muzialis said:
HallsofIvy,
many thanks for your comment.
I understand the notation, what is puzzling me is ots very audacious use.
In the paper I am reading, by a physicist, the starting point is the ODE
$$q =- G + \frac{\partial G }{\partial t}$$ (nothing else than Fourier's Law, for the curious), presented in the particular notation we discussing as
$$ q = - (1 - \frac{\partial }{\partial t}) G$$, or even
$$ \frac{q}{(1 - \frac{\partial }{\partial t}) } = -G$$
Then, by noting that
$$\frac{1}{1-\frac{\partial}{\partial t}} \approx 1 + \frac{\partial}{\partial t} $$
the following ODE is obtained
$$ \dot{q} + q = -G $$, well, I do not see how this and the initial ODE could be related in more rigorous way, the two seem very different to me. If for example the given q is identically zero, the first one is solved by an exponent, while the second one is solved by the zero constant function.
Apparently the physicist Cattaneo used this procedure in the late '40 to derive the hyperbolic heat equation.
Many thanks
What's going on, basically, is an abuse of notation in which ##\frac{1}{1-\frac{\partial}{\partial t}}## is a shorthand for the inverse operator ##L^{-1}= \left(1-\frac{\partial}{\partial t}\right)^{-1}##, where ##L = 1 - \frac{\partial}{\partial t}##.
Under certain conditions, you can write the inverse operator as a power series, so what your reference is doing is writing
$$L^{-1} = 1 + \sum_{k=1}^\infty \left(\frac{\partial}{\partial t}\right)^k = 1 + \sum_{k=1}^\infty \frac{\partial^k}{\partial t^k},$$
and then keeping only the lowest order derivative term, presumably because the function q or G vary slowly and the higher derivatives can be neglected to first order.
I believe this approximation procedure is called (or at least related to) the
Resolvent formalism, which comes up in the study of Green's functions (Here, the inverse operator ##L^{-1}## should be a Green's function, such that when you apply it to ##q = -LG \Rightarrow L^{-1}q = -L^{-1}LG##, there is actually an integration involved ##\int dt~ L^{-1} q = -\int dt~L^{-1}L G = -\int dt~\delta(t-t') G(t') = -G(t)##, so it's really ##\int dt~ L^{-1} q \approx q + \dot{q}##.)
Edit: To make this clearer, hopefully, let's do this in Fourier space. The Fourier transformed equation
$$q(t) = -G(t) - \frac{dG}{dt}(t)$$
becomes
$$\hat{q}(\omega) = -\hat{G}(\omega) + i\omega \hat{G}(\omega).$$
Note that the time derivative effectively got replace by a factor of ##i\omega##. We can solve this equation for ##\hat{G}(\omega)##,
$$-\hat{G}(\omega) = \frac{\hat{q}(\omega)}{1-i\omega}.$$
Inverse Fourier transforming this will give ##G(t)##. However, suppose the integration over the left hand side is hard to do, but, we expect that it is the low frequency components of ##q(t)## that dominate its behavior. Then, we can expand the denominator and write
$$-\hat{G}(\omega) \approx (1 + i\omega) \hat{q}(\omega).$$
This equation is much easier to inverse transform, and note that just as ##d/dt## became ##i\omega## in Fourier space, ##i\omega## will become ##d/dt## when back in the time domain, and so we find
$$-G(t) \approx q(t) + \frac{dq}{dt}.$$
So what the author did in the reference linked to in the first post was an abuse of notation which provided a shortcut to this result.