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Time-Varying Initialization and Corrected Laplace Transform of the Caputo Derivative
6th Workshop on Fractional Differentiation and Its Applications Part of 2013 IFAC Joint Conference SSSC, FDA, TDS Grenoble, France, February 4-6, 2013
Time-Varying Initialization and Corrected Laplace Transform of the Caputo Derivative Carl F. Lorenzo*, Tom T. Hartley**, Jay L. Adams** *NASA Glenn Research Center, Cleveland, Ohio, 44135 USA [email protected] **University of Akron,,Akron, Ohio, 44325-3904 USA, [email protected], [email protected] Abstract: This paper derives the time-varying initialization function for the Caputo derivative of any order u>0 where m 1 u m, m 1,2, . The derivative is redefined to include this initialization function. Then, the Laplace transform for the redefined Caputo derivative is determined which corrects (supplants) that given for the derivative in the literature and properly accounts for time-varying initialization effects. The paper generalizes previous results for order 0 u 1. Keywords: Initialization function, Caputo derivative, Laplace transform For 0 u 1, m 1,
1. INTRODUCTION - THE CAPUTO DERIVATIVE The Caputo derivative [1]-[3] has been applied as a replacement for the Riemann-Liouville derivative in many studies. The attraction for the use of the Caputo derivative is explained by Podlubny [3], “The main advantage of Caputo’s approach is that the initial conditions for fractional differential equations with Caputo derivatives take on the same form as for integer-order differential equations, i.e. contain the limit values of integer-order derivatives of unknown functions at the lower terminal…” Its use is widespread because it eliminates the singularity at time t=0. It thereby, supposedly, simplifies the initialization of fractional differential equations. This paper will show that the Caputo derivative like the Riemann-Liouville [8] derivative requires a time–varying initialization.
C u a Dt
f (t )
1 ( m u )
t
f
( m)
(t )
( )
u 1 m
t
f ( )
(t )
u
d , 0 u 1.
(2)
a
L C0 Dtu f (t ) s u F ( s) n 1
s
u k 1 ( k )
f
(3)
(0), n 1 u n,
k 0
and for 0 u 1, simplifies to
D
L
C 0
u t
n 1, and the Laplace transform
f (t ) s u F (s) s u 1 f (0), 0 u 1 .
(4)
The remainder of this paper will determine corrected equations for the Laplace transform of the Caputo derivative, that is, of Equations (3) and (4) accounting for the requirement of a time-varying initialization term. The next section determines the time-varying initialization function required to make the Caputo derivative compatible with its past.
d , m 1 u m
a
m 0,1,2 .
1 (1 u )
In the original work, [1]-[2], a was taken as a 0. The Laplace transform given in the literature for the Caputo derivative [1]-[3] is
Previous studies [4]-[7] have examined the some of the problems and issues associated with use of the Caputo derivative. Trigeassou [9] studies the initialization of the Caputo derivative based on infinite-dimensional state formulation. This paper extends the Laplace transform for the Caputo derivative from 0 u 1 case, [10], to the general case of m 1 u m, m 1,2, . The Caputo derivative [1]-[3] is defined as C u a Dt
f (t )
2.TIME VARYING-INITIALIZATION CAPUTO DERIVATIVE
(1)
FOR
THE
In this section we determine the time-varying initialization function for the Caputo derivative. Let us consider two instances of the Caputo derivative, one starting at t=-a and 978-3-902823-27-4/13/$20.00 © 2013 IFAC
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10.3182/20130204-3-FR-4032.00189
2013 IFAC FDA Grenoble, France, February 4-6, 2013
second starting at t=0, with 0>-a. Both instances will be operating on the same function f (t ) . Further, we require that f (t ) and all of its derivatives are zero for all t<-a to clearly delineate the history period, a t 0. Now since both instances of the derivatives are operating on the same function we should expect that C0 Dtu f (t ) should be a C u a Dt
continuation of
to formally include the initialization function, and where c (t ) is defined by (7) and C0 d tu f (t ) represents the uninitialized Caputo derivative. When 0 u 1 , m 1 and we have C u 0 Dt
when the appropriate
f (t )
initialization function is added to C0 Dtu f (t ) . Therefore, we wish to determine a function such that C u 0 Dt
f (t )
C u a Dt
f (t ), t 0, m 1 u m .
t
f
( m)
(t )
( )
u 1 m
f ( )
(t )
u
d c (t ), t 0 .
(10)
0
OF
THE
CAPUTO
The Laplace transform for the Caputo derivative with 0 u 1 , was derived in reference [10]. The key results are
0
t
t
3. LAPLACE TRANSFORM DERIVATIVE, 0
(5)
d
1 (m u )
1 (1 u )
where c (t ) is given by Equation (8).
