Average amplitudes analysis for a phenomenological model under hydrodynamic interactions with periodic perturbation and multiplicative trichotomous noise

Abstract

From a statistical mechanics perspective, to describe the dynamics of a tracer, a phenomenological model has been established by a generalized Langevin equation (GLE) which includes a Basset force, a periodic perturbation force, a Stokes force, an external force and a thermal noise. Using the generalized Shapiro-Loginov formula, the iterative expressions of the first moments of the system are obtained. The time series of the first moments have been extensively investigated. By analyzing the time series of the first moments of the system with different system parameters, the irregular responses of the curves are revealed and tend to be stable for a long time. Significantly, the dynamics of average amplitudes of the first moments, influenced by various system parameters, have also been addressed in detail. Especially, the monotonic and non-monotonic properties of the average amplitudes of the first moments versus the memory exponent $\alpha$ are discussed.

Appendices

Appendix A: Coefficients of Eqs. (29), (30), (31)

$$\begin{aligned} {H_1}\left( s \right)&= \frac{{\left[ {d_2^2\left( s \right) - {a^2}{k^2}} \right] {g_1}\left( s \right) + {k^2}{g_4}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.1)

$$\begin{aligned} {H_2}\left( s \right)&= \frac{{ - k{d_2}\left( s \right) {g_3}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.2)

$$\begin{aligned} {H_3}\left( s \right)&= \frac{{{k^2}{g_3}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.3)

$$\begin{aligned} {H_4}\left( s \right)&= \frac{{\left[ {d_2^2\left( s \right) - {a^2}{k^2}} \right] {g_2}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.4)

$$\begin{aligned} {H_5}\left( s \right)&= \frac{{ - Mk{d_2}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.5)

$$\begin{aligned} {H_6}\left( s \right)&= \frac{{M{k^2}}}{\sigma }, \end{aligned}$$

(A.6)

$$\begin{aligned} {I_1}\left( s \right)&= \frac{{2q{a^2}k{d_3}\left( s \right) {g_1}\left( s \right) - k{d_1}\left( s \right) {g_4}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.7)

$$\begin{aligned} {I_2}\left( s \right)&= \frac{{{d_1}\left( s \right) {d_2}\left( s \right) {g_3}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.8)

$$\begin{aligned} {I_3}\left( s \right)&= \frac{{ - k{d_1}\left( s \right) {g_3}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.9)

$$\begin{aligned} {I_4}\left( s \right)&= \frac{{2q{a^2}k{d_3}\left( s \right) {g_2}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.10)

$$\begin{aligned} {I_5}\left( s \right)&= \frac{{M{d_1}\left( s \right) {d_2}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.11)

$$\begin{aligned} {I_6}\left( s \right)&= \frac{{ - Mk{d_1}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.12)

$$\begin{aligned} {J_1}\left( s \right)&= \frac{{{d_2}\left( s \right) \left[ {{d_1}\left( s \right) {g_4}\left( s \right) - 2q{a^2}{d_3}\left( s \right) {g_1}\left( s \right) } \right] }}{\sigma }, \end{aligned}$$

(A.13)

$$\begin{aligned} {J_2}\left( s \right)&= \frac{{ - {a^2}k\left[ {{d_1}\left( s \right) - 2q{d_3}\left( s \right) } \right] {g_3}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.14)

$$\begin{aligned} {J_3}\left( s \right)&= \frac{{{d_1}\left( s \right) {d_2}\left( s \right) {g_3}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.15)

$$\begin{aligned} {J_4}\left( s \right)&= \frac{{ - 2q{a^2}{d_2}\left( s \right) {d_3}\left( s \right) {g_2}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.16)

$$\begin{aligned} {J_5}\left( s \right)&= \frac{{ - M{a^2}k\left[ {{d_1}\left( s \right) - 2q{d_3}\left( s \right) } \right] }}{\sigma }, \end{aligned}$$

(A.17)

$$\begin{aligned} {J_6}\left( s \right)&= \frac{{M{d_1}\left( s \right) {d_2}\left( s \right) }}{\sigma }, \end{aligned}$$

(A.18)

where $d_1(s),d_1(s),d_3(s)$ and $g_1(s),g_2(s),g_3(s),g_4(s)$ are given by Eqs. (27),(28), respectively.

