Fractional Vector Calculus and Fractional Special Function

0.2 MB·English

by  Ming-Fan Li

Fractional Vector Calculus and Fractional Special Function
Ming-Fan Li,∗ Ji-Rong Ren,† and Tao Zhu‡
Institute of Theoretical Physics, Lanzhou University, Lanzhou, 730000, China
Fractional vector calculus is discussed in the spherical coordinate framework. A variation of the
Legendre equation and fractional Bessel equation are solved by series expansion and numerically.
Finally, we generalize the hypergeometric functions.
PACSnumbers: 02.30.Gp,45.10.Hj,02.40.Ky
I. INTRODUCTION ter we will use Riemann-Liouville fractionalintegraland
0 Caputo fractional derivative.
1 Fractionalcalculusisthecalculusofdifferentiationand Let f(x) be a function defined on the right side of a.
0
integrationofnon-integerorders[1,2]. Duringlastthree Let α be a positive real. The Riemann-Liouville frac-
2
decades, fractional calculus has gained much attention tional integral is defined by
n
due to its demonstrated applications in various fields of
a 1 x
J science and engineering, such as anomalous diffusion [3], Iαf(x)= (x−ξ)α−1f(ξ) dξ. (1)
fractional dynamical systems [4–6], fractional quantum Γ(α)Z
7 a
1 mechanics [7], fractional statistics and thermodynamics
[8], to name a few. The integral has a memory kernel.

] Fractional vector calculus (FVC) is important in de- Let A ≡ [α]+1. The Caputo fractional derivative is
h
scribing processes in fractal media, fractional electrody- defined by
p
- namics and fractional hydrodynamics [9, 10]. But an
h effective FVC is still lacking. There are many problems dA
t Dαf(x) = IA−α f(x)
a in defining an effective FVC. One is that fractional in- dxA
m tegral and fractional derivative are defined “half” , that 1 x dA
is to say, they are defined only on the right or the left = (x−ξ)A−α−1 f(ξ) dξ.(2)
[ Γ(A−α)Z dξA
side of an initial point. And if we make fractional series a

1 expansion,functions areallexpandedonthe rightorthe
v They have the following properties:
left neighborhood of the initial point. We cannot across
9
thiscuttingpoint. Soifwewanttodescribethebehavior
8 Dα(x−a)β =0, β ∈{0,1,...,[α]};(3)
8 near the initial point, we need define both the right and
Γ(β+1)
2 the left functions. The situation will become even more Dα(x−a)β = (x−a)β−α,β otherwise;(4)
. complicated when we deal with high dimensions. In this Γ(β−α+1)
1
0 letter we will define FVC in spherical coordinate frame- Iα(x−a)β = Γ(β+1) (x−a)β+α. (5)
0 work. Since in the spherical coordinate framework the Γ(β+α+1)
1 radius is naturally bounded to the positive half, we need
v: just one fractional derivative. These properties are just fractional generalizations of
i Using this frame, we will discuss the Laplacian equa-
X tion[11]withfractionalradiusderivativein3-dspaceand dn m!
xm = xm−n, n∈N, m6=0, (6)
r the heat conduct equation [11] in 2-d space with cylin- dxn (m−n)!
a
drical symmetry. As a result, the corresponding special
functions will be generalized. Finally, we will generalize
hypergeometric functions [12–14]. x m!
dx xm = xm+1. (7)
Z (m+1)!
0
II. FRACTIONAL CALCULUS Fora‘good’function,onecandefineitsfractionalTay-
lor series
To start, let’s briefly recall some basic facts in frac-
∞
tional calculus [1, 2]. There are many ways to define
f(x)= (Dα)mf(x) ·[(Iα)m·1]. (8)
fractional integral and fractional derivative. In this let- x=a
mX=0 (cid:12)
(cid:12)
Explicitly,
∗Electronicaddress: [email protected] 1
†Electronicaddress: [email protected] (Iα)m·1= (x−a)mα. (9)
‡Electronicaddress: [email protected] Γ(mα+1)
2
A. Spherical coordinates
z
|OA|=rα/Γ(α+1)
A The spherical coordinates of 3-dimension space is a
triplet (r,θ,φ). In one-order calculus, the gradient of
vs. r in Euclidian metric
N A a scale function u(r,θ,φ) is
A=(r , θ , φ ) ∂ 1 ∂ 1 ∂
A A A Grad u= e +e +e u. (10)
r θ φ
(cid:18) ∂r r∂θ rsinθ∂φ(cid:19)
θ
A
O We generalize this definition to fractional calculus as
L φA
A’
Γ(α+1) ∂ Γ(α+1) ∂
y Gradαu= e Dα+e +e u,
(cid:18) r r θ rα ∂θ φ rαsinαθ ∂φ(cid:19)
(11)
x
where Dα u = 1 r(r − ρ)A−α−1 dA u(ρ,θ,φ)dρ.
r Γ(A−α) 0 dρA
FIG. 1: Spherical framework with fractional radius deriva- For anisotropic space, αRis a function of θ and φ. For
tive. |OA| = Γ(α1+1)rAα, NdA = Γ(α1+1)rAαθA, and LdA′ = isotropicspace,αisaconstantandindependentofθ and
Γ(α1+1)rAαsinαθA φA. φ. We will just consider the isotropic case.
