Linear Differential Operators

Writing an $n$ th order linear equation as $L [y] = f (x)$ is more than shorthand. It reveals that solving a differential equation is the same kind of problem as solving a matrix equation $A 𝐱 = 𝐛$ , because $L$ is a linear operator. Every structural result in this chapter is a consequence of that single property.

Quick Reference

Item	Statement
General linear equation of order $n$	$a_{0} (x) \frac{d^{n} y}{d x^{n}} + a_{1} (x) \frac{d^{n - 1} y}{d x^{n - 1}} + \dots + a_{n} (x) y = b (x)$
Standard form	$\frac{d^{n} y}{d x^{n}} + P_{1} (x) \frac{d^{n - 1} y}{d x^{n - 1}} + \dots + P_{n} (x) y = f (x)$
Conversion	$P_{i} = \frac{a_{i}}{a_{0}}$ and $f = \frac{b}{a_{0}}$ , valid where $a_{0} (x) \neq 0$
The operator	$L \equiv \frac{d^{n}}{d x^{n}} + P_{1} (x) \frac{d^{n - 1}}{d x^{n - 1}} + \dots + P_{n} (x)$
Compact equation	$L [y] = f (x)$
Domain of $L$	$C^{n}$ functions, those with $n$ continuous derivatives
Linearity	$L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}]$
Homogeneous case	$L [y] = 0$

Standard Form

A linear differential equation of order $n$ is an equation of the form

a_{0} (x) \frac{d^{n} y}{d x^{n}} + a_{1} (x) \frac{d^{n - 1} y}{d x^{n - 1}} + \dots + a_{n - 1} (x) \frac{d y}{d x} + a_{n} (x) y = b (x)

where $a_{0}, \dots, a_{n}$ and $b$ are functions of $x$ defined on an interval $I$ .

Suppose

a_{0} (x) \neq 0 for every x in I .

Then, dividing both sides by $a_{0} (x)$ , we get

\frac{d^{n}y}{dx^{n}} + P_{1}(x)\frac{d^{n-1}y}{dx^{n-1}} + \cdots + P_{n}(x)y = f(x) , \tag{1}

where

P_{i} (x) = \frac{a_{i} (x)}{a_{0} (x)}, (i = 1, \dots, n) and f (x) = \frac{b (x)}{a_{0} (x)} .

Equation (1) is called the standard form of a linear differential equation of order $n$ .

The hypothesis $a_{0} (x) \neq 0$ is essential and is not always satisfied. A point where $a_{0}$ vanishes is called a singular point. In x^{2}y'' + xy' + y = 0 the origin is singular, and the standard form's coefficients $1 / x$ and $1 / x^{2}$ blow up there. Solutions can behave strangely at such points, which is exactly the situation that the method of Frobenius in Chapter 6 was invented to handle.

The Differential Operator

We define $L$ , the linear differential operator of order $n$ , as

L \equiv \frac{d^{n}}{d x^{n}} + P_{1} (x) \frac{d^{n - 1}}{d x^{n - 1}} + \dots + P_{n} (x) .

According to this definition,

L [y] \equiv \frac{d^{n} y}{d x^{n}} + P_{1} (x) \frac{d^{n - 1} y}{d x^{n - 1}} + \dots + P_{n} (x) y .

In fact, $L$ is a function whose domain is the set of functions with $n$ continuous derivatives; such functions are called $C^{n}$ functions. The output $L [y]$ is another function of $x$ . So $L$ takes functions to functions, which is why it is called an operator rather than simply a function.

Using this operator, we can write the differential equation (1) as

L [y] = f (x) .

For example, if $L$ is a linear differential operator of order $2$ , then

L [y] = \frac{d^{2} y}{d x^{2}} + P_{1} (x) \frac{d y}{d x} + P_{2} (x) y

and $L [y] = f (x)$ is shorthand for a second-order linear differential equation.

Note the last term. The operator ends in $P_{n} (x)$ with no derivative attached, which should be read as $P_{n} (x) \frac{d^{0}}{d x^{0}}$ , that is, multiplication by $P_{n} (x)$ . Multiplication by a function is a perfectly respectable linear operator, and including it keeps the pattern uniform.

