Abstract

We discuss thermodynamical restrictions for a linear constitutive equation containing fractional derivatives of stress and strain of different orders. Such an equation generalizes several known models. The restrictions on coefficients are derived by using entropy inequality for isothermal processes. In addition, we study waves in a rod of finite length modelled by a linear fractional constitutive equation. In particular, we examine stress relaxation and creep and compare results with the quasistatic analysis.

1. Introduction

Fractional calculus is intensively used to describe various phenomena in physics and engineering. We mention just few: diffusion and heat conduction processes [1, 2], damping in inelastic bodies [3], thermoelasticity [4], dissipative bending of rods [5], and concrete behavior of rods [6].

Generalizations of classical equations of mathematical physics in order to include fractional derivatives, can be conducted in several ways. In the first approach, one changes the ordinary and partial integer order derivatives with the fractional ones in the relevant system of equations. Often, in this approach, the physical meaning of the newly defined terms with fractional derivatives might become unclear. In the second approach, one defines Lagrangian density function in which the integer order derivatives are replaced with fractional ones. In the next step, such a modified Lagrangian is used in the Hamilton principle (minimization of action integral) to obtain a system of the Euler-Lagrange equations, that describes the process (cf. [7, 8]). Although this variational approach have more sound physical meaning, it is not frequently used in applications since it, in principle, leads to equations with both left and right fractional derivatives. In both methods, the obtained equations have to satisfy requirements following from the Second Law of Thermodynamics, often expressed in the form of the Clausius-Duhamel inequality (see e.g., [9, 10]).

In this work we propose a new model for linear viscoelastic body (obtained from the one-dimensional model) with an arbitrary number of springs and dashpots (see e.g., [11, page 28]) by replacing integer order derivatives with the Riemann-Liouville derivatives of real order (the first approach). Thus, we consider the class of constitutive equations of the form𝑁𝑛=0𝑎𝑛0𝐷𝛼𝑛𝑡𝜎=𝑀𝑚=0𝑏𝑚0𝐷𝛽𝑚𝑡𝜀,𝑡>0.(1.1) Here 𝜎 denotes the Cauchy stress, 𝜀 the strain at the time instant 𝑡, while 0𝐷𝜂𝑡, 𝜂∈[0,1], denotes the operator of the left Riemann-Liouville fractional derivation. Recall, for 𝜂∈[0,1], the left 𝜂th order Riemann-Liouville fractional derivative 0𝐷𝜂𝑡𝑦 of a function 𝑦∈AC([0,𝑇]), 𝑇>0, is defined as 0𝐷𝜂𝑡1𝑦(𝑡)∶=𝑑Γ(1−𝜂)𝑑𝑡𝑡0𝑦(𝜏)(𝑡−𝜏)𝜂𝑑𝜏,𝑡>0,(1.2) where Γ is the Euler gamma function and AC([0,𝑇]), 𝑇>0, denotes the space of absolutely continuous functions (for a detailed account on fractional calculus see e.g., [12]). For technical purposes, the orders of the fractional derivatives in (1.1) are assumed to satisfy 0≤𝛼0<𝛼1<⋯<𝛼𝑛<⋯<𝛼𝑁≤1,0≤𝛽0<𝛽1<⋯<𝛽𝑚<⋯<𝛽𝑀≤1.(1.3) Also, in (1.1), the coefficients {𝑎𝑛}𝑛=0,…,𝑁, {𝑏𝑚}𝑚=0,…,𝑀 have the physical meaning of relaxation times and are assumed to be given. Many known constitutive equations of one-dimensional viscoelasticity are included as special cases of (1.1). For example, the choice 𝛼0=𝛽0=0, 𝛼1=𝛽1=1, 𝑎0=𝑏0=1, 𝑎1=𝑎 and 𝑏1=𝑏, and all other coefficients being equal to zero, leads to the classical Zener model (see [13]). If 𝑎0=𝑏0=1, 𝛼0=𝛽0=0, 𝑎1=𝑎, 𝑏1=𝑏, 𝛼1=𝛽1=𝛼, where 𝛼∈[0,1], and all other coefficients vanishing, then one has the generalized Zener model (see [9]). It is known that the Clausius-Duhamel inequality in both cases restricts constants 𝑎 and 𝑏 so that 𝑎≤𝑏. However, we are not aware of restrictions on coefficients and orders of derivatives in generalized constitutive equation given by (1.1). Thus, in Section 2 of this work we shall derive these restrictions.

In the second part of this work, we chose one specific equation of the type (1.1), proposed in [14]. Namely, the constitutive equation takes the form𝑎1+𝑏0𝐷𝑡𝛼−𝛽𝑎𝜎(𝑥,𝑡)=𝐸0𝐷𝛼𝑡+𝑐0𝐷𝛾𝑡+𝑎𝑐𝑏0𝐷𝑡𝛼+𝛾−𝛽[]𝜀(𝑥,𝑡),𝑥∈0,𝐿,𝑡>0,(1.4) where 𝐸 is the generalized Young modulus (positive constant having dimension of stress), 𝑎, 𝑏, and 𝑐 are given positive constants, while 0<𝛽<𝛼<𝛾<1/2. It should be stressed that in [14] there is no discussion concerning restrictions on the parameters 𝑎, 𝑏, 𝑐, 𝛼, 𝛽, and 𝛾. Instead, only 𝑎,𝑏,𝑐≥0, 0≤𝛽<𝛼≤1 and 0≤𝛾≤1 were assumed. We will show, by the use of the method presented in [10], that conditions 𝑎,𝑏,𝑐≥0, and 0<𝛽<𝛼<𝛾<1/2 guarantee that the dissipation inequality is satisfied. The constitutive equation of the form (1.4) is obtained in [14] via the rheological model that generalizes the classical Zener rheological model by substituting spring and dashpot elements by fractional elements. It is assumed that the stress-strain relation for the fractional element is given by 𝜎(𝑡)=0𝐷𝜂𝑡𝜀, 𝑡>0, where 𝜂∈[0,1]. We refer to [14] for more details of the derivation, creep compliance, and relaxation modulus of (1.4).

The constitutive equation (1.4) actually describes a viscoelastic fluid body. For such a body of finite dimension we will analyze the wave propagation as well as two characteristic properties of the material: stress relaxation and creep. In this respect, the present work is a continuation of previous investigations (cf. [15–18]), where waves in viscoelastic solid body have been studied. We will consider the equation of motion of one-dimensional continuous body as𝜕𝜕𝜕𝑥𝜎(𝑥,𝑡)=𝜌2𝜕𝑡2[]𝑢(𝑥,𝑡),𝑥∈0,𝐿,𝑡>0,(1.5) where 𝜌, 𝜎, and 𝑢 denote density, stress, and displacement of material at a point positioned at 𝑥 and time 𝑡, respectively, as well as the strain measure, defined by𝜕𝜀(𝑥,𝑡)=[]𝜕𝑥𝑢(𝑥,𝑡),𝑥∈0,𝐿,𝑡>0.(1.6) These two equations are coupled with a constitutive equation (1.4). We will impose initial and two types of boundary conditions to system (1.4), (1.5), and (1.6). The first type of boundary conditions describes the rod fixed at one end (displacement is zero during the time), while the other end is subject to a prescribed displacement. The second type describes the same rod but its other end is subject to the prescribed stress. We will obtain solutions in both cases in the convolution form of boundary conditions and the kernel of certain type. This setting is appropriate for examining stress relaxation and creep processes. If the displacement of the body's free end is prescribed as the Heaviside function then one can examine stress relaxation, while prescribing the stress of rod's free end by the Heaviside function will enable the study of creep.

Similar problems have already been studied by several authors. System (1.5) and (1.6) coupled with the constitutive law of distributed-order type10𝜙𝜎(𝜂)0𝐷𝜂𝑡𝜎(𝑥,𝑡)𝑑𝜂=𝐸10𝜙𝜀(𝜂)0𝐷𝜂𝑡[]𝜀(𝑥,𝑡)𝑑𝜂,𝑥∈0,𝐿,𝑡>0,(1.7) with 𝜙𝜎(𝜂)∶=𝑎𝜂 and 𝜙𝜀(𝜂)∶=𝑏𝜂, 𝑎≤𝑏, was considered in [15] for the case of the stress relaxation in a viscoelastic material described by (1.7), and in [16] for the case of the creep and forced vibrations in a viscoelastic material of the same type. Special cases of (1.7) were studied in [19] (with 𝜙𝜎(𝜂)∶=𝛿(𝜂)+𝜏𝛼𝜀𝛿(𝜂−𝛼) and 𝜙𝜀(𝜂)∶=𝐸∞𝜏𝛽𝜀𝛿(𝜂−𝛽), where 𝛿 is the Dirac distribution), and in [20] (with 𝜙𝜎(𝜂)∶=𝛿(𝜂)+𝜏𝛼𝜀𝛿(𝜂−𝛼) and 𝜙𝜀(𝜂)∶=𝐸0(𝛿(𝜂)+𝜏𝛼𝜎𝛿(𝜂−𝛼)+𝜏𝛽𝜎𝛿(𝜂−𝛽)). Also our constitutive law (1.4) follows from (1.7) by choosing 𝜙𝜎(𝜂)∶=𝛿(𝜂)+(𝑎/𝑏)𝛿(𝜂−(𝛼−𝛽)) and 𝜙𝜀(𝜂)∶=𝑎𝛿(𝜂−𝛼)+𝑐𝛿(𝜂−𝛾)+(𝑎𝑐/𝑏)𝛿(𝜂−(𝛼+𝛾−𝛽)).

We stress here that there is a strong connection between thermodynamical restrictions on coefficients in (1.1) and conditions for the existence of the inverse Laplace transform of equation of motion (see [17, 18]). It has been proved that the thermodynamical restrictions guarantee the existence of solutions.

2. Thermodynamical Restrictions

In this section we consider generalized linear fractional model and distributed-order fractional model of a viscoelastic body and give thermodynamical restrictions on coefficients and orders of fractional derivatives that appear in those models.

2.1. Generalized Linear Fractional Model

Our aim is to find restrictions on parameters of the model (1.1), that is, on {𝛼𝑛}𝑛=0,…,𝑁, {𝛽𝑚}𝑚=0,…,𝑀, {𝑎𝑛}𝑛=0,…,𝑁, {𝑏𝑚}𝑚=0,…,𝑀, so that the generalized linear fractional model of a viscoelastic body satisfies the requirements of the Second Law of Thermodynamics.

