ODE An Algebraic Approach (2)

In this essay, I try to discuss the basic idea of solving ODE Algebraically.

$\begin{array}{c}D:=\frac{d}{dx}\end{array}$$D:=\frac{d}{dx}\\$ is a linear map; according to the first isomorphism theory, $\begin{array}{c}\frac{C^1(\mathrm{\Omega })}{\mathbb{R}}\cong \mathrm{Im}D\end{array}$$\frac{C^1(\Omega)}{\mathbb R}\cong \mathrm{Im} D\\$

The isomorphism is $\bar{D}$$\overline D$ and $\begin{array}{c}{\bar{D}}^{-1},\text{that is,}\int \end{array}$$\overline D^{-1},\text{that is,}\int\\$

In general, let $\mathrm{\Delta }$$\Delta$ is a linear map, the domain is $C^k(\mathrm{\Omega })$$C^k(\Omega)$ , then $\begin{array}{c}\frac{C^k(\mathrm{\Omega })}{Ker\mathrm{\Delta }}\cong \mathrm{Im}\mathrm{\Delta }\end{array}$$\frac{C^k(\Omega)}{Ker\Delta}\cong \mathrm{Im}\Delta\\$

Linear ODE is $\mathrm{\Delta }:=P(D)\in C(\mathrm{\Omega })[D]$$\Delta:=P(D)\in C(\Omega)[D]$ ,

There is already a general solution for any homogeneous ordinary differential equation with constant coefficients,

$(D^n+a_{n-1}{D}^{n-1}+a_{n-2}{D}^{n-2}+a_{n-3}{D}^{n-3}+...+a_2{D}^2+a_{1}D)y=0$$(D^n+a_{n-1} D^{n-1}+a_{n-2} D^{n-2}+a_{n-3} D^{n-3}+...+a_{2} D^{2}+a_1D)y=0$

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

I use $e^{\lambda x}(f):=e^{\lambda x}\cdot f$$e^{\lambda x}(f):=e^{\lambda x}\cdot f$ is a linear automorphism, to pull $Ker(D-\lambda {)}^k$$Ker (D-\lambda)^k$ back to $Ker{D}^n$$KerD^n$

That is a good method, because $D\circ e^{\lambda x}(f)=D(e^{\lambda x}\cdot f)=({\partial }_1+{\partial }_2)(e^{\lambda x}\cdot f)=(\lambda +{\partial }_2)(e^{\lambda x}\cdot f)$$D\circ e^{\lambda x}(f)=D(e^{\lambda x}\cdot f)=(\part_1+\part_2)(e^{\lambda x}\cdot f)=(\lambda+\partial_2)(e^{\lambda x}\cdot f)$

Then $(D-\lambda )(e^{\lambda x}\cdot f)={\partial }_2(e^{\lambda x}\cdot f)=e^{\lambda x}\cdot D^{n}f$$(D-\lambda)(e^{\lambda x}\cdot f)=\part_2(e^{\lambda x}\cdot f)=e^{\lambda x}\cdot D^n f$

Thus $(D-\lambda {)}^k(e^{\lambda x}\cdot f)=0\text{ iff }{e}^{\lambda x}\cdot D^{k}f=0\text{ iff }{D}^{k}f=0$$(D-\lambda)^k(e^{\lambda x}\cdot f)=0\text{ iff }e^{\lambda x}\cdot D^kf=0\text{ iff } D^k f=0$

Now, I want to discuss some ODE like

$(D^n+a_{n-1}{D}^{n-1}+a_{n-2}{D}^{n-2}+a_{n-3}{D}^{n-3}+...+a_2{D}^2+a_{1}D)y=f$$(D^n+a_{n-1} D^{n-1}+a_{n-2} D^{n-2}+a_{n-3} D^{n-3}+...+a_{2} D^{2}+a_1D)y= f$

That is $P(D)y=f$$P(D)y=f$, and we need to find ${\overline{P(D)}}^{-1}$$\overline{P(D)}^{-1}$

We need to discuss the different behaviors when $P(D)$$P(D)$ acts on different functions

1.1 $P(D)e^{\lambda x}=P(\lambda )e^{\lambda x}$$P(D)e^{\lambda x}=P(\lambda)e^{\lambda x}$

Thus for $P(D)y=e^{ax}$$P(D)y=e^{ax}$, the solution is $\begin{array}{c}\frac{e^{ax}}{P(a)}+KerP(D)\end{array}$$\frac{e^{ax}}{P(a)}+Ker P(D)\\$

1.2

For $(D-1)y=P(x),\mathrm{deg}P=n$$(D-1)y= P(x),\deg P=n$, thus $D^{k}P(x)=0,\mathrm{\forall }k>n$$D^k P(x)=0,\forall k>n$

That is, when we limit $D$$D$ act on $P_n(x)$$P_n(x)$, $D$$D$ is nilpotent.

