Suppose we have a set of polynomials where $\deg(Q_k(x))\le k$, and consider the following differential equation,

$$W:=\sum_{k=0}^n Q_k(x)\frac{d^k}{dx^k} .$$

It is known that if there is a polynomial of degree $n$, $B_n(x)$, and a constant, $c$, such that,

$$W[B_n(x)]=c\cdot B_n(x) ,$$


$$c=\sum_{k=0}^n \binom{n}{k}Q_k^{(k)}.$$

The proof follows by simply equating the leading coefficients.

We now ask the converse. Suppose we are given that,

$$c=\sum_{k=0}^n \binom{n}{k}Q_k^{(k)}.$$

Then must there exist a polynomial of degree $n$ such that

$$W[B_n(x)]=c\cdot B_n(x)\ \ \ \ \ ?$$

An affirmative answer to the question, would provide an ``if and only if'' condition for polynomial solutions for differential equations of this nature. More importantly an affirmative answer would indicate that if you have any linear operator on the space of polynomials,

$$T:=\sum_{k=0}^\infty Q_k(x)D^k$$

where $\deg(Q_k(x))\le k$ for every $k$. Then there would exist a sequence of eigenvalues, $\{a_n\}$, and eigenvectors, $\{B_n(x)\}$, such that,

$$T[B_n(x)]=a_n\cdot B_n(x),$$

for every $n$ in the natural numbers.

A simple explanation or reference would be great.


Using the basis of monomials $x^j$, $j = 0 \ldots n$, of the space ${\mathcal P}_n$ of polynomials of degree $\le n$, the matrix representing $W$ is upper triangular. The eigenvalues of $W$ (counted by algebraic multiplicity) are the diagonal elements of this matrix. In particular, the diagonal entry corresponding to basis element $x^n$ is $\sum_{k=1}^n {n \choose k} Q_k^{(k)}$, so this is an eigenvalue. However, there is not necessarily an eigenvector of degree $n$ if some diagonal entries are repeated.

EDITED: For example, in the case $n=2$, consider $$ W(f) = f - 2 x \dfrac{df}{dx} + (x^2 + x/2) \dfrac{d^2 f}{dx^2}$$ for which the matrix is $$ \pmatrix{1 & 0 & 0\cr 0 & -1 & 1\cr 0 & 0 & -1\cr}$$ The only polynomials of degree $\le 2$ with $W(f) = cf$ are the constants (with $c=1$) and multiples of $x$ (with $c = -1$).

  • $\begingroup$ I may be misunderstanding you, but isn't $$W[x^n]=(xD+1)x^n = (n+1)x^n$$ for every $n$. Hence, for example $x^2$ with $c=2$ is a solution. $\endgroup$ Mar 5 '14 at 7:18
  • $\begingroup$ I changed my example: now it should work. $\endgroup$ Mar 5 '14 at 7:36
  • $\begingroup$ First, thank you for your responses. It is helpful to talk this out with others. :-). Now lets say $Q_0(x)=1$, $Q_1(x)=-2x$, $Q_2(x)=x^2+x/2$. We calculate $c=1(1)+2(-2)+1(2) = -1$, in other words we don't get to pick the $c$, we have to calculate if from the formula (in fact it is already known that if $c$ doesn't satisfy this formula, then we won't have solutions to our equation). So, you are saying, $$(x^2+x/2)y''-2xy'+y=(-1)y,$$ has no quadratic solutions. But consider $$p(x)=(x+1/2)^2.$$ Unless I made an terrible error, I think that is a solution. Nevermind, I did make a terrible error. $\endgroup$ Mar 5 '14 at 8:06
  • $\begingroup$ Thank you so very much for you efforts. This example seems to work fantastic, just what I was looking for. I used an arbitrary $$ax^2+bx+c$$ and precisely we must conclude that $a=0$. $\endgroup$ Mar 5 '14 at 8:15
  • $\begingroup$ Due diligence is in order and I feel I must acknowledge our personal communication. I am currently a graduate student trying to write a paper and so I ask, would you mind, if I acknowledge your help? Here is a link: arxiv.org/pdf/1402.0141.pdf . In my next update, I can now change the section concerning question 1 and provide a "personal communications" in the bibliography. :-) $\endgroup$ Mar 5 '14 at 8:34

Do I understand correctly, that $Q^{(k)}_k$ is a $k$-th derivative of $Q_k(x)$?

If not, disregard the rest.

Let $Q_k(x) = \sum_{i=0}^k q_{k,i}x^i$, then $Q^{(k)}_k = q_{k,k}k!$ and $c = \sum_{k=0}^n \binom{n}{k} q_{k,k} k!$

For polynomial $y = \sum_{i=0}^n y_ix^i$ to satisfy the equation, $y_i$ must solve the system of linear equations I am too lazy to spell out; what's important is that it has $n+1$ equation for $n+1$ unknowns. What is even more important, consider an $x^n$ term at the left hand side, which is $\sum_{k=0}^n q_{k,k}\frac{n!}{(n-k)!} = \sum_{k=0}^n q_{k,k}\binom{n}{n-k}k! = c$. In other words, the last equation in the system is tautology, making the whole system underdefined, therefore solvable.

  • $\begingroup$ I am not exactly understanding what you are saying. Your statement about n+1 equations with n+1 unknowns, makes it sound like the problem is solvable for arbitrary $c$. Which is known to not be true. Of course, more than likely I am just misunderstanding you. I will edit my response to try and make things more clear. $\endgroup$ Mar 5 '14 at 6:23
  • $\begingroup$ Also, I am not quite sure about the reasoning about the coefficients with an "underdefinement". Lets put this in to a little more perspective with a concrete example. Given $$(x^2-1)y''+(2x)y'=(6)y$$ there is only one quadratic solution (up to a constant multiple), namely the second Legendre polynomial. $\endgroup$ Mar 5 '14 at 7:11
  • $\begingroup$ Oh, I can't edit comments....did not know that. Anyway, here is the Legendre differential equation, $$(x^2-1)y''+(2x)y'=(n^2+n)y.$$ Apparently (or so I have read), this differential equation uniquely (up to a constant multiple) defines the Legendre polynomials of all degrees. Now the $n^2+n$ part, cannot be modified and must be $n^2+n$, if $W:=(x^2-1)D^2+(2x)D$ is to have a polynomial solution of degree $n$. Thus, if differential equation $W$ has a polynomial solution of degree $n$, then the solution must be $$W[L_n(x)]=(n^2+n)L_n(x)$$ up to constant multiple. $\endgroup$ Mar 5 '14 at 7:38
  • $\begingroup$ makes it sound like the problem is solvable for arbitrary $c$ - no it doesn't. The reason is that the system is homogenouos, and for most $c$ yields only a trivial zero solution. For some particular $c$s (which are eigenvalues of the system matrix, as Robert Israel pointed out) it does have a nontrivial one. $\endgroup$
    – user58697
    Mar 6 '14 at 17:35

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