theorem”, theorem 13.3 follows immediately
from this lemma, as does the theorem on Wronskians, theorem 13.6, for the case where N = 2 .
Proving the Rest of Theorem 13.3
All that remains to achieving our goal of verifying theorem 13.3 is to verify that fundamental
sets of solutions exist, and that a fundamental set of solutions must contain exactly two solutions.
Verifying these two facts will be easy.
Existence of Fundamental Sets
Let x0 be some point in (α, β) , and let y1 and y2 be, respectively, be the solutions to the
initial-value problems
and
ay ′′ + by ′ + cy = 0
with
y(x0 ) = 1
and
y ′ (x0 ) = 0
ay ′′ + by ′ + cy = 0
with
y(x0 ) = 0
and
y ′ (x0 ) = 1
.
(By lemma 13.2, we know these solutions exist.) Computing their Wronskian at x0 , we get
W (x0 ) = y1 (x0 )y2 ′ (x0 ) − y2 (x0 )y1 ′ (x0 ) = 1 · 1 − 0 · 0 6= 0
,
which, according to lemma 14.4, means that {y1 , y2 } is a fundamental set of solutions to our
differential equation.
Size of Fundamental Sets of Solutions
It should be clear that we cannot solve every initial-value problem
ay ′′ + by ′ + cy = 0
with
y(x0 ) = A
and
y ′ (x0 ) = B
using a linear combination of a single particular solution y1 of the differential equation. After
all, just try finding a constant c1 so that
y(x) = c1 y1 (x)
satisfies
when
ay ′′ + by ′ + cy = 0
A = y1 (x0 )
with
and
y(x0 ) = A
and
y ′ (x0 ) = B
B = 1 + y1 ′ (x0 ) .
So a fundamental set of solutions for our differential equation must contain at least two solutions.
Linear Differential Operators
307
On the other hand, if anyone were to propose the existence of a fundamental solution set of
more than two solutions
{ y1 , y2 , y3 , . . . } ,
then the required linear independence of the set (i.e., that no solution in this set is a linear
combination of the others) would automatically imply that the set of just the first two solutions,
{ y1 , y2 }
,
is also linearly independent. But then lemma 14.4 tells us that this smaller set is a fundamental
set of solutions for our differential equation, and that every other solution, including the y3 in
the set originally proposed, is a linear combination of y1 and y2 . Hence, the originally proposed
set of three or more solutions cannot be linearly independent, and, hence, is not a fundamental
set of solutions for our differential equation.
So, a fundamental set of solutions for a second-order, homogeneous linear differential equation cannot contain less than two solutions or more than two solutions. It must contain exactly
two solutions.
And that completes our proof of theorem 13.3.
14.2
Proving the More General Theorems on General
Solutions and Wronskians
Extending the discussion in the previous section into proofs of the more general theorems in
chapter 13 (theorem 13.5 on page 289 and theorem 13.6 on page 291) is relatively straightforward
provided you make use of some basic facts normally developed in a good introductory course
on linear algebra. Alas, at this point in the text, it is not assumed that the reader has had such a
course. So proving these theorems at this time will be left as a challenge to the interested reader
who has had a good course on linear algebra.
Besides, later in this text (at a point where a more complete knowledge of linear algebra
is assumed), we will prove analogous theorems involving “systems of differential equations”.
Theorems 13.5 and 13.6 can then be derived as corollaries.
14.3
Linear Differential Operators
The Operator Associated with a Linear Differential Equation
Sometimes, when given some N th -order linear differential equation
a0
dN y
d N −1 y
d2 y
dy
+ a1 N −1 + · · · + a N −2 2 + a N −1
+ aN y = g
N
dx
dx
dx
dx
,
it is convenient to let L[y] denote the expression on the left side, whether or not y is a solution
to the differential equation. That is, for any sufficiently differentiable function y ,
L[y] = a0
dN y
d N −1 y
d2 y
dy
+
a
+
·
·
·
+
a
+ a N −1
+ aN y
1
N
−2
N
N
−1
2
dx
dx
dx
dx
.
308
Verifying the Big Theorems and an Introduction to Differential Operators
To emphasize that y is a function of x , we may also use L[y(x)] instead of L[y] . For much
of what follows, y need not be a solution to the given differential equation, but it does need to
be sufficiently differentiable on the interval of interest for all the derivatives in the formula for
L[y] to make sense.
