Properties of the Mittag-Leffler relaxation function

Transcription

1 Journal of Mathematical Chemistry Vol. 38, No. 4, November 25 25) DOI: 1.17/s z Properties of the Mittag-Leffler relaxation function Mário N. Berberan-Santos Centro de Química-Física Molecular, Instituto Superior Técnico, Lisboa, Portugal Received 26 April 25; revised 27 May 25 The Mittag-Leffler relaxation function, E α x), with α 1, which arises in the description of complex relaxation processes, is studied. A relation that gives the relaxation function in terms of two Mittag-Leffler functions with positive arguments is obtained, and from it a new form of the inverse Laplace transform of E α x) is derived and used to obtain a new integral representation of this function, its asymptotic behaviour and a new recurrence relation. It is also shown that the fastest initial decay of E α x) occurs for α = 1/2, a result that displays the peculiar nature of the interpolation made by the Mittag-Leffler relaxation function between a pure exponential and a hyperbolic function. KEY WORDS: Mittag-Leffler function, Laplace transform, relaxation kinetics AMS MOS) classification: 33E12 Mittag-Leffler functions and generalizations, 44A1 Laplace transform 1. Introduction The Mittag-Leffler function E α z), named after its originator, the Swedish mathematician Gösta Mittag-Leffler ), is defined by [1 4] E α z) = n= z n Ɣαn + 1), 1) where z is a complex variable and α forα = the radius of convergence of equation 1) is finite, and one has by definition E z) = 1/1 z)). The Mittag-Leffler function is a generalization of the exponential function, to which it reduces for α = 1, E 1 z) = expz). For <α<1 it interpolates between a pure exponential and a hyperbolic function, E z) = 1/1 z). The precise nature of this interpolation is the subject of the present study. The Mittag-Leffler function obeys the following relations [3]: ] n 1 E 1/n z 1/n ) = e [n z Ɣ1 k/n, z) n = 2, 3,..., 2) Ɣ1 k/n) k= /5/11-629/ 25 Springer Science+Business Media, Inc.

2 63 M.N. Berberan-Santos / Mittag-Leffler relaxation function E mα z m ) = 1 m 1 E α z e 2ik/m ) m = 2, 3,.... 3) m k= It follows that E 1/2 z) = expz 2 )erfc z) and E 2 z) = cosh z). An explicit expression for any rational value of the parameter α = m/n can be obtained from equations 2) and 3). The generalized Mittag-Leffler function is [3] E α,β z) = n= z n Ɣαn + β), 4) so that E α,1 z) = E α z). In the simplest form α, β. Algorithms for the computation of the generalized Mittag-Leffler function were recently discussed [5]. There has been much recent interest in the Mittag-Leffler and related functions in connection with the description of relaxation phenomena in complex physical and biophysical systems [6 18], namely within the framework of fractional non-integer) kinetic equations. In this work, and having in view the applications of this function to relaxation phenomena, the discussion will be generally restricted to E α x) that corresponds to a relaxation function when x is a nonnegative real variable usually standing for the time) and α Basic relation Using equations 1) and 4), E α x) can be written in terms of two Mittag-Leffler functions with positive arguments, E α x) = E 2α x 2 ) xe 2α,1+α x 2 ). 5) A particular case of equation 3) follows immediately from equation 5), It also follows from equation 5) that E 2α x 2 ) = E α x) + E α x). 6) 2 E α iω) = E 2α ω 2 ) iω E 2α,1+α ω 2 ), 7) a result that will be used in the next section.

