- #1
Einj
- 470
- 59
Hi everyone! I have a question on functional derivatives. I have a function defined as:
$$
F[\{u\}]=\int d^3r \sum_{i=1}^3 \frac{\partial u_i}{\partial r_i},
$$ where [itex]u_i(\vec r)[/itex] is a function of the position. I need to compute its functional derivative. To do that I did the following:
$$
F[\{u_k+\delta u_k\}]=F[\{u_k\}]+\int d^3r\sum_{i=1}^3\delta_{ik}\frac{\partial \delta u_k}{\partial r_i}=F[\{u_k\}]+\int d^3r\frac{\partial \delta u_k}{\partial r_k}.
$$
Now take, for example, [itex]k=1[/itex]. Then we have:
$$
\int d^3r\frac{\partial \delta u_x}{dx}=\int dydz \delta u_x(x,y,z)=\int dx'dy'dz'\delta(x-x')\delta u_x(x',y',z'),
$$
is then correct to say that:
$$
\frac{\delta F[\{u\}]}{\delta u_k(r')}=\delta(r_k-r_k')
$$??
Thanks
$$
F[\{u\}]=\int d^3r \sum_{i=1}^3 \frac{\partial u_i}{\partial r_i},
$$ where [itex]u_i(\vec r)[/itex] is a function of the position. I need to compute its functional derivative. To do that I did the following:
$$
F[\{u_k+\delta u_k\}]=F[\{u_k\}]+\int d^3r\sum_{i=1}^3\delta_{ik}\frac{\partial \delta u_k}{\partial r_i}=F[\{u_k\}]+\int d^3r\frac{\partial \delta u_k}{\partial r_k}.
$$
Now take, for example, [itex]k=1[/itex]. Then we have:
$$
\int d^3r\frac{\partial \delta u_x}{dx}=\int dydz \delta u_x(x,y,z)=\int dx'dy'dz'\delta(x-x')\delta u_x(x',y',z'),
$$
is then correct to say that:
$$
\frac{\delta F[\{u\}]}{\delta u_k(r')}=\delta(r_k-r_k')
$$??
Thanks