Solutions to $f'(x)-f(\alpha x)=0$

Question

Looking at this question:

Solution of recursion $T(n) = T(n-2) + 2T(\frac{n}{2})$

I was interested to know what are/is the solution/s to the (functional) differential equation:

$f'(x)-f(\alpha x)=0, \quad f(0)=0,f:\mathrm{R}^+\rightarrow\mathrm{R},0<\alpha<1$

I am not sure that the differential equation is related to the linked problem, but I got interested in the equation by itself.

TRIAL: I just noticed that defining $g(x) \equiv f(e^x),g:[-\infty,+\infty]\rightarrow \mathrm{R}$, this satistfies $g'(x)=e^xg(x+ln(\alpha))$ so that now we have a time delay $\tau=ln(\alpha)<0$, $g(-\infty)=0$ and maybe standard techinques can be used? But actually I do not know which are the standard techniques, if they exists, for this type of equations.

Answer 1

If we assume a series solution $$ f(x) = \sum_{k=0}^\infty c_k x^k$$ we get $$ k c_k = \alpha^{k-1} c_{k-1} \ \text{for}\ k \ge 1$$ so that $$ c_k = \frac{c_0 \alpha^{k(k-1)/2}}{k!} $$ The series converges for all $z$ if $|\alpha| \le 1$.

Answer 2

The only differentiable function $f: \Bbb R^+ \to \Bbb R$ satisfying $$ f'(x) = f(\alpha x) \, , \, f(0) = 0 $$ for a given $\alpha \in (0, 1)$ is the “zero function” $f(x) = 0$.

One can proceed as in $|f'(x)|\le|f(x)|$ and $f(0)=0$, prove that $\forall x\in[0,\frac 1 2]:f(x)=0$ and prove that

$$ \tag{*} f(x_0) = 0 \implies f(x) = 0 \text{ for } x_0 \le x \le x_0 + \frac 12 \, . $$

It is clear that this – together with $f(0) = 0$ – implies that $f(x) = 0$ for all $x \ge 0$.

So it remains to prove $(*)$: Assume that $f(x_0) = 0$ and define $$ M = \max \{ |f(x) | : x_0 \le x \le x_0 + \frac 12 \} \, . $$ Then $$ f(x) = \int_{x_0}^x f'(t) \, dt = \int_{x_0}^x f(\alpha t) \, dt \\ \implies |f(x)| \le \int_{x_0}^x |f(\alpha t)| \, dt \le x M \le \frac 12 M $$ for $x_0 \le x \le x_0 + \frac 12$. It follows that $$ 0 \le M \le \frac 12 M \implies M = 0\, . $$