Then we have
1 (m u )
f (t )
f
( m)
(t )
( )
u 1 m
L
d
C u 0 Dt
f (t ) s u L f (t )
s u 1 (1 u )
a
(6)
(11)
0
f ( ) exp( s)(1 u, s) d , i
a
t 0, 0 u 1.
for t>0 and m 1 u m. Thus, for m 1 u m. and
c (t )
1 ( m u )
0
( m)
f
(t )
( )
u 1 m
d , t 0
L C0 Dtu f (t ) s u L f (t ) s u 1 fi (0)
(7)
s u 1 as e (1 u, as ) fi (a) (1 u )
a
is the required initialization function for the Caputo derivative. Then for 0 u 1 , m=1, and
c (t )
1 (1 u )
0
f ( )
(t )
u
d , t 0
u su (1 u )
f ( ) exp( s)(u, s)d , i
a
a
where the argument is f instead of f . For the special case where f i (t ) constant b, - a t 0 ,
L
C u 0 Dt
f (t ) s u L f (t )
exp( a s)(1 u, a s) b s u 1 1 , (1 u ) t 0, 0 u 1. Notice the last bracketed term is
Therefore, the derivative is re-defined as
f (t ) C0 d tu f (t ) c (t ), t 0,
or
b s u 1 s u 1 f (0) . C u 0 Dt
f (t )
1 ( m u )
(12)
t 0, 0 u 1.
(8)
Thus, for the Caputo derivative to be consistent with its past, a time-varying initialization function of the form of Equation (7) must be added to the derivative. It is seen that only when u 0 does Equation (8) reduce to an ordinary Riemann integral and is equal to a constant.
C u 0 Dt
0
t
f
( m)
(t )
( )
u 1 m
d c (t ) ,
This is to be expected, when
0
u 1
(13)
(14)
f i (t ) f (t ) on passage
through t=0, we have s f (0) s u 1 f (0) , as occurs in Equation (4). The remaining term in the bracket is the effect of the time–varying initialization. The effect of a step at t=0 is accounted for in f(t), this is made clear in the example problem to follow.
(9)
t 0, f (t ) 0 for t a, m 1 u m,
162
2013 IFAC FDA Grenoble, France, February 4-6, 2013
4. EXAMPLE Consider the initialization of the example illustrated in Figure 1. Mathematically, the functions are given by
fi (t ) b, a t 0,
(15)
f (t ) c t 0 ,
(16)
u
(23)
Here we determine the Laplace transform for the Caputo derivative for any u > 0, that is
The problem is first solved in the time domain, thus
(t )
(22)
5. GENERAL CASE
Figure 1. Example problem history function and function to be fractionally differentiated.
D
L d , 0 u 1.
a
exactly matching that given using the theory developed and shown in Equation (13).
0
f ( )
d , t 0
exp( a s)(1 u, a s) L c (t ) b s u 1 1 , (1 u ) t 0, 0 u 1.
t
t
u
Taking the Laplace transform of this result gives
fi ( t) = b
1 f (t ) (1 u )
f ( )
(t )
f (t) = c
C u a d t
0
b (t a) u (c b) (t ) u c t u (1 u ) (1 u ) (1 u ) b c (t ) (t a ) u (t ) u , t 0. (1 u )
with fi (t ) 0, t a.
-a
1 (1 u )
c (t )
C 0
u t
f (t ) L C0 d tu f (t ) c (t ) , t 0 .
(24)
(17) Then we have
a
For t 0 C u a d t
f (t )
1 (1 u )
t
b ( a) (c b) ( )
a
0 u 1, C u a d t
f (t )
u
(t ) u
d ,
L
(18)
t 0, (19)
C u 0 dt
f (t )
c t u , (1 u )
0 u 1
t
f
( m)
(t ) 0
( )
u 1 m
d
where f ( m) (t ) has been subscripted in the initialization integral to indicate the derivative over the initialization period.
t
1 c ( ) f (t ) d , 0 u 1. (1 u ) (t )u 0
1 f (t ) L ( m u)
The uninitialized Caputo derivative is based on f (t ) cU (t ), t 0 , then C u 0 dt
C u 0 Dt
0 f i( m) ( ) 1 L d , t 0, m 1 u m , (25) u (m u ) a (t ) m 1,2,
u
b(t a) (c b)(t ) (1 u ) (1 u ) . 0 u 1, t 0,
We first consider the Laplace transform of the uninitialized Caputo derivative, with m 1 u m , m 1,2, , namely
(20) (21)
t 1 f ( m ) ( ) L d u 1 m (m u ) 0 (t ) (26) t 1 f ( m ) ( ) st 0 e (m u) 0 (t )u 1m d dt , t 0
The required initialization for the Caputo derivative starting at t=0 is given by the difference between the derivative starting at t=-a and that starting at t=0, see Equation (6), thus
163
t
f ( m) ( ) 1 e st d dt ( m u ) (t ) u 1m 0 0
(27)
2013 IFAC FDA Grenoble, France, February 4-6, 2013
1 ( m u )
e st f ( m) ( )
(t )
u 1 m
(1 U ( t )) d dt
c (t )
(28)
0 0
Noting that 1 U ( t ) U (t ), and under the conditions of uniform convergence of the integrals, we may interchange the order of integration giving
1 ( m u )
f
( m)
0
e st ( ) U ( t ) dt d u 1 m ( t ) 0
1 ( m u )
f ( m) ( ) L (t ) mu 1U (t ) d
L c (t ) (29)
1 ( m u )
f
( m)
( )
( m u )
0
s
m u
exp( s)d
f
1 ( m u )
L c (t )
(30)
L c (t )
(31)
f (t ) s
( m)
( ) exp( s)d
(32)
s
a
(36)
0
f i( m) ( )
a
e st
(t )
u 1 m
dt d ,
0
1 ( m u )
0
f
i
( m)
( )L (t ) m u 1 d ,
a
(37)
t 0.
