NASA / TMm1998-208407

Insights Into the Fractional Order Initial

Value Problem via Semi-infinite Systems

Tom T. Hartley

University of Akron, Akron, Ohio

Carl F. Lorenzo

Lewis Research Center, Cleveland, Ohio

National Aeronautics and

Space Administration

Lewis Research Center

November 1998

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Insights Into the Fractional Order Initial Value Problem

Via Semi-Infinite Systems

Tom T. Hartley

Department of Electrical Engineering

The University of Akron

Akron, OH 44325-3904

Carl F. Lorenzo

NASA Lewis Research Center

2 I000 Brookpark Rd.

Cleveland, OH 44135

Abstract

This paper considers various aspects of the initial value problem for fractional order differential

equations. The main contribution of this paper is to use the solutions to known spatially

distributed systems to demonstrate that fractional differintegral operators require an initial condition

term that is time-varying due to past distributed storage of information.

Introduction

The problem to be considered here is the initial condition problem for systems with components of

fractional mathematical order. Specifically, the case when the fractional components are of

mathematical order 1/2 is considered at length. This is due to the availability of a convenient

physical system to use as reference, namely, the semi-infinite lossy line. Laplace transform

techniques are utilized throughout the study, as they are the fundamental tool of linear system

theory. The main contribution of this work is the demonstration that fractional differintegral

operators require the addition of time-dependent initial condition terms in their application. The

implication is that the solution of fractional ordinary differential equations will require that the

problem statement include the appropriate time-dependent initial condition terms along with some

information about the structure of these terms. It is also shown, by using a semi-infinite elastic

bar, that the simple derivative or integral can require these terms as well. The present section

provides some background into fractional order systems before the distributed line is addressed in

the later sections.

The idea of fractional integrals and derivatives has been known since the development of the

regular calculus, with the first reference probably being associated with Leibniz in 1695 [Oldham

and Spanier (1974), page 3]. Although not well known to most engineers, the fractional calculus

has been considered by prominent mathematicians [Courant and Hilbert (1953)] as well as the

"engineers" of the operational calculus [Heaviside (1893)], [Bush (1929)]. In fact, many

textbooks written before 1960 have a small section on fractional calculus [Goldman (1949)],

[Holbrook (1966)], [Starkey (1954)], [Carslaw and Jeager (1941)], [Scott (1955)], [Mikusinski

(1959)]. An outstanding historical survey can be found in Oldham and Spanier who also give

what is unquestionably the most readable and complete mathematical presentation of the fractional
calculus. Other bounddiscussionsof the areaaregivenby Ross,McBride, and McBride and

Roach. Unfortunately,manyof theresultsin thefractionalcalculusaregivenin the languageof

themathematicianandarenotreadilyaccessibletothegeneralengineeringcommunity.

Many systemsareknownto display fractional order dynamics. Probably the first physical system

to be widely recognized as demonstrating fractional behavior is the semi-infinite lossy (RC) line.

Looking into the line, the operational impedance is

Z(s) - V(s) _ 1

I(s)

The current is thus equal to the half-derivative of the applied voltage (assuming time begins at t--0),

which can be written in the time domain as,

i(t) - dl/2v(t)

dt 1/2

Although studied by many, Heaviside considered this system extensively by using the operational

calculus. He also states that "there is a universe of mathematics lying in between the complete

differentiations and integrations" [Heaviside (1893), p. 459] This specific system is studied

further in the next section. An equivalent system is the diffusion of heat into a semi-infinite solid.

Here the temperature at the boundary is equal to the half integral of the heat rate there. Other

systems that are known to display fractional order dynamics are viscoelastic systems [Bagley and

Calico (1991)], [Koeller (1984)], [Koeller (1986)], [Skaar, Michel, and Miller (1988)], [LopezMarcos (1990)]; colored noise [Mandelbrot (1967)]; electrode-electrolyte polarization [Ichise,

Nagayanagi, and Kojima (1971)], [Sun, Onaral, and Tsao (1984)]; dielectric polarization [Sun,

Abdelwahab, and Onaral (1984)]; boundary layer effects in ducts [Sugimoto (1991)]; and

electromagnetic waves [Heaviside (1893)]. As many of these systems depend upon specific

material and chemical properties, it is expected that a wide range of fractional order behaviors are

possible by considering different materials.

