ASET colloquium TIFR Mumbai May‐20, 2016
Fractional Calculus from mathematician’s paradox to application in physics and paradox to application in physics and engineering Shantanu Das Scientist, RCSDS E&I Group BARC Mumbai, Senior Research Professor Dept. of Phys, Jadavpur Univ (JU), Adjunct Professor DIAT‐Pune dj f UGC‐Visiting Fellow Dept. of Appl. Math Calcutta Univ.
[email protected] http://scholar.google.co.uk/citations?user=9ix9YS8AAAAJ&hl=en p // g g / www.shantanudaslecture.com
From Laplace operator the fractional derivative Classically we have
L { f ′(t )} = sF ( s ) − f (0)
where
L { f (t )} = F ( s )
0 < α < 1 Say for case of RHS if we have for s α F ( s ) − f (0)
Then we relate to as th Then we relate to L { f ( t )} = s F ( s ) − f (0 ) as α − th derivative of derivative of f ( t ) (α )
α
α negative we have order integration i.e. with α− α = −β For ‐ L
{f
(− β )
}
(t ) = s − β F ( s )
So there is possibility that we have in between operations for integration & differentiation, 1 − 1 11 like half one third etc D 1 2 f ( t ) like half, one third etc D 3 f (t ) D 2 f (t ) D 2 f (t ) We are having thus fractional Laplace variables like etc s 2 s− 2 1
1
Assume presently that we have fractional differentiation and integration then how can we use the corresponding fractional Laplace variable ? sα
CFE for approximation of fractional semi differential Laplace operator CFE is Continued Fraction Expansion CFE is defined as following 1
def
(1 + x ) α =
x
1−α 1+
( 12 ) ( α
x
+ 1) 1−
( 16 ) ( α
x
− 1) 1+
( 16 ) ( α
x
+ 2) 1−
( 110 ) ( α
− 2)
x 1 + ....
x = s −1 α = For obtaining rational approximation of put in CFE and s CFE for four number term approximation for s is CFE for four number term approximation for is
1 2
5s2 + 10s + 1 s ≈ s2 + 10s + 5
Here we have got approximation for half‐differentiation in Laplace domain‐is implementable
We can have a transfer function of a filter to realize fractional Laplace variable
CFE for approximation of fractional semi differential Laplace Operator for various number of terms S.No
No. of terms in CFE Rational approximation for approximation
1
2
2
4
3
6
4
8
s
3s + 1 s + 3 5s2 + 10s + 1 s2 + 10s + 5 7 s 3 + 3 5 s 2 + 2 1s + 1 s 3 + 2 1s 2 + 3 5 s + 7
1 1s 5 + 1 6 5 s 4 + 4 6 2 s 3 + 3 3 0 s 2 + 5 5 s + 1 s5 + 55s4 + 330s3 + 462s2 + 165s + 11
CFE gives approximation for fractional Laplace operator in terms of ratio of rational polynomials CFE gives approximation for fractional Laplace operator in terms of ratio of rational polynomials. The above is semi‐differentiation operator. The semi‐integration operator will be reciprocal of the above ratio.
Makes way to have fractional order analog and digital circuits & systems
We make use of the fractional Laplace variables to make controller C P ID ( s ) = k
p
+
ki + kd s s
C F O P ID ( s ) = k p +
α > 0
u PID (t ) = k p ε (t ) + k i D −1ε (t ) + k d D 1ε (t )
ki + kd sβ α s
u F O P ID ( t ) = k p ε ( t ) + k i D
−α
ε (t ) + k d D β ε (t )
β > 0
v
r
y sp
+−
ε
Fractional Controller
u
+
+
C (s)
Plant
r : Load disturbance
ε : u :
v:
Controller output
y
G (s)
y s p : Set-point Error
+
+
y:
Measurement Noise Out-put
PIα D β We have a fractional order PID (FO‐PID)system called as
Observe robust control i d iso‐damping i We observe for uncertain plant G W b f t i l t G (s) () Plant with gain uncertainty
G (s) =
k s
2
+ 2ξ ω 0 s + s
2
k ∈ [0 . 2 , 5 . 0 ]
y (t )
Enhanced robustness
t
Functional Fractional Calculus
Digital FO‐PID for DC Motor Speed Control hardware
Circuit Systems y & Process Springer p g DOI 10.1007/s00034-016-0262-2:2016
Courtesy: BRNS funded joint project of VNIT Nagpur and BARC to develop industrial digital fractional order PID.
