Editing Functional derivative (section)

===Formula===
Given a functional
<math display="block">F[\rho] = \int f( \boldsymbol{r}, \rho(\boldsymbol{r}), \nabla\rho(\boldsymbol{r}) )\, d\boldsymbol{r},</math>
and a function <math>\phi(\boldsymbol{r})</math> that vanishes on the boundary of the region of integration, from a previous section [[#Definition|Definition]],
<math display="block">\begin{align}
\int \frac{\delta F}{\delta\rho(\boldsymbol{r})} \, \phi(\boldsymbol{r}) \, d\boldsymbol{r}
& = \left [ \frac{d}{d\varepsilon} \int f( \boldsymbol{r}, \rho + \varepsilon \phi, \nabla\rho+\varepsilon\nabla\phi )\, d\boldsymbol{r} \right ]_{\varepsilon=0} \\
& = \int \left( \frac{\partial f}{\partial\rho} \, \phi + \frac{\partial f}{\partial\nabla\rho} \cdot \nabla\phi \right) d\boldsymbol{r} \\
& = \int \left[ \frac{\partial f}{\partial\rho} \, \phi + \nabla \cdot \left( \frac{\partial f}{\partial\nabla\rho} \, \phi \right) - \left( \nabla \cdot \frac{\partial f}{\partial\nabla\rho} \right) \phi \right] d\boldsymbol{r} \\
& = \int \left[ \frac{\partial f}{\partial\rho} \, \phi - \left( \nabla \cdot \frac{\partial f}{\partial\nabla\rho} \right) \phi \right] d\boldsymbol{r} \\
& = \int \left( \frac{\partial f}{\partial\rho} - \nabla \cdot \frac{\partial f}{\partial\nabla\rho} \right) \phi(\boldsymbol{r}) \ d\boldsymbol{r} \, .
\end{align}</math>

The second line is obtained using the [[total derivative]], where {{math|''∂f'' /''∂∇ρ''}} is a [[Matrix calculus#Scalar-by-vector|derivative of a scalar with respect to a vector]].<ref group="Note">For a three-dimensional Cartesian coordinate system,
<math display="block">\frac{\partial f}{\partial\nabla\rho} = \frac{\partial f}{\partial\rho_x} \mathbf{\hat{i}} + \frac{\partial f}{\partial\rho_y} \mathbf{\hat{j}} + \frac{\partial f}{\partial\rho_z} \mathbf{\hat{k}}\, ,</math>
where <math>\rho_x = \frac{\partial \rho}{\partial x}\, , \ \rho_y = \frac{\partial \rho}{\partial y}\, , \ \rho_z = \frac{\partial \rho}{\partial z}</math> and <math>\mathbf{\hat{i}}</math>, <math>\mathbf{\hat{j}}</math>, <math>\mathbf{\hat{k}}</math> are unit vectors along the x, y, z axes.</ref>

The third line was obtained by use of a [[Divergence#Properties|product rule for divergence]]. The fourth line was obtained using the [[divergence theorem]] and the condition that <math>\phi=0</math> on the boundary of the region of integration. Since <math>\phi</math> is also an arbitrary function, applying the [[fundamental lemma of calculus of variations]] to the last line, the functional derivative is
<math display="block">\frac{\delta F}{\delta\rho(\boldsymbol{r})} = \frac{\partial f}{\partial\rho} - \nabla \cdot \frac{\partial f}{\partial\nabla\rho} </math>

where {{math|1=''ρ'' = ''ρ''('''''r''''')}} and {{math|1=''f'' = ''f'' ('''''r''''', ''ρ'', &nabla;''ρ'')}}. This formula is for the case of the functional form given by {{math|''F''[''ρ'']}} at the beginning of this section. For other functional forms, the definition of the functional derivative can be used as the starting point for its determination. (See the example [[#Coulomb potential energy functional|Coulomb potential energy functional]].)

