Vector potential

A vector potential for a divergence-free vector field $F$ is a vector field $A$ satisfying $F=\nabla \times A$. It is not unique because adding a gradient leaves its curl unchanged.

Let $D$ be bounded, connected and simply connected, and let $u$ be a smooth divergence-free field. Suppose a sufficiently smooth vector field $A$ satisfies the displayed Poisson and boundary conditions. Divergence gives $\Delta \mathrm{div}A=0$ with zero boundary value, so harmonic uniqueness gives $\mathrm{div}A=0$. The vector Laplacian identity then implies $\nabla \times (\nabla \times A-u)=0$.

On a simply connected domain, define a scalar potential by integrating this curl-free difference along a path from a fixed point. Vanishing circulation around contractible loops proves path independence. Thus the difference is $\nabla \chi$; its divergence and boundary normal component give $\Delta \chi =0$ and ${\partial }_n\chi =0$. Green's identity yields ${\int }_D|\nabla \chi {|}^2=0$, proving the reconstruction. Additional conditions on $A$ can specify its own boundary problem; the argument assumes its existence and proves that its curl is exactly $u$. Simple connectivity is essential for eliminating possible non-gradient harmonic circulation fields.

Let $B$ be a continuously differentiable solenoidal vector field on an open star-shaped set containing the origin. Thenis a vector potential for $B$. For $D={\int }_0^{1}tB(tx)dt$, differentiation gives $\nabla \cdot D=0$ and $\nabla \times (D\times x)=2D+(x\cdot \nabla )D={\int }_0^1{\partial }_t(t^{2}B(tx))dt=B(x)$. Boundedness near the origin removes the lower endpoint. A field defined only on a punctured domain need not satisfy these hypotheses.