National SeminarApplication of Generalized CalculusApplication of Generalized Calculus in Physics and Applied Mathematics

Keynote address

Department of Physics Jadavpur UniversityApril 26‐27, 2016

Keynote address

Generalized Fractional Calculus is far from a paradox‐is a reality‐is a reality

Shantanu DasRCSDS (E&I Group) BARC Mumbai, ( p) ,

Senior Research Professor Dept. of Phys, Jadavpur Univ (JU),Adjunct Professor DIAT‐Pune

UGC‐Visiting Fellow Dept. of Appl. Math Calcutta Univ.shantanu@barc gov inshantanu@barc.gov.in

http://scholar.google.co.uk/citations?user=9ix9YS8AAAAJ&hl=enwww.shantanudaslecture.com

What is there for next 90 minutes ?Calculus is as old as three hundred plus years so is fractional calculus

Why  we  need to have                                       ?fractional calculus

The differentiation          operation is taking rate per unit (infinitesimal)          as dd x dx d 0x ↓

F l t d fi i t f i ti f( )d α d 0↓ 0 1 f ti l diff ti tiFor an element             defining rate as               for say                      gives notion of( )d x d 0x ↓ 0 1α< < fractional differentiation

Similarly integration with respect to element               gives notion of( )dx α fractional integration

Obviously ( )d dx xα > what does it say then ?

Fractional calculus is extension of the classical calculus

Is there any                                               ? conjugation or parallel

Can we apply functions?to non‐differentiable

System dynamics having memory has direct link with fractional calculus

Plus several other seemingly paradoxical views that is linked to  fractional calculus

We will discuss aspects of fractional calculus which is not covered in anymodern and still developing

Can we apply                                             functions?to non‐differentiable

We will discuss                                                        aspects of fractional calculus, which is not covered in any books on classical fractional calculus, for the later 50 % of the presentation i.e. Part‐B

modern and still developing

Hope to deal with these                                 in the next 90 minutes130 odd slides

Relax & enjoy

Part‐A

Classical fractional calculus

Birth date  of fractional calculus

I l tt t L’H it l i 1695 S t b 30 L ib i i d th ibilit f li i

L’Hospital’s query

In a letter to L’Hospital in 1695 September 30, Leibniz raised the possibility of  generalizing the operation of differentiation to non‐integer orders, and L’Hospital asked what would be the result of half‐differentiating ; that is; 

1 2 1d ( )1 2 12

1 2

d ( ) ? o r [ ] ?d x

x D xx

= =

Leibniz replied “It leads to a paradox, from which one day useful consequences will be drawn”.

We can say this query dated 30th September, 1695 gave birth to “Fractional Calculus”; therefore this subject of fractional calculus with half derivative and integrals etc, are as old as conventional Newtonian or Leibniz’s calculus. 

Date of Birth of Fractional Calculus 30th September, 1695

What is Fractional Calculus

The Fractional Calculus is a name of theory of integration and derivatives of arbitrary order

which unify and generalize the notion of integer order  n‐fold repeated differentiation 

d ( )n f x

and n‐fold repeated integration.

d nx

1 2 1( )

1 2 1 0 0( ) ( )(d ) d d ... d ( )dnx x xx

xn nn na

f x f y y x x x f x x−

−− −= ≡∫ ∫ ∫ ∫ ∫

to arbitrary order

Fractional Calculus is

1 2 1n na a a a−

; ,nα α α← ∈ ∈

Fractional Calculus isGeneralized differentiation and integration.

Generalized Differ‐integration

The Generalized CalculusComplex‐order Calculus

d ( ) ,d

i

i

f xx

α β

α β α β+

+ ∈

p

3

F ti l d l l

d ( )d

f xx

α

α α ∈

Fractional order calculus

Non‐local

1 2

Non localDistributedHistory/heredityNon‐MarkovianNon‐NewtonianNon‐differentiability

Integer order calculus

4

d ( )d

n

n

f x nx

∈NewtonianLocalPoint property

Continuous‐order Calculus

( )d ( )d

( ) db f t

xaK

α

αα α∫

Integer Order mass spring damper is 0( ) ( ) ( ) ( )m x t b x t k x t f t+ + =

Motivation‐Discrete Delta Distributed Integer Orders

20( ) ( ) ( )m s b s k X s F s+ + =Laplace of above gives

Using  the property of Delta function i.e.                                                                           with 0 0( ) ( )d ( )g gα δ α α α α− =∫ ( )g s αα ≡0 0( ) ( ) ( )g g∫

( )2

0( ) ( ) ( )

( 2 ) d ( 1 ) d ( 0 ) d ( ) ( )

m s b s k X s F s

b k X Fα α αδ δ δ

+ + =

∫ ∫ ∫

We write the following

( )( 2) ( 1) ( ) d ( ) ( )m b k s X s F sαδ α δ α δ α α∞⎛ ⎞

− + − + =⎜ ⎟∫

( )0( 2 ) d ( 1 ) d ( 0 ) d ( ) ( )m s b s s k s s X s F sα α αδ α α δ α δ α− + − + − =∫ ∫ ∫

( )00

( 2) ( 1) ( ) d ( ) ( )m b k s X s F sδ α δ α δ α α+ + =⎜ ⎟⎝ ⎠∫

Delta distributed integer order

0bk

0 1 2 α

m

Motivation‐Discrete Delta Distributed Fractional Order

We have arrived at the following for simple mass spring damper system

( )00

( 2) ( 1) ( ) d ( ) ( )m b k s X s F sαδ α δ α δ α α∞⎛ ⎞

− + − + =⎜ ⎟⎝ ⎠∫

We can generalize the above by adding few delta functions with weights to involve fractional dampersWe can generalize the above by adding few delta functions with weights to involve fractional dampers

( ) ( ) ( ) ( ) ( )( )3 11 0 22 2

0

2 1 d ( ) ( )m b b b k s X s F sαδ α δ α δ α δ α δ α α∞⎛ ⎞

− + − + − + − + =⎜ ⎟⎝ ⎠∫

t t d di t ib ti

0bk m

to get order distribution as

1b

2b

0 1 2

α

12

32

( ) ( ) ( ) ( )3 12 22 1

1 0 2( ) ( ) ( ) ( ) ( ) ( )t t t tm D x t b D x t b D x t b D x t k x t f t+ + + + =

Corresponding Fractional Order mass spring  damper system is thus

Delta distributed fractional order system

Motivation‐Continuous Order System

From distributed discrete Fractional Order system that is

( ) ( ) ( ) ( ) ( )( )3 11 0 22 22 1 d ( ) ( )m b b b k s X s F sαδ α δ α δ α δ α δ α α

∞⎛ ⎞− + − + − + − + =⎜ ⎟

⎝ ⎠∫ ( ) ( ) ( ) ( ) ( )( )0⎝ ⎠∫

We generalize the above of several delta functions with several weights as follows

( ) d ( ) ( )N

k X Fαδ∞⎛ ⎞⎛ ⎞

⎜ ⎟⎜ ⎟∑∫ ( )0

d ( ) ( )n n nn

k s X s F sαδ α α α α=−∞

− = ∈⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∑∫

When these adjacent delta functions become very‐very close the Delta distribution resembles a  continuous function ( )0

( ) Nn nK kα δ α α= −∑ ( )0

( ) n nn =∑( )K α

0 α

With this we write the Continuous Order System as                                                                         ( ) d ( ) ( )K s X s F sαα α∞⎛ ⎞

=⎜ ⎟⎝ ⎠

∫− ∞⎝ ⎠

Tough to solve requires inverse Laplace of                                   ?( ) dK s αα α∞

− ∞∫

Functional Fractional Calculus Part‐B

The main or at least the motive to investigate fractional differential manifolds is in fact that

Motivation‐ Hurst exponent and fractional calculus

The main or at least the motive to investigate fractional differential manifolds is in fact that

fractional calculus seems to be quite relevant to investigate some problems which occur in 

in fractal space‐time physics.

First elementary point of view the mathematics framework of fractal physics deals with

functions                of which the differential satisfies the condition ( )f x( ) ( )d ( ) d 0Hf( ) ( )d ( ) d 0Hf x x H∝ >

is referred as Hurst parameter

h b l h l l l l h ld

:H

This above relation means the classical equality                                  no longer hold, i.e. ( )d ( ) ( ) df x f x x′= ( )f xis not differentiable, and that instead we should be using modeling in form

( ) ( )d ( ) ( ) d Hf x g x x=( ) ( )

Which directly leads to fractional calculus ( ) ( )( )

d ( ) d ( )d d

f x f xH

x x αα← ←

Part‐B

Motivation‐Self similarity , non‐differentiability and fractional calculus

Qualitatively speaking a self‐similar function is a function which exhibits similar patterns when

one changes the scale of observation, that is –the pattern generated  by               and             

with                looks like same    

( )f ax

0a >

( )f x

Formally ,               ,                   is similar of order                , if one has equality( )f x x ∈ H

( ) ( ) 0 0Hf ax a f x a H= > >

Which means that the landscapes in the vicinity of            and                  looks alike      x ax

Such a function satisfies the condition                           (0) 0f =

Also provides relation ( via                      ,                   )a x← 1x ←

( ) (1)Hf x x f=

As a special case when ; the function is non differentiable0 1H< < ( )fAs a special case when                               ; the    function                    is non‐differentiable 0 1H< < ( )f xbut is fractionally‐differentiable for fractional order    H

Part‐B

Motivation‐ Coarse graining phenomenon and fractional calculus

In systems involving course grained phenomenon, every thing happens as if the elemental

point is not infinitesimally small (zero) rather it exhibits some thickness, what could be  

pictured by using instead of(d )x α0 1α< < dxpictured by using                                        , instead of (d )x 0 1α< < dx

As                   we will have always d 0x ↓ ( ) ( )d dx xα >

In other words we are considering rate of variation as           ( )( )

d ( )

d

f x

x α

f l d d f l l l

For fine grained phenomena rate of variation is                           that is classical derivative ( )( )

d ( )df x

x

Here again we come across fractional derivative and fractional calculus

( )dx

Part‐B

Other Motivations for Fractional Calculus 

Distributed systems behave as fractional order‐we consider non‐locality rather than point property.; thus representation of distributed system is better with fractional calculus.Almost all semi‐infinite system gets representations in half derivative.

Memory based systems are represented by fractional calculus‐relaxation of di‐electrics, Visco‐l ti l t S it h i di h i d th‐elastic elements, Super‐capacitor charging discharging and others

Transport phenomena in fractal space‐time

Transport phenomena in porous medium

...and many others

Functional Fractional Calculus

Magnetic Levitation the ball afloat via action of fractional calculus

Voltage to coil  u ( t )

ll i iBall position

Courtesy: BRNS funded joint project of VNIT Nagpur and BARC to develop digital fractional order controller for industrial applications.

Magnetic Levitation the ball afloat via action classical calculus

Voltage to coil  u ( t )

Ball position

Courtesy: BRNS funded joint project of VNIT Nagpur and BARC to develop digital fractional order controller for industrial applications.

Why this control effort less? and its repercussion‐a conjecture

Comparing these two cases we see that to do the same job‐that is to position the floating ball and slowly making its position follow sinusoidal command the output of the controller in case of second case is fluctuating severely. Many critics will say it is ‘noise’, but why the noisy output is absent in fractional calculus based systems? The reason is that fractional derivatives andis absent in fractional calculus based systems?  The reason is that fractional derivatives and fractional integrals are having inherent memory. This memory in the systems works to govern the ball position based on its previous experience‐therefore these fractional differentiation and integration gives an ideal filtering action. Whereas the classical differentiation is a point property‐does not therefore has memory, and acts instantly with no previous experience. 

Thus in the second case the maneuvering signal is going very high instantaneously and in the next moment going very low again and again giving a notion as if some ‘noise’ is getting injectednext moment going very low again and again giving a notion as if some  noise  is getting injected into the system. 

We also infer in the second case the controller is making lot of efforts to control  this ball and keep it afloat, while the first case the controller action is smooth and effort‐less. 

So we can see fractional calculus based system does the control action with a lesser effort than h i l l i l l l b d ll h f b ffi ithe conventional classical calculus based controllers, therefore are better efficient.  

So we have to have operations like ‘half derivative’

How?

We know to calculate one whole derivative 

We know to calculate one whole integration ( )We know to calculate one whole integration

We make use of this knowledge and ( )f x ( )f x′

( )12 ( )f x

1. Do one whole derivativemake use of number‐line

2. Do half‐integration

( ) ( ) ( )1 12 2 (1)( ) ( )f x D D f x−=

Here function requires to be differentiable

1. Do one whole derivative

0 112

Here function requires to be differentiable

( )f x1

2 ( )f x−( )1

2 ( )f xAlternatively we move opposite

1. Do half integration

2. Do one whole derivation( ) ( )1 12 2(1)( ) ( )f x D D f x−=

012− 11

2

Here function need not to be differentiable

We should know how to find ‘half‐integration’ for both methods

Generalizing the expression for repeated integration

[ ]1 1x

∫

What we know is following repeated integration

[ ]

( ) [ ]

( ) [ ]

0

0 1

1 10 00

2 2 21 1 0 0 00 0 0 0

3 3 3 2

( ) d ( ) ( )

( ) d ( ) d d ( ) ( ) ( ) ( ) d

1( ) d ( ) d d d ( ) ( ) ( ) ( ) d

x x

x x x x

x x

x x x x x

f y y I f x D f x

f y y f x x x I f x D f x f y x y y

f y y f x x x x I f x D f x f y x y y

−

−

−

= =

= = = = −

= = = = −

∫∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ ∫( ) [ ]

( ) 3

2 2 1 0 0 00 0 0 0 0

43 3 2 1 0 00 0

( ) d ( ) d d d ( ) ( ) ( ) ( ) d2

( ) d ( ) d d d d

x x

x x

x

f y y f x x x x I f x D f x f y x y y

f y y f x x x x x I

= = = =

= =

∫ ∫ ∫ ∫ ∫

∫ ∫[ ]

0 1 4

0 0 0

4 30 0

( )

1( ) ( ) ( ) d( 2 ) (3 )

x x x

x

x

f x

D f x f y x y y−

=

= −

∫ ∫ ∫

∫[ ]0 0( ) ( ) ( )

( 2 ) (3 ). . . . . . . . . .

x f f y y y∫

So for                       , we have following generalizationn +∈

( )

[ ]

0 2 1

3 1 2 1 0 00 0 0 0 0

10 0 0

( ) d . . . . . ( ) d d . . . . . d d ( )

1( ) ( ) ( ) ( ) d( 1 ) !

n nx x x x xn nn n x

xn n nx x

f y y f x x x x x I f x

I f x D f x f y x y yn

− −

− −

− −

= =

= = −−

∫ ∫ ∫ ∫ ∫

∫( )

What is further generalization if             is non‐integer, i.e. what to do with n ( 1)!n −

About Generalization of Factorial

1 0n =⎧F t i l 1 0!

( 1 ) ! ( ) 0n

nn n n

=⎧= ⎨ − >⎩

Factorial

Generalization of factorial

0

( !)( 1) ! lim( )

z

nk

n nzz k∞→ ∞

=

− =+∏

Euler

! 2nnn n

eπ ⎛ ⎞≈ ⎜ ⎟

⎝ ⎠

Stirling

( )1

1

0

! l n dnxn x= ∫Euler

0! dt nn e t t

∞ −= ∫

[ ]1( ) d R 0t t tα∞

− −Γ >∫Euler’s Gamma function [ ]1

0

( ) d R e 0te t tαα αΓ = >∫Euler s Gamma function

( )( 1 ) ( )α α αΓ + = Γ This gives analytic continuity to negative axis

So                               ! ( 1)α α= Γ +

Generalized Factorial to Gamma Function plots

! ( 1)α α= Γ +

( )xΓ( ) ( )( ) ( )

3 5 32 2 4

5 3 4

!

!

π

π

= Γ =

Γ

Some values are following

( ) ( )( ) ( )( ) ( )

5 3 42 2 3

31 12 2 2

1 12 2

!

!

!

π

π

π

− = Γ − =

= Γ =

− = Γ =

xPlot of Gamma‐Function

( ) ( )( ) ( )

2 2

3 12 2! 2 π− = Γ − = −

Plot of reciprocal of Gamma Function 

1

Generalizing the expression for repeated integration for positive real number to get fractional integration formula

[ ] [ ] 10 0 0

1

0

1( ) ( ) ( ) ( ) d( 1) !

1 ( ) ( ) d( )

x

x x

x

I f x D f x f y x y y

f y x y y

α α α

α

α

α

− −

−

= = −−

= −Γ

∫

∫

Riemann‐Liouville formula

E l ( )f

( )( )11 1

22 2 10 0 1

2 0

1 ( ) dx

x xI x D x y x y y−−⎡ ⎤ ⎡ ⎤= = −⎣ ⎦ ⎣ ⎦ Γ ∫( ) ( )2 sinx x xπ θθ +

To get the following result

Example ( )f x x=

( )1 22 0

1 d

1

x

x

y yxy y

y

=Γ −

∫

∫

( )( ) ( )

( ) ( ) ( )

21

2

2 2

2

2

sin2 2

0 1 22 4 4

2 21 1

cos1 dsin

1 sin d

x x x

xx x

x x

D xπ

π

θθθ

θ

θ θ

+−

−

⎡ ⎤ =⎣ ⎦ Γ −

= +Γ

∫

∫( ) ( )

( )

2 21 22 0 4 4 2

1

1 d2

1 d

yx x

x

y yy x

y y

=Γ − − +

=Γ

∫

∫

( ) ( ) ( )

( )

2

2

2

2 21 12 2

12

1 cos2 2x x

π

π

π

θ θ

−

−

Γ

⎡ ⎤= −⎢ ⎥Γ ⎣ ⎦

∫

( ) ( )21 22 0 4 2

x xy

yΓ − −∫

( ) ( )121

22

2

x

x

π π

π

= Γ =Γ

=Put s i nx xy θ+= ( )d (cos )dxy θ θ 2Put s i n

2x xy θ+= ( )2d (cos )dxy θ θ=

2πθ = − to

2πθ =New limit Rigorous indeed? 

Generalizing the expression for repeated differentiation‐Euler formula repeated differentiation

This is exactly what we would expect based on a straightforward interpolation of the derivatives of a power of  x . Recalling that the first few (whole) derivatives of are

mx

2 31 2 3

2 3

d d d; ( 1) ; ( 1)( 2)d d d

m m mm m mx x xm x m m x m m m x

x x x− − −= = − = − −

Thus we expect to find that the general form of the n‐th derivative of         ismx

d !d ( ) !

n mm n

n

x m xx m n

−=−( )

Replacing the integer  n with the general value   ,  and using the gamma function to express the factorial, this suggests that the a fractional derivative of         is simply

υmx

( ) ( )

12 1

21

2

1

312 2

d ! ( 1) d 1! 1 2d ( 1) ( 1) 1 1d

mm mx m m x xx x x

x m m x

υυ υ

υ υ υ π−− −Γ +

= = = = =Γ − + Γ − + Γ − + Γ

1which is answer to L’Hopital’s question                              , is it?

This Euler formula holds for           ; that is requires function to be better behaved than1m > − 1x −

[ ]12 2 xD x π=

k∑

Use of Euler’s formula and Fractional derivative of ‘Constant’as non‐zero!!

