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attfact vtcavftfdtfD v tc v tc allfor0)(and, ,0),,,,,()()( (5) where c a v dft v tcavf )()( )( 1 ),,,,( 1 , as defined in Eq. (4) The generalized fractional derivative, for q and p real is defined by ,,)()( cttfDDtfD p t c m tc q t c (6) where, m is an integer such that mqm 1 , and pmq . Furthermore, 0q and 0 ct . In terms of the conventional notation, ),,,,(),,,,()()( tcamhtcapf dt d tfd dt d tfD m m p t c m m q t c (7) where, ,ct and )(tfDh p t a . It is of course clear that tf may be considered to be a composite function, for example a function different than tf t remains the that for terminal initialization of the integer derivative, ,,0),,,,( cttcamh (8) and the definition in Eq. (4) is applied for ),,,,( tcapf in Eq. (7). The next section considers the extention to the Caputo fractional derivative. 3 Initialization of the Caputo Fractional Derivative The Caputo fractional derivative was introduced [8,9] to alleviate some of the difficulties associated with the RL approach to fractional differential equations when applied to the solution of physical problems and is defined by [8]: )1()()( )( 1 )( 1 t a mm t C a mmdft m tfd (9) may be used for the history period, i.e., a t c, while f function to be fractionally differintegrated, i.e., t c . It has also been shown [7] THE CAPUTO FRACTIONAL DERIVATIVE 29 30 As is well known, in the solution of fractional differential equations, the initial conditions are specified in terms of fractional derivatives in the RL approach, but, in terms of integer order derivatives with known physical interpretations in the Caputo approach [10]. In view of the popularity of the Caputo formulation in applications of physical interest, the key question to be asked is: when viewed from the LH general initialization perspective, what “history” is inferred [11,12] for the Caputo derivative? 4 Relation Between the Initialized LH and Caputo Fractional Derivatives As has been noted, the generalized initialization as applied to RL fractional 4.1 We first consider the case when 1,10 m , then )()( )1( 0 1 00 tfDDtfD t tt ,0,),0,,1,()( )1( 0 ttahtfD dt d t (10) Noting that the initialization for the integer order derivative is zero 0,0,0,),1(,)( )1( 00 ttaftfd dt d tfD t t (11) Substituting explicitly for the quantities in curly brackets in Eq. (11) yields Achar, Lorenzo, and Hartley derivative, according to LH is given by Eq. (7) and will be used in the following Simple cases : 0 1 and 1 2 i .e., examples, where for convenience, is used in the place of q, i.e., m p 0 , m is a positive integer, and as before, for terminal initialization, (h, m, a, c, t) 0 . Hereafter t c corresponds to t 0 . Three cases will be considered below. 31 0 0 0 )()( )1( 1 )()( )1( 1 a t t dft dt d dft dt d tfD (12) Recasting the convolution integral by interchanging the arguments and carrying out the differentiation of the integral using Leibnitz’ rule, yields .0,)()( )1( 1 )0( )1( )( )1( 1 0 0 0 tdft dt d f t dtftfD a t t (13) Rewriting the argument of the convolution integral as ft and using the definition of the Caputo derivative, Eq. (9) with 1m and 10 , one can write the following expression relating the Caputo derivative to the initialized LH derivative for 10 : .0,),0,),1(,( )1( )0( )()( 0 0 ttaf dt d ft tfdtfD t C t (14) where