Controller output u ( t ) for PID and Fractional PID
r(t)
Controller C t ll PID / FO-PID
+ -
u(t)
Plant Pl Mag-Lev
c(t)
ki + kd s s k C F O P ID ( s ) = k p + αi + k d s β s α >0 β >0 C P ID ( s ) = k p +
r ( t ) : set point sinusoidal c(t):p position of the levitated ball u ( t ) : out‐put of controller
We note that Mag‐Lev is inherent unstable unit
Control effort in case of PID control
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.
Control effort in case of Fractional Order PID Controls
Voltage to coil u ( t ) Ball position ll ii
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 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 first case is fluctuating severely. The reason is that fractional derivatives and fractional integrals are having . The reason is that fractional derivatives and fractional integrals are having inherent memory 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 first case the maneuvering signal is going very high instantaneously and in the next moment going very low again and again‐lot high instantaneously and in the next moment going very low again and again lot of effort? of effort?
So we can see fractional calculus based system does the control action with a lesser effort than the conventional classical calculus based controllers, therefore are better efficient.
Can we say Fuel Efficient Controls?
Liouville’s Postulation In parallel to classical approach Joseph Liouville postulated exponential approach d α e ax α ax = a e ; α dx
α ∈
α p Negative values of represent integrations (anti‐derivative) and we can even extend this to g g ( ) allow complex values of or even to a continuous distribution of this order in some interval. α dα f ( x) dxα
d −β f ( x) dx − β
d α + iβ f ( x ) d xα + iβ
∑
αn
n
an D x f ( x )
∫
b
a
⎛ dα f ( x) ⎞ ⎜ K (α ) ⎟d α α d x ⎝ ⎠
In a way this generalizes notion of integration and differentiation to arbitrary order!
Liouvelli’s class of functions and approach Any function expressible as a sum of exponential functions can then be differentiated in the same way.
d α cos( x ) d α ⎛ e ix + e − ix ⎞ ( i ) α e ix + ( − i ) α e − ix = ⎟= α α ⎜ dx dx ⎝ 2 2 ⎠
( ) ( ( ) = e
iα
π
e ix + e
2
− iα
2 = cos ( x + α
Since ( ± i )
α
(
= e
± iπ
2
)
α
=e
± iα
π
2
π
2
π
2
) (e ) = e ( − ix
i x +α
)
we have the nice result
π
2
)
+e 2
− i ( x +α
π
2
)
d α cos( x ) = cos ( x + α dxα
Is it so simple ?
π
2
)
Fractional order differentiator and integrator circuit realized α
d cos( x ) Liouville’s exponential approach allows us to simply write = c o s ( x + α π2 ) α dx
Thus the generalized differential operator simply shifts the phase of the cosine function and likewise the sine function by α × (90) 0 that is in proportion to the order of the differentiation. 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.
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.
