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We know that given the divergence and curl of a vector field (and appropriate boundary conditions) it is possible to construct a unique vector field in $\mathbb R^3$. The specific problem I am thinking about is related to the PDE $$\operatorname{div} F = g,$$ where $F \colon \mathbb R^n \to \mathbb R^n$ is a vector field and $g \colon \mathbb R^n \to \mathbb R$ is a scalar field, and $\operatorname{div}$ is the $n$-dimensional generalization of the divergence given by $$\operatorname{div} F = \frac{dF_{i}}{dx_{i}}$$ (summation implied). What additional pieces of information are necessary to uniquely specify $F$ given the function $g$ (we know the answer is the curl of $F$ in 3D)?

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The appropriate generalization you want is to consider the vector field $F$ and the scalar field $g$ as $(n-1)-$ and $n$-forms in the deRham complex. That is $$ F \in \Omega^{n-1}(\mathbb{R}^n) = \Omega^{n-1} \otimes C^\infty(\mathbb{R}^n) , $$ $$F = f_1 \ dx_2 \wedge ... \wedge dx_n + f_2 \ dx_1 \wedge dx_3 \wedge ... \wedge dx_n + $$ $$ f_3 \ dx_1 \wedge dx_2 \wedge dx_4 \wedge ... \wedge dx_n + ...+ f_n \ dx_1 \wedge ... \wedge dx_{n-1} \wedge dx_n $$ and $$ g \in \Omega^n(\mathbb{R}^n) $$ $$ g = G \ dx_1 \wedge dx_2 \wedge ... \wedge dx_n $$

Framing it in these terms the equation $$ div( F) = g $$ is $$dF = g $$ where d is the external derivative $$ d : \Omega^j(\mathbb{R}^n) \rightarrow \Omega^{j+1}(\mathbb{R}^n) $$

Now your question becomes: given $ g$, what $F$ satisfy $dF = g$. The answer is certainly there is no unique function/n-1 form $F$ which satisfies this equation. Given any such solution $F_0$, then $F = F_0 + dE$ is also a solution where $E$ is a $(n-2)-$form, because $d^2 = 0$. Further, given any such $g$ there must exist at least one $F$ because $g$ is a closed $n$-form, meaning $dg = 0$ by the Poincare theorem, every closed $n$-form is also exact, meaning there exists such a $(n-1)-$form $F$ such that $dF = g$. (In the language of deRham cohomology, $H^n_{DR}(\mathbb{R}^n) = 0$.)

The more interesting question is this: what happens if we look now not at the deRham complex on $\mathbb{R}^n$ but on subsets of $\mathbb{R}^n$ that are topologically non-trivial? Or on $n$-dimensional manifolds? Are there then $n$-forms $g$ for which there are no $F$ such that $dF = g$?

In other words, the existence of closed $n$-forms that are not exact will depend on the topology of the underlying space, and vice versa. It's a great topic.

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+1: This is a nice explanation of the setting in which the higher-dimensional generalization of $\operatorname{div} F = g$ must be understood. But I don't see how it answers the question of what additional information is needed to uniquely determine $F$. Would the codifferential $\delta F = \pm*^{-1}d*F$ suffice, at least on $\mathbb R^n$? – Rahul Jan 18 '12 at 2:02
My point is there is no such unique F, because you can always add a member of the kernel of d, and the image of d from (n-2) forms provides you with a subset of the kernel from which to choose. – Simon S Jan 18 '12 at 13:06

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