the last integral in Eq. (13) is restated as an LH initialization. For the case 21 , 2m and the initialized LH derivative given by 0,0,0,),2(,)( )2( 00 ttaftfd dt d dt d tfD t t (15) yields on substituting explicit expressions for the quantities in the curly brackets ),0,),2(,()()( )2( 1 2 2 0 1 0 taf dt d dft dt d dt d tfD t t (16) THE CAPUTO FRACTIONAL DERIVATIVE 32 Recasting the convolution integral in Eq. (16) by interchanging the arguments and carrying out the differentiation of the integral using Leibnitz’ rule yields the expression relating the Caputo derivative to the initialized LH derivative for the case 21 as [11]: .0),,0,),2(,( )1( )0( )2( )0( )( 2 2 1 0 0 ttaf dt d ftft tfdtfD t C t (17) The expressions in Eq. (14) and Eq. (17) can be generalized as shown below. Generalizing to the case when mm 1 we get 0,),0,),(,()()( )( 00 ttamf dt d tfd dt d tfD m m m t m m t , (18) ,),0,),(,( )1( )0( )()( )( 1 )( 1 0 0 1 0 tamf dt d k ft dft m tfD m m m k kk t mm t (19) or, .1,0,),0,),(,( )1( )0( )( 1 0 0 0 mmttamf dt d k ft tfdtfD m m m k kk t C t (20) Equation (20) expresses the LH order derivative )( 0 tfD t in terms of the order Caputo derivative and additional terms. The additional terms consist of a polynomial in t with coefficients given by the values of the function t k , all evaluated at 0t , and the LH initialization for a fractional derivative under the assumption of terminal initialization. The polynomial contains a term ( 0k term), which is singular at range 10 , Eq. (20) simplifies to the Eq. (14), and for the range 21 , Achar, Lorenzo, and Hartley 4.2 General case m 1 m f (t) and its integer-order derivatives f t 0 for 0 . The details of the derivation can be found in ref. [11]. For the 33 5 Inferred History of the Caputo Derivative It is important to determine the “history” inferred by use of the Caputo derivative of a function tf . This can be achieved by setting the Caputo derivative equal to the LH fractional derivative of the same order , and for the same function tf , for 0t . It follows from Eq. (14) that the two derivatives will be equal for 10 if )1( )0( ),0,),1(,( ft taf dt d t > 0. (21) For clarity of presentation we will call the initialization function, yet to be determined, 0for )( 1 tatf , to differentiate it from 0f side (RHS). For the terminal initialization considered in this note, it follows that “history” would be given by a function 1 f , satisfying the following equation: .0, )1( )0( )()( )1( 1 0 1 t ft dft dt d a (22) “ a ”. To determine the inferred history of the Caputo derivative we require a general representation for 1 f . We will consider continuous functions and assume that 1 f 0 1 i i i bf . Specifically, the Maclaurin series, or on the right-hand It is important to note that the left-hand side (LHS) of Eq. (22), which is the instant prior to t 0 . Specifically, RHS of Eq. (22) is not a function of THE CAPUTO FRACTIONAL DERIVATIVE it reduces to Eq. (17). Expressions in Eq. (14), Eq. (17), and Eq. (20) will now be used to determine the history inferred by the use of the Caputo derivative. 