ODE, An Algebraic Approach (1)

(Decomposition Theorem).

Let $V$$V$ be a vector space over a field $K$$K$ and let $T$$T$ be a linear operator on $V$$V$ . If $a_1,a_2,...,a_m$$a_1, a_2, ..., a_m$ are distinct scalars and $k_1,k_2,...,k_m\in N$$k_1, k_2, ..., k_m ∈ N$

then

$Ker((T-a_1{)}^{k_1}(T-a_2{)}^{k_2}(T-a_3{)}^{k_3}...(T-a_m{)}^{k_m}$$Ker((T-a_1)^{k_1}(T-a_2)^{k_2}(T-a_3)^{k_3}...(T-a_m)^{k_m}$

$=Ker(T-a_1{)}^{k_1}\oplus Ker(T-a_2{)}^{k_2}\oplus Ker(T-a_3{)}^{k_3}\oplus ...\oplus Ker(T-a_m{)}^{k_m}$$=Ker(T-a_1)^{k_1}\oplus Ker(T-a_2)^{k_2}\oplus Ker(T-a_3)^{k_3}\oplus...\oplus Ker(T-a_m)^{k_m}$

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$

It can be viewed as $p(D)y=0$$p(D)y=0$

And $P(D)$$P(D)$ can be factored to $(D-{\lambda }_1{)}^{k_1}(D-{\lambda }_2{)}^{k_2}(D-{\lambda }_3{)}^{k_3}...(D-{\lambda }_m{)}^{k_m}$$(D-\lambda_1)^{k_1} (D-\lambda_2)^{k_2}(D-\lambda_3)^{k_3}...(D-\lambda_m)^{k_m}$ over $\mathbb{C}$$\mathbb C$

Therefore, to solve $(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$

We just need to solve $(D-\lambda {)}^{k}y=0$$(D-\lambda)^k y=0$

An interesting analogy is Math Essays: Dual basis and Taylor series (wuyulanliulongblog.blogspot.com)

We know that the kernel of $D^n$$D^n$ is Span$\begin{array}{c}\{1,x,\frac{x^2}{2!},\frac{x^3}{3!},...,\frac{x^{n-1}}{(n-1)!}\}\end{array}$$\left\{1,x,\frac{x^2}{2!},\frac{x^3}{3!},...,\frac{x^{n-1}}{(n-1)!}\right\}\\$

But what is the kernel of $(D-\lambda {)}^n$$(D-\lambda)^n$

When $k=1$$k=1$, it is simple; the solution is $e^{\lambda x}$$e^{\lambda x}$

So, we can guess that

$\begin{array}{c}Ker(D-\lambda {)}^n=e^{\lambda x}Span\{1,x,\frac{x^2}{2!},\frac{x^3}{3!},...,\frac{x^{n-1}}{(n-1)!}\}\end{array}$$Ker(D-\lambda)^n=e^{\lambda x}Span\left\{1,x,\frac{x^2}{2!},\frac{x^3}{3!},...,\frac{x^{n-1}}{(n-1)!}\right\}\\$

$\begin{array}{c}=Span\{e^{\lambda x},x{e}^{\lambda x},\frac{x^2}{2!}{e}^{\lambda x},\frac{x^3}{3!}{e}^{\lambda x},...,\frac{x^{n-1}}{(n-1)!}{e}^{\lambda x}\}\end{array}$$=Span\left\{e^{\lambda x},xe^{\lambda x},\frac{x^2}{2!}e^{\lambda x},\frac{x^3}{3!}e^{\lambda x},...,\frac{x^{n-1}}{(n-1)!}e^{\lambda x}\right\}\\$

But why?

