- #1

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- 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

##\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?

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?