Operators and Linear Ordinary Differential Equations (Part 1)

Introduction

Recently, while studying quantum mechanics, I inadvertently encountered a lot of knowledge regarding matrices (linear algebra), group theory, and related fields. I discovered many shared underlying ideas, primarily the use of operators to represent actions and reactions on functions. For instance, if we denote D = \frac{d}{dx}, then the derivative of a function f(x) can be viewed as the result of the operator D acting on it once. The second derivative is the result of the operator acting twice, and so on. Conversely, D^{-1} represents the inverse action of this operator, which restores the acted-upon function (the image) to the original function (the pre-image). Of course, this is not a simple division of the derivative operator, but rather an integration operation. Using this conceptual framework to solve linear differential equations offers a sense of unified and concise beauty.

Linear differential equations are the foundation for solving all differential equations. Generally, their forms are relatively simple, and in most cases, we can find their general solutions. In the Schrödinger equation of non-relativistic quantum mechanics, the essence lies in solving a second-order linear partial differential equation. On the other hand, in many nonlinear systems that we cannot solve directly, the linear solution serves as a first-order approximation, which is extremely important for qualitative analysis.

First-Order Linear Ordinary Differential Equations

This is the fundamental form for the integration of all subsequent differential equations, namely the solution to \frac{dy}{dx} + g(x)y = f(x). This is solved using the method of variation of constants, an idea consistent with the "perturbation method" in celestial mechanics. First, when the original differential equation cannot be solved directly, we ignore some small terms to find an approximate solution. That is, we first solve: \frac{dy}{dx} + g(x)y = 0

Direct integration yields y = C \exp(-\int g(x)dx), where C is a constant. Now, considering the solution to the original equation, we expect the original solution to also take the form: y = C \exp\left(-\int g(x)dx\right)

However, here C is a function. Substituting this into \frac{dy}{dx} + g(x)y = f(x), we get: \frac{dC}{dx} \exp\left(-\int g(x)dx\right) = f(x)

This is also solvable, yielding: C = \int f(x) \exp(\int g(x)dx)dx. Thus, the general solution to the original equation is: y = \left[\int f(x) \exp\left(\int g(x)dx\right)dx\right] \exp\left(-\int g(x)dx\right) In particular, when g(x) = \alpha = \text{constant}, it can be quickly derived that: y = \left[\int f(x) e^{\alpha x}dx\right] e^{-\alpha x}

Linear Operators

The operator D is defined as D = \frac{d}{dx}, representing the derivative of a function of one variable. The n-th derivative is denoted as D^n = \frac{d^n}{dx^n}, representing the operator D acting repeatedly on the same function. We also denote D^0 = 1 as the identity transformation. A known function or even a constant can be treated as an operator; its action on a function is equivalent to ordinary multiplication with that function. It can be proven that all derivative operators D^n, inverse derivative (integral) operators D^{-n}, and all non-zero differentiable functions constitute an operator group, which is a non-Abelian group.

The above terminology might be a bit professional, but it can be understood this way: in a differential equation, only the function being sought is the "real object," while other operations (including the four arithmetic operations, differentiation, etc.) are actions on the function. Solving a differential equation means using these actions and inverse actions to transform the function back to its original state. This is similar to the ideas in linear algebra; or rather, the content introduced in this article can itself be categorized within the scope of linear algebra. The term "non-Abelian group" means that the composition of these actions generally does not satisfy the commutative law, which will be discussed in the examples below.

Thus, the aforementioned ordinary differential equation can be expressed as: [D + g(x)]y = f(x)

This means that y, after being acted upon by the operator [D + g(x)], becomes f(x). The operator [D + g(x)] is composed of the sum of two basic operators (note that this is addition, not composition; composition represents multiplication). Now, to find y, the function before the action, we only need to apply the inverse action to f(x): y = [D + g(x)]^{-1} f(x)

According to (1), it is easy to obtain: ^{-1} f(x) = \left[\int f(x) \exp\left(\int g(x)dx\right)dx\right] \exp\left(-\int g(x)dx\right) For the constant coefficient case (D + \alpha)y = f(x), we have: y = (D + \alpha)^{-1} f(x) = \left[\int f(x) e^{\alpha x}dx\right] e^{-\alpha x} However, although general function multiplication and the derivative operator do not satisfy the commutative law, the derivative operator and constants are commutative. This point will play an essentially important role in solving constant coefficient linear differential equations below.

Constant Coefficient Linear Differential Equations

Next, we will study general linear ordinary differential equations: (D^n + a_1 D^{n-1} + a_2 D^{n-2} + \dots + a_n)y = f(x) In general differential equation theory, the cases f(x) = 0 and f(x) \neq 0 are discussed separately, and different forms arise when the characteristic equation has repeated roots. However, from the research below, we know they can be handled using the same method (or rather, they can be written in a unified form). First, we solve the characteristic equation: x^n + a_1 x^{n-1} + a_2 x^{n-2} + \dots + a_n = 0 to find the n roots r_1, r_2, \dots, r_n (counting multiplicities). Then (5) can be rewritten as: (D - r_1)(D - r_2)\dots(D - r_n)y = f(x) This form is much clearer. It is simply n operators acting sequentially on a function. Since we already know the inverse action of each operator from (4), we can "follow the same method" and apply the inverse actions of these n operators sequentially to f(x) to obtain the original function. That is: y = (D - r_n)^{-1} \dots (D - r_2)^{-1} (D - r_1)^{-1} f(x) According to formula (4), this can be organized as: y = e^{r_n x} \int \dots e^{r_2 x} \int [e^{-r_2 x} \cdot e^{r_1 x} \int f(x) e^{-r_1 x} dx] dx \dots dx Which can be abbreviated as: y = e^{r_n x} \int \dots \int e^{(r_2 - r_3)x} \int e^{(r_1 - r_2)x} \int f(x) e^{-r_1 x} dx^n Equation (9) is the general solution for constant coefficient differential equations. In this form, almost all cases are unified. At the same time, the practicality of solving using this form is very good; it is not just for theoretical analysis. For example, when f(x) = 0, equation (9) can be integrated directly (when characteristic roots are all distinct) to become: y = C_1 e^{r_1 x} + C_2 e^{r_2 x} + \dots + C_n e^{r_n x} When repeated roots occur, without loss of generality, let r_{n-1} = r_n. Then e^{(r_{n-1} - r_n)x} = 1, and it is easy to obtain: y = C_1 e^{r_1 x} + C_2 e^{r_2 x} + \dots + (C_{n-1}x + C_n) e^{r_n x} In the final integration step, the term C_{n-1}x + C_n appears. More general cases of repeated roots can be handled by analogy and will not be detailed here. It should be noted that the integration process for repeated roots is not necessarily simpler, but it is indeed effective and achieves unification.

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