Now, since analytic functions can be expanded into power series  we can use Euler formula, applying it term by term to determine the fractional derivatives of all such functions. Furthermore, applying this formula with negative  values of           gives a plausible 

i f th f ti l i t l f f F l t fi d th h l

( ) kkk

f x a x= ∑

υexpression for the fractional‐ integral of a power of            . For example, to find the whole integral of           we set            and then compute via Euler formula as

x3x 3m =

13 3 ( 1) 4 4 4 3d 3! 3! 6 1 x−

∫( )3 3 ( 1) 4 4 4 3

10

d 3! 3! 6 1( ) dd 3 ( 1) 1 (5) 24 4

x x x x x y yx

− −− = = = = =

Γ − − + Γ ∫

Note that the above integration is valid only if the initial point be zero, else initial value isNote that the above integration is valid only if the initial point be zero, else initial value is subtracted. The unification of these two operations makes it even less surprising that generalized differentiation is non‐local, just as is integration‐has memory history hereditary!

Incidentally, the generalized derivative as developed so far gives some slightly surprising results. For example, the half‐derivative of any constant function           is0Cx

( )

12

12 1

2

d 0 ! C( C ) C 0d x xπ

= = ≠Γ Fractional derivative of constant is not zero!!

The Founders

Joseph Liouville1809‐1882

George Friedrich Bernhard Riemann1826‐1866

Michelle Caputo 1967

Liouville’s Postulation

d ;d

a x a xe a ex

υυ

υ υ += ∈

This postulate is Liouville’s proposition conjugation to normal differentiation.  

Negative values of  represent integrations (anti‐derivative) and we can even extend this to allow complex values of or even to a continuous distribution of this order in some interval.

In a way this generalizes notion of integration and differentiation to arbitrary  order!

Liouville’s class of functions and approach

Any function expressible as a sum of exponential functions can then be differentiated in the same way. 

( ) ( )2 2/ 2 / 2

d d ( ) ( )co s( )d d 2 2

( ) ( )

ix ix ix ix

i x i xi ix i ix

e e i e i exx x

e e e e e eπ π

υ υ υ υ

υ υ

υ υυ π υ π

− −

+ − +− −

⎛ ⎞+ + −= =⎜ ⎟

⎝ ⎠

+ +

( )2

( ) ( )2 2

co s

e e e e e e

x πυ

+ += =

= +

Since / 2 / 2( ) ( )i ii e eυ π υ υπ± ±± = = we have the nice result d cos( ) cosd 2

x xx

υ

υ

υπ⎛ ⎞= +⎜ ⎟⎝ ⎠

d υ

Phase Shifting

( )2d c o s ( ) c o s

dx x

xπ

υ υ= +

Thus the generalized differential operator simply shifts the phase of the cosine function and likewise the sine function by that is in proportion to the order of the differentiation0(90)υ ×likewise the sine function by                , that is in proportion to the order of the differentiation. For differentiation the process advances the phase, needless to say the integration makes the phase lagged. We test by giving a sinusoid at the input of the circuit and measure the output and record the phase lag or lead, depending on the fractional order.

0(90)υ ×

Output of fractional order differentiator circuit for ½ order

Courtesy: BRNS funded joint project of VNIT Nagpur and BARC to develop analog fractional order PID.

Fractional derivative

( )f x ( )f x′

( ) ( )f xα

Caputo Derivative

( ) ( ) ( )1 (1)( ) ( ) 0 1f x D D f xα α α− −= < < 1. Do one whole derivative0 1α

1

p

( ) (1)1( ) ( )d 0 1x

D f x x y f y yαα α−= − < <∫ 2. Do                 ‐integration1 α−( )0

( ) ( )d , 0 1(1 )xD f x x y f y y α

α= < <

Γ − ∫

1

0

1( ) ( ) ( )( ) d( )

xI g x D g x g y x y yβ β β

β− −= = −

Γ ∫We used 

( )f x(1 ) ( )f xα− −

( ) ( )f xα

( )β

Riemann‐ Liouville fractional derivative

1. Do             integration

h l d

( ) ( )(1) (1 )( ) ( )f x D D f xα α− −=0(1 )α− − 1α

(1 )α−

2. Do one whole derivation

( )0

1 d( ) ( )d , 0 1(1 ) d

x

xD f x x y f y yx

αα αα

−= − < <

Γ − ∫

For ,              we move the integer point             to a integer point 1α > 1 m

( 1)m mα− < <where

Relation between RL and Caputo fractional derivative

0 1α< ≤For ( )x a α−

when ( ) 0f a = [ ] [ ]( ) ( )CD f x D f xα α=

[ ] [ ] ( )( ) ( ) ( )(1 )

Ca x a x

x aD f x D f x f aα α

α−

= +Γ −

Proof is by employing integration by parts to RL formula and then via induction

[ ] [ ]

[ ] [ ]( )1 ( )( ) ( ) ( )km

C kf af fα α α−

∑

1m m mα +− < ≤ ∈For we have

[ ] [ ]

[ ]

0

1 1(1) ( 1)

( )( ) ( ) ( )( 1)

( ) ( ) ( )( ) ( ) ( ) ... ( )(1 ) (2 ) ( )

C ka x a x

k

mC ma x

f aD f x D f x x ak

x a x a x aD f x f a f a f am

α α α

α α αα

α

α α α

−

=

− − − −−

= + −Γ − +

⎛ ⎞− − −= + + +⎜ ⎟Γ − Γ − Γ −⎝ ⎠

∑

Kindergarten of Fractional Calculus

The fundamental theorem of Integral Calculus & RL‐Caputo relation

( ) ( ) ( )d [ , ] ( ) [ , ]x

af x f a f t t x a b f x AC a b′ ′= + ∈ ∈∫We know from our classical calculus

( ) ( ) ( )

( ) ( )( ) ( )12 2 (1) 2 2

( ) ( ) ( ) ( )d ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

x

x x x x x xaf x D I f x f a f t t f x D I f x f a I D f x

f x D I f x f a f a x a I D f x

′= = + = = +

= = + − +

∫Generalization to n‐fold is:

( ) ( )( ) ( )

( ) ( )( ) ( )(2) ( 1)1 ( ) ( )(1) 2 ( 1) ( )2! ( 1)!

( ) ( ) ( ) ( ) ( )

......

( ) ( ) ( ) ( ) ( ) .. ( ) ( )nf a f an n n n nn

f f f f f

f x D I f x f a f a x a x a x a I f x− −−= = + − + − + − +

x

( ) ( (n nx xf x D I f x=

( )

( )

1 ( ) 1 ( )1! ( 1)!0

1 ( ) 1 ( )1( 1) ( )0

)) ( ) ( ) ( )d

( ) ( ) ( ) ( ) ( )d

k

k

xn f a k n n

k nka

xn f an n k n n

k nk

x a x t f t t

f x D I f x x a x t f t t

− −−=

− −Γ + Γ=

= − + −

= = − + −

∑ ∫

∑ ∫a

( )1 ( ) 1 xkn f a−

∫

Differentiation of integration is‐constant(s) at start point plus integration of differentiation. This is also RL, Caputo fractional derivative; a very important generalization.

1 ( )

0

( )1

0

( ) 1( ) ( ) ( ) ( )d( 1 ) ( )

( )( ) ( ) ( )( 1 )

n n k n n

k akn

k Cx x

k

f aD I f x x a x t f t tk n

f aD f x x a D f xk

α α α

α α α

α α

α

− − − −

=

−−

=

= − + −Γ + − Γ −

= − +Γ + −

∑ ∫

∑( )0 1 ( ) ( ) ( )

(1 )C

x xf aD f x x a D f xα α αα

α−< < = − +

Γ −

A look at the condition of differentiabilityThe Caputo definition requires the function be differentiable, whereas there is no such restrictions on R‐L definition. Both the definitions are equal if initial point of function is zero

The Caputo and R‐L are related by 

[ ] [ ] ( )( )(1 )( ) ( ) ( ) x aC

a x a xD f x D f x f aαα α −−

Γ= +

[ ] [ ] ( )( )( )( ) ( ) x a f aC D f x D f xαα α −−= −

[ ] [ ] ( )(1 )( ) ( ) ( )a x a xf x f x f a αΓ −

We carry on following arrangement

[ ] [ ] ( )(1 )( ) ( )a x a xD f x D f x αΓ −=

[ ] ( ) ( )(1 )( ) x a f a

a xD f aααα

−−Γ −=Using fractional derivative of constant  that is                  as

[ ] [ ] ( ) ( )( ) ( )C x aD f x D f x f aα

α α−−

( )f a

we get [ ] [ ] ( )

[ ]

( ) ( )(1 )

( ) ( )

a x a x

a x

D f x D f x f a

D f x f aα

α= −

Γ −

= −

we get

Here we get a clue. That is relaxing the condition of differentiability, if we calculate the R‐L 

derivative of offset function                                           then the condition of differentiability is( ) ( ) ( )g x f x f a= −

relaxedrelaxed

Also the offset function has zero initial condition                      gives advantage. ( ) 0g a =

Fractional derivative of non‐differentiable function

( )f x x b= +( )f x x b= +

b

0 x[ ]

( )

1 12 2

1

0 0( )x xD f x D x b⎡ ⎤= +⎣ ⎦

( )( )

( )

( )

( ) ( )

12

1 12 2

12

12

1 1 12 2 2

32 1

11 1

xx b

b x

−−

−

Γ += +

Γ + − Γ −

= Γ +Γ

( ) ( )2 12Γ

[ ] ( )1 12 2

0 0( ) ( 0 )x xD f x f D x b b⎡ ⎤− = + −⎢ ⎥⎣ ⎦

If we offset the constant then we get

( ) ( )3 12 2 != Γ =

we get fractional derivative  value at the non‐differentiable points

Fractional Derivative of Exponential Function is not Exponential !!

Liouville proposed that the general  n‐ th derivative of          is simply ,  and yet if we expand the exponential function        into a power series  

a xe ( )n a xa exe

2 32 31 . . .a x a a ae x x x= + + + +

and apply  Euler formula to determine the half‐derivative, term by term, we get  (not at all what Liouville postulated earlier!), that is not‐exponential one

1 ! 2 ! 3 !

112

12

12

2 3 40

d ( ) 1 4 8 1 61 2 .. .3 1 5 1 0 5d

xx

xeD e x x x x

x xπ⎛ ⎞⎡ ⎤ = = + + + + +⎜ ⎟⎣ ⎦ ⎝ ⎠

6

8

½ Derivative of xe

0 5 1 0 1 5 2 00

2

4

xe( )1

21

2

* 10 2 ,

xx

xeD e xx

γ±

±⎡ ⎤ = − ±⎣ ⎦

* 1( ) dx txx e t tα αγ α − − −= ∫ 0 .5 1 .0 1 .5 2 .00

x

Tricomi’s incomplete Gamma function

( ) 0( , ) dxx e t tαγ α Γ= ∫

Kindergarten of Fractional Calculus

Perhaps Leibniz was confused due to        !xe

Here we see one of the paradoxes that might have intrigued Leibniz. According to a very reasonable general definition we expect any derivative (including fractional derivatives)of the exponential function to equal itself, and the exponential goes to 1 as  x goes to 

d t f ll d i d f l f th h lf d i ti f th ti lzero, and yet our carefully‐derived formulas for the half‐derivative of the exponential function goes to infinity at   .0x =

Clearly something is wrong. Must we abandon the elegant exponential approach, along with its beautiful explanation of the trigonometric derivatives as simple phase shifts, etc?

No, we can reconcile our results, provided we recognize that the derivative is non‐local, and therefore depends on the chosen range of differentiation; that is like integration it has lower and upper limits. Fractional derivatives thus have start and end point.

The lower terminal to minus infinity is Liouvelli formulation, and this refers also to  steady state. 

In these cases our simplified results of phase shifting of trigonometric  functions do hold.

Consider the two anti‐differentiations (integrations) shown below

Fractional Derivatives Require Lower & Upper Limit like Integration !

Consider the two anti differentiations (integrations) shown below

0

0 0

443 0d d

4 4

x xxu x

x x

xxu u e u e e= − = −∫ ∫

The first integral shows that when we say is the derivative of we are implicitly3 4 / 4The first integral shows that when we say          is the derivative of                       we are implicitly assuming                  . 

3x 4 / 4x0 0x =

Similarly we have for exponential function derivative with start point                      as 0x a=

( )*d

( )cx cx

cxe eD e c x

αα γ α

⎡ ⎤⎣ ⎦⎡ ⎤ = = − + ∞⎣ ⎦

( )( )*( )

, ( )C cxC cx ea x x aD e c x aα

α γ α−−−

⎡ ⎤ = − − −⎣ ⎦

For start point as  0x = −∞

This is obtained via RL formula

[ ]( )

*

* *

, ( )( )d( )

( , )lim

x

cx

y

D e c xxx

cye x yy

α α

α

γ α

γ α

−∞

→∞

⎡ ⎤ = = + ∞⎣ ⎦ + ∞+ ∞

⎛ ⎞−= + ∞ =⎜ ⎟

⎝ ⎠⎛ ⎞ ⎛ ⎞

( )

* *

211

( , ) ( , )lim lim( )

1 1 ( )( )

cx cx

y z

zcx

z

cy ze c c e cy zcy z

ec e O zz

α αα α

α αα

γ α γ α

α

→∞ →∞

−−+

+

⎛ ⎞ ⎛ ⎞− −= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞

= − − +⎜ ⎟Γ −⎝ ⎠( )cxc eα

⎝ ⎠=

We have got Liouvelli’s postulateKindergarten of Fractional Calculus

Comment about Liouville’s approach

Needless to say, this approach can be applied to the exponential, i.e. ½ derivative of  a xe be

12d 2

12

12

dd

a x a xe a ex

=

This approach is correct in some cases for derivative with lower terminal at time‐immemorial i.e.steady state case            .This approach is also called Liouville’s approach.− ∞

The exponential approach (Liouville’s) seems to give a very satisfactory way of defining fractional derivatives……. but …..

we have yet to answer L’Hopital’s question, which was to determine the  half‐derivative of

( )f x x= Here we cannot be using this Liouvlle’s postulate

though we have found via Euler formulation of repeated differentiation. [ ]12 2 xD x π=

Most fundamental approach

To get a clearer idea of the ambiguity in the concept of a generalized derivative it’s useful toTo get a clearer idea of the ambiguity in the concept of a generalized derivative, it’s useful to examine a few other approaches, and compare them with the exponential approach of Liouville.

The most fundamental approach may be to begin with the basic definition of the whole deriva‐‐tive of a function   

( )d ( ) ( ) ( )l imf x f x f x− − ε( )

0l im

d x ε ↓=

ε

Repeating  n ‐times of this operation leads to a binomial series of following type

0 0

d ( ) 1l i m ( 1 ) ( )d

n nj

n nj

nf x f x jjx ε ↓

=

⎛ ⎞= − − ε⎜ ⎟ε ⎝ ⎠

∑

Illustration

0 0

d ( ) 1lim ( 1) ( )d

n nj

n nj

nf x f x jjx ε ↓

=

⎛ ⎞= − − ε⎜ ⎟ε ⎝ ⎠

∑

To illustrate, this formula gives the second derivative ( n = 2 ) of the function 

4( )f x x=

2 4 24d 1x ∑

( )

42 20 0

4 4 4

20

d 1lim ( 1) ( )d

2( ) ( 2 )lim

j

j

x x jx

x x x

ε ↓=

ε↓

= − − εε

− − ε + − ε=

ε

∑

( )0

2 2 2

0lim 12 24 14 12x x x

ε↓

ε↓

ε= − ε + ε =

⎛ ⎞

Generalization

We can generalize  above formula for non‐integer orders ( n ), but to do this we must not only generalize the binomial coefficients !n n⎛ ⎞

0 0

d ( ) 1lim ( 1) ( )d

n nj

n nj

nf x f x jjx ε ↓

=

⎛ ⎞= − − ε⎜ ⎟ε ⎝ ⎠

∑

generalize the binomial coefficients,

we also need to determine the appropriate generalization of the upper summation limit, which we wrote as  n. 

!! ( ) !

nj

n nCj j n j

⎛ ⎞= =⎜ ⎟ −⎝ ⎠

Consider an arbitrary smooth function          . 

In addition to the point at x we’ve also take other equally‐spaced values on the interval from

( )f x

In addition to the point at  x , we ve also take other equally‐spaced values on the interval from 0 to x , each a distance      from its neighbors. 

The number k of these points is related to the values of   x and     by                                        

ε

ε k kx = ε , = 0 , 1 , 2 , 3 ...

For convenience, we define a (backward) shift operator        such that                                               Eε

d e fE ( ) ( )f x f xε = − ε

Differentiation with shift operator

( )f x

xx − ε2x − ε3x − ε4x − ε5x − ε

With this back ward shift operator we get differentiation as 

[ ] ( ) ( )0 0

d ( ) ( ) E ( )( ) ( )( ) lim limd

1 Elim ( )

f x f x f xf x f xD f xx

f x

ε

ε ↓ ε ↓

ε

−− − ε= = =

ε ε−⎛ ⎞= ⎜ ⎟0

lim ( )f xε ↓

= ⎜ ⎟ε⎝ ⎠

Differentiation & Integration via shift operator

With the help of the series expansion as  1 2 3(1 ) 1 ....x x x x−− ≅ + + + +and the shift operator as             indicating  N backward shift till              i.e.E N

ε ( 0 )f

th t t i t f th f ti E ( ) 0 1 2 3N f N Nthe start point of the functions: E ( ) ; 0,1, 2,3,....N f x N Nε = − ε =

We obtain the following

1⎛ ⎞[ ]

[ ]

11

0 0

11

0

1 E ( ) ( )( ) lim ( ) lim

1 E( ) lim ( )

f x f xD f x f x

D f x f x

ε

ε

ε

↓ ε →

−

− ε

↓

⎛ ⎞− − − ε= =⎜ ⎟ε ε⎝ ⎠

⎛ ⎞−= ⎜ ⎟ε⎝ ⎠

[ ]

( ) ( )

0

2

0 0lim 1 E E .. ( ) lim ( ) ( ) ( 2 ) ... (0 )f x f x f x f x f

ε ↓

ε εε ↓ ε ↓

ε⎝ ⎠

= ε + + + = ε + − ε + − ε +

As in limit goes to zero the operator is a simply a differentiation operator and isε 1D 1D −As in limit        goes to zero the operator        is a simply a differentiation operator and              is  a simple integration operator. We can do   n ‐ times the above and write the following: 

D D

[ ]0

1 E( ) l i m ( )n

nD f x f xε

↓

⎛ ⎞−= ⎜ ⎟

⎝ ⎠[ ]

0ε ↓⎜ ⎟ε⎝ ⎠

Contd..

1 En

⎛ ⎞Contd..

This reproduces the ordinary whole multiple derivatives. For example, the second derivative of( )f

[ ]0

1 E( ) l i m ( )nD f x f xε

ε

↓

⎛ ⎞−= ⎜ ⎟ε⎝ ⎠

is( )f x

[ ]2

22

1 E ( ) 2 ( ) ( 2 )( ) ( ) f x f x f xD f x f xε⎛ ⎞− − − ε + − ε= =⎜ ⎟ε ε⎝ ⎠

in the limit as        goes to zero, which illustrated how we recover the binomial equation ε

d ( ) 1n n nf x ⎛ ⎞∑

for any whole number of differentiations. 

0 0

d ( ) 1l i m ( 1 ) ( )d

jn n

j

nf x f x jjx ε ↓

=

⎛ ⎞= − − ε⎜ ⎟ε ⎝ ⎠

∑

However, strictly speaking, this context makes it clear that we should actually write the second derivative as

[ ]2

2 1 E (1) ( ) (2) ( ) (1) ( 2 ) (0) ( 3 ) ...(0) (0)( ) ( ) f x f x f x f x fD f fε⎛ ⎞− − − ε + − ε − − ε +⎜ ⎟[ ]2

2

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) f f f f fD f x f xε= =⎜ ⎟ε ε⎝ ⎠Contd..