Recall that $(1-D{)}^{-1}=1+D+D^2+...+D^n+...$$(1-D)^{-1}=1+D+D^2+...+D^n+...$

Thus $y=(1-D{)}^{-1}P(x)=(1+D+D^2+...+D^n)(-P(x))$$y=(1-D)^{-1} P(x)=(1+D+D^2+...+D^n)(-P(x))$

In general, for $(\alpha -D)y=P(x),\mathrm{deg}P=n$$(\alpha-D) y=P(x),\deg P=n$

$\begin{array}{c}(\alpha -D)=\alpha (1-\frac{D}{\alpha })\end{array}$$(\alpha-D)=\alpha\left(1-\frac{D}{\alpha}\right)\\$, thus $\begin{array}{c}(\alpha -D{)}^{-1}=\frac{1}{\alpha }(1+\frac{D}{\alpha }+{\left(\frac{D}{\alpha }\right)}^2+...+{\left(\frac{D}{\alpha }\right)}^n+...\end{array}$$(\alpha - D)^{-1}=\frac{1}{\alpha}(1+\frac{D}{\alpha}+\left(\frac{D}{\alpha}\right)^2+...+\left(\frac{D}{\alpha}\right)^n+...\\$

$\begin{array}{c}y=(\frac{1}{\alpha }(1+\frac{D}{\alpha }+{\left(\frac{D}{\alpha }\right)}^2+...+{\left(\frac{D}{\alpha }\right)}^n)P(x)+Ker(\alpha -D)\end{array}$$y=\left(\frac{1}{\alpha}(1+\frac{D}{\alpha}+\left(\frac{D}{\alpha}\right)^2+...+\left(\frac{D}{\alpha}\right)^n\right)P(x)+Ker(\alpha -D)\\$

More generally, we know that $(D-\lambda )$$(D-\lambda)$ is nilpotent when it acts on space like $e^{\lambda x}P(x)$$e^{\lambda x}P(x)$

And we have for any $c\ne \lambda$$c\ne\lambda$

$\begin{array}{c}(c-D)=(c-\lambda -(D-\lambda ))=(c-\lambda )(1-\frac{D-\lambda }{c-\lambda })\end{array}$$(c-D)=(c-\lambda -(D-\lambda))=(c-\lambda)(1-\frac{D-\lambda}{c-\lambda})\\$

Example 1.2.1

$(1-D)y=x^3+x^2+x$$(1-D)y=x^3+x^2+x$

$y=(1+D+D^2+D^3)(x^3+x^2+x)$$y=(1+D+D^2+D^3)(x^3+x^2+x)$

$=x^3+x^2+x+3x^2+2x+1+6x+2+6=x^3+4x^2+9x+9$$=x^3+x^2+x+3x^2+2x+1+6x+2+6=x^3+4x^2+9x+9$

Let us check that

$(1-D)(x^3+4x^2+9x+9)=x^3+4x^2+9x+9-(3x^2+8x+9)=x^3+x^2+x$$(1-D)(x^3+4x^2+9x+9)=x^3+4x^2+9x+9-(3x^2+8x+9)=x^3+x^2+x$

It is true!

Example 1.2.2

$(3-D)y=e^x{x}^2$$(3-D)y=e^x x^2$

$(3-1-(D-1))=2(1-\frac{(D-1)}{2})$$(3-1-(D-1))=2\left(1-\frac{(D-1)}{2}\right)$

Thus $\begin{array}{c}{\left(2(1-\frac{(D-1)}{2})\right)}^{-1}=\frac{1}{2}(1+\frac{(D-1)}{2}+{\left(\frac{(D-1)}{2}\right)}^2)\end{array}$$\left(2\left(1-\frac{(D-1)}{2}\right)\right)^{-1}=\frac{1}{2}\left(1+\frac{(D-1)}{2}+\left(\frac{(D-1)}{2}\right)^2 \right)\\$

