Is there a general solution for the differential equation $f''(x) = f(f(x))$?

Question

I'm currently an undergraduate studying differential equations and I've been fixated on the differential equation $f''(x) = f(f(x))$ for the past 2 days. I can't seem to crack it but it feels like it should have a general solution?

Answer 1

Remark: I had a little time to write a draft of my notes on the proofs of the claims I make below and have posted it on my home webpage here. (It would have made a very long post on MO, so I decided that it would be better to just link it to a file in my public directory.)

There are many local solutions of this equation. For example, suppose that one starts with a $C^2$ function $f$ on an interval $I\subset\mathbb{R}$ such that $f'$ is positive on $I$ and $f(I)$ is disjoint from $I$. Then an inverse $g:f(I)\to I$ of $f:I\to f(I)$ exists and is $C^2$. Now define $f$ on the interval $f(I)$ so that $f(y) = f''(g(y))$ for $y\in f(I)$. Then for $x\in I$, we will have $x = g(y)$ for some $y\in f(I)$ and, of course, $y = f(x)$. Then $f''(x) = f''(g(y)) = f(y) = f(f(x))$ for all $x\in I$.

These sorts of 'rough' solutions are constructed without any fixed points. Solutions with fixed points are much more rigid. A $C^2$ solution on an open domain $D$ such that $f(D)\subset D$ must be smooth on $D$, since $f''=f{\circ}f$, implying that if $f$ is $C^k$, then $f$ must be $C^{k+2}$. In fact, with a little effort, one can show that a $C^2$ solution with an contracting fixed point must be real-analytic in a neighborhood of the fixed point, since the equation $f''=f{\circ}f$ allows one to prove an estimate of the form $|f^{(k)}|\le C^k\,k!$ for some constant $C$ on a neighborhood of the fixed point.

Note 1: For every constant $b\in\mathbb{C}$, there is a unique formal power series with lowest order term $bz$ that satisfies $f''(z) = f(f(z))$. The first few terms are $$ f(z) = bz+\frac{{b}^{2}}{3!}\,{z}^{3} +{\frac {{b}^{3} \left( {b}^{2}{+}1 \right)}{5!}}\,{z}^{5} +{\frac {{b}^{4} \left( {b}^{6}{+}{b}^{4}{+}11\,{b}^{2}{+}1\right)}{7!}}\,{z}^{7}+\cdots.\tag1 $$ When $|b|<1$, this series converges absolutely and uniformly on the disk $|z|^2\le 6\bigl(1{-}|b|\bigr)$, and satisfies $|f(z)|\le |z|$ there. See the Addendum below for a sharper (but still not sharp) estimate of the radius of convergence.

Update (1 Mar 2021): One can show that, when $b$ is a small negative real number, the above function $f$ extends real-analytically and periodically to $\mathbb{R}$ and gives a $1$-parameter family of nontrivial solutions $f:\mathbb{R}\to\mathbb{R}$. In particular, such an $f$ extends holomorphically to a strip of fixed width about $\mathbb{R}\subset\mathbb{C}$. (Meanwhile, when $-1<b<0$, the radius of convergence of the power series (1) is only $r(|b|)\in(0,\infty)$ (see the Addendum below), which is a very different behavior from that when $0<b<1$.)

Addendum to Note 1: One can show that, when $0<b<1$, the real-analytic odd function $f$ that equals the power series (1) on its interval of convergence extends real-analytically to a bounded interval $\bigl(-r(b),r(b)\bigr)\subset\mathbb{R}$ on which $|f(x)|<|x|$ and that $\lim_{x\to r(b)^-}f(x)=r(b)$. In particular, $f:\bigl(-r(b),r(b)\bigr)\to\bigl(-r(b),r(b)\bigr)$ is a real-analytic diffeomorphism with a single contracting fixed point at $x=0$. Moreover, the formal series converges to $f$ uniformly on compact subsets of $\bigl(-r(b),r(b)\bigr)$, and $f$ cannot be extended real-analytically to any larger interval. (There are some indications that $f$ may extend smoothly beyond $x = r(b)$, in which case, $x=r(b)$ would become an expanding fixed point of $f$.) Also, $r:(0,1)\to(0,\infty)$ is a continuous, decreasing bijection, and $$ \frac{\sqrt{6\bigl(1{-}b\bigr)}}{b} > r(b)> \begin{cases} \sqrt{\displaystyle\frac3{2b}} & \text{for}\ 0<b\le\tfrac12,\\ \\ \sqrt{6(1{-}b)} & \text{for}\ \tfrac12\le b<1, \end{cases} $$ from which it follows that, for $b=1$, the radius of convergence of the series is $0$.

