# Calculation of the Relaxation Modulus in the Andrade Model by Using the Laplace Transform
###### Abstract  
In the framework of the theory of linear viscoelasticity, we derive an analytical expression of the relaxation modulus in the Andrade model \(G_{\alpha}(t)\) for the case of rational parameter \(\alpha=m/n\in(0,1)\) in terms of Mittag-Leffler functions from its Laplace transform \(\hat{G}_{\alpha}(s)\). It turns out that the expression obtained can be rewritten in terms of Rabotnov functions. Moreover, for the original parameter \(\alpha=1/3\) in the Andrade model, we obtain an expression in terms of Miller-Ross functions. The asymptotic behaviours of \(G_{\alpha}(t)\) for \(t\to 0^{+}\) and \(t\to+\infty\) are also derived applying the Tauberian theorem. The analytical results obtained have been numerically checked by solving the Volterra integral equation satisfied by \(G_{\alpha}(t)\) by using a successive approximation approach, as well as computing the inverse Laplace transform of \(\hat{G}_{\alpha}(s)\) by using [PERSON]'s method. Andrade model; relaxation modulus in linear viscoelasticity; Mittag-Leffler function; Laplace transform 2024 8439 10.3390/fractalfractal8080439 10.3390/10. useful parametrization given in [8], whose validity is discussed in [4]. Essentially, this parametrization performs the following change of variables:  
\[J_{U}=\frac{1}{\mu},\quad\beta=\frac{1}{\mu\,\tau^{\alpha}},\quad\eta=\mu\,\tau, \tag{2}\]  
thus, we obtain  
\[J_{\alpha}(t)=\frac{1}{\mu}\bigg{[}1+\bigg{(}\frac{t}{\tau}\bigg{)}^{\alpha}+ \frac{t}{\tau}\bigg{]},\quad t\geq 0, \tag{3}\]  
where the parameters \(\mu,\tau>0\) have a clear and important physical meaning. The aim of this paper is to analytically calculate in closed form the relaxation modulus \(G_{\alpha}(t)\) (i.e., the stress per unit strain) from the creep compliance \(J_{\alpha}(t)\) given in (3) by using the inverse Laplace transform. It is worth noting that the computation of \(G_{\alpha}(t)\) from \(J_{\alpha}(t)\) is also possible by numerically solving the following Volterra integral equation of the second kind ([9] [Eqn. 2.87]):  
\[G_{\alpha}(t)=\mu\bigg{[}1-\int_{0}^{t}\frac{dJ_{\alpha}(t^{\prime})}{dt^{ \prime}}G_{\alpha}\big{(}t-t^{\prime}\big{)}dt^{\prime}\bigg{]}, \tag{4}\]  
just as has been done for other models such as that of Jeffrey-Lomnitz rheological law [10]. Another numerical approach is to obtain the relaxation modulus in the Laplace domain \(\tilde{G}_{\alpha}(s)=\mathcal{L}[G_{\alpha}(t);s]\), and then numerically evaluate the inverse Laplace transform to obtain \(G_{\alpha}(t)\). However, an analytical solution is more desirable since it shows the role and weight of the model parameters explicitly. In addition, analytical approaches are generally more computationally efficient. In the present approach, we just analytically calculate the inverse Laplace transform of \(\tilde{G}_{\alpha}(s)\). This result has been applied by some of the authors in [11], and here we provide the mathematical details of the calculation. Next, we derive \(\tilde{G}_{\alpha}(s)\) in the Andrade model. For this purpose, apply the Laplace transform to (3) in order to obtain  
\[\mathcal{L}[J_{\alpha}(t);s]=\tilde{J}_{\alpha}(s)=\frac{1}{\mu\,s^{2}}\bigg{[} s+\Gamma(1+\alpha)\tau^{-\alpha}s^{1-\alpha}+\frac{1}{\tau}\bigg{]}. \tag{5}\]  
However, since the following relation is satisfied in any linear viscoelasticity rheology ([9] [Eqn. 2.8])  
\[\tilde{J}_{\alpha}(s)\tilde{G}_{\alpha}(s)=\frac{1}{s^{2}}, \tag{6}\]  
we conclude  
\[\tilde{G}_{\alpha}(s)=\frac{\mu\,\tau}{s\,\tau+\Gamma(\alpha+1)(s\,\tau)^{1- \alpha}+1}. \tag{7}\]  
As mentioned above, the range of values that are interesting for \(\alpha\) are between \(0\) and \(1\), so we restrict our study to \(\alpha\in(0,1)\). It is worth noting that (7) can also be obtained rewriting the Volterra integral Equation (4) in terms of the Laplace convolution product [12], i.e.,  
\[G_{\alpha}(t)=\mu\bigg{(}1-G_{\alpha}(t)\ast\frac{dJ_{\alpha}(t)}{dt}\bigg{)}, \tag{8}\]  
thus taking the Laplace transform in (8) and solving for \(\tilde{G}_{\alpha}(s)\), we arrive at (7).