The above equation for the functional derivative can be generalized to the case that includes higher dimensions and higher order derivatives. The functional would be,
<math display="block">F[\rho(\boldsymbol{r})] = \int f( \boldsymbol{r}, \rho(\boldsymbol{r}), \nabla\rho(\boldsymbol{r}), \nabla^{(2)}\rho(\boldsymbol{r}), \dots, \nabla^{(N)}\rho(\boldsymbol{r}))\, d\boldsymbol{r},</math>

where the vector {{math|'''''r''''' &isin; '''R'''<sup>''n''</sup>}}, and {{math|&nabla;<sup>(''i'')</sup>}} is a tensor whose {{math|''n<sup>i</sup>''}} components are partial derivative operators of order {{math|''i''}},
<math display="block"> \left [ \nabla^{(i)} \right ]_{\alpha_1 \alpha_2 \cdots \alpha_i} = \frac {\partial^{\, i}} {\partial r_{\alpha_1} \partial r_{\alpha_2} \cdots \partial r_{\alpha_i} } \qquad \qquad \text{where} \quad \alpha_1, \alpha_2, \dots, \alpha_i = 1, 2, \dots , n \ . </math><ref group="Note">For example, for the case of three dimensions ({{math|1=''n'' = 3}}) and second order derivatives ({{math|1=''i'' = 2}}), the tensor {{math|&nabla;<sup>(2)</sup>}} has components,
<math display="block"> \left [ \nabla^{(2)} \right ]_{\alpha \beta} = \frac {\partial^{\,2}} {\partial r_{\alpha} \, \partial r_{\beta}} </math>where <math>\alpha</math> and <math>\beta</math> can be <math>1,2,3</math>.</ref>

An analogous application of the definition of the functional derivative yields
<math display="block">\begin{align}
\frac{\delta F[\rho]}{\delta \rho} &{} = \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)} + \dots + (-1)^N \nabla^{(N)} \cdot \frac{\partial f}{\partial\left(\nabla^{(N)}\rho\right)} \\
&{} = \frac{\partial f}{\partial\rho} + \sum_{i=1}^N (-1)^{i}\nabla^{(i)} \cdot \frac{\partial f}{\partial\left(\nabla^{(i)}\rho\right)} \ .
\end{align}</math>

In the last two equations, the {{math|''n<sup>i</sup>''}} components of the tensor <math> \frac{\partial f}{\partial\left(\nabla^{(i)}\rho\right)} </math> are partial derivatives of {{math|''f''}} with respect to partial derivatives of ''ρ'',
<math display="block"> \left [ \frac {\partial f} {\partial \left (\nabla^{(i)}\rho \right ) } \right ]_{\alpha_1 \alpha_2 \cdots \alpha_i} = \frac {\partial f} {\partial \rho_{\alpha_1 \alpha_2 \cdots \alpha_i} }  </math>
where <math>  \rho_{\alpha_1 \alpha_2 \cdots \alpha_i} \equiv \frac {\partial^{\,i}\rho} {\partial r_{\alpha_1} \, \partial r_{\alpha_2} \cdots \partial r_{\alpha_i} } </math>, and the tensor scalar product is,
<math display="block"> \nabla^{(i)} \cdot \frac{\partial f}{\partial\left(\nabla^{(i)}\rho\right)} = \sum_{\alpha_1, \alpha_2, \cdots, \alpha_i = 1}^n \ \frac {\partial^{\, i} } {\partial r_{\alpha_1} \, \partial r_{\alpha_2} \cdots \partial r_{\alpha_i} } \ \frac {\partial f} {\partial \rho_{\alpha_1 \alpha_2 \cdots \alpha_i} } \ . </math> <ref group="Note">For example, for the case {{math|1=''n'' = 3}} and {{math|1=''i'' = 2}}, the tensor scalar product is,
<math display="block"> \nabla^{(2)} \cdot \frac{\partial f}{\partial\left(\nabla^{(2)}\rho\right)} = \sum_{\alpha, \beta = 1}^3 \ \frac {\partial^{\, 2} } {\partial r_{\alpha} \, \partial r_{\beta} } \, \frac {\partial f} {\partial \rho_{\alpha \beta} } , </math>where <math>\rho_{\alpha \beta} \equiv \frac {\partial^{\, 2}\rho} {\partial r_{\alpha} \, \partial r_{\beta} }</math>.</ref>