2
$\begingroup$

Say I have these two things: $$ A(x,y)+B(x,y)\frac{dy}{dx}=0 $$ $$A(x,y)dx+B(x,y)dy=0$$

what is the difference? I always thought they were just different ways of expressing the same things, but recently I was told it wasn't quite like that. How come?

$\endgroup$
  • $\begingroup$ math.stackexchange.com/q/21199/431814 $\endgroup$ – aidangallagher4 Sep 29 '18 at 15:52
  • $\begingroup$ Ok, I read that, and also this one: math.stackexchange.com/a/23914/520539 . But I'm still having trouble translating it to equations $\endgroup$ – Bidon Sep 29 '18 at 16:14
  • $\begingroup$ I never understood this either, until I studied differential geometry. In my undergraduate courses, mostly designed for Engineers and scientists, it all seemed like symbol pushing to me. The link aidagallagher4 posted is a great resource, but if you really want to know how it all fits together $rigorously$, then working through basic diff geometry would be well worth the effort. $\endgroup$ – Matematleta Sep 29 '18 at 16:26
2
$\begingroup$

There is a lot of good information in the comments; let me see what I can do about hacking it together into an answer.

First, as suggested by the top answer in aidangallagher's link, you can treat these equations as exactly the same if you are reading both of them as equations over some suitable nonstandard extension of the real numbers.

Along the same lines, the top answer in OP's link explains that if you look at a historical perspective, these equations used to be read as exactly the same, because of the notion of a "differential" which was thought to be an extremely (i.e. arbitrarily) small real number. Nowadays, this interpretation is generally not considered rigorous.

The contemporary, non-nonstandard view on the matter is what Matematleta is hinting at in his two comments: that the two equations are extremely different, and only related because of a nontrivial theorem.

  • The first equation is "elementary" in the sense that it deals only with functions over the real numbers. Namely, $y$ is a function $\Bbb R\to\Bbb R$, so that $\frac{dy}{dx}:=y'$ is also a function $\Bbb R\to\Bbb R$. Then $A$ and $B$ are functions $\Bbb R^2\to\Bbb R$, and the symbol $A(x,y)$ means the function $\Bbb R\to\Bbb R$ sending $x\mapsto A(x,y(x))$. This takes a few tries to digest, in my experience.

  • The second equation is not elementary; it is a statement about differential forms. As Mathematleta suggests, the definition of a differential form involves a lot of machinery, and is best learned over the course of a month or so. (Alternatively, if you google "What is a differential form", there are no shortage of people trying to explain it in simple terms. I've always found these explanations unsatisfying, even before I knew the machinery.) Suffice it to say that there exist some exotic objects that we call "$dx$" and "$dy$", and:

    • These way these objects are constructed, it is possible to multiply them by any sufficiently nice functions $\Bbb R^2\to\Bbb R$, so the symbols $Adx$ and $Bdy$ make sense.
    • The symbol $0$ is also such an object because you can multiply either of these objects by the function that takes value 0 everywhere.
    • Finally, these types of objects can be added together, so $Adx+Bdy$ also makes sense.

    ${}$

    Then the equation is saying the differential form that this sum represents, is equal to the zero form.

Theorem: The equation $Adx+Bdy=0$ holds as differential forms if and only if for every "sufficiently nice" curve $t\mapsto (X(t), Y(t))$, we have $$A\big(X(t),Y(t)\big)\frac{dX}{dt} + B\big(X(t),Y(t)\big)\frac{dY}{dt} = 0.$$

Corollary: If $Adx+Bdy=0$, then $A(x,y)+B(x,y)\frac{dy}{dx}=0$. (Proof: Use the Theorem with $X(t)=t$ and $Y(t)=y(t)$, and then rename $t$ to $x$.)

In practice, you can take this theorem as the definition of a differential form. Under this interpretation, the second equation is just a lazy way to write $A\big(X(t),Y(t)\big)\frac{dX}{dt} + B\big(X(t),Y(t)\big)\frac{dY}{dt} = 0,$ which, despite being a bit of a mouthful, is at least elementary.


Finally, my own contribution: the second equation has a last interpretation. It is a common abuse of notation to write $dx$ instead of $\Delta x$ to mean "a small change in $x$". This makes the second equation elementary, but it has the very bad side effect that the two equations no longer contain the same information.

If you assume that the new second equation holds for all $\Delta x$ (which is reasonable), then it implies the first equation, but it is incredibly more restrictive: it tells you something not just about all tangent lines of $y$, but all secant lines of $y$!

On the other hand, if it only holds for some $\Delta x$, it doesn't say anything about tangent lines at all.

$\endgroup$
  • $\begingroup$ I am not an expert in geometry. If you are, please be swift and brutal about correcting this answer— since I actually quite like it and am planning on using it elsewhere. $\endgroup$ – aleph_two Oct 19 '18 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.