Uniquely solving for a curl- and divergence-free 3D vector field What boundary conditions do I need to uniquely determine a vector field in 3D space, given that the field has zero curl and zero divergence?
I tried to start with an $xy$-plane that was curl-free in the $z$ component, but then I realized that this overdetermines the solution, since the evolution of the field in the $z$-direction can give the field a non-zero curl.
 A: Let $G(\vec r,\vec r')=\frac{1}{4\pi |\vec r-\vec r'|}$.  Helmholtz's Theorem states that a twice continuously differentiable vector field $\vec A(\vec r)$ can be written 
$$\begin{align}
\vec A(\vec r)&=-\nabla \left(\int_D  G(\vec r, \vec r')\,\nabla'\cdot\vec A(\vec r')\,dV' -\oint_{\partial D}G(\vec r;\vec r')\,\hat n'\cdot \vec A(\vec r')\,dS'\right)\\\\
&+\nabla \times\left(\int_D  G(\vec r, \vec r')\,\nabla'\times\vec A(\vec r')\,dV' -\oint_{\partial D}G(\vec r;\vec r')\,\hat n'\times \vec A(\vec r')\,dS'\right) \tag1
\end{align}$$
Therefore, if both $\nabla \cdot \vec A(\vec r)$ and $\nabla \times \vec A(\vec r)$, $\vec r\in D$ are given , then $\vec A(\vec r)$ is uniquely determined by specification of its normal and tangential components on the boundary $\partial D$.
In the special case for which $\nabla \cdot \vec A(\vec r)=0$ and $\nabla \times \vec A(\vec r)=0$, we see that 
$$\begin{align}
\vec A(\vec r)&=\nabla \oint_{\partial D}G(\vec r;\vec r')\,\hat n'\cdot \vec A(\vec r')\,dS'-\nabla \times\oint_{\partial D}G(\vec r;\vec r')\,\hat n'\times \vec A(\vec r')\,dS' \\\\
&= \oint_{\partial D}(\hat n'\cdot \vec A(\vec r'))\nabla G(\vec r;\vec r')\,dS'+ \oint_{\partial D}(\hat n'\times \vec A(\vec r'))\times \nabla G(\vec r;\vec r')\,dS'
\end{align}$$
