When is a matrix function the Jacobian matrix of another mapping Suppose $J(x)$ is a continuous matrix function $\mathbb{R}^D \to \mathbb{R}^D \times \mathbb{R}^D$. Do there always exist a mapping $f: \mathbb{R}^D \to \mathbb{R}^D$ so that $J = \nabla f$. If not, are there well-known conditions such that this mapping exists?
 A: Let $J_{1}, \ldots, J_{D}$ denote the columns of $J$. Then each
$J_{i}:\mathbb{R}^{D}\rightarrow\mathbb{R}^{D}$ and so you are trying to find
functions $f_{i}:\mathbb{R}^{D}\rightarrow\mathbb{R}$ such that $J_{i}=\nabla
f_{i}$ for every $i=1, \ldots, D$. To construct counter-examples is easy. If
$J$ is $C^{1}$ and not just continuous, since the domain is $\mathbb{R}^{D}$,
then a necessary and sufficient condition for each $J_{i}$ to be the gradient
of a function is that $J_{i}$ is irrotational, that is, $\frac{\partial
J_{i,j}}{\partial x_{k}}=\frac{\partial J_{i,k}}{\partial x_{j}}$ for all $j$,
$k$, where $J_{i}=(J_{i,1},\ldots,J_{i,D})$. In $\mathbb{R}^{2}$ take
$J_{1}(x,y)=(y,2x)$ and anything you want for $J_{2}$. Then $\frac{\partial
}{\partial y}(y)=1\neq\frac{\partial}{\partial x}(2x)=2$, and so $J_{1}$ is
not irrotational. 
If $J$ is just continuous, then a necessary and sufficient condition for each
$J_{i}$ to be the gradient of a function is that $\int_{\gamma}J_{i}=0$ for
every closed curve $\gamma$. This is not so easy to use because you have to check every closed curve, but if you find one for which the integral is nonzero, then you immediately know that $J_{i}$ cannot be the gradient of a function.
You can find all this stuff in Fleming "Functions of several variables". Look
for exact differential forms.