Why $L$ Is Called Linear

For any constants $c_{1}$ and $c_{2}$ and any $C^{n}$ functions $y_{1}$ and $y_{2}$ ,

L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}] .

Proof

The proof rests on two facts from calculus, applied term by term.

First, differentiation is additive: $\frac{d^{k}}{d x^{k}} (u + v) = \frac{d^{k} u}{d x^{k}} + \frac{d^{k} v}{d x^{k}}$ .

Second, constants pass through derivatives: $\frac{d^{k}}{d x^{k}} (c u) = c \frac{d^{k} u}{d x^{k}}$ .

Applying both to each of the $n + 1$ terms making up $L$ ,

\begin{aligned} L\left[c_1y_1 + c_2y_2\right] &= \sum_{k} P_k(x)\frac{d^{n-k}}{dx^{n-k}}\left(c_1y_1 + c_2y_2\right) \\ &= \sum_{k} P_k(x)\left(c_1\frac{d^{n-k}y_1}{dx^{n-k}} + c_2\frac{d^{n-k}y_2}{dx^{n-k}}\right) \\ &= c_1\sum_{k} P_k(x)\frac{d^{n-k}y_1}{dx^{n-k}} + c_2\sum_{k} P_k(x)\frac{d^{n-k}y_2}{dx^{n-k}} \\ &= c_1 L\left[y_1\right] + c_2 L\left[y_2\right] . \end{aligned}

Note that the coefficients $P_{k} (x)$ are not constants and cannot be pulled out of the derivatives; they simply multiply each term after the differentiation is done, which is all the argument requires.

The immediate payoff is the superposition principle.

Superposition Principle. If $y_{1}$ and $y_{2}$ are solutions of the homogeneous equation $L [y] = 0$ , then $c_{1} y_{1} + c_{2} y_{2}$ is also a solution, for any constants $c_{1}$ and $c_{2}$ .

Proof

By linearity,

L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}] = c_{1} \cdot 0 + c_{2} \cdot 0 = 0 .

In the language of linear algebra: the set of solutions of $L [y] = 0$ is closed under addition and scalar multiplication, so it is a vector space, specifically the kernel of $L$ . Section 3.3 shows that this space has dimension exactly $n$ .

Worked Examples

Write x^{2}y'' - 3xy' + 4y = x^{3} in standard form and identify the operator $L$ . Where is the standard form valid?

Solution

Here $a_{0} (x) = x^{2}$ , which vanishes at $x = 0$ . Dividing by $x^{2}$ ,

y'' - \frac{3}{x}y' + \frac{4}{x^{2}}y = x .

So $P_{1} (x) = - \frac{3}{x}$ , $P_{2} (x) = \frac{4}{x^{2}}$ , and $f (x) = x$ , with

L = \frac{d^{2}}{d x^{2}} - \frac{3}{x} \frac{d}{d x} + \frac{4}{x^{2}} .

The standard form is valid on any interval not containing the origin, so on $x > 0$ or on $x < 0$ , but not across $x = 0$ .

Let $L = \frac{d^{2}}{d x^{2}} + 4$ . Compute $L [\sin 2 x]$ , $L [e^{x}]$ , and $L [3 \sin 2 x - 5 e^{x}]$ .

Solution

L [\sin 2 x] = - 4 \sin 2 x + 4 \sin 2 x = 0,

so $\sin 2 x$ is a solution of the homogeneous equation.

L [e^{x}] = e^{x} + 4 e^{x} = 5 e^{x} .

By linearity, without any new differentiation,

L [3 \sin 2 x - 5 e^{x}] = 3 L [\sin 2 x] - 5 L [e^{x}] = 3 \cdot 0 - 5 \cdot 5 e^{x} = - 25 e^{x} .

Note the labor saved: linearity let us assemble the answer from results already computed. This is exactly how the method of undetermined coefficients works in Chapter 4.

Show that the operator N[y] = y'' + y^{2} is not linear.

Solution

Take $y_{1} = y_{2} = 1$ , so $N [1] = 0 + 1 = 1$ . Then

N [1 + 1] = N [2] = 0 + 4 = 4,

while

N [1] + N [1] = 1 + 1 = 2 .

Since $4 \neq 2$ , the additivity property fails and $N$ is not linear.

