TITLE: If the Jacobian of two functions is zero, how are the two functions related?
QUESTION [5 upvotes]: Let $ x^{1} = f^{1}(u^{1},u^{2})$ and $x^{2} = f^{2}(u^{1},u^{2}) $. If the Jacobian of $f^{1}$ and $f^{2}$ is identically equal to zero (i.e. equal to 0 for all values of $u^1$ and $u^2$), why does this mean that there must be a functional relations between $x^{1}$ and $ x^{2}$ such that there exists a function $\phi$ such that $\phi(x^{1},x^{2}) = 0$ ?

REPLY [5 votes]: The Jacobian of two functions is $J(f^1, f^2) = \partial_x f^1 \partial_y f^2 - \partial_y f^1 \partial_x f^2$. This can be re-written as the dot product of the gradient of $f^1$ with the vector orthogonal to the gradient of $f^2$.
$0 = J(f^1, f^2) = (\partial_x f^1 , \partial_y f^1) \cdot (\partial_y f^2, -\partial_x f^2).$
So these two vectors are orthogonal over the whole domain, so the gradients of $f^1$ and $f^2$ are always parallel. Now consider a contour of constant $f^1$. It is constant because the gradient of $f^1$ is always perpendicular to this contour. But then the gradient of $f^2$ is also perpendicular to this contour. Hence $f^2$ cannot change value along this contour. Along this contour $f^2$ always takes the same value. Doing this for every value / contour of $f^1$ describes a function $q$ from the image of $f^1$ to the image of $f^2$. 
That is, we can express $f^2(u^1, u^2) = q(f^1(u^1, u^2))$ for some function $q$. 
I find the above the most useful result. But we could also express it as requested: $\phi(x^1, x^2) = x^2 - q(x^1) = 0$ for $x^1 = f^1(u^1, u^2), x^2 = f^2(u^1, u^2)$ for all $(u^1, u^2)$.