The two most common definitions for the general fractional differintegral are probably the

Griinwald definition and the Riemann-Liouville definition [Oldham and Spanier (1974)]. The

Riemann-LiouviUe definition is given here:

dqf(t)

[d(t-a)]q

1 f(x)

r(-q) (t - "_)q+l

dx, q<O, (1)

where (a) represents the beginning of the domain of the differintegral. Here q can have non-integer

values, and thus the name fractional differintegral. Notice that the definition is based on

integration, and more importantly is a convergent convolution integral for q<0. For fractional

differentiation (q>0), the usual integer nth derivative is take: 1 of the fractional (q-n) th integral,

dqf_ d n [dq-nf]

dtq dt n [dtq-nj '

q>0 and n an integer > q. (2)

The basic engineering tool for analyzing linear systems i:; the Laplace transform. The Laplace

transform of Equation (1) as given by Oldham and Spanier ',¢ith a=0 is
L/dqf(t)/= sqL{f(t)} - n_ s kIdq-l-kf(t)l

[_ k=0 k dtq-l-k it-0- '

for all q, (3)

where n is an integer such that n-l<q<n. The problem with this formulation is that the initial

condition terms are fractional integrals and derivatives evaluated at time zero. Unfortunately, this

information is not readily specified in many problems. Notice that when the initial conditions are

assumed to be zero, this equation reduces to

LIdqf(t) t sqL{f(t)} I--a74-J=

(4)

It is the nature of the initial condition terms that motivates this study. The next section addresses

this problem through a study of the semi-infinite lossy line.

The Semi-Infinite Lossy Line

This section contains a classical iterated Laplace transform derivation of the operational impedance

for a semi-infinite lossy line. Of particular importance is the inclusion of the spatial initial

conditions in the formulation. These are necessary in order to specify the initial state completely,

so that a proper time response can be obtained. Thus the semi-infinite lossy line becomes the

reference case for understanding the initialization of more general fractional order differential

equations.

Assuming a lossy RC-line, the boundary value problem can be written in terms of the current or

voltage variable [Karplus (1958)]. Since a semi-infinite line is considered, the measurable inputs

or outputs are assumed to occur at the left end (x=0), while the right end (x=oo) is kept at a finite

value. The desired result is to mathematically define the behavior of the line at x=0, that is, to

determine both the input impedance and the initialization response. In terms of the voltage

variables, the problem can be written as

av(x,t) _ Ot a2v(x't) , v(O,t) = vI(t), v(oo,t) = O, v(x,O) given,

at ax 2

(5)

with

i(x,t) = - 1 av(x,t) , (6)

R ax

where v is the voltage, i is the current, vI(t) is a time-dependent input variable, and Ot is a constant

which is one over the product of resistance per unit length, R, and capacitance per unit length, C.

A classical solution using iterated Laplace transforms is used to solve this problem.

Taking the Laplace transform with respect to time and using s as the temporal Laplace variable

gives
sV(x,s) - v(x,O) = _ d2V(x's)

dx 2

with

V(0,s) = VI(S), V(°°,s) = 0, I(x,s) = -1 dV(x,s) (7)

R dx

Then taking the Laplace transform with respect to spatial position x, and using p as the spatial

Laplace variable gives

_-V(p,s) - 1V(p,O) = p2V(p,s) - pV(O,s)- [dV] .

t dx J(O,s)

(8)

Replacing dWdx with V*, this equation can be manipulated to give V(p,s) as

[l_- _-]V(p,s) =- 1V(p,O) + pV(O,s) + V*(O,s) (9)

or

(10)

Each term in this equation will be considered separately in order to inverse transform this with

respect to the p variable. The first term, the transform of the initial spatial distribution, V(p,0), is

problem-dependent A slight rearranging of the expression gives;

V(p,s) = [2_f___ (__ ___)- 23/--_ (1+ _--_-)t [_- V(p,0)]

+[p2-1_ tlpv o,s +v, o,s l. (11)

Here the first term represents the effect of any initial spatial voltage distribution, while the second

term represents the voltage and current present at the x=0 end of the line. The first term is now

inverse-Laplace-transformed with respect to the p variable nsing convolution, and the other term is

inverse-transformed using standard transform pairs;
x

V(x,s) = 1 _+./i(x-m[ 1 v(Z,0)] d_,

xl__k__ e-yr_ (x-_.)[_1__v(k,O)] dX 2¢-_ t ot

+v(0,s)cosecx)+v,(0,s)sin Cx) (12)

Equivalently,

x + i

V(x,s) = -1 e yr_ x

2C e-_ _"[-1 v(Z,0)] d_,

V*(O,s) [e+,V_X- e_,V_x] .