That we have padded all the other values of  f ( x ) to zero

0 0

2

d 1( ) lim ( 1) ( )d

1 E (1) ( ) (2) ( ) (1) ( 2 ) (0) ( 3 ) (0) (0)

n nj

n nj

nf x f x j

jx

f f f f f

ε↓=

⎛ ⎞= − − ε⎜ ⎟ε ⎝ ⎠

⎛ ⎞

∑Contd..

It just so happens that, if   n is a positive integer, all the binomial coefficients after the first         terms  are identically zero 1n +

[ ]22

1 E (1) ( ) (2) ( ) (1) ( 2 ) (0) ( 3 ) ...(0) (0)( ) ( ) f x f x f x f x fD f x f xε⎛ ⎞− − −ε + − ε − − ε += =⎜ ⎟ε ε⎝ ⎠

0 ;nj

nC j n

j⎛ ⎞

= = >⎜ ⎟⎝ ⎠

so we can truncate the series.But for any ‘negative ‘or ‘fractional positive’ values of  n , the binomial coefficients are non‐‐terminating, so we must include the entire summation over the specified range. 

Consequently, the upper summation limit in binomial formula should actually be  0( ) /x x− εwhere  0x is the lower bound on the range of evaluation

In any case, we can re‐write formula in the more correct form that does not rely on  n being a positive integer as

0

d ( ) 1x x

n nf x−ε

⎢ ⎥⎣ ⎦ ⎛ ⎞

0 0

d ( ) 1l i m ( 1 ) ( )d

jn n

j

nf x f x jjx

⎣ ⎦

ε ↓=

⎛ ⎞= − − ε⎜ ⎟ε ⎝ ⎠

∑Contd..

Generalizing the binomial coefficients to real numbers

These coefficients are defined as !

!( )!n

j

n nCj j n j

⎛ ⎞= =⎜ ⎟ −⎝ ⎠

so we need a way of evaluating the factorial function for non‐integer arguments

α ∈ ( ) ( 1) !α αΓ = −We know

! ( 1)⎛ ⎞ Γ

Now that we have a general way of expressing “factorials” for non integers we can re write

! ( 1)!( )! ( !) ( 1)jC

j j j j jα α α α

α α⎛ ⎞ Γ +

= = =⎜ ⎟ − Γ − +⎝ ⎠So we get

Now that we have a general way of expressing  factorials  for non‐integers, we can re‐write formula in generalized form, replacing each appearance of the integer  n with the real number    This gives

υ

0

d ( ) 1 ( 1)x x

jf xυ υ−ε

⎢ ⎥⎣ ⎦ Γ +∑0 0

d ( ) 1 ( 1)lim ( 1) ( )d ! ( 1 )

j

j

f x f x jx j jυ υ

υυε↓

=

Γ += − − ε

ε Γ + −∑

The Generalized Derivative Formula 

Thus ranges of integration/differentiation we have tacitly assumed for these two definitions are different. To get agreement between the interpolated binomial expansion method and the definition based on exponential functions  (Liouvelli’s approach) we must return to equation, 

⎢ ⎥

and replace the condition  with the condition               with the condition  . 

0

0 0

d ( ) 1 ( 1)lim ( 1) ( )d ! ( 1 )

x x

j

j

f x f x jx j j

υ

υ υ

υυ

−⎢ ⎥⎢ ⎥ε⎣ ⎦

ε ↓=

Γ += − − ε

ε Γ + −∑

0 0x = 0x = −∞

This is easy to do, because it simply amounts to setting the upper summation limit to infinity, i.e., we take the following formula for our generalized derivative

d ( ) 1 ( 1)lim ( 1) ( )jf x f x jυ υ∞ Γ +

= − − ε∑0 0lim ( 1) ( )

d ! ( 1 )jf x j

x j jυ υε υε ↓=

= εΓ + −∑

With this, we do indeed find that the       ‐th derivative of          is simply   , consistent i h h l i l h f i ill

υ a xe ( ) axa eυ

with the purely exponential approach of Liouville.

Kindergarten of Fractional Calculus

0

d ( ) 1 ( 1)x x

fυ−ε

⎢ ⎥⎣ ⎦ Γ +

Non‐locality of Generalized Derivative thus Fractional Derivative is has memory

If           is an integer  n , the vanishing of the binomial coefficients for all  j greater than  nimplies that we don’t really need to carry the summation beyond j = n and in the limit as

υε

0 0

d ( ) 1 ( 1)lim ( 1) ( )d ! ( 1 )

j

j

f x f x jx j j

υ

υ υ

υυ

ε⎣ ⎦

ε ↓=

Γ += − − ε

ε Γ + −∑

implies that we don t really need to carry the summation beyond  j n , and in the limit as         goes to zero the   n values of                        with non‐zero coefficients all converge on  x so the derivative is local.

H i l h bi i l i h i fi i l ffi i h

ε( )f x j− ε

However, in general, the binomial expansion has infinitely many non‐zero coefficients, so the result depends on the values of    x all the way down to       . We typically choose                  so we are effectively evaluating the “derivative” (which is not the same as the “slope”) for the entire interval from       to       .  

0x 0 0x =

0 x0

It just so happens that this non‐locality disappears for positive whole derivatives, when n is whole number like 1, 2, 3…. . But for                    the property in non‐local, thus requiring all the points of the function to construct the generalized derivative of the function

n υ= ∈points of the function to construct the generalized derivative of the function. 

We may also state that fractional derivative is a non‐local quantity, whereas the normal integer order derivative is local or point property.  The

i h if id h d fi i i f hif i b k d difffractional derivative thus depends on history or

in the sense if we consider the definition of shift operation as backward difference .memory

The fractional differ‐integration

GL Formula                                                                                                                   [ ]/1 ( 1)( ) lim ( 1) ( ) 0

xjD f x f x j xυ υε⎢ ⎥⎣ ⎦ Γ +

ε∑

Choosing                       we get Liouville’s postulate or exponential approach 0x = −∞

[ ]0 00 0

( )( ) lim ( 1) ( ) 0! ( 1 )

jx

jD f x f x j x

j jυ υε ↓=

= − − ε =ε Γ + −∑

g g p p pp

Liouville’s [ ] 00

1 ( 1)( ) lim ( 1) ( )! ( 1 )

jxD f x f x j x

j jυ

υ

υυ

∞

−∞ ε↓

Γ += − − ε = −∞

ε Γ +∑[ ]0 0 ! ( 1 )j j jυ υε↓

=ε Γ + −∑

Kindergarten of Fractional Calculus

[ ] ( )( )( )11 22

1 1

10 ! 1

1l im ( 1) ( )x

jj j

D x x jε⎢ ⎥⎣ ⎦

Γ +

Γ↓= − − ε∑

With GL method can we answer L’Hopital’s question ?

[ ] ( )( )1 12 2

0 ! 10 0( ) ( )

j jj

x

j

y

Γ + −ε ↓=

ε

ε

=⎢ ⎥⎣ ⎦

∑Let

1 1⎛ ⎞⎛ ⎞ ⎛ ⎞

then [ ] ( )( )( )31 22

32

10 !0

lim ( 1) ( )y jj jj

D x x jΓ

ε ↓ Γ −ε == − − ε∑ expanding this we have

Evaluating this expression (numerically) in the limit as        goes to zero, we find that the half‐ε

[ ]12

1 12 251 1 1

2 8 16 1280

1lim 1( ) ( ) ( 2 ) ( 3 ) ( 4 )........ (2 ) ( )2 1

D x x x x x xy yε→

⎛ ⎞⎛ ⎞ ⎛ ⎞= − −ε − − ε − − ε − − ε − ε − ε⎜ ⎟⎜ ⎟ ⎜ ⎟− −ε ⎝ ⎠ ⎝ ⎠⎝ ⎠

g p ( y) g ,derivative of   x is (almost)

[ ]12 2 xD x π∼

We can thus write the above as                               indicating that differentiation is  starting at start1

20 ( ) 2 x

xD x π=0point 0x =

In a way we answered L’Hospital’s query  by most fundamental approach

Oscillator‐ a mass spring damper system

A mass‐spring oscillator under viscoelastic damping

2 10( ) ( ) ( )t tm D x t c D x t k x t F+ + =Classical frictional damping

20( ) ( ) ( )q

t tm D x t c D x t k x t F+ + =Fractional frictional damping

Cornell Univ. Library Archive: http://arxiv.org/abs/1508.06202, 2015

( )( )2 ( ) ( ) ( )q x tm D x t c D x t k x t F+ + =

Variable fractional order for friction

2( )1 x tL

⎧ ⎛ ⎞+⎪ ⎜ ⎟⎝ ⎠⎪

0( ) ( ) ( )t tm D x t c D x t k x t F+ + =

The viscoelastic order is variable within the travel length as

( ) 2

,2( ) ( )( )1

L

q x t q tx tL

⎜ ⎟⎝ ⎠⎪

⎪= = ⎨⎛ ⎞⎪ −⎜ ⎟⎪ ⎝ ⎠⎪

if viscous ‐ viscoelastic oscillator. 

if viscoelastic ‐ viscousoscillator,2

⎪⎩

if viscoelastic ‐ viscous oscillator.

The fractional order is continuously variable

A mass‐spring oscillator sliding on a continuously variable order guide

Cornell Univ. Library Archive: http://arxiv.org/abs/1508.06202, 2015

Character of variable fractional order for friction

2 ( )( ) ( ) ( ) ( )q tt tm D x t c D x t k x t tδ+ + =

322 ( ) 2( ) 2 ( ) ( ) ( )q t

t n t nD x t D x t x t tη ω ω δ+ + =

With adjustments in constants of the above we write the following form

( ) ( ) ( ) ( )t n t nη

Cornell Univ. Library Archive: http://arxiv.org/abs/1508.06202, 2015

322 ( ) 2( ) 2 ( ) ( ) ( )q tD t D t t tδ

With viscoelastic‐viscous guide

22 ( ) 2( ) 2 ( ) ( ) ( )q tt n t nD x t D x t x t tη ω ω δ+ + =

The displacement‐time graph for fractional continuous variable order spring‐mass damper model for free oscillation with viscoelastic‐viscous damping plots

Cornell Univ. Library Archive: http://arxiv.org/abs/1508.06202, 2015

322 2( ) 2 ( ) ( ) ( )qD t D t t tδ

With viscous‐viscoelastic guide

22 2( ) 2 ( ) ( ) ( )qt n t nD x t D x t x t tη ω ω δ+ + =

The displacement‐time graph for fractional continuous order mass‐spring damper model with viscous‐viscoelastic damping plots

Cornell Univ. Library Archive: http://arxiv.org/abs/1508.06202, 2015

Expressing fractional impedance via Curie relaxation law

As in di‐electric relaxation, the relaxation of current to an impressed constant voltage stress            an step voltage applied at to an initial uncharged system the current is by following law

0U0t =voltage applied at              , to an initial uncharged system the current is by following law 0t =

0( ) 0 0 1Ui t th t α

α

α= > < <

We now get Transfer Function of a capacitor via Laplace Transform⎧ ⎫

This is Curie law

‐4

‐5

Slope = ‐ n

{ } 0( ) ( ) UI s i th t α

α

⎧ ⎫= = ⎨ ⎬

⎩ ⎭L L

Use Laplace pair1

! nn

n ts + ↔ and ! (1 )n n= Γ −

‐6

‐7

p s

To get 00 1

(1 ) (1 )( ) UI s Uh s h s sα α

α α

α α− + −

⎛ ⎞Γ − Γ − ⎛ ⎞= = ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

‐8

‐1 0 1 2

Note that the voltage excitation is a constant step input at time zero

0 0( )

0 0U t

u tt

>⎧= ⎨ ≤⎩

Laplace is 0( ) UU ss

=0 0t ≤⎩

( ) (1 ) (1 )( )( )

(1 ) ( ) 1

I sH s s C sU s h s h

U s

α ααα

α α

α α

α

−

Γ − Γ −= = = =

Γ

For unit step we have

(1 ) ( ) 1( )( )

1( )( )

U sC Z sh I s C s

ZC i

α αα α

αα

α

ωω

Γ −= = =

=

Indicates fractional derivative in current‐voltage expression

( ) 1

We have just expressed impedance in Laplace domain for capacitor as 

( ) 1( ) 0 1( )

U sZ sI s C s α

α

α= = < <

This gives current voltage expression for capacitor as

Fractional order impedance

This gives current‐voltage expression for capacitor as 

( ) 1( ) 0 1( )

U sZ sI s C s α

α

α= = < <

( ) ( )( ) ( ) ( )I s C s U s C s U sα αα α= =

As  we have for zero initial condition ( ) { }( ) ( )d ( )( ) ( )

df t

s F s s f tt

= ↔L

( ) { }( ) ( )d ( )( ) ( )

df t

s F s s f tt

αα α

α= ↔Lwe generalize to fractional order derivative and write

Our new‐capacitor expression is                                                                                                 ( )0

d ( ) 1( ) ( ) ( ) dd

tu ti t C u t it C

αα

α α τ τ= = ∫ ( )0d t C α

∫

Contrary to  classical expression( ) ( )

0

d ( ) 1( ) ( ) ( ) dd

tu ti t C u t i

t Cτ τ= = ∫

The charging time is memorizedWe charge a ultra‐capacitor to maximum limit voltage and keep for time T on float charge, then it is kept under open‐circuit. The self‐discharge plot is depending on Tthen it is kept under open circuit. The self discharge plot is depending on T

Courtesy: BRNS Funded joint Project CMET Thrissur‐BARC Development of CAG Super‐capacitors and application in electronics circuits

Self discharge with memory via fractional derivative for supercapacitor

A capacitor is charged from time –T to t with a constant voltage U0; the charging current isA capacitor is charged from time     T to    t with a constant voltage    U0;    the charging current is

0 0 0d d(1 ) (1 )( ) ( ( ))d d (1 )

t t

cT T

U U Ui t C t Tt h t h

α αα

α α αα α

α αα

−

− −

⎛ ⎞ ⎛ ⎞Γ − Γ −= = = − −⎜ ⎟ ⎜ ⎟Γ −⎝ ⎠⎝ ⎠

Using FD of constant i.e.( )

(1 )x aD C C

αα −−Γ=

0 0 1 ( ) 0( )U t T

h t T αα

α= < < + >+

At   t = 0  it is kept in open‐circuit; there will be self discharge thus a discharge current will appear depending d f h l l ( )

(1 )a xD C C αΓ −

on decaying of the terminal voltage u ( t ) 

0 0

d ( ) (1 ) d ( )( )d d

t t

du t u ti t Ct h t

α α

α α αα

αΓ −= =

0 0

We combine the charge and self discharge together and write ( ) ( ) 0c di t i t+ =

0 (1 ) d ( ) 0( ) dU u t

h t T h t

α

α α

αΓ −+ =

+( ) dh t T h tα α+

Do fractional integration of order      for above expression, to write the following

[ ]00 0

(1 ) ( ) 0( )tUD u t U

h t T hα

αα α

α− ⎡ ⎤ Γ −+ − =⎢ ⎥+⎣ ⎦

α

( )α α⎣ ⎦

We apply the formula for fractional integration i.e.                                                                     to get[ ]0 10

1 ( )d( )( ) ( )

t

t -

f x xD f tt - x

ααα

− =Γ ∫

Contd…

Contd…0

010

1 d (1 ) (1 )( ) 0( ) ( ) ( )

tU x u t Uh T x t x h hα α

α α α

α αα −

⎡ ⎤Γ − Γ −+ − =⎢ ⎥Γ + − ⎣ ⎦

∫

Rearranging the above, we write0

0 10

d( )( ) (1 ) ( ) ( )

tU xu t UT x t xα αα α −= −

Γ Γ − + −∫

Put T x τ+ = d dx τ= so for 0x = Tτ = and x t= T tτ = + We have thus

0 00 01

d( ) ( )d( ) (1 ) ( ) ( ) (1 )

T t T t

T T

U Uu t U U FT tα α

τ τ τα α τ τ α α

+ +

−= − = −Γ Γ − + − Γ Γ −∫ ∫

Now we break                         as                                                                  and call the second term as  ( )dT t

TF τ τ

+

∫0

0

( )d ( )d ( )dT t T t

T T

F F Fτ τ τ τ τ τ+ +

= +∫ ∫ ∫ ( )tI

We write in terms of convolution of two functions with substitution  T t t+ =

1 1 10 0 0

d d 1 1( ) ( )d *( ) ( )

T t T t t

t FT t t t tα α α α α α

τ ττ ττ τ τ τ

+ +

− − −⎛ ⎞ ⎛ ⎞= = = = ⎜ ⎟ ⎜ ⎟+ − − ⎝ ⎠ ⎝ ⎠∫ ∫ ∫I

Using Laplace pair                             we write                                                                       as { } { } { }(1 )( ) ( )t s t tα α− − −= = ×L I I L L1( 1)s

t ααα

+Γ +↔

( )[ ]1 (1 ) 1

(1 ) 1( 1) (1 ) ( )( )ss ssα α

αα α α− + − − +

Γ − − +Γ − + Γ − Γ= × =I

Extracting            by inverse Laplace of obtained            we get( )tI ( )sI

{ } ( ){ }1 1 1( ) ( ) (1 ) ( ) (1 ) ( )st s α α α α− −= = Γ − Γ = Γ − ΓI L I L

We used Laplace pair 11 s↔ Contd…

Contd…

Using just derived expression i.e.                                                                                                    we write theexpression for open circuit voltage             as follows

11

( )00

( )d d (1 ) ( )T t

T t

T tF α ατ τ

τ τ τ α α−

++

+ −= = Γ − Γ∫ ∫

( )u t( )

[ ]0 0

0 0 00 0

0

00 0

( ) ( )d ( )d (1 ) ( ) ( )d(1 ) ( ) (1 ) ( ) (1 ) ( )

( )d ( )d

T t

T T

T

U U Uu t U F F U F

U UF F

τ τ τ τ α α τ τα α α α α α

τ τ τ τ

+⎡ ⎤= − + = − Γ − Γ −⎢ ⎥Γ − Γ Γ − Γ Γ − Γ⎣ ⎦

= − =

∫ ∫ ∫

∫ ∫0

01

0

( )d ( )d(1 ) ( ) (1 ) ( )

d(1 ) ( ) ( )

TT

F F

UT tα α

τ τ τ τα α α α

τα α τ τ −

Γ − Γ Γ − Γ

=Γ − Γ + −

∫ ∫

∫

Therefore

01

d( )(1 ) ( ) ( )

TUu tT tα α

τα α τ τ −=

Γ − Γ + −∫0(1 ) ( ) ( )T tα α τ τΓ Γ +

is the voltage over open capacitor at self discharge. This function of time,  depends on the total time T the capacitor has been on the voltage sourcetime  T the capacitor has been on the voltage source.

In a way this capacitor is memorizing its charging history.

This explanation was possible only by usage of fractional derivative

Self Discharge Curve of Super‐capacitor

2.5

2.0

Open CircuitVoltage ( V ) T = 1000hr

1.0T = 10hr

T = 100hr

10 100 1000

i

log ( t )

timeAt this instant open circuit is done

Self Discharge voltage after open circuit is  u ( t )  is almost constant for about time  T then it  decays as  (1 )t α− −proportional to             . More time the capacitor is kept at charge the more time it takes to self‐discharge

This interesting observation we described by use of fractional derivative

proportional to . More time the capacitor is kept at charge the more time it takes to self discharge

Ending part‐Ag p

Part‐B

The modern treatment 

–’can we evolve some parallel with classical calculus ?’

