# Can a discontinuous vector field be conservative?

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.

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.