In mathematics, the Riemann–Liouville integral associates with a real function <math>f: \mathbb{R} \rightarrow \mathbb{R}</math> another function of the same kind for each value of the parameter . The integral is a manner of generalization of the repeated antiderivative of in the sense that for positive integer values of , is an iterated antiderivative of of order . The Riemann–Liouville integral is named for Bernhard Riemann and Joseph Liouville, the latter of whom was the first to consider the possibility of fractional calculus in 1832. The operator agrees with the Euler transform, after Leonhard Euler, when applied to analytic functions. It was generalized to arbitrary dimensions by Marcel Riesz, who introduced the Riesz potential.
Motivation
The Riemann-Liouville integral is motivated from Cauchy formula for repeated integration. For a function continuous on the interval [,], the Cauchy formula for -fold repeated integration states that
<math display="block">I^n f(x) = f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t.</math>
Now, this formula can be generalized to any positive real number by replacing positive integer with , Therefore we obtain the definition of Riemann-Liouville fractional Integral by
:<math>I^\alpha f(x) = \frac{1}{\Gamma(\alpha)} \int_a^x\left(x-t\right)^{\alpha-1} f(t)\,\mathrm{d}t</math>
Definition
The Riemann–Liouville integral is defined by
:<math>I^\alpha f(x) = \frac{1}{\Gamma(\alpha)}\int_a^x(x-t)^{\alpha-1}f(t)\,dt</math>
where is the gamma function and is an arbitrary but fixed base point. The integral is well-defined provided is a locally integrable function, and is a complex number in the half-plane . The dependence on the base-point is often suppressed, and represents a freedom in constant of integration. Clearly is an antiderivative of (of first order), and for positive integer values of , is an antiderivative of order by Cauchy formula for repeated integration. Another notation, which emphasizes the base point, is
:<math>{}_aD_x^{-\alpha}f(x) = \frac{1}{\Gamma(\alpha)}\int_a^x (x-t)^{\alpha-1}f(t)\,dt.</math>
This also makes sense if , with suitable restrictions on .
The fundamental relations hold
:<math>\frac{d}{dx}I^{\alpha+1} f(x) = I^\alpha f(x),\quad I^\alpha(I^\beta f) = I^{\alpha+\beta}f,</math>
the latter of which is a semigroup property. and produces a derivative that has different properties: it produces zero from constant functions and, more importantly, the initial value terms of the Laplace Transform are expressed by means of the values of that function and of its derivative of integer order rather than the derivatives of fractional order as in the Riemann–Liouville derivative. The Caputo fractional derivative with base point , is then:
:<math>D_x^{\alpha}f(y)=\frac{1}{\Gamma(1-\alpha)}\int_x^y (u-x)^{-\alpha}f'(y-u)du.</math>
Another representation is:
:<math>{}_a\tilde{D}^\alpha_x f(x)=I^{\lceil \alpha\rceil-\alpha}\left(\frac{d^{\lceil \alpha\rceil}f}{dx^{\lceil \alpha\rceil\right).</math>
Fractional derivative of a basic power function
right|thumb|320px|The half derivative (purple curve) of the function (blue curve) together with the first derivative (red curve).
right|thumb|320px|The animation shows the derivative operator oscillating between the [[antiderivative (: ) and the derivative (: ) of the simple power function continuously.]]
