Once upon a time in the fantastical realm of quantum field theory, three functional giants roamed the land: $H[f]$, $I[f]$, and $J[f]$. These functionals were peculiar entities: they did not simply evaluate a number at a point, but instead swallowed entire functions and returned a single value. Today, dear reader, we embark on an odyssey to dissect the inner workings of these giants and understand how they respond to infinitesimal changes. In the language of mathematics, we seek their functional derivatives.
The guiding idea is this: if a functional $F[f]$ changes under a small perturbation $f(y) \mapsto f(y) + \epsilon \eta(y)$ as
\[F[f + \epsilon \eta] – F[f] = \epsilon \int \frac{\delta F}{\delta f(z)}\eta(z)\,dz + O(\epsilon^2),\]
then the coefficient $\delta F/\delta f(z)$ is the functional derivative.
The functional $H[f]$ is defined as
\[H[f](x) = \int G(x,y)f(y)\,dy.\]
We want to find
\[\frac{\delta H[f](x)}{\delta f(z)}.\]
Using a delta-function probe, perturb $f(y)$ by $\epsilon\delta(z-y)$:
\[\frac{\delta H[f](x)}{\delta f(z)} = \lim_{\epsilon\to 0}\frac{H[f(y)+\epsilon\delta(z-y)](x)-H[f](x)}{\epsilon}.\]
Now compute the perturbed functional:
\[H[f(y)+\epsilon\delta(z-y)](x)
= \int G(x,y)\left[f(y)+\epsilon\delta(z-y)\right]dy.\]
Expanding gives
\[H[f(y)+\epsilon\delta(z-y)](x)
= \int G(x,y)f(y)\,dy + \epsilon\int G(x,y)\delta(z-y)\,dy.\]
Using the defining property of the delta function,
\[\int G(x,y)\delta(z-y)\,dy = G(x,z),\]
so
\[H[f(y)+\epsilon\delta(z-y)](x)=H[f](x)+\epsilon G(x,z).\]
Therefore,
\[\frac{\delta H[f](x)}{\delta f(z)}
= \lim_{\epsilon\to 0}\frac{H[f](x)+\epsilon G(x,z)-H[f](x)}{\epsilon}
= G(x,z).\]
Let
\[I[f] = \int_{-1}^{1} f(x)^3\,dx.\]
Perturb $f(x)$ by $\epsilon\delta(x-x_0)$:
\[I[f+\epsilon\delta(\cdot-x_0)]
= \int_{-1}^{1}\left[f(x)+\epsilon\delta(x-x_0)\right]^3dx.\]
To first order in $\epsilon$,
\[\left[f(x)+\epsilon\delta(x-x_0)\right]^3
= f(x)^3 + 3\epsilon f(x)^2\delta(x-x_0) + O(\epsilon^2).\]
The higher-order terms contain products of distributions and do not contribute to the first variation in this formal calculation. Keeping only the linear term,
\[I[f+\epsilon\delta(\cdot-x_0)]
= I[f] + 3\epsilon\int_{-1}^{1} f(x)^2\delta(x-x_0)\,dx + O(\epsilon^2).\]
If $x_0\in[-1,1]$, this becomes
\[I[f+\epsilon\delta(\cdot-x_0)] = I[f] + 3\epsilon f(x_0)^2 + O(\epsilon^2).\]
Thus,
\[\frac{\delta I[f]}{\delta f(x_0)}
= \lim_{\epsilon\to 0}\frac{I[f+\epsilon\delta(\cdot-x_0)]-I[f]}{\epsilon}
= 3f(x_0)^2.\]
For the second functional derivative,
\[\frac{\delta^2 I[f]}{\delta f(x_0)\delta f(x_1)},\]
we differentiate the first functional derivative with respect to $f(x_1)$:
\[\frac{\delta^2 I[f]}{\delta f(x_0)\delta f(x_1)}
= \frac{\delta}{\delta f(x_1)}\left(3f(x_0)^2\right).\]
Since
\[\frac{\delta f(x_0)}{\delta f(x_1)} = \delta(x_0-x_1),\]
we obtain
\[\frac{\delta^2 I[f]}{\delta f(x_0)\delta f(x_1)}
= 6f(x_0)\delta(x_0-x_1).\]
The functional $J[f]$ is given by
\[J[f] = \int \left(\frac{\partial f}{\partial y}\right)^2dy.\]
We want to find
\[\frac{\delta J[f]}{\delta f(x)}.\]
Using the formal definition, perturb $f(y)$ by $\epsilon\delta(x-y)$:
\[\frac{\delta J[f]}{\delta f(x)}
= \lim_{\epsilon\to 0}\frac{J[f(y)+\epsilon\delta(x-y)]-J[f(y)]}{\epsilon}.\]
First compute the perturbed functional:
\[J[f(y)+\epsilon\delta(x-y)]
= \int \left(\frac{\partial}{\partial y}\left[f(y)+\epsilon\delta(x-y)\right]\right)^2dy.\]
Expanding to first order in $\epsilon$ gives
\[J[f(y)+\epsilon\delta(x-y)]
= \int \left(\frac{\partial f}{\partial y}\right)^2dy
+ 2\epsilon\int \frac{\partial f}{\partial y}\frac{\partial}{\partial y}\delta(x-y)\,dy
+ O(\epsilon^2).\]
The $O(\epsilon^2)$ term vanishes in the functional derivative limit. For the middle term, integrate by parts, assuming the boundary term vanishes:
\[2\epsilon\int \frac{\partial f}{\partial y}\frac{\partial}{\partial y}\delta(x-y)\,dy
= -2\epsilon\int \frac{\partial^2 f}{\partial y^2}\delta(x-y)\,dy.\]
Therefore,
\[2\epsilon\int \frac{\partial f}{\partial y}\frac{\partial}{\partial y}\delta(x-y)\,dy
= -2\epsilon\left.\frac{\partial^2 f}{\partial y^2}\right|_{y=x}.\]
So the first variation is
\[\delta J[f] = -2\epsilon\left.\frac{\partial^2 f}{\partial y^2}\right|_{y=x},\]
and hence
\[\frac{\delta J[f]}{\delta f(x)}
= \lim_{\epsilon\to 0}\frac{-2\epsilon\left.\frac{\partial^2 f}{\partial y^2}\right|_{y=x}}{\epsilon}
= -2\left.\frac{\partial^2 f}{\partial y^2}\right|_{y=x}.\]
Equivalently,
\[\frac{\delta J[f]}{\delta f(x)} = -2f”(x),\]
provided the boundary term from integration by parts vanishes. This reveals that $J[f]$ is sensitive to the curvature of $f$ at the point $x$.
Thus, we have computed the functional derivatives for $H[f]$, $I[f]$, and $J[f]$ from their formal definitions:
\[\frac{\delta H[f](x)}{\delta f(z)} = G(x,z),\]
\[\frac{\delta I[f]}{\delta f(x_0)} = 3f(x_0)^2,\]
\[\frac{\delta^2 I[f]}{\delta f(x_0)\delta f(x_1)} = 6f(x_0)\delta(x_0-x_1),\]
and
\[\frac{\delta J[f]}{\delta f(x)} = -2f”(x).\]
The enigmas are solved, and the giants’ secrets are laid bare through the power of rigorous mathematical formulation.
Happy questing!
Original (English)
