Fractional Calculus

Student: Manal AL-Ali

Dr. Abdalla Obeidat

Designation

Designation means integration and differentiation of arbitraryorder, In other expressions it means dealing with operators like

, , is arbitrary real or Complex value.

Many definitions are proposed to find the fractional derivatives

and integrals , the most common one is Riemann – Liouville Definition.

υ

υ

dx

dυ

υ

−

−

dx

d υ

Riemann Riemann Riemann Riemann –Liouville DefinitionLiouville DefinitionLiouville DefinitionLiouville Definition....

Riemann – Liouville Definition is given by the following

equation :

⟨−−Γ

≤≤−⟩

=

∫ −−

−

x

a

m

xa

m

xa

duufux

nnxfIdx

d

xfD

0Re.....)()()(

1

Re1....0Re).......()(

)(1 υ

υ

υυ

υ

υ

υ (1)

When a = -∞, eq. (1) is equivalent to Riemann's definition,

and when a = 0, we have Liouville's definition.

Example:

evaluation of the fractional derivative for the functionf(x) = xb, b >-1.

We have for m-1 ≤ ≤ m , m € N

For b=0,

*

You can see that the fractional derivative of constants is notzero!

∫ −−−

−

+−Γ

+Γ=−

−Γ=

=

x

bbmmb

x

m

x

m

x

xb

bduuux

mdx

dxD

xfIdx

dxfD

0

1

0

00

)1(

)1()(

)(

1{)(

)()()(

υυυ

υυ

υυ

)1(0

α

αυ

−Γ=

−cx

cD x

(2)

υ

MittagMittagMittagMittag----LefflerLefflerLefflerLeffler FunctionFunctionFunctionFunctionMittag-Leffler function of one parameter is denoted by

∑∞

= +Γ=

0 )1()(

k

k

k

zzE

αα (3)

A two- parameter Mittag-Leffler function is defined by the series

expansion

∑∞

= +Γ=

0

,)(

)(k

k

k

zzE

βαβα , (α > 0, β > 0) (4)

Relation to some other function

It is follows from the defection that :

∑∞

=

=+Γ

=0

1,1)1(

)(k

zk

ek

zzE ∑

∞

=

−=

+Γ=

0

2,1

1

)2()(

k

zk

z

e

k

zzE

∑∞

=

=+Γ

=0

22

1,2 )cosh()12(

)(k

k

zk

zzE

∑∞

=

=+Γ

=0

22

2,2

)sinh(

)22()(

k

k

z

z

k

zzE

∑∞

=

=+Γ

−=−

0

22

1,2 )cos()12(

)1()(

k

kk

zk

zzE

∑∞

=

=+Γ

−=−

0

22

2,2

)sin(

)22(

)1()(

k

kk

z

z

k

zzE

Laplace transformLaplace transformLaplace transformLaplace transformLaplace transform of fractional differential operator is given by

∑−

==

−−−=ℑ1

0

0

1

00 )]([)(});({n

k

t

kp

t

kpp

t tfDssFsstfD where n-1<p<n. (5)

Laplace transform of Mittag-Leffler function

∫∞

+

−−+− =±

0 1

)(

,

1

)(

!))((

k

kkpt

ap

pkdtatEte

∓α

βαα

βαβα

,(Re (p) > ) , (6)α

1

a

Fractionalization of physical problem

Fractionalization of physical problem means bringing the

tools of fractional derivatives / fractional integrals into the

theory of the problem by fractionalization of some appropriate operators. Then search for physical meanings.

The question now is how we can choose those operators ???

The problem fractional multipoles

the general problem of

electrostatic potential

of a static electric chargedistribution in free space

the general problem of

electrostatic potential

of a static electric chargedistribution in free space

point monopole

Case 2Case 1

r

1αΦ

point dipole Intermediate cases

2

1

rαΦ

υL

L

υα

r

1Φ

0

2/ ερ−=Φ∇Poisson Equation

is the volume charge density of the source ρ

)(0 rq�

δρ = )(.1 rp���

δρ ∇−=Case 1 Case 2

zq

pL

∂

∂−= )(

Then the fractional operator will be υ

υυυ

zq

pL

∂

∂= )(

The intermediate “fractional” source can be defined by the

following equation: υρ

))((0 rqz

lL�

δρρυ

υυυ

υ∂

∂== (7)

