Reference for a nice proof of "undetermined coefficients"

Question

I'm teaching an honors differential equations class and have been using linear algebra heavily. I thought it would be interesting to include a proof of the method of undetermined coefficients along the lines described below, which I figured is already known. Nonetheless, Google turns up no proof other than the obvious one by integration, which is significantly messier and uses no linear algebra, so perhaps I'm wrong.

What is the (standard, best, earliest) reference to the following proof?

Note: I have made a conscious decision not to ask this on math.stackexchange since, based on my last question asking about a reference there, I don't think it will even be noticed.

Theorem: Let $L$ be a constant-coefficient $n$th order linear differential operator with characteristic polynomial $p(\lambda)$, and let $y(t) = t^m e^{qt}$ for some integer $m \geq 0$ and $q \in \mathbb{C}$. Suppose that $q$ is a $k$-fold root of $p(\lambda)$. Then the equation $Lx = y$ has a solution of the form $t^k f(t) e^{qt}$, where $f(t)$ is a polynomial of degree $m$.

Proof outline:

1. $y$ is a solution of the equation $L_2 y = 0$, where $L_2 = (\tfrac{d}{dt} - q)^m$. The pair of equations $Lx = y, L_2 y = 0$ can be written as a system of $n + m + 1$ first-order equations via the linear transformation $\vec{z} = (x, x', \dots, x^{(n - 1)}, y, y', \dots, y^{(m)})$, for which the coefficient matrix is: $$M = \left(\begin{array}{c|c} A & \begin{smallmatrix} 0 & \cdots & 0 \\ \vdots & & \vdots \\ 1 & \cdots & 0 \end{smallmatrix} \\ \hline 0 & C \end{array}\right)$$ where $A$ is the companion matrix of $L$ and $C$ is the companion matrix of $L_2$.

2. It is easily verified that:

   • The eigenvalues of $M$ are the same as those of $A$, with an additional $m + 1$ repetitions of $q$; each eigenvalue $\lambda$ has a one-dimensional eigenspace, spanned by $(1, \lambda, \dots, \lambda^{n - 1}, p(\lambda), \lambda p(\lambda), \dots, \lambda^m p(\lambda))$.

   • Every Jordan chain for $A$, namely, vectors $\vec{v}_i$ associated with an eigenvalue $\lambda$ such that $(A - \lambda I) \vec{v}_i = \vec{v}_{i - 1}$, upgrades to vectors $\vec{u}_i$ with the same property for $M$, by attaching $m + 1$ zeros to the end.

   It follows that $M$ has a generalized eigenbasis consisting of upgrades of a generalized eigenbasis for $A$, together with $m + 1$ additional generalized eigenvectors filling out a Jordan chain of length $k + m + 1$ for eigenvalue $q$.

3. A generalized eigenvector $\vec{v}_h$ of eigenvalue $\lambda$ at the top of a Jordan chain produces a fundamental solution to $\vec{z}' = M\vec{z}$ via the matrix exponential: $$e^{\lambda t} \sum_{i = 0}^{h - 1} \frac{1}{i!} t^i \vec{v}_{h - i}.$$ The general solution is therefore the general linear combination of these. Taking the first components (which we may assume are nonzero only for the eigenvectors (indexed $\vec{v}_1$ in the above sum)), we get a solution of $Lx = y$. The terms coming from "upgrades" are part of the homogeneous solution to $Lx = 0$, and the remaining terms are some linear combination of $t^k e^{qt}, \dots, t^{k + m} e^{qt}$, as claimed.

Answer 1

1. If $L$ has characteristic polynomial $\lambda \mapsto p(\lambda)$, then the conjugated differential operator $e^{-qt} L e^{qt}$ has characteristic polynomial $\lambda \mapsto p(\lambda+q)$. From this we may easily reduce the problem to the $q=0$ case.

2. If the characteristic polynomial $\lambda \mapsto p(\lambda)$ of $L$ has a zero of order $k$ at the origin, then $p$ factors as $p(\lambda) = \tilde p(\lambda) \lambda^k$, and $L$ similarly factors as $L = \tilde L \frac{d}{dt^k}$. Since every degree $m$ polynomial has a $k$-fold antiderivative that is equal to $t^k$ times a degree $m$ polynomial, we can thus reduce the problem to the $q=0, k=0$ case.

3. If $q=0$ and $k=0$, then $p(0)$ is non-vanishing; by rescaling we may take $p(0)=1$. Then $p(\lambda) = 1 + \lambda r(\lambda)$ for some polynomial $r$, so $L = 1 + R \frac{d}{dt}$ for some differential operator $R$. On the finite dimensional vector space spanned by $1,t,\ldots,t^m$, the operator $R \frac{d}{dt}$ acts nilpotently (as it always reduces the degree) and so $L$ is unipotent, hence invertible (by Neumann series), in this space, and the claim follows.

Answer 2

Hi Ryan and everybody,

Besides the very beautiful proof by Tao, a very nice and easy linear algebra approach to the undetermined coefficients method can be found in C. C. Ross ``Why the method of undetermined coefficients works'', Am. Math. Monthly 98 (1991), pp. 747-749.

I also suggest R. C. Gupta ``Linear differential equations with constant coefficients: a recursive alternative to the method of the undetermined coefficients'', Int. J. Math. Educ. Sci. Technol. 27 (1996), pp. 757-760.

A very easy formula (quite good for the students) that gives a particular solution of the equation $$L(\frac{d}{dt})y=t^m e^{q t}, \ q \in \mathbb C,$$ where $L$ is a nth order linear differential operator with constant coefficients, can be found in O. R. B. de Oliveira ``A formula substituting the undetermined coefficients and the annihilator methods'', Int. J. Math. Educ. Sci. Technol. 44 (3), 2013, pp. 462-468, http://dx.doi.org/10.1080/0020739X.2012.714496. The proof of the formula is in p. 465 and it is short and very elementary.

The idea is the following: the given ode has a particular solution $y=f(t)e^{q t}$ where $f$ is a polynomial that satisfies the following ode with constant coefficients $$\frac{p^{(n)}(q)}{n!}f^{(n)} + \cdots + \frac{p'(q)}{1!}f' + \frac{p(q)}{0!}f= t^m,$$ where $p=p(\lambda)$ is the characteristic polynomial associated to $L$. The short and elementary proof of this formula, with some examples, is over there. Using it, we find a particular solution through solving a triangular linear system.

The same formula shows that if $q$ is a root of multiplicity $k$ of $p(\lambda)=0$ then $e^{qt},\ldots, t^{k-1}e^{qt}$ are solutions of the homogeneous equation $Ly=0$, since we have $p(q)=\cdots=p^{(k-1)}(q)=0$ and $\frac{d^l}{dt^l}(t^j)=0$ if $l>j$. Furthermore, in this case the formula boils down to $$\frac{p^{(n)}(q)}{n!}f^{(n)} + \cdots + \frac{p^{(k)}(q)}{k!}f^{(k)} = t^m,$$ which shows that $f^{(k)}$ is a polynomial of order $m$. Integrating it $k$-times we are able to choose $f(t)=t^kg(t)$, where $g$ is a polynomial of order $m$.

Oswaldo R. B. de Oliveira