L c (t ) 0
1
m 1 m s k F ( mk 1) (0) . (33) s F ( s) k 0
m u
f ( m u )
i
( m)
( ) e s (m u,s)d ,
(38)
a
m 1 u m, t 0 Combining this result with that of Equation (34) yields,
L
C 0
Dtu f (t ) s u L f (t ) 0
s
m u
1 fi ( m ) ( ) e s (m u, s)d , (m u ) a
(39)
m 1 u m, t 0. An alternate form of this equation shows its correction of the Caputo derivative Laplace transform as originally published, that is Equation (1). To see this, the following integral relationship [8] is applied repeatedly to the integral term in (39)
F ( m k 1) (0) is taken to be zero. This gives the
k
d dt ,
Then from [11] p. 245 we have
In this derivation we are concerned with the Laplace transform of the uninitialized operator. Equation (3) is recognized as the original Caputo derivative Laplace transform. The initial condition terms are present to provide continuity for the function through the origin. The continuity will be correctly provided by the second term in Equation (25). Thus the summation in Equation (33) will be ignored. Said differently, the initialization is accounted for in the c (t ) term of Equation (24), thus the term m 1
u 1 m
m 1 u m,
s L
(t )
The inner integral is recognized as a Laplace transform, giving
The integral is seen to be the Laplace transform of f ( m) (t ) , thus u m
f i( m) ( )
m 1 u m, t 0
0
C u 0 dt
0
Interchanging the order of integration, yields
(35)
a
1 e st ( m u )
Thus,
L C0 d tu f (t ) s u m
d , t 0
u 1 m
m 1 u m, t 0
0
fi( m) ( )
(t )
0
Applying the transform pair Spanier & Oldham [11] p. 245 (or Erdelyi [12] p. 137) gives
0
is the required initialization function. The Laplace transform is given by
The bracketed term is recognized as a Laplace transform, thus
1 ( m u )
k 0
transform for the uninitialized operator as
L
C u 0 dt
f (t ) s L f (t ), m 1 u m . u
0
e
(34)
s
(q 1, s ) f i( )d (q 1) f i (0)
a
0
e as (q 1, as) f i (a) s q (q, s)e s f i ( )d .
We now consider the Laplace transform of the initialization function for the Caputo derivative general case. This is given by Equation (7), then for m 1 u m.
a
(40) After the first application we get 164
2013 IFAC FDA Grenoble, France, February 4-6, 2013
L c (t )
From this equation we may determine the general expression that m applications of Equation (40) will yield
(m u ) f i( m 1) (0) e as (m u, as ) f i( m 1) (a ) 1 0 , (41) m u ( m 1) s s (m u ) s (m u 1) f i ( ) e a (m u 1, s )d m 1 u m, t 0.
L c (t )
m
s u m s j 1 f i ( m j ) (0) j 1
m
s u m s j 1e as j 1
This may be written as
L c (t )
1 s m u
f i( m 1) (0) e as (m u , as ) f ( m1) (a ) i (m u ) s , (m u 1) 0 ( m 1) s ( ) e (m u 1, s )d fi a
u m m 0
s s (u )
(m u j 1, as ) ( m j ) fi (a ) (m u j 1)
(45)
fi ( m m ) ( ) e s ( u, s)d .
a
m 1 u m, t 0. This may be re-written as
L c (t )
(42)
m
s
u j m 1
j 1
m
s
m 1 u m, t 0.
f i ( m j ) (0) (46) (m u j 1, as ) ( m j ) fi (a) (m u j 1)
u j m 1 as
e
j 1
0
Applying (40) again to the integral in (42)
L c (t )
suu fi ( ) e s (u, s )d . (1 u ) a m 1 u m, t 0.
fi ( m 1) (0) e as (m u , as ) f ( m 1) (a ) i (m u ) s (43) (m u 1) 1 ( m 2) , (0) s m u (m u 1) f i e as (m u 1, as ) f i ( m 2) (a ) s m u 2 0 f ( m 2) ( ) e s (m u 2, s )d a i m 1 u m, t 0.
