On a Nonlocal Problem for Mixed-Type Equation with Partial Riemann-Liouville Fractional Derivative

Abstract

The present paper presents a study on a problem with a fractional integro-differentiation operator in the boundary condition for an equation with a partial Riemann-Liouville fractional derivative. The unique solvability of the problem is proved. In the hyperbolic part of the considered domain, the functional equation is solved by the iteration method. The problem is reduced to solving the Volterra integro-differential equation.

Full text

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Citation: Ruziev, M.; Zunnunov, R.
On a Nonlocal Problem for
Mixed-Type Equation with Partial
Riemann-Liouville Fractional
Derivative . Fractal Fract. 2022,6, 110.
https://doi.org/10.3390/
fractalfract6020110
Academic Editor: Hari Mohan
Srivastava
Received: 29 December 2021
Accepted: 2 February 2022
Published: 14 February 2022
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fractal and fractional
Article
On a Nonlocal Problem for Mixed-Type Equation with Partial
Riemann-Liouville Fractional Derivative
Menglibay Ruziev and Rakhimjon Zunnunov *
Institute of Mathematics, Academy of Sciences of the Republic of Uzbekistan, University Str. 4b,
Tashkent 100174, Uzbekistan; mruziev@mail.ru
*Correspondence: zunnunov@mail.ru
Abstract: The present paper presents a study on a problem with a fractional integro-differentiation
operator in the boundary condition for an equation with a partial Riemann-Liouville fractional
derivative. The unique solvability of the problem is proved. In the hyperbolic part of the considered
domain, the functional equation is solved by the iteration method. The problem is reduced to solving
the Volterra integro-differential equation.
Keywords:
fractional order derivative; Riemann-Liouville operator; boundary value problem; singular
coefficient; mixed-type equation
1. Introduction and Formulation of a Problem
Boundary value problems for the mixed-type equations of fractional order were in-
vestigated in [
1
–
4
]. In [
5
], the unique solvability was investigated for the problem of
an equation with the partial fractional derivative of Riemann-Liouville and a boundary
condition that contains the generalized operator of fractional integro-differentiation. A
problem, in which the boundary condition contains a linear combination of generalized
fractional operators with a Gauss hypergeometric function for a mixed-type equation with a
Riemann-Liouville partial fractional derivative, was studied in [
6
]. The nonlocal boundary
value problem for mixed-type equations with singular coefficients was considered in [
7
].
The Gellerstedt-type problem, with nonlocal boundary and integral gluing conditions for
the parabolic-hyperbolic-type equation, with nonlinear terms and Gerasimov-Caputo oper-
ator of differentiation, was studied in [
8
]. The work [
9
] is devoted to study the boundary
value problems for a mixed type fractional differential equation with Caputo operator.
A nonlocal boundary value problem for weak nonlinear partial differential equations of
mixed type, with a fractional Hilfer operator, was solved in [
10
]. The work [
11
] is con-
cerned with the existence and uniqueness of solutions for a Hilfer-Hadamard fractional
differential equation.
Let
D
be a finite domain bounded by segments
AA0
,
BB0
, and
A0B0
of lines
x=
0,
x=
1, and
y=
1, respectively, lying in the half-plane
y>
0, and characteristics
AC =
{(x
,
y):x−2
m+2(−y)m+2
2=
0
}
,
BC ={(x
,
y):x+2
m+2(−y)m+2
2=
1
}
of the following
equation:





uxx −Dγ
0,yu=0, γ∈(0, 1),y>0,
−(−y)muxx +uyy +α0
(−y)1−m
2
ux+β0
yuy=0, y<0, (1)
in the half-plane,
y<
0, and the interval,
AB
, of the straight line,
y=
0. In (1)
m>
0,
|α0|<(m+
2
)/
2, 1
<β0<(m+
4
)/
2. Here,
Dγ
0,y
is the partial fractional Riemann-
Liouville derivative [3]
(Dγ
0,yu)(x,y) = ∂
∂y
1
Γ(1−γ)
y
Z
0
u(x,t)dt
(y−t)γ,(0<γ<1, y>0).
