To show that a vector field $F (R^2)$ is conservative, is it sufficient to say that $(dQ/dx) - (dP/dy) = 0$?

Question

If i have a vector field

$$ F = (y,x) $$ is it then possible to say that $$ dQ/dx - dP/dy =0 $$ to be certain that F is conservative? Is this then related to Greens formula somehow?

Answer 1

The vector field $F=(y,x)$ is the gradient of the function $(x,y) \to xy$, so it is conservative. In more fancy terms, the 1-form $f=y dx+x dy$ is exact since it holds $f=d(xy)$. In general, it is not sufficient to check that $\partial_y F_x -\partial_x F_y=0$ the domain over which the field is defined should be simply connected, i.e. there should not be "holes" in it. The celebrated counterexample is $F=(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$. Writing the associated 1-form: $\omega=F_x dx+F_y dy$ in polar coordinates gives $d\theta$ which is not exact, despite the poor notation, since $\int_{S^1} \omega=\int_0^{2\pi} d\theta=2\pi$.