Note 2: More generally, for any two constants $a,b\in\mathbb{C}$, there is a formal power series $$ f(z) = a+b\,(z{-}a) +\frac{a}{2!}\,(z{-}a)^2 +\frac{b^2}{3!}\,(z{-}a)^3 +\frac{ab(b{+}1)}{4!}\,(z{-}a)^4 +\cdots\tag2 $$ that has $a$ as a formal fixed point, i.e., $f(a) = a$, so that the composition $f(f(z))$ makes sense as a power series centered at $z = a$ and, formally, $f'(a) = b$, that satisfies $f''(z) = f(f(z))$ as formal power series centered at $z = a$. Moreover, this is the unique power series centered at $z=a$ that has $f(a) = a$ and $f'(a) = b$ and satisfies $f''(z) = f(f(z))$ as formal power series.

As in the case $a=0$, when $|b|<1$, so that $f$ is a 'formal contraction' on a neighborhood of $a$, it turns out that the series converges absolutely and uniformly on a disc of the form $|z-a| \le r(a,b)$ for some $r(a,b)>0$, so this gives a two-parameter family of local solutions with a contracting fixed point.

Note 3: The (two) multivalued solutions described by Michael Engelhardt have fixed points and hence are (analytic continuations of) solutions of the type (2). One can see this as follows: These (multivalued) solutions can be written in the form $$ f(x) = i\sqrt{2}\,\left(\frac{x}{i\sqrt{2}}\right)^b,\qquad \text{where}\ b = \tfrac12(1\pm i\sqrt{7}). $$ Clearly, $a\in\mathbb{C}$ will be a fixed point, i.e., $f(a) = a$ if and only if $$ 1 = \left(\frac{a}{i\sqrt{2}}\right)^{b-1}, $$ and this happens (for $b = \tfrac12(1+i\sqrt7)$) when, for some integer $k$, $$ a = a_k = i\sqrt{2}\, e^{i\pi k(1+i\sqrt7)/2} = i^{k+1}\sqrt{2}\,\left(e^{-\pi\sqrt7}\right)^{k/2}. $$ Moreover, we have $$ f'(a_k) = b\left(\frac{a_k}{i\sqrt2}\right)^{b-1} = b, $$ so $|f'(a_k)| = |b| = \sqrt 2>1$, which implies that the fixed point is a repelling fixed point.

This is interesting because it implies that the formal power series (2) for $(a_k,b)$ must have a positive radius of convergence, even though $|b|>1$. This led me to speculate that maybe the formal power series (2) might have a positive radius of convergence for any $(a,b)\in\mathbb{C}$, but Will Sawin (in a comment below) pointed out that this cannot be true.

Answer 2

The equation has solutions with powers, $f(x) = ax^b$. Inserting this ansatz, one has $$ a b (b-1) x^{b-2} = a (a x^b)^b = a^{b+1} x^{b^2} \ , $$ so the requirements on $a$ and $b$ are $$ b-2 = b^2 \ \ \ \Rightarrow \ \ \ b = \frac{1\pm i\sqrt{7} }{2} $$ and $$ b(b-1) = a^b \ \ \ \Rightarrow \ \ \ a = (b(b-1))^{1/b} \ \ \ \Rightarrow \ \ \ a = (-2)^{1/b}$$

So this yields two solutions, which will have to be restricted to the complex $x$ plane with a cut to make sense of the non-integer exponents.

Answer 3

The solutions are:

$$\displaystyle f_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$ $$\displaystyle f_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$

The solution technique can be found in this paper.

For a general case, solution of the equation

$$f'(z)=f^{[m]}(z)$$

has the form

$$f(z)=\beta z^\gamma$$

where $\beta$ and $\gamma$ should be obtained from the system

$$\gamma^m=\gamma-1$$ $$\beta^{\gamma^{m-1}+...+\gamma}=\gamma$$

In your case $m=2$.