Wikipedia, the free encyclopedia

Mittag-Leffler function

Mathematical function
Not to be confused with Mittag-Leffler polynomials.
The Mittag-Leffler function can be used to interpolate continuously between a Gaussian and a Lorentzian function.

In mathematics, the Mittag-Leffler function E α , β {\displaystyle E_{\alpha ,\beta }} {\displaystyle E_{\alpha ,\beta }} is a special function, a complex function which depends on two complex parameters α {\displaystyle \alpha } {\displaystyle \alpha } and β {\displaystyle \beta } {\displaystyle \beta }. It may be defined by the following series when the real part of α {\displaystyle \alpha } {\displaystyle \alpha } is strictly positive:[1][2]

E α , β ( z ) = ∑ k = 0 ∞ z k Γ ( α k + β ) , {\displaystyle E_{\alpha ,\beta }(z)=\sum _{k=0}^{\infty }{\frac {z^{k}}{\Gamma (\alpha k+\beta )}},} {\displaystyle E_{\alpha ,\beta }(z)=\sum _{k=0}^{\infty }{\frac {z^{k}}{\Gamma (\alpha k+\beta )}},}

where Γ ( x ) {\displaystyle \Gamma (x)} {\displaystyle \Gamma (x)} is the gamma function. When β = 1 {\displaystyle \beta =1} {\displaystyle \beta =1}, it is abbreviated as E α ( z ) = E α , 1 ( z ) {\displaystyle E_{\alpha }(z)=E_{\alpha ,1}(z)} {\displaystyle E_{\alpha }(z)=E_{\alpha ,1}(z)}. For α = 0 {\displaystyle \alpha =0} {\displaystyle \alpha =0}, the series above equals the Taylor expansion of the geometric series and consequently E 0 , β ( z ) = 1 Γ ( β ) 1 1 − z {\displaystyle E_{0,\beta }(z)={\frac {1}{\Gamma (\beta )}}{\frac {1}{1-z}}} {\displaystyle E_{0,\beta }(z)={\frac {1}{\Gamma (\beta )}}{\frac {1}{1-z}}}.

In the case α {\displaystyle \alpha } {\displaystyle \alpha } and β {\displaystyle \beta } {\displaystyle \beta } are real and positive, the series converges for all values of the argument z {\displaystyle z} {\displaystyle z}, so the Mittag-Leffler function is an entire function. This function is named after Gösta Mittag-Leffler. This class of functions are important in the theory of the fractional calculus.

For α > 0 {\displaystyle \alpha >0} {\displaystyle \alpha >0}, the Mittag-Leffler function E α , 1 ( z ) {\displaystyle E_{\alpha ,1}(z)} {\displaystyle E_{\alpha ,1}(z)} is an entire function of order 1 / α {\displaystyle 1/\alpha } {\displaystyle 1/\alpha }, and is in some sense the simplest entire function of its order.

The Mittag-Leffler function satisfies the recurrence property (Theorem 5.1 of [1])

E α , β ( z ) = 1 z E α , β − α ( z ) − 1 z Γ ( β − α ) , {\displaystyle E_{\alpha ,\beta }(z)={\frac {1}{z}}E_{\alpha ,\beta -\alpha }(z)-{\frac {1}{z\Gamma (\beta -\alpha )}},} {\displaystyle E_{\alpha ,\beta }(z)={\frac {1}{z}}E_{\alpha ,\beta -\alpha }(z)-{\frac {1}{z\Gamma (\beta -\alpha )}},}

from which the following Poincaré asymptotic expansion holds : for 0 < α < 2 {\displaystyle 0<\alpha <2} {\displaystyle 0<\alpha <2} and μ {\displaystyle \mu } {\displaystyle \mu } real such that π α 2 < μ < min ( π , π α ) {\displaystyle {\frac {\pi \alpha }{2}}<\mu <\min(\pi ,\pi \alpha )} {\displaystyle {\frac {\pi \alpha }{2}}<\mu <\min(\pi ,\pi \alpha )} then for all N ∈ N ∗ , N ≠ 1 {\displaystyle N\in \mathbb {N} ^{*},N\neq 1} {\displaystyle N\in \mathbb {N} ^{*},N\neq 1}, we can show the following asymptotic expansions (Section 6. of [1]):

