Say $V$ is a smooth vector field on $\mathbb{R}^n\setminus 0$. I was wondering whether it is always possible to find a field $W = \nabla u$ such that $V + W$ is essentially bounded (I am assuming everything to be smooth here, for simplicity). I guess this is not true, for example I suspect that for the field $V = \frac{x^{\perp}}{\left|x\right|^2}$ such a $u$ doesn't exist, even though I am not able to prove it rigorously! Any hint?
Gauge vector field to a bounded one
0
$\begingroup$
analysis
multivariable-calculus
1 Answers
0
Hint Over a circle $C_{\epsilon}$ of radius $\epsilon > 0$ centered at the origin and oriented anticlockwise, computing directly gives that the vortex vector field $${\bf V} = \left(-\frac{y}{x^2 + y^2} , \frac{x}{x^2 + y^2}\right)$$ you give satsifies $$\oint_{C_{\epsilon}} {\bf V} \cdot d{\bf s} = 2 \pi .$$ Now, what is the value of the path integral $$\oint_{C_{\epsilon}} ({\bf V} + {\bf W}) \cdot d{\bf s} = \oint_{C_{\epsilon}} ({\bf V} + \nabla u) \cdot d{\bf s} ?$$ And what can we say about the value of $$\oint_{C_{\epsilon}} {\bf X} \cdot d{\bf s}$$ for (essentially) bounded $\bf X$ as $\epsilon \to 0$?