ODE and Differential Operator Ring-module

Abstract

In the previous essay, "ODE: An Algebraic Approach (1)(2)(3)(4),

Math Essays: ODE, An Algebraic Approach (1) (wuyulanliulongblog.blogspot.com)

Math Essays: ODE An Algebraic Approach (2) (wuyulanliulongblog.blogspot.com)

Math Essays: ODE, An Algebraic Approach (3) (wuyulanliulongblog.blogspot.com)

Math Essays: ODE An Algebraic Approach (4) (wuyulanliulongblog.blogspot.com)

I focused on solving ODEs algebraically.

In this essay, I will concentrate on the algebraic structure we used.

An Isomorphism and a Module

Let's consider the algebra (both ring and module) $\mathbb{C}[X]$$\mathbb C[X]$.

Now, let's look at the algebra isomorphism $h:\mathbb{C}[X]\to \mathbb{C}[D]$$h:\mathbb C[X]\to\mathbb C[D]$, which is specifically a ring isomorphism.

Since $\mathbb{C}[X]$$\mathbb C[X]$ is commutative, $\mathbb{C}[D]$$\mathbb C[D]$ is also commutative.

Now, we can consider $\mathbb{C}[D]$$\mathbb C[D]$ acting on $C^{\mathrm{\infty }}$$C^{\infty}$ since $\mathbb{C}[D]$$\mathbb C[D]$ is a subring of ${\mathrm{End}}_{\mathrm{Ab}}(H(\mathrm{\Omega }))$$\mathrm{End}_{\mathrm{Ab}}(H(\Omega))$, where $H(\mathrm{\Omega })$$H(\Omega)$ is the space of holomorphic functions on an open set $\mathrm{\Omega }$$\Omega$.

This gives us a $\mathbb{C}[D]$$\mathbb C[D]$-module. More generally, we can consider $C^{\mathrm{\infty }}[D]$$C^{\infty}[D]$ acting on ${\mathbb{C}}^{\mathrm{\infty }}$$\mathbb C^{\infty}$.

Reviewing ODEs like $P(D)y=g$$P(D)y=g$, it is similar to considering $ra=b$$ra=b$ in an $R$$R$-module.

In ODE (1), we solve for $ra=0$$ra=0$, and in ODE (2), we consider $ra=b$$ra=b$ where $a={\overline{r}}^{-1}b+\text{Ker}(r)$$a=\bar{r}^{-1}b+\text{Ker}(r)$...

Corresponding between Quotient|Module and particular solution

If we consider $\mathbb{C}[D]$$\mathbb C[D]$ acting on ${\mathbb{P}}_n$$\mathbb P_n$, where ${\mathbb{P}}_n$$\mathbb P_n$ is the polynomial space, we can view ${\mathbb{P}}_n=H(\mathrm{\Omega })/(z^{n+1})$$\mathbb P_n=H(\Omega)/(z^{n+1})$ with $0\in \mathrm{\Omega }$$0\in\Omega$.

Then, we have a natural quotient and an interesting correspondence.

Let ${\pi }^n:H(\mathrm{\Omega })⟶H(\mathrm{\Omega })/(z^{n+1})=Ker(D^{n+1})$$\pi^n:H(\Omega)\longrightarrow H(\Omega)/(z^{n+1})=Ker(D^{n+1})$ be the quotient map.

It will deduce the quotient map ${\pi }_n:\mathbb{C}[D]⟶\mathbb{C}[D]/(D^{n+1})$$\pi_n:\mathbb C[D]\longrightarrow\mathbb{C}[D]/{(D^{n+1})}$ when $\mathbb{C}[D]$$\mathbb C[D]$ acti on $Ker(D^{n+1})$$Ker(D^{n+1})$.

One way to understand this is by considering the duality between ${\displaystyle \frac{z^n}{n!}}$$\displaystyle\frac{z^n}{n!}$ and $D^n{|}_0$$D^n|_0$.