Fractional derivative for non‐differentiable functions defined

A function              is said to have a fractional derivative                 of the order          with  ( )x t ( ) ( )x tα α

0 1α< < whenever the limit ( )def

( ) (0)x x t xΔ =0 1α< < whenever the limit (                                   ) ( ) (0)x x t xΔ = −

( )( )

0

( )( ) limx

x tx tt

ααΔ ↓

Δ∝

Δexists and is finiteexists and is finite

Shortly, this amounts us to write the equality in terms of differentials 

( )d dx b t α=

d l f h d f h h( )tAs a direct result of this definition, when             is a constant, one has ( )x t d ( ) 0x t =

and thus the fractional derivative of a constant function should be zero.

In a likewise manner, when            is differentiable we have                        , and as a   ( )x t d ( )dx x t t′=

result the fractional derivative is zero i.e. ( )( ) ( )( )1( ) ( )

( ) li li li ( ) 0x t txt t t αα −′ ΔΔ ′ Δ

( )( )( )

( )( )( )( )

0 0 0( ) lim lim lim ( ) 0

t t tx t x t t

t tα

α αΔ ↓ Δ ↓ Δ ↓′= = = Δ =

Δ Δ

Clearly, the definition deals with non‐differentiable functions

Corollary for fractional derivative of non‐differentiable function

Assume that                                   where            is differentiable and           is          th( ) ( ) ( )x t f t g t= + ( )f t ( )g t α −

differentiable, then one has                                                  , therefore one still  take the limit( ) ( )0 1x b t b tαΔ = Δ + Δ(1)

condition of definition                                                        . It follows that, exactly like a ( )( )( )

0( ) l im ( ) /

tx t x t t αα

Δ ↓∝ Δ Δ

derivative defines a function up to an arbitrary additive constant, a function  of which 

l h f i l d i i i k i d fi d ddi i diff i bl f ionly the fractional derivative is known, is defined up to an additive differentiable function

( ) ( )0 1

( ) ( ) ( )x t f t g t

x b t b tα

= +

Δ = Δ + Δ 1

( ) ( )0

x t f t Cx b t

= +Δ = Δ +

( )( )1 ( )

0 1 00 0 0lim lim lim ( )

t t t

x b b t b x tt

α αα

−

Δ ↓ Δ ↓ Δ ↓

Δ= + Δ = =

Δ10

l i m ( )t

x b x ttΔ ↓

Δ ′= =Δ

More generally if one has equality                                                                             then  ( ) ( )0 1 ,x b t b tα β α βΔ = Δ + Δ <the function            is         th differentiable at    ( )x t α − 0t =

( )In definition                                  one has                  ; but            may have any value  ( ) ( )x at a x tα∝ (0) 0x = (0)x(2)

Mittag‐Leffler Function

The Mittag‐Leffler function               ,                is defined by series   ( )E yα y tα=k∞

( ) ( )0 !

k

k

yE ykα α

∞

=

= ∑Exactly like exponential function          is solution of differential equationte

( )( ) ( )D y t y t= (0) 1y =

Dαwe would like to have a fractional derivative            which should have                      as  ( )E tαα

l ( f ) h f l d ff lsolution (eigenfunction) to the fractional differential equation

( ) ( )D E t E tα α αα α=

This way we would be in a position to expand the fractional calculus in conjugation

or parallel to classical calculus, by substituting                        for            almost   ( )E tαα λ teλ

everywhere

We note                                      ( ) ( )1, expE t tααα λ λ= =

Plot of decaying Mittag‐Leffler Function

Plot of in linear scale of Mittag‐Leffler Function

( ) 0 1 0α

ty e−=

( ) , 0 1, 0y E t tαα α= − < ≤ >

Not‐differentiable at              0t =

Mittag‐Leffler function as a solution of fractional differential equationcomposed via modified RL fractional derivative

On substituting                                            into ( ) ( )!0

ktkk

E t ααα α

∞

== ∑ ( ) ( )D E t E tα α α

α α=

( ) ( )D E t E tα α α

If h fi

( ) ( )D E t E tα α αα α=

( ) ( )2 3 2 3

! (2 )! (3 )! ! (2 )! (3 )!1 ... 1 ...t t t t t tD α α α α α ααα α α α α α+ + + + = + + + +

( ) ( ) ( )2α α αIf then fine

Therefore we arrive at following ‘so called’ definition for a suitable fractional derivative

( ) ( ) ( )2

! (2 )! !1 0 1 ...t t tD D Dα α αα α αα α α= = =

( ) ( )( )

!0

!k kk

D C D x xk

α α α α ααα α

−= =−

C : Constant

Obviously this fractional derivative provides customary derivative when  1α =

This modification looks as conjugation to classical derivative

( )( )

1, 0

( 1) 1k kk

D x x D Ck

α α α α ααα

−Γ += =

Γ − +In Gamma function form we have

( ) ( ) ( )cosh sinhE +

Some aspects of Mittag‐Leffler function

( ) ( ) ( )

( ) ( ) ( ) ( )def def

1 12 2

cosh sinh

cosh ( ) ( ) sinh ( ) ( )

E x x x

x E x E x x E x E x

α α α

α α α α α α

= +

= + − = − −

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( )def def

1 12 2

cos sin

cos ( ) ( ) sin ( ) ( )i

E ix x i x

x E ix E ix x E ix E ix

α α α

α α α α α α

= +

= + − = − −

( ) ( ) ( )( )E E Eα α αλ λ λ

( ) ( )D E x E xα α αλ λ λ=

( ) ( ) ( )( )E x y E x E yα α αα α αλ λ λ+ =

( ) ( )cos sin sin cos

cosh sinh sinh cosh

D E x E x

D x x D x x

D x x D x x

α α

α α α α α αα α α α

α α α α α αα α α α

λ λ λ=

= − =

= − =

We find interesting parallel to usual exponential , hyperbolic and trigonometric functions

Only with modified R‐L derivative‐as we described as with                      & ( ) 0D Cα =( )

( )!

!kk k

kD x xαα α α αα α

−−=

Fractional Derivative via Fractional Difference

FWE ( )hLet                                          , is a continuous: , ( )f t f t→ →

Let,              represent a constant discretization element. We call a forward operator 0h >def

(but not necessarily differentiable) function

[ ]def

FWE ( ) ( ) ( )h f t f t h= +

One can write the finite difference                 of             in the form    ( )f tΔ ( )f t

( )FW( ) ( ) ( ) E 1 ( )f t f t h f t f tΔ = + − = −

and quite in natural way, this suggests to define the fractional difference of order  α0 1α< < of                 by expression( )f t

( ) ( )def

FW0

( ) E 1 ( ) ( 1) ( )k

kf t f t f t k h

kαα α

α∞

=

⎛ ⎞Δ = − = − + −⎜ ⎟

⎝ ⎠∑

( )E ( ) ( )f t f t hμ +with notation ( )FWE ( ) ( )f t f t hμ μ= +with notation

With these above notations and definition in a quite natural way, we are led to define fractional derivative of order as limitαfractional derivative of order       as limit α

( )

0

( )( ) limh

f tf th

αα

α↓

Δ=

Contd…

According to our definition ,                                        and via previous argument of  ( )( )

0

( )( ) limx

f tf tt

ααΔ ↓

Δ∝

Δ( ) ( )y t E tα

α=

Contd…

( )tΔ

be the solution of                              we have seen fractional derivative of a constant should be ( ) ( )D y t y tα =

zero. 

According to                                                  the Laplace transform of  ( )

0

( )( ) limh

f tf th

αα

α↓

Δ= { }( ) ( ) ( )f t s F sα α=L

But we have to examine that if                                           definition gives zero for constant( )

0

( )( ) limh

f tf th

αα

α↓

Δ=

( )f t C=

0h h↓

Accordingly, the                        gives    { }

{ }

( ) ( ) ( )f t s F s

C

α α=

⎛ ⎞

L

{ } { }

{ }

1

1 1 0( )! ( )!

CD C s C s s Cs

t ts C C D C C

α α α α

α αα α

α α

−

− −− −

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= = ≠

L L

L { } ( )! ( )!α α− −

Fractional derivative of constant not zero.

But how we wrote                                                                                    ? { }( )

0

( )( ) lim ( )h

f tf t s F sh

αα α

α↓

⎧ ⎫Δ= =⎨ ⎬

⎩ ⎭L L

Contd…

{ }( ) ( )dstf t h f t h t t h∞ −+ + +∫L

Contd…

{ }0

( )

( ) ( )d

( )d

st

s h

hhh h h

f t h e f t h t t h

e fτ

τ

τ τ∞ − −

∞ ∞

+ = + + =

=

∫∫

∫ ∫ ∫

L

0 0

0

( )d ( )d ( )d

( ) ( )d

sh s sh s sh s

hhsh sh s

e e f e e f e e f

e F s e e f

τ τ τ

τ

τ τ τ τ τ τ

τ τ

− − −

−

= = −

= −

∫ ∫ ∫∫

For small           we have  h ( ) ( )( 0 )

0( )d (0 ) (0 )

h st se f t t e f h h f− −≅ =∫

{ }( ) ( ) (0)sh shf t h e F s he f+ ≅LTherefore

With above derived Laplace identity we apply to our fractional difference expression i.e. 

{ }( ) ( ) (0)f t h e F s he f+ ≅ −LTherefore

α∞ ⎛ ⎞ ( )0

( ) ( 1) ( )k

kf t f t k h

kα α

α=

⎛ ⎞Δ = − + −⎜ ⎟

⎝ ⎠∑

and try to write expression for  { }( )f tαΔL { }( )f

{ } ( )( ) ( 1) ( )kf t f t k hα α∞⎧ ⎫⎛ ⎞Δ ⎨ ⎬⎜ ⎟∑L L

Proof of Laplace of Fractional Difference

{ } ( )

( ) ( ) ( ){ }( ){ } ( ){ } ( ){ }

0

( 1)2

( 1)

( ) ( 1) ( )

( 1) ( 2) ....

k

kf t f t k h

k

f t kh f t h f t h

α

α α

α

α α α α

=

−

Δ = − + −⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

= + − + − + + − +

∑L L

L

( ){ } ( ){ } ( ){ }( ) ( )

( 1)2

( 1) ( 1)

( 1) ( 2)

) ( 1) ( 2) ....

( ) (0) ( ) ( 1) (0)

(

hs hs hs hs

hs

f t kh f t h f t h

e F s he f e F s he f

e F s

α α

α α α α

α α α

α α α α

α α α α

−

− −

− −

= + − + − + + − +

= − − − −

+

L L L

( )( 1)( 2) ( 2)) (0)h hse fα α α α− − − +2 (e F s+ ( )( )

( )

2

( 1)( 1) ( 2)2

( 1)( 2 )( 1) ( 2 )2

) (0) ....

( ) ( ) ( ) ...

(0) ( 1) (0) (0) ...

hs hs hs

hhs hs hs

e f

e F s e F s e F s

he f he f e f

α αα α α

α α αα α α

α

α α α

−− −

− −− −

− +

= − + +

− − − + +( )2( ) ( ) ( ) ( )f f f

( ) ( )( 1) ( 1)( 1) ( 2) 2h h h h h h

Take the first term in bracket and we do following manipulation and approximation for small  h

( ) ( )( ) ( )( )

( 1) ( 1)( 1) ( 2) 22 2( ) ( ) ( ) ... ( ) 1 ...

( ) 1 1 ( )

hs hs hs hs hs hs

hs hs hs hs

e F s e F s e F s e F s e e

e F s e e e F s

α α α αα α α α

ααα

α α− −− − − −

− −

− + + = − + −

= − = −

( )( )

1 ( ) 1

1 1 ( ) ( )

hs hse F s e hs

hs F s h s F s

α

α α α

= − ≅ +

= + − = Contd…

Contd…

{ } ( )( 1)( 2 )( 1) ( 2)2( ) ( ) (0) ( 1) (0) (0) ...hhs hs hsf t h s F s he f he f e fα α αα α α α α αα α α − −− −Δ = − − − + +L

Therefore

{ } ( )

Divide by  hα

( )fα⎧ ⎫ ( )( 1)( 2)1 22

( ) ( ) (0) 1 ( 1) ...hs hs hsf t s F s h e f e eh

αα αα α α

α α α − −− − −⎧ ⎫Δ= − − − + +⎨ ⎬

⎩ ⎭L

k h l d h f ll0h ↓Take the limit                and we get the following0h ↓

{ }( )

0

( )( ) lim ( )h

f tf t s F sh

αα α

α↓

⎧ ⎫Δ= =⎨ ⎬

⎩ ⎭L L

0h h↓⎩ ⎭

With this fractional difference we saw that                      as   0D Cα ≠ { } 1 0D C s Cα α −= ≠L

In other words, with this definition the fractional derivative of a constant would not be zeroIn other words, with this definition the fractional derivative of a constant would not be zero what contradicts our assumption. This feature is due to presence of           in the coefficients1h α−

of (0)f

W k th t l i l L l f l f d i ti i { }( ) ( ) (0)f t F f′L

Observation with classical Laplace identity and remarks  

We remark that classical Laplace formula for derivative is { }( ) ( ) (0)f t sF s f′ = −L

and as is evident it is not equivalent to  { } ( ){ }( )( )

0( ) lim ( )f t

hhf t s F s

α

αα αΔ

↓= =L L{ } ( ){ }0 hh↓

when               .1α =

This gives rise to a query as to whether there is some mistake? 

First of the above two the consistency is complete when                  , which is condition  (0) 0f =satisfied by self‐similar functions. Therefore the definition  ( )( )

0( ) lim f t

hhf t

α

αα Δ

↓=

f f lf l f0h↓

is quite satisfactory for self‐similar functions

According to our earlier discussion, when we restrict ourselves to the class of functions

which satisfy the condition                   then we can claim that (1) if function is fractional  (0) 0f =

differentiable, then it is self similar and (2) conversely , if it is self similar then it is

fractional differentiablefractional differentiable

To cope up with the problem due to constant term                  we shall assert that  (0)f C= 0D Cα =

Fractional Derivative for a function with non‐zero initial condition

Let denote a continuous function: ( )f t f t→ → (but not necessarily differentiable)Let                                         , denote a continuous                                                             function: , ( )f t f t→ →

and define                                       . Let               denote a constant discretization element (span) def

( ) ( ) (0)f t f t f= − 0h >

The fractional derivative of order of is defined as the limitα ( )f t

(but not necessarily differentiable)

The fractional derivative of order          of               is defined as the limitα ( )f tdef

( )

0

( )( ) limh

f tf th

αα

α↓

Δ=

Alternatively, assume               is self similar function with Hurst index       ,  ( )g t H 0 1H< <

Then its fractional derivative of order           is defined by the expression H

( )

0

( )( ) limH

HHh

g tg th↓

Δ=

According to this definition the fractional derivative of constant is zero as  ( )f t C=

as we are having here                      for  ( ) 0f t = ( )f t C=

This is similar to Caputo derivative but not same they are same when is differentiable( )f tThis  is similar to Caputo derivative but not same‐they are same when            is differentiable( )f t

We call it modified R‐L fractional derivative‐Jumarie type

Laplace of modified R‐L derivative and its conjugation with ordinary classical derivative

def( )

0

( )( ) limh

f tf th

αα

α↓

Δ=

{ } ( )( )

0

( ) (0)( ) lim

( ) (0)

h

f t ff t

h

f f

αα

α

α α

↓

⎧ ⎫Δ −⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭⎧ ⎫Δ Δ

L L

{ }0 0

( ) (0)lim lim

( ) (0)

h h

f t fh h

s F s s f

α α

α α

α α

↓ ↓

⎧ ⎫Δ Δ= −⎨ ⎬

⎩ ⎭= −

L

L

1

(0)( )

( ) (0)

fs F s ss

s F s s f

α α

α α −

⎛ ⎞= − ⎜ ⎟⎝ ⎠

= −( ) ( )f

1α =For                                                                        here classical result is recovered.{ }( ) ( ) (0)f t sF s f′ = −L

We have found similarity with classical derivative in Laplace transform identity.

Modification in Riemann‐Liouvelli Derivative (via Integral )‐Jumarie

Assume that                 is a constant function       , then fractional derivative of order   ( )f t C α

0(1 )t

CD C tα α αα

−= ≤Γ −is

(1)

0 0α= >Note that first one is fractional integration of constant and second one is fractional derivative

Wh i h h ( )( ) (0) ( ) (0)f f f f(2) When               is not a constant then one can have  ( )f t ( )( ) (0) ( ) (0)f t f f t f= + −

and derivative is defined by expression                                                                        in which ( )( ) ( ) (0) ( ) (0)t tf t D f D f t fα α α= + −

(2)

negative  0α ≤ ( )11

0( ) ( ) ( )d

tf t t fα α

α τ τ τ− −Γ −= −∫

For positive  0α > ( ) ( )( ) ( 1)( ) ( ) (0) , 0 1tf t D f t f fα α α α− ′= − = < <( ) ( )( )

0

( ) ( ) ( ) ,

1 d ( ) ( ) (0) d(1 ) d

t

t

f f f f

t f ft

ατ τ τα

−= − −Γ − ∫

When we generalize that to write1≤When                          we generalize that to write  1n nα≤ < +

( )( )( ) ( )( ) ( ) , ; 1nnf t f t n n nα α α−= ≤ < ≥

Application example and proof of eigenfunction of fractional differential equation with modified RL derivative

Take For deri ati e e take offset f nction2 3( ) ( ) ( )( ) ( ) 1 x x xf E

α α αα λ λ λλTake                                                                                            . For derivative we take offset function ( ) ( ) ( )( ) ( ) 1 ...

! (2 )! (3 )!f x E xα

α λα α α

= = + + + +

2 3( ) ( ) ( )( ) (0) ( ) 1 ...( !) (2 )! (3 )!

x x xf x f E xα α α

αα

λ λ λλα α α

− = − = + + +we get

( ) ( )( ) (0 ) ( ) 1D f x f D E xα α αλ=( ) ( )

( ) ( ) ( ) ( )2 3

2 3

2 32

( ) (0 ) ( ) 1

( ) .. .( !) ( 2 ) ! (3 ) !

( !) ( 2 ) ! (3 ) ! ...( !) 0 ! ( 2 ) ! ( !) (3 ) ! ( 2 ) !

D f x f D E x

D E x D x D x D x

x x

α

α α α α α α α αα

α α

λ

λ λ λλα α α

λ α λ α λ αα α α α α

− = −

= + + +

⎛ ⎞ ⎛ ⎞⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2

( !) 0 ! ( 2 ) ! ( !) (3 ) ! ( 2 ) !

1 ... ( )( !) ( 2 ) !

x x E xα α

αα

α α α α α

λ λλ λ λα α

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞

= + + + =⎜ ⎟⎝ ⎠

So we verified that with this modified RL definition                      is solution of                      ( )y E xαα λ= D y yα λ=

For integration we are not making offset function 

( ) ( ) ( ) ( ) ( ) ( )2 3

2 3

2 3

( ) ( ) 1 ...( !) (2 )! (3 )!

1 ( !) (2 )! (3 )!

D f x D E x D D x D x D xα α α α α α α α α αα

λ λ λλα α α

λ α λ α λ α

− − − − − −= = + + + +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞2 32 3 4

2 3 42 3

1 ( !) (2 )! (3 )! ...( )! ( !) (2 )! (2 )! (3 )! (3 )! (4 )!

1( )! (2 )! (3 )! (4

x x x x

x x x

α α α α

α α α

λ α λ α λ αα α α α α α α

λ λ λ λλ α α α α

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

4

2 3 4

...)!

