User:Bob K31416/FD

.
A functional derivative relates a change in a functional to a change in a function that the functional depends on. For example, consider a functional F [ρ] that depends on the function ρ(x). Consider a small change in ρ(x) that adds to it a function $δ$ρ(x). The corresponding change in F [ρ] is,
 * $$ \delta F [\rho]=  F [\rho + \delta \rho] - F [\rho] $$.

The functional derivative $δF/δ$ρ relates $δF$ to $δ$ρ.

.
Given a manifold M representing (continuous/smooth/with certain boundary conditions/etc.) functions ρ and a functional F defined as
 * $$F\colon M \rightarrow \mathbb{R} \quad \mbox{or} \quad F\colon M \rightarrow \mathbb{C} \, ,$$

the functional derivative of $F[$ρ], denoted $δF/δ$ρ, is defined by



\begin{align} \int \frac{\delta F}{\delta\rho(x)} \ \phi(x) \ dx &= \lim_{\varepsilon\to 0}\frac{F[\rho+\varepsilon \phi]-F[\rho]}{\varepsilon} \\ &= \left [ \frac{d}{d\epsilon}F[\rho+\epsilon \phi]\right ]_{\epsilon=0}, \end{align} $$

where $εφ$ is the variation of ρ, and  $φ$ is an arbitrary function. These equations come from the expansion of the integrand of a functional in a power series. For example, consider the functional
 * $$ F[\rho] = \int \limits_a^b L[ \, x, \rho(x), \rho \, '(x) \, ] \, dx \, $$

where ρ&prime;(x) &equiv; dρ/dx. If ρ is varied by adding to it $εφ$, then the resulting integrand expanded in a Taylor series in powers of $εφ$ is,



\begin{align} L(x, \rho+\epsilon\phi, \rho'+\epsilon\phi') &= L(x,  \rho, \rho')  + \left [ \frac{d}{d(\rho+\epsilon\phi)}L(x,  \rho+\epsilon\phi, \rho'+\epsilon\phi')\right ]_{\epsilon\phi=0} \epsilon\phi +  \frac{1}{2} \left [ \frac{d^2}{d(\rho+\epsilon\phi)^2}L(x,  \rho+\epsilon\phi, \rho'+\epsilon\phi')\right ]_{\epsilon\phi=0} (\epsilon\phi)^2 + \cdots \\ &= L(x,  \rho, \rho')  + \left [ \frac{d}{d\epsilon}L(x,  \rho+\epsilon\phi, \rho'+\epsilon\phi')\right ]_{\epsilon=0} \epsilon +  \frac{1}{2} \left [ \frac{d^2}{d\epsilon^2}L(x,  \rho+\epsilon\phi, \rho'+\epsilon\phi')\right ]_{\epsilon=0} \epsilon^2 + \cdots \end{align} $$ Integrating both sides,
 * $$ F[\rho+\epsilon \phi] = F[\rho] + \int \left \{ \left [ \frac{d}{d\epsilon}L(x,  \rho+\epsilon\phi, \rho'+\epsilon\phi')\right ]_{\epsilon=0} \epsilon +  \frac{1}{2} \left [ \frac{d^2}{d\epsilon^2}L(x,  \rho+\epsilon\phi, \rho'+\epsilon\phi')\right ]_{\epsilon=0} \epsilon^2 + \cdots \right \} \ dx $$
 * $$ \frac {F[\rho+\epsilon \phi] - F[\rho]} {\epsilon} = \left [ \frac{d}{d\epsilon} F[\rho+\epsilon \phi] \right ]_{\epsilon=0}  +  \frac{1}{2} \left [ \frac{d^2}{d\epsilon^2}F[\rho+\epsilon \phi] \right ]_{\epsilon=0} \epsilon + \cdots  $$
 * $$ \lim_{\epsilon\to 0}\frac{F[\rho+\epsilon \phi]-F[\rho]}{\epsilon} = \left [ \frac{d}{d\epsilon} F[\rho+\epsilon \phi] \right ]_{\epsilon=0} $$



\begin{align} \delta F &= \int_a^b \frac {\delta F} {\delta \rho(x)}  \ \delta \rho(x) \ dx \\ &= \int_a^b \frac {\delta F} {\delta \rho(x)}  \ \epsilon \phi(x) \ dx \end{align} $$

then the change in the value of $J$ to first order in $δf$ can be expressed as
 * $$ \delta J = \int_a^b  \frac{\delta J}{\delta f(x)} {\delta f(x)} dx \, . $$

The coefficient of $δf(x)$, denoted as $δJ/δf(x)$, is called the functional derivative of $J$ with respect to $f$ at the point $x$.

