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!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


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.