1

x α

λ λ λ λ

⎛ ⎞+⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞

( )

2 3 42 3 4

1

1 1 ... 1( )! (2 )! (3 )! (4 )!

( ) 1

x x x x

E x

α α α α

αα

λ λ λ λλ α α α α

λ λ−

⎛ ⎞⎛ ⎞⎛ ⎞= + + + + + −⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= −

Exactly  similar to exponential function

Application example and generation of two parameter Mittag‐Leffler (ML) function with modified RL derivative

We do fractional deri ati e of order on one parameter ML f nction i e0β > ( ) ( )y x E xα

( )2 3

2 3

( ) ( ) 1 ..(1 ) (1 2 ) (1 3 )

x x xD y x D E x Dα α α

β β α βα

α β α β α β

α α α− − −

⎛ ⎞= = + + + +⎜ ⎟Γ + Γ + Γ +⎝ ⎠

We do fractional derivative of order                on one parameter ML function i.e. 0β > ( ) ( )y x E xα=

2 3

, 1

0 ...(1 ) (1 2 ) (1 3 )

( )

x x x

x E x

α β α β α β

α β αα α β

α β α β α β−

− +

= + + + +Γ + − Γ + − Γ + −

=

( )kxE

∞

∑Where we have two‐parameter ML function as                                                             ,0

( )( )p q

k

xE xq kp=

=Γ +∑

1( ) ( )xe x E tλ α αλ−=Let

For classical RL fractional derivative where                        the eigensolution for                         is as follows 0RLD Cα ≠ 0RL

xD y yα λ=

( ) ( )

,

( 1) 11

0 0

( ) ( )

( )( 1) ( 1)

k kk k

k k

e x E t

x xxk k

α α α

α αα

λ

λ λα α

+ −∞ ∞−

= =

= =Γ + Γ +∑ ∑

( )( 1) 1

1( )( )

0 0 ( 1)0 1

( )( )

kk

xRL x RL k kx x k

k k

x

xD e Dk

e

αα

α λ αα α

λα

λ λα

λ

+ −−∞ ∞

Γ += =

⎛ ⎞⎡ ⎤ = =⎜ ⎟⎣ ⎦ Γ⎝ ⎠=

∑ ∑

( )1 RL α λ

Note the difference

Thus we have                                        , solution of FDE with classical RL derivative                           ( )1,y x E xα α

α α λ−= 0RL

xD y yα λ=

or eigen function for                               is                                   0RL

xD y yα λ= ( )1,y x E xα α

α α λ−=Kindergarten of Fractional Calculus

One more example with modified fractional RL derivative of Mittag‐Leffler function

What is                         where             is modified RL fractional derivative ( )D E xαα λ Dα

W k th i f ti l ti f th t i( )E αλ ( )D Eα αλWe know the eigen function                    solution of                           that is  ( )E xαα λ ( )D E xα α

α λ

( )( ) ( )dd

E x E xx

αα α

α αα λ λ λ=

Let  y x α= then 1d dy x xαα −= ( ) ( ) ( )1d dy x xαα ααα −=and

With this substitution we write

( )( ) ( )dd E yE y

ααα λ

λ( )( ) ( )( ) ( )

( ) ( )( )( )( )

( )1

( 1)

1

d d

d1d

E yy x x

E x

xy α

α αα αα

α αα

α αα α

λα

λ

α

−

−

=

⎛ ⎞⎜ ⎟=⎜ ⎟⎝ ⎠( ) ( ) ( )

( )( ) ( )1 1

y

y E x y E yα α α α αα α

α

α λ λ λα λ− − − −

⎝ ⎠

= =

Therefore we have interesting relation ( )1( )D E x x E xα α αλ λα λ− −=Therefore we have interesting relation                                                             ( )( )D E x x E xα αλ λα λ=

Relation between Fractional Derivative and Derivative

In classical calculus the expression (equality)p ( q y)d ( ) , (0) 0d

x g t xt

= =

is equivalent tot

∫0( ) ( )d

tx t g τ τ= ∫

The RHS is considered as anti‐derivative, in a the sense that it is defined by the equality

( )d ( )d ( ) d ( )d ( )dt t

∫ ∫( )0 0

d ( )d ( ) d ( )d ( )dd

g g t g g t tt

τ τ τ τ= =∫ ∫On applying the same point of view to fractional derivative we have fractional differential

equality ( )d ( ) d (0) 0 0 1y g t t yα α< <equality ( )d ( ) d , (0) 0 , 0 1y g t t y α= = < <

is equivalent to fractional integral ( )0

( ) ( ) dt

y t g ατ τ= ∫

and here the RHS is  fractional anti‐derivative of order ‐ defined by the equality α

( )( ) ( ) ( )dd ( ) d ( ) d ( ) d ( )t tg g t t g g tα α ατ τ τ τ∫ ∫( )( ) ( ) ( )

0 0d ( ) d ( ) d ( ) d ( )

dg g t t g g t

t ατ τ τ τ= =∫ ∫

Completeness of approach

We started while defining fractional derivative via quotient then in further( )/x t αΔ ΔWe started while defining fractional derivative via quotient                       , then in further ( )/x tΔ Δ( )( )0

lim ( ) /t

x t t αα

Δ ↓Δ Δdefinition we defined                                         via fractional difference of zero‐corrected function. 

What is the relation between these two ; will make completeness of the approach.

With the above definitions of fractional derivative it is easy to show the solution of FDE( )

0( ) ( ) , (0)y t y t y yα λ= = is ( )0( )y t y E tαα λ=

(1)

Also we have a formal operational equality  FW FWE ( ) E ( )h tD h D hα α=

( )FWE ( ) ( ) ( )h x t x t h= +where                                                . Now look at the Laplace

(2)

{ } ( ) { } ( )( ) ( ) (0) ( ) ( ) (0)st st hs hsh tD f t h s e F s te f D f t h s e F s he fα α α α+ ≅ − + ≅ −L L

Now we see for                and             both are equal i.e.   0h ↓ 0t ↓ { } { }( ) ( )t hD f t h D f t hα α+ = +L L

Therefore with the operational identity                                            considering as FDE and takingFW FWE ( ) E ( )h tD h D hα α=

tDα as constant we write the solution as  ( )FWE ( ) th E h Dα αα=

Fractional Taylor series

( )FWE ( ) th E h Dα αα=

( )( )We got                                          as operational identity; and this we operate on             to write( )x t

( )( ) ( )

W t T l i

( ) ( )( )F WE ( ) ( ) ( )th x t E h D x tα αα= that is ( )( ) ( )2 2

! (2 )!( ) ( ) 1 ... ( )h ht t tx t h E h D x t D D x tα αα α α α

α α α+ = = + + +

writing                                   ,                                        we have following series( )( ) ( )tD x t x tα α= 2 (2 )( ) ( )tD x t x tα α=

(3) We get a Taylor series2

( ) ( 2 )

( )

( ) ( ) ( ) ( ) . . .( ) ! ( 2 ) !

( ) ( )k

h hx t h x t x t x t

hx t x t

α αα α

αα

α α∞

+ = + + +

= + ∑

(3)

Fractional Taylor series

1( ) ( )

( ) !kx t x t

kα=

= + ∑

0 ,t t h← ←Making                               gives Fractional Maclaurent series

( )( ) (0 ) (0 )k

ktx t x xα

α∞

= + ∑ ( )

1( ) (0 ) (0 )

( ) !kx t x x

kα=

= + ∑

Note:                                                                      that is ( )( ) ( ) .... ( ) ( )k k

k

x t D D D x t D x tα α α α α= ≠ 2D Dα α α+≠

For small        we consider only first two terms of Taylor and write

( )( ) ( ) ( )( )!hx t h x t x t

αα

α+ +

h(4)

( ) ( ) ( )1 ( ) ( )

( )

( ) ( !) ( ) !d ( ) dx t x t t x x t tα αα αα α−Δ Δ

Expansion of non‐differentiable function in Fractional Taylor Series

kx α∞Our Fractional Maclaurent series is( )

1( ) ( 0 ) (0 )

( ) !k

k

xf x f fk

α

α=

= + ∑

The Mittag‐Leffler function                            with                              is ( ) ( )f x E xαα= 0 1α< < not‐differentiable at                0x =

( ) ( ) ( )( )

( ) ( )

2

0 0

2

0 0

( ) ( 0 ) ( ) ( ) . . . . .! 2 !

1 ( ) ( . . . . .! 2 !

x x

x x

x xE x E D E x D D E x

x xE x D E x

α αα α α α α α

α α α α

α αα α α

α α

α α

α α

= =

= =

= + + +

= + + +( )

( )

( )

2

0

2

1 ( ) . . . .! 2 !

1 . . .! 2 !

x

x x E x

x x

α αα

α

α α

α α

α α

== + + +

= + + + We have got the series definition of  ( ) ( )f x E xαα=

( )! 2 !α α

Let 12( )f x x b= + which is not‐differentiable at                0x =

( ) ( )( )

( )( ) ( )( )1122

1 1 1 1 12 2 2 2 2

2

1 10 02 2

( ) . . . . .! 2 !x x

x xf x b D x b D D x b= =

= + + + + +( ) ( )( )

( )( ) ( )( )

12

1 12 2

2 2

12 1

212

! 2 !

!!

! 0 !xb x D b x= + + = +

We apply normal Maclaurent series for a polynomial                                       and write the following2( )f x x x c= + +We apply normal Maclaurent series for a polynomial and write the following( )f

( ) ( ) ( )( ) ( )2 2

00

2

( ) (0 ) (0 ) (0 ) .. . 2 1 2 02 2 xx

x xf x f x f f c x x

c x x

==′ ′′= + + + = + + + +

= + + We observe parallel to classical calculus 

Comment about Fractional Taylor Series and Mittag Leffler Function

Intuitively if one considers that in fractional calculus the Mittag‐Leffler function plays a similar

role as exponential function in classical calculus, one is led to extend the Taylor series

( )( ) exp ( )f x h hD f x+ = in form of

1.

( )( ) ( )f x h E h D f xα α+ =( )( ) exp ( )xf x h hD f x+ = in form of

Of course, despite the simplicity of the formula this is not sufficient and needs to supported by

a sound derivation. As a matter of fact the problem which appears is to define suitably the

( )( ) ( )xf x h E h D f xα+

concept of fractional derivative in order that it is fully consistence with the Mittag‐Leffler

function considered as the solution of the FDE, i.e. ( ) ( ) ( )x t x tα λ=

From Taylor series we have a conversion formula that is

From this we can have the relation

2. ( )( )! d ( ) dx x t t ααα

( ) ( )dd

!d d dxt

x t xα

α

α αα = d ! dx xα αFrom this we can have                                                             the  relation 

From formula of fractional derivative we have                                       which gives following ( )1d 1

d 1 !x x

x

αα

α α−=

−

( )( ) ( )1 1d 1 ! dx x x αα αα− −= −

( ) ( )d t

3.

( )( ) ( )d 1 ! dx x xα

These are conversion formula of differential              to           and                d xα dx ( )dx α

Integration with respect to fractional differentialdeft xα

∫ ∫( ) 1

0 0( ) d ( ) ( )d 0 1

t xf x fα αξ ξ α ξ ξ ξ α−= − < <∫ ∫

This relates fractional integral w.r.t.            and Riemann‐Liouvelli integral RHS( )dx α

The integral w.r.t.                  is defined as solution of fractional differential equation ( )dx α

( )d ( ) d , 0 , (0) 0y f x x x yα= ≥ =

M l i l b h id b ( )α

that is ( )0

( ) ( ) dx

y x f x x α= ∫Multiply both sides by         we get !α ( )( !)d ( !) ( ) dy f x x αα α=

Taylor series gives                                                 using this expression in above we get( )( )!d ( ) dy y x x ααα

( ) ( ) ! ( )y x f xα α=which provides

( )11( ) 0

( ) ! ( ) ! ( ) ( )dx

y x D f x x fα ααα α ξ ξ ξ− −

Γ= = −∫

or ( )( ) ! ( )D y x f xα α=

( )( )1 11

( 1)! 0 0! ( ) ( )d ( ) ( )d

x xx f x fα α

αα ξ ξ ξ α ξ ξ ξ− −+= − = −∫ ∫

x x

∫ ∫Thus we have proved                                                                                                         ( ) 1

0 0( ) d ( ) ( )d 0 1

x xf x fα αξ ξ α ξ ξ ξ α−= − < <∫ ∫

Corollaries for integration w.r.t. fractional differential

From our discussion we have                                      as a result integration of  ( ) ! ( )y x D f xαα −= ( )d ( ) dy f x x α=

that is  ( )0

( ) ( ) dx

y x f αξ ξ= ∫ So we have1 x α

∫

(1)

( )0

1( ) ( ) d!

xD f x f αα ξ ξ

α− = ∫

gives

( )( )0

1 d( ) ( ) d! d

xf x f

x

αα

α ξ ξα

= ∫

As a direct consequence of                                                                          we derive ( ) 1

0 0( ) d ( ) ( )d

x xf x fα αξ ξ α ξ ξ ξ−= −∫ ∫

( ) 1( ) d ( ) ( ) ( )d 0 1x x

f fα β α βξ ξ β ξ ξ ξ β+ + −+ < + <∫ ∫

(2)

(α

( )

( )

1

0 0

1

0

( ) d ( ) ( ) ( )d 0 1

( ) ( ) ( )dx

f x f

x x f

β α β

β α

ξ ξ α β ξ ξ ξ α β

α β ξ ξ ξ ξ

+

−

= + − < + <

= + − −

∫ ∫∫

writing that is taken from definition we get( ) 1( ) d ( )( ) ( )df x fα αξ ξ α ξ ξ ξ−= −writing                                                                          that is taken from definition we get( )( ) d ( )( ) ( )df x fξ ξ α ξ ξ ξ=

Also(3)

( ) ( ) ( )0 0

( ) d ( ) ( ) d 0 1x x

f x fα β αα β βαξ ξ ξ ξ ξ α β+ += − < + <∫ ∫

Also(3)

( ) ( ) ( )0 0

( ) d ( ) ( ) d 0 1x x

f x fα β βα β αβξ ξ ξ ξ ξ α β+ += − < + <∫ ∫

Some examples of integration w.r.t. fractional differential ( ) 1

0 0( ) d ( ) ( )d , 0 1

x xf x fα αξ ξ α ξ ξ ξ α−= − < <∫ ∫Our formula is

On making                           we get ( )f x x γ=

( ) ( )( )1 11( )0 0 0

d ( ) d ( ) ( ) dx x x

x xαγ α γ α γαξ ξ α ξ ξ ξ α α ξ ξ ξ− −

Γ= − = Γ −∫ ∫ ∫

(1)

( )( ) ( )( )( )0

! !( 1)1 ( 1) , 0 1( 1 ) !xD x x xα γ γ α γ αα γγα α αγ α α γ

− + +Γ += Γ + = Γ + = < <

Γ + + +On making                       we get( ) 1f x =

( )d 0 1x

xα αξ α< <∫(2)

( )0

d , 0 1xξ α= < <∫The coordinates are drawn in normal case via process  of integration of the differential elementi.e.                            on this drawn coordinate we write function                 w.r.t. the coordinate  

0d

xxξ =∫ ( )f x x

On making                            we get( ) ( )f x xδ=(3)

Thus with the fractional differential the drawn coordinate will be                             with functionmapped as                     w.r.t. transformed coordinate  

( )0

dx

xα αξ =∫( )f x α

, by using                                          the following( )( ) ( ) d (0)f x x x fδ =∫

x α

g g( ) ( )f x xδ

( ) 1 1

0 0( ) d ( ) ( )d , 0 1

x xx xα α αδ ξ ξ α ξ δ ξ ξ α α− −= − = < <∫ ∫

( ) , y g g( )( ) ( ) d (0)f x x x fδ∫

(4) On making                                         we get( )( )f x E x αα λ= ( )

( )( ) ( ) ( )( )( )( )

0

1

d ( )

! 1

xE D E x

E x

αα α αα α

αα

λ ξ ξ α α λ

α λ λ

−

−

= Γ

= −

∫

Leibniz’s rule for product of two functions( )( )!d d αα ( ) ( ) ( )From the fractional Taylor series we have                                     let us write ( )( )!d dy y x ααα = ( ) ( ) ( )y x u x v x=

that is product of two functions. We do the following manipulations

( ) ( ) ( )( )! d du v u v xα αα =

( )( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )

( )

( ) ( ) ( )

! ( ) d ( ) d d

d ( ) ! d ( ) ! d

d ( ) ( d ) ( ) ( d )

u x v v x u u v x

u v x u x v v x u

u v x u x v x v x u x

α α

α α

α α α α α α

α

α α

+ =

= +

= +

We get the formula                                                                                      ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x v x u xα α α= +

Dividing both sides by                   we obtain ( )dx α

Let                           and                            with application of above we have ( ) ( )u x xδ= ( ) ( )v x f x=( )( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( 0 ) 0

x f x x f x f x x

x f x f

α α α

α α

δ δ δ

δ

= +

= =

( )( ) ( )( ) ( ) ( )x f x x f xα αδ δ= −

Now integrating w.r.t. fractional differential we get 

( ) ( )( ) ( )( ) ( ) d ( ) ( ) d , 0 1f fα αα αδ ξ ξ ξ δ ξ ξ ξ α= − < <∫ ∫( ) ( )1 ( )

1 ( )

( ) ( ) ( )d

(0)

x f

x f

α α

α α

α ξ δ ξ ξ ξ

α

−

−

= − −

= −

∫ ∫∫

Using Leibniz’s product rule

( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x v x u xα α α= +The expression is ( )( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x v x u x= +The expression is

Let                            and                             with                             then by using above rule we get ( ) au x x= ( ) bv x x= 0 , 1a b< <

( ) ( ) ( ) ( ) ( )( )

! !( )! ( )!

! !

a b a b b a a b b ab ab a

a bb

D x x x D x x D x x x x xα α α α αα α

α

− −− −

+

= + = +1.

( )! !( )! ( )!

a ba ba b x α

α α+ −

− −= +

Using direct formula we get 2.

( ) ( ) ( )!( )!

a ba b a b a bx a bD x x D x xα α α

α++ + −

+ −= =

Well they are not the same yet they are Is there a contradiction? These two are( )a bK x α+ −Well they are not the same yet they are               . Is there a contradiction?  These two are

dealing with             and               , we must conclude from the viewpoint of fractional derivative a bx x a bx +

they are not same, in the sense that their fractional differentials (or differential increments)

( )K x

have not the same value.                                              required in first case (1) and                      ( ) ( )( ) (d ) , ( ) (d )a bx x x xα α α α ( )( ) (d )a bx xα α+

for second case (2).

Integration by parts

( ) ( ) ( ) ( )( ) ( ) ( )d ( ) (d ) ( ) (d )u v x u x v x v x u xα α α α α α= +We have

Integrate above ( ) ( ) ( ) ( )( ) ( ) ( )d ( ) (d ) ( ) (d )u v x u x v x v x u xα α α α α α= +∫ ∫ ∫

Also we have  ( ) ( )( )! d dy y x ααα = Integrating both sides we get ( ) ( )( ) ( )

0 0! d d

y x xy y x ααα =∫ ∫

gives ( )( )

0! ( ) d

xy x y x ααα = ∫ Using this relation

we write ( ) ( ) ( )( )( ) ( ) d ! ( ) ( )u x v x x u x v xα α α=∫

With use of the concepts developed for fractional differentials we write the following

( ) ( ) ( )( ) ( )( ) ( ) d ! ( ) ( ) ( ) ( ) db bb

aa au x v x x u x v x u x v x xα αα αα= −∫ ∫

( )d ! db bb

aa av u v u u vα αα= −∫ ∫

or

It is like we have for classical calculus i.e. 