Finally, the summations are taken from the alternate direction to yield m 1
L c (t ) s u j 1 f i ( j ) (0) j 0
m 1
s
u j 1 as
e
j 0
( j u 1, as ) ( j ) fi (a ) ( j u 1)
(47)
0
suu fi ( ) e s (u, s )d . (1 u ) a m 1 u m, t 0. Thus the Laplace transform for any u > 0 is given by
fi ( m 1) (0) e as (m u , as ) f ( m 1) (a ) i (m u ) ( m 2) (0) s fi (44) ( m u 1, as ) 1 f i ( m 2) (a ) . m u se as (m u 1) s s2 (m u 2) 0 fi ( m 2) ( ) e s (m u 2, s )d a
L
C 0
m 1
Dtu f (t ) s u L f (t )
s j 0
m 1
s
u j 1
fi ( j ) (0)
u j 1 as
e
j 0
0
( j u 1, as ) ( j ) fi (a ) ( j u 1)
suu fi ( ) e s (u , s )d . (1 u ) a
165
(48)
2013 IFAC FDA Grenoble, France, February 4-6, 2013
[3] Podlubny, Igor, Fractional Differential Equations: An Introduction to Fractional Derivatives, Fractional Differential Equations, to methods of their Solution and some of their Applications. Mathematics in Science and Engineering, Vol. 198, Academic Press, 1999.
for m 1 u m, m 1, 2, , t 0. Clearly the substitution of m=1 into Equation (48) yields Equation (12) as a point of validation. The first summation in (48) matches that found in Equation (3) with a change of sign. This has the effect of moving the starting point to t=-a which is seen in the second summation. Further, the integral term brings in the timevarying initialization effect. In the application of both Equations (12) and (48) it is important to note that
f i ( )
must
( j)
( j)
be
compatible
[4] Achar, B. N. Narahari, Lorenzo, Carl F., and Hartley, Tom T., “Initialization and the Caputo Fractional Derivative” NASA TM-2003-212482, July 2003 [5] Achar, B. N. Narahari, Lorenzo, Carl F., and Hartley, Tom T., “Initialization Issues of the Caputo Fractional Derivative”, Proceedings of DETC/CIE 2005, ASME Int. Design Engineering Technical Conference, DETC2005-84348, Long Beach Ca., Sept. 24-28, 2005.
with
f i (0) and f i (a). It is also noted that the second summation in Equation (48) may be taken as zero if such case is of interest to the application.
[6] Achar, B. N. Narahari, Lorenzo, Carl F., and Hartley, Tom T., “The Caputo Fractional Derivative: Initialization Issues Relative to Fractional Differential Equations” pp.27-42, in Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering, Ed. J. Sabatier, O. P. Agrwal, J. A. Tenreiro Machado, ISBN 978-1-4020-6041-0 (HB), Springer, 2007.
Thus, the key result of this paper is that, Equation (12) and its generalization Equation (48) provide a corrected Laplace transform for the Caputo derivative. 6. DISCUSSION AND CONCLUSIONS
[7] Lorenzo, Carl F., and Hartley, Tom T., “On Self-Consistent Operators with Application to Operators of Fractional Order”, Proceedings of ASME International Design Engineering Technical Conference, DETC2009-86730, San Diego, Ca. Aug. 30-Sept. 2, 2009.
This paper has determined the time-varying initialization function associated with the Caputo derivative. This function is required to properly continue/initialize a Caputo differentiation. Based on the time-varying initialization function the corrected Laplace transform for the Caputo derivative has been derived. That is; the Laplace transform that properly accounts for the history associated with the differentiation. The key results for the case 0
[8] Lorenzo, Carl F., and Hartley, Tom T., “Initialization of Fractional-Order Operators and Fractional Differential Equations” Special Issue on Fractional Calculus of ASME Journal of Computational and Nonlinear Dynamic, Special Issue on Discontinuous and Fractional Dynamical Systems, Vol. 3, No. 2 , 020201, April 2008.
The results given in Equation (13) identically match those derived for the same case for the RiemannLiouville derivative operator as determined in [8], suggesting an equivalence of the operators. In future works it will be required to compare results to those of Trigeassou and Maamri [9] to assure compatibility of results for this important problem.
[10] Lorenzo, Carl F., and Hartley, Tom T., “Time-Varying Initialization and Laplace Transform of the Caputo Derivative: With Order Between Zero and One” DETC2011-47396, Proceedings of the ASME 2011 International Design Engineering Technical Conferences & Computers and Information in Engineering Conference, IDETC/CIE 2011 August 28-31, 2011, Washington, DC, USA
ACKNOWLEDGEMENT
[11] Spanier, J. and Oldham, K. B., “An Atlas of Functions,” Hemisphere Publishing Corp. ISBN 0-89116-573-8, 1987.
[9] Trigeassou, Jean-Claude, and Maamri, Nezha, “The Initial Conditions of the Riemann-Liouville and Caputo Fractional Derivative: an Integrator Interpretation,” The 4th IFAC Workshop Fractional Differentiation and its Applications Badajoz, Spain, October 18-20, 2010.
The authors gratefully acknowledge the support of NASA Glenn Research Center.
[12] Erdelyi, A., Oberhettinger, F., and Tricomi, F. G., Tables of Integral Transforms, Bateman Manuscript Project, McGraw-Hill Book Co, 1954.
REFERENCES [1] Caputo, M, Linear model of dissipation whose Q is almost frequency independent-II Geophy. J. R. Astr. Soc. vol. 13, 1967 pp529-539. [2] Caputo, M. Elastica e Dissipaszione, Zanicelli, Bolgna, 1969.
