My differential equations professor gave the class an assignment introducing us to delay differential equations, and he put some questions for us to answer on it. I've been completely stuck on this problem for too long.

$$ y'(t)=\alpha y(t-\tau) $$ Re-scale the independent variable to show that that equation is equivalent to either one of the following equations: $$ \frac{dy}{ds}(s)= \pm y(s-\bar{\tau}) \quad\text{or} \quad\frac{dy}{ds}(s)= \bar{\alpha}y(s-1) $$ for some new parameters $\bar{\alpha}$ and $\bar{\tau}$ (Find what these parameters are in terms of $\alpha$ and $\tau$).

I can't for the life of me figure out how to do this. First of all, embarrassingly enough, I am not 100% sure what to make of the notation $\frac{dy}{ds}(s)$. I am assuming that it means the same thing as just $\frac{dy}{ds}$.

I talked with my professor and he said that by re-scaling $t$, he meant to multiply it by a constant and add (or subtract) another constant. The second of those last equations is super easy, but I'm stuck on the first one. $$ t=As+B\\ y'(t)=\frac{dy}{dt}=\frac{dy}{d(As+B)}=\frac1A\cdot\frac{dy}{ds}\\ \frac1A\frac{dy}{ds}=\alpha y(As+B-\tau)\implies\frac{dy}{ds}=\frac\alpha Ay(A(s+C)) \qquad C=\frac{B-\tau}2 $$ So, in order to satisfy the first of the equations, $A$ would have to be $\pm\alpha$. But it is impossible to make it match the equation, because I am left with $$ \frac{dy}{ds}=\pm y(\pm\alpha(s+C)) $$ There is nothing I can do to get rid of that alpha. So what mistake am I making? I've been at this for hours.


Take care in defining the new function $\overline y(s)$. You would like to change variables, so you'll define a new function $\overline y(s)=y(t(s))$, where $t(s) = As+B$. Now,

$$ \frac{dy}{dt}=\alpha y(t-\tau), $$

so by changing variables, the chain rule gives

$$ \frac{d\overline y}{ds}=\frac{d y}{dt}\frac{d t}{ds}=A\alpha y(t-\tau)=A\alpha \overline y\left( \frac{t-\tau-B}A \right)=A\alpha \overline y\left( s-\frac{\tau}A \right). $$

That's the core of your mistake: that the original function evaluated at $t-\tau$ cannot be replaced by the new function merely by substitution, $y(t-\tau)\neq \overline y(As+B-\tau)$. It is a common mistake when changing variables to not recognise that $\overline y$ is actually a different function to the original $y$, not helped by the confusing fact that many people don't write it with the line.

From here it is easy to see that we can choose $A$ to either make the term inside the brackets $1$, or the prefactor, but in general not both. Also, note that $B$ isn't part of the transformed equation - this is because a shift in time doesn't alter the dynamics (if you like, the system is invariant to time translation).

  • $\begingroup$ Thank you. This answer makes sense, but I'm thrown by the fact that you added a new function. None of my calculus classes that I can recall have had me introduce a new function, and I've never seen the notation with the line over the function (or maybe you just added that because it matched the rest of the problem). Also, the question did not ask for $\bar{y}$, just $y$, but I guess that might just be lazy writing. I'll stick with this because it makes more sense than anything else. $\endgroup$ – Polygon Sep 26 '17 at 22:47
  • $\begingroup$ The line notation is arbitrary. It is a new function since it's defined on a different domain, just that it's simpler to write only $y$ and hope that everyone realises which $y$ is being used when. $\endgroup$ – bjorne Sep 27 '17 at 7:58

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