Generalized Factorial to Gamma Function Factorial is for positive Integer number is 1 ⎧ n!= ⎨ ⎩ ( n − 1) !( n )
n = 0
Γ(x)
n > 0
Euler’ss Gamma function Euler Gamma function Γ (α ) =
∞
∫e 0
Γ (α + 1 ) = α
−t
tα
−1
dt
R e [α
]>
0
(Γ (α ) )
This gives analytic continuity to negative axis We write thus
α ! = Γ(α + 1)
Plot of Gamma‐Function x
Some values are following
( 32 )! = Γ ( 52 ) = 34 π ( − 52 )! = Γ ( − 23 ) = 34 π ( 12 )! = Γ ( 32 ) = 12 π ( − 12 )! = Γ ( 12 ) = π ( − 32 )! = Γ ( − 12 ) = −2 π
Plot of reciprocal of Gamma Function
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 x m are dx m = m x m −1 ; dx
d2xm m−2 = − m ( m 1) x ; 2 dx
d3xm m −3 = − − m ( m 1)( m 2 ) x dx3
xm Thus we expect to find that the general form of the n‐th derivative of is dnxm m ! x = dxn (m − n )!
m − n
α 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 xm dα x m m! Γ ( m + 1) = x m −α = x m −α α Γ ( m − α + 1) dx ( m − α )! x d1/ 2 x 1! 1 1− 12 x = = = 2 dx1/ 2 Γ (1 − 12 + 1) Γ ( 32 ) π −1
d 1/ 2 [1] 0! x 2 0 − 12 = = = x Γ ( 0 − 12 + 1) Γ ( 12 ) d x 1/ 2
Euler’s formula
1
πx
Half derivative of constant as non‐zero !
Use of Euler’s formula f ( x ) = ∑ k ak x k 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 α functions. Furthermore, applying this formula with negative values of gives a plausible expression for the fractional‐ i f th f ti l integral of a power of . For example, to find the whole i t l f f x F l t fi d th h l 3 x m=3 integral of we set and then compute via Euler formula as dα xm Γ ( m + 1) x m −α = α dx Γ ( m − α + 1) d −1 x 3 3! 3! 6 1 4 3 − ( −1) 4 4 = x = x = x = x d x −1 Γ (5 ) 24 4 Γ (3 − ( − 1 ) + 1 )
Note 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 generalized differentiation is non‐local, just as is integration‐has memory history hereditary
d ±α x m Γ ( m + 1) = x ±α dx Γ ( m ∓ α + 1)
m ∓α
Fractional Derivative of Exponential Function is not Exponential !! e a e α ‐ th derivative of is simply , and yet We previously proposed that the general ex if we expand the exponential function into a power series a a 2 a 3 a x 2 e = 1 + x + x + x 3 + ... 1 ! 2 ! 3 ! and apply Euler formula to determine the half‐derivative, term by term, we get (not at all what we postulated earlier!) , that is following Liouville’s postulate α
ax
d 1/2 (e x ) = d x1/2
ax
4 2 8 16 ⎛ ⎞ 3 4 1 2 x x x x . . . + + + + + ⎜ ⎟ 3 15 105 π x ⎝ ⎠ 1
Here we have D e x ≠ e x Here we have 1
8
½ Derivative of e
6
2
x
Was Liouville wrong or is there a contradiction? 4
2
ex 0
0 .55
1 .0 0
1 .5 5
2 .0 0
x
Kindergarten of Fractional Calculus
Most fundamental approach TTo get a clearer idea of the ambiguity in the concept of a generalized derivative, it’s useful to t l id f th bi it i th t f li d d i ti it’ f lt 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 derivative of a function d f (x) f (x) − f (x − ε) = li m ε↓ 0 dx ε
Repeating n ‐times of this operation leads to a binomial series of following type ⎛n⎞ ( − 1) j ⎜ ⎟ f ( x − j ε ) j=0 ⎝ j⎠ n Cj = 0 j>n Note that thus summation above ends at n d n f (x) 1 = li m ε↓ 0 ε n dxn
n
∑
Generalizing the binomial coefficients to real numbers we get a formula d
α
f (x) 1 = l i m ε↓ 0 ε α dxα
⎢ x − x0 ⎥ ⎢ ⎥ ε ⎣ ⎦
∑
j= 0
( − 1)
j
Γ (α + 1 ) f ( x − jε ) j ! Γ (α + 1 − j )
Kindergarten of Fractional Calculus
Non‐locality of Generalized differ‐integration α
d f (x) 1 = lim ε↓0 εα dxα
⎢ x − x0 ⎥ ⎢ ε ⎥ ⎣ ⎦
∑
j=0
( − 1) j
Γ (α + 1) f ( x − jε ) j ! Γ (α + 1 − j )
Thus, the generalized derivative is a non‐local operation, just as is integration. jε) This can be seen from the factor in the summation formula, showing that as j This can be seen from the factor f ( x − jε in the summation formula, showing that as j ( x − x0 ) / ε ranges from zero to the argument of f ranges from x down to zero (or the start point of origination of function).
