problem with a functional derivative

Question

I've the following problem with a functional derivative (I'm not a specialist). Let's start with something I know (hope!): given a functional $\mathcal{F}[\psi]$, say $$ \mathcal{F}[\psi] = \int^{T}_{t} G(\psi(s),s)ds $$ ($G$ is a function of $\psi(s)$ and $s$), I interpret the functional derivative of $\mathcal{F}$ w.r.t. the function $\psi(s)$ as the function $\frac{\delta\mathcal{F}}{\delta \psi(s)}$ of the variable $s$ such that the first variation $\delta \mathcal{F} \stackrel{def}{=} \mathcal{F}[\psi + \delta \psi] - \mathcal{F}[\psi]$ of the functional $\mathcal{F}$ verifies: $$ \delta\mathcal{F} = \int^{T}_{t}\frac{\delta\mathcal{F}}{\delta \psi(s)} \delta \psi(s) ds $$ In this simple case, the direct computation of $\delta \mathcal{F}$ gives us (at first order in $\delta \psi$): $$ \frac{\delta\mathcal{F}}{\delta \psi(s)} = \frac{\partial G}{\partial \psi}(\psi(s),s) $$ which - if remember well - should be the simplest form of the Eulero-Lagrange equations, since the integrand $G$ does not depend on the derivative $\frac{d\Psi}{ds}$ too. For example I had the following (family of) functional $\beta_{t}[\psi]$ defined as follows: $$ \beta_{t}[\psi] = \int^{T}_{t} e^{k(s-t)}\psi(s)ds $$ which depends functionally on $\psi$ and is indexed by the parameter $t$ (also $T$ is of course a parameter but unnecessary for what follows, so please forget it!). By direct computation of $\delta \beta_{t}[\psi] = \beta_{t}[\psi + \delta \psi] - \beta_{t}[\psi]$ one easily gets: $$ \delta \beta_{t} = \int^{T}_{t} e^{k(s-t)} \delta \psi(s)ds $$ and therefore, by definition above: $$ \frac{\delta\beta_{t}}{\delta \psi(s)} = e^{k(s-t)} $$ which is a function of $s$ with parametric dependence on $t$, as it should. Up to here no problem. Now consider the following rather nasty (family of) functional: $$ \begin{eqnarray} \alpha_{t}[\psi] &=& \int^{T}_{t} \beta_{s}[\psi]ds\\ &=& \int^{T}_{t} \Big(\int^{T}_{s} e^{k(u-s)} \psi(u) \Big) ds \\ &=& \int^{T}_{t} e^{-sk} \Big(\int^{T}_{s} e^{ku} \psi(u) \Big) ds \end{eqnarray} $$ The last "$=$" was just to be crystal clear about the integrating variables. For what I said before, I would be able to recognize the functional derivative $\frac{\delta\alpha_{t}}{\delta \psi(s)}$ of $\alpha_t[\psi]$ if I somehow write: $$ \delta\alpha_t = \alpha_{t}[\psi + \delta \psi] - \alpha_{t}[\psi] = \int^{T}_{t}\frac{\delta\alpha_t}{\delta \psi(s)} \delta \psi(s) ds $$ But what I was able to write up to now is only: $$ \delta \alpha_{t} = \int^{T}_{t} e^{-sk} \Big(\int^{T}_{s} e^{ku} \delta \psi(u) \Big) ds $$ In conclusion, my question is: $$ \frac{\delta\alpha_{t}}{\delta \psi(s)} = ??? $$

Answer 1

EDIT - SOLVED.

Sorry guys, didn't notice it was so easy at first glance. Fubini is the answer:

$$ \begin{eqnarray} \delta \alpha_{t} &=& \int^{T}_{t}\Big( \int^{T}_{s} e^{k(u-s)} \delta \psi(u) du \Big) ds \\ &=& \int^{T}_{t} \Big( \int^{u}_{t} e^{k(u-s)} ds \Big) \delta \psi(u) du \\ \end{eqnarray} $$ since $$\left\{(s,u): t \leq s \leq T, s \leq u \leq T \right\} = \left\{(s,u): t \leq u \leq T, t \leq s \leq u \right\}$$ and thus (renaming variables): $$ \frac{\delta\alpha_{t}}{\delta \psi(s)} = \int^{s}_{t} e^{k(s-u)} du = \frac{e^{k(s-t)} - 1}{k} $$