# $f'=e^{f^{-1}}$, 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. Commented 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. Commented Jan 3, 2017 at 10:40
• Commented Jan 3, 2017 at 21:05
• @Fan Zheng , i think you should start a bounty for this question since it's very interesting Commented 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
Commented Jan 7, 2017 at 7:55
• Yet there is a small question left: Is there a local $C^\infty$ solution at $0$ ? Commented 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 $$(f^{-n})'=\exp(-f^{-2}-f^{-3}-\dots-f^{-n+1}).$$ Commented 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.) Commented May 30, 2021 at 20:15
• Yes, Sorry, always assuming $f$ analytic on the whole real line. Fixed now. Commented May 30, 2021 at 23:17