$$\text{Find}\quad\frac{\delta F_k}{\delta G} \quad \text{given} \quad F_k=\left(\sum_{r=0}^{N-1} e^{ikr}\int_{-\infty}^{\infty} dt \ e^{i\omega t} G(r,t)\right)^{-1}$$

noting that $k$ and $r$ are discrete while $\omega$ and $t$ are continuous.


I am trying to show that $F_k[G(r,t)]$ is very sensitive to perturbations in $G(r,t)$. By numerical work I know this is (often) the case. Ultimately I'd like to argue that constructing $F_k$ by measuring $G(r,t)$ won't work well when the measurement is noisy. I figured a good way to show that was to compute the functional derivative and show it gets large (or perhaps diverges) in some places. However, I am doing something wrong.


By the definition here:

$$\frac{\delta F_k}{\delta G}:\int dt \frac{\delta F_k}{\delta G}\eta(t)=\left[\frac{d}{d\epsilon}\left(\sum_{r=0}^{N-1} e^{ikr}\int dt \ e^{i\omega t} G(r,t)+\epsilon\eta(t)\right)^{-1}\right]_{\epsilon=0}$$

Using the chain rule, $\sum_{r=0}^{N-1} e^{ikr}=N\delta_{k,0}$ i.e. the Kronecker delta, and the definition $G(k,\omega)\equiv\sum_{r=0}^{N-1} e^{ikr}\int dt \ e^{i\omega t} G(r,t)$ transforms the above to

$$\frac{-\int dt\sum_{r=0}^{N-1} e^{i(kr+\omega t)}\eta(t)}{\left(\sum_{r=0}^{N-1} e^{ikr}\int dt \ e^{i\omega t} G(r,t)\right)^{2}}=\int dt\frac{-\delta_{k,0} \ e^{i\omega t}}{G(k,\omega)^2}\eta(t)\to \frac{\delta F_k}{\delta G(r,t)}=\frac{-\delta_{k,0} \ e^{i\omega t}}{G(k,\omega)^2}$$

I know this is wrong because this numerics show this isn't $0$ for all $k\neq 0$. I suspect the problem has something to do with $\eta$ only depending on $t$ and not $r$, but I'm not sure how to account for a discrete variable like $r$.


I think my understanding of the definition above is likely wrong, and probably needs a test function that depends on both $r$ and $t$? I'd be pretty jazzed if the right way to do it looked something like

$$\frac{\delta F_k}{\delta G}:\sum_{r=0}^{N-1}\int dt \frac{\delta F_k}{\delta G}\eta(r,t)=\left[\frac{d}{d\epsilon}F_k\left[G(r,t)+\epsilon\eta(r,t)\right]\right]_{\epsilon=0}$$

but my math is too weak to assert this is the right way, even though it looks like it makes some sense (to me). I really have no idea how to handle the fact that $r$ is discrete. If the above is correct, this would give

$$ \frac{\delta F_k}{\delta G(r,t)}=\frac{-e^{i(\omega t+kr)}}{G(k,\omega)^2} $$

which is what I was hoping to show since $G(k,\omega)=0$ for at least one $\omega$ at every $k$.


Any of the following

  1. $\dfrac{\delta F_k}{\delta G}$
  2. How to approach such functional derivatives?
  3. What's wrong with my attempt?
  4. Confirm/deny the correctness of my "resolution".

Thanks in advance!

  • $\begingroup$ can you clarify what the summation variable is? also from the very beginning only one term has $k$ in it, which allows you to rewrite before you take the derivative. Sounds like this is not what you intended? $\endgroup$ – Calvin Khor Oct 7 at 2:34
  • $\begingroup$ @CalvinKhor I am not sure I follow. I am summing over $r$ so that $F$ is a function of $k$. Since $G$ is a function of $r$, I don't think this becomes trivial? $\endgroup$ – bRost03 Oct 7 at 2:40
  • $\begingroup$ Well that was one way to answer my question. What you mean is, by $\sum_r^N$ you mean $\sum_{r=\dots}^N$ where $\dots$ is presumably 0 instead of say $-17$? With so many variables, omitting details is confusing $\endgroup$ – Calvin Khor Oct 7 at 2:41
  • $\begingroup$ if there is still a nontrivial function of $r$, how can you apply the identity $\sum_r e^{ikr} = \delta_{\text{something with no r}}$? $\endgroup$ – Calvin Khor Oct 7 at 2:45
  • $\begingroup$ @CalvinKhor Ah I see. I have updated the post. If it helps this is a Fourier Transform from position space to momentum space for a finite lattice, so $k=2\pi n/N$ for $0\leq n<N$. If that doesn't help just ignore it. I can apply the identity there because $G(r,t)$ isn't present. $\endgroup$ – bRost03 Oct 7 at 2:48

I'll answer to your questions not in their former order, in order to possibly answer in a better way.

