2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
+50
$\begingroup$

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

$$a_{n+1}=(\lambda+y+n+1)a_n-(y+n)a_{n-1},$$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.