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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?

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    This is a really good question to ask, however I don't believe an easy answer exists. In fact, it has been shown by M. Gromov and Y. Eliashberg that the [existence of mapping with prescribed jacobian satisfies the $h$-principle](http://link.springer.com/article/10.1007%2FBF01075646). – 2017-02-07

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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.