5.1 Simple cases : 0 1 and 1 2 required initialization, is only related to the value of the function evaluated at t 0 , on the RHS, and not to the function or its derivatives at any may be represented by a polynomial in , that is 34 0 1 1 .0, )1( )0( )( n n n n f f (23) will be used as the desired representation. Substituting Eq. (23) into the yields .0),0()( 1 )0( 0 0 1 tftdt n f dt d a n n n (24) this result, we obtain 0),0( ! !1 ! ! )0( 0 1 1 1 1 1 1 1 1 tft j tn in n j aat n f dt d n n i n j n i j in i n (25) Differentiating with respect to t gives .0),0( 1 !1 1 )0( 0 1 1 1 11 1 1 1 tft j tn in j aati f n n i n j n i j in i n (26) the RHS, (the summation of the higher power terms, n t , cannot sum to a t term), and because all derivatives 1,00 1 nf n we have Achar, Lorenzo, and Hartley integrand of Eq. (22) and interchanging the order of integration and summation It is clear that only the n = 0 case on the LHS can match the exponent of t on A general solution for the definite integral can be derived [11] and substituting 35 .0,0 1 0 1 tftt at f (27) Because the starting point of the initialization “a” does not occur on the RHS of Eq. (26), we must have a , to force the first term of Eq. (27) to zero. Therefore, for 10 we have 0()0( 1 ff 0,0 1 af . (28) Therefore, the only history that can make the Caputo derivative the same as the LH derivative, and that is tacitly assumed when evaluating a Caputo fractional derivative, for 10 is the “constant” function of time, that is 0for,0constantt 1 tafftf . (29) The above arguments can be extended to the case when 21 as outlined below. It follows from Eq. (17) that the Caputo derivative and the LH fractional derivative of the same order and the same function )(tf would be equal to each other if .0, )1( )0( )2( )0( )()( )2( 1 1 0 1 1 2 2 t ftft dft dt d a (30) .0),0()1()0()( 1 )0( 1 0 0 1 2 2 tftftdt n f dt d a n n n (31) Substituting the result of integration and performing the differentiation operation yields ) , and from Eq. (23), f THE CAPUTO FRACTIONAL DERIVATIVE Substituting as before from the McLaurin expansion in Eq. (23), and inter- changing the order of integration and summation yields 36 .0),0()1()0( 1 1)2(! )0( !1 ! 1 )1( )0( 1 1 1 1 1 0 1 0 1 1 1 1 1 1 tftft j tnnn f in n j aatii f n i n j n n n n n i i j in i n (32) It is clear that only the 0n and match those of the terms on the RHS. It is required therefore that 0)0( )( 1 n f the RHS and because 21 we must have a . Thus we must have 0),0()1()0( )3)(2( )2)(3)(0( )2( )1)(2)(0( )1(1 1 )1( 11 tftft t f t f (33) Therefore we must have )0()0(),0()0( )1( 1 )1( 1 1 ffff . It follows from Eq. (23) that tfftf )0()0()( 0 t , (34) for the case of 21 . Both the results of Eq. (29) and Eq. (34) can be obtained as a special case of the more general result derived below. n 1 cases can allow exponents of t that will for all n 2 . Because the starting point of the initialization a does not occur on Achar, Lorenzo, and Hartley 37 In this case [11] setting the two derivatives in Eq. (20) to be equal yields .1,0, )1( )0( ),0,),(,( 1 0 mmt k ft tamf dt d m k kk m m (35) Again for clarity of presentation we set 1 ff in the initialization function. Then under the assumption of terminal initialization, this becomes .1,0 , )1( )0( 1 1 0 0 1 1 mmt k ft dft m dt d m k kk a m m m (36) Again representing 1 f as a continuous function by Eq. (23), gives .1,0, )1( )0( 1 0 1 1 0 0 0 1 1 mmt k ft d n f t dt d m m k kk a n n n m m m (37) Interchanging the order of integration and summation, and substituting the result for the definite integral, and noting that