-as | z | → + ∞ , | arg ( z ) | ≤ μ {\displaystyle \,|z|\to +\infty ,|{\text{arg}}(z)|\leq \mu } {\displaystyle \,|z|\to +\infty ,|{\text{arg}}(z)|\leq \mu }:

E α ( z ) = 1 α exp ⁡ ( z 1 α ) − ∑ k = 1 N 1 z k Γ ( 1 − α k ) + O ( 1 z N + 1 ) {\displaystyle E_{\alpha }(z)={\frac {1}{\alpha }}\exp(z^{\frac {1}{\alpha }})-\sum \limits _{k=1}^{N}{\frac {1}{z^{k}\,\Gamma (1-\alpha k)}}+O\left({\frac {1}{z^{N+1}}}\right)} {\displaystyle E_{\alpha }(z)={\frac {1}{\alpha }}\exp(z^{\frac {1}{\alpha }})-\sum \limits _{k=1}^{N}{\frac {1}{z^{k}\,\Gamma (1-\alpha k)}}+O\left({\frac {1}{z^{N+1}}}\right)},

-and as | z | → + ∞ , μ ≤ | arg ( z ) | ≤ π {\displaystyle \,|z|\to +\infty ,\mu \leq |{\text{arg}}(z)|\leq \pi } {\displaystyle \,|z|\to +\infty ,\mu \leq |{\text{arg}}(z)|\leq \pi }:

E α ( z ) = − ∑ k = 1 N 1 z k Γ ( 1 − α k ) + O ( 1 z N + 1 ) {\displaystyle E_{\alpha }(z)=-\sum \limits _{k=1}^{N}{\frac {1}{z^{k}\Gamma (1-\alpha k)}}+O\left({\frac {1}{z^{N+1}}}\right)} {\displaystyle E_{\alpha }(z)=-\sum \limits _{k=1}^{N}{\frac {1}{z^{k}\Gamma (1-\alpha k)}}+O\left({\frac {1}{z^{N+1}}}\right)},

where we used the notation E α ( z ) = E α , 1 ( z ) {\displaystyle E_{\alpha }(z)=E_{\alpha ,1}(z)} {\displaystyle E_{\alpha }(z)=E_{\alpha ,1}(z)}.

The Mittag-Leffler function, characterized by three parameters, is expressed as follows:

E α , β γ ( z ) = ( 1 Γ ( γ ) ) ∑ k = 1 ∞ Γ ( γ + k ) z k k ! Γ ( α k + β ) , {\displaystyle E_{\alpha ,\beta }^{\gamma }(z)=\left({\frac {1}{\Gamma (\gamma )}}\right)\sum \limits _{k=1}^{\infty }{\frac {\Gamma (\gamma +k)z^{k}}{k!\Gamma (\alpha k+\beta )}},} {\displaystyle E_{\alpha ,\beta }^{\gamma }(z)=\left({\frac {1}{\Gamma (\gamma )}}\right)\sum \limits _{k=1}^{\infty }{\frac {\Gamma (\gamma +k)z^{k}}{k!\Gamma (\alpha k+\beta )}},}

where α , β {\displaystyle \alpha ,\beta } {\displaystyle \alpha ,\beta } and γ {\displaystyle \gamma } {\displaystyle \gamma } are complex parameters and ℜ ( α ) > 0 {\displaystyle \Re (\alpha )>0} {\displaystyle \Re (\alpha )>0}.[3]

For γ ∈ N {\displaystyle \gamma \in \mathbb {N} } {\displaystyle \gamma \in \mathbb {N} }, the Mittag-Leffler function with three parameters is reformulated as:

E α , β γ ( z ) = ∑ k = 0 ∞ ( γ ) k z k k ! Γ ( α k + β ) , {\displaystyle E_{\alpha ,\beta }^{\gamma }(z)=\sum _{k=0}^{\infty }{\frac {(\gamma )_{k}z^{k}}{k!\Gamma (\alpha k+\beta )}},} {\displaystyle E_{\alpha ,\beta }^{\gamma }(z)=\sum _{k=0}^{\infty }{\frac {(\gamma )_{k}z^{k}}{k!\Gamma (\alpha k+\beta )}},}

where ( γ ) k := Γ ( γ + k ) Γ ( γ ) = γ ( γ + 1 ) ⋯ ( γ + k − 1 ) {\displaystyle (\gamma )_{k}:={\frac {\Gamma (\gamma +k)}{\Gamma (\gamma )}}=\gamma (\gamma +1)\cdots (\gamma +k-1)} {\displaystyle (\gamma )_{k}:={\frac {\Gamma (\gamma +k)}{\Gamma (\gamma )}}=\gamma (\gamma +1)\cdots (\gamma +k-1)} and it exhibits the following property:

E ( α , β ) γ ( z ) = 1 α γ ( E ( α , β − 1 ) γ − 1 ( z ) + ( 1 − β + α γ ) E ( α , β ) γ − 1 ( z ) ) {\displaystyle E_{(\alpha ,\beta )}^{\gamma }(z)={\frac {1}{\alpha ^{\gamma }}}\left(E_{(\alpha ,\beta -1)}^{\gamma -1}(z)+(1-\beta +\alpha \gamma )E_{(\alpha ,\beta )}^{\gamma -1}(z)\right)} {\displaystyle E_{(\alpha ,\beta )}^{\gamma }(z)={\frac {1}{\alpha ^{\gamma }}}\left(E_{(\alpha ,\beta -1)}^{\gamma -1}(z)+(1-\beta +\alpha \gamma )E_{(\alpha ,\beta )}^{\gamma -1}(z)\right)}.[4]

Additionally, a relation concerning the first parameter of the 2-parameter Mittag-Leffler function is as follows: E ( α , β ) ( r t α ) = 1 m ∑ i = 1 m E ( ρ , β ) ( s i t ρ ) , {\displaystyle E_{(\alpha ,\beta )}(rt^{\alpha })={\frac {1}{m}}\sum _{i=1}^{m}E_{(\rho ,\beta )}(s_{i}t^{\rho }),} {\displaystyle E_{(\alpha ,\beta )}(rt^{\alpha })={\frac {1}{m}}\sum _{i=1}^{m}E_{(\rho ,\beta )}(s_{i}t^{\rho }),}

where α ρ = m ∈ N {\displaystyle {\frac {\alpha }{\rho }}=m\in \mathbb {N} } {\displaystyle {\frac {\alpha }{\rho }}=m\in \mathbb {N} } and s i {\displaystyle s_{i}} {\displaystyle s_{i}} are roots of s m = r {\displaystyle s^{m}=r} {\displaystyle s^{m}=r}.[5][6]

Special cases[edit]

For α = 0 , 1 / 2 , 1 , 2 {\displaystyle \alpha =0,1/2,1,2} {\displaystyle \alpha =0,1/2,1,2} we find: (Section 2 of [1])

Error function:

E 1 2 ( z ) = exp ⁡ ( z 2 ) erfc ⁡ ( − z ) . {\displaystyle E_{\frac {1}{2}}(z)=\exp(z^{2})\operatorname {erfc} (-z).} {\displaystyle E_{\frac {1}{2}}(z)=\exp(z^{2})\operatorname {erfc} (-z).}

The sum of a geometric progression:

E 0 ( z ) = ∑ k = 0 ∞ z k = 1 1 − z , | z | < 1. {\displaystyle E_{0}(z)=\sum _{k=0}^{\infty }z^{k}={\frac {1}{1-z}},\,|z|<1.} {\displaystyle E_{0}(z)=\sum _{k=0}^{\infty }z^{k}={\frac {1}{1-z}},\,|z|<1.}