While we defined L[y] as the left side of the above differential equation, the expression for
L[y] is completely independent of the equation’s right side. Because of this and the fact that the
choice of y is largely irrelevant to the basic definition, we will often just define “ L ” by stating
L = a0
dN
d2
d
d N −1
+ a1 N −1 + · · · + a N −2 2 + a N −1
+ aN
N
dx
dx
dx
dx
where the ak ’s are functions of x on the interval of interest.1
!◮Example 14.1:
If our differential equation is
√
d2 y
dy
+ x2
− 6y = x + 1
2
dx
dx
then
L =
d
d2
− 6
+ x2
2
dx
dx
,
,
and, for any twice-differentiable function y = y(x) ,
L[y(x)] = L[y] =
dy
d2 y
+ x2
− 6y
2
dx
dx
.
In particular, if y = sin(2x) , then
d d2 L[y] = L sin(2x) =
sin(2x) + x 2
sin(2x) − 6 sin(2x)
2
dx
dx
2
= −4 sin(2x) + x · 2 cos(2x) − 6 sin(2x)
= 2x 2 cos(2x) − 10 sin(2x)
.
Observe that L is something into which we plug a function (such as the sin(2x) in the
above example) and out of which pops another function (which, in the above example, ended up
being 2x 2 cos(2x) − 10 sin(2x) ). Anything that so converts one function into another is often
called an operator (on functions), and since the general formula for computing L[y] looks like
a linear combination of differentiations up to order N ,
L[y] = a0
dN y
d N −1 y
d2 y
dy
+
a
+ aN y
+
·
·
·
+
a
+ a N −1
1
N
−2
N
N
−1
2
dx
dx
dx
dx
,
it is standard to refer to L as a linear differential operator (of order N .
We should also note that our linear differential operators are “linear” in the sense normally
defined in linear algebra:
1 If using “ L ” is just too much shorthand for you, observe that the formulas for L can be written in summation
form:
L[y] =
N
X
k=0
ak
d N −k y
d x N −k
and
L =
You can use these summation formulas instead of “ L ” if you wish.
N
X
k=0
ak
d N −k
d x N −k
.
Linear Differential Operators
309
Lemma 14.5
Assume L is a linear differentiable operator
L = a0
dN
d N −1
d2
d
+
a
+ aN
+
·
·
·
+
a
+ a N −1
1
N
−2
dx N
dx
dx N −1
dx 2
where the ak ’s are functions on some interval (α, β) . If y1 and y2 are any two sufficiently
differentiable functions on (α, β) , and c1 and c2 are any two constants, then
L[c1 y1 + c2 y2 ] = c1 L[y1 ] + c2 L[y2 ] .
To prove this lemma, you basically go through the same computations as used to derive the
principle of superposition (see the derivations just before theorem 13.1 on page 280).
The Composition Product
Definition and Notation
The (composition) product L 2 L 1 of two linear differential operators L 1 and L 2 is the differential
operator given by
L 2 L 1 [φ] = L 2 L 1 [φ]
for every sufficiently differentiable function φ = φ(x) .2
!◮Example 14.2:
Let
L1 =
d
+ x2
dx
and
L2 =
For any twice-differentiable function φ = φ(x) , we have
dφ
2
L 2 L 1 [φ] = L 2 [L 1 [φ]] = L 2
+ x φ
dx
d
=
dx
Cutting out the middle yields
L 2 L 1 [φ] =
dφ
+ x 2φ
dx
d
+ 4
dx
.
dφ
+ 4
+ x 2φ
dx
=
d 2φ
d 2 dφ
+
x φ + 4
+ 4x 2 φ
dx
dx
dx 2
=
d 2φ
dφ
dφ
+ 2xφ + x 2
+ 4
+ 4x 2 φ
dx
dx
dx 2
=
dφ
d 2φ
+ 4 + x2
+ 2x + 4x 2 φ
2
dx
dx
d 2φ
2
2 dφ
+
4
+
x
+
2x
+
4x
φ
dx
dx 2
for every sufficiently differentiable function φ . Thus
L2 L1 =
d
d2
+ 4 + x2
+ 2x + 4x 2
2
dx
dx
2 The notation L ∘ L , instead of L L would also be correct.
2
1
2 1
.
.