Fractal Fract. 2022,6, 110. https://doi.org/10.3390/fractalfract6020110 https://www.mdpi.com/journal/fractalfract

Fractal Fract. 2022,6, 110 2 of 8
Let D+=D∩(y>0),D−=D∩(y<0).
Problem 1.
Find a solution
u=u(x
,
y)
of Equation (1) in the domain
D
, satisfying the following
boundary conditions:
u(0, y) = ϕ1(y),u(1, y) = ϕ2(y), 0 ≤y≤1, (2)
x¯
αD1−¯
β
0,xx1−¯
α−¯
βu[θ0(x)] + (1−x)¯
βρ(x)D1−¯
α
x,1 (1−x)1−¯
α−¯
βu[θk(x)]
=µ1D1−¯
α−¯
β
0,xτ(x)−µ2D1−¯
α−¯
β
x,1 τ(x) + f(x),x∈[0, 1](3)
and the following conjugation conditions:
lim
y→+0y1−γu(x,y) = lim
y→−0(−y)β0−1u(x,y),x∈[0, 1],
lim
y→+0y1−γ(y1−γu(x,y))y=lim
y→−0(−y)2−β0((−y)β0−1u(x,y))y,x∈I= (0, 1). (4)
Here,
¯
α=m+2(2−β0+α0)
2(m+2)
,
¯
β=m+2(2−β0−α0)
2(m+2)
,
τ(x) = lim
y→−0(−y)β0−1u(x
,
y)
,
ϕ1(y)
,
ϕ2(y)
,
ρ(x)
,
f(x)
are given functions; moreover,
y1−γϕ1(y)
,
y1−γϕ2(y)∈C([
0, 1
])
,
ϕ1(
0
) = ϕ2(
0
) =
0,
f(
0
) =
0,
µ1
and
µ2
are constants;
θ0(x0) = x0
2,−(m+2
4x0)2/(m+2)
is a point of in-
tersection of characteristics of Equation (1), outgoing from the point
(x0
, 0
) (x0∈I)
, with
the characteristic
AC
;
θk(x0) = x0+k
1+k,−((m+2)(1−x0)
2(1+k))2
m+2
is the intersection point of the
curve
x−2k
m+2(−y)m+2
2=x0
,
k=const >
1, with the characteristic
BC
,
D1−¯
α
x,1 f(x) =
−d
dx D−¯
α
x,1 f(x) = −d
dx 1
Γ(¯
α)
1
R
x
f(t)dt
(t−x)1−¯
α
. We are looking for a solution,
u(x
,
y)
, of the problem
in the class of twice differentiable functions in the domain
D
, such that
y1−γu∈C(¯
D+)
,
u(x
,
y)∈
C(¯
D−\OB)
,
y1−γ(y1−γu)y∈C(D+∪ {(x
,
y):
0
<x<
1,
y=
0
})
,
uxx ∈C(D+∪D−)
,
uyy ∈C(D−).
Note that from Equation (1), at
m=
2,
β0=
0, we obtain the moisture transfer
Equation [12]
, and at
α0=
0,
β0=
0, Equation (1) passes to the Gellerstedt equation, which
finds application in the problem of determining the shape of the dam slot.
In this article, we study a problem with a shift for Equation (1), in which some part of
the characteristic BC is freed from nonlocal boundary conditions.
2. Main Results
We denote that
lim
y→+0y1−γu(x
,
y) = τ(x)
,
lim
y→+0y1−γ(y1−γu(x
,
y))y=ν(x)
. The solu-
tion of Equation (1) in the domain
D+
, satisfying condition (2) and condition
lim
y→+0y1−γ
u(x,y) = τ(x),x∈¯
I, has the following form [13]:
u(x,y) =
y
Z
0
∂G
∂ξ |ξ=0ϕ1(η)dη−
y
Z
0
∂G
∂ξ |ξ=1ϕ2(η)dη−
1
Z
0
G(x,y;ξ, 0)τ(ξ)dξ,
where
G(x,y;ξ,η) = Γ(γ)
2(y−η)δ−1
∞
∑
n=−∞e1,δ
1,δ−|x−ξ+2n|
(y−η)δ−e1,δ
1,δ−|x+ξ+2n|
(y−η)δ,
ep,q
b,c(z) =
∞
∑
k=0
zk
Γ(p+kb)Γ(q−kc),b>c,b>0, z∈C,δ=γ
2,γ>0,

Fractal Fract. 2022,6, 110 3 of 8
e1,δ
1,δ(z) =
∞
∑
k=0
zk
Γ(δ−kδ)k!,δ<1.
