# Fixed point of curvature operator

Let $$y=f(x)$$ the graph of a real-valued function. We define its curvature by : $$curv(f) = \frac{|f''|}{(1+(f')^2)^{3/2}}$$

I would like to know if there is any function (apart from the trivial anwser $$f(x)=0$$) whose curvature is itself. So what is the fixed point of $$curv$$ ?

• What have you tried? Surely you have some thoughts, partial results? Jan 31 at 22:09
• So it would seem (just watch the positive ones). But it looks the the maximal solutions cease to exist at some point; although I suspect it might just be graphical rendering.
– user239203
Jan 31 at 22:15
• I found some very messy closed form solutions with Wolfram Alpha. Jan 31 at 22:16

If, as @Gae. S. did, we try to solve the equation

$$y'' - y (1 + [y']^2)^{\frac 32}=0$$ I suppose that there is no explicit solution.

However, switching variables, the equation $$y \left(1+\frac{1}{[x']^2}\right)^{\frac 32}+\frac{x''}{[x']^3}=0$$ shows explicit solutions in terms of elliptic integrals.

Multiply both sides by $$2f'$$ to get $$2ff'=\frac{2f''f'}{(1+(f')^2)^{3/2}}$$ which integrates to $$f^2=a-2(1+(f')^2)^{-1/2}$$ Rearranging and substituting $$u=f$$,$$\int\frac{1}{\sqrt{4(a-u^2)^{-2}-1}}\,du=x-x_0$$ Perhaps $$y=u^2-a$$, $$F(y):=\frac{1}{2}\int\sqrt{\frac{y^2}{(4-y^2)(y+a)}}dy$$ so that $$f(x)=\sqrt{2(F^{-1}(x-x_0)+a)}$$ This explains Leibovici's answer in using inverse functions and elliptic functions.