166
Time-Varying Initialization and Corrected Laplace Transform of the Caputo Derivative Carl F. Lorenzo*, Tom T. Hartley**, Jay L. Adams** *NASA Glenn Research Center, Cleveland, Ohio, 44135 USA [email protected] **University of Akron,,Akron, Ohio, 44325-3904 USA, [email protected], [email protected] Abstract: This paper derives the time-varying initialization function for the Caputo derivative of any order u>0 where m 1 u m, m 1,2, . The derivative is redefined to include this initialization function. Then, the Laplace transform for the redefined Caputo derivative is determined which corrects (supplants) that given for the derivative in the literature and properly accounts for time-varying initialization effects. The paper generalizes previous results for order 0 u 1. Keywords: Initialization function, Caputo derivative, Laplace transform For 0 u 1, m 1,
1. INTRODUCTION - THE CAPUTO DERIVATIVE The Caputo derivative [1]-[3] has been applied as a replacement for the Riemann-Liouville derivative in many studies. The attraction for the use of the Caputo derivative is explained by Podlubny [3], “The main advantage of Caputo’s approach is that the initial conditions for fractional differential equations with Caputo derivatives take on the same form as for integer-order differential equations, i.e. contain the limit values of integer-order derivatives of unknown functions at the lower terminal…” Its use is widespread because it eliminates the singularity at time t=0. It thereby, supposedly, simplifies the initialization of fractional differential equations. This paper will show that the Caputo derivative like the Riemann-Liouville [8] derivative requires a time–varying initialization.
C u a Dt
f (t )
1 ( m u )
t
f
( m)
(t )
( )
u 1 m
t
f ( )
(t )
u
d , 0 u 1.
(2)
a
L C0 Dtu f (t ) s u F ( s) n 1
s
u k 1 ( k )
f
(3)
(0), n 1 u n,
k 0
and for 0 u 1, simplifies to
D
L
C 0
u t
n 1, and the Laplace transform
f (t ) s u F (s) s u 1 f (0), 0 u 1 .
(4)
The remainder of this paper will determine corrected equations for the Laplace transform of the Caputo derivative, that is, of Equations (3) and (4) accounting for the requirement of a time-varying initialization term. The next section determines the time-varying initialization function required to make the Caputo derivative compatible with its past.
d , m 1 u m
a
m 0,1,2 .
1 (1 u )
In the original work, [1]-[2], a was taken as a 0. The Laplace transform given in the literature for the Caputo derivative [1]-[3] is
Previous studies [4]-[7] have examined the some of the problems and issues associated with use of the Caputo derivative. Trigeassou [9] studies the initialization of the Caputo derivative based on infinite-dimensional state formulation. This paper extends the Laplace transform for the Caputo derivative from 0 u 1 case, [10], to the general case of m 1 u m, m 1,2, . The Caputo derivative [1]-[3] is defined as C u a Dt
f (t )
2.TIME VARYING-INITIALIZATION CAPUTO DERIVATIVE
(1)
FOR
THE
In this section we determine the time-varying initialization function for the Caputo derivative. Let us consider two instances of the Caputo derivative, one starting at t=-a and 978-3-902823-27-4/13/$20.00 © 2013 IFAC
161
10.3182/20130204-3-FR-4032.00189
2013 IFAC FDA Grenoble, France, February 4-6, 2013
second starting at t=0, with 0>-a. Both instances will be operating on the same function f (t ) . Further, we require that f (t ) and all of its derivatives are zero for all t<-a to clearly delineate the history period, a t 0. Now since both instances of the derivatives are operating on the same function we should expect that C0 Dtu f (t ) should be a C u a Dt
continuation of
to formally include the initialization function, and where c (t ) is defined by (7) and C0 d tu f (t ) represents the uninitialized Caputo derivative. When 0 u 1 , m 1 and we have C u 0 Dt
when the appropriate
f (t )
initialization function is added to C0 Dtu f (t ) . Therefore, we wish to determine a function such that C u 0 Dt
f (t )
C u a Dt
f (t ), t 0, m 1 u m .
t
f
( m)
(t )
( )
u 1 m
f ( )
(t )
u
d c (t ), t 0 .
(10)
0
OF
THE
CAPUTO
The Laplace transform for the Caputo derivative with 0 u 1 , was derived in reference [10]. The key results are
0
t
t
3. LAPLACE TRANSFORM DERIVATIVE, 0
(5)
d
1 (m u )
1 (1 u )
where c (t ) is given by Equation (8).
