1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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). $\endgroup$ Sep 17, 2020 at 13:41

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks. Can one obtain an explicit formula for the general solution from the above form? $\endgroup$ Sep 19, 2020 at 15:50
  • $\begingroup$ @M.H.Hooshmand: yes, of course. This is a simple exercise. $\endgroup$ Sep 19, 2020 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.