the maximum value of the exponent of t on the RHS is 1m coefficients, that is 1nfor ,00 1 mf n , yields .1,0 , )1( )0( 1 0 1 1 0 1 0 1 1 1 1 mmt k ft jm jntf m m k kk m n n j m j n n (38) In general, the equality will only hold when .1,,1,0,00 1 miff i i (39) 5.2 General case: m1 m THE CAPUTO FRACTIONAL DERIVATIVE , and hence, terms on the LHS, (after the mth order differentiation,) with exponents greater than this must have zero 38 Placing these results into Eq. (23) we have, for mm 1 1 0 0, )1( )0( )( m n n n tat n f tf , (40) as the only allowable initialization that will make the Caputo derivative equal to the LH derivative. Thus in general, we find for mm 1 , that the Caputo derivative infers a history in the form of a polynomial in t back to with maximum order of 1m . The coefficients of the polynomial are related to the + derivatives of tf 5.3 Example 2 2 tt 2 at (inferring 2for0 attf ), and with the differentiation of interest starting at 0t . The Caputo derivative is given by 0,8 1 2 1 2 2/12/3 3 8 2 1 2 0 2/1 2 1 2 2/1 0 tttdttd t t C (41) have by Eq. (29) the inferred initialization function (history for Caputo derivative) given as 0tfor40 ftf . 2 2 tt Achar, Lorenzo, and Hartley values of the integer-order derivatives evaluated at t = 0 . It is also observed that A simple example will illustrate the profound differences between the Caputo derivative and the LH initialization of the RL derivative. Consider the semi-derivative of f with a history period starting at which has removed the effect of the singularity at t = 0. Because 1/ 2 , we We now consider the LH initialization of the RL derivative for terminal initialization of the function f with order higher than order m 1 will in general be discon- tinuous at t = 0. Eq. (40) yields Eq. (29) and Eq. (34) as special cases. [...]... CAPUTO FRACTIONAL DERIVATIVE 1/ 2 0 Dt t 22 1/ 2 t 22 0 Dt 1 Dt 0 Dt 1 / 2 t 2 2 d 1/ 2 t 22 0 dt dt 1 2 d t t dt 0 1 ( 42) f , 1 / 2, 2, 0, t 2 2d 1 2 0 1 2 t 1 2 2 0,t 1 0, 2 2d ,t 0, (43) Integrating , collecting terms, and simplifying yields 1/ 2 t 22 0 Dt 40(t 2) 3 / 2 15 1 2 , t>0 (44) The results of Eq (41) and Eq (44), are shown graphically in Figure 1 It is clear from the figure that the Caputo inferred... 0.00 625 0.003 125 0.59 1.67 5.14 17. 72 65.81 25 2.64 989 .27 0.45 1 .25 3. 92 13. 52 49.61 189.53 740.64 0.75 1.77 4.88 14.77 51.06 186 .20 713.03 24 .95 25 .23 23 .39 23 .50 23 .22 23 .77 23 .55 0.13 0 .25 0.53 1.05 2. 09 4 .22 8.53 0.66 1.86 5.48 18.16 66.98 25 3.09 989.09 α D⋆ y(t) = S2 S3 S5 0.48 1. 02 0.16 1.34 2. 00 0 .25 3.97 5.47 0.63 13.70 15.53 1.13 49.59 52. 13 2 14 189.61 186 30 4 .25 7 42. 73 7 12. 84 8.63 Γ (5 + α /2) ... September 2 6, Chicago, Illinois Padovan J (1987) Computational algorithms and finite element formulation involving fractional operators Comput Mech 2: 271 28 7 Gorenflo R (1997) Fractional calculus: some numerical methods in: Carpinteri A, Maincardi, F (eds.), Fractals and Fractional Calculus in Continuum Mechanics Springer, Wein, New York, pp 27 7 29 0 17 60 18 19 20 21 22 23 24 25 26 27 28 29 30 Agrawal. .. initial conditions For example, Figure 2 shows that results obtained using S5 15 58 16 Agrawal and Kumar Table 14 Comparison of maximum errors in y(t) for example 5 for different h h S1 S2 0 .2 1 .24 e−1 7.40e 2 0.1 −9.59e 2 −8.48e 2 −7.34e 2 −6.32e 2 0.05 0. 