Is My Understanding of Higher-Order Functional Derivatives Correct?

[Hill]

Homework Statement

Consider the functional $I[f]=\int_{-1}^1 f(x) \,dx$. Find the functional derivative $\frac {\delta ^2 I[f^3]} {\delta f(x_0) \delta f(x_1)}$.

Relevant Equations

Definition of the functional derivative: $\frac {\delta G[f]} {\delta f(x_0)}=\left. \frac d {d \epsilon} G[f(x)+\epsilon \delta(x-x_0)] \right|_{\epsilon=0}$

(To moderators: although the question is mathematical, I post it in the physics forum because the definition and the notation are those used by physicists and because it comes from a QFT textbook; please move it if I'm wrong.)

My issue with this question is that the textbook has neither defined nor discussed higher-order functional derivatives, so I proceed by assuming that

$\frac {\delta ^2 I[f^3]} {\delta f(x_0) \delta f(x_1)}=\left. \frac {d^2} {d \epsilon_0 d \epsilon_1} I[(f(x)+\epsilon_0 \delta(x-x_0)+\epsilon_1 \delta(x-x_1))^3] \right|_{\epsilon_0=\epsilon_1=0}$.

In this case,

$\left. \frac {d^2} {d \epsilon_0 d \epsilon_1} \int_{-1}^1(f(x)+\epsilon_0 \delta(x-x_0)+\epsilon_1 \delta(x-x_1))^3 \,dx \right|_{\epsilon_0=\epsilon_1=0}$

$=\left.\int_{-1}^16(f(x)+\epsilon_0 \delta(x-x_0)+\epsilon_1 \delta(x-x_1))\delta(x-x_0) \delta(x-x_1)\,dx \right|_{\epsilon_0=\epsilon_1=0}$

$=6\int_{-1}^1 f(x) \delta(x-x_0) \delta(x-x_1)\,dx$

$=\begin{cases} 6f(x_0) & \text{ if } -1\leq x_0=x_1 \leq 1\\0 & \text{ otherwise}\end{cases}$

Is this a correct answer?

 

[TSny]

Consider the functional $I[f]=\int_{-1}^1 f(x) \,dx$. Find the functional derivative $\frac {\delta ^2 I[f^3]} {\delta f(x_0) \delta f(x_1)}$.

This looks like part of Exercise 1.2 in the text Quantum Field Theory for the Gifted Amateur by Lancaster and Blundell. I'm not skilled in the topic of functional derivatives, so take my comments below with a grain of salt. We have experts here who can set us straight.

My issue with this question is that the textbook has neither defined nor discussed higher-order functional derivatives

Yes, unfortunately, it looks like the textbook expects you to intuit the meaning of a second-order functional derivative.

so I proceed by assuming that

$\frac {\delta ^2 I[f^3]} {\delta f(x_0) \delta f(x_1)}=\left. \frac {d^2} {d \epsilon_0 d \epsilon_1} I[(f(x)+\epsilon_0 \delta(x-x_0)+\epsilon_1 \delta(x-x_1))^3] \right|_{\epsilon_0=\epsilon_1=0}$.

With this assumption, you get the result

$6\int_{-1}^1 f(x) \delta(x-x_0) \delta(x-x_1)\,dx$

$=\begin{cases} 6f(x_0) & \text{ if } -1\leq x_0=x_1 \leq 1\\0 & \text{ otherwise}\end{cases}$

It seems to me that the result should contain a delta function:

$=\begin{cases} 6f(x_0)\delta(x_0-x_1) & \text {if both}\, x_0 \, \text{and} \, x_1 \, \in (-1, 1) \\ \\0 &\text{otherwise}\end{cases}$

What do you think?

When I first saw this exercise, I interpreted the meaning of a second-order functional derivative as a "derivative of a derivative":
$$\frac {\delta ^2 I[f^3]} {\delta f(x_0) \delta f(x_1)}=\frac{\delta}{\delta f(x_0)}\left[\frac{\delta I[f^3]}{\delta f(x_1)}\right]$$ To evaluate the first-order functional derivative $\large \frac{\delta I[f^3]}{\delta f(x_1)}$ you can use the epsilon definition given in the text, or you can use the result of the second example on page 13 where it is shown that $$\frac{\delta}{\delta f(x)}\int \left[ f(y)\right]^p \phi(y) dy= p\left[f(x)\right]^{p-1} \phi(x). $$ Then see if you can go on and take the next functional derivative: $\large \frac{\delta}{\delta f(x_0)}$. For this, it might be helpful to note the last bulleted comment in Example 1.1 on page 12. With this approach, I get the same result $6f(x_0)\delta(x_0-x_1)$ as I get with your approach.

 

[Hill]

When I first saw this exercise, I interpreted the meaning of a second-order functional derivative as a "derivative of a derivative"

I think that both approaches lead to the same result, but I doubt the delta function gets outside the integral. $\delta(x_0-x_1)=0$ if $x_0 \neq x_1$.

