Proof of Exact differential

Question

I'm thinking of proving the exact differential equation, but don't know if I understand it quite right.

Let's say we have $\alpha(x,y) = A(x,y) \mathrm{d}x + B(x,y)\mathrm{d}y$ and know that a function $f$ it's called exact if $\mathrm{d}F = \frac{\partial F}{\partial x} \mathrm{d}x + \frac{\partial F}{\partial y} \mathrm{d}y$, so how do I prove that $\alpha$ is only exact if and only if $\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$.

My idea: Let's start by some followings from this expression $\mathrm{d}F = \frac{\partial F}{\partial x} \mathrm{d}x + \frac{\partial F}{\partial y} \mathrm{d}y$: \begin{align*} \mathrm{dd}F&= \frac{\partial \mathrm{d} F}{\partial x} \mathrm{d}x + \frac{\partial \mathrm{d} F}{\partial y} \mathrm{d}y = \frac{\partial^{2} F}{\partial x^{2}} \mathrm{d}x^{2} + \frac{\partial^{2}F}{\partial x \partial y}\mathrm{d}x \mathrm{d}y + \frac{\partial^{2}F}{\partial y \partial x} \mathrm{d}x \mathrm{d}y + \frac{\partial^{2}F}{\partial y^{2}} \mathrm{d}y \\ &= \frac{\partial^{2} F}{\partial x^{2}} \mathrm{d}x^{2} + 2 \frac{\partial^{2}F}{\partial x \partial y}\mathrm{d}x \mathrm{d}y + \frac{\partial^{2}F}{\partial y^{2}} \mathrm{d}y \end{align*} Now this expression can only be equal to $\mathrm{d}\alpha$ if $\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$ is satisfied, because \begin{align*} \mathrm{d} \alpha =& \frac{\partial^{2} A}{\partial x^{2}} \mathrm{d}x^{2} + \frac{\partial^{2}A}{\partial x \partial y}\mathrm{d}x \mathrm{d}y + \frac{\partial^{2}B}{\partial x \partial y} \mathrm{d}x \mathrm{d}y + \frac{\partial^{2}B}{\partial y^{2}} \mathrm{d}y \, . \end{align*} Is that correct?

Please help me solving that, because I think I'm running in circles.

Thanks.

Answer 1

There are two issues here:

The first is, the claim is not true! For example, computing directly shows that $$\alpha := \frac{-y \,dx + x \,dy}{x^2 + y^2}$$ satisfies $$d\alpha = 0 ,$$ but there is so $F$ such that $\alpha = dF$. If there were, then by the Fundamental Theorem of Calculus for line integrals the integral of $\alpha$ along the unit circle $\Bbb S^1$ (oriented anticlockwise) would be $\oint_{\Bbb S^1} \alpha = \oint_{\Bbb S^1} dF = 0$, but parameterizing $\Bbb S^1$ and computing directly gives $\oint_{\Bbb S^1} \alpha = 2 \pi$, a contradiction.

The claim is true locally, however: If $d \alpha = 0$, then around any point $x$ in the domain of $\alpha$ there is a neighborhood $U$ and a function $F : U \to \Bbb R$ such that $\alpha\vert_U = dF$, and in fact, we can take $U$ to be any simply connected subset of the domain of $\alpha$ containing $x$.

The second problem is that there are some evident mechanical misconceptions. First, if $A \,dx + B \,dy$ is exact, we have $$ \begin{align} 0 &= d(A \,dx + B \,dy) \\ &= dA \wedge dx + dB \wedge dy \\ &= \left(\frac{\partial A}{\partial x} dx + \frac{\partial A}{\partial y} dy\right) \wedge dx + \left(\frac{\partial B}{\partial x} dx + \frac{\partial B}{\partial y} dy\right) \wedge dy \end{align} $$ The antisymmetry of $\wedge$ gives that $dx \wedge dx = dx \wedge dy = 0$ and $dx \wedge dy = - dy \wedge dx$, so we can write the above equation as $$0 = C(x, y) \,dx \wedge dy$$ for some function $C$ given in terms of $A$ and $B$---what is it?---and hence conclude that $C(x, y) = 0$. (Note that this computation does not depend on the domain of $A \,dx + B \,dy$, so exactness implies that $\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$, but again, as we say above, the reverse in general requires a topological restriction.)

Answer 2

Well, you were nearly there at one point, then you wandered off.

If $A \text{d} x + B \text{d} y$ is exact, then, as you correctly say, $A = \frac{\partial f}{\partial x}$ and $B = \frac{\partial f}{\partial y}$ for some function $f(x,y)$.

Now just have another look at the thing you want to prove: $\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$, and the solution should be obvious.