Understanding exactness and finding a potential.$[e^x\cos(\pi y^2)+ay-1]dx+[by\; e^x\sin(\pi y^2)+x+y]dy$

Question

$$\underbrace{[e^x\cos(\pi y^2)+ay-1]}_{P}dx+\underbrace{[by\; e^x\sin(\pi y^2)+x+y]}_Q dy$$ is given

I want to check if the given equation is exact, has potential, conservative.

Question:

1. If I want to check exactness, I always begin with taking partial derivatives of $P_y$ and $Q_x$ and check whether they are equal or not. I guess if $P$ and $Q$ have continuous partials then we have conservaive and potential by Green's Theorem but what about exactness?
2. Without derivating $P$ and $Q$ just finding potential $\phi$ for the given equation how can I be sure about exactness?
3. What does exactness mean realy? Geometrically or intuitionaly ?

For $b=-2\pi$ and $a=1$ $$P_y=Q_x=-2\pi e^x\sin(\pi y^2+1)$$ Is it enough for exactness?

Note: I'm taking a course equivalent to vector calculus. Lecturer does not use a textbook and give us a refference book, he gives lectures in oldstyle. I want to learn the whole thing, any refference text would be appreciated.

Answer 1

$(e^x\cos(\pi y^2))dx=(d(e^x\cos(\pi y^2))+2\pi ye^x\sin(\pi y^2) dy$

Hence, for $ F(x,y)= e^x\cos(\pi y^2)$, $\omega = dF+(ay-1) dx+(x+y) dy+ (b+2\pi)ye^x\sin(\pi y^2) dy$,

and $\omega= dF+ ay dx+xdy+ d(-x+y^2/2)+(b+2\pi)ye^x\sin(\pi y^2) dy$

Then, for $G(x,y)=-x+y^2/2$ $\omega=dF+dG+d(xy)+ (a-1)y dx+ (b+2\pi)ye^x\sin(\pi y^2) dy$.

If one write the condition that $(a-1)y dx+ (b+2\pi)ye^x\sin(\pi y^2) dy$ is a potential, $a=1, b+2\pi=0$, and $\omega =d(F(x,y)+G(x,y)+xy)$

Answer 2

We say that a differential form $\omega$ is closed if $d\omega = 0$. We say that a differential form $\omega$ is exact if there exists a differential form $\eta$ such that $d\eta = \omega$. Notice that any exact form is closed, since $\omega = d\eta$ implies $d\omega = d(d\eta) = d^2 \eta = 0$, where we have used the fact that $d^2 = 0$ for all differential forms.

I don't know if I can provide geometric intuition, but I can provide a geometric analogy. Think about curves in the plane. Some curves are loops, i.e. they stop where they start. Every loop you can draw in the plane can be thought of as the boundary (edge) of some region (the region is just the area enclosed by the loop). But what if I require that no region contain the origin? Then there are loops you can draw, e.g. a circle around the origin, which are not the boundary of any allowed region (any such region would have to contain the origin). In this analogy, curves are like differential forms. Loops are like closed forms, i.e. forms $\omega$ for which $d\omega = 0$. Loops which can be thought of as the boundary of an allowed region are like exact forms, i.e. forms for which $\omega = d\eta$. Notice that the boundary of a region is always a loop, just like an exact form is always closed.

To see what's going on at the computational level, let's start by thinking about what it would mean for a 1-form $\omega = Pdx + Qdy$ to be closed. That means $d\omega = 0$, i.e. $d(Pdx + Qdy) = 0$. Hence $d(Pdx + Qdy) = P_y dy \wedge dx + Q_x dx \wedge dy = -P_y dx \wedge dy + Q_x dx \wedge dy = (Q_x - P_y) dx \wedge dy = 0$, which means $Q_x - P_y = 0$, i.e. $Q_x = P_y$. Thus, every closed 1-form $\omega = Pdx + Qdy$ satisfies $Q_x = P_y$. But we already said that every exact form is closed. Therefore, every exact 1-form $\omega = Pdx + Qdy$ satisfies $Q_x = P_y$.

But what about the other direction? Is it true that if a 1-form $\omega = Pdx + Qdy$ in 2 dimensions satisfies $P_y = Q_x$ then it is exact? The answer in general is no. As an example, consider the 1-form $$ \alpha = -\frac{ydx}{x^2 + y^2} + \frac{xdy}{x^2 + y^2} $$ We have $P = -\frac{y}{x^2 + y^2}$ and $Q = \frac{x}{x^2 + y^2}$. You can check that we do in fact have $P_y = Q_x$. But, it can be shown that there is no function $F$ such that $dF = \alpha$, i.e. $\alpha$ isn't exact. This happened because $\alpha$ isn't defined on the whole plane (it isn't defined at the origin). Think about the geometric picture. Once we removed a point from the plane, we were no longer able to think of a circle containing the origin as the boundary of an allowed region. The same thing has happened here. If we think about forms that aren't defined at the origin, like $\alpha$, then it can happen that $P_y = Q_x$ (the form is closed) but the form isn't exact.

The good news is that Poincare's lemma tells us that if we only think about forms defined on the whole plane, then a form being closed does in fact imply exactness. The proof of Poincare's lemma is beyond the scope of this answer, but there's a good proof in Spivak's Calculus on Manifolds. Since a form $\omega = Pdx + Qdy$ being closed is the same as the statement $P_y = Q_x$, we can conclude that it is enough to check that $P_y = Q_x$ to show that the form is exact. The $P$ and $Q$ in your question are defined on the whole plane, so it really is enough to check that $P_y = Q_x$.

The big idea here, which the geometric analogy hopefully made clear, is that the difference between closedness and exactness isn't a property of the form in question (provided the form is defined everywhere), but of the space in which it sits. In the geometric picture, whether or not a loop can be thought of as a boundary doesn't depend on the loop itself, but on the space in which the loop sits.

Finally, to make a physics connection: being exact, being conservative, and being derivable from a potential are all the same thing. Importantly, none of these are the same thing as being closed. But usually it's enough to check that something is closed to show that it's exact because it's rare in physics (especially undergraduate physics) to encounter a space where there's a difference between closed and exact forms. That said, the difference can become important, even in classical physics. One form of Maxwell's equations looks like $dF = 0, -\star d\star F = J$, where $F = dA$. The end of this sentence is the crucial point: by saying $F = dA$, we are saying that $F$ (known as the field strength tensor) is exact. But this isn't necessarily the case. In fact, if magnetic monopoles exist, then it isn't possible for $F$ to be exact, and Maxwell's equations take a slightly different form.

I strongly recommend Spivak's Calculus on Manifolds as a good introduction to some of the ideas you'll encounter in this course. If that book seems too difficult to start with, then I recommend Schey's Div, Grad, Curl and All That.