Exponential function:

E 1 ( z ) = ∑ k = 0 ∞ z k Γ ( k + 1 ) = ∑ k = 0 ∞ z k k ! = exp ⁡ ( z ) . {\displaystyle E_{1}(z)=\sum _{k=0}^{\infty }{\frac {z^{k}}{\Gamma (k+1)}}=\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=\exp(z).} {\displaystyle E_{1}(z)=\sum _{k=0}^{\infty }{\frac {z^{k}}{\Gamma (k+1)}}=\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=\exp(z).}

Hyperbolic cosine:

E 2 ( z ) = cosh ⁡ ( z ) ,  and  E 2 ( − z 2 ) = cos ⁡ ( z ) . {\displaystyle E_{2}(z)=\cosh({\sqrt {z}}),{\text{ and }}E_{2}(-z^{2})=\cos(z).} {\displaystyle E_{2}(z)=\cosh({\sqrt {z}}),{\text{ and }}E_{2}(-z^{2})=\cos(z).}

For β = 2 {\displaystyle \beta =2} {\displaystyle \beta =2}, we have

E 1 , 2 ( z ) = e z − 1 z , {\displaystyle E_{1,2}(z)={\frac {e^{z}-1}{z}},} {\displaystyle E_{1,2}(z)={\frac {e^{z}-1}{z}},}
E 2 , 2 ( z ) = sinh ⁡ ( z ) z . {\displaystyle E_{2,2}(z)={\frac {\sinh({\sqrt {z}})}{\sqrt {z}}}.} {\displaystyle E_{2,2}(z)={\frac {\sinh({\sqrt {z}})}{\sqrt {z}}}.}

For α = 0 , 1 , 2 {\displaystyle \alpha =0,1,2} {\displaystyle \alpha =0,1,2}, the integral

∫ 0 z E α ( − s 2 ) d s {\displaystyle \int _{0}^{z}E_{\alpha }(-s^{2})\,{\mathrm {d} }s} {\displaystyle \int _{0}^{z}E_{\alpha }(-s^{2})\,{\mathrm {d} }s}

gives, respectively: arctan ⁡ ( z ) {\displaystyle \arctan(z)} {\displaystyle \arctan(z)}, π 2 erf ⁡ ( z ) {\displaystyle {\tfrac {\sqrt {\pi }}{2}}\operatorname {erf} (z)} {\displaystyle {\tfrac {\sqrt {\pi }}{2}}\operatorname {erf} (z)}, sin ⁡ ( z ) {\displaystyle \sin(z)} {\displaystyle \sin(z)}.

Mittag-Leffler's integral representation[edit]

The integral representation of the Mittag-Leffler function is (Section 6 of [1])

E α , β ( z ) = 1 2 π i ∮ C t α − β e t t α − z d t , ℜ ( α ) > 0 , ℜ ( β ) > 0 , {\displaystyle E_{\alpha ,\beta }(z)={\frac {1}{2\pi i}}\oint _{C}{\frac {t^{\alpha -\beta }e^{t}}{t^{\alpha }-z}}\,dt,\Re (\alpha )>0,\Re (\beta )>0,} {\displaystyle E_{\alpha ,\beta }(z)={\frac {1}{2\pi i}}\oint _{C}{\frac {t^{\alpha -\beta }e^{t}}{t^{\alpha }-z}}\,dt,\Re (\alpha )>0,\Re (\beta )>0,}

where the contour C {\displaystyle C} {\displaystyle C} starts and ends at − ∞ {\displaystyle -\infty } {\displaystyle -\infty } and circles around the singularities and branch points of the integrand.

