Mean square calculus and random linear fractional differential equations: Theory and applications
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28 juil. 2017
À propos de cet article
Publié en ligne: 28 juil. 2017
Pages: 317 - 328
Reçu: 04 avr. 2017
Accepté: 28 juil. 2017
DOI: https://doi.org/10.21042/AMNS.2017.2.00026
Mots clés
© 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=20251006T234746Z&X-Amz-Expires=3600&X-Amz-Signature=03e151f05d54befadfd880bbbc97c48b1833c84f2ac5e8b8ec45bd6fe101d00c&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=20251006T234746Z&X-Amz-Expires=3600&X-Amz-Signature=ce376d89813267266289e7cd9c4574ab1c3b16cdd451d10b63f6d8b31d49b790&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=20251006T234746Z&X-Amz-Expires=3600&X-Amz-Signature=0fc37530e9fb98ea925f2ea04c707d24978723595110065461a9a48a9a67f445&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=20251006T234746Z&X-Amz-Expires=3600&X-Amz-Signature=7e49ec2f71d1634f02e96dc7914c80a2c11e4cb4dab14d77504e3a2b7d4dc1a2&X-Amz-SignedHeaders=host&x-amz-checksum-mode=ENABLED&x-id=GetObject)