Appendix B: The inverse Laplace transform of Eqs. (A.1)–(A.18)

From Eqs. (A.1)–(A.18), due to the following limit

$$\begin{aligned} \mathop {\lim }\limits _{s \rightarrow 0} \frac{{{H_n}\left( s \right) }}{{{Y_n}\left( s \right) }} = 1,\mathop {\lim }\limits _{s \rightarrow 0} \frac{{{I_n}\left( s \right) }}{{{U_n}\left( s \right) }} = 1,\mathop {\lim }\limits _{s \rightarrow 0} \frac{{{J_n}\left( s \right) }}{{{V_n}\left( s \right) }} = 1,n = 1,2, \ldots ,6, \end{aligned}$$

(B.1)

where

$$\begin{aligned} {Y_1}\left( s \right)&= \frac{{\mathrm{{P}}{r_h}{s^{\frac{1}{2}}} - 2q{a^2}{k^2}{\mathrm{{P}}_4}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.2)

$$\begin{aligned} {Y_2}\left( s \right)&= \frac{{ - k{\mathrm{{P}}_1}{\mathrm{{P}}_4}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.3)

$$\begin{aligned} {Y_3}\left( s \right)&= \frac{{{k^2}{\mathrm{{P}}_4}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.4)

$$\begin{aligned} {Y_4}\left( s \right)&= \frac{{M\mathrm{{P}}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.5)

$$\begin{aligned} {Y_5}\left( s \right)&= \frac{{ - Mk{\mathrm{{P}}_1}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.6)

$$\begin{aligned} {Y_6}\left( s \right)&= \frac{{M{k^2}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.7)

$$\begin{aligned} {U_1}\left( s \right)&= \frac{{ - 2q{a^2}k\left[ {{\mathrm{{P}}_2}{r_h}{s^{\frac{1}{2}}} - {\mathrm{{P}}_4}\left( {{r_\alpha }{s^\alpha } + 1} \right) } \right] }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.8)

$$\begin{aligned} {U_2}\left( s \right)&= \frac{{\left( {{r_\alpha }{s^\alpha } + 1} \right) {\mathrm{{P}}_1}{\mathrm{{P}}_4}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.9)

$$\begin{aligned} {U_3}\left( s \right)&= \frac{{ - k{\mathrm{{P}}_4}\left( {{r_\alpha }{s^\alpha } + 1} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.10)

$$\begin{aligned} {U_4}\left( s \right)&= \frac{{ - 2q{a^2}kM{\mathrm{{P}}_2}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.11)

$$\begin{aligned} {U_5}\left( s \right)&= \frac{{M{\mathrm{{P}}_1}\left( {{r_\alpha }{s^\alpha } + 1} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.12)

$$\begin{aligned} {U_6}\left( s \right)&= \frac{{ - Mk\left( {{r_\alpha }{s^\alpha } + 1} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.13)

$$\begin{aligned} {V_1}\left( s \right)&= \frac{{ - 2q{a^2}{\mathrm{{P}}_1}\left[ {{\mathrm{{P}}_4}\left( {{r_\alpha }{s^\alpha } + 1} \right) - {\mathrm{{P}}_2}{r_h}{s^{\frac{1}{2}}}} \right] }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.14)

$$\begin{aligned} {V_2}\left( s \right)&= \frac{{ - {a^2}k{\mathrm{{P}}_4}\left( {{r_\alpha }{s^\alpha } + {\mathrm{{P}}_3}} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.15)

