Solutions of homogeneous linear differential equations are a special case of structure theorem for f.g. modules over a PID

In this other post, Qiaochu Yuan comments that the solutions for the homegeneous linear differential equation with constant coefficients are a special case of the structure theorem for finitely generated modules over a PID.

He states:

the space of solutions to a homogeneous linear differential equation with constant coefficients has a $k[D]$-module structure, where $D$ is differentiation, and moreover it's finitely-generated. So the structure theorem tells you what kind of decomposition to expect, and then you explicitly construct the corresponding generalized eigenvectors $x^ne^{rx}$.

I don't quite understand this, since I don't know what a $k[D]$-module is, and I'm not at all familiar with derivations.

I'm very interested in seeing explicitely how the description of the solutions of these equations is a special case of the structure theorem.

Solution 1:

Let our homogeneous linear differential equation be $p(D) f = 0$, where $p$ is some polynomial and $D$ is differentiation. By existence and uniqueness the space of solutions is a vector space of dimension $\deg p = n$. Moreover the space of solutions is acted on by $D$, hence is a $k[D]$-module generated by $n$ elements.

The structure theorem says that all finitely-generated $k[D]$-modules are direct sums of modules of the form $k[D]/(D - \lambda)^m$. Here we know that $p(D)$ acts by zero, so any direct summand must have the property that $(D - \lambda)^k$ divides $p(D)$. Writing

$$p(D) = \prod (D - \lambda_i)^{m_i}$$

suggests that we should look for submodules isomorphic to $k[D]/(D - \lambda_i)^{m_i}$. Such a submodule consists of functions $f$ which are solutions to

$$(D - \lambda_i)^{m_i} f = 0.$$

Now, write $f = e^{\lambda_i x} g$. We compute that

$$(D - \lambda_i) f = e^{\lambda_i x} Dg$$

hence it follows by induction that

$$(D - \lambda_i)^{m_i} f = e^{\lambda_i x} D^{m_i} g = 0.$$

It follows that $g$ is a polynomial of degree less than $m_i$. So we have found our desired submodule: it consists of all $f$ of the form $f(x) = e^{\lambda_i x} g(x)$ where $g$ is a polynomial of degree less than $m_i$.

(By the way, you don't actually need the structure theorem to do any of this. If $p(D) f = 0$ then it already follows that any function $f$ satisfying $q(D) f = 0$ where $q | p$ is a solution, and then the above discussion constructs all solutions. The structure theorem is just a convenient mental heading to place this situation under, since it's a natural source of nontrivial Jordan blocks.)

Solution 2:

I can't read Qiaochu's mind, but I guess that something like the following is what he had in mind. Consider a homogeneous DE with constant coefficients $a_1,\ldots,a_n\in k$ (here $k$ is a field, either the reals or the complex numbers) $$ y^{(n)}+a_1y^{(n-1)}+\cdots+a_{n-1}y'+a_ny=0. $$

You undoubtedly know that the set of solutions forms a subspace $V$ of the function space. IOW, if $y=y_1(x)$ and $y=y_2(x)$ are solutions, then so is any linear combination $y=c_1y_1(x)+c_2y_2(x)$ for any constants $c_1,c_2\in k$.

Furthermore, if $y$ is a solution, then so is its derivative $Dy=y'$. In other words $D:V\rightarrow V$. Therefore also $D^2y, D^3y,\ldots$ are always solutions, if $y$ is. This allows us to act on any solution $y$ by any polynomial $b_n D^n+b_{n-1}D^{n-1}+\cdots+b_0$ as follows: $$ (b_n D^n+b_{n-1}D^{n-1}+\cdots+b_0)y=b_ny^{n}+b_{n-1}y^{n-1}+\cdots+b_1y'+b_0. $$ This action turns $V$ into a $k[D]$-module. Algebraically the ring $k[D]$ is just the ring of polynomials in the "unknown" $D$. So it is a PID, and we can use the structure theory of finitely generated modules over a PID. One variant of the structure theorem tells us that any finitely generated $k[D]$-module can be written as a sum of cyclic modules of the either of the form $k[D]$ or of the form $k[D]/p(D)^m$, where $p(D)\in k[D]$is an irreducible polynomial.

In this case there will be no free components. This follows from the fact that $V$ is a finite dimensional vector space over $k$. More directly it follows from the fact that the characteristic polynomial $$ \chi(D)=D^n+a_1D^{n-1}+\cdots+a_{n-1}D+a_n $$ obviously kills everything in $V$. This also implies that all the polynomials $p(D)$ that appear in the above direct sum decomposition must be factors of $\chi(D)$.

In the case $k=\mathbf{C}$ an irreducible polynomial $p(D)$ is linear by the fundamental theorem of algebra, i.e. $p(D)=D-\alpha$ for some complex constant $\alpha$. If $y$ is a generator of a cyclic $k[D]$-submodule $V_\alpha$ of the form $k[D]/(D-\alpha)^m$, then $V_\alpha$ has a basis $$\{y_0=y,y_1=(D-\alpha)y,y_2=(D-\alpha)^2y,\ldots,y_{m-1}=(D-\alpha)^{m-1}y\}$$ because any coset of the ideal generated by $(D-\alpha)^m$ has a unique representative of degree less than $m$. Here, by the properties of the module action, we have $$(D-\alpha)y_{m-1}=0\qquad\hbox{and}\qquad (D-\alpha)y_i=y_{i+1},i=0,1,\ldots,m-2.$$ From this we can deduce that $y_{m-1}=c_{m-1}e^{\alpha x}$, and (selecting $c_{m-1}=1$) then $y_{m-2}=xe^{\alpha x}$ et cetera. Yet another way of looking at this is to observe that $V_\alpha$ corresponds to a Jordan block of the endomorphism $V$.

In the real case it is possible that $p(D)$ is quadratic and has complex roots, i.e. $$ p(D)=(D-[\alpha+i\beta])(D-[\alpha-i\beta])=D^2-2\alpha D+(\alpha^2+\beta^2). $$ I leave it to you to guess what kind of cyclic $k[D]$-submodules of $V$ look like the $k[D]/p(D)^m$.

I may have overlooked some aspects of this way of looking at a DE, but this answer is long enough as it is, so I stop for now.