By this definition, the real space metric is changed to
an effective metric. This can be easily seen from Fig.
III. FRACTIONAL VECTOR CALCULUS 1. The radius |OA| now is [Iα · 1]|r=rA = 1 rα;
r r=0 Γ(α+1) A
the arc length NA = 1 rαθ and the arc length
As has been aforementioned, fractional derivative is Γ(α+1) A A
defined only on the right or the left half of the real line, LA′ = Γ(α1+1)rAαdsinαθA φA. This kind of metric is not
which gives complications in defining an effective FVC addictive since |OB| 6= |OA| + |AB| even when O, A
d
with Cartesian coordinates. So it may be more feasible and B are on the same straight line. This is due to the
to do FVC with spherical coordinates. non-locality of the fractional operations.
By the above generalization,the divergence of a vector function A=(A ,A ,A ) is
r θ φ
1 Γ(α+1) ∂ Γ(α+1) ∂
divα A= Dα(r2αsinαθA )+ (rαsinαθA )+ (rαA ). (12)
r2αsinαθ r r r2αsinαθ∂θ θ r2αsinαθ∂φ φ
The Laplacian of a scale function u(r,θ,φ) will be
△α u ≡ divαGradα u
1 Γ2(α+1) ∂ ∂ Γ2(α+1) ∂2
= Dα(r2αDα u)+ (sinαθ u)+ u. (13)
r2α r r r2αsinαθ ∂θ ∂θ r2αsin2αθ ∂φ2
The rotor operator can be defined as well. Since in The divergence is
this letter we do not deal with the rotor, its definition
will not be given.
1 Γ(α+1) ∂
divα A= Dα(rαA )+ A . (15)
rα r r rα ∂θ θ
B. Polor coordinates
The Laplacian of a scale function u(r,θ) is
Likewise, the fractional gradient operator of two di-
mensions in polor coordinates can be defined as
△α u ≡ divαGradα u
Γ(α+1) ∂ 1 Γ2(α+1) ∂2
Gradα u= e Dα+e u. (14) = Dα(rαDα u)+ u. (16)
(cid:18) r r θ rα ∂θ(cid:19) rα r r r2α ∂θ2
3
IV. FRACTIONAL SPHERICAL EQUATION A. Fractional Laplacian equation
AND FRACTIONAL CYLINDRICAL EQUATION
In this section, we consider the Laplacian equation
with the 3-d Laplacian operaror defined above and the
fractional heat conduct equation in a 2-d space with WiththeLaplacianoperatordefinedabove,theLapla-
cylindrical symmetry. cian equation becomes
1 Γ2(α+1) ∂ ∂ Γ2(α+1) ∂2
△α u= Dα(r2αDα u)+ (sinαθ u)+ u=0. (17)
r2α r r r2αsinαθ ∂θ ∂θ r2αsin2αθ ∂φ2
Thisequationcanbesolvedbyseparationofvariables. 6
Let u=R(r)Θ(θ)Φ(φ) and substitute, the result is
Legendre function
4
1 d sinαθdΘ + 1 d2Φ =−λ, (18) even, lamda=12
Θsinαθdθ(cid:18) dθ(cid:19) Φsin2αθ dφ2
2
1
RDrα(r2αDrαR)=λΓ2(α+1). (19) 0 α=1/10
α=1
The secondequationcanbe solvedby fractionalseries −2
expansion. LetR= ∞ c rmα,andsubstitute,the
m=−∞ m α=4
result is that the nonPzero components are the ones with
−4
m satisfying
Γ(mα+α+1) −6
=λΓ2(α+1). (20) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Γ(mα−α+1)
FIG. 2: Legendre even function. Since the fractional index
Otherwise, cm=0. α occurs as a multiplicative factor in thevariation version of
The firstequationis a variationofthe ordinaryspher- theLegendre equation,small difference from 1 will not make
ical harmonic equation. By further decomposition, it large changes to the profile. We calculated the function with
turns to other small difference α’s, but the results are not shown for
theircurvesare veryclose toeach other. Notice thedirect of
d2Φ thecurves.
