= Velocity reconstruction from a mixed-boundary vector Poisson problem
{title2=$\Delta A=-\nabla\times u,\quad\nabla\cdot A|_{\partial D}=0,\quad n\cdot\nabla\times A=n\cdot u\implies\nabla\times A=u$}
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 $\Delta\operatorname{div}A=0$ with zero <boundary> value, so harmonic uniqueness gives $\operatorname{div}A=0$. 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.
Back to article page