$$\begin{aligned} {V_3}\left( s \right)&= \frac{{\left( {{r_\alpha }{s^\alpha } + 1} \right) {\mathrm{{P}}_1}{\mathrm{{P}}_4}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.16)

$$\begin{aligned} {V_4}\left( s \right)&= \frac{{2q{a^2}M{\mathrm{{P}}_1}{\mathrm{{P}}_2}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.17)

$$\begin{aligned} {V_5}\left( s \right)&= \frac{{ - M{a^2}k\left( {{r_\alpha }{s^\alpha } + {\mathrm{{P}}_3}} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.18)

$$\begin{aligned} {V_6}\left( s \right)&= \frac{{M{\mathrm{{P}}_1}\left( {{r_\alpha }{s^\alpha } + 1} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}, \end{aligned}$$

(B.19)

therefore, one has

$$\begin{aligned}&{H_n}\left( s \right) \simeq {Y_n}\left( s \right) ,{I_n}\left( s \right) \simeq {U_n}\left( s \right) ,\nonumber \\&{J_n}\left( s \right) \simeq {V_n}\left( s \right) ,n = 1,2, \ldots ,6. \end{aligned}$$

(B.20)

One assumes that ${h_n}\left( t \right) ,n = 1,2, \ldots ,6$, are the inverse Laplace transform of ${H_n}\left( s \right) ,n = 1,2, \ldots ,6$, respectively. Due to the fact that ${H_n}\left( s \right) ,n = 1,2, \ldots ,6$ are equivalent to ${Y_n}\left( s \right) ,n = 1,2, \ldots ,6$, taking the inverse Laplace transform of Eqs. (A.1)-(A.6), one can obtain

$$\begin{aligned} {h_1}\left( t \right)&= {L^{ - 1}}\left[ {{H_1}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{Y_1}\left( s \right) } \right] \nonumber \\&= \frac{{{r_h}}}{{{r_\alpha }}}{t^{\alpha - \frac{3}{2}}}{E_{\alpha ,\alpha - \frac{1}{2}}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) \nonumber \\&\quad + \frac{{2q{a^2}{k^2}{\mathrm{{P}}_4}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= \frac{{{r_h}}}{{{r_\alpha }}}{t^{\alpha - \frac{3}{2}}}{E_{\alpha ,\alpha - \frac{1}{2}}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) \nonumber \\&\quad + \frac{{2q{a^2}{k^2}{\mathrm{{P}}_4}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.21)

$$\begin{aligned} {h_2}\left( t \right)&= {L^{ - 1}}\left[ {{H_2}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{Y_2}\left( s \right) } \right] \nonumber \\&= \frac{{k{\mathrm{{P}}_1}{\mathrm{{P}}_4}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= \frac{{k{\mathrm{{P}}_1}{\mathrm{{P}}_4}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.22)

$$\begin{aligned} {h_3}\left( t \right)&= {L^{ - 1}}\left[ {{H_3}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{Y_3}\left( s \right) } \right] \nonumber \\&= - \frac{{{k^2}{\mathrm{{P}}_4}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - \frac{{{k^2}{\mathrm{{P}}_4}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.23)

$$\begin{aligned} {h_4}\left( t \right)&= {L^{ - 1}}\left[ {{H_4}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{Y_4}\left( s \right) } \right] \nonumber \\&= - \frac{{M\mathrm{{P}}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - \frac{{M\mathrm{{P}}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.24)

$$\begin{aligned} {h_5}\left( t \right)&= {L^{ - 1}}\left[ {{H_5}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{Y_5}\left( s \right) } \right] \nonumber \\&= \frac{{Mk{\mathrm{{P}}_1}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= \frac{{Mk{\mathrm{{P}}_1}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.25)

$$\begin{aligned} {h_6}\left( t \right)&= {L^{ - 1}}\left[ {{H_6}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{Y_6}\left( s \right) } \right] \nonumber \\&= - \frac{{M{k^2}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - \frac{{M{k^2}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.26)

