Linear differential equations versus abstract algebra... (Q&A).

Question

Context: Lately I've been "self-learning" a little algebra, one motivation being a desire to prove that the algorithms I've taught in differential equations for years actually work. The other day I noticed something about all this that would make a nice answer to this question or this question. Both those questions are on hold, so I'm putting it here instead.

Disclaimer: Everything I say here is doubtless commonplace to experts in differential equations - I'm posting it for the benefit of readers who, like me, never actually studied differential equations as a mathematical topic, like with theorems and proofs, and who've wondered about proofs of things that are simply asserted in elementary DE texts.

Let $D$ denote the derivative operator:: $Df=f'$.

Question: Given a complex polynomial $p$, how do we show that the solutions to $p(D)y=0$ are linear combinations of the functions $t^ke^{\lambda t}$, where $p(\lambda)=0$?

Answer 1

We start with a well known lemma, in the context of linear algebra over some field:

Lemma. Suppose $p_1$ and $p_2$ are polynomials with no common factor and let $p=p_1p_2$. If $T$ is a linear operator on the vector space $V$ then $\ker(p(T))=\ker(p_1(T))\oplus\ker(p_2(T)).$

Proof: There exist polynomials $r_1$ and $r_2$ such that $$r_1p_1+r_2p_2=1.$$So for any $x\in V$ we have$$x=r_1(T)p_1(T)x+r_2(T)p_2(T)x=x_1+x_2.$$If $p(T)x=0$ this shows that $$p_2(T)x_1=0=p_1(T)x_2.$$So $\ker(p(T))=\ker(p_1(T))+\ker(p_2(T)).$The fact that this is a direct sum is also clear: If $p_1(T)x=p_2(T)x=0$ then $r_1p_1+r_2p_2=1$ implies that $x=0$.

Now by a trivial induction

Lemma. Similar result if $p=\prod_{j=1}^np_j$, where the $p_j$ are pairwise relatively prime.

Of course this is just where the Jordan canonical form comes from; it's interesting that it also has this application to differential equations.

The question asks us to identify the kernel of $p(D)$. Assume $p$ is monic. We can write $$p(t)=\prod_{j=1}^k(t-\lambda_j)^{n_j},$$ where the $\lambda_j$ are distinct complex numbers. So by the lemma we only need to identify the kernel of $(D-\lambda I)^n$. And this is very simple, starting with an observation inspired by the standard algorithm for solving $y'-\lambda y=g$:

Let $m_\lambda$ denote the operation "multiply by $e^{\lambda t}$": $$(m_\lambda f)(t)=e^{\lambda t}f(t).$$Note that

$(D-\lambda I)^n=m_\lambda D^nm_{-\lambda}$

(For $n=1$ this is just the product rule; the general case follows immediately.)

And it's more or less obvious what the kernel of $m_\lambda D^nm_{-\lambda}$ is. Suppose that $m_\lambda D^nm_{-\lambda}y=0$. Then $$D^nm_{-\lambda}y=0,$$so $$m_{-\lambda}y=q(t),$$where $q$ is a polynomial of degree less than $n$. So $$y=q(t)e^{\lambda t}.$$

The same analysis gives an algorithm for solving inhomogenous equations:

Exercise Show that the solutions to $p(D)y=g$ are given by $y=\sum y_j$, where $(D-\lambda_jI)^{n_j}y_j=r_j(D)g$.