What conditions are necessary or sufficient for this manipulation of differentials to be justified?

Question

This comes from a problem given in "Physics", Halliday-Resnick-Krane, Chapter 2, Problem 55.

It asks to study a non-uniformly accelerated motion defined by $ a(t)=-3v(t)^2 $ and derive a numerical value for the time elapsed given the initial and final velocity (the initial velocity is $1.5$ and the final velocity is $0.75$).

This becomes a differential equation, $ v'(t)=-3v(t)^2 $.

Since, of course, first-year (standard) calculus doesn't provide, as far as I know, tools to solve this, after trying in vain on my own I looked for solutions on the web, and I found the following procedure in a physics forum:

$$ \frac{dv}{dt} = -3v^2 \ \implies \ \frac{dv}{v^2}=-3dt \\ \implies \int_{v_0}^v\frac{dv}{v^2}=\int_0^t(-3dt) \implies \frac{1}{v_0}-\frac{1}{v}=-3t \ .$$ The numerical result stemming from this procedure is in perfect agreement with the numerical value given by the textbook itself (the textbook gives $0.2222$, my calculator gives $0. \bar 2 $).

There are quite a lot of things that I don't understand here.

There's the infamous multiplication by $dt$ and consequent cancellation of it in the LHS. What assumptions need to be made to justify this, if it is even possible?

Apart from that, in the same passage they also divide by $v^2$. Is this only justified in this case because they know that both the initial and final velocity are greater than zero and the velocity is strictly decreasing? If not, how?

In the next passage, they integrate the LHS as if $v$ was a variable, and I'm not sure if or how the variable substitution theorem and/or the chain rule can apply in this specific context, since we're coming from an expression (between the first and second passage) that either doesn't make sense or presents differential forms, which I haven't studied yet.

In general, is there a theorem that somehow justifies this notational manipulations?, And in what conditions would such a theorem apply?

Answer 1

In the specific case of separable ODEs, the standard manipulation of differentials is easy to justify. If you have

$$\frac{dy}{dx}=f(x) g(y)$$

you try to write

$$\frac{dy}{g(y)}=f(x) dx$$

and then integrate both sides to get

$$\int \frac{1}{g(y)} dy = \int f(x) dx.$$

A clue that things are weird is that these are indefinite integrals over different variables. These should really be definite integrals over the same variable if this equality is going to make sense. And they can be: the usual integration by substitution, which is justified by the fundamental theorem of calculus and the chain rule, tells us

$$\int_{y(a)}^{y(b)} \frac{1}{g(y)} dy = \int_a^b \frac{1}{g(y(x))} \frac{dy}{dx} dx.$$

Thus the interpretation of these steps is going from the initial ODE directly to the integral equation

$$\int_a^b \frac{1}{g(y(x))} \frac{dy}{dx} dx = \int_a^b f(x) dx$$

which can be entirely justified without separating differentials.

Something very similar works for exact equations. Note that this method is already nonsensical if there is some $x \in (a,b)$ with $g(y(x))=0$. In this case if the solution is unique then it must be constant; separation of variables "destroys" such solutions.

Answer 2

I have an example which may help clarify some things. Bear in mind, my perspective a bit more physics-oriented (and in practice, physicist rarely worry about the things you bring up here). I personally have always taken the treatment of differentials as fractions for granted, so I won't discuss that part of your question: I can't provide rigorous mathematical insight into it. But when I read your question, this example stood out in my memory:

Find all functions $v(t)$, continuous and differentiable on $\mathbb R$, such that $v'(t) = 2v/t$ everywhere, and $v(-1) = 2$, $v(1) = 3$.

Using the same technique as the one you present in your question, one gets $$ \frac{dv}{v} = \frac{2dt}{t}$$ If we pick $t=1$, $v(1) = 3$ as a base point we can integrate to get $$ \ln|v|-\ln {3} = 2 \ln|t| $$ $$ |v| = 3t^2 $$ If we naively say $v = 3t^2$, we certainly cannot satisfy the other condition, $v(-1) = 2$. We could say that we have to be careful in how we choose the sign in removing the modulus. Continuity of $v$ requires that we choose the positive sign for any $t > 0$ because $v(1) = +3$. But what if we try $v = -3t^2$ for $t < 0$? The function is still continuous and differentiable everywhere, but now we can have $v(-1) \neq v(1)$. Still can't get $v(-1) = 2$ though. That's because we started "being careful" too late.

The division to get $dv/v$ on the left is only allowed if $v \neq 0$. Same is true for the division by $t$, in fact. This should have been the first clue that something funny will be going on. The differential equation in the problem cannot possibly be satisfied everywhere, because it is meaningless at $t = 0$. The problem should have said "for every $t \neq 0$." I made the omission intentionally, to better set up the "trap."

And another (related) problem is the integration: if our integrals extend up to or across $t = 0$ and $v = 0$, they are, strictly speaking, improper integrals. Those are defined through limits, and if we evaluate them separately, there is no guarantee that the limit-taking will be consistent between them and the equality will be preserved. Here's how we could avoid that altogether:

Consider the function $v$ on the interval $t < 0$. By continuity, for some interval around $t = -1$, $v$ is also non-zero. Then we can divide, integrate (the integrals are proper), and get $v = 2t^2$ using the condition $v(-1) = 2$. We can now see that this is the only function on $(-\infty, 0)$ which meets the requirements, because any function that has zeroes for some $t < 0$ would still have to agree with this formula on some interval between those zeros which contains $-1$, and that would lead to discontinuities; and any function that does not have zeros for any $t < 0$ has to agree with this formula on the whole of $(-\infty, 0)$.

Apply the same argument for $(0, +\infty)$, and we get $v = 3t^2$ for $t > 0$. By continuity, $v(0) = 0$. We can verify that the pieced-together function does solve the problem.

So what is the lesson from this? We have to be careful when dividing by variables, in case they could be zero, and when integrating the separated variables. The integrals have to be well-defined. Provided these conditions are met, I think the method is valid.

The treatment of $v$ as a variable in its own right is something that is more closely related to your questions about treating the differential forms as independent quantities in a fraction, so as I said from the start, I won't touch that. Maybe someone with more rigorous mathematical understanding of it will be able to address it.