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I'm reading in Tenenbaum and Pollard's Ordinary Differential Equations where they introduce the concept of the differential. Suppose $y=f(x)$ is differentiable. He defines the differential by $dy(x, \Delta x)=f'(x)\Delta x$, and explains that if we think of $dx$ as the differential of the function $x\mapsto x$, then we can write $dy=f'(x)dx$, and this continues to hold when $x$ is a function of another variable, $t$. On page 51-52, he writes

The first-order differential equations we will study in this chapter can be written in the form $$Q(x,y)\frac{dy}{dx} + P(x,y)=0.\tag{6.6}$$ Written in this form, it is assumed that $x$ is the independent variable and $y$ is the dependent variable. If we multiply (6.6) by $dx$, it becomes $$P(x,y)dx + Q(x,y)dy=0.$$ Written in this form, either $x$ or $y$ may be considered as being the dependent variable. In both cases, however, $dy$ and $dx$ are differentials, and not increments.

I'm a little shaky with the notion of dividing by $dx$ in the first place. I guess $dy/dx$ has a removable singularity at $dx=0$ which we can fill in, giving us $(dy/dx)(x)=f'(x)$ for all $x$, whether $x$ depends on some other variable(s) or not. Is that the way I should think of it?

Another thing that concerns me is the potential switching of the dependency. If a solution $y(x)$ is not injective, how can we arbitrarily decide to think of $y$ as the dependent variable? Or perhaps we'll get singularities in our solution where $y'(x)=0$?

I have actually worked with differential forms on smooth manifolds before, so I'm happy with an answer where we think of them as smooth covector fields (here I guess $\Delta x$ is an element of the tangent space at $x$). I would feel wrong dividing by a smooth covector field unless I knew it was nonzero!

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  • $\begingroup$ Because you have worked on smooth manifolds, then you can assume you're working with differential forms, which should be rigorous to you. In this context, you're not dealing with increments. $\endgroup$ – DisintegratingByParts Feb 9 '14 at 0:28
  • $\begingroup$ @T.A.E. But the question is how can I justify dividing by a function which is sometimes zero. Should it be by an argument about removable singularities, as I mentioned above? $\endgroup$ – Eric Auld Feb 9 '14 at 0:49
  • $\begingroup$ When you consider this as a differential form, then you're not dividing by anything. $\endgroup$ – DisintegratingByParts Feb 9 '14 at 0:50
  • $\begingroup$ @T.A.E. I mean in the line $Q(x,y)(dy/dx)+P(x,y)=0 \implies Q(x,y)dy + P(x,y)dx=0$ we are multiplying and dividing by $dx$, right? $\endgroup$ – Eric Auld Feb 9 '14 at 0:53
  • $\begingroup$ @T.A.E. And more generally, what is the interpretation of the ratio of two differential forms? Suppose we have $Adx + Bdy + Cdz =0$. In what sense can the ratio of $dz/dx$ be the partial derivative of $z$ w.r.t. $x$ while $y$ is held constant? $\endgroup$ – Eric Auld Feb 9 '14 at 1:51
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Yeah, this seems like a common bit of voodoo. We can't really multiply and divide by differentials, but we can do something like this: imagine a vector-valued function $\ell(t)$, where $\ell: \mathbb R \mapsto \mathbb R^2$. Let $V: \mathbb R^2 \mapsto \mathbb R^2$ be a vector field such that $V(x,y) = P(x,y) \hat x + Q (x,y) \hat y$.

Clearly, if we integrate the $V$ on $\ell$, we can get

$$\int V \cdot \frac{d\ell}{dt} \, dt$$

If we write $\ell(t) = \bar x(t) \hat x + \bar y(t) \hat y$, we get

$$\int P(\bar x, \bar y) \bar x'(t) + Q(\bar x, \bar y) \bar y'(t) \, dt$$

It's still important, I think, to distinguish between the coordinates $x, y$ and the scalar functions $\bar x(t), \bar y(t)$.

Now, the paramterization is arbitrary. We can, for instance, choose as our parameterization $\bar x(t) = t$. Or rather, we can just use $x$ itself for the parameter, and as such, $\bar x' = 1$, so we get

$$\int P(x, \bar y(x)) + Q(x, \bar y(x)) \frac{d \bar y}{dx} \, dx$$

So all of this can be phrased in terms of a rigorous set of ideas. Usually, when integrating over curves, the distinction between a coordinate (like $y$) and a component function of the parameter (like $\bar y(t)$) is dropped completely. Usually, we can understand that this is exactly what's meant, and so it feels redundant, but I think from a pedantic perspective, it's helpful to maintain the distinction.

Overall, there's no need to "divide" any differentials; all we have here are notations for derivatives. Choosing to parameterize with respect to $x$ does have dangers: not when $dy/dx = 0$ (these are well-handled) but when the derivative does not exist (is infinite). Curves that might otherwise be smooth and differentiable can give problems when the derivative $dy/dx$ does not exist. Still, since the thrust of this topic is ODEs, such issues should seldom crop up.

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  • $\begingroup$ Thank you for the answer! Why can't we multiply by differentials? For instance, what is wrong with multiplying by $dx$, thought of as a function of $x$ and $\Delta x$, or perhaps of $t$ and $\Delta t$? $\endgroup$ – Eric Auld Feb 9 '14 at 0:04
  • $\begingroup$ Well, I guess I should leave issues about the algebra of differentials to people who deal with that kind of stuff regularly. I just see no need for them. In integrals, they don't mean anything; they just denote what variable to integrate over. In derivatives, $dy/dx$ notation doesn't need to be thought of as any kind of division. It's all very suggestive of the kinds of things you can do, but think of the chain rule: what looks like a "cancellation" of differentials is actually quite a bit more complicated. The notation just hides that, instead of exposing what's going on. $\endgroup$ – Muphrid Feb 9 '14 at 7:42
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Here's what is really behind the differential equation. If you can write the differential form as an exact differential form after multiplying by (mostly) non-zero $R(x,y)$, then $$ R(Qdx + Pdy) = df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y} dy = 0 $$ gives an algebraic equation $$ f(x,y) = C. $$ By the implicit function theorem, you can solve for $y=y(x)$ locally near $x=x_{0}$ and $y=y_{0}$ if $f(x_{0},y_{0})= C$ and if $$ \frac{\partial f}{\partial y}(x_{0},y_{0}) \ne 0. $$ In such a case, $$ \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}=0, $$ which gives $$ \frac{dy}{dx} = -\frac{\partial f/\partial x}{\partial f/\partial y}=-\frac{RQ}{RP}=-\frac{Q}{P}. $$ Over some regions, you'll solve for $y=y(x)$ and over other others you'll solve for $x=x(y)$. It depends on the derivatives $\partial f/\partial x$, $\partial f/\partial y$, which are propotional to $Q$, $P$, respectively. So the end result is that it appears you can just divide by $dy$ or $dx$ and solve. I think this makes it clear that you are not actually just dividing by $dx$ or $dy$, even though the final result makes it look that way.

BTW: Such ODEs are sometimes called exact differential equations because of the technique of turning the differential equation into an algebraic equation through the use of exact differential forms.

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