How to find a general solution of the following functional (recurrence) equation: $$f(x) = c_1 f(x - a_1) + \dots + c_n f(x - a_n), \tag{1}$$ where $c_i, a_i$, $i = 1, ... n$ are arbitrary real numbers? I'm looking only for real functions $f$. Without losing generality let's consider $0 < a_1 < a_2 < \dots < a_n$, $c_i \neq 0$.

More info on the problem:

The core of the problem is that $a_i$ are non rational. It leads us to the transcendental characteristic equation of the form $$\lambda^{a_n} - c_1 \lambda^{a'_{n-1}} - \dots - c_{n-1} \lambda^{a'_1} - c_n = 0, \tag{2}$$ where $a_n > a'_{n-1} > \dots > a'_1$.

Is it true that this equation might have infinite number of complex roots? Then, is it true that general solution will be represented as an infinite series? When will these series converge? Can you please suggest something for reading on this topic?

Addition to the question:

I came across an encyclopedia article that considers equation (1). Because it is what I need, I put the part from there and posted it as an answer, but I have the following questions remaining:

  1. What if some roots of the equation equation (2') are not simple? What are the partial solutions of (9) when root $\lambda_m$ has multiplicity $\mu_m$? I guess the function $x^l e^{\lambda_m}$ will be the solution for any $l = 0, 1, \dots, \mu_m$, but what is the bi-orthogonal function $\psi_{m l}$ corresponding to this solution?
  2. The encyclopedia article states: "under certain conditions the series (11) converges to the solution f." What are those conditions in particular? Is it the condition that the system of functions $\{x^l e^{\lambda_m x}\}$ must compose a Rice base? How to show it?
  3. Let's return to my equation (1). I want to show that in this case there is a simple positive root of the equation (2') that is greater than any other root of (2').

Recently, I came across the example in an Encyclopedia of Mathematics (Finite-difference calculus). This partially answers my question in a simple case, but it has some nuances that I want to be cleared, so I edited the head post to include more questions. The following is the part from the encyclopedia relevant to my question.

$$\varPhi(x, f(x), f(x+1), \dots, f(x+n)) = 0 \tag{4}$$ it's solution with respect to $f(x+n)$ is $$f(x + n) = \psi(x, f(x), f(x+1), \dots, f(x+n-1)) \tag{5}$$


$$f(x + k) + \alpha_1 f(x + k - 1) + \dots + \alpha_k f(x) = 0 \tag{8}$$


Equation (4) can be found not only when $x$ varies discretely, taking values $0, 1, \dots$, but also when $x$ varies continuously. Let $f$ be arbitrarily defined in the half-open interval $[0,n)$ . From (5) one obtains by setting $x=0$. If $f$ is continuously defined in $[0,n)$, then $f$ may prove to be discontinuous in the closed interval $[0,n]$. If one wishes to deal with continuous solutions, $f$ needs to be defined on $[0,n)$ in such a way that by virtue of (5) it proves to be continuous in $[0,n]$. Knowledge of $f$ in $[0,n]$ enables one to find from (5) $f(x)$ for $x \in (n,n+1]$, then for $x \in (n+1,n+2]$, etc.

More general than (8) is the equation $$f(x + h_k) + \alpha_1 f(x + h_{k-1}) + \dots + \alpha_k f(x) = 0 \tag{9}$$

Here $0 < h_1 < \dots < h_k$ are not necessarily integers and are not necessarily commensurable relative to one another. Equation (9) has the particular solutions $$f(x) = e^{\lambda x}$$

where $\lambda$ is a root of the equation $$L(\lambda) \equiv e^{h_k \lambda} + \alpha_1 e^{h_{k-1} \lambda} + \dots + \alpha_k = 0 \tag{2'}$$

This equation has an infinite number of roots $\lambda_1, \lambda_2, \dots$. Consequently, (9) has an infinite number of particular solutions $e^{\lambda_m x}$, $m = 1, 2, \dots$. Suppose that all the roots are simple. To express the solutions of (9) in terms of these elementary particular solutions, it is convenient to write the equation in the form: $$\int_{0}^{\alpha} f(x+t) d\sigma(t) = 0, \quad \alpha = h_k, \tag{10}$$

where $\sigma(t)$ is a step function having jumps at the points $0, h_1, \dots, h_k$, equal to $\alpha_k, \alpha_{k-1}, \dots, 1$, respectively. Let $$\psi_\nu(t) = \frac{-e^{\lambda_\nu t}}{L'(\lambda_\nu)} \int_{0}^{t} e^{\lambda_\nu \xi} d\sigma(\xi) = 0, \qquad \nu \geqslant 1.$$

The functions $\psi_\nu(t)$ have the property: $$\int_{0}^{\alpha} e^{\lambda_m t} \psi_\nu(t) dt = \delta_{m \nu}$$

($\delta_{m \nu} = 1$ if $m = \nu$, and $\delta_{m \nu} = 0$ if $m \neq \nu$), that is, they form a system that is bi-orthogonal to the system $\{e^{\lambda_m t}\}$. On this basis, the solution $f$ of (10) corresponds to the series $$ f(x) \sim \sum_{\nu = 1}^{\infty} c_{\nu} e^{\lambda_\nu x}, \tag{11}$$

$$c_\nu = \int_{0}^{\alpha} f(t) \psi_\nu(t) dt.$$

If (9) has the form $$f(x + 2\pi) - f(x) = 0 \tag{12}$$ (that is, $f$ is a periodic function with period $2\pi$); $L(x) = e^{2\pi x} - 1$; the roots of the equation $L(\lambda) = 0$ are $m i$ ($m = 0, \pm 1, \dots$); and (11) is the Fourier series of $f$ in complex form. The series (11) can be regarded as a generalization to the case of the difference equation (9) of the ordinary Fourier series corresponding to the simplest difference equation (12). Under certain conditions the series (11) converges to the solution $f$. If $f$ is an analytic function, then (9) is expressible in the form of an equation of infinite order $$\sum_{n = 0}^{\infty} \alpha_n f^{(n)} (x) = 0$$


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