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I was having trouble understanding a step pertaining to the solution to a Dirichlet problem for Laplace's equation in a cylindrical domain. However, I realized my question was fundamentally much simpler and could be restated.

Given the differential equation

$y'(t) = ty(t)$,

let us say we decide to make the change of variables $t = 2s$. Then it is clear that

$\frac{dy}{ds} = \frac{dy}{dt}\frac{dt}{ds}$. Thus $\frac{dy}{dt}$ is $\frac{1}{2}\frac{dy}{ds}$. We then have

$\frac{1}{2}\frac{dy}{ds} = 2s*y(t)$.

If we replace y(t) with y(s), then we can solve the differential equation in terms of $s$, replace $s$ with $t$ by their known relationship, and obtain the expected result. However, I do not understand the justification for the above replacement. Wouldn't it imply the false relationship that $y(t) = y(0.5t)$?

To make matters even more confusing, my book once indicated a slightly different artifact of change of variables, in which it essentially defined a new function to account for the change of variables. In the above case, it would be something like $Y(s) := y(t)$. Please help.

Please provide an answer that considers that I am still only a high school student.

Ahdhehshdjdj
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1 Answers1

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Changing variables $t=2s$ means you’re defining a new function $Y(s):=y(2s)$ (which implies $y(t)=Y(\frac{t}{2})$). Then, by the chain rule, \begin{align} Y’(s)=y’(2s)\cdot 2=[2s\cdot y(2s)]\cdot 2=4s\cdot y(2s)=4s\cdot Y(s). \end{align}

It is exactly for this very reason that I abhor introducing the chain rule via the “mnemonic” of cancelling the differentials. The $y$ on the two sides of the equation mean completely different things!

peek-a-boo
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  • Thank you. My book specifically makes the claim that with the change of variables $s = \frac{mr\pi}{b}$, we can transform the equation $r^{2}R''(r) + rR'(r) - (r^{2}({\frac{m\pi}{b}})^2 + n^{2})R(r) = 0$ into $s^{2}R''(s) + sR'(s) - (s^{2} + n^{2})R(s) = 0 $. Why are we not defining a new function here with the change of variables? Or is this just laziness on the part of the book? What is going on here? – Ahdhehshdjdj Jul 15 '23 at 02:35
  • Yes it is pure laziness brcause they know what they’re doing and would rather do mental gymnastics (but to be fair, it foes get tiring once yiu know what’s going on). It’s kind of like how we all learn grammar but in everyday life, we rarely speak or write with perfect grammar. – peek-a-boo Jul 15 '23 at 03:16
  • So, to summarize, changes of variables in differential equations like the ones above should entail defining a new function to account for the new variable; however, this step may be omitted in mathematical texts. – Ahdhehshdjdj Jul 15 '23 at 03:43
  • Yes that’s right. – peek-a-boo Jul 15 '23 at 12:01
  • Does it work with $t=2s:\frac{dy}{dt}=ty\implies \frac{dy}{d(2s)}=(2s)y$ and get $\frac{dy}{2ds}=2sy\iff\frac{dy}{ds}=4sy$? – Тyma Gaidash May 03 '24 at 00:06