 

[Hill]

It seems to me that the result should contain a delta function

Hmm. I've just got a comment in my other thread, and the article under the second link there shows an example of second order functional derivative that contains a delta function and obtained by your iterative approach:

 

[TSny]

This link gives a nice way to define higher-order functional derivatives on page 2. As an example, Page 3 gives the first-order and second-order functional derivatives of the functional $F[f] = \int_a^b [f(x)]^n dx \,$: $$\frac{\delta F[f]}{\delta f(x)} = n [f(x)]^{n-1}$$ $$\frac{\delta^2 F[f]}{\delta f(x) \delta f(x')} = n(n-1)[f(x)]^{n-2} \delta(x-x').$$ For the case $n = 3$, we have essentially the same functional as yours : $F[f] = \int_a^b [f(x)]^3 dx$. The second-order functional derivative is then $\frac{\delta^2 F[f]}{\delta f(x) \delta f(x')} = 6f(x) \delta(x-x')$.

 

[Hill]

This link gives a nice way to define higher-order functional derivatives on page 2. As an example, Page 3 gives the first-order and second-order functional derivatives of the functional $F[f] = \int_a^b [f(x)]^n dx \,$: $$\frac{\delta F[f]}{\delta f(x)} = n [f(x)]^{n-1}$$ $$\frac{\delta^2 F[f]}{\delta f(x) \delta f(x')} = n(n-1)[f(x)]^{n-2} \delta(x-x').$$ For the case $n = 3$, we have essentially the same functional as yours : $F[f] = \int_a^b [f(x)]^3 dx$. The second-order functional derivative is then $\frac{\delta^2 F[f]}{\delta f(x) \delta f(x')} = 6f(x) \delta(x-x')$.

Thank you very much. This definition resolves my problem of having delta function outside the integral: as (C.6) shows, it is in fact inside the integral.

 

[TSny]

Thank you very much. This definition resolves my problem of having delta function outside the integral: as (C.6) shows, it is in fact inside the integral.

In C.6 the expression $\delta f(x)$ does not necessarily represent a Dirac delta function. $\delta f(x)$ is just a general small variation in the function $f(x)$.

As an example, consider our functional $F[f] = \int_a^b [f(x)]^3dx$. Then we look at $F[f+\delta f]$. $$\begin{align*} F[f+\delta f] & = \int_a^b [f(x)+\delta f(x)]^3dx = \int_a^b \Bigl\{ [f(x)]^3 + 3[f(x)]^2 \delta f(x) + 3 f(x) [\delta f(x)]^2 + [\delta f(x)]^3 \Bigr\} dx \\ & = F[f] + \int_a^b dx \, 3[f(x)]^2 \delta f(x) + \int_a^b dx \, 3 f(x) [\delta f(x)]^2 + \int_a^b dx \, [\delta f(x)]^3 \,\,\,\,\,\,\,\,\,\,\,\,\,\, (1) \end{align*}$$ We want to get this into the form (C.6). We can use a Dirac delta function to write the second integral in line (1) above as $$\begin{align*}\int_a^b dx \, 3 f(x) \delta f(x) \delta f(x) & = \int_a^b dx \int_a^b dx' \, 3 f(x) \delta(x'-x) \delta f(x) \delta f(x') \\ & = \frac{1}{2!} \int_a^b dx \int_a^b dx' \, 6 f(x) \delta(x'-x) \delta f(x) \delta f(x') \end{align*}$$ We don't need the last integral in (1) if we are only interested in the first-order and second-order functional derivatives. So, we have $$ F[f+\delta f] = F[f] + \int_a^b dx \, 3[f(x)]^2 \delta f(x) +\frac{1}{2!} \int_a^b dx \int_a^b dx' \, 6 f(x) \delta(x'-x) \delta f(x) \delta f(x') + \, ... $$Comparing this with (C.6) we can identity $$\Gamma_1(x) = 3[f(x)]^2 $$ and $$\Gamma_2(x, x') = 6 f(x) \delta(x'-x)$$ But $\Gamma_1(x)$ and $\Gamma_2(x,x')$ are, by definition, the first-order and second-order functional derivatives, respectively. See (C.7) and (C.8) in the link. Note that the second-order functional derivative includes a factor of the Dirac delta function $\delta(x'-x)$.

 

[Hill]

In C.6 the expression δf(x) does not necessarily represent a Dirac delta function. δf(x) is just a general small variation in the function f(x).

Yes, I understand. When I said,

as (C.6) shows, it is in fact inside the integral.

I did not refer to $\delta f(x)$, $\delta f(x')$, etc. I've referred to $\Gamma_1$, $\Gamma_2$, etc. They are inside the integrals, and they are the functional derivatives, which include Dirac delta function that was my concern.
Thank you again.

 

[TSny]

I did not refer to $\delta f(x)$, $\delta f(x')$, etc. I've referred to $\Gamma_1$, $\Gamma_2$, etc. They are inside the integrals, and they are the functional derivatives, which include Dirac delta function that was my concern.
Thank you again.

I see. I misunderstood. Everything's good.