Mean square calculus and random linear fractional differential equations: Theory and applications
, , e
28 lug 2017
INFORMAZIONI SU QUESTO ARTICOLO
Pubblicato online: 28 lug 2017
Pagine: 317 - 328
Ricevuto: 04 apr 2017
Accettato: 28 lug 2017
DOI: https://doi.org/10.21042/AMNS.2017.2.00026
Parole chiave
© 2017 C. Burgos, J.C Cortés, L. Villafuerte, R.J. Villanueva, published by Sciendo
This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.
Figure 1
![Approximations of the mean (left) and the standard deviation (rigth) of the solution SP to the random IVP (2)α = 0.7 and λ = 3/4 using different orders of truncations M = 6, 7, 8, 9, 10 over the time interval [0, 5].](https://sciendo-parsed.s3.eu-central-1.amazonaws.com/64709e5971e4585e08aa181d/j_AMNS.2017.2.00026_fig_001.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Content-Sha256=UNSIGNED-PAYLOAD&X-Amz-Credential=AKIA6AP2G7AKOUXAVR44%2F20251006%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20251006T234008Z&X-Amz-Expires=3600&X-Amz-Signature=6732d065c89d9ac8c1bf1269327f413d1a0468d59adb20413362eb37945e3625&X-Amz-SignedHeaders=host&x-amz-checksum-mode=ENABLED&x-id=GetObject)
Figure 2
![Approximations of the mean (left) and the standard deviation (right) of the solution SP to the random IVP (2) with α = 0.7 and λ = 5/4 using different orders of truncations M = 10, 12, 14, 16, 18 over the time interval [0,5].](https://sciendo-parsed.s3.eu-central-1.amazonaws.com/64709e5971e4585e08aa181d/j_AMNS.2017.2.00026_fig_002.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Content-Sha256=UNSIGNED-PAYLOAD&X-Amz-Credential=AKIA6AP2G7AKOUXAVR44%2F20251006%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20251006T234008Z&X-Amz-Expires=3600&X-Amz-Signature=8c43610a478d99024e1a1386ca44f44f318a0e422d38edfb5c5cd6cda0bf7d71&X-Amz-SignedHeaders=host&x-amz-checksum-mode=ENABLED&x-id=GetObject)
Figure 3
![Approximations of the mean (left) and the standard deviation (right) of the solution SP to the random IVP (2) with α = 0.7, λ = 5/4, E[b0]=E[c]=−1 $\mathbb{E}[b_0]=\mathbb{E}[c]=-1$ and V[b0]=V[c]=1/4 $\mathbb{V}[b_0]=\mathbb{V}[c]=1/4$ using different orders of truncations M over the time intervals [0,5].](https://sciendo-parsed.s3.eu-central-1.amazonaws.com/64709e5971e4585e08aa181d/j_AMNS.2017.2.00026_fig_003.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Content-Sha256=UNSIGNED-PAYLOAD&X-Amz-Credential=AKIA6AP2G7AKOUXAVR44%2F20251006%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20251006T234008Z&X-Amz-Expires=3600&X-Amz-Signature=03e7ec0c304162b5cb8924d7cb32af4173ee50fb286cbf99d9b268a47639dfec&X-Amz-SignedHeaders=host&x-amz-checksum-mode=ENABLED&x-id=GetObject)
Figure 4
![Approximations of the mean (left) and the standard deviation (right) of the solution SP to the random IVP (2) with M = 20, λ = 5/4, E[b0]=E[c]=−1 $\mathbb{E}[b_0]=\mathbb{E}[c]=-1$ and V[b0]=V[c]=1/4 $\mathbb{V}[b_0]=\mathbb{V}[c]=1/4$ using different orders of the derivative α = {0.4, 0.5, 0.6, 0.7, 0.99} over the time interval [0, 5].](https://sciendo-parsed.s3.eu-central-1.amazonaws.com/64709e5971e4585e08aa181d/j_AMNS.2017.2.00026_fig_004.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Content-Sha256=UNSIGNED-PAYLOAD&X-Amz-Credential=AKIA6AP2G7AKOUXAVR44%2F20251006%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20251006T234008Z&X-Amz-Expires=3600&X-Amz-Signature=ee3be1acd75533e00310c2d498e609497f90f9b70a8b4d3e3bfb134a6ebc76e3&X-Amz-SignedHeaders=host&x-amz-checksum-mode=ENABLED&x-id=GetObject)