# Find necessary and sufficient conditions for these functions to have vector potentials

I've encountered this exercise in my textbook, which I'm not really sure how to approach. Help would be appreciated!

Let $a,b,c,d$ be constants, and $P(x,y)=(ax+by)/(x^2+y^2), Q(x,y)=(cx+dy)/(x^2+y^2)$. We define a vector field $F=(P,Q)$ and $A=\{(x,y)|(x-3)^2+(y-2)^2<1\}, B=\{(x,y)|1<x^2+y^2<2\}$. What are necessary and sufficient conditions on the constants (a,b,c,d) for F to have a vector potential in A? What about B?

Edit: since A is a "star set" (not sure what the English term is) we can find necessary and sufficient conditions by comparing the (x and y) derivatives of P and Q. I'm not sure about B though.

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If ${\bf F}(x,y)=\bigl(P(x,y),Q(x,y)\bigl)$ is a vector field in a domain $\Omega\subset{\mathbb R}^2$ then a necessary condition for the existence of a potential of ${\bf F}$, i.e., a function $f:\Omega\to{\mathbb R}$ with $\nabla f={\bf F}$, is that $${\rm rot}\,{\bf F}:={\partial Q\over\partial x}-{\partial P\over\partial y}\ \equiv\ 0$$ in $\Omega$. In your case this amounts to $${(b+c)(y^2-x^2)+2(a-d)x y\over (x^2+y^2)^2}\ \equiv\ 0\ ,$$ which implies the conditions $a=d$, $b=-c$. It follows that ${\bf F}$ has to be of the form $${\bf F}(x,y)=a\bigl({x\over x^2+y^2},{y\over x^2+y^2}\bigr) + c\bigl({-y\over x^2+y^2},{x\over x^2+y^2}\bigr)\ .\qquad(*)$$ Here the first summand is $a\,\nabla g(x,y)$ for the potential $g(x,y):={1\over2}\log(x^2+y^2)$ which is unproblematic in both your domains $A$ and $B$. It follows that $a$ may take any real value.
The second summand in $(*)$ is nothing else but $\ c\,\nabla{\rm arg}(x,y)$. The symbol ${\rm arg}$ denotes the polar angle in the $(x,y)$-plane. This "function" of $x$ and $y$ is only defined up to multiples of $2\pi$ and has no continuous representant in all of ${\mathbb R}^2\setminus\{(0,0)\}$, but it has a well defined gradient $\nabla{\rm arg}(x,y)=\bigl({-y\over x^2+y^2},{x\over x^2+y^2}\bigr)$. It is possible to choose a nice representant of ${\rm arg}$ in the domain $\Omega:=A$, e.g., the principal value $${\rm Arg}(x,y):=\arctan{y\over x}\qquad(x>0)\ ,$$ but not in the domain $B$, which is an annular domain around $(0,0)$. It follows that $c$ may take any value if $\Omega:=A$, and has to be $=0$, if $\Omega:=B$.
To put it another way: If you integrate the field ${\bf A}(x,y):=\bigl({-y\over x^2+y^2},{x\over x^2+y^2}\bigr)$ around the circle $\gamma$ of radius ${3\over2}$ in $B$ you don't get $0$ (as you should for a field with a potential), but you get the total increment of the polar angle along $\gamma$, which is $2\pi$.
Thanks for your help! I don't know what $arg$ means, could you please explain? – ro44 Jun 23 '12 at 20:26