Following the procedure analogous to the one presented in [13], we apply the Fourier transform to (1.1) and obtain𝜎(𝜔)𝑁𝑛=0𝑎𝑛(𝑖𝜔)𝛼𝑛=̂𝜀(𝜔)𝑀𝑚=0𝑏𝑚(𝑖𝜔)𝛽𝑚,𝜔∈ℝ,(2.1) where ∫𝑓(𝜔)=ℱ[𝑓(𝑡)](𝜔)=∞−∞𝑓(𝑡)𝑒−𝑖𝜔𝑡𝑑𝑡, 𝜔∈ℝ, and ℱ[0𝐷𝜂𝑡𝑓](𝜔)=(𝑖𝜔)𝜂𝑓(𝜔). Writing (2.1) in the form 𝜎(𝜔)=𝐸(𝜔)̂𝜀(𝜔),𝜔∈ℝ,(2.2) with 𝐸 being the complex modulus defined as 𝐸∑(𝜔)∶=𝑀𝑚=0𝑏𝑚(𝑖𝜔)𝛽𝑚∑𝑁𝑛=0𝑎𝑛(𝑖𝜔)𝛼𝑛,𝜔∈ℝ,(2.3) one uses the conditions (cf. [10, 13]) Re𝐸(𝜔)≥0,∀𝜔>0,(2.4)Im𝐸(𝜔)≥0,∀𝜔>0,(2.5) which follow from the Second Law of Thermodynamics in case of the isothermal process, in order to obtain restrictions on parameters 𝛼𝑛, 𝛽𝑚, 𝑎𝑛, and 𝑏𝑚.

A straightforward calculation yields 𝐸∑(𝜔)=𝑀𝑚=0𝑏𝑚𝜔𝛽𝑚𝛽cos𝑚∑𝜋/2+𝑖𝑀𝑚=0𝑏𝑚𝜔𝛽𝑚𝛽sin𝑚𝜋/2∑𝑁𝑛=0𝑎𝑛𝜔𝛼𝑛𝛼cos𝑛∑𝜋/2+𝑖𝑁𝑛=0𝑎𝑛𝜔𝛼𝑛𝛼sin𝑛.𝜋/2(2.6) Introducing 𝐸 as 𝐸|||||(𝜔)=𝐸(𝜔)𝑁𝑛=0𝑎𝑛(𝑖𝜔)𝛼𝑛|||||2,(2.7) one obtainsRe𝐸(𝜔)=𝑁𝑛=0𝑎𝑛𝜔𝛼𝑛𝛼cos𝑛𝜋2𝑀𝑚=0𝑏𝑚𝜔𝛽𝑚𝛽cos𝑚𝜋2+𝑁𝑛=0𝑎𝑛𝜔𝛼𝑛𝛼sin𝑛𝜋2𝑀𝑚=0𝑏𝑚𝜔𝛽𝑚𝛽sin𝑚𝜋2,Im𝐸(𝜔)=𝑁𝑛=0𝑎𝑛𝜔𝛼𝑛𝛼cos𝑛𝜋2𝑀𝑚=0𝑏𝑚𝜔𝛽𝑚𝛽sin𝑚𝜋2−𝑁𝑛=0𝑎𝑛𝜔𝛼𝑛𝛼sin𝑛𝜋2𝑀𝑚=0𝑏𝑚𝜔𝛽𝑚𝛽cos𝑚𝜋2.(2.8)

Since 𝛼𝑛,𝛽𝑚∈[0,1], 𝑛=0,1,…,𝑁, 𝑚=0,1,…,𝑀, we have that 𝛼𝑛𝜋/2,𝛽𝑚𝜋/2∈[0,𝜋/2], and consequently, sine and cosine of those angles are positive. Therefore, assuming 𝑎𝑛,𝑏𝑚≥0 we obtain that Re𝐸(𝜔)≥0, and hence (2.4) is satisfied. In the sequel we will restrict our attention to this case (i.e., 𝑎𝑛,𝑏𝑚≥0).

After a straightforward calculation of (2.8) one concludes that (2.5) holds if and only if Im𝐸(𝜔)=−𝑛∈{0,1,…,𝑁}𝑚∈{0,1,…,𝑀}𝜔𝛼𝑛+𝛽𝑚𝛼sin𝑛−𝛽𝑚𝜋2𝑎𝑛𝑏𝑚≥0,∀𝜔>0.(2.9)

Lemma 2.1. Let (1.3) hold and 𝑎𝑛,𝑏𝑚≥0, 𝑛=0,1,…,𝑁, 𝑚=0,1,…,𝑀. Suppose that 𝛼𝑁≠𝛽𝑀. Then a necessary condition for (2.9) is that 𝛼𝑁<𝛽𝑀.
In other words, the highest order of fractional derivatives of stress in (1.1) could not be greater than the highest order of fractional derivatives of strain.

Proof. We observe that for large 𝜔 the sign of Im𝐸(𝜔) coincides with the sign of the term in the sum on the right-hand side of (2.9) with the largest power of 𝜔. It follows from (1.3) that the latter is achieved for 𝛼𝑁 and 𝛽𝑀. Therefore, in order that Im𝐸(𝜔)>0, 𝛼𝑁 has to be less than 𝛽𝑀, as claimed.

Remark 2.2. (i) Similarly as in Lemma 2.1 one can prove that if 𝛼𝑁=𝛽𝑀 then a necessary condition becomes 𝛼𝑛<𝛽𝑚, for the largest 𝛼𝑛 and 𝛽𝑚 which do not coincide.
(ii) A similar conclusion for a particular problem with the constitutive equation of the form (1.1) has been obtained in [21].

Example 2.3. Equation of the form 𝑎0𝜎+𝑎𝛼0𝐷𝛼𝑡𝜎=𝑏0𝜀, 0<𝛼<1, cannot be a constitutive equation of a viscoelastic body. It does not obey the Second Law of Thermodynamics. Indeed, (2.9) reduces to −𝜔𝛼sin(𝛼𝜋/2)𝑎𝛼𝑏0, which is strictly less than zero for all 𝜔>0.

For practical purposes it can happen that the condition (2.9) is too general and thus hardly applicable to concrete problems. Therefore, it will be useful to extract from (2.9) particular conditions on parameters 𝑎𝑛,𝑏𝑚,𝛼𝑛 and 𝛽𝑚 which guarantee that (2.5) is satisfied. We will separately consider several possibilities:

(1) 𝛼𝑛≠𝛽𝑚, for all 𝑛,𝑚, that is, there are 𝑁+1 and 𝑀+1 terms of different order in (1.1).

Since we assumed that 𝑎𝑛,𝑏𝑚≥0, we can choose the orders of fractional derivatives as 𝛼𝑛≤𝛽𝑚, forall𝑛,𝑚, so that all sin((𝛼𝑛−𝛽𝑚)𝜋/2) are negative, and consequently Im𝐸(𝜔)≥0. This, together with (1.3), further leads to the following condition: 0≤𝛼0<𝛼1<⋯<𝛼𝑛<⋯<𝛼𝑁<𝛽0<𝛽1<⋯<𝛽𝑚<⋯<𝛽𝑀≤1.(2.10) In other words, 𝑎𝑛,𝑏𝑚≥0 and (2.10) are sufficient conditions for (2.5).

Remark 2.4. Note that the constitutive equation (1.4) belongs to this class. Indeed, by setting 𝑁=1, 𝑎0=1, 𝛼0=0, 𝑎1=𝑎/𝑏, 𝛼1=𝛼−𝛽, 𝑀=2, 𝑏0=𝐸𝑎, 𝛽0=𝛼, 𝑏1=𝐸𝑐, 𝛽1=𝛾, 𝑏2=𝐸𝑎𝑐/𝑏, 𝛽2=𝛼+𝛾−𝛽 in (1.1) we obtain (1.4). Thermodynamical restrictions are satisfied since constants 𝐸, 𝑎, 𝑏, 𝑐 are positive, while (2.10) in this case reads 0≤𝛼−𝛽<𝛼<𝛾<𝛼+𝛾−𝛽≤1,(2.11) and is satisfied according to assumption 0<𝛽<𝛼<𝛾<1/2.

(2) 𝑀>𝑁 and 𝛼𝑖=𝛽𝑖, 𝑖=0,1,…,𝑁, that is, there are 𝑁+1 first terms of the same order and 𝑀−𝑁 terms left in (1.1).

Then (2.9) reduces to Im𝐸(𝜔)=𝑖,𝑗∈{0,1,…,𝑁},𝑖<𝑗𝜔𝛼𝑖+𝛼𝑗𝛼sin𝑗−𝛼𝑖𝜋2𝑎𝑖𝑏𝑗−𝑎𝑗𝑏𝑖−𝑖∈{0,1,…,𝑁},𝑚>𝑁𝜔𝛼𝑖+𝛽𝑚𝛼sin𝑖−𝛽𝑚𝜋2𝑎𝑖𝑏𝑚≥0∀𝜔>0.(2.12) We are interested in a special case when each term in the above sum is nonnegative. Then the following should hold: 𝑎𝑖𝑏𝑖≥𝑎𝑗𝑏𝑗𝛼,∀𝑖,𝑗∈{0,1,…,𝑁}∶𝑖<𝑗,𝑖≤𝛽𝑚,∀𝑖∈{0,1,…,𝑁},𝑚>𝑁,(2.13) which implies that 𝑎0𝑏0≥𝑎1𝑏1𝑎≥⋯≥𝑁𝑏𝑁≥0,0≤𝛼0<𝛼1<⋯<𝛼𝑁<𝛽𝑁+1<⋯<𝛽𝑀≤1.(2.14)

(3) 𝑁>𝑀 and 𝛼𝑁−𝑀+𝑖=𝛽𝑖, 𝑖=0,1,…,𝑀, that is, there are 𝑀+1 last terms of the same order and 𝑁−𝑀 terms left in (1.1).