  1. How to approach such functional derivatives?

The approach you adopted (though with some slips as we'll see below) is the correct one, i.e. $$ \frac{\delta F_k }{\delta G}[G](\eta) \triangleq DF_k[G](\eta) \triangleq \left.\frac{\mathrm{d}}{\mathrm{d \epsilon}} F_k(G+\epsilon \eta)\right|_{\epsilon=0}\label{1}\tag{1} $$ The calculations have to be done as the right hand were a function of the real variable $\epsilon$: the definition does not depend on how the functional $F_k$ acts on its argument $G(r,t)$, thus is irrespective of the discreteness and / or the continuity of the involved variables. We'll see below that the discreteness and finiteness of the values of the variable $r= 0, 1, \ldots, N-1$, in this particular case, can be efficiently dealt by representing the argument $G$ and the variation $\eta$ as vectors in an $N$-dimensional space, but this is by no means implied by the structure of equation \eqref{1} itself.
The structure of functional derivatives is illustrated in some detail in this Q&A: for the reasons given there, you'll see that it is not safe to trust the Wikipedia entry on the topic.

  1. What's wrong with my attempt?

The main slip which is usually done (as in this case and in this other one) is to expect that this kind of derivative has the structure of a function: it is almost never so. Customarily, functional derivatives are operators that map a variation $\eta$ of the argument $G$ of the functional to the variation of the corresponding value of the functional itself. Answering more extensively this point of the question question,

  1. you are considering variations of the single $t$ variable: this is not an absolute error, since you can licitly do so if you are sure that the input function $G=G(r,t)$ cannot change its value "randomly" if the discrete variable $r$ is kept constant. However, since you are trying to analyze the performance of $F_k$ on noisy inputs, you should consider the possibility to have $\eta=\eta(r,t)$ for $r=0, 1,\ldots,{N-1}$, and
  2. you expect the functional derivative to be a function, while it is an integral operator: seeing this will help you in evaluating the variation of your functional when subjected to noisy inputs.
  1. $\dfrac{\delta F_k}{\delta G}$

Formally, by putting $$ \begin{split} &\mathbf{r} =(0, 1, \ldots, {N-1}),\\ & e^{ik\mathbf{r}} =\big(1, e^{ik}, \ldots, e^{ik{(N-1)}}\big), \\ & \eta(\mathbf{r}, t) =\big(\eta(0,t), \eta(1,t), \ldots, \eta({N-1},t)\big), \end{split} $$ we can write the functional derivative of $F_k$ as $$ \begin{split} \frac{\delta F_k}{\delta G}[G](\eta) &= DF_k[G](\eta)\\ & =-\frac{{1}}{G(k,\omega)^2}\sum_{r=0}^{N-1} e^{ikr} \int \eta(r, t)\;e^{i\omega t} \mathrm{d}t\\ & = -\frac{{1}}{G(k,\omega)^2} \int \left\langle e^{ik\mathbf{r}}, \eta(\mathbf{r}, t)\right\rangle e^{i\omega t} \mathrm{d}t \end{split}\label{2}\tag{2} $$ where $\langle\,\cdot\,,\,\cdot\,\rangle:\Bbb R^N\times\Bbb R^N \to\Bbb R$ is the ordinary scalar product. By using equation \eqref{2} and a suitable choice of the variation $\eta(r,t)$, you will be able to analyze the sensitivity of $F_k$ to particular variations of its argument function $G$.

  1. Confirm/deny the correctness of my "resolution".

Your resolution is "almost correct" in the sense that, after performing correctly the derivative, you should chose a suitable variation $\eta(\mathbf{r}, t) =\big(\eta(0,t), \eta(1,t), \ldots, \eta({N-1},t)\big)$ in order to get your result. And obviously your choice should be consistent with the problem you are dealing with.

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