The functional relation between
τ=τ(x)
and
ν=ν(x)
, transferred from the parabolic
part, D+, to the line, y=0, has the following form [1]:
ν(x) = 1/(Γ(1+γ))τ00 (x). (5)
Applying the Darboux formula, given in the domain
D−
, the solution of the modified
Cauchy problem with the initial data lim
y→0−(−y)β0−1u(x,y) = τ(x),
lim
y→−0(−y)2−β0((−y)β0−1u(x,y))y=ν(x),x∈I, has the following form:
u(x,y) = ¯
γ1(−y)1−β0
1
R
0
τx+2
m+2(2t−1)(−y)m+2
2t¯
β−1(1−t)¯
α−1dt
+¯
γ2
1
R
0
νx+2
m+2(2t−1)(−y)m+2
2t−¯
α(1−t)−¯
βdt,
(6)
where ¯
γ1=Γ(¯
α+¯
β)
Γ(¯
α)Γ(¯
β),¯
γ2=−Γ(2−¯
α−¯
β)
(β0−1)Γ(1−¯
α)Γ(1−¯
β).
From (6) we have the following:
u[θ0(x)] = ¯
γ1m+2
4¯
α+¯
β−1
Γ(¯
α)D−¯
α
0,xx¯
β−1τ(x) + ¯
γ2x¯
α+¯
β−1Γ(1−¯
β)D¯
β−1
0,xx¯
−αν(x). (7)
Multiplying both sides of (7) by x1−¯
α−¯
β, we have the following:
x1−¯
α−¯
βu[θ0(x)] = ¯
γ1Γ(¯
α)m+2
4¯
α+¯
β−1x1−¯
α−¯
βD−¯
α
0,xx¯
β−1τ(x)
+Γ(1−¯
β)¯
γ2D¯
β−1
0,xx¯
−αν(x).(8)
Applying the operator D1−¯
β
0,xto both sides of relation (8) we obtain the following:
D1−¯
β
0,xx1−¯
α−¯
βu[θ0(x)] = ¯
γ1((m+2)/4)¯
α+¯
β−1Γ(¯
α)D1−¯
β
0,xx1−¯
α−¯
βD−¯
α
0,xx¯
β−1τ(x)
+Γ(1−¯
β)¯
γ2D1−¯
β
0,xD¯
β−1
0,xx¯
−αν(x).(9)
Equalities are true, as follows:
D1−¯
β
0,xx1−¯
α−¯
βD−¯
α
0,xx¯
β−1τ(x) = x−¯
αD1−¯
α−¯
β
0,xτ(x)(10)
D1−¯
β
0,xD¯
β−1
0,xx¯
−αν(x) = x¯
−αν(x). (11)
Let us show Relation (10).
Denoting the left-hand side of Equality(10) by g1(x), we obtain the following:
g1(x) = D1−¯
β
0,xx1−¯
α−¯
βD−¯
α
0,xx¯
β−1τ(x) = d
dx D−¯
β
0,xx1−¯
α−¯
βD−¯
α
0,xx¯
β−1τ(x)
=1
Γ(¯
β)Γ(¯
α)
d
dx
x
Z
0
ξ1−¯
α−¯
βdξ
(x−ξ)1−¯
β
ξ
Z
0
t¯
β−1τ(t)dt
(ξ−t)1−¯
α.
Changing the order of integration, we obtain the following:
g1(x) = 1
Γ(¯
β)Γ(¯
α)
d
dx
x
Z
0
τ(t)t¯
β−1dt
x
Z
t
ξ1−¯
α−¯
β(x−ξ)¯
β−1(ξ−t)¯
α−1dξ.