By the help of Riemann – Liouville definition of fractional

operator eq.(7) can be written as

})()()()()2(

1{)(

1

2

2

∫∞−

−−−Γ

=z

duyxuuzdx

d

q

pq δδδ

υρ υυ

υ (8)

−Γ= −υυ

υυ

δδρ 1

2

2

)2(

1)()()( zzUyx

dz

dql (9)

The scalar potential of this factional source is found to be

R

qDlLzyx z

πεψψ υυυ

υ4

),,( 0 ∞−==

)cos(4

)1(1

0

θπε

υυυ

υ

−+Γ

=+

PR

ql

Figure.2. The Fractional potential

(10)

Fractional problems using MittagFractional problems using MittagFractional problems using MittagFractional problems using Mittag----Leffler functionLeffler functionLeffler functionLeffler function

Now we list here some solved fractional physical problemsusing series solution. and then rewrite the solutions of these

problems in terms of Mittag-Leffler function.

1) The fractional LC-RC circuit

A.A.Rousan, N.Y.Ayoub, etal suggested a fractional differential

equation that combines the simple harmonic oscillation of anLC circuit with discharging of an RC circuit.

0)(1

1

1

=+ +

+

+

Qdt

Qd α

α

α

αω where 0 ≤ α ≤ 1

ω(α) = ω1αω2

1-α(11)

When α = 0, equation (11) goes over to the equation of RC circuit

0)()(

2 =+ tQdt

tdQω where ω2 = 1/RC (12)

and, when α = 1, equation (11) become the LC circuit equation

0)()(

12

2

=+ tQdt

tQdω where ω1

2 = 1/LC (13)

Apply laplace transform to eq.11,and use the following initial

conditions

00 )())(

( Qdt

tQdt

α

α

α

αω== 0))(

( 01

1

==−

−

tdt

tQdα

α

We get

αα

α

αω

αω++ +

=ℑ110

)(

)()(

sQtQ

(14)

compare eq. (14) by eq. (6), we get

))(()()(1

1,10

ααα

α ωω +++ −= tEtQtQ (15)

Now when α = 0

And when α = 1

teQtEQtQ 2

021,10 )()(ωω −=−= Solution of eq.12

)sin())(()()( 10

2

2,20 tQtiEtQtQ ωωω ==

)cos()cos()(

)( 10110 tItQdt

tdQtI ωωω === Solution of eq.13

2) Fractional Simple Harmonic Oscillator

The equation of motion that covers a dynamic system is given

by the following equation:

0)(

2

2

=++ kxdt

dxb

dt

txdm (16)

Akram. A. Rousan, Nabil.Y.Ayoub and Khetam Khasawinah

suggested a modified equation of eq. (16) using fractionalized second term as follows:

0)(

2

2

=++ kxdt

dxb

dt

txdm

α

α

α (17)

Define the damping ratio αωη

−=

22m

band

m

k=2ω

Then take laplace transform of eq.17

0)(])(

[2])(

[22

2

2

=ℑ+ℑ+ℑ −tx

dt

txd

dt

txdωαηω

α

αα

0)(}][)({2])(

[)]([)(2

01

12

00

2 =+−+−− =−

−−

== txdt

xdsxs

dt

tdxtxssxs ttt ωαηω

α

ααα

Use the initial conditions

0][ 0 ==tdt

dx0][ 0 ==t

dt

xdα

α

cdt

xdt ==−

−

01

1

][α

α

, ,

We get

0)(})({2)(22

0

2 =+−+− −txcsxssxsxs ωαηω αα

(18)

Rearranging the equation, one can write it as follows

21222

2

0

2

2)( xx

ss

csxsx +=

++

+=

−

−

ωαηω

αηωαα

α

Let

222

01

2)(

ωαηω αα ++=

− ss

sxsx

222

2

22

2)(

ωαηω

αηωαα

α

++=

−

−

ss

csx

One can rewrite x1 and x2 as follows

∑∞

=+−−

−−

+−=

0122

12

01)2(

)1()(p

p

ppp

s

sxsx

αα

αα

αηω

ω

∑∞

=+−−

−−−

+−=

0122

22

2)2(

)1(2)(p

p

ppp

s

scsx

αα

ααα

αηω

ωαηω

(19)

(20)

Comparing eq.19 and 20 by eq(6) , we get

∑∞

=

−+− −

−=

0

2

1,2

2

01))(2()(

!