3 3. Inverse Laplace transform M.N. Berberan-Santos / Mittag-Leffler relaxation function 631 A simple form of the inverse Laplace transform of E α x) can be obtained by the method outlined in [17]. Briefly, the three following equations can be used for the direct inversion of a function Ix) to obtain its inverse Hk), Hk) = eck Hk) = 2eck [Re[Ic+ iω) ] coskω) Im[Ic+ iω)] sinkω)]dω, 8) Re[Ic+ iω)] coskω)dω k >, 9) Hk) = 2eck Im[Ic+ iω)] sinkω)dω k >. 1) where c is a real number larger than c, c being such that Iz) has some form of singularity on the line Rez) = c but is analytic in the complex plane to the right of that line, i.e., for Rez)>c. Using equation 7), application of equation 9) to E α x) with c =, implying α 1, yields a general relation for its inverse Laplace transform H α k), H α k) = 2 hence, for instance H 1 k) = 2 E 2α ω 2 ) coskω)dω k >, α 1, 11) coshiω) coskω)dω = 2 cosω) coskω)dω = δk 1), 12) H 1/2 k) = 2 e ω2 coskω)dω = 1 e k2 /4, 13) H 1/4 k) = 2 e ω4 erfcω 2 ) coskω)dω, 14) H k) = 2 coskω) 1 + ω dω = 2 e k. 15) Another integral representation of H α k), based on the Lévy one-sided distribution L α k) [8], was obtained by Pollard [19] see also [17,18]), H α k) = 1 ) α k 1+ 1 α ) Lα k 1 α. 16)

4 632 M.N. Berberan-Santos / Mittag-Leffler relaxation function If equation 8) is used instead of equation 9), and taking into account equation 7), H α k) = 1 [ E2α ω 2 ) coskω) + ωe 2α,1+α ω 2 ) sinkω) ] dω α 1. 17) This form is less simple than equation 11), but is formally) necessary for finding the asymptotic expansion of E α x), as will be done in Section Complete monotonicity It is known [19,2] that E α x) is completely monotonic for x if α 1, i.e., that 1) n dn E α x), x, α 1. 18) dx n We remark here that this result follows immediately from 1) n dn E α x) = k n H dx n α k)e kx dk, 19) by noting that H α k) > fork if α 1H α k) is a probability density function), as could be conjectured simply by plotting the function H α k) for several values of α. Demonstration that H α k) > fork is direct for α 1/2, using equation 11) and knowing that E α x) α 1) is always positive and decreases monotonically. General demonstrations for α 1weregiven by Feller [2] and Pollard [19]. 5. Behaviour near the origin Any relaxation function Ix) can be written as x ) Ix) = exp ku)du, 2) where kx) is a x-dependent rate coefficient. When the relaxation is exponential, kx) is obviously constant. For the Mittag-Leffler relaxation function, kx) = d dx ln E 1 n + 1) x) n α x) = E α x) Ɣ1 + α + αn), 21) whose initial value is finite and close to unity, k) = kh α k)dk = n= 1 Ɣ1 + α). 22)

5 M.N. Berberan-Santos / Mittag-Leffler relaxation function 633 It follows nevertheless from equation 22) that the fastest initial decay occurs for α = 1 / 2, a result hitherto unnoticed, and that shows the peculiar nature of the interpolation between a pure exponential and a hyperbolic function performed by the Mittag-Leffler relaxation function. 6. Asymptotic behaviour with Expansion of equation 17) in a power series gives H α k) = 1 a n α) k n, α<1, 23) n= a α) = E 2α ω 2 )dω. 24) The Laplace transform of equation 23) is the asymptotic expansion of E α x), E α x) = 1 n= a n α), α<1. 25) xn+1 Since a α) for α < 1, the Mittag-Leffler relaxation function has a hyperbolic x 1 ) asymptotic decay for α < 1, and an exponential decay only for α = 1. The cross-over between the initial exponential-like decay and the asymptotic hyperbolic decay occurs at a value of x that is the shorter, the smaller the α. It follows from equation 25) that E α x 2 ) asymptotically decays as x 2, hence equation 24) is clearly convergent for α<1/2. 7. A recurrence relation By taking the Laplace transform of equation 11), a recurrence relation is obtained, E α x) = 2x E 2α ω 2 ) dω, α 1. 26) x 2 + ω2 This relation works in the opposite way of equation 6), and allows the direct calculation of E α x) from E 2α x 2 ). In this way, it follows, for instance, that E 1 x) = 2x E 1/2 x) = 2x coshiω) x 2 + ω 2 dω = e x, 27) e ω2 x 2 + ω 2 dω = ex2 erfcx), 28)