Related to the Laplace transform and Mittag-Leffler summation is the expression (Eq (7.5) of [1] with m = 0 {\displaystyle m=0} {\displaystyle m=0})

∫ 0 ∞ e − t z t β − 1 E α , β ( ± r t α ) d t = z α − β z α ∓ r , ℜ ( z ) > 0 , ℜ ( α ) > 0 , ℜ ( β ) > 0. {\displaystyle \int _{0}^{\infty }e^{-tz}t^{\beta -1}E_{\alpha ,\beta }(\pm r\,t^{\alpha })\,dt={\frac {z^{\alpha -\beta }}{z^{\alpha }\mp r}},\Re (z)>0,\Re (\alpha )>0,\Re (\beta )>0.} {\displaystyle \int _{0}^{\infty }e^{-tz}t^{\beta -1}E_{\alpha ,\beta }(\pm r\,t^{\alpha })\,dt={\frac {z^{\alpha -\beta }}{z^{\alpha }\mp r}},\Re (z)>0,\Re (\alpha )>0,\Re (\beta )>0.}

Applications of Mittag-Leffler function[edit]

One of the applications of the Mittag-Leffler function is in modeling fractional order viscoelastic materials. Experimental investigations into the time-dependent relaxation behavior of viscoelastic materials are characterized by a very fast decrease of the stress at the beginning of the relaxation process and an extremely slow decay for large times. It can even take a long time before a constant asymptotic value is reached. Therefore, a lot of Maxwell elements are required to describe relaxation behavior with sufficient accuracy. This ends in a difficult optimization problem in order to identify a large number of material parameters. On the other hand, over the years, the concept of fractional derivatives has been introduced to the theory of viscoelasticity. Among these models, the fractional Zener model was found to be very effective to predict the dynamic nature of rubber-like materials with only a small number of material parameters. The solution of the corresponding constitutive equation leads to a relaxation function of the Mittag-Leffler type. It is defined by the power series with negative arguments. This function represents all essential properties of the relaxation process under the influence of an arbitrary and continuous signal with a jump at the origin.[7][8]

See also[edit]

Notes[edit]

References[edit]

  1. ^ a b c d e f Saxena, R. K.; Mathai, A. M.; Haubold, H. J. (2009-09-01). "Mittag-Leffler Functions and Their Applications". arXiv:0909.0230 [math.CA].
  2. ^ Weisstein, Eric W. "Mittag-Leffler Function". mathworld.wolfram.com. Retrieved 2019-09-11.
  3. ^ Gorenflo, Rudolf; Kilbas, Anatoly A.; Mainardi, Francesco; Rogosin, Sergei V. (2014). Mittag-Leffler Functions, Related Topics and Applications: Theory and Applications. Springer Monographs in Mathematics. Berlin, Heidelberg: Springer Berlin Heidelberg. doi:10.1007/978-3-662-43930-2. ISBN 978-3-662-43929-6.
  4. ^ T. R., Prabhakar (1971). "A Singular Integral Equation with a Generalized Mittag-Leffler Function in the Kernel". Yokohama Mathematical Journal. 19: 7–15.
  5. ^ Erman, Sertaç; Demir, Ali (2020-12-01). "On the construction and stability analysis of the solution of linear fractional differential equation". Applied Mathematics and Computation. 386: 125425. doi:10.1016/j.amc.2020.125425. ISSN 0096-3003.
  6. ^ Erman, Sertaç (2023-05-31). "Undetermined Coefficients Method for Sequential Fractional Differential Equations". Kocaeli Journal of Science and Engineering. 6 (1): 44–50. doi:10.34088/kojose.1145611. ISSN 2667-484X.
  7. ^ Pritz, T. (2003). Five-parameter fractional derivative model for polymeric damping materials. Journal of Sound and Vibration, 265(5), 935-952.
  8. ^ Nonnenmacher, T. F., & Glöckle, W. G. (1991). A fractional model for mechanical stress relaxation. Philosophical magazine letters, 64(2), 89-93.
This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (May 2021) (Learn how and when to remove this message)

External links[edit]

This article incorporates material from Mittag-Leffler function on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

Authority control databases: National Edit this at Wikidata