+ V(O,s)2 [e+'_x + e-'V_x] + 2ff-_-

(13)

Collecting like exponentials then gives

+ ,,/c_ s x

V(x,s) = e re,

2 V(0,s) V*(0,s) + ¢__

° $ x +e-Ca-x

2

V(0,s)- V*(0,s) + 1 (14)

It should be recognized that the coefficients multiplying the two exponential functions are

unknowns. Although the integral and either V(0,s) or V*(0,s) are given in the problem statement,

the other condition (V* or V, respectively) at x=0 is determined as a response to these two given

terms. Imposing the boundary condition at x=oo, allows the determination of a relationship

between these three terms at x=O, and thus allows the impedance and initial condition response of

the system.

It is shown in the Appendix that in the limit as x--+oo, the second term in Equation (14) goes to zero

through application of L'Hospital's rule and Leibniz's rule, and by the fact that V(oo,s)=0 from the

problem statement. Then
V(oo,s) = e V

2

V(0,s) + V*(0,s) Lira 1 fxe_ v(_,O) d)_ 1= O. (15)

x---_ 0_3_--_

Dropping the limit notation, by necessity we have

V(0,s) + V*(0,s)

_0 _ . S

1 e 3/_ 7. v(_,,0) d)_ = 0. (16)

Remembering that the current anywhere in the line is related to the voltage, then at x---0

I(0,s) = - 1 dV(0,s) = V*(0,s)

R dx R

(16')

and solving for voltage in terms of source current in Equation (16) gives

R I(0,s) 1 r1oo

V(0,s)- +

J0e-Yr_ _"v(Z,0) dZ. (17)

In evaluating the integral on the right, it is now recognized that this term is equivalent to a Laplace

transform integral with the Laplace operator _f--_. Thus, Laplace transform tables can simplify the

evaluation of this term as follows:

_0 _

1 e-3/-_ x v(_,,O) d_. -

1 [V(p,O)]r=_/_ •

(18)

The notation on the right side of this equation is used to indicate the evaluation procedure. First the

initial spatial distribution v(x,0) is Laplace transformed with respect to the spatial Laplace variable

p to give V(p,0). The integral on the left of Equation (18) is then easily evaluated by replacing the

spatial Laplace variable p with s_. Using this, Equation (17) becomes

V(O,s) - R I(O,s) + ! [V(p,O)Jw_ _

(19)

Notice that this contains the driving point impedance functic_n Z(s), which is obtained by setting the

initial condition terms to zero,

Z(s)= V(0,s) _ R

I(0,s)
or, as _ = 1/RC,

Z(s) - v(O,s) _ _ 1

I(O,s)

(20)

Notice from Equation (19) that the voltage V(0,s) is seen to be composed of two parts, the forced

response due to I(0,s) and the initial condition response due to the initial voltage distribution in the

lossy line.

Applying the other boundary condition of Equation (7) at x---0, allows the solution of the voltage

transfer function V(x,s)/V(0,s) and has no effect on the terminal impedance. This is now done for

completeness. At x=0, the boundary condition is

V(O,s)=VI(s). (21)

Now, inserting Equations (21), (19), and (16') into Equation (14), gives

+ A/--if-s x

V(x,s_= _ v_

_ s__ x

2

iV(p,O)]p=3/- _ - fxe-Y_ z v(_,,O) d)_ ]

2Vi(s) [V(p,O)]p_/-_ 1 e+_/-_v(X,o)d_..

+

(22)

This equation is the general expression for the voltage anywhere inside the line as a function of the

applied voltage at the terminal Vi(s) and the initial conditions on the line. Furthermore, the current

at any point in the line can be determined directly from Equation (22) as

I(x,s) = - _1 dV(x,s) (7)

R dx

We will now discuss the voltage-current behavior of the lossy line. Referring back to the

definition of the Laplace transform from Equation (4), and applying it to Equation (19), it can be

seen that the input-output properties (with initial conditions equal to zero) of the lossy line satisfy

i(t) - 1 dl/2v(t) or v(t) = Rq-_- d-1/2i(t)

R'¢-_ dt 1/2 dt -1/2

More completely, from Equation (17) or (19),

v(t) = R4-_- d-1/2i(t) + ¢1(t)

dt-_/2

(23)
where Ol(t) is obtained asthe inverse Laplacetransform of either term in Equation (18).