+m2Φ=0, (m=0,1,2,3,...) (21)
dφ2
c · p (x) + c · p (x). The relation of successive
0 even 1 odd
sinαθ d dΘ coefficients is
sinαθ =m2−λsin2αθ. (22)
Θ dθ(cid:18) dθ(cid:19)
n2+nα−λ
c =c . (24)
The first equation is simple. The second equation n+2 n(n+2)(n+1)
can be transformed by changing variables x = cosθ and
p(x)=Θ(θ) to Letα=1andλ=l(l+1),wewillgettheLegendrepoly-
nomials. We calculated numerically the functions with
d2p dp m2 some different values of the parameters and displayed
(1−x2) −(1+α)x + λ− p=0. (23)
dx2 dx (cid:20) (1−x2)α(cid:21) the results in Fig. 2 and Fig. 3.
This is a variation of the ordinary associated Legendre
equation [11, 12]. By setting α=1, we recover the ordi- B. Fractional cylindrical equation
nary associated Legendre equation.
When m = 0 it is the Legendre equation. Make the
The heat conduct equation in 2-d space is
ansatz p(x) = ∞ c xn and substitute, we find that
n=0 n
the terms withPeven powersof x and the terms with odd ∂u
powers of x are independent, so we can write p(x) = ∂t =a2△2u. (25)
4
0.4
0.2
30
0 25
ν
20
α=1/4
−0.2 15
−0.4 Legendre function α=1 10
5
odd, lamda=12
−0.6 α=1.3 0
−5
4
−0.8
3 2
2 1.5
−1 ρ 1
1 0.5 α
0 0
−1.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
FIG.4: ParametersoffractionalBesselfunction. ν asafunc-
FIG. 3: Legendre odd function. Notice the direct of the tion of α and ρ.
curves.
1
Here △ is the two dimensional Laplacian operator. Us-
2
0.8 Fractional Bessel functions
ing the fractional cylindrical Laplacian operator defined
above, this equation becomes ν=3, k=1
0.6
∂u 1 Γ2(α+1) ∂2
= Dα(rαDα u)+ u. (26)
∂t rα r r r2α ∂θ2 0.4
Byseparationofvariablesu(t,r,θ)=R(r)Θ(θ)T(t), it α=3/2
0.2
can be decomposed to α=1
T′+a2k2T =0, (27) 0 α=2/3
−0.2 α=1/2
∂2Θ
+ν2Θ=0, (28)
∂θ2
−0.4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1 rαDα(rαDα R)+k2 r2α R−ν2R=0. FIG. 5: Fractional Bessel function. The fractional index α
Γ2(α+1) r r Γ2(α+1) showsupintheexponentialsofthefractional Bessel series; a
(29) small difference from 1 changes the profile largely. For a big
Thefirsttwoequationsaresimple. Thethirdequation α, ρ is small, so thechange is suppressed.
is a fractional generalization of the Bessel equation [11,
12]. Itcanbesolvedbyfractionalseriesexpansion. Since
Bessel equation is singular at r = 0. We must use such and c1 =0.
ansatz: R = rαρ ∞ c rαm. Substitute it into the By the recursive relation, a solution is implied
m=0 m
above equation, wePget ∞
Γ(αρ+1) 2 Rρ(r)=rαρ (−1)ndnk2nrα·2n, (34)
c0(cid:20)(cid:18)Γ(αρ−α+1)(cid:19) −ν2Γ2(α+1)(cid:21)=0, (30) nX=0
where d =1,
0
Γ(αρ+α+1) 2
c1(cid:20)(cid:18) Γ(αρ+1) (cid:19) −ν2Γ2(α+1)(cid:21)=0, (31) dn =dn−1 1 2 , (35)
Γ(αρ+α·2n+1) −ν2Γ2(α+1)
(cid:20)(cid:18)Γ(αρ+α·2n−α+1)(cid:19) (cid:21)
Γ(αρ+αm+1) 2
c −ν2Γ2(α+1) +c k2 =0.
m(cid:20)(cid:18)Γ(αρ+αm−α+1)(cid:19) (cid:21) m−2 and ρ satisfies Eq.(33).
(32) Eq.(33)inone-ordercalculus(α=1)issimplyρ=±ν.
To have a starting term, c 6=0, so We meshedin Fig. 4 the surface defined by the equation
0
(33). After solvingthe equationwith ν =3forρ whenα
Γ(αρ+1) 2 varies, we plotted in Fig. 5 R (r) belonging to different
−ν2Γ2(α+1) =0, (33) ρ
(cid:20)(cid:18)Γ(αρ−α+1)(cid:19) (cid:21) values of α.