( ) ( ) ( )( ) ( ) d ( ) ( ) ( ) ( ) db bb

u x v x x u x v x u x v x x′ ′= −∫ ∫( ) ( ) ( )( ) ( ) d ( ) ( ) ( ) ( ) d

d ( ) d

aa ab bb

aa a

u x v x x u x v x u x v x x

v u v u u v

=

= −

∫ ∫∫ ∫

Chain Rule

( )( ) ( )( ) ( )( ) ( ) ( )f f αα α α⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ Δ ⎛ ⎞ ⎛ ⎞

Assume that               is          differentiable w.r.t.          and                is differentiable w.r.t( )f u α − u ( )u x x

then we have following

(1)

( )( ) ( )( )( )

( )( )( )

( )( )

( )

0 0

( ) ( ) ( ) d ( ) d( ) lim limd d( )x x

f u x f u x u x f u uf u xu xx u x x

αα α ααα

α α α αΔ ↓ Δ ↓

⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ Δ Δ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠Δ Δ Δ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )( ) ( )( )( )( ) ( )( ) ( )uf u x f u u xα αα ′=

Assume that               is  differentiable w.r.t.          and                is           differentiable w.r.t( )f u u ( )u x α − x

( ) ( )1

1d 1d 2 1 !

u uuu

α αα

α α α

−−= =

Γ − −We have fractional derivative of as from here( )f u u=

(2)

( )

( ) ( )d 2 1 !u α αΓWe have fractional derivative of                     as                                                             from here

we get a conversion formula                                                       and like wise we also write( )( ) ( )1 1d 1 ! du u u αα αα− −= −

( )( ) ( )1 1d 1 ! df f fαα αα

− −= − This being the case one has( )( ) ( )

( )( ) ( )( )( ) ( )

1 1( )

1

1 ! dd d d d ( ) ( )d d d d1 ! d

u

f ff f u u f f u xx u x x uu u

αα αα α α αα α

αα α α αα

α

α

− −

−

− ⎛ ⎞ ′= = = ⎜ ⎟⎝ ⎠−( )( ) ( )

( )( ) ( )1

( ) ( )( ( ))( ) ( ( )) ( )( ) u

f u xf u x f u x u xu x

ααα α

−⎛ ⎞ ′= ⎜ ⎟⎝ ⎠

Chain Rule when both are fractionally differentiableWe have the following(3)

d d dd d d

f f ux u x

α α α

α α α=

( )( ) ( )1 1d 1 ! du u u αα αα− −= −use                                                        for the denominator on the RHS and get following

g( )

( )( ) ( )

( )( )

1 1

d d d d dd d d d1 ! d

d d

f f u f ux u x xu u

f u

α α α α α

α α α ααα

α α

α− −

= =−

⎛ ⎞ ⎛ ⎞( )( )

( ) ( )( )( ) ( )( )1 1 ( ) ( )d d1 ! 1 ! ( ) ( )

d du

f uu u f u u xu x

α α α αα αα α− −

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )( ) ( )( ) ( )( )( ) 1 ( ) ( )α

In other form we write 

( )( ) ( )( ) ( )( )( ) 1 ( ) ( )( ) 1 ! ( ) ( )uf u x u f u u xα α α αα −= −

( )( ) 1 ( ) ( )( ) ( 2 ) ( ) ( )x u xf u x u f u u xα α α αα −= Γ −

Corollary:  The         derivative of        provides the equality ( ) ( )1 1d (1 )! dx x x αα αα − −= −α − x

which allows us to write successively                                      from this we get( )( )d dxf f x αα α=

( )( ) 1d ( ) (1 ) ! df fα α α α( )( ) 1d ( ) (1 ) ! dxf f x x xα α α αα −= −

About the three chain rule formulas

Let us have  ( ) ( ) ( )( ) ( )1 1( ) ;f u x E x E x E u u xα ααα α

α α αλ λ λ= = = =

With                                                              we have( )( ) ( )( )( ) ( )u xD f u x f u u αα α ′=

( )( ) ( )( ) ( )1D E x E u E xααα α α α

α α αλ λ λ λ= = known result

1

With                                                                                     we get( )( ) ( ) ( )1 ( )( ) ( ) ( )f

uuD f u x f u u xα αα α−

′=

( )( ) ( ) ( )( )( ) ( )11

1E xα α αλ−

⎛ ⎞

2

( )( ) ( ) ( )( )( ) ( )( )( ) ( )( )( )

1

11

(1 )!

1

(1 )!

E xux

u

D E x D E u x

E x D E u

α α

α

λα α α αλα α αλ

ααα αλ

α αα

λ

λ

−−

−

−

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

But the                          is not differentiable w.r.t.      in such a manner that it fails( )E u αα

u

With                                                                                              we get( )( ) 1 ( ) ( )( ) (1 )! ( ) ( )uD f u x u f u u xα α α αα −= −3 ( )( )( ) ( ) ( )( )( )

( )

11 1 1(1 )!(1 )!D E x x E x x

E x

αααα α α αλ

α α α

αα

λ α λ λ

λ λ

− −−= −

= known result

We note that                             is not differentiable and                is differentiable and also         differentiable( )E uαα

1u xαλ= α −

Contd…

L t ( ) ( )( ) 0 1bab a b af u x x x u a b u x= = = < < =

Contd…

Let  ( ) ( )( ) , 0 , 1,f u x x x u a b u x= = = < < =

According to standard R‐L formula we have 

( ) ( ) !a ba b a bD x xα α−=( ) ( ) !x a bD x xα−=

(1) ( ) ( ) ( ) ( ) ( )1 ( )! ! !

! ! !( ( ) ) b a a b a a bb b b ab b bD f u x u a x x a x xααα α α α α α α

α α α− − − − −

− − −= = =

(2) ( ) ( ) ( ) ( )( )( ) ( ) ( )

1 1 !!

1 ( ! )( 1 )

( ( ) ) b

a

b au au a

bb b b

D f u x b u xα αα α

α

α α

− − −−

−

=

( ) ( ) ( )( ) ( )1 ( ! )( 1 ) !

! !

ab aa b a a b a a baa ax b x x x

α αα α αα α

− − − −− −= =

( ) ( ) (1 ) ! ! !( 1 ) ! !( ( ) ) (1 ) !b a ba a a a bb aD f

α αα α α α− −− − −(3) ( ) ( ) ( ) ( )

(1 ) ! ! !( 1 ) ! !( ) !( ) !! !( ( ) ) (1 ) ! a ba a a a bb a

a bb aD f u x x x x xαα α α αα αα αα − − −

− −− −= − =

We note that  and           both are not‐differentiableax bu so the (1) and (2) disqualified

thus the derivation with R‐L formula and the third chain rule (3) are qualified however are

involving two functions with different fractional increments

Coarse grained space (CGS) and coarse grained time (CGT)

CGT case d x

According to fractional Taylor series we have                                            ,using this we have

( )( )

( )1( )1 dd (d ) d ! d! d

xx x t x tt

ααα α

ααα

−= =

CGT case( )d

ut

α α=

using this we have 

( )( )

( )( )1 1 ( )d d( ) ! ! ( ) 0 1 , d 0

d dx xu t x t t

t t

αα

α α αα α α− −= = = < < >

CGS case                                                                              ( )d( ) d 0 , 0 1

dx

v t xt

α

α α= > < <

Given a function                       and its inverse                         , their fractional derivative of order ( )y f x= ( )x g y=α satisfy the condition0 1α< < ( ) ( )2 1( ) ( )( ) ( ) (1 ) !y x x y x y αα α α − −= −

( ) ( ) d d d d( ) ( )d d d d

y x y xy x x yx y y x

α α α αα α

α α α α

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

and proof is: 

( ) ( )

( )( )

11 1

21 1

(1 )! (1 )! (1 )!

y y

xyy x

αα α

α α α

−− −

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠ Contd…

( ) ( )( )

( )( )

d ! d d !( ) !d ! d d ( )x x x

v tt t t t x

α α α

α α α

α ααα

= = = =

Contd…

For CGS we have ( ) ( )d ! d d ( )t t t t xα

Above is expressing                 in terms of fractional derivative of time w.r.t space( )v tα

2 1

( ) d ( )( )d

t xt xx

αα

α=

( ) ( )2 1( ) ( )( ) ( ) (1 )!y x x y xy αα α α − −= −Using                                                                              and manipulating above we have following

( )

( ) ( ) ( )

! ! ( )( )( ) ( ) ( )

x tv tt t t

α

α α α α

α α= =

( ) ( )( )( )

( ) ( )

( ) ( ) ( )

( )2 1 ( )

2 1

1 2( )

( ) ( ) ( )! ( ) ! (1 ) ! ( ) ( )

(1 ) !

t x x t t xx t xt x t

xt

α α α

αα α

α

α

α α αα

−− −

−

= = −−

( ) ( )1 2( )( ) ( ) ( ) ! (1 ) !k xt x t kα αα α α α−= = −

So we have fractional velocity as

Fractional velocity in CGT as                                                                                     ( ) 1 ( )( ) ! ( ) , 0 1, d 0u t x t tαα α α−= < < >

Fractional velocity in CGS as ( ) ( )1 2( )( ) ( ) ( ) ( ) ! (1 )!v t k xt x t kα αα α α α−= = −Fractional velocity in CGS as                                                                                               ( ) ( )( ) ( ) ( ) , ( ) ! (1 )!v t k xt x t kα α α α α

Lorentz transformation in coarse grained system

1. CGS system : The differential displacement is approximated by ,d x ( )d x α 0 1α< <1. CGS system : The differential displacement           is approximated by               ,  d x ( )d x 0 1α< <

The velocity of the particle is defined as                        and if we state that this is bounded by ( )d / dx tα c

i.e. ( )d / dx t cα <

We consider pseudo‐geodesic as ( ) ( ) ( )2 2 22d d ds c t x α= − We have Lorentz relation as

( ) ( )( ) ( )( ) ( )1

2 2

2 2

defd d d d d d 1u u

c cx k x u t t k t x kα α α −

= + = + = −

On integrating in sense                                                                                       we get ( ) 1

0 0( ) d ( ) ( )d

x xf x fα αξ ξ α ξ ξ ξ−= −∫ ∫

( ) ( )2u

cx k x u t t k t xα α α= + = +

This is case with coarse grain space and continuous time

2. CGT system:  The particle velocity will be                        and pseudo‐geodesic is( )d / dx t α

( )

( ) ( ) ( )2 2 22d d ds c t xα= −

Here we will have

( ) ( )2uc

x k x ut t k t xα α α= + = +

Solution of Fractional Differential Equation with modified RL fractional derivative

( )0( ) ( ) , (0)x t b t x xα = =

( )d ( ) d ( ) dx b t b t tα

αα ( )( ) d ( ) dd

b t x b t tt

αα = =

One has successively

( )0

( )

0 0d ( ) ( ) d

x t t

xx t x b ααξ τ τ= − =∫ ∫

Therefore                                                                 10 0

( ) ( ) ( )dt

x t x t bαα τ τ τ−= + −∫

( )

Solution of Fractional Differential Equation with modified fractional RL derivative 

( )0( ) ( ) ( ) ( ) (0)x t a t x t b t x xα = + =

( ) ( ) ( ) ( )x t a t x tα =The solution of homogeneous equation                                     is   ( )( )0( ) ( ) d

tx t C E a α

α τ τ⎛ ⎞= ⎜ ⎟⎝ ⎠∫

This being the case, let us look for a special solution of the complete equation in the form

( )( )0( ) ( ) ( ) d

tx t C t E a α

α τ τ= ∫Use                                                                               to write the following( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x u x v xα α α= +

( )( ) ( ) ( )( )( ) ( )

0 0 0

d( ) ( ) ( ) d ( ) ( ) d ( ) dd

t t tx t C t E a C t a E a

t

αα α αα α

α αατ τ τ τ τ τ⎛ ⎞

= + ⎜ ⎟⎝ ⎠

∫ ∫ ∫( )( ) ( ) ( )( )0 0 0d t α⎜ ⎟⎝ ⎠

∫ ∫ ∫

Use the above expression in                                                   to get  ( ) ( ) ( ) ( ) ( )x t a t x t b tα = +

( )( )( ) 1( ) ( ) ( ) dt

C t b t E a αα τ τ−= ∫ integrating this we have( )( )0( ) ( ) ( ) dC t b t E aα τ τ= ∫ integrating this we have

( )( )1

0 0( ) ( ) ( ) (d ) d

t tC t b u E a u αα

α τ τ−= ∫ ∫

and general solution is                                                                                                      ( ) ( )( ) ( )( )10 0 0 0

( ) ( ) ( ) d , ( ) ( ) ( )(d ) dt t t

x t x C t E a C t b u E a uα ααα ατ τ τ τ−= + =∫ ∫ ∫

Fractional Differential Equation with constant coefficients( )

0( ) ( ) (0)x t ax t b x xα = + =

( ) ( )( )( )( ) ( )( )1

0 0 0 0( ) d d d

t t tx t x bE a u E aα α α

α ατ τ−⎛ ⎞= +⎜ ⎟⎝ ⎠∫ ∫ ∫

Using the result  obtained  ( ) ( )( ) ( )( )10 0 0 0

( ) ( ) ( ) d , ( ) ( ) ( )(d ) dt t t

x t x C t E a C t b u E a uα ααα ατ τ τ τ−= + =∫ ∫ ∫

we write

( )( )( ) ( )

( ) ( ) ( )( )

10 0

10 0

d

d

t

t

x b E at u E at

x E at bE at E at u

αα αα α

αα α αα α α

−

−

= +

= +

∫

∫We simplify the second term as follows 

( ) ( ) ( ) ( ) ( )( )1

0 0d ( ) d

t tb E a t E a b E a t E aα α αα α α

α α α ατ τ τ τ− = −∫ ∫( )( )

( ) ( )( )0

0

( ) d

( ) 1

t

t

b E a t

b bE a t E a ta a

ααα

α αα α

τ τ

τ

= −

⎡ ⎤= − − = −⎣ ⎦

∫

Combining with                         we write                                                        ( )0x E atαα ( ) ( )0( ) b b

a ax t x E atαα= + −

Solution of  ( ) ( ) ( ) ( )x t ka t x tα =

Fractional Differential Equation with non‐constant coefficients

( ) ( )( ) ( )( )1( ) ( ) ( ) d ( ) ( ) ( )(d ) dt t t

x t x C t E a C t b u E a uα αατ τ τ τ−= + =∫ ∫ ∫We have

( )( )( ) ( )( )( )( )

10 0 0 0

( ) ( ) ( ) d d ( ) d

( ) d

t t t

t

x t x b u E k a u E k a

x E k a

α α αα α

α

τ τ τ τ

τ τ

−⎛ ⎞= +⎜ ⎟⎝ ⎠

=

∫ ∫ ∫

∫

here ( ) 0b t =

( ) ( )( ) ( )( )0 0 0 0( ) ( ) ( ) d , ( ) ( ) ( )(d ) dx t x C t E a C t b u E a uα ατ τ τ τ= + =∫ ∫ ∫We have

( )( )0 0( ) dx E k aα τ τ= ∫

As a special case when                            we have the following 1( )a t t α−=

( ) ( ) 11 1t tα α −

∫ ∫( ) ( )

( )

11 1

0 0

1 1

0

d d

1( ) ( ) d( )

t t

t

t

t

α αα α

α α

τ τ α τ τ τ

α α τ τ τα

−− −

− −

= −

⎛ ⎞= Γ −⎜ ⎟Γ⎝ ⎠

∫ ∫

∫

( )( )( ) ( )( )( )( )( )( ) ( )( )( )( )

(1 1)1 10 (1 1 )( ) ( )

( ) (2 ) ( 1) (2 )

! (1 ) !

tD t t

t t

t

αα α α αα αα α α α

α α α α α

α α

Γ − +− − − +Γ − + +⎡ ⎤= Γ = Γ⎣ ⎦

= Γ Γ − = Γ + Γ −

= −( )( )

Therefore                                                        ( )0( ) !(1 )!x t x E ktα α α= −

( ) ( )(2 ) ( )( ) ( ) 0a x t b x t cα α+ + =

Fractional Differential Equation with constant coefficients & two orders

( ) ( )( ) ( ) 0a x t b x t c+ + =

Can be re‐written as 

( ) ( ) ( )( )( ) ( ) 0 0a D D x t b D x t c D r D rα α α α α+ + = − − =( ) ( ) ( )( )1 2( ) ( ) 0 0a D D x t b D x t c D r D r+ + = − − =

Where the           and              are the roots of  1r 2r 2 0ar br c+ + =

then                                                                    ( ) ( )1 1 2 2( )x t C E r t C E r tα αα α= +1 2r r≠

then ( ) ( )ˆ( )x t C C t E r tα α= +1 2 ˆr r r= =

The constants         ,           depends on initial condition1C 2C

then                                                                    ( ) ( )1 2( )x t C C t E r tα= +1 2r r r

Application to dynamics close to equilibrium position

Consider the dynamics                                with equilibrium point as          i.e. d ( )dx f x t= 0x 0( ) 0f x =

Close to this equilibrium we have variation equation                                               . We assume( ) 0d ( ) dxx f x x tδ δ=

this is disturbed by and external force (input)                                            ; that it turns out as ( )( ) dw t t α 0 1α< <

( ) ( )0d ( ) d ( ) d , 0 1xx f x x t w t t αδ δ α= + < <

With                  we have  :y xδ= ( )0d ( ) d ( ) d , 0 1xy f x y t w t t α α= + < <

In order to get solution of above we write ( ) ( ) ( )y t y t y t=In order to get solution of above, we write  1 2( ) ( ) ( )y t y t y t=

( )( )

1 2 0 1 2d ( ) ( ) d ( ) d

d d ( ) d ( ) d

xy y f x y y t w t t

y y y y f x y y t w t t

α

α

= +

+ + ( )1 2 2 1 0 1 2d d ( ) d ( ) dxy y y y f x y y t w t t+ = +

( )d ( ) d d ( ) df t t t α

Comparing the coefficients of          and          in LHS and RHS of above we write following   1y 2y

( )2 1 0 1 2 1 2d ( ) d d ( ) dxy y f x y y t y y w t t= =

Contd…

Contd…

We have two equations ( ) ( )11 0 1 2 1d ( ) d d ( ) dxy f x y t y y w t t α−= =

The first one directly gives  ( )1 1 0( ) (0) exp ( )xy t y f x t=y g ( )1 1 0( ) ( ) p ( )xy y f

To solve second we multiply both sides by             and notice !α ( ) 2 2! d dy yαα =

( )( ) ( )( ) ( )

12 1

12 1

!d ! ( ) d

d ! ( ) ( ) d

y y w t t

y y t w t t

α

αα

α α

α

−

−

=

=

( ) 1( )2 1( ) ! ( ) ( )y t y t w tα α −=

We write solution as

( ) ( )1( ) (0) ! ( ) ( ) dt

y t y w y αα τ τ τ−= + ∫ ( ) ( )2 2 10( ) (0) ! ( ) ( ) dy t y w yα τ τ τ= + ∫

What happens when                            is subjected to coarse graining in time CGT and( ) ( )x t f x=

coarse graining in space CGS ?

