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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.

Thank you for your assistance.

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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 $2\pi$.

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Thanks for your help! I don't know what $arg$ means, could you please explain? – ro44 Jun 23 '12 at 20:26
I suppose the entire argument about B is kind of unclear to me--how does B being annular come into play? Is there a more elementary way to do this? – ro44 Jun 23 '12 at 20:32
@ro44: See my edit. The nonexistence of a continuous polar angle in the punctured plane is one of the most basic facts of all of mathematics and can not be described in a "more elementary way". – Christian Blatter Jun 24 '12 at 8:59

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