Waves in fractional Zener type viscoelastic media

0.22 MB·English

by  Sanja Konjik

Waves in fractional Zener type viscoelastic media
Sanja Konjik ∗
1
Ljubica Oparnica †
1
0 Duˇsan Zorica ‡
2

n
a
J
Abstract
5
1 Classical wave equation is generalized for the case of viscoelastic materials obeying frac-
tional Zener model instead of Hooke’s law. Cauchy problem for such an equation is studied:

] existenceand uniquenessof the fundamentalsolution is proven and solution is calculated.
P
A MathematicsSubjectClassification(2000): Primary: 26A33;Secondary: 46F12,74D05,
74J05
.
h
Keywords: fractional derivatives, Laplace and Fourier transforms, fundamental solution,
t
a fractional Zener model
m
[
1 Introduction

1
v
The need for a use of derivatives of arbitrary real (or even complex) order, also called fractional
6
derivatives, is permanently underlined by various applied problems. Although, the idea about
6
9 differentiating with respect to noninteger order originates from the work of Leibnitz and Euler at
2 theturnofthe18thcentury,thelast30yearshavebroughtarealexpansionoffractionalcalculus.
. Certainly,itisduetogreatnumberofpossibilitiesforitsapplicationinmanydiversefieldssuchas
1
0 mechanics,viscoelasticity,automaticcontrol,signalprocessing,stochasticandfinance,biomedical
1 engineering, etc. For a detailed exposition of the theory of fractional calculus we refer to [16] or
1 [19].
:
v In this paper we study existence and uniqueness of solutions for the system
i
X ∂ ∂2
σ(x,t) = u(x,t),
r ∂x ∂t2
a
σ(x,t)+τ Dασ(x,t) = ε(x,t)+ Dαε(x,t), (1)
0 t 0 t
∂
ε(x,t) = u(x,t),
∂x
wherex R,t>0,σ,uandεarestress,displacementandstrain,respectively,consideredasfunc-
∈
tions of x and t, 0<τ <1 is a constant and Dα, 0 α<1, denotes the left Riemann-Liouville
0 t ≤
operatoroffractionaldifferentiationoforderα. Thissystemdescribeswavesinviscoelasticmedia.
System (1) is subjected to initial conditions
∂
u(x,0)=u (x), u(x,0)=v (x),
0 ∂t 0 (2)
σ(x,0)=0, ε(x,0)=0,
∗Faculty of Agriculture, Department of Agricultural Engineering, University of Novi Sad, Trg Dositeja
Obradovi´ca8,21000NoviSad,Serbia. Electronicmail: [email protected]
†Institute of Mathematics, Serbian Academy of Science and Art, Kneza Mihaila 35, 11000 Belgrade, Serbia.
Electronicmail: [email protected]
‡Faculty of Civil Engineering, University of Novi Sad, Kozaraˇcka 2a, 24000 Subotica, Serbia. Electronic mail:
[email protected]
1
as well as boundary conditions
lim u(x,t)=0, lim σ(x,t)=0. (3)
x x
→±∞ →±∞
As we shall show, system (1) can be reduced to the equation
∂2 ∂2
u(x,t)=L(t) u(x,t), x R,t>0, (4)
∂t2 ∂x2 ∈
which will be called wave equation for fractional Zener type viscoelastic media or shortly FZWE
(Fractional Zener Wave Equation). Here L denotes a linear operator (of convolution type) acting
on (R), whose explicit form is given by
′
S
1+sα
L(t)=L−1 1+τsα ∗t.
(cid:18) (cid:19)
In fact, equation (4) will be the subject of our consideration. System (1), and in particular (4),
generalize classical wave equation.