Then (2.9) reduces to Im𝐸(𝜔)=−𝑛<𝑁−𝑀,𝑖∈{0,1,…,𝑀}𝜔𝛼𝑛+𝛼𝑁−𝑀+𝑖𝛼sin𝑛−𝛼𝑁−𝑀+𝑖𝜋2𝑎𝑛𝑏𝑖+𝑖,𝑗∈{0,1,…,𝑀},𝑖<𝑗𝜔𝛽𝑖+𝛽𝑗𝛽sin𝑗−𝛽𝑖𝜋2𝑎𝑁−𝑀+𝑖𝑏𝑗−𝑎𝑁−𝑀+𝑗𝑏𝑖≥0,∀𝜔>0.(2.15) Again, we concentrate on a special case when each term in the above sum is nonnegative. By assumption (1.3) it follows that all terms in the first part of the sum are positive. Thus, 𝑎𝑁−𝑀+𝑖𝑏𝑖≥𝑎𝑁−𝑀+𝑗𝑏𝑗,∀𝑖,𝑗∈{0,1,…,𝑀}∶𝑖<𝑗,(2.16)

which implies 𝑎𝑁−𝑀𝑏0≥𝑎𝑁−𝑀+1𝑏1𝑎≥⋯≥𝑁𝑏𝑀≥0.(2.17)

(4) 𝑁=𝑀 and 𝛼𝑖=𝛽𝑖, 𝑖=0,1,…,𝑁, that is, there are 𝑁+1 terms of the same order on both sides of (1.1).

In this case (2.9) becomes Im𝐸(𝜔)=𝑖,𝑗∈{0,1,…,𝑁},𝑖<𝑗𝜔𝛼𝑖+𝛼𝑗𝛼sin𝑗−𝛼𝑖𝜋2𝑎𝑖𝑏𝑗−𝑎𝑗𝑏𝑖≥0,∀𝜔>0.(2.18) As in the previous cases, we look for those parameters for which all terms in the above sum are nonnegative. Hence, we have to restrict our attention only to those parameters 𝑎𝑛,𝑏𝑚≥0 which in addition satisfy 𝑎𝑖𝑏𝑖≥𝑎𝑗𝑏𝑗,∀𝑖,𝑗∶𝑖<𝑗.(2.19) This implies that 𝑎0𝑏0≥𝑎1𝑏1𝑎≥⋯≥𝑁𝑏𝑁≥0.(2.20)

Remark 2.5. It can be shown that in other cases which are not listed above (e.g., when there are 𝐾<min{𝑁,𝑀} terms of the same order in (1.1)) it is not possible to make all terms in (2.9) nonnegative simultaneously, thus it is more difficult to find some particular conditions on parameters 𝑎𝑛, 𝑏𝑚, 𝛼𝑛, and 𝛽𝑚 which implies (2.5).

2.2. Distributed-Order Fractional Model

The constitutive equation (1.1) can be generalized by (1.7), and further, if the integrals in (1.7) are interpreted in the distributional setting as supp𝜙𝜙(𝛼)0𝐷𝛼𝑡ℎ𝑑𝛼,𝜑(𝑡)∶=𝜙(𝛼),0𝐷𝛼𝑡ℎ,𝜑(𝑡),𝜑∈𝒟(ℝ),(2.21) then (1.1) is generalized by 𝜙𝜎(𝛼),0𝐷𝛼𝑡=𝜙𝜎,𝜑(𝑡)𝜀(𝛼),0𝐷𝛼𝑡𝜀,𝜑(𝑡),𝜑∈𝒟(ℝ)(2.22) (cf. [22]). Here 𝜙𝜎 and 𝜙𝜀 are positive integrable functions of 𝛼, 𝛼∈[0,1], or distributions with compact support in [0,1], and 𝜑 denotes a test function belonging to the space 𝒟(ℝ) of compactly supported smooth functions on ℝ.

Notice that by setting 𝜙𝜎∑(𝛼)∶=𝑁𝑛=0𝑎𝑛𝛿(𝛼−𝛼𝑛), and 𝜙𝜀∑(𝛼)∶=𝑀𝑚=0𝑏𝑚𝛿(𝛼−𝛽𝑚) one obtains (1.1).

In the sequel, we will consider 𝜙𝜎(𝛼)∶=𝑎𝛼 and 𝜙𝜀(𝛼)∶=𝑏𝛼 and look for the restrictions on 𝑎 and 𝑏. Then constitutive equation (1.7) becomes ∫10𝑎𝛼0𝐷𝛼𝑡∫𝜎𝑑𝛼=10𝑏𝛼0𝐷𝛼𝑡𝜀𝑑𝛼, 𝑡>0. Interpreting these integrals as the Riemann sums, one obtains 𝑁𝑛=0𝑎𝛼𝑛0𝐷𝛼𝑛𝑡𝜎Δ𝛼𝑛=𝑁𝑛=0𝑏𝛼𝑛0𝐷𝛼𝑛𝑡𝜀Δ𝛼𝑛,(2.23) where 𝑁→∞ and Δ𝛼𝑛→0. Putting 𝑎𝑛∶=𝑎𝛼𝑛Δ𝛼𝑛 and 𝑏𝑛=𝑏𝛼𝑛Δ𝛼𝑛, one obtains (1.1) with 𝑁=𝑀 and all terms of the same order.

Taking 𝑎,𝑏>0 we have that (2.4) holds, while, as in the case (4) above (with equal number of terms of the same order on both sides of (1.1)), condition 𝑎𝑏𝛼𝑖≥𝑎𝑏𝛼𝑗,∀𝑖,𝑗∶𝑖<𝑗,(2.24) provides validity of (2.5). Since 𝛼𝑖≤𝛼𝑗, for all 𝑖,𝑗, 𝑖<𝑗, the latter is equivalent to 𝑎𝑏<1,thatis,𝑎<𝑏.(2.25)

3. A Model of Viscoelastic Body of Finite Length

In this section we analyze wave propagation in a viscoelastic rod of finite length. The rod is made of a viscoelastic material described by a fractional-type constitutive equation (1.4), which is a special case of generalized linear fractional model (1.1). In fact, we will study an initial boundary value problem for system (1.4), (1.5), and (1.6).

3.1. Convolution Form of Solutions

Consider system (1.4), (1.5), and (1.6) supplied with initial conditions 𝜕𝑢(𝑥,0)=0,[],𝜕𝑡𝑢(𝑥,0)=0,𝜎(𝑥,0)=0,𝜀(𝑥,0)=0,𝑥∈0,𝐿(3.1) as well as with two types of boundary conditions:𝑢(0,𝑡)=0,𝑢(𝐿,𝑡)=Υ(𝑡),𝑡∈ℝ,𝑢(0,𝑡)=0,𝜎(𝐿,𝑡)=Σ(𝑡),𝑡∈ℝ.(3.2) Functions Υ and Σ are locally integrable functions supported in [0,∞). Note that if Υ=Υ0𝐻 we have the case of stress relaxation, while if Σ=Σ0𝐻 we have the case of creep, where 𝐻 is the Heaviside function.

Introducing dimensionless quantities 𝑥𝑥=𝐿,𝑡𝑡=𝐿√,𝜌/𝐸𝑢𝑢=𝐿,𝜎𝜎=𝐸,ΥΥ=𝐿,ΣΣ=𝐸,𝑎𝑎=𝐿√𝜌/𝐸𝛼,𝑏𝑏=𝐿√𝜌/𝐸𝛽,𝑐𝑐=𝐿√𝜌/𝐸𝛾,(3.3) and using the fact that fractional derivatives are transformed as 0𝐷𝜂𝑡𝑢(√𝑡)=(𝐿𝜌/𝐸)𝜂0𝐷𝜂𝑡𝑢(𝑡) (cf. [17, 23]), after omitting the bar over dimensionless quantities, we obtain the following system: 𝑥∈[0,1], 𝑡>0, 𝜕𝜕𝜕𝑥𝜎(𝑥,𝑡)=2𝜕𝑡2𝜕𝑢(𝑥,𝑡),𝜀(𝑥,𝑡)=𝑎𝜕𝑥𝑢(𝑥,𝑡),1+𝑏0𝐷𝑡𝛼−𝛽𝑎𝜎(𝑥,𝑡)=𝐸0𝐷𝛼𝑡+𝑐0𝐷𝛾𝑡+𝑎𝑐𝑏0𝐷𝑡𝛼+𝛾−𝛽𝜀(𝑥,𝑡).(3.4) System (3.4) is subject to initial𝜕𝑢(𝑥,0)=0,[],𝜕𝑡𝑢(𝑥,0)=0,𝜎(𝑥,0)=0,𝜀(𝑥,0)=0,𝑥∈0,1(3.5) and two types of boundary conditions 𝑢(0,𝑡)=0,𝑢(1,𝑡)=Υ(𝑡),𝑡∈ℝ,(3.6)𝑢(0,𝑡)=0,𝜎(1,𝑡)=Σ(𝑡),𝑡∈ℝ(3.7) as described above.

Solutions of the above system will be determined by the Laplace transform method. Recall, the Laplace transform of 𝑓∈𝐿1loc(ℝ), 𝑓≡0 in (−∞,0), and |𝑓(𝑡)|≤𝑀𝑒𝑎𝑡, for some 𝑀>0, 𝑎∈ℝ and 𝑡≥0, is defined by []𝑓(𝑠)=ℒ𝑓(𝑡)(𝑠)∶=∞0𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡,Re𝑠>𝑎.(3.8) Thus, applying the Laplace transform to (3.4) and (3.5), one obtains 𝜕𝜕𝑥𝜎(𝑥,𝑠)=𝑠2𝜕̃𝑢(𝑥,𝑠),̃𝜀(𝑥,𝑠)=𝑎𝜕𝑥̃𝑢(𝑥,𝑠),1+𝑏𝑠𝛼−𝛽𝜎(𝑥,𝑠)=𝑎𝑠𝛼+𝑐𝑠𝛾+𝑎𝑐𝑏𝑠𝛼+𝛾−𝛽̃𝜀(𝑥,𝑠).(3.9) System (3.9) reduces to 𝜕2𝜕𝑥2̃𝑢(𝑥,𝑠)−(𝑠𝑀(𝑠))2[]̃𝑢(𝑥,𝑠)=0,𝑥∈0,1,𝑠∈𝐷,(3.10) where𝑀(𝑠)∶=1+(𝑎/𝑏)𝑠𝛼−𝛽𝑎𝑠𝛼+𝑐𝑠𝛾+(𝑎𝑐/𝑏)𝑠𝛼+𝛾−𝛽=1√𝑎𝑠𝛼1+(𝑎/𝑏)𝑠𝛼−𝛽1+(𝑐/𝑎)𝑠𝛾−𝛼+(𝑐/𝑏)𝑠𝛾−𝛽.(3.11) It has a solutioñ𝑢(𝑥,𝑠)=𝐶1(𝑠)𝑒𝑥𝑠𝑀(𝑠)+𝐶2(𝑠)𝑒−𝑥𝑠𝑀(𝑠)[],𝑥∈0,1,𝑠∈𝐷,(3.12) where 𝐶1 and 𝐶2 are functions of 𝑠 which will be determined from the boundary conditions. Since the power function 𝑠𝜂 is analytic on the complex plane except the branch cut along the negative axis (including the origin) we take 𝐷∶=ℂ⧵(−∞,0] to be the domain for variable 𝑠 in (3.9). Applying either (3.6)1 or (3.7)1, we obtain 𝐶1=−𝐶2=∶𝐶, and thus𝑒̃𝑢(𝑥,𝑠)=𝐶(𝑠)𝑥𝑠𝑀(𝑠)−𝑒−𝑥𝑠𝑀(𝑠)[]].,𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.13) From (3.9) and (3.11) it follows that1𝜎(𝑥,𝑠)=𝑀2𝜕(𝑠)[]].𝜕𝑥̃𝑢(𝑥,𝑠),𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.14)

As announced in Section 1 we will separately seek solutions in different cases: displacement 𝑢 and stress 𝜎 in the case of prescribed displacement (such as e.g., stress relaxation), and displacement 𝑢 in the case of prescribed stress (e.g., creep). For the former we supply to the system boundary conditions (3.6), while for the latter we assume (3.7). In the sequel we derive convolution forms of solutions in all these cases.

