Showing a differential equation is non-exact on R² minus the origin

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 5K views
Settembrini
Messages
5
Reaction score
0
I try to show, that equation
[itex]\frac{-y}{ x^{2}+y ^{2} } + \frac{x}{ x^{2}+y ^{2}}y'=0[/itex]
is not exact in [itex]\mathbb{R^{2}} \setminus \{(0,0)\}[/itex].
It's obvious that I have to use the fact, that the set is not simply connected, but I don't know how to do it.
 
Physics news on Phys.org
I would try this:
Assume that it is exact and that there is a potential function F. Integrate its derivative (as given by your problem) in a circle around (0,0). If such a function F exists, the result has to be 0.
 
I'm not sure, if we can use the line integral here. We are trying to show, that there doesn't exist function F such that F is exact differential, that is
[tex]\frac{ \partial F}{ \partial x}=\frac{-y}{ x^{2}+y ^{2} }[/tex] and [tex]\frac{ \partial F}{ \partial y}=\frac{x}{ x^{2}+y ^{2}}[/tex]
Existence of such function is equivalent to [tex]\int_{L}\frac{-y}{ x^{2}+y ^{2} } \mbox{d}x +\frac{x}{ x^{2}+y ^{2}} \mbox{d}y=0[/tex] for every closed curve L. Although, this statement is true only if the domain is simply connected; and our domain isn't of this kind.
 
I know how to solve this equation and how to find F in any "regular" domain, for example in real plane [tex]\mathbb{R^{2}}[/tex]. Problems appear in the neighbourhood of point (0,0) in our domain, because all methods of solving this kind of equation, I know are valid only in simply connected domain.
 
I do not know why you could not use integrals.

http://en.wikipedia.org/wiki/Atan2

Suppose
F=C+Arctan2(y,x)
F'=0
but consider the nonexistence of the limit (by inequality of directional limits)

[itex]\lim_{(x,y) \rightarrow (0,0)} F(x,y)[/itex]

F cannot be continuous and the equation is not exact

Note that if we had excluded a path to infinity along with 0 we would have a simply connected region and an exact equation