Introduction Some problems Fractional Calculus and Some Problems Ewa Girejko [email protected] Faculty of Computer Science Bialystok University of Technology, Poland 15 March, 2013 Rutgers University, Camden, USA Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals Background of fractional derivatives Its origins go back more to the year 1695 and the conversation of L'Hopital and Leibniz (throught the letters): Leibniz: Can the meaning of derivatives with integer order be generalized to derivatives with non-integer orders? De L'Hospital: What if n=1/2? Leibniz: It will lead to a paradox, from which one day useful consequences will be drawn. Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals After that, many famous mathematicians, like J. Fourier, N. H. Abel, J. Liouville, B. Riemann and others, contributed to the development of the Fractional Calculus. The theory of derivatives and integrals of arbitrary order took more or less finished form by the end of the XIX century. Fractional differentiation may be introduced in several different ways { fractional derivatives of Riemann-Liouville; Gr¨unwald-Letnikov; Caputo; Miller-Ross. Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals For three centuries the theory of fractional derivatives was developed as a pure theoretical field of mathematics, useful only for mathematicians. In the last few decades, fractional differentiation turned out to be very useful in various fields: physics (classic and quantum mechanics, thermodynamics, etc.), chemistry, biology, economics, engineering, signal and image processing, and control theory. The web site of Podlubny: http: // people. tuke. sk/ igor. podlubny/ fc_ resources. html Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals How to get fractional derivatives of exponential function We know already: D1eλx = λeλx ; D2eλx = λ2eλx ;::: Dneλx = λneλx ; when n is an integer. Why not to replace n by 1=2 and write D1=2eλx = λ1=2eλx ? p Why not to go further and put 2 instead of n? Let us write Dαeλx = λαeλx (exp) for any value of α, integer, rational, irrational, or complex. Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals We naturally want eλx = D1 D−1 eλx : λx 1 1 λx Since e = D λ e , we have 1 Z D−1 eλx = eλx = eλx dx : λ Similarly, ZZ D−2 eλx = eλx dxdx : So it is reasonable to interpret Dα when α is a negative integer −n as the nth iterated integral. Dα represents a derivative if α is a positive real number, and an integral if α is a negative real number. Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals Examples Let f (x) = e2x , 0 < α < 1. Then Dα(e2x ) = 2αe2x . We see that D0 (f (x)) = f (x) ≤ Dα (f (x)) ≤ D1 (f (x)) 1 α 1 3 x α 1 1 3 x Let f (x) = e , 0 < α < 1. Then D ( 3 x) = 3 e . We see that D1 (f (x)) ≤ Dα (f (x)) ≤ D0 (f (x)) Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals dα x−α 1 = dxα Γ(1 − α) where we replace n! = 1 · 2 · 3 · ::: (n − 1)n by more general R +1 −t z−1 Gamma function Γ(z) = 0 e t dt ; t > 0: Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals The fractional derivative of a constant is not equal to zero for 0 < α < 1. However, there is an agreement with the classical calculus: α ! 1 ) Γ(1 − α) ! +1 ) D1(1) = 0 Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals First trial to define fractional derivative We can extend the idea of a fractional derivative to a large number of functions. Given any function that can be expanded in a Taylor series in powers of x, +1 X n f (x) = anx ; n=0 and assuming we can differentiate term by term we get +1 +1 X X Γ(n + 1) Dαf (x) = a Dαxn = a xn−α : (Liouville) n n Γ(n − α + 1) n=0 n=0 Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals Do we get a contradiction? We wrote Dαex = ex : (1) Let us now compare this with our definition (Liouville) to see if they agree. From the Taylor series, +1 X 1 ex = xn n! n=0 and our definition (Liouville) gives +1 X xn−α Dαex = (2) Γ(n − α + 1) n=0 The expressions (1) and (2) do not match! We have discovered a contradiction that historically has caused great problems (controversy between 1835 and 1850). Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals Iterated Integrals We could write D−1f (x) = R f (x)dx, but the right-hand side is indefinite. Instead, we will write Z x D−1f (x) = f (t)dt : 0 The second integral will then be Z x Z t2 −2 D f (x) = f (t1)dt1dt2 : 0 0 The region of integration is a triangle, and if we interchange the order of integration, we can write the iterated integral as a single integral (method of Dirichlet, 1908): Z x Z x Z x Z x Z x −2 D f (x) = f (t1)dt2dt1 = f (t1) dt2dt1 = f (t1)(x−t1)dt1 : 0 t1 0 t1 0 Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals Dirichlet: writing iterated integrals as a single integral Using the same procedure we can show that 1 Z x D−3f (x) = f (t)(x − t)2dt ; 2 0 1 Z x D−4f (x) = f (t)(x − t)3dt 2 · 3 0 and, in general, 1 Z x D−nf (x) = f (t)(x − t)n−1dt : (n − 1)! 0 Now, as we have previously done, let us replace the n with an arbitrary α and the factorial with the gamma function... Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals Second trial to define fractional derivative ... replacing n by α, factorial by gamma function: 1 Z x D−αf (x) = f (t)(x − t)α−1dt : (Riemann) Γ(α) 0 Remark As t ! x, x − t ! 0. The integral diverges for every α ≤ 0; when 0 < α < 1 the improper integral converges. Since (Riemann) converges only for positive α, it is truly a fractional integral. Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals The choice of zero for the lower limit was arbitrary. It could have been a. In general, people who work in the field use the notation α aDx f (x) indicating limits of integration going from a to x. With this notation we have: Riemann-Liouville integral Z x −α 1 α−1 aDx f (x) = f (t)(x − t) dt : Γ(α) a Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals We are not surprised that fractional integrals involve limits, because integrals involve limits. Since ordinary derivatives do not involve limits of integration, we were not expecting fractional derivatives to involve such limits! We think of derivatives as local properties of functions. The fractional derivative symbol Dα incorporates both derivatives (positive α) and integrals (negative α). Integrals are between limits, it turns out that fractional derivatives are between limits too. Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals What was the reason of contradiction? The reason for the contradiction is that we used two different limits of integration! We have Z x −1 λx λt 1 λx 1 λa aDx e = e dt = e − e a λ λ 1 λa We get the form we want when λ e = 0, i.e., when λa = −∞. If λ is positive, then a = −∞: Weyl fractional derivative α λx α λx −∞Dx e = λ e Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals We have Z x p+1 p+1 −1 p p x a aDx x = t dt = − : a p + 1 p + 1 ap+1 Again, we want p+1 = 0. This will be the case when a = 0. We conclude that Γ(p + 1) Dαxp = xp−α : 0 x Γ(p − α + 1) Explanation α λx First we calculated −∞Dx e ; the second time we calculated α λx 0Dx e . Ewa Girejko Seminar in Rutgers University, Camden March 2013 Introduction The fractional calculus Some problems Derivatives and integrals Riemann-Liouville: Fractional Integrals and Derivatives Fractional integral of f of order α: Z x −α 1 α−1 aDx f (x) = f (t)(x − t) dt ; α > 0 : Γ(α) a Let α > 0 and let m be the smallest integer exceeding α. Then we define the fractional derivative of f of order α as dm h i Dαf (x) = D−(m−α)f (x) a x dxm a x m Z x 1 d m−α−1 = m f (t)(x − t) dt : Γ(m − α) dx a FD of integer order is the ordinary derivative n dn If α = n 2 N, then m = n + 1 and aDx f (x) = dxn f (x).