Inverse Laplace Transform of $\frac{1}{s+s^p/a}$

Question

I am analyzing creep functions for some rock mechanics problem, and I found that the function: \begin{equation} J(t)=1/E +Bt^p, \end{equation} is the best choice for my dataset, where $E$, $B$ and $p$ are positive constants, and additionally $p$ is between 0 and 1. Now, I'd like to invert it to get the relaxation modulus. Thus first, I take the usual viscoelastic inversion formula: \begin{equation} M(t) = \mathcal{L}^{-1}\bigg[\frac{1}{\widetilde{J}(s)s^2} \bigg]; \end{equation} in my case, the function to invert equals: \begin{equation} \widetilde{M}(s)=\frac{Es^{p-1}}{s^p+a}=E\bigg[\frac{1}{s}-\frac{1}{s+s^p/a}\bigg], \end{equation} where $a=EB\Gamma(p+1)$. The $1/s$ part is obvious, but I wonder if there is a way to invert $1/(s+s^p/a$). It looks like a mixture of a power law with exponential function, but I've got no idea how to proceed to get some result. The $p$ values I obtained experimentally were between 0.1 and 0.4 approximately. I found out that inversion is possible (of course) for $p=0$ and $p=1$, and also for $p=1/2$. Maybe, there are some other fixed $p$ values with existing inverse transforms.

I will appreciate any help or hints on this problem. Thanks!

Answer 1

In case if the topic is still of interest: The expression

\begin{equation} \tilde{M}(s) = E \frac{s^{p-1}}{s^p+a} \end{equation}

can be inverted using the following formula: \begin{equation} \mathcal{L} \left\{ E_{\alpha} \left( \lambda t^{\alpha} \right) \right\} = \frac{s^{\alpha-1}}{s^\alpha - \lambda} \end{equation} where $E_\alpha(z)$ is the one-parametric Mittag-Leffler function.

Setting $\alpha = p$ and $\lambda=-a$, the above expression can be inverted as: \begin{equation} M(t) = E \cdot E_p\left( -a t^{\alpha} \right) \end{equation}

Detailed information about the Laplace-transforms involving the one-parametric and the generalized Mittag-Leffler function can be found in:

• Rogosin, Sergei, The role of the Mittag-Leffler function in fractional modeling, Mathematics 3, No. 2, 368-381 (2015). ZBL1318.33036.

• Podlubny, Igor, Fractional differential equations. An introduction to fractional derivatives, fractional differential equations, to methods of their solution and some of their applications, Mathematics in Science and Engineering. 198. San Diego, CA: Academic Press. xxiv, 340 p. (1999). ZBL0924.34008.

• Mainardi, Francesco, Fractional calculus and waves in linear viscoelasticity. An introduction to mathematical models, Hackensack, NJ: World Scientific (ISBN 978-1-84816-329-4/hbk). xx, 347 p. (2010). ZBL1210.26004.

Answer 2

This is a half answer and little more.

Not that:

$$\sum _{n=0}^{\infty } (-x)^n=\frac{1}{1+x}$$

so, $$\color{red}{\mathcal{L}_s^{-1}\left[\frac{1}{s+\frac{s^p}{a}}\right](t)}=\mathcal{L}_s^{-1}\left[\frac{1}{s \left(1+\frac{s^{-1+p}}{a}\right)}\right](t)=\mathcal{L}_s^{-1}\left[\frac{\sum _{n=0}^{\infty } \left(-\frac{s^{-1+p}}{a}\right)^n}{s}\right](t)=\sum _{n=0}^{\infty } \mathcal{L}_s^{-1}\left[(-1)^n a^{-n} s^{-1+n (-1+p)}\right](t)=\color{red}{\sum _{n=0}^{\infty } \frac{(-1)^n a^{-n} t^{-n (-1+p)}}{\Gamma (1-n (-1+p))}}$$

The answer is a $\color{red}{red}$ Sum,but I can't find closed form it.

Only for given parameter $\color{red}{p}$ CAS can find solution.

Mathematica code:

for p=1/10:

  Sum[((-1)^n a^-n t^(-n (-1 + p)))/Gamma[1 - n (-1 + p)] /. 
  p -> 1/10, {n, 0, Infinity}] // Simplify