Let us assume that is a monomial of the form
:<math>f(x) = x^k\,.</math>
The first derivative is as usual
:<math>f'(x) = \frac{d}{dx}f(x) = k x^{k-1}\,.</math>
Repeating this gives the more general result that
:<math>\frac{d^a}{dx^a}x^k = \dfrac{k!}{(k-a)!}x^{k-a}\,,</math>
which, after replacing the factorials with the gamma function, leads to
:<math>\frac{d^a}{dx^a}x^k = \dfrac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}, \quad\ k > 0.</math>
For and , we obtain the half-derivative of the function <math>x \mapsto x</math> as
:<math>\frac{d^\frac12}{dx^\frac12}x=\frac{\Gamma(1+1)}{\Gamma\left(1-\frac12 + 1\right)} x^{1-\frac12}=\frac{\Gamma(2)}{\Gamma\left(\frac{3}{2}\right)}x^\frac12 = \frac{1}{\frac{\sqrt{\pi{2x^\frac12.</math>
To demonstrate that this is, in fact, the "half derivative" (where ), we repeat the process to get:
:<math>\dfrac{d^\frac12}{dx^\frac12} \dfrac{2x^{\frac12{\sqrt{\pi
=\frac{2}{\sqrt{\pi\dfrac{\Gamma(1+\frac12)}{\Gamma(\frac12-\frac12+1)}x^{\frac12-\frac12}
=\frac{2}{\sqrt{\pi \frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma(1)} x^0
=\frac{2 \frac{\sqrt{\pi{2} x^0}{\sqrt{\pi=1\,,</math>
(because <math display="inline">\Gamma\!\left(\frac{3}{2}\right) = \frac{\sqrt{\pi{2}</math> and ) which is indeed the expected result of
:<math>\left(\frac{d^\frac12}{dx^\frac12} \frac{d^\frac12}{dx^\frac12}\right)\!x = \frac{d}{dx} x = 1\,.</math>
For negative integer power , 1/<math display="inline">\Gamma</math> is 0, so it is convenient to use the following relation:
:<math>\frac{d^a}{dx^a}x^{-k} = \left(-1\right)^a\dfrac{\Gamma(k+a)}{\Gamma(k)}x^{-(k+a)} \quad\text{ for } k \ge 0.</math>
This extension of the above differential operator need not be constrained only to real powers; it also applies for complex powers. For example, the -th derivative of the -th derivative yields the second derivative. Also setting negative values for yields integrals.
For a general function and , the complete fractional derivative is
:<math>D^\alpha f(x)=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dx} \int_0^x \left(x-t\right)^{-\alpha}f(t) \, dt.</math>
For arbitrary , since the gamma function is infinite for negative (real) integers, it is necessary to apply the fractional derivative after the integer derivative has been performed. For example,
:<math>D^\frac32 f(x) = D^\frac12 D^1 f(x)=D^\frac12 \frac d {dx} f(x).</math>
Laplace transform
We can also come at the question via the Laplace transform. Knowing that
:<math>\mathcal L \left\{Jf\right\}(s) = \mathcal L \left\{\int_0^t f(\tau)\,d\tau\right\}(s) = \frac1 s \bigl(\mathcal L\left\{f\right\}\bigr)(s)</math>
and
:<math>\mathcal L \left\{J^2f\right\}=\frac1s\bigl(\mathcal L \left\{Jf\right\} \bigr)(s)=\frac1{s^2}\bigl(\mathcal L\left\{f\right\}\bigr)(s)</math>
and so on, we assert
:<math>J^\alpha f=\mathcal L^{-1}\left\{s^{-\alpha}\bigl(\mathcal L\{f\}\bigr)(s)\right\}</math>.
For example,
:<math>J^\alpha(t^k) = \mathcal L^{-1}\left\{\frac{\Gamma(k+1)}{s^{\alpha+k+1\right\} = \frac{\Gamma(k+1)}{\Gamma(\alpha+k+1)} t^{\alpha+k} </math>
as expected. Indeed, given the convolution rule
:<math>\mathcal L\{f*g\}=\bigl(\mathcal L\{f\}\bigr)\bigl(\mathcal L\{g\}\bigr)</math>
and shorthanding for clarity, we find that
:<math>\begin{align}
\left(J^\alpha f\right)(t) &= \frac{1}{\Gamma(\alpha)}\mathcal L^{-1}\left\{\bigl(\mathcal L\{p\}\bigr)\bigl(\mathcal L\{f\}\bigr)\right\}\\
&=\frac{1}{\Gamma(\alpha)}(p*f)\\
&=\frac{1}{\Gamma(\alpha)}\int_0^t p(t-\tau)f(\tau)\,d\tau\\
&=\frac{1}{\Gamma(\alpha)}\int_0^t\left(t-\tau\right)^{\alpha-1}f(\tau)\,d\tau\\
\end{align}</math>
which is what Cauchy gave us above.
Laplace transforms "work" on relatively few functions, but they are often useful for solving fractional differential equations.
See also
- Caputo fractional derivative