Equation that describes waves occurring in the elastic media are obtained by using basic
equationsofelasticity(see[2]). Ourinterestisrestrictedtoaninfiniteelasticrod(onedimensional
body), positioned along x axis, that is not under influence of body forces. Equations of elasticity
then read
∂ ∂2
σ(x,t) = ρ u(x,t) (5)
∂x ∂t2
σ(x,t) = Eε(x,t) (6)
∂
ε(x,t) = u(x,t) (7)
∂x
where x R, t > 0, σ, u and ε are stress, displacement and strain, as above, ρ = const. is the
∈
density of the media and E = const. is the Young modulus of elasticity. Equation (5) is the
equilibrium equation and it is the consequence of the Second Newton Law. Equation (6) is the
constitutive equation and in the case of elastic media it is known as the Hooke law. Equation (7)
isthestrainmeasureforthelocalsmalldeformations,i.e. connectionbetweenstrainanddisplace-
ment. Wave equation is now obtained by substituting equation (7) into (6) and subsequently (6)
into (5). It reads
∂2 ∂2 E
u(x,t)=c2 u(x,t), c= . (8)
∂t2 ∂x2 sρ
Newly introduced parameter c can be physically interpreted as the wave speed.
The firstauthorswhointroducedthe fractionalZenermodelwere M.Caputo andF.Mainardi
[6, 7]. Such model was also considered in [17] but without using fractional calculus. A similar
problem of stress waves in a viscoelastic medium was investigated in [9] by using the inversion
formulaoftheclassicalLaplacetransform. Waveequation(8)canbegeneralizedintheframework
of fractional calculus in severalways. One of the possibilities is to substitute second order partial
derivative with respect to time by fractional one, as it has been done in [13] for one dimension,
and in [10] and [11] for d dimensions, d 1,2,3 . Time partial derivative can also be replaced
∈ { }
witheither twofractionalderivativesofdifferentorder,asin[3],orbydistributedorderfractional
derivative (integral over given range of order of fractional derivatives), as in [12] and [14] for
α (0,1), and in [4] and [5] for α [0,2]. In each of these papers the intention was to stress
∈ ∈
similaritybetweenwaveanddiffusionequation,sincebyallowingtheorderoffractionalderivative
to be α [0,2] both equations can be obtained as special cases.
∈
In this article wave equation (8) is generalized for the case of viscoelastic media described by
fractional Zener model. Generalization is conducted so that constitutive equation (6) is changed,
since Hooke’s law is inappropriate to describe viscoelastic media. Other two equations, namely
equation of motion of deformable body (5) and relation between strain and displacement (7),
2
are valid for any type of deformable body, with mentioned restrictions, therefore they remain
unchanged. Similar generalization was done in [18] in case of bounded domain.
The paper is organized as follows. In Section 2 we recall basic notions and fix notation which
will be used throughoutthis paper. In Section 3 we introduce dimensionless quantities, derive (1)
and (4) and set up the Cauchy problem (4), (2) in the setting of distribution theory. Section 4
is devoted to examination of the Cauchy problem (4), (2) where we prove a theorem on existence
and uniqueness of its fundamental solution. In Section 5 we give a numerical example, which
illustratesthebehaviorofthefundamentalsolutionforthespecificchoiceoftheparametersαand
τ, as well as initial conditions.
2 Mathematical preliminaries
Let Ω be an open set in Rn. The space of distributions will be denoted by (Ω) and the space
′
of Schwartztempered distributions by (Rn). (R) and (R) are the subDspaces of (R) and
(R) respectively,containing distributSio′ns suppDo+′rted on [0S,+′ ). Since we shall mostlyDw′ork with
′
fSunctions and distributions depending on two variables, u =∞u(x,t), we introduce (R R ) to
′ +
be the space of all distributions u (R2), which vanish for t<0. S ×
′
We shall also need the set C =∈Sz C Rez >0 .
+
{ ∈ | }
Let t [0,a], a > 0, and y AC([0,a]). The left Riemann-Liouville fractional derivative of
∈ ∈
order α [0,1), Dαy, is defined as
∈ 0 t
1 d t y(ζ)
Dαy(t)= dζ,
0 t Γ(1 α)dt (t ζ)α
− Z0 −
where Γ is the Euler gamma function.