One way to consider this is $e^{\lambda x}(f)$$e^{\lambda x}(f)$ is a linear automorphism, ($e^{\lambda x}(f)=e^{\lambda x}\cdot f$$e^{\lambda x}(f)=e^{\lambda x}\cdot f$) and $Ker(e^{\lambda x})=0$$Ker(e^{\lambda x})=0$

$D\circ e^{\lambda x}(f)=D(e^{\lambda x}f)=({\partial }_1+{\partial }_2)(e^{\lambda x}f)=(\lambda +{\partial }_2)(e^{\lambda x}f)$$D\circ e^{\lambda x}(f)=D(e^{\lambda x}f)=(\partial_1+\part_2) (e^{\lambda x}f)=(\lambda+\partial_2)(e^{\lambda x}f)$

Thus $(D-\lambda )\circ e^{\lambda x}(f)={\partial }_2(e^{\lambda x}f)=e^{\lambda x}{f}^{\prime }$$(D-\lambda)\circ e^{\lambda x}(f)=\part_2(e^{\lambda x} f)=e^{\lambda x} f'$

Thus $(D-\lambda {)}^k\circ e^{\lambda x}(f)={\partial }_2^k(e^{\lambda x}f)=e^{\lambda x}{D}^{k}f$$(D-\lambda)^k\circ e^{\lambda x}(f)=\partial_2^k (e^{\lambda x}f)=e^{\lambda x}D^kf$

Therefore $(D-\lambda {)}^k\circ e^{\lambda x}(f)=(D-\lambda {)}^k(e^{\lambda x}\cdot f)=0\iff e^{\lambda x}{D}^{k}f=0\iff D^{k}f=0$$(D-\lambda)^k\circ e^{\lambda x}(f)=(D-\lambda)^k(e^{\lambda x }\cdot f)=0\Leftrightarrow e^{\lambda x} D^k f=0\Leftrightarrow D^k f=0$

Thus the $\begin{array}{c}Ker(D-\lambda {)}^k=Span\{e^{\lambda x},x{e}^{\lambda x},\frac{x^2}{2!}{e}^{\lambda x},\frac{x^3}{3!}{e}^{\lambda x},...,\frac{x^{n-1}}{(n-1)!}{e}^{\lambda x}\}\end{array}$$Ker (D-\lambda)^k=Span\left\{e^{\lambda x},xe^{\lambda x},\frac{x^2}{2!}e^{\lambda x},\frac{x^3}{3!}e^{\lambda x},...,\frac{x^{n-1}}{(n-1)!}e^{\lambda x}\right\}\\$

In this case, $e^{\lambda x}$$e^{\lambda x}$ pull $Ker(D-\lambda {)}^k$$Ker (D-\lambda)^k$ back to $Ker{D}^n$$Ker D^n$.

The idea is just like if you try to find the Kernel of matrix $A$$A$, but it is hard.

However, $AB=0$$AB=0$ is easy, (B is invertible) thus you can consider $v\in Ker(AB)$$v\in Ker(AB)$, then $KerA=B(Ker(AB))$$Ker A= B(Ker(AB))$

Because $ABv=0$$ABv=0$ iff $A(Bv)=0$$A(Bv)=0$

By the way, the Gaussian elimination is similar, but it is through $BAv=0$$BAv=0$ to find $Av=0$$Av=0$

$B$$B$ is invertible, so $BAv=0$$BAv=0$ iff $Av=0$$Av=0$

But the most amazing thing is $(D-\lambda )=e^{\lambda x}\circ D\circ e^{-\lambda x},$$(D-\lambda)=e^{\lambda x}\circ D\circ e^{-\lambda x},$

And we will see that $D$$D$ act on ${\mathbb{P}}_n(x)\text{that is all the polynomias with }\mathrm{deg}p\le n$$\mathbb P_n(x)\text{that is all the polynomias with }\deg p\le n$

is nilpotent, thus we could find formal power series to find the inverse,

and because $D$$D$ is nilpotent, thus we have $(D-\lambda {)}^n=e^{\lambda x}\circ D^n\circ e^{-\lambda x}$$(D-\lambda)^n=e^{\lambda x}\circ D^n\circ e^{-\lambda x}$

Therefore $(D-\lambda )$$(D-\lambda)$is also nilpotent