Functional Equation $f(f'(x))=-f(x)$

Question

Assume $f:\mathbb{R}_{>0}\to\mathbb{R}$ is differentiable and satisfies $\forall x>0:f(f'(x))=-f(x)$. What is $f(x)$?

I know that $f(x)=\ln x$ is a solution, but I don't know if there is another one, or how can I prove that.

Answer 1

It seems the following.

Differentiate both parts of the equation:

$$f’(f’(x))f’’(x)=-f’(x).$$

Then $$f’(f’(x))=-\frac{f’(x)}{ f’’(x)}.$$

Take $f$ of both parts:

$$f(x)=-f(f’(x))=f(f’(f’(x)))=f\left(-\frac{f’(x)}{ f’’(x)}\right).$$

Since the function $f$ is injective,

$$x=-\frac{f’(x)}{ f’’(x)}.$$

This is a differential equation. Solve it

$$xf''(x)+f'(x)=0$$

$$\bigl(xf'(x)\bigr)'=0$$

$$xf'(x)=C_{1}$$

$$f(x)=C_{1}\ln x+C_{2}$$ The verification:

$$f(f’(x))=C_1\ln\left(\frac {C_1}x\right)+C_2=-(C_1\ln (x)+C_2).$$

So $C_2=-\frac{C_1\ln C_1}2$ and

$$f(x)=\frac{\ln Kx}{K^2}.$$

PS. In general, we cannot a priori assume that $f$ is twice differentiable. For instance, consider a similar situation. Let $g:\Bbb R\to\Bbb R$ be a function such that $g(x)=x^2$ if $x\ge 0$ and $g(x)=-x^2$ if $x< 0$. Then $g’(x)=|x|$, $g(g’(x))=x^2$ is a differentiable function, but $g’’(0)$ does not exist. So a question remains open for a not twice differentiable function $f$.