025 −6 .26 e 2 −5.87e 2 0.0 125 −5.83e 2 −5.67e 2 0.00 625 −5.65e 2 −5.58e 2 S3 S5 7.39e 2 3 .28 e−1 −8.04e 2 1.92e−1 −6.35e 2 1.13e−1 −5.95e 2 −8.09e 2. .. Substituting Eq 14 into Eq 13, we obtain y2m +2 in terms of F2m+1 , and F2m +2 Note that F2m is not included here as it can be computed directly from y2m To compute the second integral in Eq 12, f (t, y(t)) is approximated over [2mh, (2m + 1)h] in terms of F2m , F2m+1 /2 and F2m+1 using QIPs similar to the one used in Eq 14 Using Eq 14, F2m+1 /2 is expressed in terms of F2m , F2m+1 , and F2m +2 This leads... Applications of Fractional Calculus in Physics World Scientific, New Jersey West BJ, Bologna M, Grigolini P (20 03) Physics of Fractal Operators Springer, New York Magin RL (20 04) Fractional calculus in bioengineering Crit Rev Biomed Eng 32( 1):1–104 Magin RL (20 04) Fractional calculus in bioengineering – Part 2 Crit Rev Biomed Eng 32( 2):105–193 Magin RL (20 04) Fractional calculus in bioengineering – Part... 0.30 0.75 2. 83 11 .20 43. 02 1 72. 06 695.05 0.47 1.08 3. 52 11.64 44.44 170.47 668.06 21 .48 21 .59 21 .98 22 .27 22 .17 22 .25 20 .53 0.03 0. 02 0.03 0.06 0.14 0.33 0.78 0.61 1.66 4.59 15.34 56.08 21 2.45 836.39 0 .23 0.70 2. 86 10.64 42. 39 168.61 674.67 0.53 1.13 3.38 11.45 43. 42 167.41 659.41 0.03 0.05 0. 02 0.06 0.14 0 .28 0.75 5 .2 Example 2 As the second example we consider the following linear FDE and the inhomogeneous... −5.95e 2 −8.09e 2 −5.70e 2 −6.80e 2 −5.59e 2 −6.15e 2 Table 15 Comparison the CPU times in seconds for example 5 h S1 S2 S3 S5 0 .2 0.1 0.05 0. 025 0.0 125 0.00 625 29 .97 99.73 384.89 1 524 .59 6 128 .41 24 7 42. 19 19.58 73.97 28 7.53 1140.45 4555.50 18340.91 22 .63 76.17 28 7 .23 11 12. 77 4 422 . 42 176 52. 16 0 .28 0. 42 1.31 3.56 12. 61 46.34 have significant error For example 3, results obtained using S5 for α > 1 were divergent... −4.74e−4 0.003 125 −4.53e−4 −3.02e−4 2. 42e−4 S4 S5 1.34e−4 1.18e−4 1.36e−4 2. 38e−5 1.89e−4 9.17e−5 1.93e−4 9 .26 e 2 8.54e 2 7.95e 2 7.51e 2 7.18e 2 6.95e 2 6.79e 2 COMPARISON OF FIVE NUMERICAL SCHEMES 53 11 Table 6 Comparison the CPU times in seconds for example 2 Set1 Set2 h S1 S2 S3 S4 S5 S1 S2 S3 S5 0 .2 0.1 0.05 0. 025 0.0 125 0.00 625 0.003 125 0.70 1.64 5 .25 18.13 66.47 26 0.98 9 92. 20 0 .27 0.86 2. 94 11.88... 1.35e−7 − 2. 00e−7 0.00 625 −1.63e−6 − 2. 38e−8 1.65e−6 0.003 125 −4.10e−7 − 1.66e−8 1.77e−5 S4 S5 6.12e−6 − 1.60e−1 6.12e−6 −8 .25 e 2 6.12e−6 −4.19e 2 6.12e−6 2. 11e 2 6.12e−6 − 1.06e 2 6.12e−6 − 5.30e−3 6.12e−6 − 2. 65e−3 Table 3 Comparison the CPU times in seconds for example 1 α = 1.5 α = 0.75 h S1 S2 S3 S4 S5 S1 S2 S3 S5 0 .2 0.1 0.05 0. 025 0.0 125 0.00 625 0.003 125 0.47 1.30 4.31 15. 02 56. 02 218.39 856.41 . ,0,0,0 ,2, 2/1 ,2 22 2 2/ 1 0 2 2/1 0 1 2 2/1 0 ttftd dt d tDDtD t ttt ( 42) ,0 ,22 2 0 2 2 2 1 1 2 0 2 1 1 2 2/1 0 2 1 2 1 td t d t dt d tD t t . τ) α−1 f(τ,y(τ))dτ ( 12) and y 2m +2 = g 2m +2 + 1 Γ (α) 2mh 0 ((2m +2) h − τ) α−1 f(τ,y(τ))dτ + 1 Γ (α) (2m +2) h 2mh ((2m +2) h − τ) α−1 f(τ,y(τ))dτ (13) Since y j , j =0, ···, 2m are known, the first integrals in both. one used in Eq. 14. Using Eq. 14, F 2m+1 /2 is expressed in terms of F 2m , F 2m+1 , and F 2m +2 . This leads to y 2m+1 in terms of F 2m+1 , and F 2m +2 . Thus, we obtain two equations in terms
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