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

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

Harry Peter
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Andrew
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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 $$f''(0) = \frac 2{e^5},\, f'''(0) = \frac2{e^4}\left(\frac 1{e^4}+ \frac{2}{e^5}\right),\, f(x) = 1 + \frac{x}{e^2} + \frac{x^2}{e^5} + \cdots.$$

abel
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    The first few derivatives seem to be (orders $0,1,2,3,4,5,6$): $1$, $e^{-2}$, $2e^{-5}$, $4e^{-9}+2e^{-8}$, $8e^{-14}+4e^{-13}+12e^{-12}$, $16e^{-20}+8e^{-19}+24e^{-18}+32e^{-17}+40e^{-16}$ and $32 e^{-27} + 16 e^{-26} + 48 e^{-25} + 64 e^{-24} + 160 e^{-23} + 40 e^{-22} + 280 e^{-21} + 80 e^{-20}$. Pretty strange, but it looks the series should converge for all $x$. – Dejan Govc Apr 15 '15 at 18:42
  • @DejanGovc, thanks for confirming and computing higher order derivatives. do you see any pattern? – abel Apr 16 '15 at 13:49
  • Not really, but proving convergence probably shouldn't be too hard. If I come up with something useful, I'll let you know. – Dejan Govc Apr 16 '15 at 17:19
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    By the way, this method shows that $f(x)=e^x$ is the unique analytic solution of the similar equation $f'(x)=f(\frac x2)^2, f(0)=1$. – Dejan Govc Apr 16 '15 at 22:12
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    Well the coefficients $f_n = [z^n]f(z)$ satisfy $f_0=1,$ and for $n>0,$ $f_n = \frac{e^{-n-1}}{n} \sum_{j=1}^n f_{j-1} f_{n-j}$. One can see $0 < f_n < 1$ for all $n$. – Andrew Apr 17 '15 at 14:46
  • (I should say $0 < f_n \leq 1$) – Andrew Apr 18 '15 at 02:29
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    @AndrewMacFie: In fact, it is easy to see by induction that $f_n\leq\frac1{n!}$. For $n=0$, we have $f_0=1$, so this holds. Suppose the statement holds for $n$. Then we have $f_{n+1}=\frac{e^{-n-2}}{n+1}\sum_{j=0}^n f_j f_{n-j}\leq\frac{e^{-n-2}}{n+1}\sum_{j=0}^n \frac1{j!}\frac1{(n-j)!}=\frac{e^{-n-2}}{(n+1)!}\sum_{j=0}^n \binom{n}{j}=e^{-2}(\frac2e)^n\frac1{(n+1)!}\leq\frac1{(n+1)!}$, where the first inequality is obtained by the induction hypothesis and in the end we use $\sum_{j=0}^n \binom{n}{j}=2^n$. So the series converges for all $x\in\mathbb R$. – Dejan Govc Apr 18 '15 at 07:56
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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 $x\in[1,2]$, $[2,3]$,…

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.

Lutz Lehmann
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  • This is a really nice idea, but shouldn't $g$ be increasing by the right hand side of the equation, restricting the valid choices for $x\in[0,1]$ a bit? – Dejan Govc Apr 15 '15 at 16:33
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    Not only that, to go $n$ units backwards by iterating $g(x)=\sqrt{e·g'(x+1)}$ you also need derivatives of any order and that all derivatives are non-negative. There is a name for this, hyper-convex or something? I think I've seen this in copula theory. – Lutz Lehmann Apr 15 '15 at 16:54
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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

enter image description here

JJacquelin
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