Editing Functional derivative (section)

===Examples===

====Thomas–Fermi kinetic energy functional====
The [[Thomas–Fermi model]] of 1927 used a kinetic energy functional for a noninteracting uniform [[free electron model|electron gas]] in a first attempt of [[density-functional theory]] of electronic structure:
<math display="block">T_\mathrm{TF}[\rho] = C_\mathrm{F} \int \rho^{5/3}(\mathbf{r}) \, d\mathbf{r} \, .</math>
Since the integrand of {{math|''T''<sub>TF</sub>[''ρ'']}} does not involve derivatives of {{math|''ρ''('''''r''''')}}, the functional derivative of {{math|''T''<sub>TF</sub>[''ρ'']}} is,<ref name=ParrYangP247A.6>{{harvp|Parr|Yang|1989|loc=p. 247, Eq. A.6}}.</ref>
<math display="block">\frac{\delta T_{\mathrm{TF}}}{\delta \rho (\boldsymbol{r}) }
= C_\mathrm{F} \frac{\partial \rho^{5/3}(\mathbf{r})}{\partial \rho(\mathbf{r})}
= \frac{5}{3} C_\mathrm{F} \rho^{2/3}(\mathbf{r}) \, .</math>

====Coulomb potential energy functional====
The '''electron-nucleus''' potential energy is
<math display="block">V[\rho] = \int \frac{\rho(\boldsymbol{r})}{|\boldsymbol{r}|} \ d\boldsymbol{r}.</math>

Applying the definition of functional derivative,
<math display="block">\begin{align}
\int \frac{\delta V}{\delta \rho(\boldsymbol{r})} \ \phi(\boldsymbol{r}) \ d\boldsymbol{r}
& {} = \left [ \frac{d}{d\varepsilon} \int \frac{\rho(\boldsymbol{r}) + \varepsilon \phi(\boldsymbol{r})}{|\boldsymbol{r}|} \ d\boldsymbol{r} \right ]_{\varepsilon=0} \\[1ex]
& {} = \int \frac {\phi(\boldsymbol{r})} {|\boldsymbol{r}|} \ d\boldsymbol{r} \, .
\end{align}</math>
So,
<math display="block"> \frac{\delta V}{\delta \rho(\boldsymbol{r})} = \frac{1}{|\boldsymbol{r}|} \ . </math>

The functional derivative of the classical part of the '''electron-electron interaction''' (often called Hartree energy) is 
<math display="block">J[\rho] = \frac{1}{2}\iint \frac{\rho(\mathbf{r}) \rho(\mathbf{r}')}{| \mathbf{r}-\mathbf{r}' |}\, d\mathbf{r} d\mathbf{r}' \, .</math>
From the [[#Functional derivative|definition of the functional derivative]],
<math display="block">\begin{align}
\int \frac{\delta J}{\delta\rho(\boldsymbol{r})} \phi(\boldsymbol{r})d\boldsymbol{r}
& {} = \left [ \frac {d \ }{d\varepsilon} \, J[\rho + \varepsilon\phi] \right ]_{\varepsilon = 0} \\
& {} = \left [ \frac {d \ }{d\varepsilon} \, \left ( \frac{1}{2}\iint \frac {[\rho(\boldsymbol{r}) + \varepsilon \phi(\boldsymbol{r})] \, [\rho(\boldsymbol{r}') + \varepsilon \phi(\boldsymbol{r}')] }{| \boldsymbol{r}-\boldsymbol{r}' |}\, d\boldsymbol{r} d\boldsymbol{r}' \right ) \right ]_{\varepsilon = 0} \\
& {} = \frac{1}{2}\iint \frac {\rho(\boldsymbol{r}') \phi(\boldsymbol{r}) }{| \boldsymbol{r}-\boldsymbol{r}' |}\, d\boldsymbol{r} d\boldsymbol{r}' + \frac{1}{2}\iint \frac {\rho(\boldsymbol{r}) \phi(\boldsymbol{r}') }{| \boldsymbol{r}-\boldsymbol{r}' |}\, d\boldsymbol{r} d\boldsymbol{r}' \\
\end{align}</math>
The first and second terms on the right hand side of the last equation are equal, since {{math|'''''r'''''}} and {{math|'''''r&prime;'''''}} in the second term can be interchanged without changing the value of the integral. Therefore,
<math display="block"> \int \frac{\delta J}{\delta\rho(\boldsymbol{r})} \phi(\boldsymbol{r})d\boldsymbol{r} = \int \left ( \int \frac {\rho(\boldsymbol{r}') }{| \boldsymbol{r}-\boldsymbol{r}' |} d\boldsymbol{r}' \right ) \phi(\boldsymbol{r}) d\boldsymbol{r} </math>
and the functional derivative of the electron-electron Coulomb potential energy functional {{math|''J''}}[''ρ''] is,<ref name=ParrYangP248A.11>{{harvp|Parr|Yang|1989|loc=p. 248, Eq. A.11}}.</ref>
<math display="block"> \frac{\delta J}{\delta\rho(\boldsymbol{r})} = \int \frac {\rho(\boldsymbol{r}') }{| \boldsymbol{r}-\boldsymbol{r}' |} d\boldsymbol{r}' \, . </math>

