18

A student of mine made a silly mistake while taking derivatives: $$ \color{red}{\partial_x \sqrt{1-x}=\sqrt{\partial_x(1-x)}}=\sqrt{-1}=i. $$ This has led me to wonder about the problem of finding pairs of functions $f, g\in C^1(\mathbb{R})$ with $$ (f\circ g)'=f \circ g'. $$ So far I found the following pairs of solutions:

  • $f(0)=0$ and $g\equiv c$ constant.
  • $f(x)=a x$ is linear and $g\in C^1(\mathbb R)$ arbitrary.
  • Thanks to Moe: $f(x)=x^n$ and $g(x)= e^{n^{1/(n-1)}x}$ for $n\geq 2.$
  • For $a>0$, let $f(x)=e^{a x}$, and $g(x)=g'(x)-\frac 1 a \log(g'(x))-\frac 1 a \log(a)$ with $g'(x)\neq 0$. (Although I´m not sure if such $g$ exists.)

Are there other such pairs, or is this the full solution set? If the latter, can we characterize the set of solutions more precisely?

Idividedbyzero
  • 1,767
  • What about $f'=f$ ? – zkutch Aug 13 '25 at 11:03
  • @Prem this case is already known to the OP. – psl2Z Aug 13 '25 at 11:06
  • Another family would be $g\in C_c^1(\mathbb{R})$ and $f\in C^1(\mathbb{R})$ with $f=0$ on the range of $g$. – Severin Schraven Aug 13 '25 at 12:48
  • 1
    $f(x) = x^2$ with $g(x) = e^{2x}$ works. I just set $f=x^2$ and solved $(g^2)' = (g')^2$ – Moe Aug 13 '25 at 13:10
  • @SeverinSchraven I´m afraid that doesn't work, if $g'$ is not in the range of $g$, because: $$ \underbrace{f'(g(x))}_{=0}\cdot g'(x) \neq f(g'(x)) $$ Take e.g. $f(x)=\begin{cases}0, x\in (0, 1)\(x-1)^2, x \geq 1\x^2, x\leq 0\end{cases}$ and $g(x)=\exp(-x^2).$ – Idividedbyzero Aug 13 '25 at 13:16
  • 1
    @Moe You mean $a=n$? I´m afraid that only works for $n=2:$ $\frac{\partial f(g(x))}{\partial x} = \frac{n^2 \left(e^{\frac{n x}{n-1}}\right)^n}{n-1}$ and $f\left(\frac{\partial g(x)}{\partial x}\right) = \left(\frac{n e^{\frac{n x}{n-1}}}{n-1}\right)^n$ – Idividedbyzero Aug 13 '25 at 13:22
  • 2
    I am sorry. I meant $f(x) = x^a$ gives $g(x) = e^{a^{\frac{1}{a-1}}x}$ – Moe Aug 13 '25 at 13:28
  • Nice work Moe! Also $f(x)=x^n$ and $g(x)=c e^{-n^{\frac{1}{n-1}} x}$ seems to be working for odd $n\geq 3$. – Idividedbyzero Aug 13 '25 at 13:35
  • 2
    @Idividedbyzero Sure, then take $f=0$ on the union of the range of $g$ and $g'$ (they both compact anyways, which leaves you with plenty functions). – Severin Schraven Aug 13 '25 at 14:51
  • 3
    I'd say the equation is surprisingly rigid (not as in "hard to solve", but as in "hard to disturb a solution once you have one"). That being said, though, for any solution $(f,g)$ we have a family of solutions $(ax+f,g)$. – Ivan Neretin Aug 13 '25 at 18:00

1 Answers1

5

Given a function $f(x)$, if we can invert $f(x)/x$ or $f'(x)$ we can find (reduce to quadrature) the corresponding function $g(x)$ which satisfies \begin{align} f(g'(x))=f'(g(x))g'(x). \end{align}

Suppose that $f(x)/x$ has inverse $F(x)$, we can rearrange to get \begin{align} g'(x)=F(f'(g(x)))\quad\implies\quad \int\frac{\mathrm dg}{F(f'(g))}=x+C \end{align} For example: \begin{align} f(x)=x^n\quad&\implies\quad g(x)=C\exp\left(n^{1/(n-1)}x\right),\\\\ f(x)=x\log x\quad&\implies\quad g(x)=C\exp(ex), \end{align} and \begin{align} f(x)=xW(x)\quad \implies\quad \exp\left(\frac{2+W(g)}{1+W(g)}\right)(1+W(g))-\frac12 \mathrm{Ei}\left(\frac{2+W(g)}{1+W(g)}\right)+\frac{e^2}{2}\mathrm{Ei}\left(\frac{-W(g)}{1+W(g)}\right)-e\mathrm{Ei}\left(\frac{1}{1+W(g)}\right)=e^2x+C. \end{align}

If we can't find an inverse for $f(x)/x$ but $f'(x)$ is invertible, we can rearrange to get \begin{align} g(x)=f'^{-1}\left(\frac{f(g'(x))}{g'(x)}\right). \end{align} Let \begin{align} \varphi(x)=f'^{-1}\left(\frac{f(x)}{x}\right)\quad \implies\quad g(x)&=\varphi(g'(x)), \end{align} a d'Alembert ODE with general parametric solution \begin{align} g(t)=\varphi(t),\quad x(t)=\int\frac{\varphi'(t)}{t}\mathrm dt+C. \end{align} In terms of $f$ we have \begin{align} g(t)=f'^{-1}\left(\frac{f(t)}{t}\right),\quad x(t)=\int\frac{tf'(t)-f(t)}{t^3f''\left\{f'^{-1}\left(f(t)/t\right)\right\}}\mathrm dt+C, \end{align} I haven't been able to find a case with known quadrature that can't be solved using the first method.

If we instead consider $g(x)$ as a given function the problem is reducible to the form \begin{align} f'(\xi)=f(h(\xi));\quad\left[\xi=\int\frac{\mathrm dx}{g'(x)},\ \ h(\xi)=g'(g^{-1}(x(\xi)))\right], \end{align} a functional-differential equation with a couple of known cases, e.g. $h(\xi)=\xi$ (trivial case), $h(\xi)=\xi-\tau$ (delay differential equation), $h(\xi)=q\xi$ (pantograph equation), but in general this strikes me as a hopeless endeavor.

Eli Bartlett
  • 2,669