recurrence relations - How to solve the delay algebraic equation $xf(x) + alpha f(x - {x_0}) - alpha f(x + {x_0}) = 0$?

In the process of solving a problem, I am faced with the problem of finding a non-zero function $f:\mathbb{R} \to \mathbb{R}$ which satisfies the equation $$xf(x) + \alpha f(x - {x_0}) - \alpha f(x + {x_0}) = 0$$ for a known $x_0$ and $\alpha$. Unfortunately all I could find searching online is the topic of delay differential equation which seems to be more general than my question. Could anyone help me with some references, keywords or hints? Thanks in advance.

Answer

Let $f(x)=\int_a^be^{xs}K(s)~ds$ ,

Then $x\int_a^be^{xs}K(s)~ds+\alpha\int_a^be^{(x-x_0)s}K(s)~ds-\alpha\int_a^be^{(x+x_0)s}K(s)~ds=0$

$\int_a^bK(s)~d(e^{xs})+\alpha\int_a^be^{-x_0s}e^{xs}K(s)~ds-\alpha\int_a^be^{x_0s}e^{xs}K(s)~ds=0$

$[e^{xs}K(s)]_a^b-\int_a^be^{xs}~d(K(s))+\alpha\int_a^be^{-x_0s}e^{xs}K(s)~ds-\alpha\int_a^be^{x_0s}e^{xs}K(s)~ds=0$

$[e^{xs}K(s)]_a^b-\int_a^be^{xs}K'(s)~ds+\alpha\int_a^be^{-x_0s}e^{xs}K(s)~ds-\alpha\int_a^be^{x_0s}e^{xs}K(s)~ds=0$

$[e^{xs}K(s)]_a^b-\int_a^b(K'(s)+\alpha(e^{x_0s}-e^{-x_0s})K(s))e^{xs}~ds=0$

$\therefore K'(s)+\alpha(e^{x_0s}-e^{-x_0s})K(s)=0$

$K'(s)=\alpha(e^{-x_0s}-e^{x_0s})K(s)$

$\dfrac{K'(s)}{K(s)}=-2\alpha\sinh x_0s$

$\int\dfrac{K'(s)}{K(s)}~ds=-2\alpha\int\sinh x_0s~ds$

$\ln K(s)=-\dfrac{2\alpha\cosh x_0s}{x_0}+c_1$

$K(s)=ce^{-\frac{2\alpha\cosh x_0s}{x_0}}$

$\therefore f(x)=\int_a^bce^{xs-\frac{2\alpha\cosh x_0s}{x_0}}~ds$

But since the above procedure in fact suitable for any complex number $s$ ,

$\therefore f_n(x)=\int_{a_n}^{b_n}c_ne^{xk_nt-\frac{2\alpha\cosh x_0k_nt}{x_0}}~d(k_nt)=k_nc_n\int_{a_n}^{b_n}e^{k_nxt-\frac{2\alpha\cosh k_nx_0t}{x_0}}~dt$

For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $k_n$ such that:

$\lim\limits_{t\to a_n}e^{k_nxt-\frac{2\alpha\cosh k_nx_0t}{x_0}}=\lim\limits_{t\to b_n}e^{k_nxt-\frac{2\alpha\cosh k_nx_0t}{x_0}}$

$\int_{a_n}^{b_n}e^{k_nxt-\frac{2\alpha\cosh k_nx_0t}{x_0}}~dt$ converges

One of the choice involving

$\int_{-\infty}^\infty e^{\frac{xt}{x_0}-\frac{2\alpha\cosh t}{x_0}}~dt$

$=\int_{-\infty}^0e^{\frac{xt}{x_0}-\frac{2\alpha\cosh t}{x_0}}~dt+\int_0^\infty e^{\frac{xt}{x_0}-\frac{2\alpha\cosh t}{x_0}}~dt$

$=\int_\infty^0e^{-\frac{xt}{x_0}-\frac{2\alpha\cosh(-t)}{x_0}}~d(-t)+\int_0^\infty e^{\frac{xt}{x_0}-\frac{2\alpha\cosh t}{x_0}}~dt$

$=\int_0^\infty e^{-\frac{xt}{x_0}-\frac{2\alpha\cosh t}{x_0}}~dt+\int_0^\infty e^{\frac{xt}{x_0}-\frac{2\alpha\cosh t}{x_0}}~dt$

$\propto\int_0^\infty e^{-\frac{2\alpha\cosh t}{x_0}}\cosh\dfrac{xt}{x_0}~dt$

$\propto K_\frac{x}{x_0}\left(\dfrac{2\alpha}{x_0}\right)$

You will find that this functional equation if fact is very similar to that of modified Bessel functions

In fact the general solution can consider as $f(x)=\Theta_1(x)I_\frac{x}{x_0}\left(\dfrac{2\alpha}{x_0}\right)+\Theta_2(x)K_\frac{x}{x_0}\left(\dfrac{2\alpha}{x_0}\right)$, where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with period $|x_0|$ .