)1()(

p

p

p

pp

tEtp

xtx ααα ωαηω

∑∞

=

−+−

− −−

=0

2

2,2

22

2))(2()(

!

)1(2)(

p

p

p

pp

ttEtp

ctx ααα

α ωαηωαηω

Now x(t) = x1 + x2

∑∞

=+−

−+− +

−=

0

2,2

2

1,20

2)}(2)({)(

!

)1()(

p

p

p

p

p

pp

ztEzExtp

tx ααα

αα εηωω (21)

Where and c taken to be αωαη −−= 2

)(2 tz1

0

−αωx

a) Simple harmonic oscillator ( )0=α)cos()()( 01,20 txtExtx ωω ==

b) Damped oscillator( ) 1=α

∑∞

=++ −+−

−=

0

2,11,1

2

0 )}2(2)2({)(!

)1()(

p

p

p

p

p

pp

ttEtEtp

xtx µωηωηωω

3) Fractional Domain Wall MotionWesam Al–Sharo’a employed the idea of fractionalization of the second term in the equation of motion suggested in the problem

of simple harmonic oscillator to study domain wall motionfor two cases: 1)

2)

)()( ttf δ=

0)( fconstf ==

The fractional domain wall motion equation is:

)()()(

2)( 22

2

2

tftxdt

txd

dt

txd=++ − ωαηω

α

αα

(22)

Use the same initial conditions in S.H.O problem, and follow the

Same procedure, one get :

a) Case1: )()( ttf δ=

∑∞

=+−+− ++

−=

0

2,201,20

2)}()12()({)(

!

)1()(

p

p

p

p

p

pp

ztExzExtp

tx αααα αηωω (23)

1) For α = 0 and f (t) = 0 eq.(23) reduced to

∑ ∑∞

=

∞

=

=+Γ

−=

−=

0 0

0

2

01,2

2

0 )cos()12(

!)(

!

)1()0()(

!

)1()(

p p

pp

ppp

txp

pt

pxEt

pxtx ωωω

(24)

2) For α = 1

∑∞

=++ −+−

−=

0

2,11,1

2

0 )}2(2)2({)(!

)1()(

p

p

p

p

p

pp

ttEtEtp

xtx µωηωηωω

αωαη −−= 2)(2 tz and

1

0

−= αωxc

(25)

2) Case 2 : 0)( fconstf ==

∑∞

=+−+−+− ++

−=

0

3,2

2

02,211,20

2)}()()({)(

!

)1()(

p

p

p

p

p

p

p

pp

zEtfztExzExtp

tx ααααααω (26)

Where: αωαη −−= 2

)(2 tz and αηω21 =x

1) For α = 0 eq.(25) reduced to

)cos()cos()(2

0

2

0

0 tff

txtx ωωω

ω −+=

2) For α = 1

∑∞

=+++

′+′′+′

−=

0

3,1

2

02,111,10

2)}()()({)(

!

)1()(

p

p

p

p

p

p

p

pp

zEtfztExzExtp

tx ω

(27)

(28)

)(2 tz ωη−=′Where: and ηω21 =′

x

4- Conclusion

We were able to rewrite some solved problems from JUST

which are published in scientific journals in terms of Mettag-Leffler function, which is a demand from most prestigious

journals.

5. Reference:

[1] I. Podlubny, Fractional Differential Equations, Volume 198,ACADEMIC PRESS, 1999.

[2] R.Hilfer, application of fractional calculus in physics, WORLD SCIENTIFIC, 2000.

[3] D.H.Werner and R.Mittra, Frontiers in Electromagnetic, IEEE Press, chapter 12.

[4] A.Rousan and N.Ayoub, A Fractional LC-RC Circuit, An International Journal for

Theory and Applications.

[5] A.Rousan, N.Ayoub and K.Khasawenah, Fractional Simple Harmonic Oscillator,

International Journal of Theoretical Physics.

.[6] Wesam Al–Sharo’a, Fractional Domain Wall Motion,