Proving the existence of a potential function $f$ if(f) $\partial_{xy}f=\partial_{yx}f$.

Question

Consider the first order ODE of the form $$ A(x,y)dx +B(x,y)dy = 0 \ . $$

It is considered exact if there exists a so-called potential function $f$, with $A := \frac{\partial f}{\partial x}$ and $B := \frac{\partial f}{\partial y}$.

I want to proof

$f$ exists if $\frac{\partial A }{\partial y} = \frac{\partial B }{\partial x}$.

I also want to see if the converse is true or not.

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I guess that we should able to show that $f$ can be uniquely constructed (up to a constant factor) by noticing something about its differential? $$ f = \int df = \int A dx + \int B dy \ . $$

We have $\partial_x B= \partial_y A $, if we require the second partial derivates of $f$ to be continuous.

But, if we start out by using $0 = df = Adx+Bdy$, we already assume the existence of $f$; so this doesn't seem the right approach.

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I'm looking for a hint.

Answer 1

Braun proves this, as a theorem, in his fourth edition of Differential Equations and Their Applications, along the following lines.

If $f$ exists such that $A=\partial_x f$ and $B=\partial_y f$, then we have $\partial_y A = \partial_x B$.

Observe that $A = \partial_x f$ for some function $f$, if, and only if, $$ f(x,y) = \int A(x,y)dx + h(y) \ . $$ Taking partial derivatives of both sides with respect to $y$, the left-hand side $h'(y)$ is a function of $y$ alone if, and only if, $$ \frac{\partial}{\partial t}\bigg[B-\int \partial_y A dt\bigg] = 0 \ . $$ From which we conclude that $\partial_y A = \partial_x B$.

Conversely,

if $\partial_y A = \partial_x B$, then we can solve for $h(y)$ in the previous equation from which we obtain $f(x,y)$, proving its existence.