In the case of prescribed displacement Υ, substituting (3.13) into (3.12) and using (3.6), one obtains[]],̃𝑢(𝑥,𝑠)=Υ(𝑠)𝑃(𝑥,𝑠),𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.15) where𝑃(𝑥,𝑠)∶=sinh(𝑥𝑠𝑀(𝑠))[]].sinh(𝑠𝑀(𝑠)),𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.16) Clearly, 𝑃(1,𝑡)=𝛿(𝑡), 𝑡∈ℝ.

Since Υ and 𝑃 are supported in [0,∞), displacement 𝑢 is given by []𝑢(𝑥,𝑡)=Υ(𝑡)∗𝑃(𝑥,𝑡),𝑥∈0,1,𝑡∈ℝ,(3.17)[]𝑢(𝑥,𝑡)=0,𝑥∈0,1,𝑡<0,(3.18) where ∗ denotes the convolution with respect to 𝑡. Recall, if 𝑓,𝑔∈𝐿1loc(ℝ), supp𝑓,𝑔⊂[0,∞), then ∫(𝑓∗𝑔)(𝑡)∶=𝑡0𝑓(𝜏)𝑔(𝑡−𝜏)𝑑𝜏, 𝑡∈ℝ. Explicit calculation of (3.17) will be done by the use of the Laplace inversion formula applied to (3.15).

Further, from (3.14), (3.15), and (3.16) it follows that[]],𝜎(𝑥,𝑠)=𝑠Υ(𝑠)𝑇(𝑥,𝑠),𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.19) where𝑇(𝑥,𝑠)=cosh(𝑥𝑠𝑀(𝑠))[]].𝑀(𝑠)sinh(𝑠𝑀(𝑠)),𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.20) Applying the Laplace inversion formula to (3.19) we obtain𝑑𝜎(𝑥,𝑡)=[]𝑑𝑡(Υ(𝑡)∗𝑇(𝑥,𝑡)),𝑥∈0,1,𝑡∈ℝ,(3.21) where the derivative is understood in the sense of distributions. Again, 𝜎(𝑥,𝑡)=0 for 𝑥∈[0,1], 𝑡<0.

In the case of prescribed stress Σ, using (3.14) at 𝑥=1 and (3.7), we obtain𝜕𝜕𝑥̃𝑢(1,𝑠)=Σ(𝑠)𝑀2].(𝑠),𝑠∈ℂ⧵(−∞,0(3.22) This combined with (3.13) gives[]],̃𝑢(𝑥,𝑠)=Σ(𝑠)𝑄(𝑥,𝑠),𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.23) where1𝑄(𝑥,𝑠)=𝑠𝑀(𝑠)sinh(𝑥𝑠𝑀(𝑠))[]].cosh(𝑠𝑀(𝑠)),𝑥∈0,1,𝑠∈ℂ⧵(−∞,0(3.24) Again, applying the Laplace inversion formula to (3.23), the displacement reads []𝑢(𝑥,𝑡)=Σ(𝑡)∗𝑄(𝑥,𝑡),𝑥∈0,1,𝑡>0,(3.25)[]𝑢(𝑥,𝑡)=0,𝑥∈0,1,𝑡<0.(3.26)

Remark 3.1. As in [15] one can show that 𝑃, 𝑇, and 𝑄 are real-valued, locally integrable functions on ℝ, supported in [0,∞) and smooth for 𝑡>0.

3.2. Explicit Forms of Solutions

This subsection is devoted to the calculation of inverse Laplace transforms of certain distributions and functions introduced above. We begin with examining some basic properties of 𝑀 given in (3.11), which will be important for further investigations. In the sequel we shall write 𝐴(𝑥)∼𝐵(𝑥) if lim𝑥→∞(𝐴(𝑥)/𝐵(𝑥))=1.

Proposition 3.2. Let 𝑀 be the function defined by (3.11). Then (i)𝑀 is an analytic function on ℂ⧵(−∞,0] if  0<𝛽<𝛼<𝛾<1. (ii)For 𝑠∈ℂ⧵(−∞,0], lim|𝑠|→0𝑀(𝑠)=∞, lim|𝑠|→0𝑠𝑀(𝑠)=0, lim|𝑠|→∞𝑀(𝑠)=0, and lim|𝑠|→∞𝑠𝑀(𝑠)=∞.

Proof. (i) Since 1+(𝑎/𝑏)𝑠𝛼−𝛽≠0 and 1+(𝑐/𝑎)𝑠𝛾−𝛼+(𝑐/𝑏)𝑠𝛾−𝛽≠0 if arg𝑠∈(−𝜋,𝜋) and 0<𝛽<𝛼<𝛾<1, it follows that the function 𝑀 given in (3.11) is analytic on the complex plane except the branch cut along the negative axis, that is, on ℂ⧵(−∞,0].
(ii) Limits in (ii) can easily be calculated.

3.2.1. Determination of Displacement 𝑢 in the Case of Prescribed Displacement Υ

To begin with, we examine properties of 𝑃 given by (3.16). 𝑃 has isolated singularities at 𝑃𝑠𝑛(±), 𝑛∈ℕ, where 𝑃𝑠𝑛(±) denotes solutions of the equation sinh(𝑠𝑀(𝑠))=0,thatis,𝑠𝑀(𝑠)=±𝑛𝑖𝜋.(3.27)

Let us examine their position and multiplicity.

Proposition 3.3. There are infinitely many complex conjugated solutions 𝑃𝑠𝑛(±), 𝑛∈ℕ, of (3.27), which all lie in the left complex half plane. Moreover, each 𝑃𝑠𝑛(±), 𝑛∈ℕ, is a simple pole.