The fractional differ‐integration is having memory f ( x)
ε x x0
x
x − 2ε x −ε Kindergarten of Fractional Calculus
Fractional Derivatives Require Lower & Upper Limit like Integration ! Consider the two anti‐differentiations (integrations) shown below x
∫
x0
x
x 04 x4 u du = − 4 4
∫
3
e u d u = e x − e x0
x0 3
x The first integral shows that when we say is the derivative of we are implicitly x4 / 4 assuming x 0 = 0 which is consistent with our derivation of equation assuming , which is consistent with our derivation of equation . dα f ( x) 1 = lim ε↓ 0 εα dxα
⎢⎣ x / ε ⎥⎦
∑
( − 1) j
j=0
Γ (α + 1) f ( x − jε ) j ! Γ (α + 1 − j )
x0 = 0
ex ex However, the second integral shows that, by saying is the derivative of , we are implicitly x0 = −∞. assuming dα f ( x) 1 = lim ε↓ 0 ε α dxα
∞
∑ ( − 1) j=0
j
Γ (α + 1) f ( x − jε ) j ! Γ (α + 1 − j )
We represent the operation with lower and higher terminals as: a
Dαx f ( x)
0
Dαx f ( x)
For any arbitrary α ∈
−∞
Dαx f ( x)
x
Dαb f ( x)
x
Dα∞ f ( x)
Kindergarten of Fractional Calculus
Repeated integration approach d −2 f ( x ) = d x −2
x u1
∫ ∫ f (u )d u d u 1
2
x
1
0 0
d −nf ( x) = dx −n
= ∫ ( x − u )f ( u )d u ; 0
u n −1
x u1 u 2
∫ ∫ ∫ ............. ∫ 0 0
0
0
d −3 f ( x ) = d x −3
x u1 u 2
∫∫∫ 0 0 0
x
1 f ( u1 )d u 3 d u 2 d u1 = ( x − u ) 2 f ( u )d u ∫ 2! 0 x
1 f ( u 1 ) d u n d u n − 1 .....d d u1 = ( x − u ) n −1 f ( u ) d u ∫ ( n − 1) ! 0
n
1 = Γ (n)
x
∫ (x − u)
n −1
f (u )d u
0
Thus we have for repeated integrals, which is Cauchy’s expression x u1 u 2
u n −1
0 0 0
0
x
1 f ( u1 )d u n d u n −1 .....d d u1 = ( x − u ) n −1 f ( u )d u ∫ ( n − 1) ! 0
∫ ∫ ∫ ............. ∫ n
α which we can express using the gamma function instead of factorials, for fractional order as d −α f ( x ) = d x −α
0
D
−α x
1 f (x) Γ (α )
x
∫ (x − u)
α −1
f (u )d u
This is Riemann fractional integral formula.