The differential (or variation or first variation) of the functional $F$[ρ] is,
 * $$ \delta F = \int   \frac {\delta F} {\delta \rho(x)} \ \delta \rho(x)   \ dx \, $$

where $δ$ρ$(x) = εϕ(x)$ is the variation of ρ$(x)$. This is similar in form to the total differential of a function $F$(ρ1, ρ2, ..., ρn),
 * $$ dF =  \sum_{i=1} ^n  \frac {\partial F} {\partial \rho_i} \ d\rho_i  \ ,$$

where ρ1, ρ2, ..., ρn are independent variables. Comparing the last two equations, the functional derivative $δF/δ$ρ$(x)$ has a role similar to that of the partial derivative $&part;F/&part;$ρi, where the variable of integration $x$ is like a continuous version of the summation index $i$.

Definition
There are several approaches to defining the functional derivative: coefficient in a first order term; derivative of a function; ratio to an area.

.
03:02, 21 May 2006‎ 66.41.8.178:

Given a functional of the form
 * $$F[\rho] = \int f( \mathbf{r}, \rho, \nabla\rho, \nabla^2\rho, \cdots ) d^3r,$$

the functional derivative can be written as
 * $$\frac{\delta F[\rho]}{\delta \rho} = \frac{\partial f}{\partial \rho} - \nabla\cdot\frac{\partial f}{\partial (\nabla \rho)} + \nabla^2 \frac{\partial f}{\partial (\nabla^2 \rho)} - \cdots.$$

19:55, 20 August 2006‎ Md2perpe: Removed

05:59, 29 September 2006‎ DrF:

In the more general case that the functional depends on higher order derivatives, i.e.,
 * $$F[\rho] = \int f( \mathbf{r}, \rho(\mathbf{r}), \nabla\rho(\mathbf{r}), \nabla^2\rho(\mathbf{r}), \cdots, \nabla^N\rho(\mathbf{r}))\, d^3r,$$

an analogous application of the definition yields

\frac{\delta F[\rho]}{\delta \rho(\mathbf{r})} = \frac{\partial f}{\partial\rho} - \nabla \cdot \frac{\partial f}{\partial\nabla\rho} + \nabla^2 \cdot \frac{\partial f}{\partial\left(\nabla^2\rho\right)} - \cdots \nabla^N \cdot \frac{\partial f}{\partial\left(\nabla^N\rho\right)} = \sum_{i=0}^N (-1)^{i}\nabla^i \cdot \frac{\partial f}{\partial\left(\nabla^i\rho\right)} $$

00:53, 7 November 2006‎ GuidoGer:

In the more general case that the functional depends on higher order derivatives, i.e.,
 * $$F[\rho] = \int f( \mathbf{r}, \rho(\mathbf{r}), \nabla\rho(\mathbf{r}), \nabla^2\rho(\mathbf{r}), \cdots, \nabla^N\rho(\mathbf{r}))\, d^3r,$$

where $$\nabla^i$$ is a vector whose $$N*N$$ components are all partial derivative operators of order $$i$$, i.e. $$\partial^i/(\partial r^{i_1}_1 \partial r^{i_2}_2 ... \partial r^{i_N}_N)$$ with $$i_1+i_2+...+i_N = i$$, an analogous application of the definition yields

\frac{\delta F[\rho]}{\delta \rho(\mathbf{r})} = \frac{\partial f}{\partial\rho} - \nabla \cdot \frac{\partial f}{\partial(\nabla\rho)} + \nabla^2 \cdot \frac{\partial f}{\partial\left(\nabla^2\rho\right)} - \cdots + (-1)^N \nabla^N \cdot \frac{\partial f}{\partial\left(\nabla^N\rho\right)} = \sum_{i=0}^N (-1)^{i}\nabla^i \cdot \frac{\partial f}{\partial\left(\nabla^i\rho\right)}. $$

a&#770; i&#770; e&#770;  j&#770;

$F [ f + δf ] &minus; F [ f ]$

$$ \hat{\mathbf{i}}              $$

where

In terms of unit vectors $$ \mathbf{\hat{i}}, \, \mathbf{\hat{j}}, \, \mathbf{\hat{k}} $$       along the axes of a cartesian coordinate system and partial derivatives ρx = &part;ρ/&part;x, ρy = &part;ρ/&part;y, ρz = &part;ρ/&part;z,