Then we have
1 (m u )
f (t )
f
( m)
(t )
( )
u 1 m
L
d
C u 0 Dt
f (t ) s u L f (t )
s u 1 (1 u )
a
(6)
(11)
0
f ( ) exp( s)(1 u, s) d , i
a
t 0, 0 u 1.
for t>0 and m 1 u m. Thus, for m 1 u m. and
c (t )
1 ( m u )
0
( m)
f
(t )
( )
u 1 m
d , t 0
L C0 Dtu f (t ) s u L f (t ) s u 1 fi (0)
(7)
s u 1 as e (1 u, as ) fi (a) (1 u )
a
is the required initialization function for the Caputo derivative. Then for 0 u 1 , m=1, and
c (t )
1 (1 u )
0
f ( )
(t )
u
d , t 0
u su (1 u )
f ( ) exp( s)(u, s)d , i
a
a
where the argument is f instead of f . For the special case where f i (t ) constant b, - a t 0 ,
L
C u 0 Dt
f (t ) s u L f (t )
exp( a s)(1 u, a s) b s u 1 1 , (1 u ) t 0, 0 u 1. Notice the last bracketed term is
Therefore, the derivative is re-defined as
f (t ) C0 d tu f (t ) c (t ), t 0,
or
b s u 1 s u 1 f (0) . C u 0 Dt
f (t )
1 ( m u )
(12)
t 0, 0 u 1.
(8)
Thus, for the Caputo derivative to be consistent with its past, a time-varying initialization function of the form of Equation (7) must be added to the derivative. It is seen that only when u 0 does Equation (8) reduce to an ordinary Riemann integral and is equal to a constant.
C u 0 Dt
0
t
f
( m)
(t )
( )
u 1 m
d c (t ) ,
This is to be expected, when
0
u 1
(13)
(14)
f i (t ) f (t ) on passage
through t=0, we have s f (0) s u 1 f (0) , as occurs in Equation (4). The remaining term in the bracket is the effect of the time–varying initialization. The effect of a step at t=0 is accounted for in f(t), this is made clear in the example problem to follow.
(9)
t 0, f (t ) 0 for t a, m 1 u m,
162
2013 IFAC FDA Grenoble, France, February 4-6, 2013
4. EXAMPLE Consider the initialization of the example illustrated in Figure 1. Mathematically, the functions are given by
fi (t ) b, a t 0,
(15)
f (t ) c t 0 ,
(16)
u
(23)
Here we determine the Laplace transform for the Caputo derivative for any u > 0, that is
The problem is first solved in the time domain, thus
(t )
(22)
5. GENERAL CASE
Figure 1. Example problem history function and function to be fractionally differentiated.
D
L d , 0 u 1.
a
exactly matching that given using the theory developed and shown in Equation (13).
0
f ( )
d , t 0
exp( a s)(1 u, a s) L c (t ) b s u 1 1 , (1 u ) t 0, 0 u 1.
t
t
u
Taking the Laplace transform of this result gives
fi ( t) = b
1 f (t ) (1 u )
f ( )
(t )
f (t) = c
C u a d t
0
b (t a) u (c b) (t ) u c t u (1 u ) (1 u ) (1 u ) b c (t ) (t a ) u (t ) u , t 0. (1 u )
with fi (t ) 0, t a.
-a
1 (1 u )
c (t )
C 0
u t
f (t ) L C0 d tu f (t ) c (t ) , t 0 .
(24)
(17) Then we have
a
For t 0 C u a d t
f (t )
1 (1 u )
t
b ( a) (c b) ( )
a
0 u 1, C u a d t
f (t )
u
(t ) u
d ,
L
(18)
t 0, (19)
C u 0 dt
f (t )
c t u , (1 u )
0 u 1
t
f
( m)
(t ) 0
( )
u 1 m
d
where f ( m) (t ) has been subscripted in the initialization integral to indicate the derivative over the initialization period.
t
1 c ( ) f (t ) d , 0 u 1. (1 u ) (t )u 0
1 f (t ) L ( m u)
The uninitialized Caputo derivative is based on f (t ) cU (t ), t 0 , then C u 0 dt
C u 0 Dt
0 f i( m) ( ) 1 L d , t 0, m 1 u m , (25) u (m u ) a (t ) m 1,2,
u
b(t a) (c b)(t ) (1 u ) (1 u ) . 0 u 1, t 0,
We first consider the Laplace transform of the uninitialized Caputo derivative, with m 1 u m , m 1,2, , namely
(20) (21)
t 1 f ( m ) ( ) L d u 1 m (m u ) 0 (t ) (26) t 1 f ( m ) ( ) st 0 e (m u) 0 (t )u 1m d dt , t 0
The required initialization for the Caputo derivative starting at t=0 is given by the difference between the derivative starting at t=-a and that starting at t=0, see Equation (6), thus
163
t
f ( m) ( ) 1 e st d dt ( m u ) (t ) u 1m 0 0
(27)
2013 IFAC FDA Grenoble, France, February 4-6, 2013
1 ( m u )
e st f ( m) ( )
(t )
u 1 m
(1 U ( t )) d dt
c (t )
(28)
0 0
Noting that 1 U ( t ) U (t ), and under the conditions of uniform convergence of the integrals, we may interchange the order of integration giving
1 ( m u )
f
( m)
0
e st ( ) U ( t ) dt d u 1 m ( t ) 0
1 ( m u )
f ( m) ( ) L (t ) mu 1U (t ) d
L c (t ) (29)
1 ( m u )
f
( m)
( )
( m u )
0
s
m u
exp( s)d
f
1 ( m u )
L c (t )
(30)
L c (t )
(31)
f (t ) s
( m)
( ) exp( s)d
(32)
s
a
(36)
0
f i( m) ( )
a
e st
(t )
u 1 m
dt d ,
0
1 ( m u )
0
f
i
( m)
( )L (t ) m u 1 d ,
a
(37)
t 0.