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Verifying the Big Theorems and an Introduction to Differential Operators
When we have formulas for our operators L 1 and L 2 , it will often be convenient to replace
the symbols “ L 1 ” and “ L 2 ” with their formulas enclosed in parentheses. We will also enclose
any function φ being “plugged into” the operators with square brackets, “ [φ] ”. This will be
called the product notation.3
!◮Example 14.3:
Using the product notation, let us recompute L 2 L 1 for
L1 =
d
+ x2
dx
and
d
+ 4
dx
L2 =
.
Letting φ = φ(x) be any twice-differentiable function,
d
d
d
dφ
+ 4
+ x 2 [φ] =
+ 4
+ x 2φ
dx
So,
L2 L1 =
dx
d
+ 4
dx
dx
dx
dφ
+ x 2φ
dx
dφ
+ x 2φ
dx
=
d
dx
=
d2φ
d 2 dφ
+
x φ + 4
+ 4x 2 φ
2
dx
dx
dx
=
d2φ
dφ
dφ
+ 2xφ + x 2
+ 4
+ 4x 2 φ
dx
dx
dx 2
=
dφ
d2φ
+ 4 + x2
+ 2x + 4x 2 φ
2
dx
dx
d
+ x2
dx
=
+ 4
d
d2
+ 4 + x2
+ 2x + 4x 2
2
dx
dx
.
,
just as derived in the previous example.
Algebra of the Composite Product
The notation L 2 L 1 [φ] is convenient, but it is important to remember that it is shorthand for
compute L 1 [φ] and plug the result into L 2
.
The result of this can be quite different from
compute L 2 [φ] and plug the result into L 1
,
which is what L 1 L 2 [φ] means. Thus, in general,
L 2 L 1 6= L 1 L 2
.
In other words, the composition product of differential operators is generally not commutative.
3 Many authors do not enclose “the function being plugged in” in square brackets, and just write L L φ . We are
2 1
avoiding that because it does not explicitly distinguish between “ φ as a function being plugged
in” and “ φ as an
operator, itself”. For the first, L 2 L 1 φ means the function you get from computing L 2 L 1 [φ] . For the second,
L 2 L 1 φ means the operator such that, for any sufficiently differentiable function ψ ,
L 2 L 1 [φ[ψ]] = L 2 L 1 [φψ]
,
The two possible interpretations for L 2 L 1 φ are not the same.
Linear Differential Operators
!◮Example 14.4:
d
dx
311
In the previous two examples, we saw that
d
d
d2
+ 4
+ x2 =
+ 4 + x2
2
dx
dx
dx
+ 2x + 4x 2
.
On the other hand, switching the order of the two operators, and letting φ be any sufficiently
differentiable function gives
d
d
dφ
d
2
2
+ x
+ 4 [φ] =
+ x
+ 4φ
dx
dx
dx
d
=
dx
Thus,
d
+ x2
dx
d
+ 4
dx
dx
dφ
+ 4φ
dx
+ x
2
dφ
+ 4φ
dx
=
dφ
d 2φ
dφ
+ 4
+ x2
+ 4x 2 φ
dx
dx
dx 2
=
dφ
d 2φ
+ 4 + x2
+ 4x 2 φ
2
dx
dx
=
d
d2
+ 4 + x2
+ 4x 2
2
dx
dx
.
.
After comparing this with the first equation in this example, we clearly see that
d
d
d
d
2
2
+ x
+ 4 6=
+ 4
+ x
.
dx
?◮Exercise 14.1:
dx
Let
L1 =
dx
d
dx
and
dx
L2 = x
,
and verify that
L2 L1 = x
d
dx
while
L1 L2 = x
d
+ 1
dx
.
Later (in chapters 17 and 20) we will be dealing with special situations in which the composition product is commutative. In fact, the material we are now developing will be most useful
verifying certain theorems involving those situations. In the meantime, just remember that, in
general,
L 2 L 1 6= L 1 L 2 .
Here are a few other short and easily verified notes about the composition product:
1.
In the above examples, the operators L 2 and L 1 were all first order differential operators.
This was not necessary. We could have used, say,
L2 = x3
√
d2
d
d3
+
sin(x)
− xe3x
+ 87 x
3
2
dx
dx
dx
and
L1 =
3
d 26
3d
−
x
dx 3
dx 26
,
though we would have certainly needed many more pages for the calculations.
312
2.