Likewise,

i(t)- 1 dl/2v(t) + ¢2(t)

RV-_ dt 1/2

(24)

where _2(t) is effectively the negative integral of _l(t) weighted by l/Rq-_. From these

expressions it should be clear that the fractional differintegral of any function must include a timedependent initial condition term.

The Semi-Infinite Lossless Line

This section considers the semi-inf'mite lossless line whose impedance is a constant or an operator

of order zero. A classical iterated Laplace transform is again used for the derivation of the

operational impedance for the semi-infinite lossless line. The inclusion of the spatial initial

conditions in the formulation is of particular importance. These are necessary in order to specify

the initial state completely, so that a proper time response can be obtained. The results will aid in

the understanding of the initial condition terms for integer order systems in a fractional ordinary

differential equation setting.

Assuming a lossless LC-line, the boundary value problem can be written in terms of the cun'ent or

voltage variable [Karplus (1958)]. Since a semi-infinite line is considered, the measurable inputs

or outputs are assumed to occur at the left end (x=0), while the right end (x=oo) must be kept at a

finite value. The desired result is an impedance at the left end of the line as well as the response

due to any initial conditions on the line. In terms of the voltage variables, the problem can be

written as

_2v(x, t) - 1 _2v(x, t)

Ot2 LC Ox 2

v(0,t) = vi(t) , v(oo,t) = 0 , v(x,0) and v'(x,0) given, (25)

with

3i(x,t) _ C 3v(x,t) and 3v(x,t) = _ L 3i(x,O , (26)

3X at 3X at

where v is the voltage, i is the current, vi(t) is a time-dependent input variable, L is inductance per

unit length, and C is capacitance per unit length. A classi.'al solution to this problem is obtained

using iterated Laplace transforms.
Taking theLaplacetransformwith respectto time andusings asthetemporalLaplacevariable

gives

s2V(x,s)- sv(x,O) dv(x,O)_ 1 d2V(x, s)

dt LC dx 2

with boundary conditions

g(O,s) = VI(S), V(oo,s) = O,

dV(x,s) _ L (Lt dI(x't)//=- L [sI(x,s) - I(x,O)]. (27)

dx _ I dt /l

Then taking the Laplace transform with respect to spatial position, x, and using p as the spatial

Laplace variable gives

LCs2V(p,s) - LCsV(p,O) - LCV'(p,O)= p2V(p,s) - pV(O,s)- [d-_--x] ,

(O,s)

(28)

where dv(x,0)/dt is replaced with V'. Further replacing dV(0,s)/dx with V*, this equation can be

manipulated to give V(p,s) as

[p2 _ LCs2]V(p,s) = _ LCsV(p,0) - LC V'(p,0) + pV(0,s) + V*(0,s) (29)

or

V(p,s) = [ 1 ] [-LCs V(p,O)-LC V'(p,O)+ pV(O,s) + V*(O,s)].

Lp2 _ LCs 2J

(30)

Each term in this equation will be considered separately in order to inverse-Laplace- transform this

with respect to the p variable. The first term, the initial spatial distribution V(p,0), is problemdependent. A slight rearranging of the expression gives

V(p,s) = 2s LfL(_ 1 _ 1 ] [-LCs V(p,O) - LC V'(p,O)] (p- sl/-L-c ) 2s_ (p + sl(-_-)

1

+[p2_ LCs2] [pV(O's) + V*(O,s)].

