Page 664 - 8th European Congress of Mathematics ∙ 20-26 June 2021 ∙ Portorož, Slovenia ∙ Book of Abstracts
P. 664
PARTIAL DIFFERENTIAL EQUATIONS AND APPLICATIONS
Some problems for a mixed type equation fractional order with non-linear
loaded term
Obidjon Abdullaev, obidjon.mth@gmail.com
Institute of Mathematics named after V.I.Romanovsky,
Uzbekistan Academy of Science, Uzbekistan
Note, that with intensive research on problems of optimal control of the agro-economical sys-
tem, regulating the label of ground waters and soil moisture, it has become necessary to in-
vestigate BVPs for a loaded partial differential equations. Integral boundary conditions have
various applications in thermo-elasticity, chemical engineering, population dynamics, etc. In
this work we consider parabolic-hyperbolic type equation fractional order involving non-linear
loaded term:
0 = uxx − cDoαyu + f1(x, y; u(x, 0)), x > 0, y > 0
0 = uxx − uyy + f2(x, y; u(x + y, 0)), x > 0, y < 0, (1)
0 = uxx − uyy + f3(x, y; uy(0, x + y)), x < 0, y > 0
where fi(x, y, u), (i = 1, 2, 3) are given functions, CDoαyu Caputo operator fractional order α
(0 < α < 1) (see [1].p.92):
x
(C Daαxf ) x = 1 α) f (t)
Γ(1 − (x − t)α dt, x > a.
a
Let Ω ⊂ R2, be domain bounded with segments B2A2, A2A1 on the lines x = l, y = h at
x > 0, y > 0; and A1C2, C2B1 on the characteristics x − y = l, x + y = 0 of the Eq. (1) at
x > 0, y < 0, also with the segments B1C1, C1B2 on the characteristics y − x = h, x + y = 0
of the Eq. (1) at x < 0, y > 0.
We denote asΩ0 parabolic part of the mixed domain Ω, and hyperbolic parts through Ω1 at
x > 0 and Ω2 at x < 0.
In the domain Ω (Ω = Ω+ ∪ Ω− ∪ (A1B1),) we will investigate following
Problem I. To find a solution u(x, y) of Eq. (1) from the class of functions: u(x, y) ∈ C(Ω¯ ) ∩
C1 {Ω2 \ A1C2} ∪ {Ω1 \ C1B2} ∩ C2(Ω2 ∪ Ω1); uxx, CDoαyu ∈ C (Ω0) ; satisfying boundary
conditions:
u(l, t) = ϕ1(y), 0 ≤ y ≤ h; (2)
d 0 ≤ x < l; (3)
dx u (θ1(x)) = a1(x)uy(x, 0) + a2(x)ux(x, 0) + a3(x)u(x, 0) + a4(x),
d (θ2(y)) = b1(y)ux(0, + b2 (y )uy (0, ≤ y < h;
u y) y) + b3(y)u(0, y) + b4(y), 0 (4)
dy
and integral gluing condition:
lim y 1−α uy (x, y) = λ1 (x)uy (x, −0) + λ2(x)ux(x, −0)+
y→+0
x (3)
+ λ3(x) r1(t)u(t, 0)dt + λ3(x)u(x, 0) + λ4(x), 0 < x < l
0
662
Some problems for a mixed type equation fractional order with non-linear
loaded term
Obidjon Abdullaev, obidjon.mth@gmail.com
Institute of Mathematics named after V.I.Romanovsky,
Uzbekistan Academy of Science, Uzbekistan
Note, that with intensive research on problems of optimal control of the agro-economical sys-
tem, regulating the label of ground waters and soil moisture, it has become necessary to in-
vestigate BVPs for a loaded partial differential equations. Integral boundary conditions have
various applications in thermo-elasticity, chemical engineering, population dynamics, etc. In
this work we consider parabolic-hyperbolic type equation fractional order involving non-linear
loaded term:
0 = uxx − cDoαyu + f1(x, y; u(x, 0)), x > 0, y > 0
0 = uxx − uyy + f2(x, y; u(x + y, 0)), x > 0, y < 0, (1)
0 = uxx − uyy + f3(x, y; uy(0, x + y)), x < 0, y > 0
where fi(x, y, u), (i = 1, 2, 3) are given functions, CDoαyu Caputo operator fractional order α
(0 < α < 1) (see [1].p.92):
x
(C Daαxf ) x = 1 α) f (t)
Γ(1 − (x − t)α dt, x > a.
a
Let Ω ⊂ R2, be domain bounded with segments B2A2, A2A1 on the lines x = l, y = h at
x > 0, y > 0; and A1C2, C2B1 on the characteristics x − y = l, x + y = 0 of the Eq. (1) at
x > 0, y < 0, also with the segments B1C1, C1B2 on the characteristics y − x = h, x + y = 0
of the Eq. (1) at x < 0, y > 0.
We denote asΩ0 parabolic part of the mixed domain Ω, and hyperbolic parts through Ω1 at
x > 0 and Ω2 at x < 0.
In the domain Ω (Ω = Ω+ ∪ Ω− ∪ (A1B1),) we will investigate following
Problem I. To find a solution u(x, y) of Eq. (1) from the class of functions: u(x, y) ∈ C(Ω¯ ) ∩
C1 {Ω2 \ A1C2} ∪ {Ω1 \ C1B2} ∩ C2(Ω2 ∪ Ω1); uxx, CDoαyu ∈ C (Ω0) ; satisfying boundary
conditions:
u(l, t) = ϕ1(y), 0 ≤ y ≤ h; (2)
d 0 ≤ x < l; (3)
dx u (θ1(x)) = a1(x)uy(x, 0) + a2(x)ux(x, 0) + a3(x)u(x, 0) + a4(x),
d (θ2(y)) = b1(y)ux(0, + b2 (y )uy (0, ≤ y < h;
u y) y) + b3(y)u(0, y) + b4(y), 0 (4)
dy
and integral gluing condition:
lim y 1−α uy (x, y) = λ1 (x)uy (x, −0) + λ2(x)ux(x, −0)+
y→+0
x (3)
+ λ3(x) r1(t)u(t, 0)dt + λ3(x)u(x, 0) + λ4(x), 0 < x < l
0
662
