# A certain nonlinear second order ODE

I am trying to find a solution to the following ODE for $f:(0,1)\to \mathbb{R}$ $$\sqrt{1+(f')^2}\frac{\partial}{\partial x} \left(\frac{f'}{\sqrt{1+(f')^2}}\right)=c$$ for some constant $c>0$. Equivalently, $$\lambda f''-(f')^2-1=0$$ for $\lambda=1/c$.

How is this equation related to the minimal surface equation $$\mathrm{div}\left(\frac{\nabla F}{1+|\nabla F|^2}\right)=0$$ for a function $F:\Omega\to \mathbb{R}$ with $\Omega\subset \mathbb{R}^{n-1}$? Minimal surfaces have zero mean curvature. Does the above ODE describe constant mean curvature curves?

• If you remove the $\sqrt{1+(f')^2}$ at the front, then the graph of $f$ is a plane curve of constant curvature $c$. Feb 16, 2018 at 4:32
• @rpf : I don't agree with the equation : $\lambda f''-(1+(f')^2)^2=0$. Feb 16, 2018 at 9:35
• @JJacquelin: Thanks for pointing out my error. I have made the correction.
– rpf
Feb 16, 2018 at 15:08

$$\sqrt{1+(f')^2}\frac{d}{dx} \left(\frac{f'}{\sqrt{1+(f')^2}}\right)=c \tag 1$$ Let $f'(x)=g(x)\quad\text{thus :}\quad \sqrt{1+g^2}\frac{d}{dx} \left(\frac{g}{\sqrt{1+g^2}}\right)=c$ $$\frac{d}{dx} \left(\frac{g}{\sqrt{1+g^2}}\right)=\frac{g'}{\sqrt{1+g^2}}-\frac{(2gg')g}{2(1+g^2)^{3/2}}$$ After simplification, Eq.$(1)$ is transformed to :

$$g'-\frac{g^2}{1+g^2}g'=c$$ $$\frac{1}{1+g^2}g'=c$$ $$\int \frac{g'}{1+g^2}=\tan^{-1}(g)=cx+\text{constant}$$ $$g(x)=\tan(cx+c_1)$$ $$f(x)=\int \tan(cx+c_1)dx = -\frac{1}{c}\ln\left|\cos(cx+c_1)\right|+c_2$$

• Hi Jean ! Good to see you here. Feb 16, 2018 at 9:39
• Jean : you won ! I was typing the same ! Feb 16, 2018 at 10:03
• @Claude : The moral of this story is that never one can trust the preliminary calculus from the OP. Feb 16, 2018 at 10:10

This is not an answer but it is too long for a comment.

I think that getting the analytical solution of equation $$\lambda f''-(1+(f')^2)^2=0$$ could be difficult (not to say impossible - at least to me).

First, let $g=f'$ to get $$\lambda g'-(1+g^2)^2=0$$ which is separable leading to $$\frac 1 \lambda x +k=\frac{1}{2} \left(\frac{g}{1+g^2}+\tan ^{-1}(g)\right)$$ that is to say $$\frac 1 \lambda x +k=\frac{1}{2} \left(\frac{f'}{1+(f')^2}+\tan ^{-1}(f')\right)$$

How could we get $f'$ ?

Have you tried numerical integration ?

By the way, do you have the initial conditions ?

• Hi Claude! I fully agree with your solution of $\lambda f''-(1+(f')^2)^2=0$. But I am afraid, rpf made a mistake and this is not the correct transformed equation. Cheers! Feb 16, 2018 at 9:31