Green's theorem exercise

I am trying to solve the following problem:

Show functions $P,Q:\mathbb R^2 \setminus \{(0,0)\} \to \mathbb R$ of class $C^1$ that verify $P_y=Q_x$ but $$\int_\gamma P(x,y)dy+Q(x,y)dy \neq 0$$ where $\gamma(t)=(\cos(t),\sin(t))$ with $0 \leq t \leq 2\pi$.

First I have some general doubts:

If the exercise asks to show that the integral is not $0$, this means that one cannot apply Green's theorem in the region enclosed by $\gamma$. I wonder if this is because this region is not simply connected, however in Tromba's textbook it only asks for the region to be simple (x-simple and y-simple region), from what I've said above, is it true that $D$ $x$ and $y$ simple $\implies$ $D$ is simply connected?

I couldn't think of two functions that satisfy the conditions asked, I would appreciate if someone could tell me how can I construct $P$ and $Q$.

As my other answer shows, the question is not well posed.

On the other hand, if, say, $(0,0)$ is excluded from the domain, then we have a shot at it, since greens theorem does not apply.

Consider the domain $D_1 = \{ (x,y) \in \mathbb{R}^2: x>0 \}$. This set could be described as the union of the quadrants $I$ and $IV$.

Define the function $\theta_1 : D_1 \to \mathbb{R}$ by the rule: $\theta_1(x,y)$ is the measure (in radians) of the angle formed by the positive $x$-axis and the ray passing through the origin and the point $(x,y)$.

$\theta_1=\arctan(y/x)$

The exterior derivative of $\theta_1$ is $\frac{-y}{x^2+y^2} dx + \frac{x}{x^2+y^2} dy$

Define the functions

1. $\theta_2$ on $D_2 = \{(x,y): y>0\}$ by $\theta_2(x,y) = \theta_1(-y,x)+\frac{\pi}{2}$
2. $\theta_3$ on $D_3 = \{(x,y): x<0\}$ by $\theta_3(x,y) = \theta_1(-x,-y)+\pi$
3. $\theta_4$ on $D_4 = \{(x,y): y<0\}$ by $\theta_4(x,y) = \theta_1(-y,x)+\frac{3\pi}{2}$

Notice that $\theta_1 = \theta_2$ on their common overlap, $\theta_2=\theta_3$ on their common overlap, $\theta_3 = \theta_4$ on their common overlap, but $\theta_4=\theta_1+2\pi$ on their common overlap. All of this work is just really to express that "$\theta$" is not a well defined function on the entire plane: it is only well defined "up to" a multiple of $2\pi$.

On the other hand, since a constant displacement does not effect the derivative, we have that $d\theta_4 = d\theta_1$ on their common overlap, despite the functions $\theta_4$ and $\theta_1$ not agreeing there.

Notice that no matter which of these $4$ functions we are looking at, we have $d\theta_i = \frac{-y}{x^2+y^2} dx + \frac{x}{x^2+y^2} dy$.

So even though there is no "global" function $\theta$ with this differential, we will call the one form $\frac{-y}{x^2+y^2} dx + \frac{x}{x^2+y^2} dy$, defined on $\mathbb{R}^2 - \{(0,0)\}$, $d\theta$. You should realize that this is a slight abuse of notation, but it is sensible, and very standard.

You can confirm that $d\theta$ is a closed one form by direct computation, but given all of the work we have done above, that would not be the most conceptual way to see that $d\theta$ is closed. At any point $(x,y) \in \mathbb{R}^2-\{(0,0)\}$, we can find at least one domain $D_i$ with $(x,y) \in D_i$. On this domain, $d\theta = d\theta_i$. So locally, around this point, $d\theta$ is exact. But exact forms are closed, so $d\theta$ is closed around this point.

Now to see that

$\int_\gamma \frac{-y}{x^2+y^2} dx + \frac{x}{x^2+y^2} dy \neq 0$, you can plug and chug. You will find that the answer is $2\pi$. More conceptually, integrating $d\theta$ along a path should tell you how much the angle $\theta$ changes along the path. In other words, $\frac{1}{2\pi} \int_\gamma d\theta$ is the winding number of $\gamma$ around the origin.

Instead of just plugging and chugging, you can also obtain this result by applying Green's theorem carefully in each of the domains $D_i$. I will leave this as a challenge to you.

• This answer is really helpful but how did you know to look at the angle function? – a student Jan 1 '15 at 1:00
• @astudent, really just because someone told me to a long time ago. It is the "classic" example of such a function. In hindsight, with all of the machinery you develop later, it is obvious to look at this example, but from the perspective of a beginner it is somewhat mysterious. From a certain perspective, you know you want a "multivalued function", and the angle function is the first example most people meet. – Steven Gubkin Jan 1 '15 at 3:32

What you ask is impossible. $\gamma$ parameterizes the unit circle, and the entire unit disk is contained in your domain, so Green's theorem does apply, and you can conclude that the integral is equal to $0$. Are you sure it was only the point $(1,2)$ which was excluded from the domain of these functions? If it were a point in the unit disk, then we could construct examples.

• You're right, I didn't even check that the point was inside the disk, I took it for granted because if not, as you've said, the theorem does apply. This was the only point excluded, I've just checked the exercise, I suppose they must have meant to exclude a point in the disk. I'll modify the problem so at least it makes sense. – user16924 Oct 15 '14 at 18:29
• If you exclude the origin then "$d\theta$" is a classic example of the sort of phenomena you are looking for. – Steven Gubkin Oct 15 '14 at 18:36
• I am not following your suggestion, how $P$ and $Q$ connect to "$d\theta$"? – user16924 Oct 15 '14 at 18:40
• I will add a new answer. – Steven Gubkin Oct 15 '14 at 18:44