6 634 M.N. Berberan-Santos / Mittag-Leffler relaxation function E 1/4 x) = 2x e ω4 erfc ω 2 ) dω. 29) x 2 + ω 2 The asymptotic behaviour of E α x) also follows directly from equation 26) for large xα < 1). 8. Integral representations The starting point is the known Laplace transform of E α x α ), J α α s), which can be obtained in closed form directly from the definition, Jα α s) = E α x α ) e sx dx = Application of inversion equation 9) to equation 3) yields sα s α. 3) E α x α ) = 2 sinα/2) ω α 1 cosxω) dω, 31) 1 + 2ω α cos α/2) + ω2α hence a new integral representation for the Mittag-Leffler relaxation function is E α x) = 2 sinα/2) ω α 1 cosx 1/α ω) dω. 32) 1 + 2ω α cos α/2) + ω2α Analogous representations are obtained with equations 8) and 1). The previously known integral representation for E α x α ), E α x α ) = sinα) k α k α cosα) + k 2α e xk dk, 33) was obtained from the Bromwich inversion integral [5]. It follows from equation 33) that E α x) = sinα) k α k α cosα) + k 2α e x1/αk dk. 34) Performing an integration by parts, equation 34) can be rewritten as E α x) = 1 1 2α + x1/α k α ) + cosα) arctan e x1/αk dk. 35) sinα) Equation 34) can be used to compute the numerical coefficient of the leading term of the asymptotic expansion of E α x). Equation 24) becomes a α) = α Ɣα) sin2α) k α k 2α cos2α) + k 4α dk, α < )

7 9. Conclusions M.N. Berberan-Santos / Mittag-Leffler relaxation function 635 The Mittag-Leffler relaxation function, E α x), with α 1, which arises in the description of complex relaxation processes, was studied. From equation 5) that gives the relaxation function in terms of two Mittag-Leffler functions with positive arguments, the inverse Laplace transform of E α x) was obtained, equation 11), and used to derive a new integral representation of this function, equation 32), its asymptotic behaviour, equations 25) and 36), and a new recurrence relation, equation 26). It was also shown that the fastest initial decay of E α x) occurs for α = 1/2, a result hitherto unnoticed, and that shows the peculiar nature of the interpolation between a pure exponential and a hyperbolic function performed by the Mittag-Leffler relaxation function. References [1] G.M. Mittag-Leffler, C. R. Acad. Sci. Paris Ser. II) ) 7. [2] G.M. Mittag-Leffler, C. R. Acad. Sci. Paris Ser. II) ) 554. [3] A. Érdelyi, W. Magnus, F. Oberhettinger and F.G. Tricomi, Higher Transcendental Functions, Vol. III McGraw-Hill, New York, 1955). [4] K.S. Miller, Integral Transform. Spec. Funct ) 41. [5] R. Gorenflo, J. Loutchko and Y. Luchko, Fract. Calc. Appl. Anal. 5 22) 491. [6] W.G. Glöckle and T.F. Nonnenmacher, Biophys. J ) 46. [7] R. Hilfer and L. Anton, Phys. Rev. E ) R848. [8] K. Weron and M. Kotulski, Physica A ) 18. [9] K. Weron and A. Klauzer, Ferroelectrics 236 2) 59. [1] R. Metzler and J. Klafter, Phys. Rep ) 1. [11] R. Hilfer, Chem. Phys ) 399. [12] R.K. Saxena, A.M. Mathai and H.J. Haubold, Astrophys. Space Sci ) 281. [13] R. Metzler and J. Klafter, J. Non-Cryst. Solids 35 22) 81. [14] R. Metzler and J. Klafter, Biophys. J ) [15] L. Sjögren, Physica A ) 81. [16] D.S.F. Crothers, D. Holland, Y.P. Kalmykov and W.T. Coffey, J. Mol. Liquids ) 27. [17] M.N. Berberan-Santos, J. Math. Chem ) 165. [18] M.N. Berberan-Santos, J. Math. Chem ) 265. [19] H. Pollard, Bull. Am. Math. Soc ) [2] W. Feller, Trans. Am. Math. Soc ) 98.