Is This Differential Equation Non-Exact in Non-Simply Connected Regions?

Click For Summary
SUMMARY

The differential equation \(\frac{-y}{x^{2}+y^{2}} + \frac{x}{x^{2}+y^{2}}y'=0\) is proven to be non-exact in the region \(\mathbb{R^{2}} \setminus \{(0,0)\}\) due to the non-simply connected nature of the domain. The discussion highlights that if a potential function \(F\) existed, the line integral \(\int_{L}\frac{-y}{x^{2}+y^{2}} \mbox{d}x + \frac{x}{x^{2}+y^{2}} \mbox{d}y\) would equal zero for every closed curve \(L\), which is not the case here. The failure of continuity of \(F\) near the origin further confirms the non-exactness of the equation.

PREREQUISITES
  • Understanding of differential equations and exactness
  • Familiarity with line integrals and their properties
  • Knowledge of simply connected and non-simply connected domains
  • Basic concepts of potential functions in multivariable calculus
NEXT STEPS
  • Study the properties of line integrals in non-simply connected domains
  • Learn about the implications of exact differential equations
  • Explore the concept of potential functions and their continuity
  • Investigate the role of the Arctan2 function in multivariable calculus
USEFUL FOR

Mathematicians, students studying differential equations, and anyone interested in the properties of non-exact equations in multivariable calculus.

Settembrini
Messages
5
Reaction score
0
I try to show, that equation
\frac{-y}{ x^{2}+y ^{2} } + \frac{x}{ x^{2}+y ^{2}}y'=0
is not exact in \mathbb{R^{2}} \setminus \{(0,0)\}.
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
\frac{ \partial F}{ \partial x}=\frac{-y}{ x^{2}+y ^{2} } and \frac{ \partial F}{ \partial y}=\frac{x}{ x^{2}+y ^{2}}
Existence of such function is equivalent to \int_{L}\frac{-y}{ x^{2}+y ^{2} } \mbox{d}x +\frac{x}{ x^{2}+y ^{2}} \mbox{d}y=0 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.
 
consider
(y/x)'
or
(Arctan(y/x))'
 
I know how to solve this equation and how to find F in any "regular" domain, for example in real plane \mathbb{R^{2}}. 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)

\lim_{(x,y) \rightarrow (0,0)} F(x,y)

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
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
5K