Thus $\begin{array}{c}y=\frac{1}{2}{e}^x{x}^2+\frac{1}{2}{e}^{x}x+\frac{1}{4}{e}^x\end{array}$$y=\frac{1}{2}e^xx^2+\frac{1}{2}e^{x}x+\frac{1}{4}e^x\\$

Check

$(2-(D-1))(\frac{1}{2}{e}^x{x}^2+\frac{1}{2}{e}^{x}x+\frac{1}{4}{e}^x)=e^x{x}^2+e^{x}x+\frac{1}{2}{e}^x-e^{x}x-\frac{1}{2}{e}^x=e^x{x}^2$$(2-(D-1))(\frac{1}{2}e^xx^2+\frac{1}{2}e^{x}x+\frac{1}{4}e^x)=e^x x^2+e^x x+\frac{1}{2} e^x-e^xx-\frac{1}{2}e^x=e^x x^2$

 

For $P(D)=(a-D)(b-D),((a-D)(b-D){)}^{-1}$$P(D)= {(a-D)(b-D)},((a-D)(b-D))^{-1}$

$\begin{array}{c}=\frac{1}{(a-b)}\frac{(a-D)-(b-D)}{(a-D)(b-D)}=\frac{1}{(a-b)}(\frac{1}{(b-D)}-\frac{1}{(a-D)})\end{array}$$=\frac{1}{(a-b)} \frac{(a-D)-(b-D)}{(a-D)(b-D)}=\frac{1}{(a-b)}\left(\frac{1}{(b-D)}-\frac{1}{(a-D)}\right)\\$

And according to

$\begin{array}{c}(\alpha -D{)}^{-1}=\frac{1}{\alpha }(1+\frac{D}{\alpha }+{\left(\frac{D}{\alpha }\right)}^2+...+{\left(\frac{D}{\alpha }\right)}^n+...\end{array}$$(\alpha - D)^{-1}=\frac{1}{\alpha}(1+\frac{D}{\alpha}+\left(\frac{D}{\alpha}\right)^2+...+\left(\frac{D}{\alpha}\right)^n+...\\$

$((a-D)(b-D){)}^{-1}$$((a-D)(b-D))^{-1}$

$\begin{array}{c}=\frac{1}{(a-b)}(\frac{1}{b}(1+\frac{D}{b}+...+{\left(\frac{D}{b}\right)}^n+...-\frac{1}{a}(1+\frac{D}{a}+...+{\left(\frac{D}{a}\right)}^n+...)\end{array}$$=\frac{1}{(a-b)}\left(\frac{1}{ b}(1+\frac{D}{ b}+...+\left(\frac{D}{b}\right)^n+...- \frac{1}{a}(1+\frac{D}{a}+ ...+\left(\frac{D}{a}\right)^n+...\right)\\$

That is $\begin{array}{c}\frac{1}{(a-b)}(\sum D^i(\frac{1}{b^{i+1}}-\frac{1}{a^{i+1}}))\end{array}$$\frac{1}{(a-b)}\left(\sum D^i\left(\frac{1}{b^{i+1}}-\frac{1}{a^{i+1}}\right)\right)\\$

Example 1.2.3

$(1-D)(2-D)y=x^2$$(1-D)(2-D) y=x^2$

$\begin{array}{c}y=(1-\frac{1}{2}+(1-\frac{1}{4})D+(1-\frac{1}{8})D^2)x^2=(\frac{1}{2}+\frac{3}{4}D+\frac{7}{8}{D}^2)x^2=\frac{1}{2}{x}^2+\frac{3}{2}x+\frac{7}{4}\end{array}$$y=(1-\frac{1}{2}+(1-\frac{1}{4}) D+(1-\frac{1}{8})D^2)x^2=\left(\frac{1}{2}+\frac{3}{4}D+\frac{7}{8} D^2\right)x^2=\frac{1}{2}x^2+\frac{3}{2}x+\frac{7}{4}\\$

Check

$(1-D)(2-D)=2-3D+D^2$$(1-D)(2-D)=2-3D+D^2$

Thus $\begin{array}{c}(2-3D+D^2)(\frac{1}{2}{x}^2+\frac{3}{2}x+\frac{7}{4})=x^2+3x+\frac{7}{2}-(3x+\frac{9}{2})+1=x^2\end{array}$$(2-3D+D^2)\left( \frac{1}{2}x^2+\frac{3}{2}x+\frac{7}{4}\right)=x^2+3x+\frac{7}{2}-(3x+\frac{9}{2})+1=x^2\\ $