5
180 Thuswegetasolutionoftheabovedifferentialequation,
160 Fractional confluent
hypergeometric function ∞ (a)α 1
140 y(z)= k zα·k. (40)
a=1, c=1 (c)α Γ(kα+1)
120 Xk=0 k
100 α=1/3
Here (a)α is defined as
k
80 α=1/2
60 α=2/3 (a)α0 = 1, (a)α1 =a,
40 Γ(α+1) Γ(kα−α+1)
α=1 (a)αk = (a)α1(cid:18)a+ Γ(1) (cid:19)...(cid:18)a+ Γ(kα−2α+1)(cid:19),
20
k ≥2. (41)
0
0 0.5 1 1.5 2 2.5 3 3.5 4 This can be seen as a fractional generalization of the
rising factorial
FIG. 6: Fractional confluent hypergeometric function.
(a) =a(a+1)...(a+k−1). (42)
k
V. FRACTIONAL HYPERGEOMETRIC
FUNCTION
Andtheseries(40)canbeseenasafractionalgeneraliza-
tiontheconfluenthypergeometricfunction. Ifα=1,itis
Thereareothertypesofspecialfunctionsinmathemat-
exactly the confluent hypergeometric function. Profiles
ical physics. A most famous one is the hypergeometric
ofthis series(40)with differentvalues ofαaredisplayed
function [12–14]. In this section, we will try to define a
in Fig. 6.
fractionalgeneralizationofthehypergeometricfunctions.
Let’s first consider the generalization of the confluent ForthefractionalGausshypergeometricfunction,con-
hypergeometric differential equation: sider the following series
zα(Dα)2y+(c−zα)Dαy−ay =0. (36)
∞ (a)α(b)α 1
Here a and c are complex parameters. When α=1, this y(z)= (kc)α k Γ(kα+1)zα·k, (43)
is the ordinary confluent hypergeometric equation. Xk=0 k
Introducing the fractional Taylor series
which reduces to the Gauss hypergeometric series when
∞
y(z)= c zα·k, (37) α=1.
k
Xk=0 The ratio of successive coefficients is
and substituting, we get the ratio of successive coeffi-
cients
a+ Γ(kα+1) b+ Γ(kα+1)
c ·Γ(kα+α+1) (cid:18) Γ(kα−α+1)(cid:19)(cid:18) Γ(kα−α+1)(cid:19)
ck+1·Γ(kα+α+1) = a+ Γ(Γk(αk−α+α+1)1), (38) k+c1k·Γ(kα+1) = c+ Γ(kα+1) ,
ck·Γ(kα+1) c+ Γ(kα+1) (cid:18) Γ(kα−α+1)(cid:19)
Γ(kα−α+1)
(44)
c ·Γ(α+1) a
1 = . (39) or
c c
0
Γ(kα+1) Γ(kα+α+1)
c ·c +c ·
k+1 Γ(kα−α+1) k+1 Γ(kα−α+1)
Γ(kα+1) Γ(kα+1) Γ(kα+1)
=c ·ab+c ·(a+b) +c · . (45)
k k k
Γ(kα−α+1) Γ(kα−α+1)Γ(kα−α+1)
∞
Since Γ(kα+1)
zαDαy(z)= c zα·k, (47)
k
Γ(kα−α+1)
∞ Xk=1
y(z)= c zα·k, (46)
k
Xk=0
6
6 Whenα=1,thisequationreducestotheordinaryGauss
hypergeometricequation. Wedrawcurvesofsomeexam-
5.5
Fractional Gauss ple functions in Fig. 7.
5 hypergeometric function
4.5
a=1, b=1, c=1
4 α=1 VI. SUMMARY
3.5
In this letter, we defined fractional vector calculus in
3
thesphericalcoordinateframework. WediscussedLapla-
α=4/3
2.5 cian equation and heat conduct equation in this kind of
α=3/2
framework. Some special functions are generalized.
2
α=5/3 The geometryinduced byfractionaloperationsis non-
1.5 addictive and nonlocal. This kind of geometry is not
Riemannian; since fractional calculus has found applica-
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
tions in many areas of science and engineering, it is the
effective geometry of such physics processes. It will be
FIG. 7: Fractional Gauss hypergeometric function. Gauss
hypergeometricfunctionisdivergentat1. Fromthefigurewe meaningful to investigate further this kind of geometry.
can see this is not the case for fractional equivalences with a For a complete fractional vector calculus, fractional
bigger α. vector integral is indispensable. But in this letter we re-
strainedourselvesfromthis direction. Itwillbe interest-
ingto discussfractionalvectorintegralinourframework
the equation(45)canbe translatedto afractionaldiffer- and generalize Green’s, Stokes’ and Gauss’s theorem.
ential equation
Special functions show up in different areas of physics
and engineering. A fractional generalization of these
ab·y(z)+(a+b)zαDαy(z)+zαDα zαDαy(z)
functionsmayfindapplicationsinsimilarsituationswith
=c·Dαy(z)+zα(Dα(cid:2))2y(z). ((cid:3)48) anomalous tailing behaviors and/or nonlocal properties.
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