In other words, considering $1,z,z^2,...,z^n,...$$1,z,z^2,...,z^n,...$ as the basis, we have $\begin{array}{c}{D}^j{|}_0\frac{z^i}{n!}=\{\begin{array}{l}1,i=j\\ 0,i\ne j\end{array}\end{array}$$D^{j}|_0\frac{z^{i}}{n!}=\begin{cases}1,i=j\\0,i\ne j\end{cases}$.

Thus, $D^j{|}_0$$D^j|_0$ forms the dual basis.

We can denote an $R$$R$-module as a pair $(R,M)$$(R,M)$,

therefore we have $({\pi }_n,{\pi }^n):(\mathbb{C}[D],H(\mathrm{\Omega }))⟶(\mathbb{C}[D]/(D^{n+1}),H(\mathrm{\Omega })/(z^{n+1}))$$(\pi_n,\pi^n):(\mathbb C[D],H(\Omega))\longrightarrow(\mathbb C[D]/(D^{n+1}),H(\Omega)/(z^{n+1}))$.

This module relate to the solution of $P(D)y=q(x),\mathrm{deg}q\le n,p_0\ne 0$$P(D)y=q(x),\deg q\le n,p_0\ne 0$.

Since the particular solution of $\begin{array}{c}P(D)y=q(x),y_p=\frac{1}{P(D)}q(x)\in (\mathbb{C}[D]/(D^{n+1}),H(\mathrm{\Omega })/(z^{n+1}))\end{array}$$P(D)y=q(x),y_p=\frac{1}{P(D)}q(x)\in (\mathbb C[D]/(D^{n+1}),H(\Omega)/(z^{n+1}))\\$

Using the identity $D-\lambda =e^{\lambda x}\circ D\circ e^{-\lambda x}$$D-\lambda=e^{\lambda x}\circ D\circ e^{-\lambda x}$, we get:

${\pi }^n:e^{\lambda z}H(\mathrm{\Omega })⟶e^{\lambda z}H(\mathrm{\Omega })/(e^{\lambda x}{z}^{n+1})$$\pi^n:e^{\lambda z}H(\Omega)\longrightarrow e^{\lambda z}H(\Omega)/(e^{\lambda x}z^{n+1})$

${\pi }_n:\mathbb{C}[D]⟶\mathbb{C}[D]/((D-\lambda {)}^{n+1})$$\pi_n:\mathbb C[D]\longrightarrow\mathbb{C}[D]/{((D-\lambda)^{n+1})}$

Module $(\mathbb{C}[D]/((D-\lambda {)}^{n+1},e^{\lambda z}H(\mathrm{\Omega })/(e^{\lambda x}{z}^{n+1}))$$(\mathbb{C}[D]/{((D-\lambda)^{n+1},e^{\lambda z}H(\Omega)/(e^{\lambda x}z^{n+1}))}$

relate to the particular solution of $P(D-\lambda )y=e^{\lambda x}q(x),\mathrm{deg}q\le n,p_0\ne 0$$P(D-\lambda)y=e^{\lambda x}q(x),\deg q\le n,p_0\ne 0$

Similarly, $\begin{array}{c}\frac{1}{P(D-\lambda )}{e}^{\lambda x}q(x)=(a_0+...+a_n(D-\lambda {)}^n)e^{\lambda x}q(x)\in (\mathbb{C}[D]/((D-\lambda {)}^{n+1},e^{\lambda z}H(\mathrm{\Omega })/(e^{\lambda x}{z}^{n+1}))\end{array}$$\frac{1}{P(D-\lambda)}e^{\lambda x}q(x)=(a_0+...+a_n(D-\lambda)^n)e^{\lambda x}q(x)\in (\mathbb{C}[D]/{((D-\lambda)^{n+1},e^{\lambda z}H(\Omega)/(e^{\lambda x}z^{n+1}))}\\$

Something really interesting

An intriguing discovery is that $C{e}^{ix}=Ker(D^2+1)$$Ce^{ix}=Ker(D^2+1)$.

Thus, $A\cos (x)+B\sin (x)=Ker(D^2+1)\subset C^{\mathrm{\infty }}(\mathbb{R})$$A\cos(x)+B\sin(x)=Ker(D^2+1)\subset C^{\infty}(\mathbb R)$.