Fractal Fract. 2022,6, 110 4 of 8
Setting in the inner integral, ξ=t+ (x−t)σ, we have the following:
g1(x) = 1
Γ(¯
β)Γ(¯
α)
d
dx
x
Z
0
τ(t)t−¯
α(x−t)¯
α+¯
β−1dt
1
Z
0
σ¯
α−1(1−σ)¯
β−1
×1−t−x
tσ1−¯
α−¯
β
dσ.
Using the Euler hypergeometric integral [14], as follows:
1
Z
0
tµ−1(1−t)k−µ−1(1−zt)−λdt =Γ(µ)Γ(k−µ)
Γ(k)F(µ,λ,k;z), 0 <µ<k,
we obtain the following:
g1(x) = 1
Γ(¯
β)Γ(¯
α)
d
dx
x
Z
0
τ(t)t−¯
α(x−t)¯
α+¯
β−1Γ(¯
α)Γ(¯
β)
Γ(¯
α+¯
β)F¯
α,¯
α+¯
β−1, ¯
α+¯
β;t−x
tdt.
From here, taking into account [14], as follows:
F(µ,λ,k;z) = (1−z)−λFk−µ,λ,k;z
z−1
we have the following:
g1(x) = 1/(Γ(¯
α+¯
β)) d
dx
x
Z
0
τ(t)t¯
β−1x−t
x¯
α+¯
β−1
F¯
β,¯
α+¯
β−1, ¯
α+¯
β;x−t
xdt.
Consider the following function:
gε(x) = 1/(Γ(¯
α+¯
β)) d
dx
x−ε
Z
0
τ(t)t¯
β−1x−t
x¯
α+¯
β−1
F¯
β,¯
α+¯
β−1, ¯
α+¯
β;x−t
xdt.
Differentiating the right-hand side of this equality and using the formulas from [
14
],
as follows:
d
dz [zµF(µ,λ,k;z)] = µzµ−1F(µ+1, λ,k;z),F(µ,λ,λ;z) = (1−z)−µ,
we obtain the following:
gε(x) = 1
Γ(¯
α+¯
β)(x−ε)¯
β−1ε
x¯
α+¯
β−1F¯
β,¯
α+¯
β−1, ¯
α+¯
β;ε
xτ(x−ε)
+¯
α+¯
β−1
Γ(¯
α+¯
β)x−¯
α
x−ε
Z
0
(x−t)¯
α+¯
β−2τ(t)dt.
Now, taking into account that
(¯
α+¯
β−1)
x−ε
Z
0
(x−t)¯
α+¯
β−2τ(t)dt =d
dx
x−ε
Z
0
(x−t)¯
α+¯
β−1τ(t)dt −ε¯
α+¯
β−1τ(x−ε),

Fractal Fract. 2022,6, 110 5 of 8
we find the following:
gε(x) = ε¯
α+¯
β−1
Γ(¯
α+¯
β)x−¯
α"x
x−ε1−¯
β
F¯
β,¯
α+¯
β−1, ¯
α+¯
β;ε
x−1#τ(x−ε)
+x−¯
α
Γ(¯
α+¯
β)
d
dx
x−ε
Z
0
(x−t)¯
α+¯
β−1τ(t)dt.
Passing in this equality to limit as
ε→
0, by virtue of the formula
F(µ
,
λ
,
k
; 0
) =
1
we obtain the following:
g1(x) = gε(x) = x−¯
α
Γ(¯
α+¯
β)
d
dx
x
Z
0
τ(t)dt
(x−t)1−¯
α+¯
β=x−¯
αd
dx D−¯
α−¯
βτ(x) = x−¯
αD1−¯
α−¯
βτ(x).
Thus, let us be convinced of the validity of Equality (10).
