Solve differential equation $f'(z) = e^{-2} (f(z/e))^2$

Question

I'm curious if there's a simple closed solution to the following DE and, if so, what it is.

$$\begin{align} f'(z) &= e^{-2} (f(z/e))^2 \\ f(0) &= 1. \end{align}$$

Answer 1

This is no differential equation at all since it connects data from two different positions. If you use $g(x)=f(e^x)$ then $$ g'(x)=f'(x)e^x=e^{-2}f(e^x/e)^2e^x=e^{x-2}g(x-1)^2 $$ is a delay-differential equation. These have in general a difficult to non-existent existence theory and equally incomplete numerical methods.

In this special case, you can prescribe a solution almost freely for $x\in[0,1]$ or $z\in[1,e]$ with the only restriction that $g'(1)=e^{-1}g(0)^2$. All other values follow from integration, $$ g(x)=g(1)+\int_1^x g'(s)\,ds=g(1)+\int_1^x e^{s-2}g(s-1)^2\,ds $$ successively for the intervals $x\in[1,2]$, $[2,3]$,…

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So far, this will construct a solution on $x\in[0,\infty)$. To keep the option for a solution on $x\in\Bbb R$ open, take any infinitely differentiable, totally (in all derivatives) increasing function $h(x)$ with $h(0)>0$ and $h'(1)>0$ and set $g(x)=a·h(x)$ with $a$ determined by $h'(1)=a/e·h(0)^2$. Even more fun can be had by setting $g(x)=a·h(x)+b$ so that a one-parameter family results.

Example for the most simple solution: $g(x)=e·(1+x)$ on $[0,1]$. Then for $x\in[1,2]$ $$ g(x)=2·e+\int_1^x e·(1+2s+s^2)\,ds = 2·e+\frac{e}3·(x-1)·(7+4·x+x^2) $$ On $[2,3]$ the degree will be $7$ etc.

Answer 2

i am wondering if we can compute the derivatives of $f$ of all order at $x = 0.$ we have $$f(0) = 1, f'(0) = \frac1{e^2}. \tag 1$$ we can take the differential equation $$f'(x) = \frac 1{e^2} f^2\left( \frac xe\right) $$ and we can differentiate repeatedly to get $$f''(x) = \frac 2{e^3} f\left( \frac xe\right) f'\left( \frac xe\right), \,f'''(x)=\frac2{e^4}\left(f'\left( \frac xe\right)f'\left( \frac xe\right)+f\left( \frac xe\right)f''\left( \frac xe\right)\right), \cdots$$

so that we have $$f''(0) = \frac 2{e^5},\, f'''(0) = \frac2{e^4}\left(\frac 1{e^4}+ \frac{2}{e^5}\right),\, f(x) = 1 + \frac{x}{e^2} + \frac{x^2}{e^5} + \cdots.$$

Answer 3

The recurrence Relationship for the coefficients of the series expension of $f(x)$ is shown below. The values of the coefficients decreasse very quickly. The figure shows that only a few terms are sufficient to obtain a good approximate of the function $f(x)$.

On the figure, the curves which represent the series made of 20 and 100 terms are identical (drawn in red)

Note: For the formula of squaring the series, see : https://en.wikipedia.org/wiki/Cauchy_product