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As we know, general solution of the linear functional equation $f(x+1)-f(x)=g(x)$ ($g$ is a known function) is $f=f_0+\phi$, where $f_0$ is an its special solution and $\phi$ any 1-periodic function.

Now, does any one know general solution of the linear functional equation: $$rf(ax+b)+sf(cx+d)=g(x),$$ where $a,b,c,d, r$ and $s$ are real constants with $rasc\neq0$?

($g$ is a given real function, the case $g=0$ and the equation $f(ax+b)=cf(x)$ is of special interest).

Are there any papers or books regarding the problem?

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  • $\begingroup$ It is the same as you wrote in the first paragraph: the general solution is the sum of a single solution plus the general solution of homogeneous equation (with g=0). $\endgroup$ Commented Sep 17, 2020 at 13:41

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As in the special case, the general solution is the sum of a particular solution and the general solution of the homogeneous equation.

To describe he general solution of the homogeneous equation $$rf\phi_1+sf\circ\phi_2=0,$$ where $\phi_1(x)=ax+b$ and $\phi_2(x)=cx+d$ are affine functions, you use the fact that affine functions form a group. So there is an affine function $\phi_3$ such that $\phi_2=\phi_3\circ\phi_1$, and making the change of the variable $t=\phi_1(x)$ we obtain the equation $$rf(t)+sf\circ\phi_3(t)=0.$$ To further simplify this, consider two cases: a) $\phi_3(t)=at+b,\; a\neq 1$, and b) $\phi_3(t)=t+b$. In the first case, by a conjugation in the affine group, the equation is reduced to $$rg(t)+sg(at)=0.$$ In the second case, it is reduced to $$rg(t)+sg(t+1)=0.$$ The general solutions of these two standard equations must be clear.

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  • $\begingroup$ Thanks. Can one obtain an explicit formula for the general solution from the above form? $\endgroup$ Commented Sep 19, 2020 at 15:50
  • $\begingroup$ @M.H.Hooshmand: yes, of course. This is a simple exercise. $\endgroup$ Commented Sep 19, 2020 at 19:46

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