where involving two-parameter Mittag-Leffler function [56, 57] is given by $E_{\mu , v}(z)=\sum _{n=0}^{\infty } \frac{z^{n}}{\varGamma (\mu n+v)}, \mathcal {R} e(\mu )>0, v \in \mathbb {C}, z \in \mathbb {C}$, and $E_{\mu , 1}(z)=E_{\mu }(z)$ turns to Mittag-Leffler function [56, 57], $E_{\mu }(z)=\sum _{n=0}^{\infty } \frac{z^{n}}{\varGamma (\mu n+1)}$, $\alpha >0, z \in \mathbb {C}$.

Letting ${i_n}\left( t \right) ={L^{ - 1}}\left[ {I_n}\left( s \right) \right] ,n = 1,2, \ldots ,6$, Since ${I_n}\left( s \right) ,n = 1,2, \ldots ,6$, are equivalent to ${U_n}\left( s \right) ,n = 1,2, \ldots ,6$, taking the inverse Laplace transform of Eqs (A.7)–(A.12), one can obtain

$$\begin{aligned} {i_1}\left( t \right)&= {L^{ - 1}}\left[ {{I_1}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{U_1}\left( s \right) } \right] \nonumber \\&= - \frac{{2q{a^2}k{\mathrm{{P}}_2}{r_h}}}{{\mathrm{{P}}{r_\alpha }}}{t^{\alpha - \frac{3}{2}}}{E_{\alpha ,\alpha - \frac{1}{2}}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) \nonumber \\&\quad + 2q{a^2}k{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - \frac{{2q{a^2}k{\mathrm{{P}}_2}{r_h}}}{{\mathrm{{P}}{r_\alpha }}}{t^{\alpha - \frac{3}{2}}}{E_{\alpha ,\alpha - \frac{1}{2}}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) \nonumber \\&\quad + 2q{a^2}k{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.27)

$$\begin{aligned} {i_2}\left( t \right)&= {L^{ - 1}}\left[ {{I_2}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{U_2}\left( s \right) } \right] \nonumber \\&= {\mathrm{{P}}_1}{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= {\mathrm{{P}}_1}{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.28)

$$\begin{aligned} {i_3}\left( t \right)&= {L^{ - 1}}\left[ {{I_3}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{U_3}\left( s \right) } \right] \nonumber \\&= - k{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - k{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.29)

$$\begin{aligned} {i_4}\left( t \right)&= {L^{ - 1}}\left[ {{I_4}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{U_4}\left( s \right) } \right] \nonumber \\&= \frac{{2q{a^2}kM{\mathrm{{P}}_4}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= \frac{{2q{a^2}kM{\mathrm{{P}}_4}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.30)

$$\begin{aligned} {i_5}\left( t \right)&= {L^{ - 1}}\left[ {{I_5}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{U_5}\left( s \right) } \right] \nonumber \\&= M{\mathrm{{P}}_1}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= M{\mathrm{{P}}_1}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.31)

$$\begin{aligned} {i_6}\left( t \right)&= {L^{ - 1}}\left[ {{I_6}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{U_6}\left( s \right) } \right] \nonumber \\&= - Mk\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - Mk\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) . \end{aligned}$$

(B.32)

Letting ${j_n}\left( t \right) ={L^{ - 1}}\left[ {J_n}\left( s \right) \right] ,n = 1,2, \ldots ,6$, using similar technique and taking the inverse Laplace transform of Eqs. (A.13)-(A.18), one can obtain