(31)

Here the first term represents the effect of any initial spatial distribution, while the second term

represents the voltage and current at the x=0 end of the line. The first term is now inverseLaplace-transformed with respect to the p variable using convolution, and the other terms are

inverse-transformed using standard transform pairs;
x

V(x,s) = 1

2¢L-e

+

e+LfL--ff s(x-_,)[_v@,O)] d_

fo x2s¢-L-c 1

__ e+Lffffffs(x'_'_- v'(_,,O)] dk

__ e-_-C-s(x-_'_- v(_,O)] d_,

X...: s(x-_O[_v'(_,,O)] d_,

1 e_L_/-L-_--

+ V(O,s) cosh( L_-C-sx)+ V*(O,s) sinh( L_-L-csx)

LfL--6s

Collecting like exponentials then gives

(32)

V(x,s) - e+L(CC-sxUV(0,s) + V*(0,s)

2 k I/LCs

x

_ 1 e-LCL-EsZ v0_,0) d_,

+ e-V-C6=x[V(0,s) - V*(0,s)

2 k L'(L--C-s

+ _ 1 Ix e+LqL-EsZ v@,0) d_, + s_--C1 ;0xe+Lq-£-6s_"v'@,0) d_ J(33)

It should be recognized that the coefficients multiplying the two exponential functions are

unknowns. Although the integrals and either V(0,s) or V*(J,s) are given in the problem statement,

the other condition (V* or V, respectively) at x=0 is determined as a response to the given terms.

Imposing the boundary condition at x=oo, allows the deterntination of a relationship between these

three terms at x--0, and thus the impedance and initial condition response of the system.

As was shown for Equation (14), in the limit as x--->_, the second term in Equation (33) goes to

zero through application of L'Hospital's rule and Leibniz's rule, and by the fact that V(oo,s)=0

from the problem statement. Then

V(oo,s) - 0 - e+l-Lc-s= [V(0,s) + V*(0,s)

2 k I/-L-Cs

/-

_ _k[ e-ffL--tsZv(_.,O) d_.

L f" l

e-Lq-_sk v'0_,0) d)_

sL_C , (34)

10
and by necessity,

V(0,s) + V*(0,s)

(fits 1 e -LfffCsz v(E,0) dE fL--6

1 e-L/ffC s_, v'(E,O) dE = O.

_LfLf_ (35)

Solving for the voltage in terms of the voltage gradient in Equation (35) gives

V(0,s) -

V*(0,s) + 1 c|-

lifts (if6 L

e-LfffC s_. v(E,0) dE

+ 1 e -LfL--tsx v'(E,0) dE (36)

sLfL-6

In evaluating the integrals, it is now recognized that these terms are equivalent to a Laplace

transform integral with the Laplace operator L_ s. Thus Laplace transform tables can simplify

the evaluation of the third term as follows:

1 e -LfL-Es_- v(E,0) dE - 1 [V(p,O)]p=LV'L-Cs, (L-6

(37)

and similarly the fourth term. The notation on the right side of this equation is used to indicate the

evaluation procedure. First the initial spatial distribution, v(x,0), is Laplace transformed with

respect to the spatial Laplace variable p to give V(p,0). The integral on the left of Equation (37) is

then easily evaluated by replacing the spatial Laplace variable p with _---C s. Using this, Equation

(36) becomes

1

V(O,s) =-V*(O,s) + 1 [V(p,O)]r,=fffff s + 1 [V'(p,O)]r,='iE-c s/. (38) (LYs --_ ¢LCL_ s

/

This contains the driving point transfer function (not an impedance),

V(0,s) _ 1

V*(0,s) (L--C-s

(39)

I1
which is obtained by setting the initial condition terms to zero. Notice from Equation (38) that the

voltage V(0,s) is seen to be composed of two parts, the forced response due to V*(0,s) and the

initial condition response due to the initial voltage distribution in the lossless line. Mathematically

this is expressed in the time domain as

v(0,t) = f dv(0,t) dx dt + _3(t)

(40)

where 0p3(t) is the time-dependent initial condition response due to the initial voltage and current

distributions on the line, which can be obtained by inverse-Laplace-transforming the two integrals

in Equation (36). Thus it is seen that a simple integral operator can also have time varying initial

condition terms.

More importantly, notice that Equation (38) contains V*(0,s). From the boundary conditions

given in Equation (27), this term can be replaced with

idV(x,s)[ _ =- L [sI(O,s) - I(O,O)] .

dx Jx=O

(41)

Equation (38) then becomes

V(0,s) = L[sI(0,s)- I(0,0)] + 1 [V(p,0)]p=,LC s + 1 [V'(p,0)]p= LffiEs LfL- s

or

V(0,s) = _f_ I(0,s) - _/_ [I(0,0)]s

+ L-_C [V(p'O)]p=fff_s + _I-_-----[V'(p'O)]P=f_]/LC s

(42)