0
Kindergarten of Fractional Calculus
Illustration of semi‐integration by Riemann formula α =1/ 2 x Now let us do semi integration of , take then and apply Riemann formula
0
D
−1/ 2 x
x= =
x
1
u
Γ ( 12 ) ∫0 ( x − u)( − 2 )+1 1
x
1
Γ( ) ∫ 1 2
0
du
udu u ( x − u) = = =
1
1
x
0
xu − u
u = 0; θ = −π / 2
u = x ; θ = +π / 2
0
2
D
−1/ 2 x
+π / 2
⎡ x⎤ = 1 ⎣ ⎦ Γ ( 1 ) ∫ dθ 2 −π / 2
x2 4
du
− u 2 − x4 + 2 x ( u2 ) 2
udu 2
x 4
− (u −
( θ ) ( x + x2sin θ ) ( 2x ) (cos x2 4
−
x2 4
sin 2 θ
x x + x sin θ ( ) 2 ) (cos θ ) ( 2 = d θ Γ ( 12 ) −π∫/ 2 ( 2x ) 1 − sin 2 θ
udu
x
Γ( ) ∫ 1 2
u
x
Γ ( 12 ) ∫0 1
du = ( 2x ) (cos θ )dθ
0
Γ( ) ∫ 1 2
x + x sin θ 2
u=
Put
)
+π / 2
1
+π / 2
dθ ( x + x2sin θ ) =
=
Γ ( 12 ) −π∫/ 2
=
1 ⎛πx⎞ ⎜ ⎟ Γ ( 12 ) ⎝ 2 ⎠ =
x 2 2
1
π 2
1
Γ ( 12 )
⎡⎣ x2θ −
x cos θ 2
θ =+ π / 2
⎤⎦θ =−π / 2
x
Rigorous indeed d −α x m Γ ( m + 1) x = −α dx Γ ( m + α + 1)
we get the same answer as from Euler formula
0
D
−1/ 2 x
⎡⎣ x
1/2
⎤⎦ = Γ
Γ
(
( 12 1 2
+
+ 1) 1 2
+ 1)
x
1 2
+
1 2
=
Γ
( 32 ) x
Γ (2)
= Γ
( 32 ) x
=
m +α
π 2
x
Riemann‐Liouvelli and Caputo Fractional Derivative d −α f ( x ) 1 −α = = D f ( x ) ( x − u ) α −1 f ( u ) d u Riemann fractional integral formula is x 0 −α ∫ dx Γ (α ) 0 x
There are two different ways in which this formula might be applied. For example, if we wish to μ = 7/3 d7/3f(x)/ dx7/3 find the (7/3)‐rd ( )derivative of a function (i.e. ), we could begin by μ g ( g g y jjust greater than g differentiating the function three whole times (taking nearest integer say m α = ( m − μ ) = ( 3 − ( 73 ) ) = 32 that is ), and then apply the above formula with m=3 α = 2/3 to “deduct” two thirds i.e. of a anti‐differentiation C a
D 7x / 3 [ f ( x ) ] = a D −x 2 / 3 ⎣⎡ f
C a
D μx [ f ( x ) ] = a D −x α ⎡⎣ f
(m)
( 3)
( x ) ⎦⎤
( x ) ⎤⎦
α = 3 − ( 73 )
Caputo 1967
α =m−μ
By Caputo we get fractional derivative of constant as zero but f needs be differentiable By Caputo we get fractional derivative of constant as zero‐but f needs be differentiable α = 2/3 Alternatively we could begin by applying the above formula with and then differentiate the resulting function three whole times ( ). m = 3
d3 ⎡ D −x 2 / 3 f ( x ) ⎤⎦ D [ f ( x)] = 3 ⎣a dx dm μ RL ⎡ D −x α f ( x ) ⎤⎦ a D x [ f ( x)] = m ⎣a dx
RL a
7/3 x
α = 3 − ( 73 )
Riemann‐Liouville 1872
α =m−μ Kindergarten of Fractional Calculus
Integral Representation of RL and Caputo Derivative Caputo Derivative f
(α )
( x) = D
C a
− (1− α )
(D
D αx f ( x ) =
(1)
f ( x) )
1 Γ (1 − α )
0
