# Uniqueness of the vector in $\mathbb{R}^n$ specified by the curl, divergence and the normal component

If I know the curl, and divergence of a n-component vector in a region, and its normal component around its boundary, is the vector uniquely specified? If yes, how do I prove it? Also, is there a straightforward way to solve these equations, and get the vector?

$\partial_i V_i=s \,\,\,\,\, \partial_iV_j-\partial_jV_i=c_{ij}$

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mathematical pedantry: You know the curl and divergence of a n-component vector field. Please. –  dmckee May 30 '13 at 14:56
Though no doubt used in physics a lot, this is most certainly a pure math question (with a pure math answer). –  Chris White May 31 '13 at 1:53

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When you know the divergence and curl of a function, it's ripe for using the Green's function to solve. However, I think you're missing one piece of information.

The Green's function $G$ for the derivative $\nabla$ in free space is given by

$$G(r) = \frac{\hat r}{4\pi|r^2|} \implies \nabla \cdot G = \delta(r), \quad \nabla \wedge G = 0$$

Let me stop for a second here and tell you about a lovely thing called geometric calculus. It lets us unify the divergence $\nabla \cdot G$ and the "curl" $\nabla \wedge G$ into a single expression $\nabla G$. This is a useful thing for the following derivation. We still keep track of both the scalar divergence and the pseudovector curl as separate "components" of the expression.

We use the Green's function and the generalized Stokes theorem to reconstruct a vector field from boundary information and information about its divergence and curl. Let $V$ be your vector field such that $\nabla V = s + c$, where $c$ is the pseudovector curl and $s$ is the scalar divergence.

The result is then

$$\oint_{\partial M} |dS'| G(r-r') \hat n' V(r') = - V(r) + \int_M |dV'| G(r-r') \nabla' V(r')$$

where $G \hat n' V = G (\hat n' \cdot V) + G \cdot (\hat n' \wedge V)$ and $G \nabla' V = Gs + G \cdot c$

The bottom line: this means that, unfortunately, you still need $\hat n' \wedge V$, which uses tangential information about the vector field $V$, to fully reconstruct $V$ in a given volume.

Now, just using the volume integral, you will be able to reconstruct $V$ to within the addition of some divergence-free and curl-free function (which come from the surface integral).

If you do ignore the surface integral as I suggest, note that $G \cdot c = G^i c_{ij}$.

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+1 for geometric calculus! :) –  Mike May 30 '13 at 21:22
@Murphid: Please could you help me prove the uniqueness of the vector field? –  ramanujan_dirac Jun 3 '13 at 12:21
With only the information about the vector field's normal component, the decomposition is not unique, as I said. The normal component effectively specifies a scalar potential, but there is still a bivector potential that you have not specified (and must be specified with the tangential component). –  Muphrid Jun 4 '13 at 3:28