Operators and Linear Ordinary Differential Equations (Part 1)

算子与线性常微分方程(上) 苏剑林 · 2012-11-30 · Mathematics differential equation linear operator

Introduction

While studying quantum mechanics recently, I found myself running into a lot of matrix (linear algebra) and group theory concepts, and noticed that they share quite a few common ideas — chief among them being the use of operators to represent an action on a function and its inverse action. For instance, we can write $D=\frac{d}{dx}$, so that the derivative of the function $f(x)$ can be regarded as the result of applying the operator D to it once, the second derivative as applying it twice, and so on. Conversely, $D^{-1}$ represents the inverse action of this operator: it restores the function after the action (the image) back to the original function (the pre-image). Of course, this is not simply "dividing out" the derivative operator, but rather an integration operation. Using this way of thinking to solve linear differential equations gives a certain unified and elegant beauty.

Linear differential equations are the foundation for solving all differential equations. Generally speaking, their form is fairly simple, and in most cases we can find their general solution explicitly. In non-relativistic quantum mechanics, the Schrödinger equation is essentially a second-order linear partial differential equation to be solved. On the other hand, in many nonlinear systems that we cannot solve exactly, the linear solution, as a first-order approximation, is extremely important for qualitative analysis.

First-order linear ordinary differential equations

This is the basic form underlying all the integrations that follow, namely solving $\frac{dy}{dx}+g(x)y=f(x)$. This is done via the method of variation of constants, whose idea is the same as the "perturbation method" in celestial mechanics: when we cannot solve the original differential equation directly, we first drop some of the small terms to obtain an approximate solution. That is, we first solve

$$\frac{dy}{dx}+g(x)y=0$$more

Direct integration gives $y=C exp(-\int g(x)dx)$, where C is a constant. Now, to find the solution of the original equation, we expect the solution to take the same form

$$y=C \exp(-\int g(x)dx)$$

but here C is a function. Substituting into $\frac{dy}{dx}+g(x)y=f(x)$, we get

$$\frac{dC}{dx} \exp(-\int g(x)dx)=f(x)$$

which can also be solved, giving

$C=\int f(x) exp(\int g(x)dx)dx$, so the general solution of the original equation is

$$y=[\int f(x) \exp(\int g(x)dx)dx]\exp(-\int g(x)dx)\tag{1}$$

In particular, when $g(x)=\alpha=Const.$, we can quickly obtain

$$y=[\int f(x) e^{\alpha x}dx]e^{-\alpha x}\tag{2}$$

Linear operators

The operator D is defined as $D=\frac{d}{dx}$, i.e., differentiation of a function of one variable, and the n-th derivative is written as $D^n=\frac{d^n}{dx^n}$, i.e., the operator D applied repeatedly to the same function. We also write $D^0=1$ for the identity transformation. A known function, or even a constant, can also be regarded as an operator: acting on some function, it is equivalent to ordinary multiplication by that function. It can be shown that all the differentiation operators $D^n$ and the antiderivative (integral) operators $D^{-n}$, together with all nonzero differentiable functions, form an operator group — and this group is non-Abelian.

The above terminology may sound a bit too technical, so let's put it this way: in a differential equation, only the function we are looking for is the "real" object; all other operations (including the four arithmetic operations, differentiation, etc.) are actions performed on that function. Solving a differential equation means using these actions and their inverse actions to bring the function back to its original image. This is analogous to the ideas of linear algebra — indeed, one could say that the content introduced in this article essentially falls under the umbrella of linear algebra. As for what "non-Abelian" means here: in general, the composition of these actions does not satisfy commutativity, which we'll see in the examples below.

With this, the above ordinary differential equation can be written as

$$[D+g(x)]y=f(x)$$

This means that y, after being acted on by the operator $[D+g(x)]$, becomes $f(x)$, and the operator $[D+g(x)]$ is formed by adding two basic operators (note: not by superposition/composition, which would correspond to multiplication). Now, to recover y — i.e., the function before the action was applied — we just need to apply the inverse action to f(x). That is,

$$y=[D+g(x)]^{-1} f(x)$$

From (1), we readily obtain

$$[D+g(x)]^{-1} f(x)=[\int f(x) \exp(\int g(x)dx)dx]\exp(-\int g(x)dx)\tag{3}$$

For a constant coefficient $(D+\alpha)y=f(x)$, we have

$$y=(D+\alpha)^{-1} f(x)=[\int f(x) e^{\alpha x}dx]e^{-\alpha x}\tag{4}$$

Although ordinary function multiplication and the differentiation operator do not in general commute, the differentiation operator does commute with a constant — a fact that will play a fundamentally important role below when solving constant-coefficient linear differential equations.

Constant-coefficient linear differential equations

We now turn to the general linear constant-coefficient ordinary differential equation

$$(D^n+a_1 D^{n-1}+a_2 D^{n-2}+...+a_n)y=f(x)\tag{5}$$

In the standard theory of differential equations, the cases f(x)=0 and f(x)≠0 are usually treated separately, and repeated roots of the characteristic equation are handled with yet another special form. However, as we'll see below, all of these cases can actually be handled with the same method (or, put another way, written in a unified form). First we solve the characteristic equation

$$x^n+a_1 x^{n-1}+a_2 x^{n-2}+...+a_n=0$$

to find its n roots $r_1,r_2,...,r_n$ (repeated roots counted with multiplicity), after which (5) can be rewritten as

$$(D-r_1 )(D-r_2 )...(D-r_n)y=f(x)\tag{6}$$

This form is much nicer to look at. Note that this is nothing more than n operators applied successively to a function, and we already know the inverse action of each individual operator — it's exactly the form given in (4). So we can "apply the same trick again": stacking up the inverse actions of these n operators and applying them successively to $f(x)$, we recover the original function. That is,

$$y=(D-r_n)^{-1}...(D-r_2)^{-1} (D-r_1)^{-1} f(x) \tag{7}$$

Using equation (4), this can be rearranged to give:

$$y=e^{r_n x}\int {...e^{r_2 x}\int[ e^{-r_2 x}\cdot e^{r_1 x} \int f(x)e^{-r_1 x}dx]dx}dx\tag{8}$$

which can be written more compactly as

$$y=e^{r_n x}\int...\int e^{(r_2-r_3 )x}\int e^{(r_1-r_2 )x}\int f(x)e^{-r_1 x} dx^n\tag{9}$$

Equation (9) is the general solution of the constant-coefficient differential equation. In this form, almost all cases are unified, and this form is also quite practical for actually solving equations, not merely useful for theoretical analysis. For instance, when f(x)=0, equation (9) can be integrated directly to give (assuming the characteristic roots are all distinct)

$$y=C_1 e^{r_1 x}+C_2 e^{r_2 x}+...+C_n e^{r_n x} \tag{10}$$

When there are repeated roots, without loss of generality suppose $r_{n-1}=r_n$, so that $e^{(r_{n-1}-r_n)x}=1$; then we readily obtain

$$y=C_1 e^{r_1 x}+C_2 e^{r_2 x}+...+(C_{n-1}x+C_n) e^{r_n x}\tag{11}$$

In the last integration step, a term of the form $C_{n-1} x+C_n$ appears. More general cases of repeated roots can be handled analogously, and we won't spell them out here. It should be noted that the integration process for handling repeated roots is not necessarily any simpler than the standard approach, but it is indeed valid, and it does achieve unification.

It's natural to wonder whether this method can be extended to solving variable-coefficient differential equations — but doing so runs into a fatal flaw related to the non-Abelian property mentioned above, which we'll discuss in the next section.