I consider a closed curve within a 2-dimensional vector field, passing through two points $A$ and $B$. Going clockwise, along the path from $A$ to $B$ the field is constant $\vec{v}(\vec{x}) = \vec{k}$, and clockwise from $B$ to $A$ the field is $\vec{v}(\vec{x}) = \nabla \phi$ where $\phi$ is some scalar field. The vector field has discontinuities at both $A$ and $B$.
Both sections of the path are locally conservative, because within each I can represent the field as the gradient of a scalar field. However, the line integral $\oint \vec{v} \cdot d\vec{l}$around this whole loop can be non-zero, because of the discontinuities at $A$ and $B$ (i.e. $\vec{v}$ might be $\vec{0}$ clockwise from A to B but $3\hat{x}$ from $B$ to $A$).
I wondered then, can the vector field still be described as conservative? I am inclined to say no, and that the field is locally conservative in two regions but not globally conservative, although I am not sure if this is correct.
Best Answer
Another requirement for conservatism for a vector field $\mathbf{F}$ in a region $\Omega$ is that $\nabla \times \mathbf{F}=0$. If $\mathbf{F}$ has discontinuities in $\Omega$, this won't be true for all points.