Could we solve the vector value ODE in a approximation way? It seems I am too fast to ask the question without thinking carefully.
The equation I consider is find $w:\Omega \to R$ satisfying 
$$\frac{Du}{\sqrt{1+|\nabla u|^2}}=Dw. \tag{*}$$
Where $u:\Omega \to R$ is given and $u=0$ on $\partial \Omega$.
It is obvious this equation could not be solved for general $u$, the central obstacle is the Frobenius condition (integrable condition). But I still wish to get some more information about this equation, in particular my eager could divide into two parts: 


*

*Could the $u$ for which $(*)$ is solvable be dense in some suitable space, maybe $C^{2}(\Omega)$?

*For the $u$ which make $(*)$ locally solvable, what information of $w$ could we get from $u$?
Background: I am try to get some geometric explanation for certain PDE similar to mean curvature equation.
I will be appreciate to any related answer and remark.
 A: The Frobenius condition which is necessary and sufficient for local integrability is that $Du$ and $D(|\nabla u|^2)$ are parallel. So going back through my answer to this question you see that $u$ must be a function whose gradient descent curves are geodesics. 
This also tells you that such functions cannot be dense in $C^2$, answering in the negative your question (1). For example, if you start with a function like
$$ f(x,y) = x + y^2 $$
then $Df = (1,2y)$ and $D |\nabla f|^2 = (0, 8y)$ can never be made to be parallel with a small $C^2$ perturbation. 
Now, supposing we have a function $u$ such that $D u$ is parallel to $D |\nabla u|^2$. Define $v = |\nabla u|^2$. Then $v$ and $u$ have the same level sets, and thus away from the level sets of $u$ (where $v = 0$) we can write $v = v(u)$ (at least locally). 
Letting $F(s)$ be a real valued function such that $F'(s) = \frac{1}{\sqrt{1 + v^2(s)}}$, you see then that the function $F\circ u$ satisfies
$$ D(F\circ u) = F'(u) Du = \frac{Du}{\sqrt{1 + v^2(u)}} $$
and so away from the critical points of $u$ this gives you the solution to your equation (*).  