$$\begin{aligned} {j_1}\left( t \right)&= {L^{ - 1}}\left[ {{J_1}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{V_1}\left( s \right) } \right] \nonumber \\&= - \frac{{2q{a^2}{\mathrm{{P}}_1}{\mathrm{{P}}_4}}}{{\mathrm{{P}}{r_\alpha }}}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&+ 2q{a^2}{\mathrm{{P}}_1}{\mathrm{{P}}_2}{t^{\alpha - \frac{3}{2}}}{E_{\alpha ,\alpha - \frac{1}{2}}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) \nonumber \\&= - \frac{{2q{a^2}{\mathrm{{P}}_1}{\mathrm{{P}}_4}}}{{\mathrm{{P}}{r_\alpha }}}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) \nonumber \\&+ 2q{a^2}{\mathrm{{P}}_1}{\mathrm{{P}}_2}{t^{\alpha - \frac{3}{2}}}{E_{\alpha ,\alpha - \frac{1}{2}}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.33)

$$\begin{aligned} {j_2}\left( t \right)&= {L^{ - 1}}\left[ {{J_2}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{V_2}\left( s \right) } \right] \nonumber \\&= - {a^2}k{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{{{\mathrm{{P}}_3}}}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - {a^2}k{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{{{\mathrm{{P}}_3}}}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.34)

$$\begin{aligned} {j_3}\left( t \right)&= {L^{ - 1}}\left[ {{J_3}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{V_3}\left( s \right) } \right] \nonumber \\&= {\mathrm{{P}}_1}{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= {\mathrm{{P}}_1}{\mathrm{{P}}_4}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.35)

$$\begin{aligned} {j_4}\left( t \right)&= {L^{ - 1}}\left[ {{J_4}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{V_4}\left( s \right) } \right] \nonumber \\&= - \frac{{2q{a^2}M{\mathrm{{P}}_1}{\mathrm{{P}}_2}}}{Q}\frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - \frac{{2q{a^2}M{\mathrm{{P}}_1}{\mathrm{{P}}_2}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.36)

$$\begin{aligned} {j_5}\left( t \right)&= {L^{ - 1}}\left[ {{J_5}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{V_5}\left( s \right) } \right] \nonumber \\&= - M{a^2}k\left( {\frac{1}{\mathrm{{P}}} - \frac{{{\mathrm{{P}}_3}}}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= - M{a^2}k\left( {\frac{1}{\mathrm{{P}}} - \frac{{{\mathrm{{P}}_3}}}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(B.37)

$$\begin{aligned} {j_6}\left( t \right)&= {L^{ - 1}}\left[ {{J_6}\left( s \right) } \right] \approx {L^{ - 1}}\left[ {{V_6}\left( s \right) } \right] \nonumber \\&= M{\mathrm{{P}}_1}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) \frac{\mathrm{{d}}}{{\mathrm{{d}}t}}\left[ {{E_\alpha }\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) } \right] \nonumber \\&= M{\mathrm{{P}}_1}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) . \end{aligned}$$

(B.38)

Appendix C: Relaxation functions

Using properties of Laplace trasform, from Eqs.(B.2)- (43), the relaxation functions ${{\hat{h}}_n}\left( t \right)$, ${{\hat{i}}_n}\left( t \right)$, ${{\hat{j}}_n}\left( t \right) (n=1,2,3)$ are given by

$$\begin{aligned} {{{\hat{h}}}_1}\left( t \right)&= {L^{ - 1}}\left[ {\frac{{ - \mathrm{{P}}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] = \frac{\mathrm{{P}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.1)

$$\begin{aligned} {{{\hat{h}}}_2}\left( t \right) \mathrm{{ }}&= {L^{ - 1}}\left[ {\frac{{k{\mathrm{{P}}_1}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] = - \frac{{k{\mathrm{{P}}_1}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.2)

$$\begin{aligned} {{{\hat{h}}}_3}\left( t \right)&= {L^{ - 1}}\left[ {\frac{{ - {k^2}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] = \frac{{{k^2}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.3)

$$\begin{aligned} {{{\hat{i}}}_1}\left( t \right) \mathrm{{ }}&= {L^{ - 1}}\left[ {\frac{{2q{a^2}k{\mathrm{{P}}_2}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] \nonumber \\&= - \frac{{2q{a^2}k{\mathrm{{P}}_2}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.4)