In the distributional setting, one introduces a family {fα}α∈R ∈D+′ (R) as
H(t)tα−1, α>0,
f (t)= Γ(α) ,
α ( ddtNNfα+N(t), α≤0,α+N >0,N ∈N
where H is the Heaviside function. Then f is a convolution operator acting on (R) (also,
f : (R) (R)). For α<0 it is calledα∗the operator of fractional differentiatioDn.+′ Moreover,
α∗ S+′ →S+′
for y AC([0,a]) it coincides with the left Riemann-Liouville fractional derivative, i.e.
∈
Dαy =f y.
0 t −α∗
For y (R) the Fourier transform is defined as
′
∈S
y,ϕ = y, ϕ , ϕ (R),
hF i h F i ∈S
where for ϕ (R)
∈S
ϕ(ξ)=ϕˆ(ξ)= ∞ ϕ(x)e−iξxdx, ξ R.
F ∈
Z−∞
Let y (R) such that e ξty (R), for all ξ > a > 0. Then the Laplace transform of y is
∈ D+′ − ∈ S′
defined by
y(s)=y(s)= (e ξty)(η), s=ξ+iη.
−
L F
It is well known that the function y is holomorphic in the half plane Res > a (see e.g. [8] or
[21]). In particular, for y L1(R) seLuch that y(t) = 0, for t < 0, and y(t) Aeat (a,A > 0) the
∈ | | ≤
Laplace transform is
y(s)= ∞y(t)e−stdt, Res>0.
L
Z0
3
We shall need the following properties: y,y ,y (R)
1 2 ′
∈S
dn
y (ξ) = (iξ)n y(ξ),
F dxn F
(cid:20) (cid:21)
[ Dαy](s) = sα y(s),
L0 t L
[y y ](s) = y (s) y (s),
1 2 1 1
L ∗ L ·L
δ(s) = 1.
L
In order to introduce the inverse Laplace transform we recall from [21] the following: Let Y be
a holomorphic function in the half plane Res > a such that |Y(s)| ≤ A(1R+e|ss|mk), m,k ∈ R. Then
there exists a distribution y (R) such that y =Y, and | |
∈S+′ L
y(t)= 1Y(t)= 1 a+i∞Y(s)estds, t>0.
−
L 2πi
Za−i∞
The notion of fundamental solution is introduced as follows (see e.g. [21]). Let P be a linear
partial integro-differential operator with constant coefficients. A fundamental solution of P, de-
noted by E, is a solution to the equation Pu = δ. Once the fundamental solution is determined
one finds a solution to Pu=f as u=E f.
∗
The Cauchy problem for the second order linear partialintegro-differentialoperator with con-
stant coefficients P is given by
Pu(x,t)=f(x,t) (9)
∂
u(x,0)=u (x), u(x,0)=u (x), (10)
0 ∂t 1
where f is continuous for t 0, u C1(R) and u C(R). A classical solution u(x,t) to the
0 1
≥ ∈ ∈
Cauchy problem (9-10) is of class C2 for t > 0, of class C1 for t 0, satisfies equation (9) for
≥
t>0,andinitialconditions(10)whent 0. Iffunctions uandf arecontinuedbyzerofort<0,
then the following equation is satisfied in→ (R2):
′
D
Pu=f(x,t)+u (x)δ (t)+u (x)δ(t). (11)
0 ′ 1
The classical solutions of the Cauchy problem (9-10) are among those solutions of equation
(11) that vanish for t <0. Therefore, the problem of finding generalized solutions (in (R2)) of
′
D
the equation (11) that vanish for t<0 will be called generalized Cauchy problem for the operator
P. If there is a fundamental solution E of the operator P and if f (R2) vanishes for t < 0
′
∈ D
then there exists a unique solution to corresponding generalized Cauchy problem and is given by
u=E (f(x,t)+u (x)δ (t)+u (x)δ(t)).
0 ′ 1
∗
Again, we refer to [8], [20] or [21] for more details.