Proof. Let us square (3.27) and define Φ(𝑠,𝑛)∶=(𝑠𝑀(𝑠))2+(𝑛𝜋)2=𝑠21+(𝑎/𝑏)𝑠𝛼−𝛽𝑎𝑠𝛼+𝑐𝑠𝛾+(𝑎𝑐/𝑏)𝑠𝛼+𝛾−𝛽+(𝑛𝜋)2.(3.28) Writing 𝑠=𝑅𝑒𝑖𝜑 it follows that Φ𝑅𝑒𝑖𝜑,𝑛=𝑅2𝑒2𝑖𝜑1+(𝑎/𝑏)𝑅𝛼−𝛽𝑒𝑖(𝛼−𝛽)𝜑𝑎𝑅𝛼𝑒𝑖𝛼𝜑+𝑐𝑅𝛾𝑒𝑖𝛾𝜑+(𝑎𝑐/𝑏)𝑅𝛼+𝛾−𝛽𝑒𝑖(𝛼+𝛾−𝛽)𝜑+(𝑛𝜋)2=𝑅2(cos(2𝜑)+𝑖sin(2𝜑))𝐴+𝑖𝐵𝐶+𝑖𝐷+(𝑛𝜋)2,(3.29) where 𝑎𝐴∶=1+𝑏𝑅𝛼−𝛽𝑎cos((𝛼−𝛽)𝜑),𝐵∶=𝑏𝑅𝛼−𝛽sin((𝛼−𝛽)𝜑),𝐶∶=𝑎𝑅𝛼cos(𝛼𝜑)+𝑐𝑅𝛾cos(𝛾𝜑)+𝑎𝑐𝑏𝑅𝛼+𝛾−𝛽cos((𝛼+𝛾−𝛽)𝜑),𝐷∶=𝑎𝑅𝛼sin(𝛼𝜑)+𝑐𝑅𝛾sin(𝛾𝜑)+𝑎𝑐𝑏𝑅𝛼+𝛾−𝛽sin((𝛼+𝛾−𝛽)𝜑).(3.30) Real and imaginary parts of Φ are then given by ReΦ𝑅𝑒𝑖𝜑=𝑅,𝑛2𝐶2+𝐷2[](𝐴𝐶+𝐵𝐷)cos(2𝜑)+(𝐴𝐷−𝐵𝐶)sin(2𝜑)+(𝑛𝜋)2,ImΦ𝑅𝑒𝑖𝜑=𝑅,𝑛2𝐶2+𝐷2[],(𝐴𝐶+𝐵𝐷)𝑠𝑖𝑛(2𝜑)−(𝐴𝐷−𝐵𝐶)𝑐𝑜𝑠(2𝜑)(3.31)𝐴𝐶+𝐵𝐷=𝑎𝑅𝛼cos(𝛼𝜑)+𝑐𝑅𝛾𝑎cos(𝛾𝜑)+2𝑏𝑅2𝛼−𝛽+𝑎cos(𝛽𝜑)2𝑐𝑏2𝑅2(𝛼−𝛽)+𝛾cos(𝛾𝜑)+2𝑎𝑐𝑏𝑅𝛼+𝛾−𝛽cos((𝛼−𝛽)𝜑)cos(𝛾𝜑),𝐴𝐷−𝐵𝐶=𝑎𝑅𝛼sin(𝛼𝜑)+𝑐𝑅𝛾𝑎sin(𝛾𝜑)+2𝑏𝑅2𝛼−𝛽+𝑎sin(𝛽𝜑)2𝑐𝑏2𝑅2(𝛼−𝛽)+𝛾sin(𝛾𝜑)+2𝑎𝑐𝑏𝑅𝛼+𝛾−𝛽cos((𝛼−𝛽)𝜑)sin(𝛾𝜑).(3.32)
Let (𝑅,𝜑) be a solution to Φ(𝑠,𝑛)=0 (or equivalently, ReΦ=ImΦ=0). Then changing 𝜑→−𝜑, we again obtain that ReΦ=ImΦ=0, which implies that solutions of (3.27) are complex conjugated.
Further, we will prove that the function Φ has no zeros in the half plane arg𝑠∈[0,𝜋/2]. For that purpose we will use the argument principle. Recall, if Φ is an analytic function inside and on a regular closed curve 𝑐 and nonzero on 𝑐, then the number of zeros of Φ is given by 𝑁𝑍=(1/2𝜋)ΔargΦ(𝑠). Let 𝛾=𝛾𝑎∪𝛾𝑏∪𝛾𝑐 be parametrized as 𝛾𝑎[],𝛾∶𝑠=𝑥,𝑥∈0,𝑅𝑏∶𝑠=𝑅𝑒𝑖𝜑𝜋,𝜑∈0,2,𝛾𝑐∶𝑠=𝑥𝑒𝑖𝜋/2[],,𝑥∈0,𝑅(3.33) and let 𝑅→∞. Along 𝛾𝑎 function Φ becomes a real-valued function, hence ΔargΦ(𝑠,𝑛)=0. Along 𝛾𝑏 we have that 𝐴𝐶+𝐵𝐷,𝐴𝐷−𝐵𝐶≥0 for 𝜑∈[0,𝜋], since sin(𝜂𝜑),cos(𝜂𝜑)>0, for 𝜂∈{𝛼,𝛽,𝛾,𝛼−𝛽}⊂(0,1/2), and 𝜑∈[0,𝜋]. Therefore, (3.31) imply that ReΦ𝑅𝑒𝑖𝜑𝜋,𝑛>0,𝜑∈0,4,ImΦ𝑅𝑒𝑖𝜑𝜋,𝑛>0,𝜑∈4,𝜋2(3.34) (inequalities at boundary points are easily checked by inserting them into (3.31)). Along 𝛾𝑐, we obtain ReΦ𝑥𝑒𝑖𝜋/2𝑅,𝑛=−2𝐶2+𝐷2||||𝑅=𝑥,𝜑=𝜋/2||(𝐴𝐶+𝐵𝐷)𝑅=𝑥,𝜑=𝜋/2+(𝑛𝜋)2,(3.35)ImΦ𝑥𝑒𝑖𝜋/2=𝑅,𝑛2𝐶2+𝐷2||||𝑅=𝑥,𝜑=𝜋/2||(𝐴𝐷−𝐵𝐶)𝑅=𝑥,𝜑=𝜋/2>0.(3.36) From (3.34) and (3.36), we may now conclude that along 𝛾𝑏 and 𝛾𝑐, ΔargΦ(𝑠,𝑛)=0.(3.37) Indeed, this follows from the following. Along 𝛾𝑏, in the case 𝜑∈[0,𝜋/4], ImΦ(𝑅𝑒𝑖𝜑,𝑛) can change its sign, but ReΦ(𝑅𝑒𝑖𝜑,𝑛)>0, while for 𝜑∈[𝜋/4,𝜋/2], ImΦ(𝑅𝑒𝑖𝜑,𝑛)>0. Along 𝛾𝑐, ImΦ(𝑥𝑒𝑖𝜋/2,𝑛)>0. Therefore, there is no change in the argument of Φ along the whole 𝛾, which implies that Φ has no zeros for 𝜑∈[0,𝜋/2]. This further implies that (3.27) has no solutions in the right complex half plane, since its solutions are complex conjugated.
In order to prove that for fixed 𝑛∈ℕ (3.27) has one solution (and its complex conjugate), we again use function Φ and the argument principle. Consider now the contour Γ=Γ𝐴∪Γ𝐵∪Γ𝐶 parametrized by Γ𝐴∶𝑠=𝑥𝑒𝑖𝜋/2[],Γ,𝑥∈0,𝑅𝐵∶𝑠=𝑅𝑒𝑖𝜑𝜋,𝜑∈2,Γ,𝜋𝐶∶𝑠=𝑥𝑒𝑖𝜋[],,𝑥∈0,𝑅(3.38) and let 𝑅→∞. Along Γ𝐴 real and imaginary parts of Φ are given by (3.35) and (3.36). Along Γ𝐵, using (3.31), we conclude that ReΦ𝑅𝑒𝑖𝜑𝜋,𝑛<0,𝜑∈2,3𝜋4,for𝑅⟶∞,and𝑛fixed,ImΦ𝑅𝑒𝑖𝜑,𝑛<0,𝜑∈3𝜋4.,𝜋(3.39) Along Γ𝐶 we have ReΦ𝑅𝑒𝑖𝜑=𝑅,𝑛2𝐶2+𝐷2||||𝑅=𝑥,𝜑=𝜋||(𝐴𝐶+𝐵𝐷)𝑅=𝑥,𝜑=𝜋+(𝑛𝜋)2>0,ImΦ𝑅𝑒𝑖𝜑𝑅,𝑛=−2𝐶2+𝐷2||||𝑅=𝑥,𝜑=𝜋||(𝐴𝐷−𝐵𝐶)𝑅=𝑥,𝜑=𝜋<0.(3.40) Along Γ𝐴 we have that ImΦ>0, while ReΦ changes its sign (since ReΦ(0,𝑛)=(𝑛𝜋)2 and lim𝑥→∞ReΦ(𝑥𝑒𝑖𝜋/2,𝑛)=−∞, for fixed 𝑛∈ℕ). Along the part of Γ𝐵 where 𝜑∈[3𝜋/4,𝜋], and along Γ𝐶, ImΦ<0. Also, lim𝑅→∞ReΦ(𝑅𝑒𝑖3𝜋/4,𝑛)=−∞ and ImΦ(𝑅𝑒𝑖3𝜋/4,𝑛)<0. This implies that the argument of Φ changes from 0 to 2𝜋.
As a conclusion one obtains that along ΓΔargΦ(𝑠,𝑛)=2𝜋,(3.41) which implies, by the argument principle, that function Φ has exactly one zero in the upper left complex plane, for each fixed 𝑛∈ℕ. Since the zeros of Φ are complex conjugated, it follows that Φ also has one zero in the lower left complex plane, for each fixed 𝑛∈ℕ.

In the next proposition we examine behavior of simple poles 𝑃𝑠𝑛(±), 𝑛∈ℕ.

Proposition 3.4. Solutions 𝑃𝑠𝑛(±), 𝑛∈ℕ, of (3.27), are such that Re𝑃𝑠𝑛(±)=𝑅cos𝜑∼2−𝛾𝑐(𝑛𝜋)2𝜋cos2−𝛾<0,Im𝑃𝑠𝑛(±)=𝑅sin𝜑∼±2−𝛾𝑐(𝑛𝜋)2𝜋sin,2−𝛾(3.42) as 𝑛→∞.

Proof. Let us square (3.27) and insert 𝑃𝑠𝑛(±)=𝑅𝑒𝑖𝜑, 𝜑∈(−𝜋,𝜋). Then, after separation of real and imaginary parts, we obtain 𝑅2𝑀cos(2𝜑)Re2𝑅𝑒𝑖𝜑−𝑅2𝑀sin(2𝜑)Im2𝑅𝑒𝑖𝜑=−(𝑛𝜋)2,(3.43)𝑅2𝑀sin(2𝜑)Re2𝑅𝑒𝑖𝜑+𝑅2𝑀cos(2𝜑)Im2𝑅𝑒𝑖𝜑=0.(3.44) Using notation from the proof of Proposition 3.3 we can write 𝑀Re2𝑅𝑒𝑖𝜑=𝐴𝐶+𝐵𝐷𝐶2+𝐷2,𝑀Im2𝑅𝑒𝑖𝜑=𝐵𝐶−𝐴𝐷𝐶2+𝐷2.(3.45) Letting 𝑅→∞, one has 𝑀Re2𝑅𝑒𝑖𝜑∼𝑎2𝑐/𝑏2𝑅2(𝛼−𝛽)+𝛾cos(𝛾𝜑)𝑎2𝑐2/𝑏2𝑅2(𝛼−𝛽)+2𝛾=1𝑐𝑅𝛾𝑀cos(𝛾𝜑),Im2𝑅𝑒𝑖𝜑𝑎∼−2𝑐/𝑏2𝑅2(𝛼−𝛽)+𝛾sin(𝛾𝜑)𝑎2𝑐2/𝑏2𝑅2(𝛼−𝛽)+2𝛾1=−𝑐𝑅𝛾sin(𝛾𝜑).(3.46) It now follows from (3.44) and (3.46) that tg(2𝜑)=−Im𝑀2𝑅𝑒𝑖𝜑Re𝑀2(𝑅𝑒𝑖𝜑)∼𝑡𝑔(𝛾𝜑)⟹sin((2−𝛾)𝜑)𝜋cos(2𝜑)cos(𝛾𝜑)∼0⟹𝜑∼±.2−𝛾(3.47) Inserting (3.47) into (3.46), and subsequently into (3.43), we obtain 𝑅2−𝛾𝑐cos2𝜋2−𝛾cos𝛾𝜋+𝑅2−𝛾2−𝛾𝑐sin2𝜋2−𝛾sin𝛾𝜋2−𝛾∼−(𝑛𝜋)2,𝑅∼2−𝛾𝑐(𝑛𝜋)2.(3.48) Thus, real and imaginary parts of 𝑃𝑠𝑛(±), as 𝑅→∞, are as claimed.

Proposition 3.5. Let 𝑝∈(0,𝑠0), 𝑠0>0. Then 1𝑀(𝑝±𝑖𝑅)∼√𝑐𝑅𝛾𝑒∓𝑖𝛾𝜋/4(3.49) as 𝑅→∞.

Proof. Set √𝜇∶=𝑝2+𝑅2 and 𝜈∶=arctan(±𝑅/𝑝). Then 𝜇∼𝑅 and 𝜈∼±𝜋/2, as 𝑅→∞. By (3.46), we have 𝑀𝜇𝑒𝑖𝜈∼1√𝑐𝜇𝛾𝑒∓𝑖𝛾𝜈/4,𝜇⟶∞,(3.50) as claimed.