By virtue of (10) and (11), Equality (9) can be written in the following form:
D1−¯
β
0,xx1−¯
α−¯
βu[θ0(x)] = ¯
γ1((m+2)/4)¯
α+¯
β−1Γ(¯
α)x−¯
αD1−¯
α−¯
β
0,xτ(x)
+Γ(1−¯
β)¯
γ2x¯
−αν(x).(12)
Now, from (6) we obtain the following:
u[θk(x)] = a1−¯
α−¯
β¯
γ1m+2
2(1+k)¯
α+¯
β−1Γ(¯
β)D−¯
β
ax+b,1 (1−x)¯
α−1τ(x)
+Γ(1−¯
α)¯
γ2a¯
α+¯
β−1(1−x)¯
α+¯
β−1D¯
α−1
ax+b,1 (1−x)¯
−βν(x).
(13)
Multiplying both sides of (13) by
(
1
−x)1−¯
α−¯
β
and applying the operator
D1−¯
α
x,1
we obtain the following:
D1−¯
α
x,1 (1−x)1−¯
α−¯
βu[θk(x)] = Γ(¯
β)¯
γ1m+2
2(1+k)¯
α+¯
β−1a1−¯
α−¯
βD1−¯
α
x,1 (1−x)1−¯
α−¯
β
×D−¯
β
ax+b,1 (1−x)¯
α−1τ(x) + ¯
γ2a¯
α+¯
β−1Γ(1−¯
α)D1−¯
α
x,1 D¯
α−1
ax+b,1 (1−x)−¯
βν(x).
(14)
It is easy to show the following:
D1−¯
α
x,1 (1−x)1−¯
α−¯
βD−¯
β
ax+b,1 (1−x)¯
α−1τ(x) = (1−x)−βD1−¯
α−¯
β
ax+b,1 τ(x),
D1−¯
α
x,1 D¯
α−1
ax+b,1 (1−x)−¯
βν(x) = a1−¯
α−¯
βν(ax +b)(1−x)−¯
β.
Then, from (14) we obtain the following:
D1−¯
α
x,1 (1−x)1−¯
α−¯
βu[θk(x)] = a1−¯
α−¯
β¯
γ1m+2
2(1+k)¯
α+¯
β−1Γ(¯
β)
×(1−x)−¯
βD1−¯
α−¯
β
ax+b,1 τ(x) + ¯
γ2Γ(1−¯
α)ν(ax +b)(1−x)−¯
β.
(15)
Now, substituting (12) and (15) into (3), we obtain the following:
¯
γ1((m+2)/4)¯
α+¯
β−1Γ(¯
α)D1−¯
α−¯
β
0,xτ(x) + Γ(1−¯
β)¯
γ2ν(x)
+a1−¯
α−¯
βm+2
2(1+k)¯
α+¯
β−1Γ(¯
β)¯
γ1ρ(x)D1−¯
α−¯
β
ax+b,1 τ(x)
+¯
γ2Γ(1−¯
α)ρ(x)ν(ax +b) = µ1D1−¯
α−¯
β
0,xτ(x)−µ2D1−¯
α−¯
β
x,1 τ(x) + f(x),
x∈[0, 1].
(16)

Fractal Fract. 2022,6, 110 6 of 8
Let µ1=Γ(¯
α+¯
β)
Γ(¯
β)m+2
4¯
α+¯
β−1. Then, from (16) we obtain the following:
¯
γ2Γ(1−¯
β)ν(x) + m+2
2(1+k)¯
α+¯
β−1a1−¯
α−¯
β¯
γ1Γ(¯
β)ρ(x)D1−¯
α−¯
β
ax+b,1 τ(x)
+ρ(x)¯
γ2Γ(1−¯
α)ν(ax +b) + µ2D1−¯
α−¯
β
x,1 τ(x) = f(x),x∈[0, 1].
(17)
Dividing both sides of (17) by ¯
γ2Γ(1−¯
β), by virtue of a=2
1+k, we obtain the following:
ν(x) + Γ(¯
α+¯
β)
¯
γ2Γ(1−¯
β)Γ(¯
α)m+2
4¯
α+¯
β−1ρ(x)D1−¯
α−¯
β
ax+b,1 τ(x)
+Γ(1−¯
α)
Γ(1−¯
β)ρ(x)ν(ax +b) + µ2D1−¯
α−¯
β
x,1 τ(x)
¯
γ2Γ(1−¯
β)=f(x)
¯
γ2Γ(1−¯
β).
