I would like to find all functions $f_{\lambda}: \mathbb R \rightarrow \mathbb C$ and scalars $\lambda \in \mathbb C$ that satisfy the following equation:

\begin{equation} \lambda f_{\lambda}(x) = x (f_{\lambda}(x-1) - f_{\lambda}(x)) + (f_{\lambda}(x+1) - f_{\lambda}(x)), \ \ \ \ \ \forall x \in \mathbb R, \end{equation} or equivalently \begin{equation} (\lambda + x + 1) f_{\lambda}(x) = x f_{\lambda}(x-1) + f_{\lambda}(x+1), \ \ \ \ \ \forall x \in \mathbb R. \end{equation}

For example, one solution is $\lambda = 0$ and $f_{\lambda}(x) = 1$ for all $x \in \mathbb R$.

I don't know how to systematically approach this problem. Any guidance would be appreciated.

  • $\begingroup$ The operator you're considering has the action $T: f \mapsto g$ where $g(x)=-(x+1)f(x) + xf(x-1) + f(x+1)$ (as Eigenfunctions of this would satisfy your equation above)? Which kind of domain are you considering? A subdomain of $L^2$? Polynomials? Continuous functions? $\endgroup$ – Roland Sep 16 '17 at 19:21
  • $\begingroup$ Yep, that's the operator. It's original form is actually $g(x) = x(f(x-1) - f(x)) + (f(x+1) - f(x))$, which is just a rearrangement of what you've written. $\endgroup$ – Garrett Sep 17 '17 at 5:04
  • $\begingroup$ As for the "domain", to be honest, I don't know what you are asking. But I suspect that there is a rational eigenfunction. $\endgroup$ – Garrett Sep 17 '17 at 5:07
  • $\begingroup$ I say this because I recently considered a similar problem in which $g(x) = x(f(x-1) - f(x))$, and found the following class of eigenfunctions: $f_{\lambda}(x) = \frac{x!}{(x+\lambda)!}$ $\endgroup$ – Garrett Sep 17 '17 at 5:09

My answer is partial, but I’ll try to extend it and it may be useful as initial general look and idea.

Given $\lambda\in\Bbb C$ the equation defines the function $f_\lambda$ on each coset $[x]=x+\Bbb Z$ of the group $\Bbb R$ with respect to a subgroup $\Bbb Z$. Namely, given $y\in\Bbb R$ for each $n\in\Bbb Z $ put $a_n=f(y+n)$. Then the sequence $\{a_n\}$ satisfies the recurrence


which uniquely defines the values of the function $f_\lambda$ on the coset $[y]$ provided we are given values of $f_\lambda(y+m)$ and $f_\lambda(y+m+1)$ for some integer $m$ and the number $y$ is non-integer or $m\le -1$ or we are given also the value of $f_\lambda(-1)$.

If $\lambda=0$ then the recurrence has a partial solution $a_n=c$. If $\lambda=-1$ then it has a partial solution $a_n=c(y+n-1)$. I guess that if $\lambda$ is a non-positive integer then the recurrence has a solution $a_n=p(n)$, where $p$ is a polynomial of degree $-\lambda$. At least, it can be proved the converse: if the recurrence has a solution $a_n=p(n)$, where $p\not\equiv 0$ is a polynomial then its degree equals $-\lambda$.

We may try to find other partial solutions and, maybe, even a complete solution. But in order to find it we may need to deal with complex powers like $n^{-\lambda}$.

In order to correspond the solutions for different cosets we need to impose additional conditions on the function $f_\lambda$ (that is the domain question, which also imposes restrictions on possible solutions of the recurrence). For instance, we may assume that the function $f_\lambda$ is continuous, rational or polynomial.

  • $\begingroup$ Thanks for weighing in, Alex. I'm actually only interested in the values of $f_\lambda(x)$ for $x \in \mathbb Z_+$, so I'm OK with "partial solutions". $\endgroup$ – Garrett Nov 15 '17 at 15:30
  • $\begingroup$ Am I right in thinking that the the condition "$y$ is non-integer or $m \leq -1$ or we are given also the value of $f_\lambda(-1)$" is to ensure we don't end up in the situation where we have zero times an unknown, bringing the recurrence to a grinding halt? $\endgroup$ – Garrett Nov 15 '17 at 15:32
  • $\begingroup$ I'm intrigued by your suggestion that the eigenvalue $\lambda$ might correspond to a polynomial of degree $-\lambda$ for $\lambda \leq 0$. Can you help me understand how you arrived at the polynomial corresponding to $\lambda = -1$? $\endgroup$ – Garrett Nov 15 '17 at 15:39
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    $\begingroup$ It seems that the eigenfunction corresponding to $\lambda = -2$ is $f_{-2} = n^2 -3n +1$. (Note, I have set $y = 0$.) Similarly, the eigenfunction corresponding to $\lambda = -3$ is $f_{-3} = n^3 -6 n^2 + 8 n -1$. $\endgroup$ – Garrett Nov 15 '17 at 18:37
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    $\begingroup$ With the help of Mathematica, I think I could determine the polynomial eigenfunction corresponding to any $\lambda \in \mathbb Z_-$. $\endgroup$ – Garrett Nov 15 '17 at 18:41

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