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Create a process from random vectors and processes

The objective is to create a process defined from a random vector and a process.

We consider the following limit state function, defined as the difference between a degrading resistance r(t) = R - bt and a time-varying load S(t):

\begin{align*} g(t)= r(t) - S(t) = R - bt - S(t) \end{align*}

We propose the following probabilistic model:

  • R is the initial resistance, and R \sim \mathcal{N}(\mu_R, \sigma_R);

  • b is the deterioration rate of the resistance; it is deterministic;

  • S(t) is the time-varying stress, which is modeled by a stationary Gaussian process of mean value \mu_S, standard deviation \sigma_S and a squared exponential covariance model;

  • t is the time, varying in [0,T].

First, import the python modules:

import openturns as ot
from openturns.viewer import View
import math as m

1. Create the gaussian process (\omega, t) \rightarrow S(\omega,t)

Create the mesh which is a regular grid on [0,T], with T=50, by step =1:

b = 0.01
t0 = 0.0
step = 1
tfin = 50
n = round((tfin - t0) / step)
myMesh = ot.RegularGrid(t0, step, n)

Create the squared exeponential covariance model:

C(s,t) = \sigma^2e^{-\frac{1}{2} \left( \dfrac{s-t}{l} \right)^2}

where the scale parameter is l=\frac{10}{\sqrt{2}} and the amplitude \sigma = 1.

ll = 10 / m.sqrt(2)
myCovKernel = ot.SquaredExponential([ll])
print("cov model = ", myCovKernel)

cov model =  SquaredExponential(scale=[7.07107], amplitude=[1])

Create the gaussian process S(t):

S_proc = ot.GaussianProcess(myCovKernel, myMesh)

2. Create the process (\omega, t) \rightarrow R(\omega)-bt

First, create the random variable R \sim \mathcal{N}(\mu_R, \sigma_R), with \mu_R = 5 and \sigma_R = 0.3:

muR = 5
sigR = 0.3
R = ot.Normal(muR, sigR)

The create the Dirac random variable B = b:

B = ot.Dirac(b)

Then create the process (\omega, t) \rightarrow R(\omega)-bt using the FunctionalBasisProcess class and the functional basis \phi_1 : t \rightarrow 1 and \phi_2: -t \rightarrow t:

R(\omega)-bt = R(\omega)\phi_1(t) + B(\omega) \phi_2(t)

with (R,B) independent.

const_func = ot.SymbolicFunction(["t"], ["1"])
linear_func = ot.SymbolicFunction(["t"], ["-t"])
myBasis = ot.Basis([const_func, linear_func])

coef = ot.ComposedDistribution([R, B])

R_proc = ot.FunctionalBasisProcess(coef, myBasis, myMesh)

3. Create the process Z: (\omega, t) \rightarrow R(\omega)-bt + S(\omega, t)

First, aggregate both processes into one process of dimension 2: (R_{proc}, S_{proc})

myRS_proc = ot.AggregatedProcess([R_proc, S_proc])

Then create the spatial field function that acts only on the values of the process, keeping the mesh unchanged, using the ValueFunction class. We define the function g on \mathbb{R}^2 by:

g(x,y) = x-y

in order to define the spatial field function g_{dyn} that acts on fields, defined by:

\forall t\in [0,T], g_{dyn}(X(\omega, t), Y(\omega, t)) = X(\omega, t) - Y(\omega, t)

g = ot.SymbolicFunction(["x1", "x2"], ["x1-x2"])
gDyn = ot.ValueFunction(g, myMesh)

Now you have to create the final process Z thanks to g_{dyn}:

Z_proc = ot.CompositeProcess(gDyn, myRS_proc)

4. Draw some realizations of the process

N = 10
sampleZ_proc = Z_proc.getSample(N)
graph = sampleZ_proc.drawMarginal(0)
graph.setTitle(r"Some realizations of $Z(\omega, t)$")
view = View(graph)
Some realizations of $Z(\omega, t)$

5. Evaluate the probability that Z(\omega, t) \in \mathcal{D}

We define the domaine \mathcal{D} = [2,4] and the event Z(\omega, t) \in \mathcal{D}:

domain = ot.Interval([2], [4])
print("D = ", domain)
event = ot.ProcessEvent(Z_proc, domain)

D =  [2, 4]

We use the Monte Carlo sampling to evaluate the probability:

MC_algo = ot.ProbabilitySimulationAlgorithm(event)
MC_algo.setMaximumOuterSampling(1000000)
MC_algo.setBlockSize(100)
MC_algo.setMaximumCoefficientOfVariation(0.01)
MC_algo.run()

result = MC_algo.getResult()

proba = result.getProbabilityEstimate()
print("Probability = ", proba)
variance = result.getVarianceEstimate()
print("Variance Estimate = ", variance)
IC90_low = proba - result.getConfidenceLength(0.90) / 2
IC90_upp = proba + result.getConfidenceLength(0.90) / 2
print("IC (90%) = [", IC90_low, ", ", IC90_upp, "]")
view.ShowAll()

Probability =  0.7514705882352942
Variance Estimate =  5.5479048319547984e-05
IC (90%) = [ 0.7392190178861351 ,  0.7637221585844534 ]