2022
DOI: 10.1134/s1995080222040229
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On a Mixed Problem for Hilfer Type Fractional Differential Equation with Degeneration

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Cited by 15 publications
(4 citation statements)
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“…These operators play a pivotal role in formulating and solving the fractional partial differential equation under consideration. To establish the groundwork, we present fundamental properties associated with these fractional operators, drawing upon existing literature [9,10,11,12,13,14].…”
Section: Introductionmentioning
confidence: 99%
“…These operators play a pivotal role in formulating and solving the fractional partial differential equation under consideration. To establish the groundwork, we present fundamental properties associated with these fractional operators, drawing upon existing literature [9,10,11,12,13,14].…”
Section: Introductionmentioning
confidence: 99%
“…Recently, in [14] the initial-boundary value problems of Dirichlet and Neumann for the Caputo type time-fractional diffusion equation are considered. Also, the regular solution of a mixed problem for the Hilfer type nonlinear partial differential equation in three-dimensional domain is studied in [15].…”
Section: Introductionmentioning
confidence: 99%
“…В работе [8] для линейного обыкновенного дифференциального уравнения дробного порядка вида (3) с производными Римана-Лиувилля была сформулирована и решена начальная задача. Краевые и начальные задачи для вырождающихся уравнений с дробным производным Хилфера исследовались в работах [9][10][11][12], а с дробными производными Римана-Лиувилля и Капуто в работах [2], [13][14]. В данной работе в терминах функции Килбаса-Сайго строится явное представление фундаментальной системы решений уравнения (1).…”
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“…,γm} 0y u (y) = λy s u, y > 0, λ ∈ C, s ≥ 0. limy→0 D α 0 0y u (y) = A 0 , lim y→0 D α 1 0y u (y) = A 1 , ... lim y→0 D α m−1 0y u (y) u (y) = A m−1 . (11) здесь A i = const, i = 0, 1, ..., m − 1, D α 0 0y = D γ 0 −1 0y , D α k 0y = D γ k −1 ≤ k ≤ s − 1, Γ (1 + α k ) , k = s, Γ(1+α k ) Γ(1+α k −αs) y α k −αs , s < k ≤ m.Подставив представление (9) в начальные условия(11) , получимlim y→0 D α 0 0y u (y) = d 0 Γ (1 + α 0 ) = A 0 , lim y→0 D α 1 0y u (y) = d 1 Γ (1 + α 1 ) = A 1 , ... lim y→0 D α m−1 0y u (y) = d m−1 Γ (1 + α m−1 ) = A m−1 .Значит решение задачи Коши будет иметь вид u (y) =…”
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