On the solutions to an $n$-th order linear homogeneous constant coefficient ODE

Question

There have been many posts on this site regarding the linear space of solutions to a linear, homogeneous, constant coefficient ODE of order $n$, in particular about the $n=2$ case. See for example here.

The technique that I knew for proving that the solution space has dimension $n$ (and has a particular basis) involves an appeal to the uniqueness theorem for the initial value problem (IVP). I have checked several elementary textbooks and found this proof in some, and no proof at all in others.

This post was prompted by a recent answer of David C. Ulrich, who pointed out for the $n = 2$ case it is completely unnecessary to appeal to the uniqueness theorem for the IVP in order to justify that the solution space has dimension 2.

That made me wonder about the following problem: Show that the solution space of $$ Lf = \prod_{j = 1}^s (D-a_j)^{r_j} f = 0 \tag{L} $$ is $n = \sum_j r_j$ dimensional with basis $$ \{x^\ell e^{a_j x} | 1 \le j \le s, 0 \le \ell \le r_j -1\}, $$ and do so (1) without appeal to the uniqueness theorem for the IVP and (2) in a manner pedagogically palatable to at least some potential audience of students. (For convenience, I assume that the characteristic polynomial of the differential operator has only linear irreducible factors; i.e. we work over the complex numbers if necessary.)

I'm going to post an answer to this question. It's a matter of taste whether this is worthwhile doing at all, but I found the question interesting because I hadn't thought before to make the connection to some relevant abstract or linear algebra. Maybe someone else will also find it interesting.

Answer 1

Since you talk about presenting this to undergraduates it seems appropriate to mention that one can do more or less the same thing without mentioning any algebra.

Say we're talking about $$\prod_{j=1}^n(D-a_j)y=g,$$ where the $a_j$ may not be distinct. Say $$z=\prod_{j=2}^n(D-a_j)y;$$then $$(D-a_1)z=g.$$So we get the general solution if we first solve $(D-a_1)z=g$, then solve $\prod_{j=2}^n(D-a_j)y=z.$

So one could formulate an algorithm: First solve $(D-a_1)z_1=g$, then solve $(D-a_2)z_2=z_1$, etc; finally solving $(D-a_n)y=z_{n-1}$.

Since the standard method of solving $(D-a)y=g$ by integrating factors does more or less contain its own proof of correctness, it's clear that this does give the general solution to the original equation. Bonus: It works the same for homogeneous or inhomogeneous equations; seems better motivated than variation of parameters or undetermined coefficients. (Undetermined coefficients is fine when the $a_j$ are distinct, but when there are repeated roots where does that $t$ come from? Here that doesn't arise, it's clear the algorithm does give the right solution, and the $t$'s come in simply when we happen to integrate a constant at one of the steps.)

So there's a simple algorithm, and it's clear that the algorithm actually gives the right answer, with no need to prove a general theorem that the solution to the hoogeneous equation is wht it is. (In a classorrom of course instead of a general proof that the algorithm works one could just show that it's right for a few specific equations, and very plausibly claim that the same thing is going to happen in general.)

If cone does want to prove that the solution to a homogeneous equation is what it is, that would be an easy induction, using

Lemma The general solution to $(D-a)z=t^ne^{bt}$ is [what it is].

Proof: Simply apply the standard method for solving first-order linear equations. At one point you'll be integrating $t^ne^{(b-a)t}$; note that if $b-a$ then the integral is $p_n(t)e^{(b-a)t}$ by integration by parts and induction on $n$, while if $b=a$ the integral is $t^{n+1}/(n+1)$.

Hmm Or if one wanted one could use this point of view to give a simple proof by induction of existence and uniqueness.

Answer 2

Recall that we are trying to describe the solution space of $$ Lf = \prod_{j = 1}^s (D-a_j)^{r_j} f = 0 \tag L $$

Any answer will have to at least deal with the case $s = 1$.

Lemma 1: The solution space of $$(D-a)^r f = 0 \tag{*} $$ has basis $B = \{x^\ell e^{a x} : 0 \le l \le r-1\}$.

Proof: By induction on $r$. The case $r = 1$ requires solving $f' = a f$, whose general solution is $f = C e^{a x}$. For $r > 1$, $(D - a)^r f = (D-a)^{r-1} (D-a) f$, so if $f$ is a solution to (*), then $g = (D - a)f $ is a solution to $ (D-a)^{r-1} g = 0$. By the induction hypothesis $g$ is in the span of $\{ x^\ell e^{a x} : 0 \le \ell \le r-2\}$. So now we have to solve the inhomogeneous first order equation $(D - a) f = g$ for $g$ in the span of $\{ x^\ell e^{a x} : 0 \le \ell \le r-2\}$. By inspection one can find a particular solution in the span of $B$, and the general solution is the sum of the particular solution and $C e^{a x}$, which completes the proof that $B$ spans the solution space. Linear independence of $B$ is left to the reader.