Notice from Equation (42) that the voltage V(0,s) is seen to be composed of two parts, the forced

response due to I(0,s) and the initial condition response due to the initial voltage distribution in the

lossless line. Considering only the first term, it can be seen that the impedance looking into this

line is then

Z(s) - v(0,s) _

I(0,s)

(43)

which is simply a constant. Mathematically, Equation (42) :s expressed in the time domain as

12
I v(O,t) = _ i(0,t) + _4(t) (44)

where this _14(t) is the time-dependent initial condition response due to the initial voltage and

current distributions on the line, which can be obtained by inverse-Laplace- transforming the three

initial condition terms in Equation (42). Thus it is seen that even a simple constant gain operator

can also have time-varying initial condition terms.

The Semi-Infinite Vibrating Bar

This section considers longitudinal vibrations in the semi-infinite elastic bar which has a transfer

function that is a pure integral of order unity. A classical iterated Laplace transform is again used

for the derivation of the operational impedance for the semi-infinite elastic bar. The inclusion of

the spatial initial conditions in the formulation is of particular importance. These are necessary in

order to specify the initial state completely, so that a proper time response can be obtained. The

results will aid in the understanding of the initial condition terms for integer order systems in a

fractional ordinary differential equation setting.

For the elastic bar, the boundary value problem can be written in terms of the longitudinal

displacement and the longitudinal force [Churchill (1958)]. Since a semi-infinite bar is considered,

the measurable inputs or outputs are assumed to occur at the left end (x=0), while the right end

(x=_) must be kept stationary. The desired result is an impedance at the left end of the bar as well

as the response due to any initial conditions on the bar. In terms of the displacement y(x,t), the

problem can be written as

32y(x, t) - E 32y(x, t)

at 2 D aX 2

, y(oo,t) = 0 , y(x,0) and y'(x,0) given, (45)

with

- E 3y(0,t) _ f(t), (46)

3x

where E is Young's modulus of elasticity, 13 is the mass per unit volume, f(t) is the force at the left

end of the bar, and _y/_x is the strain which is proportional to the force. The force can be the

applied input with the displacement as the output, or the displacement can be the input with the

force as the output, as desired by the user. In what follows we will let a 2 = E/p.

A classical solution using iterated Laplace transforms will be used to solve this problem. The

results from this solution will provide further guidance into fractional ordinary differential

equations.

Taking the Laplace transform with respect to time and using s as the temporal Laplace variable

gives

13
s2y(x,s) - sy(x,O) - dy(x,O)_ a2 d2y(x,s)

dt dx 2

with

Y(oo,s) = 0, dY(0,s) = _ 1 F(s) = Y*(0,s). (47)

dx E

Then taking the Laplace transform with respect to spatial position x and using p as the spatial

Laplace variable gives

dY

a_2S2y(p,s)- a_SY(p,O)- _Y'(p,O)= p2y(p,s)- pY(O,s)- [-d-_-x](0,s) (48)

Replacing dY(x,s)/dx with Y* and dy(x,0)/dt by y', this equation can be manipulated to give

Y(p,s) as

p2_ __ s2]y(p,s) sY(p,O) - Y'(p,O) + pY(O,s) + Y*(O,s)

"! 1

I_L 1__

a _ 3 a 2 a 2

(49)

or

Y ps, :Ip2

_,1_s2t j isa 2 Y p0, a 2Y p0, pY 0s, +Y, 0s, (50)

Each term in this equation will be considered separately in order to inverse-Laplace- transform this

with respect to the p variable. The first term, the initial spatial distribution Y(p,0), is problemdependent. A slight rearranging of the expression gives

Y(p,s) = 2s _s) 2s (p +

a

1 s Y(p,O - 1__ Y'(p,O)]

a 2 . a2 J

+ I ! s2-1 [pY(O,s) + Y*(O,s)] . (51) p2

Here the first term represents the effect of any initial spa!ial distribution, while the second term

represents the relationship of the displacement and force at he x=0 end of the bar. The fn'st term is

now inverse-Laplace-transformed with respect to the p va:iable using convolution, and the other

term is inverse-transformed using standard transform pairs;

