Suppose we are in $\mathbb R^3$ and let $\nu\in S^2$ be fixed. We have a regular function $u:\mathbb R^3\to \mathbb R$ with the following property:
- $u$ is decreasing along the integral lines of the constant vector field $\nu$, i.e. $\frac{\partial u}{\partial \nu}\leq 0$;
- $u$ is constant along the integral lines of any constant vector field $\nu'$ orthogonal to $\nu$, that is $\frac{\partial u}{\partial \nu'}= 0$.
We want to show that there exists a regular function $f:\mathbb R\to \mathbb R$ such that $f'\leq 0$ and $u(P)=f(P\cdot \nu)$.
I think this is done if one can show that:$$\forall P,P'\in \mathbb R^n: P\cdot \nu=P'\cdot \nu \implies u(P)=u(P') \quad (*).$$ I've seen that $(*)$ is true in the easier case where $\nu=(0,0,1)$, for example, because we can connect two points $P=(x,y,z), P'=(x',y',z)$ following the integral lines of the vector fields $(1,0,0)$ and $(0,1,0)$, where $u$ is constant by hypothesis. I think we can do something similar with $\nu$ general too, so my idea would be to connect $P,P'$ using integral lines of vector fields orthogonal to $\nu$, but I get stuck because I can't formalize the argument.
Could anyone help me, please?
If you (as you claimed) know how this is done for $\nu=(1,0,0)$ you are done, since, by an orthogonal transformation, you can always choose your coordinate system in such a way that $\nu$ takes this form, and the problem is, quite obviously, invariant under this kind of transformation.