$$\begin{aligned} {{{\hat{i}}}_2}\left( t \right)&= {L^{ - 1}}\left[ {\frac{{ - {\mathrm{{P}}_1}\left( {{r_\alpha }{s^\alpha } + 1} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] \nonumber \\&= - {\mathrm{{P}}_1}\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.5)

$$\begin{aligned} {{{\hat{i}}}_3}\left( t \right) \mathrm{{ }}&= {L^{ - 1}}\left[ {\frac{{k\left( {{r_\alpha }{s^\alpha } + 1} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] \nonumber \\&= k\left( {\frac{1}{\mathrm{{P}}} - \frac{1}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.6)

$$\begin{aligned} {{{\hat{j}}}_1}\left( t \right)&= {L^{ - 1}}\left[ {\frac{{2q{a^2}{\mathrm{{P}}_1}{\mathrm{{P}}_2}}}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] \nonumber \\&= - \frac{{2q{a^2}{\mathrm{{P}}_1}{\mathrm{{P}}_2}}}{Q}{t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.7)

$$\begin{aligned} {{{\hat{j}}}_2}\left( t \right)&= {L^{ - 1}}\left[ {\frac{{{a^2}k\left( {{r_\alpha }{s^\alpha } + {\mathrm{{P}}_3}} \right) }}{{\mathrm{{P}}{r_\alpha }{s^\alpha } + Q}}} \right] \nonumber \\&= {a^2}k\left( {\frac{1}{\mathrm{{P}}} - \frac{{{\mathrm{{P}}_3}}}{Q}} \right) {t^{ - 1}}{E_{\alpha ,0}}\left( { - \frac{Q}{{\mathrm{{P}}{r_\alpha }}}{t^\alpha }} \right) , \end{aligned}$$

(C.8)

$$\begin{aligned} {{{\hat{j}}}_3}\left( t \right)&= {{{\hat{i}}}_2}\left( t \right) . \end{aligned}$$

(C.9)

Appendix D: commutative property between average and differentiation

Supposed that x(t) is a random process with probability density function p(x, t). Its average or expectation is defined as $\langle x\rangle =\int _{-\infty }^{+\infty }xp(x,t)\text {d}x$. Distinctly, the integral in the definition of average converges uniformly. Using differential properties and integral properties, one could yield

$$\begin{aligned} \left\langle { }_0^C D_t^\alpha x(t)\right\rangle&=\left\langle \frac{1}{\varGamma (1-\alpha )} \int _0^t \frac{1}{(t-u)^\alpha } \dot{x}(u) d u\right\rangle \nonumber \\&=\int _{-\infty }^{+\infty } \frac{1}{\varGamma (1-\alpha )} \int _0^t \frac{1}{(t-u)^\alpha } \dot{x}(u) d u p(x, u) d x \nonumber \\&=\frac{1}{\varGamma (1-\alpha )} \int _0^t \frac{1}{(t-u)^\alpha } \int _{-\infty }^{+\infty } \dot{x}(u) p(x, u) d x d u \nonumber \\&=\frac{1}{\varGamma (1-\alpha )} \int _0^t \frac{1}{(t-u)^\alpha }\langle \dot{x}(u)\rangle d u \nonumber \\&={ }_0^C D_t^\alpha \left\langle x(t)\right\rangle . \end{aligned}$$

(D.1)

For $\alpha \rightarrow 1$ one could find that $\left\langle \frac{d}{dt}x(t)\right\rangle =\frac{d}{dt}\left\langle x(t)\right\rangle$. Analogously, for $\alpha \rightarrow 2$, $\left\langle \frac{d^2}{dt^2}x(t)\right\rangle =\frac{d^2}{dt^2}\left\langle x(t)\right\rangle$. For commutativity of integer order, one can also see Refs. [50,51,52,53,54].