(18)
We write Equality (18) in the following form:
ν(x) + µ2
¯
γ2Γ(1−¯
β)D1−¯
α−¯
β
x,1 τ(x) = −Γ(1−¯
α)
Γ(1−¯
β)ρ(x)
×"ν(ax +b) + (4
m+2)1−¯
α−¯
β
¯
γ2Γ(1−¯
β)
Γ(¯
α+¯
β)Γ(1−¯
β)
Γ(¯
α)Γ(1−¯
α)D1−¯
α−¯
β
ax+b,1 τ(x)#+f(x)¯
γ0,(19)
where ¯
γ0=1
¯
γ2Γ(1−¯
β).
Let µ2=4
m+21−¯
α−¯
β(Γ(¯
α+¯
β)Γ(1−¯
β))/(Γ(¯
α)Γ(1−¯
α)).
We introduce the following notation:
Φ(x):=ν(x) + Γ(¯
α+¯
β)Γ(1−¯
β)
Γ(¯
α)Γ(1−¯
α)4
m+21−¯
α−¯
β
¯
γ0D1−¯
α−¯
β
x,1 τ(x). (20)
Then, from (19), we obtain the following functional equation:
Φ(x) = Φ(ax +b)ω(x) + f1(x), (21)
where ω(x) = −Γ(1−¯
α)
Γ(1−¯
β)ρ(x),f1(x) = ¯
γ0f(x).
Applying the iteration method to (21), after simple calculations, we obtain the following:
Φ(x) = ω(x)An−1(x)Φ(anx+1−an)
+ω(x)∑n−1
j=1f1(ajx+1−aj)Aj−1(x) + f1(x),(22)
where
Aj−1(x) = ω(ax +1−a)ω(a2x+1−a2)···ω(aj−1x+1−aj−1),A0=1.
By
|ω(x)| ≤
1, for
x∈[
1
−ε
, 1
]
, here,
ε
is a sufficiently small positive number;
therefore, we obtain the following:
|Aj−1(x)| ≤ Mn0
0, (23)
where
M0=max
[1−ε,1]ω(x)
, and
n0= [loga(aε)]
. We seek a solution of Equation (21), in the
class of functions bounded at the point x=1.
After passing to the limit,
n→∞
, in (22), and considering that, by virtue of
ρ(x)
,
f(x)∈
C(¯
I)∩C2(I)
and (23), the series on the right-hand side of (22) is uniformly converging on
the interval [0, 1], we obtain the following:
Φ(x) = P(x), (24)

Fractal Fract. 2022,6, 110 7 of 8
P(x) = ω(x)
∞
∑
j=1
f1(ajx+1−aj)Aj−1(x) + f1(x).
Thus, according to (20) from (24), we have
ν(x) + ¯
γ0(Γ(¯
α+¯
β)Γ(1−¯
β))/(Γ(¯
α)Γ(1−¯
α))4
m+21−¯
α−¯
β
D1−¯
α−¯
β
x,1 τ(x) = P(x). (25)
Thus, we obtained the functional relation between the functions
τ=τ(x)
and
ν=
ν(x), on I, from the domain D−.
By virtue of (4), substituting (5) into (25) we obtain the following:
1/(Γ(1+γ))τ00(x) + ¯
γ3D1−¯
α−¯
β
x,1 τ(x) = P(x), (26)
where ¯
γ3=¯
γ0(4/(m+2))1−¯
α−¯
βΓ(¯
α+¯
β)Γ(1−¯
β)/(Γ(¯
α)Γ(1−¯
α)).
Equation (26) can be written as follows:
1/(Γ(1+γ))τ00(x)−¯
γ3/(Γ(¯
α+¯
β)) d
dx
1
Z
x
τ(t)dt
(t−x)1−¯
α−¯
β=P(x).
We have obtained the Volterra integro-differential equation, which is uniquely solvable,
see [15].