This is why nonlinear equations have no superposition principle, and why adding two solutions of a nonlinear equation almost never produces a third.

Exercises

Write \left(1-x^{2}\right)y'' - 2xy' + 6y = 0 in standard form and state where it is valid.

Solution

Dividing by $1 - x^{2}$ ,

y'' - \frac{2x}{1-x^{2}}y' + \frac{6}{1-x^{2}}y = 0 .

The coefficient $a_{0} = 1 - x^{2}$ vanishes at $x = \pm 1$ , so the standard form is valid on $(- \infty, - 1)$ , on $(- 1, 1)$ , or on $(1, \infty)$ , but not across the endpoints. This is Legendre's equation with parameter $2$ , and the interval $(- 1, 1)$ is the one of interest in applications.

Let $L = \frac{d^{2}}{d x^{2}} - \frac{d}{d x} - 6$ . Verify that $e^{3 x}$ and $e^{- 2 x}$ are solutions of $L [y] = 0$ , and use superposition to write down infinitely many more.

Solution

For $y = e^{3 x}$ : $L [y] = 9 e^{3 x} - 3 e^{3 x} - 6 e^{3 x} = 0$ . ✓

For $y = e^{- 2 x}$ : $L [y] = 4 e^{- 2 x} + 2 e^{- 2 x} - 6 e^{- 2 x} = 0$ . ✓

By superposition, $y = c_{1} e^{3 x} + c_{2} e^{- 2 x}$ solves $L [y] = 0$ for every choice of $c_{1}$ and $c_{2}$ . Section 3.3 shows this two-parameter family is in fact all the solutions.

Suppose $L [y_{1}] = f_{1} (x)$ and $L [y_{2}] = f_{2} (x)$ . What equation does $y_{1} + y_{2}$ satisfy?

Solution

By linearity, $L [y_{1} + y_{2}] = L [y_{1}] + L [y_{2}] = f_{1} (x) + f_{2} (x)$ .

So $y_{1} + y_{2}$ solves the equation with the combined forcing function. This is the superposition principle for nonhomogeneous equations, and it is genuinely useful: to solve $L [y] = e^{x} + \sin x$ , solve $L [y] = e^{x}$ and $L [y] = \sin x$ separately and add the answers.

Frequently Asked Questions

What is a linear differential operator?

It is the rule $L$ that takes a function $y$ and returns the left-hand side of the differential equation, $y^{(n)} + P_{1} y^{(n - 1)} + \dots + P_{n} y$ . Writing the equation as $L [y] = f$ separates the machinery (the operator) from the data (the forcing function), which is what makes the general theory possible.

Why must

a_{0} (x)

be nonzero?

Because the standard form is obtained by dividing by $a_{0}$ . Where $a_{0}$ vanishes, the leading coefficient of the equation disappears and the equation effectively drops in order at that point. Such singular points require special treatment, and they are exactly where power series methods become necessary.

What does

C^{n}

mean?

It is the set of functions having $n$ continuous derivatives. This is the natural domain for an operator of order $n$ , since applying $L$ requires differentiating $n$ times and the result should be a continuous function. Polynomials, exponentials, sines, and cosines all belong to $C^{n}$ for every $n$ .

Does linearity require the coefficients

P_{i}

to be constants?

No. Linearity is a statement about how $L$ treats $y$ , not about the coefficients. The operator $L = \frac{d^{2}}{d x^{2}} + \sin (x) \frac{d}{d x} + e^{x^{2}}$ is perfectly linear. Constant coefficients matter for solving the equation, not for the theory in this chapter.

What is the difference between

L [y] = 0

and

L [y] = f

The first is homogeneous, the second nonhomogeneous. Their solution sets differ in an important way: the homogeneous solutions form a vector space closed under addition, while the nonhomogeneous solutions do not (adding two of them gives $L [y] = 2 f$ ). The relationship between the two is the subject of Section 3.4.

Quick Reference

Standard Form

The Differential Operator

Why $L$ Is Called Linear

Worked Examples

Exercises

Frequently Asked Questions

Quick Links

Social Media

Linear Differential Operators

Quick Reference

Standard Form

The Differential Operator

Why L Is Called Linear

Worked Examples

Exercises

Frequently Asked Questions

Quick Links

Social Media

Why $L$ Is Called Linear