3 Set up of the Cauchy problem
In order to obtain (1), we introduce dimensionless coordinates in the system (5-7), where (6) is
modified by the fractional Zener model of the viscoelastic body:
σ(x,t)+τ Dασ(x,t)=E[ε(x,t)+τ Dαε(x,t)], x R,t>0, (12)
σ0 t ε0 t ∈
where τ , τ are relaxation times satisfying τ > τ > 0. The latter condition follows from the
σ ε ε σ
Second Law of Thermodynamics (see [1]).
We subject to the system (5), (7) and (12) initial conditions
∂
u(x,0)=u (x), u(x,0)=v (x),
0 ∂t 0
σ(x,0)=0, ε(x,0)=0,
4
where x R, u and v are initial displacement and velocity. Note that there are no initial stress
0 0
∈
and strain. We also supply boundary conditions
lim u(x,t)=0, lim σ(x,t)=0, t 0.
x x ≥
→±∞ →±∞
We will need the following lemma.
Lemma 3.1. Let y AC([0,a]), a > 0, 0 α < 1 and T > 0. Let t and y be transformed
∗
as (t¯,y¯)= t ,y , th∈en the left Riemann-Lio≤uville fractional derivative Dαy(t), t 0, is trans-
T∗ 0 t ≥
formed as follows
(cid:0) (cid:1) Dαy¯(t¯)=(T )α Dαy(t).
0 t¯ ∗ 0 t
Proof. Since y¯(t¯)=y(t)=y(t¯T ), we have
∗
1 d t¯ y¯(ξ)
Dαy¯(t¯) = dξ
0 t¯ Γ(1 α)dt¯ (t¯ ξ)α
− Z0 −
t
= 1 T d T∗ y(ξT∗) dξ
Γ(1 α) ∗dt ( t ξ)α
− Z0 T∗ −
1 d t y(z)
= (T∗)α dz
Γ(1 α)dt (t z)α
− Z0 −
= (T )α Dαy(t).
∗ 0 t
2
Let0 α<1, τ >τ >0 and E be constants appearingin (12), and let ρ be the constantin
ε σ
≤
(5). Set T∗ = √ατε and X∗ = √ατε Eρ. Then X∗ and T∗ are constants (measured in meters and
seconds respectively) which satisfyqX∗ 2 ρ =1 and (T )α = 1.
T∗ E ∗ τε
We introduce dimensionless qua(cid:16)ntitie(cid:17)s in (5), (12) and (7) as
x t u σ
x¯= ; t¯= ; u¯= ; σ¯ =
X T X E
∗ ∗ ∗
and
u v τ
u¯ = 0 ; v¯ = 0 T ; τ¯= σ
0 X 0 X ∗ τ
∗ ∗ ε
and obtain system (1-3). Note that ε is already dimensionless quantity.
In order to simplify notation, a bar ¯ will be dropped and dimensionless quantities: x¯, t¯, σ¯,
u¯, τ¯, u¯ and v¯ will be written as: x,t,σ,u,τ,u and v . Obviously,condition 0<τ <τ implies
0 0 0 0 σ ε
0<τ <1.
In our work an appropriate setting for studying system (1-3) will be distributional one. In
fact, we will look for a fundamental solution to the generalized Cauchy problem for FZWE (4) in
(R R ). This suffices to obtain a solution to (1-3), since (1) and (4) are equivalent. Indeed,
′ +
S ×
by applying the Laplace transform with respect to time variable t to the second equation in (1),
one obtains
(1+τsα)σ(x,s)=(1+sα)ε(x,s).