Equilibrium position of two dimensional system 

We consider a particle with mass            and subjected to the potential function  m ( )V xdriven by  ( ) ( )xmx t V x= −

or in vector form (the state variable form) we may formulate the above equation of motion as ( ) y q1( ) ( ) ( )xx t y y t m V x−= = −

We assume the particle is at rest at            that is the equilibrium position is  0x 0( , 0)x

We then have fractional differential equations for the purturbations( ) ( ) 1

0( )xxx y y m V x yα α −= = −

The second one gives 

( )( ) (0) (0) ( )t D tα−

( )10( ) (0) ( )xxy t y E m V x tα

α−= −

From first one we get ( )( )

0

10

( ) (0) (0) ( )

(0) ( )(0)

t

xxxx

x t x y D y t

my E m V x tV

α

αα

−

− =

= −

From first one we get

CGT and CGS comparison with normal case 

What happens to equilibrium position           in  dynamics  of                           with CGT, CGS   d ( ) dx f x t=0x

For continuously differentiable deviation, in                           one has  1 0 1( ) ( ) ( )xy t f x y t=

( ) ( )x t f x= is subjected to CGT, we have following

( ) ( )x t f x=

This being the case assume 

( )( )

( )( )1 1 ( )d d( ) ! ! ( ) 0 1 , d 0

d dx xu t x t t

t t

αα

α α αα α α− −= = = < < > using this

i h f CGT d ( )(d )f αd ( )dx f x t=in                               we have for CGT  d ( )(d )x f x t α=

( ) ( ) ! ( )x t f xα α=

gives

Therefore the deviation equation for CGT case is

( )0 00 0

d ( ) d ! ( ) ( )d dx x x f x x f x

t t

α α

α α

δ α δ+− = + −

( ) ( )

( )

0 0( ) ( )0 2

2

d dd ( ) ! ! ( ) :

dd ! ( ) ( )

f x x f xxx

t tx x f x x y x

ty f x y t

αδδα

α

δ α δ α δ δ

α

+ −= = =

= ( )0 2! ( ) ( )d xf x y t

tα α=

That is                                                      ( )2 0 2( ) ! ( ) ( )xy t f x y tα α=

Contd….

Contd….

Assume now                             subjected to CGS. For CGS we have the followingd ( )dx f x t=

( ) ( ) ( )1 2( )d( ) ( ) ! (1 ) !

dx

v t k xt x t kt

αα α

α α α−= = = −

using this and substituting in                               we get( )1

( ) ( ) ( )xt

x t f xk

αα

−

=

D i i i i

(d ) ( )dx f x tα =

Deviation equation is

( ) ( )1 10 0( ) ( )

0 0 0 0 3

1

( )( ) ( ) ( ) :

x x t x tx x x f x x f x x y

k k

α αα α δ

δ δ δ− −+

+ − = + − =

⎛ ⎞ ( )( )1

( ) 1 13 0 3 0 0 3 0 0

1 10 3 0 0 3

1

( ) ( ) ( ) ( )

use ( ) (1 ) .....

xty x y f x f x y x f xk

x y x x y

t

αα α α

α α α

α

α

−− −

− − −

−

⎛ ⎞= + + −⎜ ⎟

⎝ ⎠+ ≅ + − +

⎛ ⎞( ) 1 13 0 0 0 0 3 0 0( ) (1 ) ( ) (x

ty x f x x f x y x f xk

α α α αα− − −⎛ ⎞= + − +⎜ ⎟

⎝ ⎠( )

( ) ( )

2 13 0 0 3 0 0

23

1( ) 1 13 0 0 0 0 3

) (1 ) ( ) .... ( )

put 0

(1 ) ( ) ( )

xy x f x y x f x

y

y k t x f x x f x y

α α

α α α α

α

α

− −

− − − −

+ − + −

≈

= − + suggest is special case in CGS0 0x =( ) ( )3 0 0 0 0 3(1 ) ( ) ( )xy k t x f x x f x yα + suggest                  is special case in CGS  0 0x

That is                                                                                                     ( ) ( )1( ) 1 13 0 0 0 0 3(1 ) ( ) ( )xy k t x f x x f x yα α α αα− − − −= − +

Comparison of coarse graining

( )1 0 1 1 1 0( ) ( ) ( ) ( ) ( ) (0) exp ( )x xx t f x y t f x y y t y f x t= = =

Fine grain system

Coarse graining in time  CGT

( ) ( )( ) ( )2 0 2 2 2 0( ) ! ( ) ( ) ! ( ) ( ) ( ) (0) ! ( )x xx t f x y t f x y t y t y E f x tα α α

αα α α= = =

( )( ) 1 1 ( ) 1 1 13 0 0 0 0 3( ) ( ) ( ) ( ) (1 ) ( ) ( )x t k xt f x y t k t x f x x f x yα α α α α αα− − − − − −= = − +

Coarse graining in space  CGS

( )

( )10 0 0 0

3 0 0 0 0 3

(1 ) ( ) ( )3 3 (1 )!

( ) ( ) ( ) ( ) (1 ) ( ) ( )

( ) (0) x

x

x f x x f x

x t k xt f x y t k t x f x x f x y

y t y E tα αα

α α

α

− −− +−

+

= ( )

Coarse graining in both  space and time

d ( )dx f x t=

(d , d ) (d , d )x t x tμ γ← 0 1 0 1μ γ< < < <( , ) ( , )

( ) ( )d ( ) dx f x tμ λ=

Depending on whether                        or                        we shall be gettingμ γ> μ γ<

( ) ( )1

d ( ) df tλ

μμ λ>( ) ( )( ) ( )1

d ( ) d

d ( ) d

x f x t

x f x t

μμ

μλ λ

μ λ

μ λ

= >

= <

Which is related to either CGT case or CGS case as discussed in earlier 

Differential geometry of fractional order‐arc lengths with modified RL fractional derivative

Let a f nction ( )f hich is b t has fractional deri ati enot differentiable α 0 1α< <Let a function  ( )f x which is                                      but has fractional derivative         ;not‐differentiable α 0 1α< <

The arc length of the curve defined by point                                from         to      ,                   as     ( ), ( )n n nx y f x= 0 nx 0nx >

( ) ( )22(1 ) ( )11( ) ( ) dnx

f αα α⎛ ⎞∫

This is obtained from classical formula for distance i.e.                                                    then using the ( ) ( ) ( )2 2 2d d ds x y= +

( ) ( ) ( )22 (1 ) ( )1

(1 )!0

1( , ) ( ) d!

n

ns x x f x x αα αα

αα

−

−

⎛ ⎞= +⎜ ⎟⎝ ⎠∫

g( ) ( ) ( )y

relations                                                                                    got from                         and with ( ) ( ) ( )1 1! d d (1 ) ! dx x x x αα αα α − −= = −( )

1d(1 ) !d

x xx

α α

α α

−

=−

Taylor series relation ( ) ( )1 ( )d ! ( ) dy f x x ααα −=

If one defines distance as                                                                and applying conversion formulas for( ) ( ) ( )2 2 2d d ds x yα α α= +

these fractional differentials one getsg

( ) ( ) ( )2

22 (1 ) ( )1(1 )!0

( , ) ( ) dnx

ns x x f x x αα αα

α −

−

⎛ ⎞= +⎜ ⎟⎝ ⎠∫

In parallel to classical arc length formula for smooth function                                                              ( )2

0( , ) 1 ( ) dnx

ns x f x xα ⎛ ⎞′= +⎜ ⎟⎝ ⎠∫

Application of fractional arc length formula to a constant function

( ) ( ) ( )2

22 (1 ) ( )1(1 ) !0

1( , ) ( ) d!

xs x f αα α

αα ξ ξ ξ

α−

−

⎛ ⎞= +⎜ ⎟⎝ ⎠∫

Assume which amounts to is a constant (with modified RL derivative)( ) ( ) 0f xα ( )f xAssume                              which amounts to                is a constant (with modified RL derivative) ( ) ( ) 0f x = ( )f x

( ) ( ) ( ) (1 )

2

22 (1 ) ( )1 1 1! ! (1 ) !(1 ) !0 0

( , ) ( ) d (d )x x

s x fαα ξα α α

α α ααα ξ ξ ξ ξ

−−−−

⎛ ⎞= + =⎜ ⎟⎝ ⎠∫ ∫

Using                                       and 1d

d (1 ) !

α α

α

ξ ξξ α

−

=− we get

(1 )x

∫

( ) 1d ! d αξ α ξ−=

( )( )

(1 )1! (1 )!0

d1! ( d )0

( , ) (d )

d

x

x

x x

s xα

α

α

ξ αα α

αξα ξ

α ξ

ξ

−

−=

=

∫∫

∫ ∫1! 0 0

d dx x

x

αα ξ ξ= =

=∫ ∫

This looks quite consistence with what we would expect. Indeed when                is a constant we come  ( )f xThis looks quite consistence with what we would expect. Indeed when is a constant we come( )f x

across an horizontal straight line and the length provided by the formula is equal to  x , in natural way 

Application of fractional arc length formula to white noise

( ) ( ) ( ) ( ) ( )2 2 2

22 (1 ) ( )1 1! (1 ) ! !0

( , ) ( ) dt

s t y αα αα α α

α τ τ τ−

−

⎛ ⎞= +⎜ ⎟⎝ ⎠∫ ( ) ( ) ( )⎝ ⎠

Assume that                                          , where                denotes ‘white‐noise’ with zero‐mean and  variance( )d ( ) ( ) dy t w t t α= ( )w t 2σ

Multiplying this equation with           gives                                              and we get distance as!α( ) ( ) ! ( )y t w tα α=

t ⎛ ⎞∫ ( ) ( ) ( )2 ( 1 )

2 22

! (1 ) !0( , ) ( ) d

ts t wα ατ

α αα τ τ−

−

⎛ ⎞= +⎜ ⎟⎝ ⎠∫

As a special case with                we get more or less a model of Brownian Motion, we then find12α =

( )12t

∫( ) ( )12 221 4

2 160, ( ) d

ts t wπ

π τ τ τ= +∫and using                                          we get( ) ( )312 2 2! π= Γ =

For a case with zero variance ( ) ( )1

21 42 0

, dt

s t π τ τ= ∫Starting with half derivative of  f ( t ) = t i.e.                                          we get ( )

( )

1 12 21

21

2 12

d!d

t ttt

= = ( ) ( )1122 1

2 !t t=

Therefore ( ) ( ) ( ) ( ) ( ) ( )( )( )

( )11 1 11 22 2 22

12

d1 4 4 1 4 12 2 20 0 0 d

, d ! d ! dt t t

s t τπ π π τ

τ τ τ τ τ= = =∫ ∫ ∫ ( )

( )( )0 0 0 d

24 12 0

! dt

t

τ

π τ= =

∫ ∫ ∫

∫Quite realistic  ( )d ( ) ( ) dy t w t t α=

( )0

( ) ( ) dy t C w tτ ατ= + ∫

With zero mean and zero variance we get a constant function Contd….( )y t C=

Contd….

For a case of non‐zero variance we write by expanding as following( )1

2 221 6 ( )wπτ τ+For a case of non zero variance we write by expanding                                       as following

( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( )

1 11 12 22 22 2

11 2 22

2 1 21 4 42 16 160 0

24

, ( ) d 1 ( ) d

1 d

t t

kt k k

s t w w

C

π ππ π

π

τ τ τ τ τ τ τ−

∞ −

= + = +

⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟

∫ ∫

∑∫

( )1 6 ( )

( ) ( )( ) ( )

( ) ( )( )

( )

22

11 2 22

1 2

241610

24161 0

( )1 2

1 d

d

k kkk

k t k kkk

t t

C w

t C w

ππ

ππ

τ τ τ τ

τ τ τ

=

∞ −=

⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= +

∑∫

∑ ∫∫ ∫( ) 2

2 ( )1 28 160 0

( ) d dwt

t w t τπ πτ τ

τ τ τ τ−−

≅ + = + = ∞∫ ∫Note for a finite non‐zero variance case we have                             and RL ½ integration of          is  2 2( ) 0w t σ ≠∼ 1t − ∞

Dealing with curves which are  continuous everywhere and nowhere differentiable like white‐noise

involves trap. Their arc length following customary Euclidian sense  i.e. ( ) ( ) ( )2 2 2d d ds x y= +

is infinite These functions cannot be replicated in such a manner that they exhibit random featuresis infinite. These functions cannot be replicated, in such a manner that they exhibit random features

It follows that                           at least when ( )12,s t = ∞ (0 ) 0w ≠

This is still developing concept

Application of fractional arc length formula to self‐similar function

A self similar function with Hurst exponent          with                          , is                                     ,H 0 1H< < ( ) ( )Hy a t a y t= 0a >

This also provides ( ) (1)Hy t t y=

In the neighborhoods of zero, we have                                        and applying the conversion formulas( )d d (1)Hy t y=

for fractional differentials we get ( ) ( ) ! (1)Hy t H y=

( ) ( ) ( )2

22 (1 ) ( )1(1 ) !0

1( , ) ( ) d!

nx

ns x x f x x αα αα

αα

−

−

⎛ ⎞= +⎜ ⎟⎝ ⎠∫Using the formula                                                                                                             we get⎝ ⎠

( ) ( )22 (1 ) 2 21

(1 ) !0

1( , ) ( !) (1) d!

t HHH

s t H H yH

τ τ−

−

⎛ ⎞= +⎜ ⎟⎝ ⎠∫ ( )( )!H ⎝ ⎠

For large values of   t we get

( )( )(1 )11 t HH∫ ( )( )( )(1 )1(1 ) !0

1( , ) d!

HHHs t H t

Hτ τ−

−≅ =∫

This result is quite similar to our Brownian motion case with zero mean and zero variance.

Brownian motion can be thought as self similar random process

This is still developing concept

Differential geometry of fractional order‐’radius of curvature for fractional curves’ with modified RL fractional derivative

Curve is defined by  ( ) ( )y t f x=

Like for differentiable continuous paths, the angle       of the tangent line of the curve at point  θ ( ),x y

with horizontal axis by tangent                                              here in this framework we have( )( ) 1tan d dy xθ −=

1.

( )( )( ) ( )1 ( )d ! ( ) dy y x x ααα −= and ( ) ( )1 1d !(1 ) ! dx x x ααα α − −= −

Thus we have tangent in this case as

F l f h h( )( ) E α ( ) ( ) 1α

1 ( )dtan (1 ) ! ( )d

y x y xx

α αθ α −= = − 0 1α< <

tan θ ∞For example for                                           we have                                    we have   ( )( )y x E x αα= ( )

0( ) 1

xy xα

==

For example                                             with white‐noise of zero mean and variance           from this we ( )d ( ) ( ) dy t w t t α=2σ

this is not‐differentiable at zero

0tan

xθ

== ∞

p ( )( ) ( )ywrite                                                   and the tangent as( )( ) ( ) ! ( )y t w tα α=

We observe here as time grows i.e.                     the angle t ↑ ∞ 0θ ↓

The shows oscillations around zero but as time increases but on average it is zeroθ

( ) ( ) 1tan 1 ! ! ( )t w tαθ α α −= −

( ) ( )1 ( ) α ( ) ( )1 ( ) α

The          shows oscillations around zero, but as time increases, but on average it is zeroθ

Loosely speaking by the fact one has                                         with probability 99.7% , and qualitative speaking3 ( ) 3w tσ σ− < <

everything happens as if               were more or less constant( )w t

This is still developing concept

For parametric curves                             we have( )( ) , ( )x t y t ( ) ( )1 ( )d ! ( ) dy y t t ααα −= & ( ) ( )1 ( )d ! ( ) dx x t t ααα −=

Thus for this case tangent is 

2.( )

( )

d ( )ta nd ( )

y y tx x t

α

αθ = =

Fractional Laplace Transforms via‐Mittag‐Leffler Function

If               , has important role in Fractional Calculus then can we have integral transforms ( )E xαα p g( )α

like Laplace and Fourier with this Mittag‐Leffler function and develop suitable laws?

Then this type of new transforms will be apt for non‐differentiable  dynamics with fractional

diff ti l ti d f ti hi h f ti ll diff ti bldifferential equations, and functions which are fractionally differentiable

Classical Laplace Transform is  { }0

( ) ( ) : l i m ( ) dT s t

Tf t F s e f t t−

↑ ∞= = ∫L

Fractional Laplace Transform is                                                                                            { } ( ) ( )0

( ) ( ) : lim ( ) ( ) dT

Tf t F s E s t f t t αα α

α α α↑ ∞= = −∫L

We will have following operational identity for fractional Laplace transform

{ } ( )( ) ( )f D Fα αL { } ( )

{ } ( ) ( ){ } ( )

1

( ) ( )

( )

( ) ( ) ( )

s

sa a

d d

t f t D F s

f a t F

f t E s F s

α αα α

αα α

α αα α ατ τ

= −

=

− = −

L

L

LConvolution of  fractional order

t

∫( ){ }( ){ } ( )

{ }

1

0

( ( ) ( )

( ) d ! ( )t

E a t f t F s a

f u u s F s

α αα α α

α αα αα − −

− = +

=∫

L

L

( ) ( )

( ){ } ( )( )0

( ) * ( ) ( ) ( ) d

( ) * ( ) ( ) ( )

tf t g t f t g

f t g t F s G s

α

α

α α αα

τ τ τ= −

=

∫L

{ } ( )( ) ( ) ( ) ! ( 0 )f t s F s fα αα α α= −L

Very similar to classical Laplace identities

Simple example of Fractional Laplace Transform for Heaviside Step function

Standard classical Laplace for and for is( ) 1f t 0t ≥ ( ) 0f t 0tStandard classical Laplace                  for            and                for            is:  ( ) 1f t = 0t ≥ ( ) 0f t = 0t <

{ }0 0

0

1( ) ( ) ( )d d( )

stst st eF s f t e f t t e t

s s

∞−∞ ∞− −= = = = =−∫ ∫L

F ti l L l iFractional Laplace is

{ } ( ) ( )

( )( ) ( ) ( )( )0

1

( ) ( ) ( ) d

d d

F s f t E s t f t t

E s t t t E s

αα αα α α

α αα α α αα αα ξ ξ ξ

∞

∞ ∞ −

= = −

= − = − −

∫∫ ∫

L

( )( ) ( ) ( )( )

( ) ( )( )

( ) ( ) ( )( )

0 0

0

d d

( )

( )

t

E s t t t E s

D E s t

E s t

α α

α α αα

α αα

α ξ ξ ξ

α α

α α

∞−

∞

⎛ ⎞= Γ −⎜ ⎟⎝ ⎠⎛ ⎞− Γ⎜ ⎟Γ

∫ ∫

( ) ( )( )

( )( )0

( )

!

ss

s

ααα

α

α α

α

⎜ ⎟= Γ =⎜ ⎟−⎜ ⎟⎝ ⎠

=

This type of Fractional Laplace Table for useful functions does not exist

What about inverse Fractional Laplace Transform ?