L c (t ) 0
1
m 1 m s k F ( mk 1) (0) . (33) s F ( s) k 0
m u
f ( m u )
i
( m)
( ) e s (m u,s)d ,
(38)
a
m 1 u m, t 0 Combining this result with that of Equation (34) yields,
L
C 0
Dtu f (t ) s u L f (t ) 0
s
m u
1 fi ( m ) ( ) e s (m u, s)d , (m u ) a
(39)
m 1 u m, t 0. An alternate form of this equation shows its correction of the Caputo derivative Laplace transform as originally published, that is Equation (1). To see this, the following integral relationship [8] is applied repeatedly to the integral term in (39)
F ( m k 1) (0) is taken to be zero. This gives the
k
d dt ,
Then from [11] p. 245 we have
In this derivation we are concerned with the Laplace transform of the uninitialized operator. Equation (3) is recognized as the original Caputo derivative Laplace transform. The initial condition terms are present to provide continuity for the function through the origin. The continuity will be correctly provided by the second term in Equation (25). Thus the summation in Equation (33) will be ignored. Said differently, the initialization is accounted for in the c (t ) term of Equation (24), thus the term m 1
u 1 m
m 1 u m,
s L
(t )
The inner integral is recognized as a Laplace transform, giving
The integral is seen to be the Laplace transform of f ( m) (t ) , thus u m
f i( m) ( )
m 1 u m, t 0
0
C u 0 dt
0
Interchanging the order of integration, yields
(35)
a
1 e st ( m u )
Thus,
L C0 d tu f (t ) s u m
d , t 0
u 1 m
m 1 u m, t 0
0
fi( m) ( )
(t )
0
Applying the transform pair Spanier & Oldham [11] p. 245 (or Erdelyi [12] p. 137) gives
0
is the required initialization function. The Laplace transform is given by
The bracketed term is recognized as a Laplace transform, thus
1 ( m u )
k 0
transform for the uninitialized operator as
L
C u 0 dt
f (t ) s L f (t ), m 1 u m . u
0
e
(34)
s
(q 1, s ) f i( )d (q 1) f i (0)
a
0
e as (q 1, as) f i (a) s q (q, s)e s f i ( )d .
We now consider the Laplace transform of the initialization function for the Caputo derivative general case. This is given by Equation (7), then for m 1 u m.
a
(40) After the first application we get 164
2013 IFAC FDA Grenoble, France, February 4-6, 2013
L c (t )
From this equation we may determine the general expression that m applications of Equation (40) will yield
(m u ) f i( m 1) (0) e as (m u, as ) f i( m 1) (a ) 1 0 , (41) m u ( m 1) s s (m u ) s (m u 1) f i ( ) e a (m u 1, s )d m 1 u m, t 0.
L c (t )
m
s u m s j 1 f i ( m j ) (0) j 1
m
s u m s j 1e as j 1
This may be written as
L c (t )
1 s m u
f i( m 1) (0) e as (m u , as ) f ( m1) (a ) i (m u ) s , (m u 1) 0 ( m 1) s ( ) e (m u 1, s )d fi a
u m m 0
s s (u )
(m u j 1, as ) ( m j ) fi (a ) (m u j 1)
(45)
fi ( m m ) ( ) e s ( u, s)d .
a
m 1 u m, t 0. This may be re-written as
L c (t )
(42)
m
s
u j m 1
j 1
m
s
m 1 u m, t 0.
f i ( m j ) (0) (46) (m u j 1, as ) ( m j ) fi (a) (m u j 1)
u j m 1 as
e
j 1
0
Applying (40) again to the integral in (42)
L c (t )
suu fi ( ) e s (u, s )d . (1 u ) a m 1 u m, t 0.
fi ( m 1) (0) e as (m u , as ) f ( m 1) (a ) i (m u ) s (43) (m u 1) 1 ( m 2) , (0) s m u (m u 1) f i e as (m u 1, as ) f i ( m 2) (a ) s m u 2 0 f ( m 2) ( ) e s (m u 2, s )d a i m 1 u m, t 0.
