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This is an excerpt from my textbook: Consider the general differential containing two variables, where $f = f(x,y)$, $$ d f=A(x, y) d x+B(x, y) d y $$ We see that $$ \frac{\partial f}{\partial x}=A(x, y), \quad \frac{\partial f}{\partial y}=B(x, y) $$ and, using the property $f_{x y}=f_{y x},$ we therefore require $$ \frac{\partial A}{\partial y}=\frac{\partial B}{\partial x} $$ This is in fact both a necessary and a sufficient condition for the differential to be exact.

I see why this is a necessary condition, but why is it a sufficient condition?

Arctic Char
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Albert
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  • In general it is not sufficient. Example – Arctic Char Aug 24 '20 at 10:53
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    It is sufficient if the domain is simply connected. You can compute a value of $f$ via line integrals, the condition makes sure that the integral value does not depend on the path, only on the end points. This is only valid if all paths in the domain connecting two points are homotopic. – Lutz Lehmann Aug 24 '20 at 11:01
  • First impression you can take from https://en.wikipedia.org/wiki/Exact_differential and then Tom M. Apostol - Calculus, Volume II from page 346. – zkutch Aug 24 '20 at 11:35

2 Answers2

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Assume that $A,B $ and its first order partial derivatives are continuous on a simply connected open set $D$.

Given $\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$, if there exists a function $h(x,y)$ such that $d(h(x,y))=Adx+Bdy\tag{A}$, then we are done.
Let's denote $h(x,y)$ by $h$.
Consider, $\frac{\partial h }{\partial x}=A$ and $\frac{\partial h }{\partial y}=B$. Let $(a,b)$ and $(x,y)\in D$
From $\frac{\partial h }{\partial x}=A$, we have : $h=\int_{x=a}^{x} A\partial x+g(y)\tag{1}$
Therefore, by $\frac{\partial h }{\partial y}=B$, we get $\frac{\partial }{\partial y}(\int_{x=a}^{x} A\partial x)+g'(y)=B\implies \int_{x=a}^{x}\frac{\partial }{\partial y}A \partial x+g'(y)=B\implies g'(y)=B-\frac{\partial }{\partial x}(\int_{x=a}^{x}B\partial x)=B-B(x,y)+B(a,y)=B(a,y)\tag{2}$


So, we have now shown that $g'(y)$ is free of $x$, that is we can find $g(y)$ from $(2)$ using FTC. $g(y)=g(b) +\int_{y=b}^{y} g'(y) dy=g(b) +\int_{y=b}^{y} B(a,y) dy$. So now $g(y)$ is known.
We'll put this $g(y)$ into $(1)$ and we'll have known $h$. And clearly the way $h$ was constructed implies that $(A)$ is satisfied by $h$.
Koro
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  • So, where did you use the fact $D$ is simply connected? – peek-a-boo Aug 24 '20 at 13:31
  • @peek-a-boo: Please note $(1)$ above. The conditions on $D$ ensure that the integral doesn't change depending upon the path. – Koro Aug 24 '20 at 16:43
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    What I'm saying is that you need to make that much clearer in your argument, because the way you currently phrased it it's unclear where you're using the hypotheses. Also, it's not clear which path you're referring to, because none of the integrals make explicit the path you're choosing. – peek-a-boo Aug 24 '20 at 16:58
  • @peek-a-boo: I have revised the answer. – Koro Aug 24 '20 at 20:44
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I think the misunderstanding here comes from the language used, and not so much the differential calculus (might want to tag this as "logic" or something). This is a good resource to understand what necessary and sufficient mean in logic, separately and together.

The most useful thing you can do to better understand these is think of counter examples: what is something that is necessary but not sufficient? A good example from that page is the following.

P: Oxygen exists in the atmosphere. Q: Humans exist.

Clearly P is necessary for Q. Having oxygen in the earth's atmosphere is a necessary condition for human life. Crucially, though, having oxygen will not guarantee human life – there are many other conditions needed for human life other than oxygen in the atmosphere. In this way, P is necessary but not sufficient for Q.

Now consider this example.

P: All men are mortal. Q: Socrates was mortal.

In this case, P being true always means Q is true (we all know Socrates was a man). There is no possible way P could be true without Q being true. Equally, Q couldn't be true without P being true. P is necessary and sufficient for Q.

TLDR:

  • P necessary and sufficient for Q = P $\Leftrightarrow$ Q,
  • P necessary but not sufficient for Q = Q $\Rightarrow$ P,
  • P not necessary but sufficient = logically invalid.

What your textbook is saying, then, is that that condition implies necessarily the exactness of the differential.