The second functional derivative is
<math display="block">\frac{\delta^2 J[\rho]}{\delta \rho(\mathbf{r}')\delta\rho(\mathbf{r})} = \frac{\partial}{\partial \rho(\mathbf{r}')} \left ( \frac{\rho(\mathbf{r}')}{| \mathbf{r}-\mathbf{r}' |} \right ) = \frac{1}{| \mathbf{r}-\mathbf{r}' |}.</math>

====von Weizsäcker kinetic energy functional====
In 1935 [[Carl Friedrich von Weizsacker|von Weizsäcker]] proposed to add a gradient correction to the Thomas-Fermi kinetic energy functional to make it better suit a molecular electron cloud:
<math display="block">T_\mathrm{W}[\rho] = \frac{1}{8} \int \frac{\nabla\rho(\mathbf{r}) \cdot \nabla\rho(\mathbf{r})}{ \rho(\mathbf{r}) } d\mathbf{r} = \int t_\mathrm{W}(\mathbf{r}) \ d\mathbf{r} \, ,</math>
where
<math display="block"> t_\mathrm{W} \equiv \frac{1}{8} \frac{\nabla\rho \cdot \nabla\rho}{ \rho } \qquad \text{and} \ \ \rho = \rho(\boldsymbol{r}) \ . </math>
Using a previously derived [[#Formula|formula]] for the functional derivative,
<math display="block">\begin{align}
\frac{\delta T_\mathrm{W}}{\delta \rho}
& = \frac{\partial t_\mathrm{W}}{\partial \rho} - \nabla\cdot\frac{\partial t_\mathrm{W}}{\partial \nabla \rho} \\
& = -\frac{1}{8}\frac{\nabla\rho \cdot \nabla\rho}{\rho^2} - \left ( \frac {1}{4} \frac {\nabla^2\rho} {\rho} - \frac {1}{4} \frac {\nabla\rho \cdot \nabla\rho} {\rho^2} \right ) \qquad \text{where} \ \ \nabla^2 = \nabla \cdot \nabla \ ,
\end{align}</math>
and the result is,<ref name=ParrYangP247A.9>{{harvp|Parr|Yang|1989|loc= p. 247, Eq. A.9}}.</ref>
<math display="block"> \frac{\delta T_\mathrm{W}}{\delta \rho} = \ \ \, \frac{1}{8}\frac{\nabla\rho \cdot \nabla\rho}{\rho^2} - \frac{1}{4}\frac{\nabla^2\rho}{\rho} \ . </math>

====Entropy====
The [[information entropy|entropy]] of a discrete [[random variable]] is a functional of the [[probability mass function]].