If we consider the space $Ker(D^2+1)=\mathrm{Span}(\cos (x),\sin (x))$$Ker(D^2+1)=\mathrm{Span}(\cos(x),\sin(x))$.

Let $\mathbb{R}[D]$$\mathbb R[D]$ act on $Ker(D^2+1)$$Ker(D^2+1)$, we deduce $\pi :\mathbb{R}[D]⟶\mathbb{R}[D]/(D^2+1)\cong \mathbb{C}$$\pi:\mathbb R[D]\longrightarrow\mathbb R[D]/(D^2+1)\cong\mathbb C$.

But if we consider $C{e}^{ix}$$Ce^{ix}$,or $C{e}^{x+iy}$$Ce^{x+iy}$ Let $\mathbb{R}[D]$$\mathbb R[D]$ or $$\begin{gathered} \begin{array}{c}\mathbb{R}\left[\frac{\partial }{\partial y \\ }\right]\end{array} \end{gathered}$$$\mathbb R\left[\frac{\partial}{\partial y\\}\right]$ act on it, we get $\begin{array}{c}\frac{\mathbb{R}[D]}{(D^2+1)}\end{array}$$\frac{\mathbb R[D]}{(D^2+1)}\\$ or $$\begin{gathered} \begin{array}{c}\mathbb{R}\left[\frac{\partial }{\partial y \\ }\right]/({\left(\frac{\partial }{\partial y}\right)}^2+1)\end{array} \end{gathered}$$$\mathbb R\left[\frac{\partial}{\partial y\\}\right]/\left(\left(\frac{\partial}{\partial y}\right)^2+1\right)$

Now we see that $(a+bi)(\cos x+i\sin x)=(a+bD)(\cos x+i\sin x)$$(a+bi)(\cos x+i\sin x)=(a+bD)(\cos x+i\sin x)$

or $\begin{array}{c}(a+bi)(e^x(\cos y+i\sin y))=(a+b\frac{\partial }{\partial y})(e^x(\cos y+i\sin y))\end{array}$$(a+bi)(e^x(\cos y+i\sin y))=\left(a+b\frac{\partial}{\partial y}\right)(e^x(\cos y+i\sin y))\\$

There are no difference for $\mathbb{C}$$\mathbb C$ act on it or $\begin{array}{c}\frac{\mathbb{R}[D]}{(D^2+1)}\end{array}$$\frac{\mathbb R[D]}{(D^2+1)}\\$ $$\begin{gathered} \begin{array}{c}/\mathbb{R}\left[\frac{\partial }{\partial y \\ }\right]/({\left(\frac{\partial }{\partial y}\right)}^2+1)\end{array} \end{gathered}$$$\big/\mathbb R\left[\frac{\partial}{\partial y\\}\right]/\left(\left(\frac{\partial}{\partial y}\right)^2+1\right)$ act on it !

Remember how to find the particular solution of $P(D)y=e^{\lambda x}$$P(D)y=e^{\lambda x}$ ? $P(D)e^{\lambda x}=P(\lambda )e^{\lambda x}$$P(D)e^{\lambda x}=P(\lambda)e^{\lambda x}$

Consider $A\cosh x+B\sinh (x)=Ker(D^2-1)$$A\cosh x+B\sinh(x)=Ker(D^2-1)$

Let $\mathbb{R}[D]$$\mathbb R[D]$ act on $Ker(D^2-1)$$Ker(D^2-1)$, we deduce $\pi :\mathbb{R}[D]⟶\mathbb{R}[D]/(D^2-1)\cong \mathbb{H}$$\pi:\mathbb R[D]\longrightarrow\mathbb R[D]/(D^2-1)\cong\mathbb H$

Similarly, we can consider $C{e}^{jx}=C(\cosh x+j\sinh x)$$Ce^{jx}=C(\cosh x+j\sinh x)$, Let $\mathbb{R}[D]$$\mathbb R[D]$ act on it, we deduce that $\mathbb{R}[D]/(D^2-1)$$\mathbb R[D]/(D^2-1)$

We have$(a+bj)(\cosh x+j\sinh x)=(a+bD)(\cosh x+j\sinh x)$$(a+bj)(\cosh x+j\sinh x)=(a+bD)(\cosh x+j\sinh x)$ as well.