According to [15], 1 1+sα is well-defined element in (R), hence
L− 1+τsα e Se+′
(cid:16) (cid:17)
1+sα
σ = 1 ε. (13)
L− 1+τsα ∗t
(cid:18) (cid:19)
Inserting ε from the third equationin(1) into (13) andthen σ into the first equationin(1) yields
∂2 1+sα ∂2
u(x,t)= 1 u(x,t). (14)
∂t2 L− 1+τsα ∗t ∂x2
(cid:18) (cid:19)
5
Setting L(t) = 1 1+sα we come to (4). Therefore we have proved that (1) and (4) are
L− 1+τsα ∗t
equivalent. Notice th(cid:16)at equa(cid:17)tion (14) is of the form Pu=0, with
∂2 1+sα ∂2
P := 1 . (15)
∂t2 −L− 1+τsα ∗t ∂x2
(cid:18) (cid:19)
4 The existence and uniqueness of a solution to the Cauchy
problem (4), (2)
The aim of this section is to find a solution to the generalized Cauchy problem to (4), i.e.
∂2 1+sα ∂2
u(x,t)= 1 u(x,t)+u (x)δ (t)+v (x)δ(t), (16)
∂t2 L− 1+τsα ∗t ∂x2 0 ′ 0
(cid:18) (cid:19)
or equivalently
Pu(x,t)=u (x)δ (t)+v (x)δ(t),
0 ′ 0
where P is given by (15).
Remark 4.1. We havealreadyexplainedthat initial conditions areincluded into the generalized
Cauchy problem (see Section 2). For functions in boundary conditions (3) are automatically
′
S
fulfilled.
We state the main theorem.
Theorem 4.2. Let u ,v (R). Then there exists a unique solution u (R R ) to (16)
0 0 ′ ′ +
∈ S ∈ S ×
given by
u(x,t)=S(x,t) (u (x)δ (t)+v (x)δ(t)), (17)
x,t 0 ′ 0
∗
where
S(x,t) = 1+ 1 ∞ 1+τqαeiαπe|x|qr11++τqqααeeiiααππ
4πiZ0 s 1+qαeiαπ
1+τqαe−iαπe|x|qr11++τqqααee−−iiααππ e−qt dq, (18)
−s 1+qαe−iαπ ! q
is the fundamental solution of operator P, S (R R ) with support in the cone x < t .
∈S′ × + | | √τ
We will need the following lemma.
Lemma 4.3. Let f (R). Then the equation
′
∈S
v ωv = f (19)
′′
− −
has a solution v (R) for all ω C ( ,0], which is of the form
′
∈S ∈ \ −∞
e √ωx
v = − | | f,
2√ω ∗
where √ω is the main branch.
Proof. We first apply Fourier transform to (19):
1
v = f. (20)
ξ2+ω
b b
6
In order to find v we need to calculate
1 1 1
1 (x)= eixξ dξ.
−
F ξ2+ω 2π R ξ2+ω
(cid:18) (cid:19) Z
Letx 0. LetΓ =Γ Γ ,whereΓ = z =Reiϕ R>0,0<ϕ<π andΓ = z =ξ R<
+ 1 2 1 2
≥ ∪ { | } { | −
ξ <R . By the Cauchy residual theorem we have that
}
k
1
eixz dz =2πi Res(z ),
z2+ω i
IΓ i=1
X
where z are poles of the function z eixz lying inside of Γ. For ω C ( ,0] there is only
i 7→ z2+ω ∈ \ −∞
one such pole z = i√ω and Res(z ) = eixz = e−x√ω. Letting R , the integral over Γ
1 1 2z z=z1 2i√ω → ∞ 1
tends to zero and eixz 1 dz = eixξ (cid:12)1 dξ, thus
Γ2 z2+ω R ξ2(cid:12)(cid:12)+ω
R R
1 1 1 e x√ω
1 (x)= eixz dz = − .
−
F ξ2+ω 2π z2+ω 2√ω
(cid:18) (cid:19) ZΓ2
For x < 0 and Γ = Γ Γ , where Γ = z = Reiϕ R > 0, π < ϕ < 0 and Γ =
3 eq:4 3 4
− ∪ { | − }
z = ξ R < ξ < R we can apply the same arguments as for the case x 0 to obtain
{ − | − } ≥
1 1 (x)= ex√ω.
F− ξ2+ω 2√ω
T(cid:16)hus fo(cid:17)r x R we can write
∈
1 e √ωx
1 (x)= − | |.