Verifying the Big Theorems and an Introduction to Differential Operators
There is no need to limit our selves to composition products of just two operators. Given
any number of linear differential operators — L 1 , L 2 , L 3 , . . . — the composition
products L 3 L 2 L 1 , L 4 L 3 L 2 L 1 , etc. are defined to be the differential operators satisfying,
for each and every sufficiently differentiable function φ ,
L 3 L 2 L 1 [φ] = L 3 L 2 L 1 [φ]
,
L 4 L 3 L 2 L 1 [φ] = L 4 L 3 L 2 L 1 [φ]
,
..
.
Naturally, the order of the operators is still important.
3.
Any composition product of linear differential operators is, itself, a linear differential
operator. Moreover, the order of the product
L K · · · L2 L1
is the sum
(the order of L K ) + · · · + (the order of L 2 ) + (the order of L 1 )
4.
.
Though not commutative, the composition product is associative. That is, if L 1 , L 2 and
L 3 are three linear differential operators, and we ‘precompute’ the products L 2 L 1 and
L 3 L 2 , and then compute
(L 3 L 2 )L 1
,
L 3 (L 2 L 1 )
and
,
L3 L2 L1
we will discover that
(L 3 L 2 )L 1 = L 3 (L 2 L 1 ) = L 3 L 2 L 1
5.
.
Keep in mind that we are dealing with linear differential operators and that their products
are linear differential operators. In particular, if α is some constant and φ is any
sufficiently differentiable function, then
L K · · · L 2 L 1 [αφ] = αL K · · · L 2 L 1 [φ] .
And, of course,
L K · · · L 2 L 1 [0] = 0
.
Factoring
Now suppose we have some linear differential operator L . If we can find other linear differential
operators L 1 , L 2 , L 3 , . . . , and L K such that
L = L K · · · L2 L1
,
then, in analogy with the classical concept of factoring, we will say that we have factored the
operator L . The product L N · · · L 2 L 1 will be called a factoring of L , and we may even refer
to the individual operators L 1 , L 2 , L 3 , . . . and L N as factors of L . Keep in mind that, since
composition multiplication is order dependent, it is not usually enough to simply specify the
factors. The order must also be given.
Linear Differential Operators
!◮Example 14.5:
313
In example 14.3, we saw that
d
d2
+ 4 + x2
+ 2x + 4x 2 =
2
dx
dx
So
d
+ 4
dx
d
+ 4
dx
d
+ x2
dx
d
+ x2
dx
.
is a factoring of
d
d2
+ 2x + 4x 2
+ 4 + x2
2
dx
dx
with factors
d
+ 4
dx
d
+ x2
dx
and
.
In addition, from example 14.4 we know
d
d2
+ 4 + x2
+ 4x 2 =
2
dx
dx
Thus
d
+ x2
dx
are also factors for
but the factoring here is
d
+ x2
dx
d
+ 4
dx
and
d
d2
+ 4 + x2
+ 4x 2
2
dx
dx
d
+ x2
dx
d
+ 4
dx
d
+ 4
dx
.
.
,
.
Let’s make a simple observation. Assume a given linear differential operator L can be
factored as L = L K · · · L 2 L 1 . Assume, also, that y1 = y1 (x) is a function satisfying
L 1 [y1 ] = 0
Then
.
L[y1 ] = L K · · · L 2 L 1 [y1 ] = L K · · · L 2 L 1 [y1 ] = L K · · · L 2 [0] = 0
.
This proves the following theorem:
Theorem 14.6
Let L be a linear differential operator with factoring L = L K · · · L 2 L 1 . Then any solution to
is also a solution to
L 1 [y] = 0
L[y] = 0
.
Warning: On the other hand, if, say, L = L 2 L 1 , then solutions to L 2 [y] = 0 will usually
not be solutions to L[y] = 0 .
314
Verifying the Big Theorems and an Introduction to Differential Operators
!◮Example 14.6:
Consider
d2 y
2 dy
+
4
+
x
+ 4x 2 y = 0
dx
dx 2
.
As derived in example 14.4,
d
d2
+ 4 + x2
+ 4x 2 =
2
dx
dx
d
+ x2
dx
d
+ 4
dx
.
So our differential equation can be written as
That is,
d
+ x2
dx
d
+ x2
dx
d
+ 4 [y] = 0
dx
dy
+ 4y
dx
= 0
.
.
(14.7)
Now consider
dy
+ 4y = 0
dx
.
