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

Solution 1:

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.

Solution 2:

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 $a=\phi$.

Solution 3:

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