<math display="block">H[p(x)] = -\sum_x p(x) \log p(x)</math>
Thus,
<math display="block">\begin{align}
\sum_x \frac{\delta H}{\delta p(x)} \, \phi(x)
& {} = \left[ \frac{d}{d\varepsilon} H[p(x) + \varepsilon\phi(x)] \right]_{\varepsilon=0}\\
& {} = \left [- \, \frac{d}{d\varepsilon} \sum_x \, [p(x) + \varepsilon\phi(x)] \ \log [p(x) + \varepsilon\phi(x)] \right]_{\varepsilon=0} \\
& {} = -\sum_x \, [1+\log p(x)] \ \phi(x) \, .
\end{align}</math>
Thus,
<math display="block">\frac{\delta H}{\delta p(x)} = -1-\log p(x).</math>

==== Exponential ====

Let
<math display="block"> F[\varphi(x)]= e^{\int \varphi(x) g(x)dx}.</math>

Using the delta function as a test function,
<math display="block">\begin{align}
\frac{\delta F[\varphi(x)]}{\delta \varphi(y)}
& {} = \lim_{\varepsilon\to 0}\frac{F[\varphi(x)+\varepsilon\delta(x-y)]-F[\varphi(x)]}{\varepsilon}\\
& {} = \lim_{\varepsilon\to 0}\frac{e^{\int (\varphi(x)+\varepsilon\delta(x-y)) g(x)dx}-e^{\int \varphi(x) g(x)dx}}{\varepsilon}\\
& {} = e^{\int \varphi(x) g(x)dx}\lim_{\varepsilon\to 0}\frac{e^{\varepsilon \int \delta(x-y) g(x)dx}-1}{\varepsilon}\\
& {} = e^{\int \varphi(x) g(x)dx}\lim_{\varepsilon\to 0}\frac{e^{\varepsilon g(y)}-1}{\varepsilon}\\
& {} = e^{\int \varphi(x) g(x)dx}g(y).
\end{align}</math>

Thus,
<math display="block"> \frac{\delta F[\varphi(x)]}{\delta \varphi(y)} = g(y) F[\varphi(x)]. </math>

This is particularly useful in calculating the [[Correlation function (quantum field theory)|correlation functions]] from the [[Partition function (quantum field theory)|partition function]] in [[quantum field theory]].

====Functional derivative of a function====
A function can be written in the form of an integral like a functional. For example,
<math display="block">\rho(\boldsymbol{r}) = F[\rho] = \int \rho(\boldsymbol{r}') \delta(\boldsymbol{r}-\boldsymbol{r}')\, d\boldsymbol{r}'.</math>
Since the integrand does not depend on derivatives of ''ρ'', the functional derivative of ''ρ''{{math|('''''r''''')}} is,
<math display="block">\frac {\delta \rho(\boldsymbol{r})} {\delta\rho(\boldsymbol{r}')} \equiv \frac {\delta F} {\delta\rho(\boldsymbol{r}')}
= \frac{\partial \ \ }{\partial \rho(\boldsymbol{r}')} \, [\rho(\boldsymbol{r}') \delta(\boldsymbol{r}-\boldsymbol{r}')]
= \delta(\boldsymbol{r}-\boldsymbol{r}').</math>

==== Functional derivative of iterated function====
The functional derivative of the iterated function <math>f(f(x))</math> is given by:
<math display="block">\frac{\delta f(f(x))}{\delta f(y) } = f'(f(x))\delta(x-y) + \delta(f(x)-y)</math>
and
<math display="block">\frac{\delta f(f(f(x)))}{\delta f(y) } = f'(f(f(x))(f'(f(x))\delta(x-y) + \delta(f(x)-y)) + \delta(f(f(x))-y)</math>

In general:
<math display="block">\frac{\delta f^N(x)}{\delta f(y)} = f'( f^{N-1}(x) ) \frac{ \delta f^{N-1}(x)}{\delta f(y)} + \delta( f^{N-1}(x) - y ) </math>

Putting in {{math|1=''N'' = 0}} gives:
<math display="block"> \frac{\delta f^{-1}(x)}{\delta f(y) } = - \frac{ \delta(f^{-1}(x)-y ) }{ f'(f^{-1}(x)) }</math>