I was looking at Haberman's book on ODEs, and it is pretty good. But one confusing point has to do with this example of using the Laplace transform on a Heavyside function. I am not clear on how the $\tau$ versus $t$ substitution occurs. Here is the example from the book.
So I marked the question area in red. I understand that the original function must be expressed in the form of $f(t-c)H(t-c)$. So that means converting the function $t^2$ into a function in terms of $t-1$ where $c=1$. But the substitutions seem a bit so in the first red line, the function is setup at $f(t-1) = t^2$, but that is not accurate since $f(t-1) = (t-1)^2$. So this might be a notational lack of clarity, but it seems confusing to me. The issue carries forward with Haberman sets $f(\tau) = (\tau + 1)$, when $\tau = t-1$. So technically $f(\tau) = (\tau - 1)$.
Again, so I understand the intents here, but the notation is really confusing. Can anyone explain this notation a bit better or provide a better/clearer way to solve this same problem. Thanks.
We need to have $t^2 H(t - 1) = f(t - c) H(t - c)$. Thus, comparing the two expressions, we see that we must take $c = 1$ and $f(t - 1) = t^2$. Note that $f(t) \neq t^2$. Rather, $f(t)$ is defined so that $f(t - 1) = t^2$. To find $f(t)$, we must perform a substitution. We take $\tau = t - 1$ so that $t = \tau + 1$. Then, substituting the expressions involving $\tau$ above, we find $f(\tau) = (\tau + 1)^2$. Since this is true for all $\tau$, this is the same as saying $f(t) = (t + 1)^2$. In any case, you can check that $$ f(t-1) = ((t - 1) +1)^2 = (t)^2 = t^2 $$ which was the property we wanted $f$ to have.
Hopefully, this is convincing, but even if it isn't, you can see that $f(t - 1) = t^2$ when we take $f(t) = (t + 1)^2$. This is the property we needed the function $f$ to have, so we are done.
I think the confusion here arises from the fact that we began by defining $f$ as a function such that $f(t - 1) = t^2$, when you are probably used to seeing $f(x) = $ "some expression involving $x$". There is no issue with writing $f(t - 1) = t^2$ as long as we check that $f$ is well defined and has the desired domain and range. It just means we have to do a little work to find out how to write $f(t)$ as an expression involving $t$.