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.

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    $\begingroup$ If you are interested in approximative solutions then one can do as follows: Take $x_0\to 0$ with $\beta = 2x_0\alpha$ fixed to obtain the differential equation $xf(x) - \beta f'(x) = 0$ with solution $f(x) = A e^{\frac{x^2}{2\beta}}$. This solution should be a good approximation for small $x_0$ at least in the region around where you set the initial condition. $\endgroup$ – Winther Mar 21 '15 at 18:04

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

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





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


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

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

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

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

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

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 $|x_0|$ .

  • $\begingroup$ Thanks! It would be nice if you could comment on the existence of other bounds for the integral (or otherwise the uniqueness of -inf and +inf). Also, I would be thankful if you could expand the integration in the last line. $\endgroup$ – Seyed Mohsen Ayyoubzadeh Aug 18 '17 at 21:47

For fixed $x$, it is a difference equation with known solution methods. This defines your function (a two-dimensional set of possibilities) on one of the cosets of $x_0\mathbb Z$. It is defined independently on each coset. (Of course, such a definition is likely not continuous. But the problem does not require continuity.)


what could have been done if the delay was $x_0$ but the advance was $y_0$ (in the case that these didn't have a common factor, e.g. $\sqrt{2}$ and $\pi$)

Incommensurable intervals: that is harder. Now the cosets, although still countable, are dense in $\mathbb R$. So (with no requirement of continuity) you can still solve on each coset separately. And the cosets are group-isomorphic to $\mathbb Z^2$. So now you need the theory of two-parameter difference equations.

a plug

Suppose we do require continuity, or measurability, and integrability or boundedness. I have a paper with Rosenblatt on this...

G. A. Edgar & J. M. Rosemblatt, "Difference equations over locally compact abelian groups." Trans. Amer. Math. Soc. 253 (1979) 273--289. MR 80i:39001

  • $\begingroup$ Seems an interesting answer. But now, as an extension, what could have been done if the delay was $x_0$ but the advance was $y_0$ (in the case that these didn't have a common factor, e.g. $\sqrt 2 $ and $\pi$) $\endgroup$ – Seyed Mohsen Ayyoubzadeh Mar 21 '15 at 18:08

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