If the Jacobian of two functions is zero, how are the two functions related? 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 $\phi(x^{1},x^{2}) = 0$ ?
 A: 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 $(u^1, u^2)$. 
A: The functions $f^1$ and $f^2$ define a map into $\mathbb R^2$. For sufficiently smooth maps (continuously differentiable is enough), the Jacobian is identically zero if and only if the image has area zero. The image can still be pretty rough, since the derivative is allowed to vanish. For example, a polygonal tree with infinitely many branches can be the image of a line segment (hence, of a planar domain) under a $C^1$ map. 
When does there exist $\phi$ that vanishes on the image of $f$, and only there? A silly answer is: always, just let $\phi=0$ on the image of $f$, and $1$ elsewhere. If the continuity of $\phi$ is required, then the image of $f$ must be closed. And this is the only restriction: for every closed subset $A\subset \mathbb R$ there is a $C^\infty$ smooth function $\phi:\mathbb R^2\to\mathbb R$ such that $\phi =0$ on $A$ and only there.
So, I think you need to clarify what you really want from the "functional relation" concept.