$\, _0F_8\left(;\frac{1}{9},\frac{2}{9},\frac{1}{3},\frac{4}{9},\frac{5}{9},\frac{2}{3},\frac{7}{9},\frac{8}{9};\frac{t^9}{387420489 a^{10}}\right)-\frac{10 t^{9/10} \, _1F_9\left(1;\frac{19}{90},\frac{29}{90},\frac{13}{30},\frac{49}{90},\frac{59}{90},\frac{23}{30},\frac{79}{90},\frac{89}{90},\frac{11}{10};\frac{t^9}{38 7420489 a^{10}}\right)}{9 a \Gamma \left(\frac{9}{10}\right)}-\frac{1000000000 t^{81/10} \, _1F_9\left(1;\frac{91}{90},\frac{101}{90},\frac{37}{30},\frac{121}{90},\frac{131}{90},\frac{47}{30},\frac{151}{90},\frac{161}{90},\frac{19}{10};\frac{t^ 9}{387420489 a^{10}}\right)}{5252921480961 a^9 \Gamma \left(\frac{1}{10}\right)}+\frac{390625 t^{36/5} \, _1F_9\left(1;\frac{41}{45},\frac{46}{45},\frac{17}{15},\frac{56}{45},\frac{61}{45},\frac{22}{15},\frac{71}{45},\frac{76}{45},\frac{9}{5};\frac{t^9}{3874 20489 a^{10}}\right)}{643458816 a^8 \Gamma \left(\frac{1}{5}\right)}-\frac{10000000 t^{63/10} \, _1F_9\left(1;\frac{73}{90},\frac{83}{90},\frac{31}{30},\frac{103}{90},\frac{113}{90},\frac{41}{30},\frac{133}{90},\frac{143}{90},\frac{17}{10};\frac{t^9 }{387420489 a^{10}}\right)}{4250022777 a^7 \Gamma \left(\frac{3}{10}\right)}+\frac{15625 t^{27/5} \, _1F_9\left(1;\frac{32}{45},\frac{37}{45},\frac{14}{15},\frac{47}{45},\frac{52}{45},\frac{19}{15},\frac{62}{45},\frac{67}{45},\frac{8}{5};\frac{t^9}{3874 20489 a^{10}}\right)}{1696464 a^6 \Gamma \left(\frac{2}{5}\right)}-\frac{32 t^{9/2} \, _1F_9\left(1;\frac{11}{18},\frac{13}{18},\frac{5}{6},\frac{17}{18},\frac{19}{18},\frac{7}{6},\frac{23}{18},\frac{25}{18},\frac{3}{2};\frac{t^9}{38742048 9 a^{10}}\right)}{945 \sqrt{\pi } a^5}+\frac{625 t^{18/5} \, _1F_9\left(1;\frac{23}{45},\frac{28}{45},\frac{11}{15},\frac{38}{45},\frac{43}{45},\frac{16}{15},\frac{53}{45},\frac{58}{45},\frac{7}{5};\frac{t^9}{3874 20489 a^{10}}\right)}{5616 a^4 \Gamma \left(\frac{3}{5}\right)}-\frac{1000 t^{27/10} \, _1F_9\left(1;\frac{37}{90},\frac{47}{90},\frac{19}{30},\frac{67}{90},\frac{77}{90},\frac{29}{30},\frac{97}{90},\frac{107}{90},\frac{13}{10};\frac{t^9}{3 87420489 a^{10}}\right)}{3213 a^3 \Gamma \left(\frac{7}{10}\right)}+\frac{25 t^{9/5} \, _1F_9\left(1;\frac{14}{45},\frac{19}{45},\frac{8}{15},\frac{29}{45},\frac{34}{45},\frac{13}{15},\frac{44}{45},\frac{49}{45},\frac{6}{5};\frac{t^9}{38742 0489 a^{10}}\right)}{36 a^2 \Gamma \left(\frac{4}{5}\right)}$

for p=2/10:

  Sum[((-1)^n a^-n t^(-n (-1 + p)))/Gamma[1 - n (-1 + p)] /. 
  p -> 2/10, {n, 0, Infinity}] // Simplify  

$-\frac{5 t^{4/5} \, _1F_4\left(1;\frac{9}{20},\frac{7}{10},\frac{19}{20},\frac{6}{5};-\frac{t^4}{256 a^5}\right)}{4 a \Gamma \left(\frac{4}{5}\right)}+\frac{625 t^{16/5} \, _1F_4\left(1;\frac{21}{20},\frac{13}{10},\frac{31}{20},\frac{9}{5};-\frac{t^4}{256 a^5}\right)}{1056 a^4 \Gamma \left(\frac{1}{5}\right)}-\frac{125 t^{12/5} \, _1F_4\left(1;\frac{17}{20},\frac{11}{10},\frac{27}{20},\frac{8}{5};-\frac{t^4}{256 a^5}\right)}{168 a^3 \Gamma \left(\frac{2}{5}\right)}+\frac{25 t^{8/5} \, _1F_4\left(1;\frac{13}{20},\frac{9}{10},\frac{23}{20},\frac{7}{5};-\frac{t^4}{256 a^5}\right)}{24 a^2 \Gamma \left(\frac{3}{5}\right)}+\cos \left(\frac{t}{\sqrt{2} a^{5/4}}\right) \cosh \left(\frac{t}{\sqrt{2} a^{5/4}}\right)$

for p=3/10:

  Sum[((-1)^n a^-n t^(-n (-1 + p)))/Gamma[1 - n (-1 + p)] /. 
  p -> 3/10, {n, 0, Infinity}] // Simplify

