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The following question is asked without malicious intentions - it's not intended as a flamebait!

In my physics textbooks (Young & Freedman in particular) I have often seen derivations of equations that use multiplication of, say, $dx$ on both sides, and then integrating.

Taking courses both at the math and physics department I've noticed that real mathematicians often frown at this. Why is that?

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    Why, mathematicians do it too. They just prefer to understand [what](http://en.wikipedia.org/wiki/Differential_form) exactly they're doing before doing it.2011-06-23
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    Could you please give an example of a derivation in your textbook upon which you think "real mathematicians" frown? I can offer one explanation that is relevant: mathematicians often love to ensure that everything they do is completely rigorous. Although rigor alone is not mathematics, it is necessary. For example, a good number of students learn about the change of variables rule in integration without knowing the formal proof of this rule. And certainly many people do not stop and look up the hypotheses of the relevant theorem every time they apply the rule ...2011-06-23
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    ... However, I am fairly certain that many mathematicians would frown upon a student applying the change of variables rule all the time without really knowing that there is a theorem (with proof) behind the rule. Unfortunately, this is how one is taught about the rule in schools these days.2011-06-23
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    Hm, I have never seen mathematicians frown upon multiplying both sides of an equation by $dx$. However, I have seen a few frown upon "canceling" the $dx$ term in expressions like $$\frac{dy}{dx}\,dx.$$ See http://math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio.2011-06-23
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    I suppose, the question asks about [separation of variables](http://en.wikipedia.org/wiki/Separation_of_variables) method for solving PDE2011-06-23
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    I second Amitesh Datta, without a concrete example it is hard to guess what the physicists have done. There are several different contexts and explanations possible.2011-06-23
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    The true scandal is that physicist are doing it all the time, mathematicians when nobody is looking, but in 300 years of calculus have not brought up an accepted paradigm of handling these things in everyday situations.2011-06-23
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    @Grigory M, that's just plain old uconstructive.2011-06-23
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    @Amitesh Datta, this is shorthand from Y&F 12th ed. pp. 459-460. We are looking at pressure in a fluid with density $\rho$, $g$ is the gravitational acceleration, $+y$ is the "up"-direction. Carve out an infinitesimal element of the fluid with area $A$ and thicknes $dy$. We examine the forces acting upon this volume. Call the pressure on the bottom $p$ and the pressure on the top $p + dp$. As the fluid is in equilibrium, $pA - (p + dp)A - \rho g dy = 0$. Divide out the area, and rearrange to get $ dp/dy = -\rho g$. Integrate with respect to height and get $p_2 - p_1 = -\rho g (y_2 - y_1)$.2011-06-23
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    That is ... $−\rho g A dy = 0.$ $p$ is pressure and the element of area is taken to be perpendicular to the height.2011-06-24
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    http://math.stackexchange.com/questions/852394/failure-of-differential-notation/1651264#16512642016-02-27

2 Answers 2

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Suppose you have the equation $$f(x) = g(x).$$ You could "multiply both sides by $dx$" $$f(x) \, dx = g(x) \, dx$$ and then integrate over some interval $$\int_a^b f(x) \, dx = \int_a^b g(x) \, dx.$$ However, unless you have learned about differential forms, the second equation above is meaningless. This can easily be avoided by simply integrating both sides of $f(x) = g(x)$ with respect to $x$ just as you learned in calculus.

Now consider the differential equation $$f(y) \frac{dy}{dx} = g(x).$$ Here, many people will "multiply by $dx$" to get $$f(y) \, dy = g(x) \, dx$$ and then integrate both sides. Again, without the machinery of differential forms, this statement is meaningless. Here we can avoid this by integrating the original equation with respect to $x$: $$\int f(y) \frac{dy}{dx} \, dx = \int g(x) \, dx$$ or to use different notation, $$\int f(y(x))y'(x)\, dx = \int g(x) \, dx.$$ Then as long as certain hypotheses are satisfied, the change-of-variables (substitution) theorem says that this is just $$\int f(y) \, dy = \int g(x) \, dx$$ so we are back to where the first method lead us.

The point of this is that manipulations involving "multiplying by $dx$" are usually shorthand for some more "rigorous" method. However, if you are aware of (and comfortable with) the mathematics going on behind the scenes, then there is no loss of rigor. "Real mathematicians" don't frown upon using these kinds of manipulations themselves, they just frown upon students using them without knowing why they work.

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    Great answer so far. What exactly are those "certain hypotheses?"2011-06-23
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    If you are referring to the change of variables part, there are a few different COV theorems depending on the integral (Riemann, Lebesgue, etc.) and the types of functions you are dealing with. A weak version for the Riemann integral is: If $\phi\in C^1$ and strictly increasing on $[a,b]$, and if $f$ is integrable on $[\phi(a),\phi(b)]$, then $\int_{\phi(a)}^{\phi(b)} f(t) \, dt = \int_a^b f(\phi(x))\phi'(x) \, dx$.2011-06-24
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    What about the methods of "homogeneous equations" and "exact equations"? In these methods they often multiply by $dx$ to get to the "diferential form" of the DE. How can we understand this?2017-09-25
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Because it is considered as non rigorous. Calculating with infinitesimals is something physicists and mathematicians have done since at least the time of Fermat, even long before that. But it always felt awkward to work with infinitesimals as they seemed to lead to contradictory statements if not manipulated carefully.

In the 19th century, under the impulse of Cauchy, calculus started to be formalized and the calculus of infinitesimals was replaced by the more precise and rigorous $\epsilon-\delta$ methodology. Physicists didn't take much notice however since calculating with differentials still worked and was more practical.

In the 1960's however, Abraham Robinson developed a rigorous approach to manipulating infinitesimals now called non-standard analysis. At the time it was controversial, nowadays, I think it is more accepted. That doesn't mean physicists really care that much, but in a way, Robinson vindicated their approach. Although you still need some work to make things rigorous.

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    I don't quite see how that's relevant — differential forms have nothing to do with infinitesimals (in modern approach, at least)2011-06-23
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    I don't see how differential forms are involved with the OP? Because he is speaking about integration? Then I guess the theory of differential forms is another approach towards treating expressions with $dx$.2011-06-23
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    Even the title of OP mentions differential form ($dx$)!.. Well, frankly speaking, it's not very clear what OP is asking about. But multiplication both sides (of PDE, say) by $dx$ and integrating (almost verbatim from OP) makes perfect sense in the context of differential forms...2011-06-23
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    OK, he uses $dx$, but doesn't mention differential forms. But I agree that the question makes sense in the context of differential forms. I just thought it also makes sense from the point of view of the theory of infinitesimals, which is how $dx$ was first conceived by Leibniz.2011-06-23