First order differential equation involving inverses

My question is to find the solutions to the following

$\frac{df(x)}{dx} = f^{-1} (x)$

where $f^{-1} (x)$ refers to the inverse of the function f. The domain really isn't important, though I am interested in either (-inf, inf) or (0, inf), so if any solutions are known for more restricted domains then they are welcome.

I cannot find any material relating to this type of question in any of my calculus and differential equations textbooks and references; it seems quite unorthodox. Any material which covers this type of diff equation would be wlecome

• Let $g(x)=f^{-1}(x)$, then $$g'(x)=f'(g(x))^{-1}=g(g(x))^{-1},$$ As such, it connects values and derivatives of the function $g$ at two different locations and is thus loosely related to delay-differential equations. – Lutz Lehmann Sep 16 '15 at 21:34
• @LutzL To be clear for readers, the term $g(g(x))^{-1}$ is the reciprocal of $g(g(x))$, and not the inverse of $g(g(x))$. – Mark Viola Sep 16 '15 at 21:37
• @LutzL Interesting comparison (will have to read up on delay differential equations). – Cicero Sep 16 '15 at 23:23

Look for solutions of the form $f(x)=A\,x^a$. The equation reads then $$A\,a\,x^{a-1}=\Bigl(\frac{x}{A}\Bigr)^{1/a}.$$ You can find $A$ and $a$ fron here.

• And the Golden ratio appears magically again! – Mark Viola Sep 16 '15 at 22:06
• Thanks for the interesting example (the golden ratio does appear!) – Cicero Sep 16 '15 at 23:22
• Is this solution unique (by a variation of the uniqueness theorem for certain differential eqs)? – Cicero Sep 16 '15 at 23:26
• This answer assumes a form without proof that it's the only possible form. Are there other forms that might satisfy this diff eq? – Jerry Guern Sep 17 '15 at 6:38
• I do not know if there are other solutions or if there is uniqueness. – Julián Aguirre Sep 17 '15 at 8:57

Here is a partial solution using power series composition.

To solve $$f(f'(x))=x\tag{1}$$ take the derivative of $(1)$ $$f'(f'(x))\,f''(x)=1\tag{2}$$ If we can find a point $a\gt0$ so that $f'(a)=a$, we can expand $f'(a+x)$ around $x=0$. Let $$g(x)=f'(a+x)-a\tag{3}$$ Applying $f'(a+x)=a+g(x)$ twice gives \begin{align} f'(f'(a+x)) &=f'(a+g(x))\\ &=a+g(g(x))\tag{4} \end{align} Differentiating $(3)$ yields $$g'(x)=f''(a+x)\tag{5}$$ Plugging $(4)$ and $(5)$ into $(2)$, we get $$g'(x)=\frac1{a+g(g(x))}\tag{6}$$ Iterating $(6)$ gives \begin{align} g(x) &=\frac xa-\frac{x^2}{2a^4}+\frac{(1+3a)x^3}{6a^8}-\frac{(1+3a+10a^2+15a^3)x^4}{24a^{13}}\\ &+\frac{\left(1+3a+10a^2+30a^3+55a^4+105a^5+105a^6\right)x^5}{120a^{19}}+\dots\tag{7} \end{align} Therefore, \begin{align} f(x) &=a+a(x-a)+\frac{(x-a)^2}{2a}-\frac{(x-a)^3}{6a^4}+\frac{(1+3a)(x-a)^4}{24a^8}\\[6pt] &-\frac{(1+3a+10a^2+15a^3)(x-a)^5}{120a^{13}}\\ &+\frac{\left(1+3a+10a^2+30a^3+55a^4+105a^5+105a^6\right)(x-a)^6}{720a^{19}}+\dots\tag{8} \end{align}

Comparison with Julián Aguirre's Answer \begin{align} f(x) &=\phi^{-1/\phi}x^\phi\\ &=\phi+\phi(x-\phi)+\frac{\phi-1}2(x-\phi)^2-\frac{5-3\phi}6(x-\phi)^3+\frac{18\phi-29}{24}(x-\phi)^4\\ &-\frac{217-134\phi}{120}(x-\phi)^5+\frac{1219\phi-1972}{720}(x-\phi)^6+\dots\tag{9} \end{align} which matches $(8)$ with $a=\phi$.

It's a bit contrived, but you could take a power series approach. First, write the original ODE as $$f(f'(x)) = x. \tag{1}$$ Then, assume a solution to $(1)$ exists, and has a power series expansion around $x=0$. That is, we write $$f(x) = \sum_{n=0}^\infty a_n x^n\quad\text{and}\quad g(x):= f'(x) = \sum_{m=0}^\infty (m+1)\, a_{m} x^{m}.$$ Then, using Faà di Bruno's formula, we can write the power series of the composition $$f(g(x)) = \sum_{n=0}^\infty c_n x^n,$$ where $c_n$ is expressed in terms of $a_n$ and $n$ through combinatorial expressions. Now, using $(1)$, we see that $$c_0 = 0,\; c_1 = 1,\;\text{and}\; c_n = 0\;\text{for}\;n\geq 2.$$ This yields a system of equations for $a_n$. Again, it's a bit ugly, but might give some insight.