I am having a bit of a confusion with some claims I keep finding on a book of Fluid Dynamics. Let's say we have two functions in 3-D space, $f(\mathbf{x})$ and $g(\mathbf{x})$, with the following constraint:
$\nabla f \times \nabla g = 0$
From my understanding of mathematics (I'm not mathematician) is that the level sets of both functions coincide. What I don't understand is why can I make the claim that I can write one function in terms of the other, that is:
$f = f(g)$
Thanks in advance
I think I've understood your question, correct me If I'm wrong. You have two scalar fields $f,g$ and know that $$\nabla f \times\nabla g = \mathbf 0$$ and you want to show that there is some function $h$ such that $f(\mathbf x) = h(g(\mathbf x))$ for all $\mathbf x.$
The first thing is that the condition on the gradients gives you that $$\nabla f = \lambda \nabla g,$$ where $\lambda$ is some scalar field. In fact if $\lambda(\mathbf x) = h'(g(\mathbf x))$ then we are done. I think however, that there are some extra conditions you have missed as your assertion is false in general.
Here is a counterexample: Define scalar fields $f,g$ as: \begin{align} f(x,y,z)&=x+y+z\\ g(x,y,z)&=(x+y+z)^2 \end{align} Then we can find \begin{align} \nabla f &=\begin{pmatrix}1\\1\\1\end{pmatrix}\\ \nabla g &=2(x+y+z)\begin{pmatrix}1\\1\\1\end{pmatrix} \end{align} Then certainly $\nabla f \times\nabla g=\mathbf 0$ but we clearly can't have $f$ as a function of $g$ as then we would have $f(1,1,1) = f(-1,-1,-1)$ as $g(1,1,1)=g(-1,-1,-1)$ which clearly isn't true.