3. Discussion
The properties of solutions of Equation (1) at
y<
0 essentially depend on the co-
efficients
α0
and
β0
, at the lowest terms of Equation (1). If
β0<
1, then the solution
of
Equation (1)
on the parabolic degeneration line is bounded. In this case, a problem
for an elliptic-hyperbolic-type equation with singular coefficients was studied in [
7
]. If
β0=
1, then the solution to Equation (1) on the parabolic degeneracy line has a logarithmic
singularity.
β0=m+4
2
is the limiting case. In these cases, boundary value problems for
Equation (1) are studied with different conditions.
Author Contributions:
Conceptualization, M.R. and R.Z.; investigation, M.R. and R.Z.; methodol-
ogy, M.R. and R.Z.; validation, M.R. and R.Z.;
writing—original
draft preparation, M.R. and R.Z.;
writing—review
and editing, M.R. and R.Z. All authors have read and agreed to the published
version of the manuscript.
Funding: This research received no external funding.
Institutional Review Board Statement: Not applicable.
Informed Consent Statement: Not applicable.
Data Availability Statement: Not applicable.
Acknowledgments: Authors would like to thank anonymous referees.
Conflicts of Interest: The authors declare no conflict of interest.
References
1.
Gekkieva, S.K. Analogue of Tricomi problem for mixed-type equation with a fractional derivative. Rep. Adyg. Inter. Acad. Sci.
2001,5, 18–22.
2.
Gekkieva, S.K. The Cauchy problem for the generalized transport equation with a fractional time derivative. Rep. Adyg. Inter.
Acad. Sci. 2000,5, 16–19.
3.
Repin, O.A. Boundary-value problem with Saigo operators for mixed type equation with fractional derivative. Russ. Math.
2018
,
62, 70–75. [CrossRef]
4.
Pskhu, A.V. Solution of the first boundary value problem for a fractional-order diffusion equation. Diff. Equ.
2003
,39, 1359–1363.
[CrossRef]

Fractal Fract. 2022,6, 110 8 of 8
5.
Tarasenko, A.V.; Egorova, I.P. On nonlocal problem with fractional Riemann-Lioville derivatives for a mixed-type equation.
J. Samara State Tech. Univ. Ser. Phys. Math. Sci. 2017,21, 112–121.
6. Repin, O.A. On a problem for mixed-type equation with fractional derivative. Russ. Math. 2018,62, 38–42. [CrossRef]
7.
Ruziev, M.K. A problem with conditions given on inner characteristics and on the line of degeneracy for a mixed-type equation
with singular coefficients. Bound. Value Probl. 2013,2013, 210 . [CrossRef]
8.
Yuldashev T.K.; Abdullaev O.K. Unique solvability of a boundary value problem for a loaded fractional parabolic-hyperbolic
equation with nonlinear terms. Lobachevskii J. Math. 2021,42, 1113–1123. 008283. [CrossRef]
9.
Yuldashev T.K.; Islomov B.I.; Ubaydullaev, H. On boundary value problems for a mixed type fractional differential equation with
Caputo operator. Bull. Karaganda Univ. Math. Ser. 2021,47, 127–137. [CrossRef]
10.
Yuldashev T.K.; Kadirkulov B.J. Boundary value problem for weak nonlinear partial differential equations of mixed type with
fractional Hilfer operator. Axioms 2020,9, 68. [CrossRef]
11.
Ahmad B.; Ntouyas S.K. Hilfer-Hadamard fractional boundary value problems with nonlocal mixed boundary conditions. Fractal
Fract. 2021,5, 195. [CrossRef]
12.
Lykov, A.V. Application of the methods of thermodynamics of irreversible processes to the investigation of heat and mass transfer.
J. Eng. Phys. 1965,9, 189–202. [CrossRef]
13. Pskhu, A.V. Partial Differential Equations of Fractional Order; Nauka: Moscow, Russia, 2005; 199p.
14.
Samko S.G.; Kilbas A.A.; Marichev O.I. Fractal Integrals and Derivatives: Theory and Applications; Gordon and Breach Science
Publishers: Basel, Switzerland, 1993; 688p.
15. Kilbas, A.A. Integral Equations: A Course of Lectures; BSU: Minsk, Belarus, 2005; 143p.