I'm trying to solve what seems to be a simple problem:
$$ f'(x) = f(x/a) $$
for $f$ real-valued and $0<a$. So far I couldn't get anything reasonable so I was wondering if there you know of any references that could help or can provide a hint.
Thanks!
The equation can be solved for all $x>0$ (or similarly $x<0$) as a usual DDE by taking a "fundamental solution" on $x\in(c,ac]$ for some $c\in\mathbb R$ and recursively by defining $f$ for $x\in(ca^n,ca^{n+1}]$ and $n\in\mathbb{Z}$ $$ f(x) = \begin{cases}f(ca^b) + \int_{ca^n}^x f(t/a)dt & n\ge 1\\ f'(xa) &n<0 \end{cases} $$
If you're interested in a solution which is analytic at $0$, you can see that $$ f(x) = \sum_{n=0}^\infty \frac{a^{-\binom{n}2}x^n}{n! } $$ works, but it only converges if $|a|\ge 1$. If you want a smooth solution with $0<a<1$ we can take something like this $$ f(x)=\sum_{n=-\infty}^\infty (-1)^n a^{\frac{n(n-1)}2} e^{-x a^n} $$ This series converges for all $x>0$. We also check:$$ f'(x)= \sum_{n=-\infty}^\infty (-1)^{n+1} a^{\frac{n(n-1)}2+n} e^{-x a^n} = \sum_{n=-\infty}^\infty (-1)^{n+1} a^{\frac{n(n+1)}2} e^{-\frac{x}a a^{n+1}} = f\left(\frac x a\right) $$ as desired.