$\, _0F_6\left(;\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7};\frac{t^7}{823543 a^{10}}\right)-\frac{10 t^{7/10} \, _1F_7\left(1;\frac{17}{70},\frac{27}{70},\frac{37}{70},\frac{47}{70},\frac{57}{70},\frac{67}{70},\frac{11}{10};\frac{t^7}{823543 a^{10}}\right)}{7 a \Gamma \left(\frac{7}{10}\right)}-\frac{10000000 t^{63/10} \, _1F_7\left(1;\frac{73}{70},\frac{83}{70},\frac{93}{70},\frac{103}{70},\frac{113}{70},\frac{123}{70},\frac{19}{10};\frac{t^7}{823543 a^{10}}\right)}{4250022777 a^9 \Gamma \left(\frac{3}{10}\right)}+\frac{15625 t^{28/5} \, _1F_7\left(1;\frac{33}{35},\frac{38}{35},\frac{43}{35},\frac{48}{35},\frac{53}{35},\frac{58}{35},\frac{9}{5};\frac{t^7}{823543 a^{10}}\right)}{3616704 a^8 \Gamma \left(\frac{3}{5}\right)}-\frac{100000 t^{49/10} \, _1F_7\left(1;\frac{59}{70},\frac{69}{70},\frac{79}{70},\frac{89}{70},\frac{99}{70},\frac{109}{70},\frac{17}{10};\frac{t^7}{823543 a^{10}}\right)}{9476649 a^7 \Gamma \left(\frac{9}{10}\right)}+\frac{3125 t^{21/5} \, _1F_7\left(1;\frac{26}{35},\frac{31}{35},\frac{36}{35},\frac{41}{35},\frac{46}{35},\frac{51}{35},\frac{8}{5};\frac{t^7}{823543 a^{10}}\right)}{22176 a^6 \Gamma \left(\frac{1}{5}\right)}-\frac{16 t^{7/2} \, _1F_7\left(1;\frac{9}{14},\frac{11}{14},\frac{13}{14},\frac{15}{14},\frac{17}{14},\frac{19}{14},\frac{3}{2};\frac{t^7}{823543 a^{10}}\right)}{105 \sqrt{\pi } a^5}+\frac{125 t^{14/5} \, _1F_7\left(1;\frac{19}{35},\frac{24}{35},\frac{29}{35},\frac{34}{35},\frac{39}{35},\frac{44}{35},\frac{7}{5};\frac{t^7}{823543 a^{10}}\right)}{504 a^4 \Gamma \left(\frac{4}{5}\right)}-\frac{1000 t^{21/10} \, _1F_7\left(1;\frac{31}{70},\frac{41}{70},\frac{51}{70},\frac{61}{70},\frac{71}{70},\frac{81}{70},\frac{13}{10};\frac{t^7}{823543 a^{10}}\right)}{231 a^3 \Gamma \left(\frac{1}{10}\right)}+\frac{25 t^{7/5} \, _1F_7\left(1;\frac{12}{35},\frac{17}{35},\frac{22}{35},\frac{27}{35},\frac{32}{35},\frac{37}{35},\frac{6}{5};\frac{t^7}{823543 a^{10}}\right)}{14 a^2 \Gamma \left(\frac{2}{5}\right)}$

for p=4/10:

  Sum[((-1)^n a^-n t^(-n (-1 + p)))/Gamma[1 - n (-1 + p)] /. 
  p -> 4/10, {n, 0, Infinity}] // Simplify

$-\frac{5 t^{3/5} \, _1F_3\left(1;\frac{8}{15},\frac{13}{15},\frac{6}{5};-\frac{t^3}{27 a^5}\right)}{3 a \Gamma \left(\frac{3}{5}\right)}+\frac{125 t^{12/5} \, _1F_3\left(1;\frac{17}{15},\frac{22}{15},\frac{9}{5};-\frac{t^3}{27 a^5}\right)}{168 a^4 \Gamma \left(\frac{2}{5}\right)}-\frac{25 t^{9/5} \, _1F_3\left(1;\frac{14}{15},\frac{19}{15},\frac{8}{5};-\frac{t^3}{27 a^5}\right)}{36 a^3 \Gamma \left(\frac{4}{5}\right)}+\frac{25 t^{6/5} \, _1F_3\left(1;\frac{11}{15},\frac{16}{15},\frac{7}{5};-\frac{t^3}{27 a^5}\right)}{6 a^2 \Gamma \left(\frac{1}{5}\right)}+\frac{1}{3} e^{-\frac{t}{a^{5/3}}}+\frac{2}{3} e^{\frac{t}{2 a^{5/3}}} \cos \left(\frac{\sqrt{3} t}{2 a^{5/3}}\right)$

EDITED:

p = 1/3(* ssumed parameter *);
a = 1(* assumed parameter *); 
f[t_] := NSum[((-1)^n a^-n t^(-n (-1 + p)))/Gamma[1 - n (-1 + p)], {n,
0, Infinity}, Method -> "AlternatingSigns"]
ListLinePlot[Table[{t, f[t]}, {t, 0.01, 5, 0.1}], 
AxesLabel -> {t, "f[t]"}, PlotLabels -> "Inverse Laplace Transform for: p=1/3,a=1"]