14
_0 x

Y(x,s) = a e+ s(x-_.)/a[_y(_.,0)] d_

2

+

fo xa--a- 2s e+ s(x-_')/a[- y'(k,O)] dZ

_a2 e-s(x4.)/a[_ y(_,,O)] d_,

' a__a_ 2s e-s(x-_.)/a[_y'(_,,0)] d)_

+ Y(O,s)cosh(_-)+ aY*(O,s)sinh(_-)

s

(52)

Collecting like exponentials then gives

r

e+sx/a aY*(0,s) Y(x,s) - /Y(0,s) +

2 L- - S

-a e- sZ/a y(k,0) d)_ - a e- sMa y'(_,0) dk

s

+ e- sra_.__La [Y(0,s) aY*(0,s)

2 L s

+ a e+ sMa V(_.,0) dL + a e+ s_./a y'(_,,0) dk. (53)

S

It should be recognized that the coefficients multiplying the two exponential functions are

unknowns. Although the integrals and either Y(0,s) or Y*(O,s) are given in the problem statement,

the other condition (Y* or Y, respectively) at x=0 is determined as a response to the given terms.

Imposing the boundary condition at x=oo, allows the determination of a relationship between these

three terms at x=0, and thus allows the impedance and initial condition response of the system.

As was shown for Equation (14), in the limit as x--_oo, the second term in Equation (53) goes to

zero through application of L'Hospital's rule and Leibniz's rule, and by the fact that Y(oo,s)=O

from the problem statement. Then

F

e +

s_/a aY*(O,s) Y(oo,s) - 0 - /Y(0,s) +

2 L - - s

- a f"e-s_/a y(L,0) dk - a ;e-sX/a y'O,,0) d_,], (54)

s

15
and by necessity,

eo

Y(0,s) + aY*(0,s) _ a e- sMa y(_,,0) d_ - a f" e- sL/a y'(%,0) d_ = 0. (55)

s s J0

Solving for the displacement in terms of the strain (or force) in Equation (55) gives

I" I" Y(0,s) - aY*(0,s) + a e- s_/a y(%,0) d_ + a e- sk/a y'(_,0) d_, (56)s s "

In evaluating the integrals, it is now recognized that these terms are equivalent to a Laplace

transform integral with the Laplace operator s/a. Thus, Laplace transform tables can simplify the

evaluation of the third term as follows:

a e- sMa y(_,,0) d_, = a [Y(p,0)]p-_a, (57)

and similarly, the fourth term. The notation on the right side of this equation is used to indicate the

evaluation procedure. First the initial spatial distribution, y(x,0), is Laplace transformed with

respect to the spatial Laplace variable p to give Y(p,0). The integral on the left of Equation (57) is

then easily evaluated by replacing the spatial Laplace variable p with s/a. Using this, Equation (56)

becomes

Y(0,s) -

- aY*(0,s)

a y, ,0 + a[Y(p,0)]p_--s/a + s[ (p)]p=s/a (58)

and this contains the transfer function, which is not an impedance,

Y(0,s)_ a

Y*(O,s) s

(59)

Notice from Equation (58) that the displacement Y(0,s) is seen to be composed of two parts, the

forced response due to the applied strain and the initial condition response due to the initial

displacement distribution in the elastic bar. Mathematically this is expressed in the time domain as

I y(O,t) =- af dx

dy(0,t) dt + _5(t) (60)

where _5(t) is the time dependent initial condition response due to the initial displacement

16
distribution in the elastic bar, which can be obtained by inverse-Laplace-transforming the two

integrals in Equation (56). Thus it is seen that a simple integral operator can also have timevarying initial condition terms.

Likewise, if the input at the left end was displacement, the force (strain) response can also be

determined from Equation (58);

Y*(O,s) _- s Y(O,s) + [V(p,O)]p=s/a + [Y'(p,O)]p=s/a- a

(61)

The strain response Y*(0,s) is then composed of two parts the forced response due to the applied

displacement Y(0,s) and the initial condition response due to any displacement distribution on the

bar at time zero. In the time domain, this is

y*(0,t) = - _1 dy(0,t) + q_6(t)

a dt

(62)

Again qb6(t) can be obtained from the given initial displacements.

Summary

This paper has demonstrated that when specific differintegral operators (q= 1/2, q= 1, and q=O) are

considered as semi-infinite systems, a time-dependent term resulting from the initial spatial

conditions should be added to the to the forced response. This is an important result with respect

to the solution of fractional ordinary differential equations that can have rather arbitrary initial

conditions.

