Both the creep function and the relaxation function are connected by the convolution integral. The usual method for calculating the relaxation function from a creep function or vice versa is to transfer the creep function into the Laplace domain. In the Laplace domain, the creep function is equal to 1/relaxation. In my case, my material model is ill-conditioned in the Laplace domain, so I am not able to transfer it back from the Laplace domain to the time domain. Does anyone know of a direct method to determine the relaxation function from a known creep function? I could not find an answer in the literature, maybe there is a paper or a way to transfer it directly. The link of my main question. Numerical instability in the inverse Laplace transform
As a friend here sed im going to write the math problem im facing now. This is my problem again Does anyone know a suitable way to transfer this function from the Laplace domain to the time domain? Working with python and Mathematica did not help me to find the correct answer.
import mpmath as mp
import numpy as np
import matplotlib.pyplot as plt
def Laplace_Max(s):
#mp.dps = 100
A= s**(5/4)/ (
0.00480931 +
0.0244077*s**(1/4) +
0.0129056*mp.exp(-35*s)*s**(1/4) +
0.00329167*mp.exp(0.707997*s)*s**(1/4) * mp.gammainc(0.0, 35.708*s,mp.inf, regularized=True) -
0.00530593*mp.gammainc(1.25, 35*s,mp.inf, regularized=True)
)
return A
t_R_inv = np.linspace(1,1200,12000)
Max_R=np.zeros(len(t_R_inv))
for i in range(len(t_R_inv)):
Max_R[i]=mp.invertlaplace(Laplace_Max, t_R_inv[i], method = 'stehfest', dps = 100, degree = 1000)
plt.plot(Max_R)
Update The inverse laplace of this function!
$\frac{1}{s}^{\frac{5}{4}}\cdot (0.00480931 + 0.02440766 \cdot s^{\frac{1}{4}} + 0.0129056 \cdot e^{-35 \cdot s} \cdot s^{\frac{1}{4}}+ 0.003291670 \cdot e^{0.7079967\cdot s} \cdot s^{\frac{1}{4}} \cdot \Gamma[0, 35.7079967 \cdot s] - 0.00530593 \cdot\Gamma[1.25, 35\cdot s])$
