# $f'=e^{f^{-1}}$, again

This question is a spin-off of this one, 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 $x\to+\infty$.

• The convergence is not hard to demonstrate. For instance, if $f,g$ are two increasing functions with $f(0)=g(0)=0$ and $f(x),g(x)\ge x$ for all $x\ge 0$, and $F,G$ are their images under the Picard map, then for every $T>0$, the functional $\Phi(f,g,T)=\int_0^T|f(t)-g(t)|\,dt$ satisfies $\Phi(F,G,T)\le \int_0^T e^t\Phi(f,g,t)\,dt$ and it follows that on every finite interval $[0,T]$, we have convergence in $L^1$ and, therefore, in $C^\infty$. The question about the analyticity at $0$ seems to be less obvious. Jan 3, 2017 at 3:33
• You might like to look at an alternate differential eqn in the inverse series alone as presented in my comments to the related earlier question. Jan 3, 2017 at 10:40
• Jan 3, 2017 at 21:05
• @Fan Zheng , i think you should start a bounty for this question since it's very interesting Jan 6, 2017 at 13:52

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.)

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).

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, as also remarked here. 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 $$0, and $$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, since $$e^{-g(g(t))}g'(t)=1$$ and $$g(t)\ge t$$, we have for any $$x\ge0$$ $$x=\int_0^{x}e^{-g(g(t))}g'(t)dt=\int_0^{g(x)}e^{-g(s)}ds\le \int_0^{+\infty}e^{-s}ds=1 ,$$

$$*$$

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.

(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. 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 $$F(0)=1$$.

• It took me a while to get all of that, but... yeah, I think that puts an end to the question. Damn that was slick.
– user78249
Jan 7, 2017 at 7:55
• Yet there is a small question left: Is there a local $C^\infty$ solution at $0$ ? Jan 7, 2017 at 19:57
• btw, the derivatives of the $n$-fold iterates $f^n$ of $f$ have a curious formula: for any $n\in\mathbb{N}$ $$(f^n)'=\exp(f^{-1}+f^0+f^{1}+\dots+f^{n-2})$$ and $$(f^{-n})'=\exp(-f^{-2}-f^{-3}-\dots-f^{-n+1}).$$ Jan 11, 2017 at 22:12
• Regarding the recent edit: for example $(1-x)^{-1}$ has all Taylor coefficients at $0$ positive, but it is not entire; you must have meant something else. (But anyway "any $x\geqslant x_0$ is reached in just one step" remains true.) May 30, 2021 at 20:15
• Yes, Sorry, always assuming $f$ analytic on the whole real line. Fixed now. May 30, 2021 at 23:17