$δf$ρ] — $δf (x)$ρ]

$δf$ρ] — $δF / δf (x)$ρ]

$x$[ρ] — $δf (x)$[ρ]

$δF (x)$ — $F (f_{1}, f_{2}, ... ):$


 * $$\rho = \rho(\mathbf{r}) $$


 * $$T_W[\rho] = \frac{1}{8} \int \frac{\nabla\rho \cdot \nabla\rho} {\rho} \ d\,^3r $$


 * $$\begin{align}

T_W[\rho + \delta\rho]  & = \frac{1}{8} \int \frac{\nabla(\rho + \delta\rho) \cdot \nabla(\rho + \delta\rho)} {\rho + \delta\rho} \, d\,^3r  \\ & = \frac{1}{8} \int \frac{\nabla \rho \cdot \nabla\rho + 2 \nabla \rho \cdot \nabla\delta\rho + \nabla\delta\rho \cdot \nabla\delta\rho} {\rho + \delta\rho} \ d\,^3r \\ \end{align}$$ Expanding the integrand in a Taylor series in powers of $dF = Σ _{i} (&part;F / &part;f_{i} ) df_{i}.$ρ and $J[ y]$ρ,
 * $$T_W[\rho + \delta\rho] = \frac{1}{8} \int \left ( \frac{\nabla\rho \cdot \nabla\rho}{\rho} \ - \ \frac{\nabla\rho \cdot \nabla\rho}{\rho^2} \, \delta\rho \ + \ 2 \, \frac{\nabla\rho \cdot \nabla\delta\rho}{\rho} \ + \ R_2 \right ) d\,^3r $$

where $h = h(x)$ is the remainder after the first order terms. Using an identity for $$\frac {\nabla\delta\rho}{\rho} \,, $$ the integral of the third term in the integrand is,
 * $$I_3 \equiv \int 2 \, \frac{\nabla\rho \cdot \nabla\delta\rho}{\rho} \  d\,^3r =  \int 2 \, \nabla\rho \cdot \left [ \nabla \left ( \frac{\delta\rho}{\rho} \right ) +  \frac{\delta\rho \ \nabla\rho}{\rho^2} \right ] \  d\,^3r $$

Using Green's first identity and the condition $y = y(x)$ρ = 0 on the boundary of the region of integration,
 * $$I_3 = \int \left ( - \, 2 \, \frac{\nabla^2\rho}{\rho} \, \delta\rho \ + \ 2 \, \frac {\nabla\rho \cdot \nabla\rho}{\rho^2} \, \delta\rho \right )  d\,^3r $$

Substituting the integrand of $y$ for the third term in the integrand of $ΔJ[h]$[ρ + $h$ρ],
 * $$ \begin{align}

T_W[\rho + \delta\rho] & = \frac{1}{8} \int \left ( \frac{\nabla\rho \cdot \nabla\rho}{\rho} \ - \ \frac{\nabla\rho \cdot \nabla\rho}{\rho^2} \, \delta\rho \ - \, 2 \, \frac{\nabla^2\rho}{\rho} \, \delta\rho \ + \ 2 \, \frac {\nabla\rho \cdot \nabla\rho}{\rho^2} \, \delta\rho \ + \ R_2 \right ) d\,^3r  \\ & = T_W[\rho] \ + \  \frac {1}{8} \int \left ( - \, 2 \, \frac{\nabla^2\rho}{\rho} \, \delta\rho \  + \  \frac {\nabla\rho \cdot \nabla\rho}{\rho^2} \,   \delta\rho \right )  d\,^3r \ + \ \frac {1}{8} \int R_2 \ d\,^3r \end{align}$$ Then the difference $φ[h]$[ρ+$ε &rarr; 0$ρ] &minus; $||h|| &rarr; 0$[ρ] to first order in $J[ y]$ρ is,
 * $$ \delta T_W =   \int \left ( - \, \frac {1}{4} \, \frac{\nabla^2\rho}{\rho}  \ + \  \frac {1}{8} \, \frac {\nabla\rho \cdot \nabla\rho}{\rho^2}     \right ) \delta\rho \  d\,^3r  $$

The functional derivative is the coefficient of $ΔJ[h]$ρ in the integrand,
 * $$ \frac {\delta T_W}{\delta\rho} =   - \, \frac {1}{4} \, \frac{\nabla^2\rho}{\rho}  \ + \  \frac {1}{8} \, \frac {\nabla\rho \cdot \nabla\rho}{\rho^2} \ . $$