In order to obtain the explicit form of solution 𝑢 to initial-boundary value problem (3.4), (3.5), and (3.6), it remains to calculate function 𝑃.

Theorem 3.6. The solution 𝑢 to initial-boundary value problem (3.4), (3.5), and (3.6) is given by (3.17), that is, 𝑢(𝑥,𝑡)=Υ(𝑡)∗𝑃(𝑥,𝑡), where 𝑃 takes the form 1𝑃(𝑥,𝑡)=2𝜋𝑖∞0sinh𝑥𝑞𝑀𝑞𝑒−𝑖𝜋sinh(𝑞𝑀(𝑞𝑒−𝑖𝜋−))sinh𝑥𝑞𝑀𝑞𝑒𝑖𝜋sinh(𝑞𝑀(𝑞𝑒𝑖𝜋𝑒))−𝑞𝑡+𝑑𝑞∞𝑛=1Res𝑃(𝑥,𝑠)𝑒𝑠𝑡,𝑃𝑠𝑛(+)+Res𝑃(𝑥,𝑠)𝑒𝑠𝑡,𝑃𝑠𝑛(−),𝑡>0.(3.51) The residues at simple poles 𝑃𝑠𝑛(±), 𝑛∈ℕ, are given by Res𝑃(𝑥,𝑠)𝑒𝑠𝑡,𝑃𝑠𝑛(±)=sinh(𝑥𝑠𝑀(𝑠))[]𝑒(𝑑/𝑑𝑠)sinh(𝑠𝑀(𝑠))𝑠𝑡𝑠=𝑃𝑠𝑛(±).(3.52)

Proof. Function 𝑃(𝑥,𝑡), 𝑥∈[0,1], 𝑡>0, will be calculated by integration over a suitable contour. The Cauchy residues theorem yields Γ𝑃(𝑥,𝑠)𝑒𝑠𝑡𝑑𝑠=2𝜋𝑖∞𝑛=1Res𝑃(𝑥,𝑠)𝑒𝑠𝑡,𝑃𝑠𝑛(+)+R𝑒𝑠𝑃(𝑥,𝑠)𝑒𝑠𝑡,𝑃𝑠𝑛(−),(3.53) where Γ=Γ1∪Γ2∪Γ3∪Γ𝜀∪Γ4∪Γ5∪Γ6∪𝛾0 is such a contour that all poles lie inside the contour Γ (see Figure 1). First we show that the series of residues in (3.51) is convergent. By Proposition 3.4, 𝑃𝑠𝑛(±) are simple poles of 𝑃, and therefore also simple poles of 𝑒st𝑃. The residues in (3.53) can be calculated as it is given in (3.52), so Res𝑃(𝑥,𝑠)𝑒st,𝑃𝑠𝑛(±)=1(𝑑/𝑑𝑠)(𝑠𝑀(𝑠))sinh(𝑥𝑠𝑀(𝑠))𝑒cosh(𝑠𝑀(𝑠))st𝑠=𝑃𝑠𝑛(±).(3.54) Using (3.27) we have that sinh(𝑥𝑠𝑀(𝑠))=±𝑖sin(𝑥𝑛𝜋), while cosh(𝑠𝑀(𝑠))=(−1)𝑛. Also, a calculation gives that (𝑑/𝑑𝑠)(𝑠𝑀(𝑠))=𝑀(𝑠)+𝑀(𝑠)⋅𝐴(𝑠), where 𝐴(𝑠)∶=1+(𝑎/𝑏)(𝛼−𝛽)𝑠𝛼−𝛽21+(𝑎/𝑏)𝑠𝛼−𝛽−𝑎𝛼𝑠𝛼+𝑐𝛾𝑠𝛾+(𝑎𝑐/𝑏)(𝛼+𝛾−𝛽)𝑠𝛼+𝛾−𝛽2𝑎𝑠𝛼+𝑐𝑠𝛾+(𝑎𝑐/𝑏)𝑠𝛼+𝛾−𝛽.(3.55) Take now that 𝑃𝑠𝑛(±)=𝑅𝑒±𝑖𝜑. Then Res𝑃(𝑥,𝑠)𝑒st,𝑃𝑠𝑛(±)=(−1)𝑛sin(𝑛𝜋𝑥)𝑛𝜋ReRtcos𝜑𝑒±𝑖(𝜑+Rtsin𝜑)𝐴(𝑅𝑒±𝑖𝜑).(3.56) Let 𝑛→∞. Then also |𝑃𝑠𝑛(±)|→∞, that is, 𝑅→∞, and ||𝐴𝑅𝑒±𝑖𝜑||𝛾⟶1−2.(3.57) Further, it follows from Proposition 3.4 that Re𝑃𝑠𝑛(±)=𝑅cos𝜑∼2−𝛾𝑐(𝑛𝜋)2𝜋cos2−𝛾≤−𝐶𝑛,forsome𝐶>0,(3.58) and by (3.48), 𝑅/𝑛∼2−𝛾√𝑐𝜋2⋅𝑛𝛾/(2−𝛾). Therefore, as 𝑛→∞, |||Res𝑃(𝑥,𝑠)𝑒st,𝑃𝑠𝑛(+)+Res𝑃(𝑥,𝑠)𝑒st,𝑃𝑠𝑛(−)|||≤||||sin(𝑛𝜋𝑥)𝑛𝜋𝑅𝑒𝑅𝑡cos𝜑𝐴(𝑅𝑒+𝑖𝜑)||||+||||sin(𝑛𝜋𝑥)𝑛𝜋𝑅𝑒𝑅𝑡cos𝜑𝐴(𝑅𝑒−𝑖𝜑)||||≤1𝜋𝑅𝑛𝑒−𝐶𝑛𝑡1||𝐴(𝑅𝑒+𝑖𝜑)||+1||𝐴(𝑅𝑒−𝑖𝜑)||∼4𝜋(2−𝛾)2−𝛾√𝑐𝜋2⋅𝑛𝛾/(2−𝛾)𝑒−𝐶𝑛𝑡,(3.59) which implies the convergence of the sum of residues in (3.53).
It remains to calculate the integral over Γ in (3.53). Consider the integral along contour Γ1∶𝑠=𝑝+𝑖𝑅, 𝑠0>𝑝>0. Then ||||Γ1𝑃(𝑥,𝑠)𝑒𝑠𝑡||||≤𝑑𝑠𝑠00||𝑃||||𝑒(𝑥,𝑝+𝑖𝑅)(𝑝+𝑖𝑅)𝑡||𝑑𝑝.(3.60) Let 𝑅→∞. In order to estimate |𝑃(𝑥,𝑝±𝑖𝑅)|, using Proposition 3.5, we write 1𝑀(𝑝±𝑖𝑅)∼𝑣±𝑖𝑤,𝑣=√𝑐𝑅𝛾cos𝛾𝜋41,𝑤=−√𝑐𝑅𝛾sin𝛾𝜋4.(3.61) Then ||𝑃||∼||||[](𝑥,𝑝±𝑖𝑅)sinh𝑥(𝑝𝑣−𝑅𝑤)±𝑖𝑥(𝑝𝑤+𝑅𝑣)[]||||≤𝑒sinh(𝑝𝑣−𝑅𝑤)±𝑖(𝑝𝑤+𝑅𝑣)𝑥(𝑝𝑣−𝑅𝑤)+𝑒−𝑥(𝑝𝑣−𝑅𝑤)||𝑒𝑝𝑣−𝑅𝑤−𝑒−(𝑝𝑣−𝑅𝑤)||=𝑒−(1−𝑥)(𝑝𝑣−𝑅𝑤)1+𝑒−2𝑥(𝑝𝑣−𝑅𝑤)||1−𝑒−2(𝑝𝑣−𝑅𝑤)||⟶0,as𝑅⟶∞.(3.62) The above convergence is valid since 1𝑝𝑣−𝑅𝑤=𝑝√𝑐𝑅𝛾cos𝛾𝜋41+𝑅√𝑐𝑅𝛾sin𝛾𝜋4⟶∞,as𝑅⟶∞.(3.63) Therefore, according to (3.62), we have lim𝑅→∞||||Γ1𝑃(𝑥,𝑠)𝑒st||||𝑑𝑠=0.(3.64) The similar argument is valid for the integral along Γ6, thus lim𝑅→∞||||Γ6𝑃(𝑥,𝑠)𝑒st||||𝑑𝑠=0.(3.65) Next, we consider the integral along contour Γ2∶𝑠=𝑅𝑒𝑖𝜑, 𝜋/2<𝜑<𝜋: ||||Γ2𝑃(𝑥,𝑠)𝑒st||||≤𝑑𝑠𝜋𝜋/2𝑅||𝑒−𝑅(1−𝑥)𝑒𝑖𝜑𝑀(𝑅𝑒𝑖𝜑)||||||1−𝑒−2𝑥𝑅𝑒𝑖𝜑𝑀(𝑅𝑒𝑖𝜑)1−𝑒−2𝑅𝑒𝑖𝜑𝑀(𝑅𝑒𝑖𝜑)||||𝑒𝑅𝑡cos𝜑𝑑𝜑.(3.66) Since 𝑠𝑀(𝑠)→∞ as |𝑠|→∞ (see Proposition 3.2 (ii)) and cos𝜑≤0 for 𝜑∈[𝜋/2,𝜋], we have lim𝑅→∞||||Γ2𝑃(𝑥,𝑠)𝑒st||||𝑑𝑠≤lim𝑅→∞𝜋𝜋/2𝑅||𝑒−𝑅(1−𝑥)𝑒𝑖𝜑𝑀(𝑅e𝑖𝜑)||𝑒𝑅𝑡cos𝜑𝑑𝜑=0.(3.67) The similar argument is valid for the integral along Γ5, thus lim𝑅→∞||||Γ5𝑃(𝑥,𝑠)𝑒st||||𝑑𝑠=0.(3.68) Since 𝑠𝑀(𝑠)→0 as |𝑠|→0 (see Proposition 3.2 (ii)), the integration along contour Γ𝜀∶𝑠=𝜀𝑒𝑖𝜑, 𝜋>𝜑>−𝜋, gives lim𝜀→0||||Γ𝜀𝑃(𝑥,𝑠)𝑒st||||𝑑𝑠≤lim𝜀→0𝜋−𝜋𝜀||||sinh𝑥𝜀𝑒𝑖𝜑𝑀𝜀𝑒𝑖𝜑sinh(𝜀𝑒𝑖𝜑𝑀(𝜀𝑒𝑖𝜑||||𝑒))𝜀𝑡cos𝜑𝑑𝜑=0.(3.69) Integrals along parts of contour Γ3∶𝑠=𝑞𝑒𝑖𝜋, 𝑅>𝑞>𝜀, Γ4∶𝑠=𝑞𝑒−𝑖𝜋, 𝜀<𝑞<𝑅, and 𝛾0∶𝑠=𝑠0+𝑖𝑟, −𝑅<𝑟<𝑅, give lim𝑅→∞𝜀→0Γ3𝑃(𝑥,𝑠)𝑒st𝑑𝑠=∞0sinh𝑥𝑞𝑀𝑞𝑒𝑖𝜋sinh(𝑞𝑀(𝑞𝑒𝑖𝜋𝑒))−𝑞𝑡𝑑𝑞,lim𝑅→∞𝜀→0Γ4𝑃(𝑥,𝑠)𝑒st𝑑𝑠=−∞0sinh𝑥𝑞𝑀𝑞𝑒−𝑖𝜋sinh(𝑞𝑀(𝑞𝑒−𝑖𝜋𝑒))−𝑞𝑡𝑑𝑞,lim𝑅→∞𝛾0𝑃(𝑥,𝑠)𝑒st𝑑𝑠=2𝜋𝑖𝑃(𝑥,𝑡).(3.70) Equation (3.51) now follows from (3.53).


