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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.

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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.

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    $\begingroup$ Thanks! So the field I described was not conservative over the whole domain since it was not continuous over the whole domain, whilst it is conservative over either of the restricted domains. $\endgroup$ – James Wirth Jun 3 at 13:51
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    $\begingroup$ Well said. It is exactly the same notion as piecewise continuity. $\endgroup$ – K.defaoite Jun 3 at 13:54

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