This question is a spin-off of [this one](https://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1), in which the OP asks whether there is a solution $f:\mathbb R\to\mathbb R$ of the functional equation (not exactly an ODE) $f'=e^{f^{-1}}$, where $f^{-1}$ is the compositional inverse of $f$. The posted answer exploits the growth of $f(x)$ when $x\to-\infty$ and obtains a contradiction, which resolves the question nicely, but also invites the following question: what if we restrict to $f:\mathbb R\_{\ge0}\to\mathbb R\_{\ge0}$ and impose $f(0)=0$? This idea has been explored in the comments, where a formal power series expansion is obtained for $f$ which does not seem to converge for any $x\ne0$.


Taking another approach, we can use an iteration scheme starting from $f\_1(x)=x$ and inductively solve the ODE $f\_{n+1}'=e^{f\_n^{-1}}$ with the initial condition $f\_{n+1}(0)=0$ to obtain $f\_{n+1}$, much in the spirit of Picard iteration. Explicitly, for example, we have


$f\_2'=e^x$ and $f\_2=e^x-1$;


$f\_3'=e^{\ln(x+1)}=1+x$ and $f\_3=x+x^2/2$;


$f\_4'=e^{\sqrt{1+2x}-1}$ and $f\_4=e^{\sqrt{1+2x}-1}(\sqrt{1+2x}-1)$


and the next iteration produces non-elementary functions. It is clear that the sequence $(f\_{2k-1})\_{k\ge1}$ is increasing, $(f\_{2k})\_{k\ge1}$ is decreasing, and $f\_{2k-1}<f\_{2k}$, so there are respective limits $f\_-=\lim\_{k\to\infty} f\_{2k-1}$ and $f\_+=\lim\_{k\to\infty} f\_{2k+1}$, with $f\_-\le f\_+$. It is also clear that from $n\ge2$ on the function $f\_n'=e^{f\_{n-1}^{-1}}$ is positive and increasing, so $f\_n$ is increasing and convex, which can be passed to the limit to show that both $f\_-$ and $f\_+$ are also increasing and convex. As such they are continuous, and by Dini's theorem $f\_{2k-1}$ converges to $f\_-$ locally uniformly and similarly for $f\_+$. Furthermore, the inequality $|x-y|\le |f\_n(x)-f\_n(y)|$ (as $f\_n'=e^{f\_{n-1}^{-1}}\ge1$) can also be passed to the limit. Then the following chain of inequalities:


$|f\_-^{-1}(x)-f\_{2k-1}^{-1}(x)|\le |x-f\_-(f\_{2k-1}^{-1}(x))|=|f\_{2k-1}(f\_{2k-1}^{-1}(x))-f\_-(f\_{2k-1}^{-1}(x))|$


shows that $f\_{2k-1}^{-1}$ converges locally uniformly to $f\_-^{-1}$, which then implies $f\_{2k}'$ converges locally uniformly to $e^{f\_-^{-1}}$. Hence $f\_+'=e^{f\_-^{-1}}$, and similarly $f\_-'=e^{f\_-^{-1}}$. From this it can be shown that $f\_{2k-1}$ converges to $f\_-$ locally in $C^\infty$, so both $f\_-$ and $f\_+$ are smooth functions, and they form an orbit of order at most 2 of the above iteration scheme. Moreover it can be shown that the first $n$ terms of the Taylor expansion of $f\_n$ agrees with what have been calculated formally in the previous comments, so both $f\_-$ and $f\_+$ have the same Taylor expansion as calculated using formal power series expansion.


In light of the above, *a priori* the following three scenarios can happen:


1. $f\_-\neq f\_+$ and we have a genuine orbit of order 2, consisting of two functions having the same Taylor expansion at 0 but not being identical.
2. $f\_-=f\_+$ is an actual solution to the equation $f'=e^{f^{-1}}$, but it is merely $C^\infty$ but not analytic, having a divergent power series expansion at 0.
3. $f\_-=f\_+$ is an actual solution to the equation $f'=e^{f^{-1}}$, and it is analytic on a neighborhood of 0; we are just misled by the first 100 or so terms of the Taylor expansion.


Now finally comes the question: which of the above scenario is the reality? In the first two scenarios, one can also ask what is the growth rate of $f\_-(x)$ and $f\_+(x)$ as $x\to+\infty$.