Now it remains to consider the general problem $s > 1$.

The first method is going to look quite abstract and therefore pedagogically impossible because of fancy terminology, but I will remark that in fact it is very concrete, or could be made very concrete if one wanted to expend the time and effort.

Method 1. Let $s > 1$ and let $V$ be the solution space of $(\rm L)$. Also write $V_{a, r}$ for the solution space of $(*)$. Then $V_{a, r} \subseteq V$. Write $p(t) = \prod_{j = 1}^s (t -a_j)^{r_j}$.

$V$ is a torsion module over $\mathbb C[t]$ with period $p$, where polynomials act on $V$ by $$ q \cdot h = q(D) h. $$ By the primary decomposition theorem for torsion modules, $$ V = V_{a_1, r_1} \oplus V_{a_2, r_2} \oplus \cdots \oplus V_{a_s, r_s}, $$ and thus we are done.

Discussion: the proof of the primary decomposition theorem is actually quite concrete and computational and depends on the following fact, which would be a useful thing for even undergraduates in science and engineering to see. For each $k$, let $q_k(t) = \prod_{j \ne k} (t -a_j)^{r_j}$. Then the collection of polynomials $q_k$, $1 \le k \le s$ has no non trivial common factor. It follows that one can compute, by repeated use of the Euclidean algorithm (Smith Normal Form for a one row matrix) polynomials $s_k(t)$ such that $$ 1 = \sum_k s_k(t) q_k(t). $$ Thus for $f \in V$, $$ f = \sum_k s_k(D) q_k(D) f. $$ Since $q_k(D) f \in V_{a_k, r_k}$, this already shows that $V$ is spanned by $\{ V_{a_k, r_k}\}$. Linear independence is shown by a variation of the same idea.

Nevertheless, it seems quite unlikely that anyone is going to actually do this in an elementary course, so this observation is mostly for the enjoyment of people who are already comfortable with the primary decomposition.

Method 2:

For this method, let us prepare the following lemma.

Lemma 2: Consider $a \ne b$. The space $V_{b, r}$ is invariant under $D -a$, and $D-a$ is invertible on $V_{b, r}$, with an explicitly computable inverse.

Proof: $V_{b, r}$ is invariant under $D-a$ since $D-a$ commutes with $D -b$. \begin{equation} D- a = (D - b) + (b-a) = (b - a) \left [ 1 - (a-b)^{-1} (D - b) \right ] \end{equation} Now $N = (a-b)^{-1} (D - b) $ is nilpotent of order $r$ on $V_{b, r}$, and therefore $D-a$ is invertible with inverse $$ (b-a)^{-1}(1 + N + N^2 + \cdots + N^{r-1}). $$

Remark: One can recover this result by the more naive approach of assuming, for given $g \in V_{b, r}$ a solution $h \in V_{b, r}$ to the equation $(D-a) h = g$, and solving by matching coefficients.

Now we continue with the 2nd method. Suppose that $f$ is a solution of $(L)$. Write $a = a_1$ and $r = r_1$ Then $g = (D-a)^r f$ is a solution to $$ \prod_{j > 1} (D - a_j)^{r_j} g = 0. $$ By the appropriate induction hypothesis, it follows that $g \in V_{a_2, r_2} \oplus \cdots \oplus V_{a_s, r_s}$. Therefore $f$ satisfies the inhomogeneous DE $$ (D-a)^r f = g, $$ with $g \in V_{a_2, r_2} \oplus \cdots \oplus V_{a_s, r_s}$. Using Lemma 2, a particular solution $f_1$ can be found with $f_1 \in V_{a_2, r_2} \oplus \cdots \oplus V_{a_s, r_s}$. The general solution to $(L)$ is $f = f_0 + f_1$, where $f_0$ is a solution to the homogenous equation $(D-a)^r f_0 = 0$, i.e. $f_0 \in V_{a_1, r_1}$. The conclusion is that $f$ is in the span of $\{V_{a_k, r_k} : 1 \le k \le s\}$.

For linear independence, suppose that $f_k \in V_{a_k, r_k}$ for each $k$ and $\sum_k f_k = 0$. Again writing $a = a_1$ and $r = r_1$, apply $(D-a)^r$ to get $\sum_{k \ge 2} (D-a)^r f_k = 0$. By an appropriate induction hypothesis, the spaces $V_{a_k, r_k}$ for $k \ge 2$ are linearly independent, so for each $k \ge 2$, $ (D-a)^r f_k = 0$. By the invertibility of $(D-a)^r$ on each $V_{a_k, r_k}$ for $k \ge 2$, it follows that $f_k = 0$ for $k \ge 2$. Hence $f_1 = 0$ as well.