3
$\begingroup$

This is my first post. :)

I'm learning about solving differential equations by separation of variables and I have questions about the rigourousness of the method. As stated, I'm solving

$${dy\over dx} = y$$

The method that I've been taught goes like this.

Case 1: Assume $y(x) = 0$ for all $x$.

  Check if $y(x) = 0$ is a solution. It is.

Case 2: Assume $y(x) = 0$ for no $x$.
  We divide by $y$ on both sides to convert into separable form. This division is justified because $y \not= 0$. We get $dy/y = dx$. "Integrating" both sides, we get $\ln|y| = x+c$. Rearrange to isolate $y$, we get $y = \pm e^xe^c$, and realize that $\pm e^c$ is a nonzero free constant. $\square$

The above is the approach I've been taught. The problem is that we're missing a case. How about this?

Case 3: Assume $y(x) = 0$ for some $x$.

I feel that a completely rigourous solution also needs to prove that there are no solutions under this case. But I don't know how to prove it. If you possess a proof, I invite you to present your answer here. I'd be delighted to read it.

I've been reading a textbook that tried to prove this but I can't fill in all the missing steps myself. It goes something like this: Suppose we're not in Case 1, so that $y(x) \not= 0$ for some $x$. Since $y$ is continuous, there is some open interval around $x$ such that $y(x) \not= 0$ for all $x$ within the interval. Then all the steps in case 2 apply within this interval. But then I don't understand why this implies that $y(x) \not= 0$ outside of the interval as well. Anyway, this is only one possible proof method and I'm open to hearing others. Ideally, I'm interested a proof method that easily generalizes to other separable differential equations as well, because dividing by $y$ or a function of $y$ is quite common during the separation process.

$\endgroup$
  • $\begingroup$ So, have you already read about existence and uniqueness theorem? $\endgroup$ – Evgeny Sep 21 '15 at 17:44
  • $\begingroup$ Related. $\endgroup$ – Workaholic Sep 21 '15 at 17:47
  • $\begingroup$ I haven't heard of the existence and uniqueness theorems. What's a good reference for these? $\endgroup$ – HazySmoke Sep 21 '15 at 17:54
1
$\begingroup$

You did a nice job with that close reading of the text.

In this particular case, there's a solution to your problem:

Suppose that $y$ is any solution to the equation, and let $y'$ denote $\frac{dy}{dx}$. Let $$ u = \frac{y}{e^x} $$ Then $$ \frac{du}{dx} = \frac{y' e^x - y e^x}{e^{2x}} = \frac{y' - y}{e^{x}} = 0 $$ because $y$ solves the original equation $y' = y$.

Thus $\frac{y}{e^x}$ is a constant $c$, to $y = ce^x$.

Now this technique worked only because the known solution, $e^x$, happened to never be zero, so dividing by it was safe. For more general equations, you have to be a good deal more clever.

Post-commment addition OP asks whether there's a way to show that this stuff works, going from the original ideas in the text. Following the idea from a response to the linked question, let me work through that:

Suppose that $f$ is a solution to the equation, and that $f(a) \ne 0$ for some $a$. Then by continuity (e.g., the intermediate value theorem), we know that there's an interval $[c, d]$ with $c < a < d$ such that $f$ is nonzero on the interval. And the solution given in the "$f$ nonzero everywhere" part of the book shows that on the interval $[c, d]$, we have that $f$ is a nonzero multiple of $e^x$.

Now knowing that $f(a) \ne 0$, let's suppose that $f(b) = 0$ for some $b > a$. (The case $b < a$ is very similar).

The set of zeroes-of-$f$-above-$a$ is bounded below (by $a$, and indeed, by $d > a$) and nonempty (because $b$ is in it), hence has an infimum; call that $A$.

There are two possibilities:

  1. $f(A) \ne 0$.

  2. $f(A) = 0$.

In the first case, we know that for some interval around $A$, we have $f(x) \ne 0$, so $A$ cannot be the infimum of the zeros of $f$.

So we must be in the second case. To summarize so far:

  • We have a function $f$ that satisfies the differential equation.

  • $f(a) \ne 0$.

  • for some $A > a$, $f(a) = 0$

  • for $a < x < A$, $f(x) \ne 0$.

Now let $Q$ be any number strictly between $a$ and $A$.

For each $x \in $[a, Q], there's an interval $(c_x, d_x)$ on which $f$ is a nonzero multiple $w_x e^x$ of $e^x$. These intervals cover $Q$, which, by the Heine Borel theorem, is compact; hence some finite number of these intervals cover $[a, Q]$. For any two that overlap, the corresponding constants $w_x$ and $w_{x'}$ must be equal, for the two nonzero exponentials agree on the (open) interval of overlap.

This shows that on the interval $[a, A)$, the solution $f$ has the form $f(x) = w e^x$ for some single nonzero constant $w$. But we also have $f(A) = 0$. But since $f$ is continuous, we know that $$ f(A) = \lim_{x \to A} f(x) = \lim_{x \to A} w e^x = w e^A \ne 0, $$ which is a contradiction. Hence the assumption -- that $f(b) = 0$ for some $b > a$ -- must be false. QED.

$\endgroup$
  • $\begingroup$ Hi John. Thanks for the answer. The accepted answer of the related post is basically this. But your proof is not a based on the proof in my textbook, correct? It's a different idea entirely. I'm wondering if there is a way to complete the proof in the textbook: That is, show that if $y(x) \not= 0$ in the interval then that's also true outside the interval. This response tried to address this concern but also didn't flesh out the details. $\endgroup$ – HazySmoke Sep 21 '15 at 18:45
  • $\begingroup$ Details added. I did have to resort to compactness, alas. $\endgroup$ – John Hughes Sep 21 '15 at 19:22
  • $\begingroup$ Answer accepted. The proof does seem to work, though this route appears to be the longer way to go compared to your first approach. $\endgroup$ – HazySmoke Sep 21 '15 at 22:36
  • $\begingroup$ Oh, yeah. Definitely. It could probably be shortened and cleaned up a good deal, but it'd still be more complex. That's probably why so many textbooks follow my first approach. :) $\endgroup$ – John Hughes Sep 21 '15 at 23:11

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.