Figure 1 
Integration contour Γ .

Corollary 3.7. In the case of stress relaxation, that is, when Υ(𝑡)=Υ0𝐻(𝑡), Υ0>0, 𝑡∈ℝ, the solution takes the form 𝑢𝐻(𝑥,𝑡)=Υ0[]𝐻(𝑡)∗𝑃(𝑥,𝑡),𝑥∈0,1,𝑡∈ℝ.(3.71) We will numerically examine it in the sequel.

3.2.2. Determination of Stress 𝜎 in the Case of Prescribed Displacement Υ

In Section 3.1 we determined stress 𝜎 (cf. (3.21)) that is a solution to (3.4), (3.5), and (3.6). In order to obtain an explicit form of 𝜎 we need to calculate function 𝑇. As in previous Section 3.2.1 it will be done by inversion of the Laplace transform of 𝑇.

Function 𝑇, which is given by (3.20), is analytic on the complex plane except the branch cut (−∞,0], and has simple poles at the same points as 𝑃, that is, 𝑃𝑠𝑛(±), 𝑛∈ℕ.

Using the similar arguments as in the proof of Theorem 3.6, one can prove the following theorem.

Theorem 3.8. The solution 𝜎 to initial-boundary value problem (3.4), (3.5), and (3.6) is given by (3.21), that is, 𝜎(𝑥,𝑡)=𝑑/𝑑𝑡(Υ(𝑡)∗𝑇(𝑥,𝑡)), where 𝑇 takes the form 1𝑇(𝑥,𝑡)=2𝜋𝑖∞0cosh𝑥𝑞𝑀𝑞𝑒𝑖𝜋𝑀(𝑞𝑒𝑖𝜋)sinh(𝑞𝑀(𝑞𝑒𝑖𝜋−))cosh𝑥𝑞𝑀𝑞𝑒−𝑖𝜋𝑀(𝑞𝑒−𝑖𝜋)sinh(𝑞𝑀(𝑞𝑒−𝑖𝜋𝑒))−𝑞𝑡+𝑑𝑞∞𝑛=1Res𝑇(𝑥,𝑠)𝑒st,𝑃𝑠𝑛(+)+Res𝑇(𝑥,𝑠)𝑒st,𝑃𝑠𝑛(−),𝑡>0.(3.72) The residues at simple poles 𝑃𝑠𝑛(±), 𝑛∈ℕ, are given by Res𝑇(𝑥,𝑠)𝑒st,𝑃𝑠𝑛(±)=cosh(𝑥𝑠𝑀(𝑠))[]𝑒𝑀(𝑠)(𝑑/𝑑𝑠)sinh(𝑠𝑀(𝑠))𝑠𝑡𝑠=𝑃𝑠𝑛(±).(3.73)

Corollary 3.9. Similarly, in the case of stress relaxation Υ=Υ0𝐻 we obtain the solution 𝜎𝐻(𝑥,𝑡)=Υ0[]𝑇(𝑥,𝑡),𝑥∈0,1,𝑡>0.(3.74)

In order to check our results for large times, we compare them with the quasistatic case. In the quasistatic case one uses only the constitutive equation (3.4)3, that is, the dynamics of the process is neglected. Taking the Laplace transform of the constitutive equation we obtain (3.9)2, and define the relaxation modulus 𝐺 via its Laplace transform, as follows: 𝐺(𝑡)∶=ℒ−1𝐺(𝑠)(𝑡),𝑡>0,𝐺(𝑠)∶=𝜎(QS)(𝑠)̃𝜀(QS)(𝑠)∶=𝑎𝑠𝛼+𝑐𝑠𝛾+(𝑎𝑐/𝑏)𝑠𝛼+𝛾−𝛽1+(𝑎/𝑏)𝑠𝛼−𝛽].,𝑠∈ℂ⧵(−∞,0(3.75) Then the stress in the quasistatic case is 𝜎(QS)=𝐺∗𝜀(QS).(3.76) Following the proof of Theorem 3.6, we obtain that 1𝐺(𝑡)=2𝜋𝑖∞0𝐺𝑞𝑒−𝑖𝜋−𝐺𝑞𝑒𝑖𝜋𝑒−𝑞𝑡𝑑𝑡.(3.77)

In the quasistatic case it holds that []𝑢(𝑥,𝑡)=𝑥⋅𝑢(1,𝑡),𝑥∈0,1𝑡>0,(3.78) and consequently by (3.4), 𝜀(𝑥,𝑡)=𝑢(1,𝑡)=∶𝜀(QS)[](𝑡),𝑥∈0,1𝑡>0.(3.79) Since according to boundary condition (3.6), 𝑢(1,𝑡)=Υ(𝑡), it follows from (3.76) that 𝜎(QS)=Υ∗𝐺,(3.80) which, in the case of stress relaxation, that is, when Υ=Υ0𝐻, Υ0>0, becomes𝜎𝐻(QS)=Υ0𝐻∗𝐺.(3.81)

3.2.3. Determination of Displacement 𝑢 in the Case of Prescribed Stress Σ

In Section 3.1 we determined displacement 𝑢 (cf. (3.25)) that is a solution to (3.4), (3.5), and (3.7). As above, we now want to find 𝑢 explicitly, by calculating the inverse Laplace transform of 𝑄.

Function 𝑄 given by (3.24) is analytic on the complex plane except the branch cut (−∞,0] and has isolated singularities at solutions 𝑄𝑠𝑛(±) of the equation cosh(𝑠𝑀(𝑠))=0thatis,𝑠𝑀(𝑠)=±2𝑛+12𝑖𝜋,𝑛∈ℕ0.(3.82)

We state a proposition that is analogous to Propositions 3.3 and 3.4. The proof is omitted since it follows the same lines as those of Propositions 3.3 and 3.4.

Proposition 3.10. (i) There are infinitely many complex conjugated solutions 𝑄𝑠𝑛(±), 𝑛∈ℕ0, of (3.82), which all lie in the left complex half plane. Moreover, each 𝑄𝑠𝑛(±), 𝑛∈ℕ0, is a simple pole.
(ii) Solutions 𝑄𝑠𝑛(±), 𝑛∈ℕ0, of (3.82), are such that Re𝑄𝑠𝑛(±)=𝑅cos𝜑∼2−𝛾𝑐2𝑛+12𝜋2𝜋cos2−𝛾<0,Im𝑄𝑠𝑛(±)=𝑅sin𝜑∼±2−𝛾𝑐2𝑛+12𝜋2𝜋sin,2−𝛾(3.83) as 𝑛→∞.

In the following theorem we calculate explicitly displacement 𝑢.

Theorem 3.11. The solution 𝑢 to initial-boundary value problem (3.4), (3.5), and (3.7) is given by (3.25), that is, 𝑢(𝑥,𝑡)=Σ(𝑡)∗𝑄(𝑥,𝑡), where 𝑄 takes the form 1𝑄(𝑥,𝑡)=2𝜋𝑖∞0𝑀𝑞𝑒−𝑖𝜋sinh𝑥𝑞𝑀𝑞𝑒−𝑖𝜋cosh(𝑞𝑀(𝑞𝑒−𝑖𝜋))−𝑀𝑞𝑒𝑖𝜋sinh𝑥𝑞𝑀𝑞𝑒𝑖𝜋cosh(𝑞𝑀(𝑞𝑒𝑖𝜋𝑒))−𝑞𝑡𝑞+𝑑𝑞∞𝑛=0Res𝑄(𝑥,𝑠)𝑒st,𝑄𝑠𝑛(+)+R𝑒𝑠𝑄(𝑥,𝑠)𝑒st,𝑄𝑠𝑛(−),𝑡>0.(3.84) The residues at simple poles 𝑄𝑠𝑛(±), 𝑛∈ℕ0, are given by Res𝑄(𝑥,𝑠)𝑒st,𝑄𝑠𝑛(±)=1𝑠𝑀(𝑠)sinh(𝑥𝑠𝑀(𝑠))[]𝑒(𝑑/𝑑𝑠)cosh(𝑠𝑀(𝑠))st𝑠=𝑄𝑠𝑛(±).(3.85)

Corollary 3.12. The case of creep is described by the boundary condition Σ(𝑡)=Σ0𝐻(𝑡), Σ0>0, 𝑡∈ℝ, in which displacement 𝑢, given by (3.25), reads 𝑢(𝑡)=Σ0[]𝐻(𝑡)∗𝑄(𝑥,𝑡),𝑥∈0,1,𝑡∈ℝ.(3.86)

