Can we solve recurrence relations indexed by R?

Question

Recurrence relations are equations with functions from $\mathbb N\rightarrow \mathbb N$. For example given $a:\mathbb N\rightarrow \mathbb N$ (write $a(n)$ as $a_n$) solving the recurrence relation

$a_n = A\cdot a_{n-1}+B\cdot a_{n-2}$

means finding the function $a$. There are methods for doing this.

Question: Are there methods for solving similar equations when the function is $a:\mathbb R\rightarrow \mathbb N$? For example,

$a(x) = A\cdot a(x-\alpha) + B\cdot a(x-\beta)$ for some fixed $\alpha,\beta\in\mathbb R$?

Answer 1

In general such equations are called functional equations.

In this case, assuming $A$ and $B$ are constants, for at least an open set of parameters $(\alpha, \beta)$ we can show that there is a solution of the ansatz form $$a(x) = \lambda \exp(\mu x)$$ just by substituting in the functional equation. Here, $\lambda$ can be prescribed freely and $\mu$ satisfies some transcendental equation involving $A, B, \alpha, \beta$.

On the other hand, the family does not necessarily exhaust the solutions of the equation: If, e.g., $\beta = 2 \alpha$, it seems to me that we can construct a solution by prescribing $a(x)$ arbitrarily on the interval $[0, \beta)$, after which the functional equation determines the value of $a$ elsewhere.

Answer 2

I know that one common strategy is to first consider the equation on the integers $x\in\mathbb{Z}$. If you manage to figure out what the values of $a(n)$ is for all integers $n$, usually the next step is to consider the equation for all rationals. When you know $a(q)$ for all rational numbers $q\in\mathbb{Q}$, you usually argue that you can extend this to all reals by continuity (but notice that this large argument only finds the continuous solutions).