$f'=e^{f^{-1}}$, again
There is no analytic local solution at $0$ to $f'=e^{f^{-1}}$, $f(0)=0$, that is, the formal power series solution is diverging. Together with the solution given in comments by fedja, this means the actual scenario is **2.** For convenience of notation, I shall consider the equivalent equation
$$\begin{cases} g' =e^{g\circ g}, \\ g(0)=0, 
 \end{cases}$$
satisfied by $g(x):=-f^{-1}(-x)$ (Indeed, by the rule of the derivative of an inverse, $(f^{-1})'(x)={1\over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so that $g'(x) =e^{g(g(x))}$; see also Tom Copeland's previous answer [here](https://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1/258695#258695).)


Indeed, assume by contradiction the formal power series solution $x+{1\over2}x^2+{1\over2}x^3+{2\over3}x^4+\&c.$ to the above equation has a positive radius of convergence. Then, it extends uniquely by analytic continuation to a maximally-defined analytic function, still denoted $g$ (that is, defined on the largest positive interval $[0,a)$, for some $0<a\le+\infty$).


Note that the Taylor series of $g$ at $0$ has non-negative coefficients. This follows immediately by induction, equating the coefficients of $g'$ and $e^{g\circ g}$; incidentally, this series is the EGF of the positive integer sequence [OEIS A214645](http://oeis.org/A214645), as also remarked [here](https://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1/258695#258695). As a consequence (check the details below), $g$ is totally monotonic on $[0,a)$; in particular $g'(x)>g'(0)=1$ and $g(x)>x$ for all $0<x<a$, and $g$ is invertible.


Then observe that $\log( g'( g^{-1}(x))$ is a well-defined analytic function on the interval $g[0,a)$, and coincides with $g$ locally at $0$. By the maximality of $[0,a)$ we have thus $g[0,a)\subset[0,a)$, but, due to the inequality $g(x)>x$ on $(0,a)$, this inclusion is only possible if $a=+\infty$, so that $g$ is unbounded. On the other hand, arguing as in Christian Remling's previous [answer](https://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1/258544#258544), since $e^{-g(g(t))}g'(t)=1$ and $g(t)\ge t$, we have for any $x\ge0$
$$x=\int\_0^{x}e^{-g(g(t))}g'(t)dt=\int\_0^{g(x)}e^{-g(s)}ds\le \int\_0^{+\infty}e^{-s}ds=1 ,$$


a contradiction.


$$\*$$


**Rmk 1.** To justify the total monotonicity of $g$, note that, as a general elementary fact, *a real analytic function on an interval $I$, whose Taylor series at some point $x\_0\in I$ has non-negative coefficients, has Taylor series with non-negative coefficients ay any point $x\in I$, $x\ge x\_0$.* Indeed, this is clear for $x\_1\ge x\_0$ within the radius of convergence of $x\_0$, and since there is a uniform radius of convergence at any $y\in [x\_0,x]$, one reaches $x$ by finitely many steps $x\_0<x\_1<\dots<x\_n=x$.


*(edit 5/29/21)* In fact more is true: *a real analytic function on $\mathbb R$, whose Taylor series at some point $x\_0\in\mathbb R$ has non-negative coefficients is an entire function*, so that any $x\ge x\_0$ is reached in just one step.


$$\*$$


**Rmk 2.** The very same argument works for other differential-functional equations like e.g.


$$\begin{cases} g' =1 + {g\circ g}, \\ g(0)=0, 
 \end{cases}$$
that generates the sequence [OEIS A001028](http://oeis.org/A001028).
As before, a maximally-defined analytic solution $g$, if any, must be totally monotonic and defined for all $x\ge0$, for otherwise $ g'\circ g^{-1} -1$ would be a proper extension of it. Then we reach a contradiction as before, with one more step needed: since we have ${ g'(t)\over 1+g(g(t))}=1$ and $g(t)\ge t$ for any $t\ge0$, we also have, for any $x\ge0$
$$x=\int\_0^{x}{ g'(t)dt\over 1+g(g(t))}=\int\_0^{g(x)}{ dt\over 1+g(t)}\le\int\_0^{g(x)}{ dt\over 1+t}=\log(1+g(x)) ,$$
whence $e^x\le 1+ g(x)$; if we plug this into the latter inequalities again, we get
$$x=\int\_0^{g(x)}{ dt\over 1+g(t)}\le \int\_0^{g(x)}e^{-t}dt\le 1 ,$$
as before. By comparison, the same conclusion also holds for $g'=F( {g\circ g})$ with any $F$ analytic and totally monotonic on $(-\epsilon,+\infty)$, and with $F(0)=1$.