Similarly as in Section 3.2.2, we examine the quasistatic case that corresponds to displacement 𝑢. Again, using only the constitutive equation (3.4)3 and its Laplace transform (3.9)2, we define the creep compliance 𝐽 via its Laplace transform as 𝐽(𝑡)∶=ℒ−1𝐽(𝑠)(𝑡),𝑡>0,𝐽(𝑠)∶=̃𝜀(𝑄𝑆)(𝑠)𝜎(𝑄𝑆)(𝑠)∶=1+(𝑎/𝑏)𝑠𝛼−𝛽𝑎𝑠𝛼+𝑐𝑠𝛾+(𝑎𝑐/𝑏)𝑠𝛼+𝛾−𝛽].,𝑠∈ℂ⧵(−∞,0(3.87) Strain measure in the quasistatic case now equals𝜀(QS)=𝐽∗𝜎(QS).(3.88) Following the proof of Theorem 3.6 one obtains 1𝐽(𝑡)=2𝜋𝑖∞0𝐽𝑞𝑒−𝑖𝜋−𝐽𝑞𝑒𝑖𝜋𝑒−𝑞𝑡𝑑𝑡.(3.89) In the quasistatic case it holds that 𝑢(𝑥,𝑡)=𝑥⋅𝑢(1,𝑡)=𝑥⋅𝑢(QS)[](𝑡),𝑥∈0,1,𝑡>0,(3.90) and consequently by (3.4), 𝜀(QS)=𝑢(QS).(3.91) Also, 𝜎(QS)(𝑡)∶=𝜎(1,𝑡)=Σ(𝑡), 𝑡>0, which is the boundary condition (3.7), hence by (3.88) and (3.91), 𝑢(QS)=Σ∗𝐽.(3.92) In the case of creep (cf. Corollary 3.12) we have𝑢𝐻(QS)=Σ0𝐻∗𝐽.(3.93)

3.3. Numerical Examples

In this subsection we give several numerical examples of displacement 𝑢𝐻 and stress 𝜎𝐻, given by (3.71) and (3.74), respectively, which correspond to the case of stress relaxation, and examine solutions (3.86), which correspond to displacement 𝑢 in the case of creep. In addition, we investigate solutions 𝑢𝐻, 𝜎𝐻, and 𝑢 for different orders of fractional derivatives.

Figure 2 presents displacements in a stress relaxation experiment, determined according to (3.71), for three different positions. Parameters in (3.71) are chosen as follows: Υ0=1, 𝑎=0.2, 𝑏=0.6, 𝑐=0.45, 𝛼=0.3, 𝛽=0.1, and 𝛾=0.4. From Figure 2, one sees that the displacements in the case of stress relaxation show damped oscillatory character and that they tend to a constant value for large times, namely, lim𝑡→∞𝑢𝐻(𝑥,𝑡)=𝑥, 𝑥∈[0,1]. Figure 3 presents the same displacements as Figure 2, but close to initial time instant. It is evident that there is a delay in displacement that increases as the point is further from the end where the prescribed displacement is applied. This is a consequence of the finite wave propagation speed.


Figure 2 
Displacements 𝑢 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 2 5 , 0 . 5 , 0 . 7 5 } for 𝑡 ∈ ( 0 , 1 0 ) .

Figure 3 
Displacements 𝑢 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 2 5 , 0 . 5 , 0 . 7 5 } for 𝑡 ∈ ( 0 , 5 ) .

Figure 4 presents the displacements 𝑢𝐻 of the points close to the end of the rod where the sudden but afterwards constant displacement is applied (i.e., 𝑢(1,𝑡)=𝐻(𝑡), 𝑡>0). One sees that the amplitudes of these points deform in shape so that they do not exceed the prescribed value of the displacement of the rod's free end.


Figure 4 
Displacements 𝑢 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 7 5 , 0 . 8 , 0 . 8 5 , 0 . 9 , 0 . 9 5 } for 𝑡 ∈ ( 0 , 5 ) .

In order to examine the influence of the orders of fractional derivatives in the constitutive equation (1.4) on the displacement 𝑢𝐻 and stress 𝜎𝐻 in a stress relaxation experiment, we plot the displacement 𝑢𝐻 and stress 𝜎𝐻 obtained by (3.71) and (3.74) for the following sets of parameters (𝛼,𝛽,𝛾)∈{(0.1,0.05,0.15),(0.3,0.1,0.4),(0.45,0.4,0.49)}, while we fix 𝑥=0.5 and leave other parameters as before. In the case of the first set, the constitutive equation (1.4) describes a body in which the elastic properties are dominant, since the orders of the fractional derivatives of stress and strain are close to zero, that is, the fractional derivatives of stress and strain almost coincide with the stress and strain. This is also evident from Figure 5, since the oscillations of the point 𝑥=0.5 for the first set of parameters vanish quite slowly comparing to the second set and in particular comparing with the third set of parameters. Note that the third set of parameters describes a body in which the fluid properties of the fractional type dominate, since in the constitutive equation (1.4), we have the low-order derivative of stress (almost the stress itself), while almost all the derivatives of strain are of order 0.5. Figure 5 also shows that the dissipative properties of a material grow as the orders of the fractional derivatives increase. Figure 6 shows that the delay in displacement depends on the order of the fractional derivative, so that a material which has dominant elastic properties (the first set) has the longest delay, compared to a material with the dominant fluid properties (the third set).


Figure 5 
Displacements 𝑢 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 = 0 . 5 for 𝑡 ∈ ( 0 , 2 0 ) .

Figure 6 
Displacements 𝑢 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 = 0 . 5 for 𝑡 ∈ ( 0 , 8 ) .

Figure 7 presents stresses 𝜎𝐻 in the case of stress relaxation, determined according to (3.74), for different points of the rod. Parameters are the same as in the previous case. Also, Figure 7 presents the quasistatic curve 𝜎𝐻(QS) obtained by (3.81).


Figure 7 
Stresses 𝜎 𝐻 ( 𝑥 , 𝑡 ) and 𝜎 𝐻 ( Q S ) ( 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 1 , 0 . 2 , 0 . 3 , 0 . 4 , 0 . 5 , 0 . 6 , 0 . 7 } for 𝑡 ∈ ( 0 , 5 ) .

Stresses, as it can be seen from Figure 7, show damped oscillatory character and for large times, in each point 𝑥∈[0,1], tend to the quasistatic curve, that is, to the same value. Eventually, the stresses in all points of the rod tend to zero, namely, lim𝑡→∞𝜎𝐻(𝑥,𝑡)=0, 𝑥∈[0,1]. From Figure 7 it is evident that as further the point is from the rod-free end the greater is the delay. This is again the consequence of the finite wave speed.

Figures 8 and 9 present the stresses of the points close to the free end. One notices from Figures 7 and 8 that as the point is closer to the end whose displacement is prescribed, the peak of stress is higher and the peak's width is smaller. Figure 9 presents the compressive phase in the stress relaxation process and the quasistatic curve as well.


Figure 8 
Stresses 𝜎 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 7 , 0 . 8 , 0 . 9 } for 𝑡 ∈ ( 0 , 0 . 5 ) .

Figure 9 
Stresses 𝜎 𝐻 ( 𝑥 , 𝑡 ) and 𝜎 𝐻 ( 𝑄 𝑆 ) ( 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 7 , 0 . 8 , 0 . 9 } for 𝑡 ∈ ( 0 , 6 ) .

Again, we fix the midpoint of the rod and investigate the influence of the change of orders of the fractional derivatives (for the same three sets as before) on the stress 𝜎𝐻 in a stress relaxation experiment. From Figure 10 one notices that there is a stress relaxation in a material regardless of the order of fractional derivatives. However, the relaxed stress depends on the order of the derivatives, as expected from the analysis of the stress 𝜎𝐻(QS) in the quasistatic case, given by (3.81). The material with the dominant elastic properties (the first set) relaxes to the highest stress, and as the fluid properties of the material become more and more dominant, the relaxed stress decreases. Again, the oscillations of the value of the stress for the first set (material with dominant elastic properties) are the least damped compared to the second and third set. Figure 11 shows that the conclusion about the dependence of delay on the orders of fractional derivatives drown earlier holds.


Figure 10 
Stresses 𝜎 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 = 0 . 5 for 𝑡 ∈ ( 0 , 1 0 ) .

Figure 11 
Stresses 𝜎 𝐻 ( 𝑥 , 𝑡 ) in a stress relaxation experiment as functions of time 𝑡 at 𝑥 = 0 . 5 for 𝑡 ∈ ( 0 , 4 . 5 ) .

Figure 12 presents displacements 𝑢 in the creep experiment, determined according to (3.86), for four different points. Parameters are the same as in the previous cases (except that in this case instead of Υ0=1 we have Σ0=1). For large times, as it can be seen from Figure 12, the displacement curves are monotonically increasing. This indicates that we deal with the viscoelastic fluid. Figure 12 also shows good agreement between the displacements obtained by the dynamic model (displacement is given by (3.86)) and the quasistatic model (displacement is given by (3.93)). Figure 13 presents the same displacements as Figure 12, but close to initial time instant and it is evident that, again, there is a delay, due to the finite speed of wave propagation.


Figure 12 
Displacements 𝑢 ( 𝑥 , 𝑡 ) and 𝑢 𝐻 ( Q S ) ( 𝑡 ) in a creep experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 2 5 , 0 . 5 , 0 . 7 5 , 1 } for 𝑡 ∈ ( 0 , 4 0 ) .

Figure 13 
Displacements 𝑢 ( 𝑥 , 𝑡 ) in creep experiment as functions of time 𝑡 at 𝑥 ∈ { 0 . 2 5 , 0 . 5 , 0 . 7 5 , 1 } for 𝑡 ∈ ( 0 , 1 0 ) .

In order to examine the dependence of the displacement 𝑢 in the case of creep experiment, given by (3.86), on the orders of fractional derivatives in the constitutive equation (1.4), we again fix the point 𝑥=0.5 of the rod and plot the displacement 𝑢 for the same set of values of 𝛼, 𝛽, and 𝛾 as before. Figure 14 clearly shows that all of the materials exhibits creep, but the displacements do not tend to a constant value. However, the material which has the dominant elastic properties (the first set) creeps slower comparing to the material with the dominant fluid properties (the third set). Figure 15 again supports the conclusion that the more elastic the material is, the greater the delay is.


Figure 14 
Displacements 𝑢 ( 𝑥 , 𝑡 ) in creep experiment as functions of time 𝑡 at 𝑥 = 0 . 5 for 𝑡 ∈ ( 0 , 4 0 ) .

Figure 15 
Displacements 𝑢 ( 𝑥 , 𝑡 ) in creep experiment as functions of time 𝑡 at 𝑥 = 0 . 5 for 𝑡 ∈ ( 0 , 1 0 ) .

Acknowledgment

This work is partially supported by Projects 174005 and 174024 of the Serbian Ministry of Science, APV Project 114-451-2167/2011-01, and START-Project Y-237 of the Austrian Science Fund.