Dirac’s delta distribution of fractional order and Inverse Fractional Laplace Transform

Like normal delta‐function with property we define fractional delta function( )xδ ( ) ( )d (0)f x x x fδ =∫Like normal delta function             with property                                     we define  fractional delta function( )xδ ( ) ( ) ( )f f∫( )( ) ( ) d (0)f x x x fα

αδ α=∫with property

Define the function                                                  then one has limit i.e.[ ]0, 0,( )

xx

εδ ε

⎧ ∉= ⎨ lim ( , ) ( )x xα αδ ε δ

↓=( , )

, 0x

xα αδ εε ε−

= ⎨< ≤⎩

0( , ) ( )α αε ↓

We note when            we recover classical Dirac’s delta function1α =

A l ti ith d l l d Mitt L ffl i it bl t f F ti l F i t fA relation with             and complex valued Mittag‐Leffler is suitable to form Fractional Fourier transform ( )xαδthat is (in conjugation to standard Fourier integral that is 

( ) ( ) ( )( ) d ( )M

E i x xαα

αααα αω ω δ

+∞

−∞− =∫

( )12 d ( )i xe xω

π ω δ+∞ −

−∞=∫

Here             is the ‘period’ of complex valued Mittag‐Leffler function                                like        forMα ( )( ) 1E i M αα α =

Inverse Fractional Laplace is

2π ( )2 1ie π =

p

( ) ( ) ( )1( ) ( ) di

M if t E s t F s sα

α

αα αα α

+ ∞

− ∞= ∫

Where we defined Fractional Laplace as

{ } ( ) ( )0

( ) ( ) : lim ( ) ( ) dT

Tf t F s E s t f t t αα α

α α α↑∞= = −∫L

Similarly developments in Fractional Fourier is there

Fractional order Gamma function

Classically we have                                                                             ( )( )( )( 1)( ) exp( ) dxx t t t∞ −Γ = −∫ integral representation of Euler Gamma

With Mittag‐Leffler function we get alpha‐order Gamma function as

( ) ( )( )( )1 ( 1)( ) ( ) dx αα α∞−

∫

( )( )( )0

( ) exp( ) dx t t tΓ ∫

( ) ( )( )( )1 ( 1)

0( ) ! ( ) dxx E t t t αα α

α αα −Γ = −∫

Some interesting conjugation with classical Gamma or Beta functions are following

( ) ( )( )( ) ( )

( 1 ) ! ( )

( 1 ) ! !n

x x x

n n

α αα

α

Γ + = Γ

Γ + ( ) ( )

( )1 ( 1 ) ( 1 )

0

( 1 ) ! !

B ( , ) (1 ) d

( ) ( )B ( )

x y

n n

x y t t t

x yx y

α

αα αα

α α

α− −

Γ + =

= −

Γ Γ=

∫B ( , )

( )x y

x yαα

=Γ +

These are very new developments and still developing

The inverse Mittag‐Leffler or Fractional Logarithmic function

In erse f nction of Mittag Leffler is

( ) ( )d ln ln 0xC

x xx E xx C

α

α α α⎛ ⎞= = >⎜ ⎟⎝ ⎠∫

Inverse function of Mittag‐Leffler is

( ) ( ) ( )( )( )( ) ( ) ( )

1 1 1

ln ln

ln ln ln

xyx y x E x y E y

uv u vα α α

α α α αα α α α

α α α

= =

= +

( )d

0

d ln ! 1d (1 ) !

1ln(1 ) !

x

xx x

x

αα

α α

αξα ξ

αα

α

=−

=− ∫

( )( )( )

0

d1(1 )! 0

1

(1 ) !

d 1

xx E

x x

αξα α ξ

α α

α

−

−

−

=

−=

∫

∫ ( )( )

( )( )( ) ( ) ( )( )

22

2

3

2

1 (1 ) !

!d 2

2 !x

x

E x αα α αα σ

α α

ασ π

α

=− −

⎛ ⎞⎜ ⎟− =⎜ ⎟⎝ ⎠

∫

∫ ( )⎝ ⎠

These are very new developments and still developing

Some of  the formulas of fractional order re‐listed with reference to modified RL fractional derivative

( ) ( )( )

( )1( )

D E x E x

D E x x E x

α α αα α

α α αα α

λ λ λ

λ λ α λ− −

=

=

( ) ( )( )( )

( )( ) ( ) ( )( )( ) ( ) ( )

1

0

1

( ) ( ) ( )

d ! 1

d ! d

x

D E u x E u x u x

E E x

αα α αα α

αα αα α

α α α

λ ξ ξ λ α λ−

−

′=

= −∫( ) ( ) ( )

( ) ( ) ( )( ) ( )

1

1 1

d ! d

! 1 ! d d

d (1 ) ! d ! d !

x x

x x x

x x x x x

α α

αα

αα α

α

α α

α α α

−

− −

=

− =

= − = =∫ ∫ ∫( ) ( )

( )( )

( ) ( ) d d ! ( )

c o s s i n

i

f x x f f x

E i x x i x

D D

αα α

α α α

α α α α

α= =

= +

∫ ∫ ∫∫ ∫

i α αc o s s i n sD x x Dα α α αα α= − i n c o s

( ) c o s h s i n h

s i n h c o s h s i n h c o s h

x xE x x x

D x x D x x

α αα α

α α α

α α α α α αα α α α

== +

= =

Formulas are applicable for modified RL Derivative

Continuous order system

Here we again come from part‐A where we introduced continuous order system

[ ]0 1 2

0 1 20 1 2 0 1d ( ) d ( ) d ( ) d ( )... ( ) , , ... ,

d d d d

m

mm mf t f t f t f ta a a a g t a b

t t t t

α αα α

α αα α α α α+ + + = ∈

( )K α

( )K k hα α= +

αa b

0 1a b< < <

d ( ) ( )nm f tα

∑ The summation is changed to∑ ∫

We will take continuous order distributed as linear function between  a and  b

⎛ ⎞

0

( ) ( )d nn

n

fa g tt α

=

=∑ The summation       is changed to          

We get Continuous Order Differential Equation 

∑ ∫

d ( )( ) d ( )d

b

a

f tK g tt

α

αα α⎛ ⎞

=⎜ ⎟⎝ ⎠

∫

Laplace Transforming the Continuous Order Differential equation

d ( )( ) d ( )d

b f tK g tt

α

αα α⎛ ⎞

=⎜ ⎟⎝ ⎠

∫To evaluate this type of FDE                                                                       we generalize this as: 

( 1 )

0

d ( ) ( ) ( )( ) d d d ( )d (1 ) ( )

b b tm m

ma a

f t K f uK u g tt t u

α

α α

αα α αα

+ +

+

⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎜ ⎟Γ − −⎝ ⎠⎝ ⎠ ⎝ ⎠

∫ ∫ ∫

da t α⎜ ⎟⎝ ⎠

∫

Used definition of FD ( 1 )

0

d ( ) 1 ( ) dd (1 ) ( )

tm m

m

f t f u ut t u

α

α αα

+ +

+ =Γ − −∫ ( ) ( 1)m m mα< + < +

m +∈ α ∈Caputo type used here

1 ( )( )d ( ) (0) ( )b m

m n m nK F f Gα α+ − −⎛ ⎞⎜ ⎟∑∫

Taking Laplace Transform

p yp

1 ( )

0

( )d ( ) (0) ( )m n m n

na

K s F s s s f G sα αα α + − −

=

⎛ ⎞− =⎜ ⎟

⎝ ⎠∑∫

For initial states zero the solution in Laplace domain is

( )( ) ( )( )

( )( ) dbm

a

G s G sF sss K s αα α

= =Φ∫

b( ) ( ) d

bm

as s s Kα α αΦ = ∫Where we define

First we need to evaluate                                        ( ) db

as Kα α α∫

Evaluating the integral                           first ( )db x

ay K x x∫

( ) ( ) lnd db bx x yk h k h+ +∫ ∫

we write                         and do thelnx x yy e= integration by partsLet us take                                    and ( )K x kx h= +

( ) ( )

( ) ( )( )

ln

d ( )ln lnd

ln

d d

d d d

x x y

a ab b kx hx y x y

xa a

bx y b

y kx h x kx h e x

kx h e x e x x

e

+

+ = +

= + −

∫ ∫∫ ∫ ∫

( ) ( )

( )

ln

ln

ln ln

( )dln

x yy b

eya

a

b bx y x y

ekx h k xy

kx h e k e

= + −

+ ⎛ ⎞= − ⎜ ⎟

∫

( )

ln ln ln

ln ln ln

aax b x bx x

y y y

kx h s k sy y y

= =

⎜ ⎟⎝ ⎠

+ ⎛ ⎞= − ⎜ ⎟

⎝ ⎠ x ax ay y y

== ⎝ ⎠

( )b a

Thus we have                                                                                                                 ( ) ( ) ( ) ( )ln

dln ln

b ay yb ab yx

a

kk b h y k a h yy k x h x

y y

−+ − +

+ = −∫

Frequency Response of Continuous order system

For a order distribution                                  in the interval                the transfer function

looks like by using the derived integral                                as following  

( )K k hα α= + [ , ]a b

( )db x

ay K x x∫

( ) ( )db

m

ab

s s s Kα α αΦ = ∫

( ) ( ) ( )ln

( )d

l

b a

m

a

k s sb asm

s s k h

kb h s ka h ss

α α α

−

= +

⎛ ⎞+ − + −⎜ ⎟=⎜ ⎟

∫

ln s⎜ ⎟⎝ ⎠

For a uniform order distribution case                         the transfer function is0; 1k h= =

( )( )b a

m s s−Φ

( )K α( )( )

l nms s

sΦ = 1.00

a b

The frequency response plot of  continuous order transfer function

O i i t h it d l t f th l t f f ti th t i ( )sΦ

( )( ) ( ) ( )( )

ln( )

m b ai i ii

iω ω ω

ωω

−Φ =Putting  we will gets iω=

Our aim is to have magnitude plot of the complex transfer function that is  ( )sΦ

(1) Magnitude of is:ln( )iω

( ) ( )22ln ( ) ln ln

ii e i

ππω ω ω⎡ ⎤= = +⎢ ⎥⎣ ⎦

giving ( ) ( )222l n l ns πω= +

(1) Magnitude of                 is:               ln( )iω

(2) Magnitude of is:( ) ( )b ai iω ω−

( )2( ) bb b bA s i πω ω= = = ∠ ( )2( ) aa a aB s i πω ω= = = ∠

( )2 2 2 cosA B A B A B A B− = + − ∠ − ∠

(2) Magnitude of                             is:               ( ) ( )i iω ω

Let and

Using vector formula

( )2 222 cos ( )b a a b b aπω ω ω ++ − −

( )2 22( ) ( ) 2 c o s ( )b a b a a bi i b aπω ω ω ω ω +− = + − −we get

From all above we write magnitude of                 is( )sΦ( )

( )2

222

2 cos ( )( ) ( )

(ln )mb a

iπ

ω ω ωω ω

ω

+Φ =

+

Solution of continuous order differential equation in frequency domain is thus

( )( )

222

2 22

(ln )( )( ) ( )

( ) 2 cos ( )m b a a b

G iF i G i

i b a

π

π

ωωω ω

ω ω ω ω ω +

+= =

Φ + − −

The time domain solution of continuous order differential equation

( )( )b a

m s ss s −Φ =With the transfer function as for continuous uniformly distributed

( ){ }1 1 1 1 ln( ) ln( ) { ( )} ( ) ( )G s sf F G G⎧ ⎫ ⎧ ⎫⎨ ⎬ ⎨ ⎬L L L L

( )l n sWith the transfer function as                                                 for continuous uniformly distributed  

order between                with unity strength for all, we write the following [ , ]a b

( ){ }{ } { } { }

1 1 1 1 ln( 1)

1 1 1ln 1 11

( )( ) { ( )} ( ) ( )( ) ( )

( )*

a b a m

a b a m

sb a m s s s

ss s s

f t F s G s G ss s s s

g t * *

−

−

− − − −−

− − −−

⎧ ⎫ ⎧ ⎫= = = =⎨ ⎬ ⎨ ⎬Φ −⎩ ⎭ ⎩ ⎭

=

L L L L

L L L

{ } 11 ln( ) ( )( )* * (1, )* m

as t

b a msg t F t −−

− Γ= L

{ } ( 1)1 n nqa t∞∑h k l d d{ }1 1m −

that is Robotnov Hartley function; a variant of Mittag‐Leffler function

{ } ( 1)0

1( , ) , ( , )( )

qq qq n

a tF a t F a t ts a nq q

∞−=

= =− Γ +∑LThe known Laplace identity are                         and  { }1 1

( )m

mt

m s

−

Γ =L

we land into problem with                       no standard tables give this.1 lna

ss

− ⎧ ⎫⎨ ⎬⎩ ⎭

L

Required to compute numerically here difficult one{ }1 ln s−LRequired to compute                               numerically here , difficult one.  { }1 lnas

sL

Regarding Laplace inverting the functions with ln (s)b b

We have                                                  , with   m = 0 we have ( )( )l n

b am s ss s

s−

Φ =( )( )

l n

b as sss

−Φ =

with                                     . This                does not converge when  0 1a b< < < ( )sΦ s ↑ ∞

We modify as                                                              Here  ( )( ) , 0 1

ln

b as ss a bs s

−Φ = < < < lim ( ) 0s s↑∞ Φ = therefore  ( )sΦ

is suited to for Laplace inversion. We note that             has pole at               and ( )sΦ 0s = 1s =p p( )sΦ

No pole in the “Left Half Plane” (LHP)

Also by applying L’Hospital rule we getd( ) 1 1

dli ( ) li lib as s b abs as b

− − −−ΦAlso by applying L Hospital rule we get  d

d( ln )1 1 1d

lim ( ) lim lim1 ln

ss ss s s

s

bs ass b as→ → →

Φ = = = −+

( )

2 1 1

2

2

d ( ) d ( ) 2 2d d

d(1 ln )d ( ln ) 10 0 0 0d

( 1) ( 1)lim ( ) lim lim lim 0b a b as s bs as b as s

ss ss s s s

b b s a a ss− −− − − −

+→ → → →

− − −Φ = = = =

( )2 dd s ss

that is the limit value of               at the poles are finite. ( )sΦ

Inverse Laplace of function with no roots in Left Half Plane

b as s ( )( )ln

s sss s

−Φ = 0 1a b< < < ( )( )( )( ) ( ) ( ) ( )s

sG s sF s s F sΦ ′= = Φ

Why ? So as to make at The inspection says that due to ln(s) there is no( ) 0sΦ → s = ∞

With                                                            we are doing

Why ? So as to make                    at             . The inspection says that due to ln(s) there is no possibility of roots in negative plane! The logarithm is also multi‐valued function,  so as the power functions of non‐integer orders; they have standard branch cut in complex plane that is 

( ) 0sΦ → s = ∞

( ,0]−∞We can use the Laplace inverse formula{ }1 1( ) lim ( )d

iT sts e s sσ +− ⎛ ⎞Φ = Φ⎜ ⎟∫L

I m ( )s

We can use                                                                                 the Laplace inverse formula{ }( ) lim ( )d2 iTT

s e s si σπ −↑ ∞

Φ = Φ⎜ ⎟⎝ ⎠∫L

BC

E D π∠ = + R e ( )sStandard Branch Cut in s‐plane

σD E

FH

E D π∠ = +

FH π∠ = − Integration contourA B C D E

−→ → → →

R e ( )s

AK

F H K A→ → → →

Apply residue theorem

B E H

∑∫ ∫ ∫ ∫

Since the integration on large arc BCD and HKA is zero at large radius, and as the radius goes to zero the integration on the small arc EF is zero, we are left with:

, , ,... ,

( )d ( )d ( )d ( )d 2 residue of polesst st st st

A B C K A A D F

e s s e s s e s s e s s iπΦ = Φ + Φ + Φ = ∑∫ ∫ ∫ ∫

Rearranging above and recognizing that contour does not enclose any poles we get

1 1 1( )d ( )d ( )d2 2 2

1 ( )d residues of poles 0

i E Hst st st

i D F

st

e s s e s s e s si i i

e s s

σ

σπ π π

+ ∞

− ∞

Φ + Φ + Φ

Φ

∫ ∫ ∫

∑∫ ( )d residues of poles 02

e s siπ

= Φ = =∑∫

From above we have Laplace inverse as

{ }1 1( ) ( )d2

ist

i

s e s si

σ

σπ

+ ∞−

− ∞

Φ = Φ∫L

1 ( )d ( )d2

E Hst st

D F

e s s e s siπ

⎛ ⎞= − Φ + Φ⎜ ⎟

⎝ ⎠∫ ∫

The result in form of Bestimmte Integral 

Use              on                we have                       and                     , that gives the followingis re θ= θ π= ± s r= − d ds r= −

{ }1

0

1( ) ( )d ( )d2

1

E Hst st

D F

s e s s e s siπ

−

∞

⎛ ⎞Φ = − Φ + Φ⎜ ⎟

⎝ ⎠⎛ ⎞

∫ ∫

∫ ∫

L

g g

( )

0

1 ( , )d ( , )d2

1 ( , ) ( , d2

rt rt

rt

e r r e r ri

e r r ri

θ π θ ππ

θ π θ ππ

− −

∞

∞−

⎛ ⎞= − − Φ = − Φ = −⎜ ⎟

⎝ ⎠⎛ ⎞

= Φ = − Φ = −⎜ ⎟⎝ ⎠

∫ ∫

∫02 iπ ⎝ ⎠∫

In polar form the TF is                                                               , using this in above and doing lot of algebra we obtain Laplace inverse of continuous order transfer function in form of 

( )( )

( 1) ( 1)

( , )ln

b ib a ia ir e r e er

r i

θ θ θ

θθ

− − −−Φ =

+g p

“Bestimmte integral”; the Green’s function is thus obtained for getting the time domain solution (by changing          to          ), the integral form is followingr x

That is { } ( )( )( ) ( )( )( )1 11

2 20

ln sin( ) cos( ) ln sin( ) cos( )1( ) d(ln )

b axt x x b b x x a a

s e xx

π π π π π ππ π

− −∞− −

− + + −Φ =

+∫L

Numerical evaluation required !!Numerical evaluation required !!

Int. J. of Appl. Mathematics and Statistics Vol. 29 Issue 5 pp 6‐16, 2012

The inverse Laplace of Low Pass Filter of Continuous Order

b as s

ln( ) b a

sss s

Ω =−

As we did analysis for High Pass Filter                            or in that case for  ( )ln

s sss

−Φ = ( )sΦ

Consider                                     , surely it converges at            , that is                    , this is got via s ↑ ∞ lim ( ) 0s

s↑∞

Ω =

L’H it l l H th i t l t W l h E if1 ( ) 1li ( ) b −ΩL’Hospital rule. Here there exists a pole at             . We also have                                    . Even if   1s = ( )

1lim ( )s

s b a→

Ω = −

there is pole at                  that is getting excluded due to “Branch cut contour”.  0s =

No pole in the “Left Half Plane” (LHP)

We can use                                                                                 the Laplace inverse formula{ }1 1( ) lim ( )d2

iT st

iTTs e s s

iσ

σπ+−

−↑ ∞

⎛ ⎞Φ = Φ⎜ ⎟⎝ ⎠∫L

No pole in the  Left Half Plane  (LHP)

and write the result as                                                                                                      { }( ) ( )( )( )( ) ( )

12 2

0

cos cos ln sin sin1( ) dcos sin sin sin

b a b a

xt

b a b a

x b x a x x b x as e x

x b x a x b x a

π π π π π

π π π π π

∞− −

− − −Ω =

− + −∫L

Further advancement required

Int. J. of Appl. Mathematics and Statistics Vol. 29 Issue 5 pp 6‐16, 2012

Closing comments

The fractional calculus is generalized differ‐integrationThe fractional calculus is generalized differ integration

Fractional calculus is as rigorous as conventional classical calculus

There are paradoxes some believe and some carry on with those paradoxesp y p

These fractional differ‐integrals have inherent memory

There are evidences in nature about fractional order dynamics

The dynamic behavior changes from classical differential equations to fractionaldifferential equations with coarse grained infinitesimal element

h d d l d h f l d ffThere are circuits and systems developed with fractional differ‐integrations

Still there exist scope to develop the fractional calculus in conjugation to classical one

Thus we today are far from ‘once considered a ‘paradoxical ‘situationThus we today are far from  once considered a  paradoxical  situation.

Fractional calculus is a reality

Still several miles to go to understandStill several miles to go to understand which 

fractional calculus does the Nature obeys

……….yet a developing subject………