User:Repliedthemockturtle

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In mathematics and theoretical physics, the functional derivative is a generalization of the directional derivative. The difference is that the latter differentiates in the direction of a vector, while the former differentiates in the direction of a function. Both of these can be viewed as extensions of the usual calculus derivative.

Two possible, restricted definitions suitable for certain computations are given here. There are more general definitions of functional derivatives.

For a functional F mapping (continuous/smooth/with certain boundary conditions/etc.) functions φ from a manifold M to R or C, the functional derivative of F, denoted δF is a distribution such that for all test functions f,

$\delta F[\phi]=\left.\frac{d}{d\epsilon}F[\phi+\epsilon f]\right|_{\epsilon=0}.$

Another definition is in terms of a limit and the Dirac delta function, δ:

$\frac{\delta F[\phi(x)]}{\delta \phi(y)}=\lim_{\varepsilon\to 0}\frac{F[\phi(x)+\varepsilon\delta(x-y)]-F[\phi(x)]}{\varepsilon}.$

[edit] Formal description

The definition of a functional derivative may be made much more mathematically precise and formal by defining the space of functions more carefully. For example, when the space of functions is a Banach space, the functional derivative becomes known as the Fréchet derivative, while one uses the Gâteaux derivative on more general locally convex spaces. Note that the well-known Hilbert space is a special case of a Banach space. The more formal treatment allows many theorems from ordinary calculus and analysis to be generalized to corresponding theorems in functional analysis, as well as numerous new theorems to be stated.

[edit] Examples

It is worthwhile to briefly discuss functional derivatives beyond their formal, mathematical definition. Functional derivatives occur regularly in physical problems which obey a variational principle, therefore, it is useful to show how functional derivatives are performed through physically-relevant examples.

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

$\frac{\delta F[\rho]}{\delta \rho} = \frac{\partial f}{\partial \rho} - \nabla\cdot\frac{\partial f}{\partial (\nabla \rho)} + \nabla^2 \frac{\partial f}{\partial (\nabla^2 \rho)} - \cdots.$

Consider the Coulomb energy functional, $J [ρ]$ ,

$J[\rho] = \int \left(\frac{1}{2}\int \frac{\rho(\mathbf{r}) \rho(\mathbf{r}')}{\vert \mathbf{r}-\mathbf{r}' \vert} d^3r'\right) d^3r.$

$J [ρ]$ depends on the charge density $ρ$ only and does not depend on it's gradient, Laplacian, or higher-order derivatives. Therefore,

$\frac{\delta J[\rho]}{\delta \rho} = \frac{\partial j}{\partial \rho} = \int \frac{\rho(\mathbf{r}')}{\vert \mathbf{r}-\mathbf{r}' \vert} d^3r'$

where

$j = \frac{1}{2}\int \frac{\rho(\mathbf{r}) \rho(\mathbf{r}')}{\vert \mathbf{r}-\mathbf{r}' \vert} d^3r'$

The second functional derivative of the Coulomb energy functional is

$\frac{\delta^2 J[\rho]}{\delta \rho^2} = \frac{\delta}{\delta \rho}\int \frac{\rho(\mathbf{r}')}{\vert \mathbf{r}-\mathbf{r}' \vert} d^3r' = \frac{\partial}{\partial \rho} \frac{\rho(\mathbf{r}')}{\vert \mathbf{r}-\mathbf{r}' \vert} = \frac{1}{\vert \mathbf{r}-\mathbf{r}' \vert}$

As a second example, consider the Weizsacker kinetic energy functional

$T[\rho]=\int \frac{1}{8} \frac{\nabla\rho(\mathbf{r}) \cdot \nabla\rho(\mathbf{r})}{ \rho(\mathbf{r}) } d^3r$

$T [ρ]$ depends on the charge density and it's gradient, therefore,

$\frac{\delta T[\rho]}{\delta \rho} = \frac{\partial t}{\partial \rho} - \nabla\cdot\frac{\partial t}{\partial (\nabla \rho)} = -\frac{1}{8} \frac{\nabla\rho(\mathbf{r}) \cdot \nabla\rho(\mathbf{r})}{ \rho(\mathbf{r})^2 } - \nabla\cdot\left(\frac{1}{4} \frac{\nabla\rho(\mathbf{r})}{ \rho(\mathbf{r}) }\right)$

where

$t = \frac{1}{8} \frac{\nabla\rho(\mathbf{r}) \cdot \nabla\rho(\mathbf{r})}{ \rho(\mathbf{r}) }$

Finally, note (although, a rather obscure note) that any function can be written in terms of a functional. For example,

$\rho(\mathbf{r}) = \int \rho(\mathbf{r}') \delta(\mathbf{r}-\mathbf{r}') d^3r'.$

Therefore,

$\frac{\delta \rho(\mathbf{r})}{\delta\rho}=\frac{\delta \int \rho(\mathbf{r}') \delta(\mathbf{r}-\mathbf{r}') d^3r'}{\delta \rho} = \frac{\partial \left(\rho(\mathbf{r}') \delta(\mathbf{r}-\mathbf{r}')\right)}{\partial \rho} = \delta(\mathbf{r}-\mathbf{r}')$

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Category: Functional analysis

User:Repliedthemockturtle

From Wikipedia, the free encyclopedia

[edit] Formal description

[edit] Examples

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