Solve $y(2x)=(y(x))^2$

Question

So, I'm working on a research project and am having issues with tackling equations that are of mixed composition. In particular, equations of the form $y(2x)=[y(x)]^2$.

My first thought to solve was attempting the following, basically finding the derivative and seeing if I could solve it analytically by taking the derivative of both sides.

$y(2x)=(y(x))^2$

$\implies y(2x)-(y(x))^2=0$

$\implies 2y'(2x)-2y(x)y'(x)=0$, by the chain rule.

$\implies 2(y'(2x)-y(x)y'(x))=0$

$\implies y'(2x)-y(x)y'(x)=0$

$\implies y'(2x)=y(x)y'(x)$

At this point how would I go about solving this equation? because it seams like all I can really do is integrate both sides but that doesn't seam to be getting me any closer.

Well, the form you've presented suggests that the function is of the form $y(x)=a^x$ for some $a$. — Rushabh Mehta, Aug 31 '19 at 01:48
One way to handle this is to take logs of both sides and introduce the function $g(x)$. Then the functional equation becomes $g(2x)=2g(x)$, which is altogether suggestive... — Semiclassical, Aug 31 '19 at 02:15
Set $x=0$ and get $y(0)=y(0)^2$ or $y(0)=0$ or $y(0)=1$. Note that the third line of the derivation is wrong, so your last line should be $y'(2x)=y'(x)y(x)$. — herb steinberg, Aug 31 '19 at 02:32
I think you can prove by induction that if $y(0)=0$ then $y^{(n)}(0)=0$ for any $n\in\mathbb{N}$ — Alessandro Cigna, Aug 31 '19 at 02:46
One thing to also be aware of is that, if you don't assume continuity, then you can create some simple but strange-looking examples of such functions. As an example, see the second part of this answer: https://math.stackexchange.com/a/873786/137524. (In that example, one cooks up examples where $$S(x)^2+C(x)^2=1, S(0)=0, C(0)=1, S(2x)=2S(x)C(x)$$ but nevertheless $S(x)$ and $C(x)$ are not sine/cosine.) — Semiclassical, Aug 31 '19 at 04:45
I think you need to write $$\implies y'(2x)-y(x)y'(x)=0$ as $\implies $y'(2x)- (1/2) y(x)y'(x)=0$ — NoChance, Sep 05 '19 at 09:36

score 1 · Answer 1 · answered Sep 05 '19 at 09:33

Let $y(x)=a^{u(x)}$ , where $a\in\mathbb{R}^+$ and $a\neq1$ ,

Then $a^{2u(x)}=a^{u(2x)}$

$u(2x)=2u(x)$

Let $x=2^t$ ,

Then $u(2^{t+1})=2u(2^t)$

$u(2^t)=\Theta(t)2^t$ , where $\Theta(t)$ is an arbitrary periodic function with unit period

$u(x)=\Theta(\log_2x)x$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

$y(x)=a^{\Theta(\log_2x)x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period, $a\in\mathbb{R}^+$ and $a\neq1$

Allawonder · Answer 2 · 2019-08-31T06:15:50.500

First, note that the function is everywhere nonnegative and even. Also, focusing on $f(1),$ we see that if $f(1)<1,$ then the function decreases for $|x|\ge 0.$ If $f(1)>1,$ it increases in that range. If $f(1)=1,$ it's the constant function $1$ (which we disregard henceforth). Of course I have assumed continuity, which is OK by you since you can differentiate.

Then by the above conditions, we see that for $f(1)>1$ we can choose either of $0$ or $1$ for $f(0).$ However, if $f(1)<1,$ we can only choose $f(0)=1.$ So, we see that the function is strictly positive if $f(1)<1$ but may vanish at the origin if $f(1)$ is chosen to be greater than $1.$ I would suggest to look at the case with $f(1)<1$ first as it seems more definite. In that case one sees that $f$ is bounded, and such functions like $$\frac{1}{1+x^2}$$ and $$e^{-x^2}$$ immediately come to mind. At this point I can't think of something more, but perhaps this may be something to go on for now? :)

PS. For the case when $f(1)>1,$ it's also easy to think of similar forms like $ax^2+b,$ with $a$ and $b$ positive, and $e^{x^2}.$ Indeed, it soon becomes apparent that the cases $f(1)<1$ and $f(1)>1$ are reciprocal, so that it suffices to focus on just one. Also, it becomes clear again why we cannot have $f(0)=0$ for the decreasing case if we want continuity everywhere; and we also see why the decreasing cases must all be bounded.

Solve $y(2x)=(y(x))^2$

2 Answers2