−
F ξ2+ω 2√ω
(cid:18) (cid:19)
Now, the inverse Fourier transform applied to (20) yields the result. 2
Proof of the main theorem. Applying Laplace transform with respect to t to (16) we
obtain:
∂2 1+τsα 1+τsα
u(x,s) s2 u(x,s)= (su (x)+v (x)). (21)
∂x2 − 1+sα − 1+sα 0 0
Setω(s):=s21+τsα andf(x,s):= 1+τsα(su (x)+v (x)),s C . InordertoapplyLemma4.3to
(21)wehaveto1+shsαowfirestthatf(,s1)+sα (eR0),which0follows∈from+ assumptionsthatu ,v (R)
′ 0 0 ′
and second, that ω(s) C ( ·,0],∈foSr all s C . The latter can be verified in the fo∈lloSwing
+
∈ \ −∞ ∈
way. Take arbitrary s = ρeiϕ, ρ > 0, π < ϕ < π. Then ω(s) = ρ2e2iϕ1+τραeiαϕ, and after a
−2 2 1+ραeiαϕ
straightforwardcalculation we arrive to
ρ2
Reω(s)= 1+ρα(1+τ)cos(αϕ)+τρ2α cos(2ϕ)+ρα(1 τ)sin(αϕ)sin(2ϕ) , (22)
A −
(cid:2)(cid:0) (cid:1) (cid:3)
ρ2
Imω(s)= 1+ρα(1+τ)cos(αϕ)+τρ2α sin(2ϕ) ρα(1 τ)sin(αϕ)cos(2ϕ) , (23)
A − −
where A=(1+ρα(cid:2)c(cid:0)os(αϕ))2+ρ2αsin2(αϕ)>0.(cid:1)Suppose that ω(s) ( ,0], for som(cid:3)e s C .
+
∈ −∞ ∈
Then Imω(s)=0, and (23) yields
cos(2ϕ)
1+ρα(1+τ)cos(αϕ)+τρ2α =ρα(1 τ)sin(αϕ) . (24)
− sin(2ϕ)
(cid:0) (cid:1)
From (22) we now obtain
ρ2 cos2(2ϕ)
Reω(s)= ρα(1 τ)sin(αϕ)sin(2ϕ) +1 0.
A − sin2(2ϕ) ≤
(cid:18) (cid:19)
7
where we have used (24) and the assumption ω(s) ( ,0], for all s C .
+
∈ −∞ ∈
Since ρ2ρα(1 τ)sin(αϕ) cos2(2ϕ) +1 >0,forρ>0, π <ϕ< π, itfollowsthatsin(2ϕ)<
A − sin2(2ϕ) −2 2
0, and therefore ϕ π, π(cid:16) π,π . H(cid:17)owever, (24) can not be satisfied for those ϕ. (Indeed,
∈ −2 −4 ∪ 2
for ϕ π, π , cos(αϕ),cos(2ϕ),sin(αϕ) > 0, but sin(2ϕ) < 0 and similarly for ϕ π,π .)
Hence,∈ω(s2)−4C ( (cid:0) ,0], for(cid:3)all(cid:2)s C(cid:1) . ∈ 4 2
+
(cid:0) ∈ (cid:3)\ −∞ ∈ (cid:2) (cid:1)
Now we can apply Lemma 4.3 to obtain a solution of (21):
e √ω(s)x ω(s)
− | |
u(x,s) = (su (x)+v (x))
2 ω(s) ∗x s2 0 0
e ωp(s)e−√ω(s)|x|
= (su (x)+v (x)). (25)
2s2 ∗x 0 0
p
To obtain a solution u to (4) we need to calculate inverse Laplace transform of (25). For that
purpose set
S(x,s):= ω(s)e2−s2√ω(s)|x| = 21s 11++τssααe−|x|sq11++τssαα, x∈R,s∈C+, (26)
p r
and e
S(x,t)= 1 S(x,s) (t), x R,t>0.
