# Calculation of the Relaxation Modulus in the Andrade Model by Using the Laplace Transform
## 4 Asymptotic Behaviour via Tauberian Theorem
### Asymptotic Behaviour for \(T\to 0^{+}\)  
Rewrite the Laplace transform of the relaxation modulus (61), as follows:  
\[\tilde{G}_{\alpha}(s)=\frac{\mu}{s}\Bigg{[}\frac{1}{1+\Gamma(\alpha+1)(s\,\tau )^{-\alpha}+(s\,\tau)^{-1}}\Bigg{]}. \tag{71}\]  
Note that, for \(a\in(0,1)\) (thus \(1-a\in(0,1)\)), and \(A\in\mathbb{R}\), we have that  
\[\lim_{y\to+\infty}\frac{1+A\,y^{-a}+y^{-1}}{1+A\,y  
Apply (73) to (71), in order to obtain  
\[\tilde{G}_{\alpha}(s)\approx\frac{\mu}{s}\Bigg{[}\frac{1}{1+\Gamma(\alpha+1)(s \,\tau)^{-\alpha}}\Bigg{]},\quad s\to+\infty. \tag{74}\]  
Now, perform the change of variables \(x=1/z\) in (63),  
\[\frac{1}{1+\frac{1}{z}}\approx 1-\frac{1}{z},\quad z\to+\infty, \tag{75}\]  
and take \(z=\frac{(s\,\tau)^{\alpha}}{\Gamma(\alpha+1)}\to+\infty\) (i.e., \(s\to+\infty\)), to arrive at  
\[\tilde{G}_{\alpha}(s)\approx\mu\bigg{[}\frac{1}{s}-\frac{\Gamma(\alpha+1)}{ \tau^{\alpha}s^{\,\alpha+1}}\bigg{]},\quad s\to+\infty. \tag{76}\]  
According to (57), the asymptotic behaviour of \(G_{\alpha}(t)\) as \(t\to 0^{+}\) is calculated as  
\[G_{\alpha}(t)\approx\mu\,\mathcal{L}^{-1}\bigg{[}\frac{1}{s}-\frac{\Gamma( \alpha+1)}{\tau^{\alpha}s^{\,\alpha+1}};t\bigg{]}, \tag{77}\]  
i.e.,  
\[G_{\alpha}(t)\approx\mu\bigg{[}1-\bigg{(}\frac{t}{\tau}\bigg{)}^{\alpha} \bigg{]},\quad t\to 0^{+}. \tag{78}\]  
Again, taking more terms in the expansion of \(\tilde{G}_{\alpha}(s)\) as \(s\to+\infty\), we can calculate more terms of \(G_{\alpha}(t)\) as \(t\to 0^{+}\) by using (57). Thereby, we obtain  
\[G_{\alpha}(t)\approx\mu\Bigg{[}1-\bigg{(}\frac{t}{\tau}\bigg{)}^{\alpha}+\frac {\Gamma^{2}(1+\alpha)}{\Gamma(2\alpha-1)}\bigg{(}\frac{t}{\tau}\bigg{)}^{2 \alpha}+\ldots\Bigg{]},\quad t\to 0^{+}. \tag{79}\]  
Note that the particular case \(\alpha=\frac{1}{3}\) in (78) yields  
\[G_{1/3}(t)\approx\mu\Bigg{[}1-\bigg{(}\frac{t}{\tau}\bigg{)}^{1/3}\Bigg{]}, \quad t\to 0^{+}. \tag{80}\]  
As a consistency test, we can obtain (80) from the the expression given in (50) for \(G_{m/n}(t)\) with \(m=1\) and \(n=3\), and the definition of the Mittag-Leffler Function (45). Furthermore, the asymptotic formula given in (78) can be obtained from the Volterra integral Equation (4). Indeed, taking into account (2) and (3) and performing the change of variables \(x=t-t^{\prime}\), this integral equation reads as  
\[G_{\alpha}(t)=\mu-\frac{1}{\tau}\int_{0}^{t}\Bigg{[}1+\alpha\bigg{(}\frac{t- x}{\tau}\bigg{)}^{\alpha-1}\Bigg{]}G_{\alpha}(x)\,dx, \tag{81}\]  
thus,  
\[\lim_{t\to 0^{+}}G_{\alpha}(t)=\mu. \tag{82}\]  
According to (82), we can take the approximation \(G_{\alpha}(x)\approx\mu\) as \(t\to 0^{+}\) in (81), thus  
\[G_{\alpha}(t) \approx \mu-\frac{\mu}{\tau}\int_{0}^{t}\Bigg{[}1+\alpha\bigg{(}\frac{t- x}{\tau}\bigg{)}^{\alpha-1}\Bigg{]}\,dx\] \[= \mu\bigg{[}1-\frac{t}{\tau}-\bigg{(}\frac{t}{\tau}\bigg{)}^{ \alpha}\bigg{]}\]Recalling that \(\alpha\in(0,1)\), we recover (78), i.e.,  
\[G_{\alpha}(t)\approx\mu\bigg{[}1-\left(\frac{t}{\tau}\right)^{\alpha}\bigg{]}, \quad t\to 0^{+}. \tag{84}\]  
Figure 1 presents the graph of \(G_{1/3}(t)\) for \(\mu=1\) and different values of \(\tau\). Figure 2 shows the asymptotic behaviours given in (80) and (69) for \(G_{1/3}(t)\) with \(\mu=1\) and \(\tau=\frac{1}{2}\).