Finally, the summations are taken from the alternate direction to yield m 1
L c (t ) s u j 1 f i ( j ) (0) j 0
m 1
s
u j 1 as
e
j 0
( j u 1, as ) ( j ) fi (a ) ( j u 1)
(47)
0
suu fi ( ) e s (u, s )d . (1 u ) a m 1 u m, t 0. Thus the Laplace transform for any u > 0 is given by
fi ( m 1) (0) e as (m u , as ) f ( m 1) (a ) i (m u ) ( m 2) (0) s fi (44) ( m u 1, as ) 1 f i ( m 2) (a ) . m u se as (m u 1) s s2 (m u 2) 0 fi ( m 2) ( ) e s (m u 2, s )d a
L
C 0
m 1
Dtu f (t ) s u L f (t )
s j 0
m 1
s
u j 1
fi ( j ) (0)
u j 1 as
e
j 0
0
( j u 1, as ) ( j ) fi (a ) ( j u 1)
suu fi ( ) e s (u , s )d . (1 u ) a
165
(48)
2013 IFAC FDA Grenoble, France, February 4-6, 2013
[3] Podlubny, Igor, Fractional Differential Equations: An Introduction to Fractional Derivatives, Fractional Differential Equations, to methods of their Solution and some of their Applications. Mathematics in Science and Engineering, Vol. 198, Academic Press, 1999.
for m 1 u m, m 1, 2, , t 0. Clearly the substitution of m=1 into Equation (48) yields Equation (12) as a point of validation. The first summation in (48) matches that found in Equation (3) with a change of sign. This has the effect of moving the starting point to t=-a which is seen in the second summation. Further, the integral term brings in the timevarying initialization effect. In the application of both Equations (12) and (48) it is important to note that
f i ( )
must
( j)
( j)
be
compatible
[4] Achar, B. N. Narahari, Lorenzo, Carl F., and Hartley, Tom T., “Initialization and the Caputo Fractional Derivative” NASA TM-2003-212482, July 2003 [5] Achar, B. N. Narahari, Lorenzo, Carl F., and Hartley, Tom T., “Initialization Issues of the Caputo Fractional Derivative”, Proceedings of DETC/CIE 2005, ASME Int. Design Engineering Technical Conference, DETC2005-84348, Long Beach Ca., Sept. 24-28, 2005.
with
f i (0) and f i (a). It is also noted that the second summation in Equation (48) may be taken as zero if such case is of interest to the application.
[6] Achar, B. N. Narahari, Lorenzo, Carl F., and Hartley, Tom T., “The Caputo Fractional Derivative: Initialization Issues Relative to Fractional Differential Equations” pp.27-42, in Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering, Ed. J. Sabatier, O. P. Agrwal, J. A. Tenreiro Machado, ISBN 978-1-4020-6041-0 (HB), Springer, 2007.
Thus, the key result of this paper is that, Equation (12) and its generalization Equation (48) provide a corrected Laplace transform for the Caputo derivative. 6. DISCUSSION AND CONCLUSIONS
[7] Lorenzo, Carl F., and Hartley, Tom T., “On Self-Consistent Operators with Application to Operators of Fractional Order”, Proceedings of ASME International Design Engineering Technical Conference, DETC2009-86730, San Diego, Ca. Aug. 30-Sept. 2, 2009.
This paper has determined the time-varying initialization function associated with the Caputo derivative. This function is required to properly continue/initialize a Caputo differentiation. Based on the time-varying initialization function the corrected Laplace transform for the Caputo derivative has been derived. That is; the Laplace transform that properly accounts for the history associated with the differentiation. The key results for the case 0
[8] Lorenzo, Carl F., and Hartley, Tom T., “Initialization of Fractional-Order Operators and Fractional Differential Equations” Special Issue on Fractional Calculus of ASME Journal of Computational and Nonlinear Dynamic, Special Issue on Discontinuous and Fractional Dynamical Systems, Vol. 3, No. 2 , 020201, April 2008.
The results given in Equation (13) identically match those derived for the same case for the RiemannLiouville derivative operator as determined in [8], suggesting an equivalence of the operators. In future works it will be required to compare results to those of Trigeassou and Maamri [9] to assure compatibility of results for this important problem.
[10] Lorenzo, Carl F., and Hartley, Tom T., “Time-Varying Initialization and Laplace Transform of the Caputo Derivative: With Order Between Zero and One” DETC2011-47396, Proceedings of the ASME 2011 International Design Engineering Technical Conferences & Computers and Information in Engineering Conference, IDETC/CIE 2011 August 28-31, 2011, Washington, DC, USA
ACKNOWLEDGEMENT
[11] Spanier, J. and Oldham, K. B., “An Atlas of Functions,” Hemisphere Publishing Corp. ISBN 0-89116-573-8, 1987.
[9] Trigeassou, Jean-Claude, and Maamri, Nezha, “The Initial Conditions of the Riemann-Liouville and Caputo Fractional Derivative: an Integrator Interpretation,” The 4th IFAC Workshop Fractional Differentiation and its Applications Badajoz, Spain, October 18-20, 2010.
The authors gratefully acknowledge the support of NASA Glenn Research Center.
[12] Erdelyi, A., Oberhettinger, F., and Tricomi, F. G., Tables of Integral Transforms, Bateman Manuscript Project, McGraw-Hill Book Co, 1954.
REFERENCES [1] Caputo, M, Linear model of dissipation whose Q is almost frequency independent-II Geophy. J. R. Astr. Soc. vol. 13, 1967 pp529-539. [2] Caputo, M. Elastica e Dissipaszione, Zanicelli, Bolgna, 1969.
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