1

Suppose I have a vector field $f: A \subset R^2 \to R^2 $. Write $f = (M,N) $. Does it follow this?

$$ \frac{ \partial M}{\partial y} = \frac{ \partial N}{ \partial x } \quad \iff \quad f \text{ is conservative}$$

Vincenzo Tibullo
  • 10,955

1 Answers1

0

No. It does not. If $A$ is the punctured plane then the vector field with $M = -y/(x^2+y^2)$ and $N = x/(x^2+y^2)$ is not conservative. This can be seen by integrating around any closed contour which contains the origin. It will give a nontrivial result depending on how many times the curve winds around the origin. On the other hand, if $A$ is simply connected then there is an equivalence between satisfying the criteria $M_y=N_x$ and $(M,N)$ being conservative.

James S. Cook
  • 16,755
  • what is the definition of simply connected? Also, I assume the same applies if I change $2$ with $n$, correct? –  May 03 '14 at 13:58
  • A space is simply connected if all loops can be continuously deformed to a point. Like a lasso shrinking, but never getting stuck on something. For $\mathbb{R}^3$ the same example works, but $A$ is missing all $(x,y,z)$ such that $x^2+y^2=0$ (the $z$-axis). Much more generally, this question eventually turns to the question of de Rahm cohomology where the condition $M_y=N_x$ is traded for the condition that a form $\alpha$ be closed $d\alpha=0$ and conservative amounts to asking if there exists $\beta$ for which $d\beta=\alpha$. – James S. Cook May 03 '14 at 14:31