I have trouble reconciling these two statements.
1) It is known that if $\nabla \times\mathbf{v}(\mathbf{x}) = 0$ on $D$, a simply connected domain, then there exists a scalar $\phi(\mathbf{x})$ such that $\mathbf{v}(\mathbf{x}) = \nabla \phi(\mathbf{x})$.
2) On the other hand, Helmholtz decomposition theorem on a finite domain states that we can decompose $\mathbf{v}(\mathbf{x})$ as $\mathbf{v}(\mathbf{x})=-\nabla \phi+\nabla\times\mathbf{A}$.
\begin{align} \phi(\mathbf{x}) & =\frac 1 {4\pi} \int_V \frac{\nabla'\cdot\mathbf{v} (\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} \, \mathrm{d}V' -\frac 1 {4\pi} \oint_S \mathbf{\hat{n}}' \cdot \frac{\mathbf{v} (\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} \, \mathrm{d}S' \\[8pt] \mathbf{A}(\mathbf{x}) & =\frac 1 {4\pi} \int_V \frac{\nabla' \times \mathbf{v}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} \, \mathrm{d}V' -\frac 1 {4\pi} \oint_S \mathbf{\hat{n}}'\times\frac{\mathbf{v} (\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} \, \mathrm{d}S' \end{align} If we apply Helmholtz decomposition to vector field $\mathbf{v}(\mathbf{x})$ such that $\nabla \times\mathbf{v}(\mathbf{x}) = 0$, then we have $\mathbf{A}(\mathbf{x})= -\frac 1 {4\pi} \oint_S \mathbf{\hat{n}}'\times\frac{\mathbf{v} (\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} \, \mathrm{d}S'$. From Helmholtz theorem perspective, it seem to me that we need an extra boundary BC in order to make $\mathbf{A}(\mathbf{x})$ vanishes. If this is the case, then we can represent irrotational vector field as a gradiant of some scalar.
This extra boundary BC, however, to my knowledge, is not needed in the proof of the first statement. My question is that how do I reconcile these statements. Of course, if the domain $D$ is unbounded,the surface term is zero.