Differintegral

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Fractional integration redirects here. Not to be confused with Autoregressive fractionally integrated moving average

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In fractional calculus, an area of applied mathematics, the differintegral is a combined differentiation/integration operator. Applied to a function ƒ, the q-differintegral of f, here denoted by

\mathbb{D}^qf

is the fractional derivative (if q > 0) or fractional integral (if q < 0). If q = 0, then the q-th differintegral of a function is the function itself. In the context of fractional integration and differentiation, there are several legitimate definitions of the differintegral.

Standard definitions

The three most common forms are:

This is the simplest and easiest to use, and consequently it is the most often used. It is a generalization of the Cauchy formula for repeated integration to arbitrary order.

\begin{align}
{}_a\mathbb{D}^q_tf(t) & = \frac{d^qf(t)}{d(t-a)^q} \\
& =\frac{1}{\Gamma(n-q)} \frac{d^n}{dt^n} \int_{a}^t (t-\tau)^{n-q-1}f(\tau)d\tau
\end{align}
The Grunwald–Letnikov differintegral is a direct generalization of the definition of a derivative. It is more difficult to use than the Riemann–Liouville differintegral, but can sometimes be used to solve problems that the Riemann–Liouville cannot.

\begin{align}
{}_a\mathbb{D}^q_tf(t) & = \frac{d^qf(t)}{d(t-a)^q} \\
& =\lim_{N \to \infty}\left[\frac{t-a}{N}\right]^{-q}\sum_{j=0}^{N-1}(-1)^j{q \choose j}f\left(t-j\left[\frac{t-a}{N}\right]\right)
\end{align}
This is formally similar to the Riemann–Liouville differintegral, but applies to periodic functions, with integral zero over a period.

Definitions via transforms

Recall the continuous Fourier transform, here denoted  \mathcal{F} :

 F(\omega) =  \mathcal{F}\{f(t)\} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t) e^{- i\omega t}\,dt

Using the continuous Fourier transform, in Fourier space, differentiation transforms into a multiplication:

\mathcal{F}\left[\frac{df(t)}{dt}\right] = i \omega \mathcal{F}[f(t)]

So,

\frac{d^nf(t)}{dt^n} = \mathcal{F}^{-1}\left\{(i \omega)^n\mathcal{F}[f(t)]\right\}

which generalizes to

\mathbb{D}^qf(t)=\mathcal{F}^{-1}\left\{(i \omega)^q\mathcal{F}[f(t)]\right\}.

Under the Laplace transform, here denoted by  \mathcal{L}, differentiation transforms into a multiplication

\mathcal{L}\left[\frac{df(t)}{dt}\right] = s\mathcal{L}[f(t)].

Generalizing to arbitrary order and solving for Dqf(t), one obtains

\mathbb{D}^qf(t)=\mathcal{L}^{-1}\left\{s^q\mathcal{L}[f(t)]\right\}.

Basic formal properties

Linearity rules

\mathbb{D}^q(f+g)=\mathbb{D}^q(f)+\mathbb{D}^q(g)
\mathbb{D}^q(af)=a\mathbb{D}^q(f)

Zero rule

\mathbb{D}^0 f=f \,

Product rule

\mathbb{D}^q_t(fg)=\sum_{j=0}^{\infty} {q \choose j}\mathbb{D}^j_t(f)\mathbb{D}^{q-j}_t(g)

In general, composition (or semigroup) rule

\mathbb{D}^a\mathbb{D}^{b}f = \mathbb{D}^{a+b}f

is not satisfied.[1]

A selection of basic formulæ

\mathbb{D}^q(t^n)=\frac{\Gamma(n+1)}{\Gamma(n+1-q)}t^{n-q}
\mathbb{D}^q(\sin(t))=\sin \left( t+\frac{q\pi}{2} \right)
\mathbb{D}^q(e^{at})=a^q e^{at}

See also

References

  1. See Lua error in package.lua at line 80: module 'strict' not found.

External links