−
L ∈
S(x,t) is well-defined since |S(x,s)| ≤ Ch(e1+R|es|sm)i, m ∈ N, so its inverse Laplace transform exists
| |
(cf. Section 2).
Note also that S is a fundeamental solution of P. Then the solution u is given by (17).
Multiform function S, given by (26), has branch points at s = 0 and s = and has no
∞
singularities. Hence, it can be evaluated by the use of the Cauchy integral formula
e
S(x,s)estds=0, x R,t>0, (27)
∈
IΓ
where Γ=Γ1 Γ2 Γε Γ3 Γ4e γ0, is a contour given in Figure 1.
∪ ∪ ∪ ∪ ∪
Im s
G
1
g
0
G G
2 e
Re s
G
3
G
4
Figure 1: Integration contour Γ
More precisely, for arbitrarily chosen R>0, 0<ε<R and a>0, Γ is defined by
Γ : s=Reiϕ,ϕ <ϕ<π;
1 0
Γ : s=qeiπ, R< q < ε;
2
− − −
Γ : s=εeiϕ, π <ϕ<π;
ε
−
Γ : s=qe iπ,ε<q <R;
3 −
Γ : s=Reiϕ, π <ϕ< ϕ ;
4 0
− −
γ : s=a(1+itanϕ), ϕ <ϕ<ϕ ,
0 0 0
−
8
where ϕ = arccos(a). Note that lim ϕ = π. In the limit when R , integral along
contour0Γ reads (xR R,t>0) R→∞ 0 2 → ∞
1
∈
1 π 1+τRαeiαϕ xReiϕ 1+τRαeiαϕ+Rteiϕ
lim S(x,s)estds= lim e−| | r 1+Rαeiαϕ idϕ. (28)
R→∞ZΓ1 2R→∞Zϕ0r 1+Rαeiαϕ
Evaluating the eabsolute value of S(x,s)estds one obtains (x R,t>0)
Γ1 ∈
R
1 e π 1+τRαeiαϕ xReiϕ 1+τRαeiαϕ
lim S(x,s)estds lim e−| | r 1+Rαeiαϕ eRtcosϕdϕ.
In thRe→li∞mi(cid:12)(cid:12)(cid:12)(cid:12)tZΓw1heen R t(cid:12)(cid:12)(cid:12)(cid:12)h≤e e2xRpr→e∞ssiZoϕn0(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)r1+1τR+αReiααϕeiαteϕnd(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)s(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) to √τ and therefore(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)
→∞ 1+Rαeiαϕ
q
√τ π
lim S(x,s)estds lim eRcosϕ(t x√τ)dϕ=0, if t> x√τ,
−| |
R→∞(cid:12)ZΓ1 (cid:12)≤ 2 R→∞Zϕ0 | |
(cid:12) (cid:12)
since cosϕ<0(cid:12)(cid:12) foreϕ∈ π2,π .(cid:12)(cid:12)Similar argument is valid for the integral along Γ4.
In the limit when ε 0, the integral along Γ is given by the formula similar to (28) and it is
ε
→(cid:0) (cid:1)
calculated as
lim S(x,s)estds= 1 lim −π 1+τεαeiαϕe−|x|εeiϕr11++τεεααeeiiααϕϕ+εteiϕidϕ= iπ.
ε→0ZΓε 2ε→0Zπ r 1+εαeiαϕ −
Integrals alongecontours: Γ , Γ and γ , in the limit when R , ε 0, give
2 3 0
→∞ →
lim S(x,s)estds = 1 ∞ 1+τqαeiαπe−q(cid:18)t−|x|r11++τqqααeeiiααππ(cid:19)dq
Rε→∞0 ZΓ2 −2Z0 s 1+qαeiαπ q
→
lim Se(x,s)estds = 1 ∞ 1+τqαe−iαπe−q(cid:18)t−|x|r11++τqqααee−−iiααππ(cid:19)dq
Rε→∞0 ZΓ3 2Z0 s 1+qαe−iαπ q
→
e
lim S(x,s)estds = 2πiS(x,t).