If we consider $\mathbb{Q}[D]$$\mathbb Q[D]$, let it act on $e^{\sqrt{d}x}$$e^{\sqrt {d}x}$,d is a square free natural number. (it is so natural, from $e^{ix},e^{jx},e^{\sqrt{d}x}...$$e^{ix},e^{jx},e^{\sqrt {d}x}...$)

$C{e}^{\sqrt{d}x}=Ker(D^2-d)$$Ce^{\sqrt{d}x}=Ker(D^2-d)$, we dedeuce $\begin{array}{c}\frac{\mathbb{Q}[D]}{(D^2-d)}\cong \mathbb{Q}[\sqrt{d}]\end{array}$$\frac{\mathbb Q[D]}{(D^2-d)}\cong\mathbb Q[\sqrt d]\\$, and $(a+b\sqrt{d})e^{\sqrt{d}x}=(a+bD)e^{\sqrt{d}x}$$(a+b\sqrt d)e^{\sqrt{d}x}=(a+bD)e^{\sqrt{d}x}$

And a convenient way to count P(D)y, y belong to KerH(D)

If you need count sth like $P(D)y,y\in KerH(D)$$P(D)y,y\in KerH(D)$, then we can consider $P(D)=Q(D)H(D)+R(D)$$P(D)=Q(D)H(D)+R(D)$

It just consider this ring $\mathbb{R}[D]/(H(D)),$$\mathbb R[D]/(H(D)),$ $P(D)\equiv R(D)\mathrm{mod}H(D)$$P(D)\equiv R(D)\mod H(D)$

Then $P(D)y=R(D)y$$P(D)y=R(D)y$, that is what we do in ODE (4)!

Why we can consider Euclidean division? Since $\mathbb{R}[D],\mathbb{C}[D]$$\mathbb R[D],\mathbb C[D]$ is Euclidean Domain !

And we already know what is the kernel of $H(D)$$H(D)$ in ODE(1)

For example, if you want to count$(D^3+D^2+D)(3\cos x+4\sin x)$$(D^3+D^2+D)(3\cos x+4\sin x)$

Actually you can stop at here $D^3+D^2+D=D(D^2+1)+D^2$$D^3+D^2+D=D(D^2+1)+D^2$

$D^3+D^2+D\equiv D^2\mathrm{mod}{D}^2+1$$D^3+D^2+D\equiv D^2\mod D^2+1$

since we already know that $D^2(3\cos x+4\sin x)=-(3\cos x+4\sin x)$$D^2(3\cos x+4\sin x)=-(3\cos x+4\sin x)$

But you can continue to do that, $(D^3+D^2+D)=(D+1)(D^2+1)-1$$(D^3+D^2+D)=(D+1)(D^2+1)-1$

Thus $(D^3+D^2+D)(3\cos x+4\sin x)=-(3\cos x+4\sin x)$$(D^3+D^2+D)(3\cos x+4\sin x)=-(3\cos x+4\sin x)$

Is that smart? In partuclualr, I mean in this example, it is stupid.

Why? We can substitute $i$$i$ in $D^3+D^2+D$$D^3+D^2+D$ directly! $i^3+i^2+i=-1$$i^3+i^2+i=-1$...

If $P(i)$$P(i)$ is not real number, substute inversely, $i\mapsto D$$i\mapsto D$ back.

Another example is, if you want to count $(D^3+2D^2+3D)e^{x}x$$(D^3+2D^2+3D)e^xx$,

since $D^3+2D^2+3D=(D+4)(D^2-2D+1)+10D-4$$D^3+2D^2+3D=(D+4)(D^2-2D+1)+10D-4$

Thus $(D^3+2D^2+3D)e^{x}x=(10D-4)e^{x}x=(10(D-1)+6)e^{x}x=10e^x+6e^{x}x$$(D^3+2D^2+3D)e^xx=(10D-4)e^xx=(10(D-1)+6)e^x x=10e^x+6e^x x$