# ODE with inverse function. Solve $f^{-1}(x)=f'(x)$.

Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that $$f^{-1}(x)=f'(x)$$

I think one such function would be of the form $f(x)=ax^b$. But then $b$ would be irrational and when $x\lt0$ this causes problems. So I guess letting $f(x)=-a(-x)^b$ for $x<0$ might work. But that's only one possibble solution, what are all the solutions?

Edit: With the comment of Joey Zou, the domain and range has been changed to $(0,\infty)$.

Edit: I already know the solution of the form $ax^b$. However, what I'm really asking is whether it is the unique solution.

• Hint: Use the Inverse Function Theorem – Moya Sep 6 '16 at 4:33
• @Moya Can you please elaborate on that? – Colescu Sep 6 '16 at 4:47
• If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable and $f'$ takes on both positive and negative values, then $f$ can't be invertible. So if an invertible differentiable function $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfied $f^{-1}(x) = f'(x)$, then $f'$ must take on both positive and negative values as the range of $f^{-1}$ is all of $\mathbb{R}$. This is a contradiction. If you change your domain and range to $f:(0,\infty)\rightarrow(0,\infty)$ you can circumvent this issue. – Joey Zou Sep 6 '16 at 4:56
• @JoeyZou I see. Let's change the domain and range to $(0,\infty)$. – Colescu Sep 6 '16 at 5:11
• See this question at MO: mathoverflow.net/questions/34052/function-satisfying-f-1-f/… – Christian Blatter Sep 6 '16 at 8:02

Partial answer, in order to confirm the Omnomnomnom's assumption that some solutions on the form $f(x)=ax^b$ do exist.

$$f^{-1}(x)=f'(x)\qquad \text{on }x>0$$ Search of particular solutions on the form $\quad f(x)=ax^b$ :

$f=ax^b \quad\to\quad x=\left(\frac{f}{a}\right)^{1/b}$

Hence the inverse fonction of $f(x)$ expressed as a function of $x$ is : $$f^{-1}(x)=\left(\frac{x}{a}\right)^{1/b}$$ to be not confused with $x$ expressed as a function of $f$.

If a solution of the form $f(x)=ax^b$ exists, it must satisfy the equation : $$f^{-1}(x)=f'(x)=\left(\frac{x}{a}\right)^{1/b}=abx^{b-1}$$ First condition : $x^{1/b}=x^{b-1}\quad\to\quad b^2-b-1=0$ $$b=\frac{1\pm\sqrt{5}}{2}$$ Second condition : $\left(\frac{1}{a}\right)^{1/b}=ab \quad\to\quad a=b^{-\frac{b+1}{b}}$

So, two different particular solutions are obtained :

First : $\quad f(x)=\left(\frac{2\:x}{1+\sqrt{5}}\right)^{\frac{1+\sqrt{5}}{2}}$

Second : $\quad f(x)=\left(\frac{2\:x}{1-\sqrt{5}}\right)^{\frac{1-\sqrt{5}}{2}}$

• How do you deal with the irrational power of $-1$ in your second partial answer? – Colescu Sep 6 '16 at 9:20
• OK. If restricted to $x>0$ AND $f(x)$ real, reject the second solution. If restricted to $x<0$ AND $f(x)$ real, reject the first solution. If $f(x)$ not restricted to real (so, can be complex), both solutions are convenient. – JJacquelin Sep 6 '16 at 9:28

Make the substitution $y = f(x)$ to rewrite this equation as $$y = f'(f(y))$$ Rewrite this as $$y\,f'(y) = f'(f(y)) \,f'(y)$$ Now, integrate both sides with respect to $y$. Using integration by parts, note that $$\int y\,f'(y) \,dy = y\,f(y) - \int f'(y)\,dy = (y - 1)f(y) + C$$ Using substitution, note that $$\int f'(f(y)) \,f'(y)\,dy = \int f'(u)\,du = f(u) + C = f(f(y)) + C$$ Conclude that any $f$ satisfying our original equation must satisfy $$(y - 1)f(y) = f(f(y)) + C$$ In terms of $x$, this says $$(f^{-1}(x) - 1)x = f(x) + C$$ This isn't quite a solution, but at least we don't have any derivatives here.

Things to be said from here:

Note that with $y = 1$, we have $$0 \cdot f(1) = f(f(1)) + C \implies C = -f(f(1))$$ Note also that if $f(y) = 0$ for any $y$, that $y$ must satisfy $$f(f(y)) = f(f(1))$$ but if $f$ is "invertible" and therefore $1$ to $1$, then this should never happen.

• I think you goofed on your integration by parts. – Mike Sep 6 '16 at 7:29