This is a simple first-order linear and separable differential equation, whose general solution
is easily found to be y = c1 e−4x . In particular, e−4x is a solution. According to the above
theorem, e−4x is also a solution to our original differential equation. Let’s check to be sure:
d
+ x2
dx
=
d
+ x2
dx
=
=
=
d0
+ x2 · 0
dx
d −4x d 2 −4x e
+ 4 + x2
e
+ 4x 2 e−4x =
2
dx
dx
= 0
d
+ 4 e−4x
dx
d −4x e
+ 4e−4x
dx
d
+ x 2 −4e−4x + 4e−4x
dx
d
+ x 2 [0]
dx
.
Keep in mind, though, that e−4x is simply one of the possible solutions, and that there will be
solutions not given by c1 e−4x .
Unfortunately, unless it is of an exceptionally simple type (such as considered in chapter
17), factoring a linear differential operator is a very nontrivial problem. And even with those
simple types that we will be able to factor, we will find the main value of the above to be in
deriving even simpler methods for finding solutions. Consequently, in practice, you should not
expect to be solving many differential equations via “factoring”.
Additional Exercises
315
Additional Exercises
14.2 a. State the linear differential operator L corresponding to the left side of
d2 y
dy
+ 6y = 0
+ 5
dx
dx 2
.
b. Using this L , compute each of the following:
ii. L[e4x ]
i. L[sin(x)]
iv. L[x 2 ]
iii. L[e−3x ]
c. Based on the answers to the last part, what is one solution to the differential equation
in part a?
14.3 a. State the linear differential operator L corresponding to the left side of
d2 y
dy
− 5
+ 9y = 0
2
dx
dx
.
b. Using this L , compute each of the following:
ii. L[sin(3x)]
i. L[sin(x)]
iii. L[e2x ]
iv. L[e2x sin(x)]
14.4 a. State the linear differential operator L corresponding to the left side of
x2
dy
d2 y
+ 5x
+ 6y = 0
dx
dx 2
.
b. Using this L , compute each of the following:
ii. L[e4x ]
i. L[sin(x)]
iii. L[x 3 ]
14.5 a. State the linear differential operator L corresponding to the left side of
dy
d3 y
− sin(x)
+ cos(x) y = x 2 + 1
dx
dx 3
,
b. and then, using this L , compute each of the following:
ii. L[cos(x)]
i. L[sin(x)]
iii. L[x 2 ]
14.6. Several choices for linear differential operators L 1 and L 2 are given below. For each
choice, compute L 2 L 1 and L 1 L 2 .
d
+ x
dx
d
=
+ x2
dx
d
= x
+ 3
dx
a. L 1 =
b. L 1
c. L 1
d. L 1 =
d2
dx 2
and
and
d
− x
dx
d
=
+ x3
dx
d
=
+ 2x
dx
L2 =
and
L2
and
L2
L2 = x
316
Verifying the Big Theorems and an Introduction to Differential Operators
e. L 1 =
d2
dx 2
and
L2 = x3
f. L 1 =
d2
dx 2
and
L 2 = sin(x)
14.7. Compute the following composition products:
d
d
d
d
+2
+3
b. x
+2
x
+3
a.
dx
dx
c.
d
x
+4
dx
e.
1
d
+
dx
x
g.
d
+ x2
dx
dx
d
1
+
dx
x
d
+ 4x
dx
d2
d
+
dx 2
dx
dx
d.
d
+ 4x
dx
f.
d
+ 5x 2
dx
2
h.
d
d2
+
dx 2
dx
d
1
+
dx
x
d
+ x2
dx
14.8. Verify that
d2
d
+ [sin(x) − 3]
− 3 sin(x) =
2
dx
dx
d
+ sin(x)
dx
d
−3
dx
and, using this factorization, find one solution to
d2 y
dy
+ [sin(x) − 3]
− 3 sin(x)y = 0
2
dx
dx
.
14.9. Verify that
d
d2
+ x
+ 2 − 2x 2 =
2
dx
dx
d
−x
dx
d
+ 2x
dx
and, using this factorization, find one solution to
d2 y
dy
+ x
+ 2 − 2x 2 y = 0
2
dx
dx
14.10. Verify that
x2
d2
d
− 7x
+ 16 =
dx
dx 2
x
d
−4
dx
.
2
and, using this factorization, find one solution to
x2
d2 y
dy
− 7x
+ 16y = 0
dx
dx 2
.
,
,
,