Interpreting Volterra Series Correctly

Question

I was reading this and found on page 6 was a description of a generalization of Taylor series to linear functionals. Reproduced below as

$$ F[\phi + \lambda] = \sum_{n=0}^{\infty} \left[\frac{1}{n!} \int\int...\int \frac{\delta ^n F[\phi] }{\delta \phi(x^0) \delta \phi(x^1) ... \delta \phi(x^{n-1}) } dx^0 dx^1 ... dx^{n-1} \right] $$

I don't understand exactly what the term $$ \frac{\delta ^n F[\phi] }{\delta \phi(x^0) \delta \phi(x^1) ... \delta \phi(x^{n-1}) }$$

Means.

And let me be extremely concrete. Consider an operator $F: \phi \rightarrow \phi + x^2\phi'$, quite literally what this means is that $F: (\mathbb{R} \rightarrow \mathbb{R}) \rightarrow (\mathbb{R} \rightarrow \mathbb{R}) $

And $F:= \phi(x) \rightarrow \phi(x) + x^2 \frac{d}{dx}\left[ \phi(x)\right] $ Can be viewed as an explicit "form".

So when we compute $$\frac{\delta F}{\delta \phi(x)}$$

Everything makes sense (since the argument of the $\phi$ we are differentiating w.r.t matches the argument of the $\phi$ in the original function and the corresponding $\frac{d}{dx}$ is compatible with the $x$ arguments in both cases.) And we use our typical euler lagrange equations.

$$ \frac{\delta F}{\delta \phi(x)} = \frac{\partial F}{\partial \phi(x)} - \frac{d}{dx} \left[ \frac{\partial F}{\partial \left( \frac{d}{dx}[\phi(x)] \right) } ...\right] = 1+2x $$

If you make an expression like

$$\frac{\delta F}{\delta \phi(x^0)}$$

Now it's unclear what you mean, but we can guess. Let us substitute every instance of the string "$x$" with "$x^0$" in $F$ and then this statement can be still well defined. And we conclude that it is equal to

$$ 1 + 2x^0 $$

Now consider:

$$\frac{\delta^2 F}{\delta \phi(x^0) \delta \phi(x^1) }$$

This is now getting us into bad territory. Not only do we not have a compatible variable $x$ in either of arguments but we cannot implicitly "guess" what string substitution needs to be made here since any global level string subtitution will render the other derivative to 0. So I am completely lost as to what this could mean.

Answer 1

I'll run through your example and hopefully that makes it clear.

$$\phi\mapsto F[\phi]= \phi+x^2\phi'$$ Vary this $$\phi+\delta\phi\mapsto F[\phi]+\delta\phi+x^2\delta\phi'$$ So we have formally $$\frac{F[\phi+\delta\phi](x)-F[\phi](x)}{\delta\phi(y)}=\frac{\delta\phi(x)}{\delta\phi(y)}+x^2\frac{\delta\phi'(x)}{\delta\phi(y)}$$

Then we take as our definition that (and more generally would have ignored $\mathcal O(\delta\phi)$ terms if we have things like $F[\phi]=\phi^2$) $$ \frac{\delta\phi(x)}{\delta\phi(y)}:=\delta(x-y), $$ the Dirac delta function. Similarly, $$ \frac{\delta\phi'(x)}{\delta\phi(y)}:=\delta'(x-y). $$ One can make this rigorous with compactly supported test functions and the like.

Answer 2

FWIW, in OP's example OP is considering the functional $$ F[\phi]~:=~ \phi(x_0) + x_0^2\phi^{\prime}(x_0), $$ which depends on the parameter $x^0\in\mathbb{R}$.

Then the functional/variational derivative is $$ \frac{\delta F[\phi]}{\delta \phi(x_1)}~=~\delta(x_0\!-\!x_1) + x_0^2\delta^{\prime}(x_0\!-\!x_1), $$ which doesn't depend on $\phi$. Therefore the higher functional derivatives vanish.