Algebraic invariant of linear constant coefficient ODE

Introduction

In this article, we explore the concept of invariants in the context of a generalization of ordinary differential equations (ODEs), denoted as ODE(3). We will discuss invariants of particular solutions and general solutions, which provide valuable insights into the behavior of these equations.

Make sure you already ready my previous work before.

Math Essays: ODE and Differential Operator Ring-module (wuyulanliulongblog.blogspot.com)

Invariant of Particular Solutions

Consider a particular type of ODE in the form of $P(D)y=h$$P(D)y = h$, where $h\in \text{Ker}H(D)$$h \in \text{Ker}H(D)$.

Two ODEs $(P(D)y=h_1)$$(P(D)y=h_1)$ and $(Q(D)y=h_2)$$(Q(D)y=h_2)$ are considered equivalent, denoted as $(P(D)y=h_1)\equiv (Q(D)y=h_2)$$(P(D)y=h_1)\equiv (Q(D)y=h_2)$, if their particular solutions come from the same space.

For $P(D)y=h$$P(D)y=h$, the solution comes from $\text{Ker}(H(D)P(D))$$\text{Ker}(H(D)P(D))$. We can define the divisor set of $H(D)P(D)$$H(D)P(D)$ as $d_{HP}$$d_{HP}$, which corresponds to the poset $(d_{HP},|)$$(d_{HP},|)$.

Definition

$S(D)$$S(D)$ is the least monic polynomial in $(d_{HP},|)$$(d_{HP},|)$ for which there exists a $y_p\in \text{Ker}S(D)$$y_p \in \text{Ker} S(D)$ such that $P(D)y_p=h$$P(D)y_p = h$. We refer to $S(D)$$S(D)$ as the particular polynomial since the particular solution comes from $\text{Ker}S(D)$$\text{Ker} S(D)$.

Remark

The definition of $S(D)$$S(D)$ is based on a universal property, making it clearer to understand its construction.

Examples

1. For the ODE $(D-1{)}^{2}y=e^x$$(D-1)^2y=e^x$, we have $S(D)=H(D)P(D)=(D-1{)}^3$$S(D) = H(D)P(D) = (D-1)^3$.

2. For the ODE $(D-1)(D-2)y=e^x$$(D-1)(D-2)y=e^x$, we find $S(D)=(D-1{)}^2$$S(D)=(D-1)^2$.

Proposition

If $S_i(D)|S_j(D)$$S_i(D)|S_j(D)$, then $\text{Ker}{S}_i(D)\subseteq \text{Ker}{S}_j(D)$$\text{Ker}S_i(D)\subseteq \text{Ker}S_j(D)$. As a result, we can say that $(P(D)y=h)\equiv (Q(D)y=h)$$(P(D)y=h)\equiv (Q(D)y=h)$ if and only if their particular polynomials are the same, making $S(D)$$S(D)$ an invariant under the equivalence relation $\equiv$$\equiv$.

Example

Consider the following ODEs:

1. $(D-1{)}^{3}y=e^x\equiv (D-1{)}^{2}y=e^{x}x\equiv (D-1)y=e^x{x}^2\equiv y=e^x{x}^3$$(D-1)^3y=e^x \equiv (D-1)^2y=e^xx \equiv (D-1)y=e^xx^2 \equiv y=e^xx^3$

Invariant of General Solutions

In ODE(3), we transformed the non-homogeneous equation $P(D)y=h$$P(D)y=h$ into the homogeneous equation $P(D)H(D)y=0$$P(D)H(D)y=0$. Solving homogeneous equations is often easier. But why we can not do sth inversely?

Since if $P_i(D)H_i(D)=\lambda P_j(D)H_j(D)$$P_i(D)H_i(D)=\lambda P_j(D)H_j(D)$, where $\lambda \in \mathbb{C}$$\lambda\in\mathbb{C}$, then they have the same type of solution.

Example

$(D-1)(D-2)(D-3)y=e^x\equiv (D-2)(D-3)y=e^{x}x\equiv y=A{e}^{x}x+B{e}^{2x}+C{e}^{3x}$$(D-1)(D-2)(D-3)y=e^x\equiv(D-2)(D-3)y=e^x x\equiv y=Ae^xx+Be^{2x}+Ce^{3x}$.

Or trans it to $gcd=1$$\gcd=1$, such as

$(D-3{)}^2(D-2{)}^2(D-1)y=e^{x}x\equiv (D-3{)}^2(D-2{)}^{2}y=e^x{x}^2$$(D-3)^2(D-2)^2(D-1)y=e^xx\equiv (D-3)^2(D-2)^2y=e^xx^2$.

And you can use long divison to help you find the particular solution. $(D^2-6D+9)(D^2-4D+4)=D^4-10D^3+37D^2-60D+36$$(D^2-6D+9)(D^2-4D+4)=D^4-10D^3+37D^2-60D+36$(you can use matrix to count this product.) Math Essays: Using matrix to count the product of two polynomias (wuyulanliulongblog.blogspot.com)

And, by the way, it just like integrating both sides.

I mean, how do you solve $Dy=x?$$Dy=x?$ integrating both sides. And it just that the solution comes from $y=A{x}^2+Bx+C=Ker{D}^3$$y=Ax^2+Bx+C=Ker D^3$

It is really stupid to use this method when you deal with $Dy=f$$Dy=f$, But you can not integraing both sides for $(D-2)y=e^x$$(D-2)y=e^x$

But as you see, you can do this for this $\mathrm{O}\mathrm{D}\mathrm{E}$$\rm ODE$ $(D-2)y=e^x\equiv y=A{e}^x+B{e}^{2x}$$(D-2)y=e^x\equiv y=Ae^x+Be^{2x}$.

Conclusion

Invariants in ODE(3) provide a powerful tool to simplify and understand the behavior of differential equations. They allow us to transform challenging problems into easier ones and relate different solutions to each other. Understanding these invariants can lead to significant insights into the study of ordinary differential equations.