To apply this result to a specific fractional order differential equation, the initial conditions, if any,

must be given along with some structural information on how they belong in the problem. For

example, if the following system were given,

d3/2f(t) + df(t____)_) + dl/2f(t) + f(t) = u(t),

dt 3/2 dt dt 1/2

any initial conditions must specify how they are imposed. Specifically, a physical implementation

of this system could contain three half-integrators, or two integrators and two half integrators. The

initial condition specification must then also specify something structural. Also, this simple

example clearly demonstrates that further study into the meaning of minimality for fractional order

representations is necessary.

It should be apparent from this study that there is a clear connection between the fractional calculus

and spatially distributed systems. Not only can spatially distributed system representations allow a

better understanding of the fractional calculus, but the fractional calculus also allows a more

compact representation and problem solution for some spatially distributed systems.

17
Appendix

Equation (14) is repeated here for reference.

V(x,s) = e +_ s x V(0,s) + V*(0,s) 1 e-_ _"v()_,0) d

2

_ s x +e< S-x

2

V(0,s) V*(0,s) + 1 J°x l

+ s

e r_ _. v()_,O) d)_ (14)

To get to Equation (16), which is the first term of Equation (14), set equal to zero, it is required to

evaluate Equation (14) in the limit as x-->oo. In this limit, the second term in Equation (14) should

go to zero due to the exponential, however, the integral inside the bracket will diverge. That is,

lim e÷ V0_,0) d_, = 0.o_

X---.-)oo

vO_

We are thus left with an indeterminate form. This can be evaluated by using L'Hospital's rule and

rearranging as follows

lim

x+:

{2 o_ fx e+_r_ _"V(_-,0) d_,)

L'Hospital's rule says that this ratio has the same value as t ae ratio of the derivatives (with respect

to x) of the numerator and denominator. Differentiating the denominator is easy, but differentiating

the numerator with respect to x requires the use of Leibniz_ s rule. Performing the differentiations

gives

2ct

Combining the leading constants, and applying the Leibniz'; rule to the numerator, gives

18
It can now be seen that the exponential terms cancel, which leaves the result

V(_,0)

lim 2-_ V(x,0) = 2s

x---ffoo

The problem statement, however, requires that the boundary condition

V(_,t) = 0

be satisfied for all time. Thus it is shown that the first term of Equation (14) is equal to zero in the

limit as x----) o°.

19
References

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1941.

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Courant, R. and Hilbert, D., Methods of Mathematical Physics, Interscience Publishers, New

York, 1953.

Goldman, S., Transformation Calculus and Electrical Transients, Prentice-Hall, New York, 1949.

Heaviside, O., Electromagnetic Theory, Chelsea, New York, 1893/1971.

Holbrook, J.G., Laplace Transforms for Electronic Engineers, 2nd, Pergamon, Oxford, 1966.

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253-265, 1971.

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San Francisco, 1979.

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Skaar, S.B., Michel, A.N., and Miller, R.K., "Stability of Viscoelastic Control Systems," IEEE

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Sugimoto, N., "Burgers Equation with a Fractional Derivative; Hereditary Effects on Nonlinear

Acoustic Waves," J. Fluid Mech., vol 225, pp 631-653, 1991.

Sun, H.H., Abdelwahab, A.A., and Onaral, B., "Linear Approximation of Transfer Function with

a Pole of Fractional Order," IEEE Trans. Auto. Cont., vol AC-29, no 5, pp 441-444, May,

1984.

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pp 664-674, Oct. 1984.

21
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Insights Into the Fractional Order Initial Value Problem via Semi-infinite Systems

6. AUTHOR(S)

Tom T. Hartley and Carl F.Lorenzo

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National Aeronautics and Space Administration

Lewis Research Center

Cleveland, Ohio 44135-3191

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Tom T. Hartley, University of Akron, Akron, Ohio 44325--0001 and Cat[ F. Lorenzo, NASA Lewis Research Center.

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This paper considers various aspects of the initial value problem for fractional order differential equations. The main

contribution of this paper is to use the solutions to known spatially distributed systems to demonstrate that fractional

differintegral operators require an initial condition term that is time-vaE¢ing due to past distributed storage of information.

14. SUBJECT TERMS

Fractional calculus; Fractional order systems; Laplace transform; Diffusion equation;

Lossy line; Initial value problem; Semi-infinite systems

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