R→∞Zγ0
Now, by the Cauchyeintegral formula (27), we obtain S as in (18). Therefore we have proved the
theorem. 2
As a consequence of what we proved in Sections 3 and 4, we have the following corollary.
Corollary 4.4. Let u be given by (17). Then
∂ 1+sα ∂
(u,ε,σ)(x,t)= u(x,t),∂xu(x,t),L−1 1+τsα ∗t ∂xu(x,t) ∈(S′(R×R+))3,
(cid:18) (cid:18) (cid:19) (cid:19)
is a unique solution to the system (1-3).
In a similar way one can also consider nonhomogeneous case.
Remark 4.5. Let the first equation in (1) be replaced by
∂ ∂2
σ(x,t)= u(x,t)+f(x,t),
∂x ∂t2
where f (R R ). This is a case of a rod under the influence of body forces. Then, the
′ +
∈ S ×
solution of the generalized Cauchy problem
∂2 1+sα ∂2
u(x,t)= 1 u(x,t)+f(x,t)+u (x)δ (t)+v (x)δ(t),
∂t2 L− 1+τsα ∗t ∂x2 0 ′ 0
(cid:18) (cid:19)
9
is
u(x,t)=S(x,t) (f(x,t)+u (x)δ (t)+v (x)δ(t))
x,t 0 ′ 0
∗
where S is as in (18).
Remark 4.6. Dimensionless condition x < t , or x < ρ τσt inthe dimensionalform,can
| | √τ | | E τε
be physically interpreted as the wave property. Namely,pin tqhe moment t, wave caused by the
initial disturbance u (x) = δ(x), x R, has reached the point which is at distance x from the
0
∈ | |
origin,wheretheinitialdisturbancewasapplied. Thus,constantcZ = Eρ ττσε canbeinterpreted
as the wave speed in the fractional viscoelastic media of Zener type. p q
5 A numerical example
Let u =δ and v =0 in the solution (17) to the Cauchy problem of wave equation for fractional
0 0
Zener type viscoelastic media. Then solution reads
∂
u(x,t)= S(x,t), x R,t>0, (29)
∂t ∈
i.e. the fundamental solution represents the solution itself. Figure 2 presents the plots of u(x,t),
x (0,3), t 0.5,1,1.5 ,given by (29), for the set of parameters: α=0.23 and τ =0.004.
∈ ∈{ }
uHx,tL
1.4
1.2 t=0.5
1
0.8
0.6 t=1
0.4 t=1.5
0.2
x
0.5 1 1.5 2 2.5 3
Figure 2: Solution u(x,t), x (0,3), t 0.5,1,1.5
∈ ∈{ }
Inordertoseetheeffectofachangeintheorderofafractionalderivative,wepresentfollowing
figures. Figure 3 presents plots of u(x,t), x (0,3), t 0.5,1,1.5 , given by (29), for different
∈ ∈ { }
values ofparameter α. Plotof u, denoted by the dashed line correspondsto α=0.25,dot-dashed
line is used for α=0.5, while the solid line denotes the plot for α=0.75. Parameter τ =0.004 is
the same in all three figures.
uHx,tL uHx,tL uHx,tL
1.2 t=0.5 0.5 t=0.5 0.35 t=0.5
1 0.3
0.4
0.8 0.25 t=1
0.3 t=1 0.2
0.6 t=1
0.4 t=1.5 0.2 t=1.5 0.15 t=1.5
0.1
0.2 0.1 0.05
x x x
0.5 1 1.5 2 2.5 3 3.5 1 2 3 4 5 6 1 2 3 4 5 6
Figure 3: Solution u(x,t), x (0,6), t 0.5,1,1.5 ,α 0.25,0.5,0.75
∈ ∈{ } ∈{ }
From figure 3 one can see that for each value of the order of fractional derivative α, as time
increases,the maximumvalue ofu decreases,whichis the consequenceof the dissipative modelof
a media.
10