In general, for solve $P(D)y=p_n(x)$$P(D) y=p_n(x)$, $P(D)=a_0+a_{1}D+a_2{D}^2+...+a_n{D}^k$$P(D)=a_0+a_1 D+a_2 D^2+...+a_n D^k$

We need to find ${\overline{P(D)}}^{-1}=b_0+b_{1}D+b_2{D}^2+....b_n{D}^m$$\overline{P(D)}^{-1}=b_0+b_1 D+b_2 D^2+....b_n D^m$

$P(D)(b_0+b_{1}D+b_2{D}^2+...+b_n{D}^m)=1$$P(D)(b_0+b_1D+b_2 D^2+...+b_n D^m)=1$

We can consider the formal power series $\begin{array}{c}\frac{1}{P(X)}=\sum _{i=0}^{\mathrm{\infty }}{a}_i(X-\lambda {)}^i\end{array}$$\frac{1}{P(X)}=\sum_{i=0}^\infty a_i (X-\lambda)^i\\$

For $a_0\ne 0$$a_0\ne 0$, we can always find a inverse Formal power series - Wikipedia

Denote $\begin{array}{c}f=\frac{1}{P(X)}\end{array}$$f=\frac{1}{P(X)}\\$

Each coefficient is $\begin{array}{c}\frac{f^n(\lambda )}{n!}\end{array}$$\frac{f^n(\lambda)}{n!}\\$,

$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}.$$\rm Proof.$

Consider $\mathbb{R}[[X]]$$\mathbb R[[X]]$ as a Linear space over $\mathbb{R}$$\mathbb R$

$\begin{array}{c}\frac{(D_{\lambda }{)}^n}{n!}\end{array}$$\frac{(D_\lambda)^n}{n!}\\$ is dual basis for $(X-\lambda {)}^n$$(X-\lambda)^n$

Thus for $P(D)y=P_n(x)$$P(D)y=P_n(x)$, $\begin{array}{c}{a}_0\ne 0,f(x)=\frac{1}{P_n(x)}\end{array}$$a_0\ne 0,f(x)=\frac{1}{P_n(x)}\\$, we have $\begin{array}{c}y=\sum _{i=1}^n\frac{f^i(\lambda )}{i!}{D}^i{P}_n(x)\end{array}$$y=\sum_{i=1}^n \frac{f^i(\lambda)}{i!} D^iP_n(x)\\$

Example 1.2.4

Consider $(D-1)(D-2{)}^2(D-3{)}^{3}y=x^2$$(D-1)(D-2)^2(D-3)^3 y= x^2$

$(D-1)(D-2{)}^2(D-3{)}^3=D^6-14D^5+80D^4-238D^3+387D^2-324D+108$$(D-1)(D-2)^2(D-3)^3=D^6 - 14D^5 + 80D^4 - 238D^3 + 387D^2 - 324D + 108$

According to Maple

$\begin{array}{c}y=(1/108+1/36\ast D+65/1296\ast D^2)x^2=\frac{x^2}{108}+\frac{x}{18}+\frac{65}{648}\end{array}$$y=(1/108 + 1/36*D + 65/1296*D^2)x^2=\frac{x^2}{108}+\frac{x}{18}+\frac{65}{648}\\ $

Check

Using Maple

$387/54-324\ast 2/108\ast x-324/18+x^2+108/18\ast x+108\ast 65/64=x^2$$387/54 - 324*2/108*x - 324/18 + x^2 + 108/18*x + 108*65/64=x^2$

Example 1.2.5

$(D^2-2D+3)y=e^{x}x$$(D^2-2D+3)y=e^x x$

$\begin{array}{c}y=\left(\frac{1}{2}\right)e^{x}x=\frac{1}{2}{e}^{x}x\end{array}$$y=\left(\frac{1}{2}\right) e^xx=\frac{1}{2}e^xx\\$

Check

$\begin{array}{c}\frac{1}{2}(e^{x}x+2e^x)-(e^{x}x+e^x)+\frac{3}{2}{e}^{x}x=e^{x}x\end{array}$$\frac12(e^x x+2e^x)-(e^xx+e^x)+\frac32e^xx=e^xx\\$

To be continued...