Conditions for Conservative Vector Field?
Is it sufficient for a vector field to be conservative, to say that it’s image is simply connected (no holes) and the partials are equivalent?? If not then what can you do if you have a situation where you can’t test every possible loop or paths? Thanks!
On
A vector field $\mathbf F\in \mathcal C^1$ is said to be conservative if exists a scalar field $\varphi$ such that:
$$\mathbf{F}=\boldsymbol{\nabla}\varphi$$ $\varphi$ it is called a scalar potential for the field $\mathbf{F}$.
In general, a vector field does not always admit a scalar potential. A necessary condition for a field to be conservative is that the equalities are satisfied:
$${\frac {\partial F_{x}}{\partial y}(x,y,z)=\frac {\partial F_{y}}{\partial x}}(x,y,z),$$$$ {{\frac {\partial F_{y}}{\partial z}}(x, y,z)={\frac {\partial F_{z}}{\partial y}}(x,y,z),\, {\frac {\partial F_{x}}{\partial y}}(x, y,z)=\frac {\partial F_{y}}{\partial x}}(x,y,z)$$ or
$$\boldsymbol \nabla \times \mathbf F = \mathbf 0$$
One way is to find a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ whose gradient is that vector field. A vector field is conservative if and only if it is a gradient.