Is it possible to solve $f'(x)=f^{-1}(x)$?

Question

I am interested in proving what family of functions have the property $$f'(x)=f^{-1}(x)$$ I've never dealt with a differential equation of this form, hence I could only go as far as to gather a little data:

$$f'(x)=f^{-1}(x)\implies f(f'(x))=x$$ $$\implies f''(x)f'(f'(x))=1$$ $$\implies f'(f'(x))=\frac{1}{f''(x)}$$

Let $f'=g$

$$\implies g(g(x))=\frac{1}{g'(x)}$$

Is this generally solvable?

Any and all information would be much useful.

Answer 1

This is a partial answer.

In what follows we seek for a solution of the form $f(x)=ax^b$.

Then $f'(x)=abx^{b-1}$ while $f^{-1}(x)=(x/a)^{1/b}=a^{-1/b}x^{1/b}$.

So, setting, $f'(x)=f^{-1}(x)$, we obtain $$ abx^{b-1}=a^{-1/b}x^{1/b}, $$ in which case $b-1=1/b$ equivalently $b^2-b-1=0$ or $b=\frac{1}{2}(1 \pm \sqrt{5})$, while $ab=a^{-1/b}$ or $a^{b}b=1$ or choosing the positive value for $b$, $$ a=b^{-1/b}=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}. $$ Altogether, $$ f(x)=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}} x^{\frac{1 + \sqrt{5}}{2}} $$ possesses the required property: $f'=f^{-1}$.

Note that this function defines a global $C^1-$solution, if we define $x^b=|x|^{b-1}x$.

Answer 2

Well, we know that:

$$\text{y}\left(\text{y}^{-1}\left(x\right)\right)=x\tag1$$

Where $\text{y}^{-1}\left(x\right)$ is the inverse of the function $\text{y}\left(x\right)$.

So, using your problem we can write:

$$\text{y}'\left(x\right)=\text{y}^{-1}\left(x\right)\tag2$$

Note that a power of $x$ fits the bill for the differential equation, given in $(2)$. So, let's set:

$$\text{y}\left(x\right)=\text{A}x^\text{r}\tag3$$

We see that:

$$\text{y}'\left(x\right)=\text{r}\text{A}x^{\text{r}-1}\tag4$$

Now the inverse of the function is given by:

$$\text{y}^{-1}\left(x\right)=\left(\frac{x}{\text{A}}\right)^\frac{1}{\text{r}}=\left(\frac{1}{\text{A}}\right)^\frac{1}{\text{r}}\cdot x^\frac{1}{\text{r}}\tag5$$

So, we need to look at:

$$\text{r}\text{A}x^{\text{r}-1}=\left(\frac{1}{\text{A}}\right)^\frac{1}{\text{r}}\cdot x^\frac{1}{\text{r}}\space\Longleftrightarrow\space x^{\text{r}-1-\frac{1}{\text{r}}}=\frac{1}{\text{r}}\left(\frac{1}{\text{A}}\right)^{1+\frac{1}{\text{r}}}\tag6$$

Now, to finish note that the RHS is a constant, so the LHS is a constant which means that $\text{r}-1-\frac{1}{\text{r}}=0$, which means that $\text{r}=\frac{1\pm\sqrt{5}}{2}$. So the LHS gives $x^0=1$, which means that $\frac{1}{\text{r}}\left(\frac{1}{\text{A}}\right)^{1+\frac{1}{\text{r}}}=1$, and that gives $\text{A}=\left(\frac{2}{1+\sqrt{5}}\right)^\frac{2}{1+\sqrt{5}}$.

Concluding, we see that this is indeed true for $\text{r}=\frac{1+\sqrt{5}}{2}$ and $\text{A}=\left(\frac{2}{1+\sqrt{5}}\right)^\frac{2}{1+\sqrt{5}}$:

$$\text{y}\left(\text{y}'\left(x\right)\right)=\text{A}\left(\text{r}\text{A}x^{\text{r}-1}\right)^\text{r}=x\tag7$$

Note that when $\text{r}=\frac{1-\sqrt{5}}{2}$ there is no solution.