Mixed problem for higher-order equations with fractional derivative and degeneration in both variables
UDC 517.9 We consider an initial-boundary-value problem for a higher-order equation with fractional Riemann – Liouville derivative in a rectangular domain degenerating in both variables. The solution to the problem is constructed in the explicit form by the method of separation of...
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2022
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| author | Irgashev, B. Yu. irgashev, bahrom Irgashev, B. Yu. |
| author_facet | Irgashev, B. Yu. irgashev, bahrom Irgashev, B. Yu. |
| author_sort | Irgashev, B. Yu. |
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UDC 517.9
We consider an initial-boundary-value problem for a higher-order equation with fractional Riemann – Liouville derivative in a rectangular domain degenerating in both variables. The solution to the problem is constructed in the explicit form by the method of separation of variables. Uniqueness is proved by the spectral method. |
| doi_str_mv | 10.37863/umzh.v74i10.6298 |
| first_indexed | 2026-03-24T03:26:59Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v74i10.6298
UDC 517.9
B. Yu. Irgashev1 (Namangan Engineering-Construction Inst., V. I. Romanovsky Inst. Math. Acad. Sci. Republic
Uzbekistan)
MIXED PROBLEM FOR HIGHER-ORDER EQUATIONS
WITH FRACTIONAL DERIVATIVE AND DEGENERATION
IN BOTH VARIABLES
МIШАНА ЗАДАЧА ДЛЯ РIВНЯНЬ ВИЩОГО ПОРЯДКУ
З ДРОБОВОЮ ПОХIДНОЮ, ЩО МАЄ ВИРОДЖЕННЯ
ЗА ОБОМА ЗМIННИМИ
We consider an initial-boundary-value problem for a higher-order equation with fractional Riemann – Liouville derivative
in a rectangular domain degenerating in both variables. The solution to the problem is constructed in the explicit form by
the method of separation of variables. Uniqueness is proved by the spectral method.
Pозглянуто початково-крайову задачу для рiвняння вищого порядку з дробовою похiдною Рiмана – Лiувiлля в пря-
мокутнiй областi, що вироджується за обома змiнними. Розв’язок задачi отримано в явному виглядi методом
вiдокремлення змiнних. Єдинiсть доводиться за допомогою спектрального методу.
1. Introduction. In the domain \Omega = \Omega x \times \Omega y, \Omega x = \{ x : 0 < x < 1\} , \Omega y = \{ y : 0 < y < 1\} ,
consider the equation
( - 1)k+1D\alpha
0xu(x, y) - xsym
\partial 2ku
\partial y2k
= 0, (1)
where 0 < \alpha < 1, 0 \leq m < k, m /\in N, s \in N \cup \{ 0\} , k \in N, D\alpha
0x is the operator of Riemann –
Liouville fractional differentiation of order \alpha ,
D\alpha
0xu(x, y) =
1
\Gamma (1 - \alpha )
\partial
\partial x
x\int
0
u(\tau , y)d\tau
(x - \tau )\alpha
.
For equation (1), consider the problem.
Problem A. Find a solution to equation (1) from the class
D\alpha
0xu(x, y) \in C(\Omega ), x1 - \alpha u(x, y) \in C
\bigl(
\Omega x \times \Omega y
\bigr)
,
(2)
\partial k - 1u(x, y)
\partial yk - 1
\in C
\bigl(
\Omega x \times \Omega y
\bigr)
,
\partial 2ku(x, y)
\partial y2k
\in C(\Omega x \times \Omega y),
satisfying the conditions
\partial ju(x, 0)
\partial yj
=
\partial ju(x, 1)
\partial yj
= 0, 0 < x \leq 1, j = 0, 1, . . . , k - 1, (3)
\mathrm{l}\mathrm{i}\mathrm{m}
x\rightarrow 0
D\alpha - 1
0x u(x, y) = \varphi (y). (4)
Here, \varphi (y) is sufficiently smooth and satisfies the natural concordance conditions.
1 e-mail: bahromirgasev@gmail.com.
c\bigcirc B. YU. IRGASHEV, 2022
1328 ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
MIXED PROBLEM FOR HIGHER-ORDER EQUATIONS WITH FRACTIONAL DERIVATIVE . . . 1329
Fractional differential equations arise in mathematical modeling of various physical processes and
phenomena [1]. Second-order equations of the form (1) with partial derivatives of fractional order
\alpha \in (0, 2) were studied in [1 – 8]. In these papers, the Cauchy problem was considered, the first, the
second and mixed boundary-value problems, a fundamental solution is found, a general representation
of solutions is constructed. Mixed equations and higher-order equations with a fractional derivative
were studied in [9 – 12]. Degenerate fractional-order equations were studied in [1, 13]. The research
will be carried out by the Fourier method. Previously, by the Fourier method, boundary-value
problems for equations with a fractional derivative were studied in [6 – 9, 12].
Based on work [5], we will make some comments. Let x1 - \alpha u(x, y) = u0(x, y) \Rightarrow u = x\alpha - 1u0,
from condition (4) we have
\mathrm{l}\mathrm{i}\mathrm{m}
x\rightarrow 0
x1 - \alpha u(x, y) =
\varphi (y)
\Gamma (\alpha )
. (5)
2. Existence of a solution. We are looking for a solution in the form
u(x, y) = X(x)Y (y).
Then with respect to the variable y, taking into account condition (3), we obtain the following
spectral problem:
Y (2k)(y) = ( - 1)k\lambda y - mY (y),
(6)
Y (j)(0) = Y (j)(1) = 0, j = 0, 1, . . . , k - 1.
Notice that \lambda = 0 is not an eigenvalue. Using the results of [14], we can write the solution to
problem (6), satisfying the conditions at the point x = 0, in the form
Yi(y) = yi \cdot 0F2k - 1
\Biggl[
i
2k - m
+ 1, . . . ,
i - (i - 1)
2k - m
+ 1,
i - (i+ 1)
2k - m
+ 1, . . .
. . . ,
i - (2k - 1)
2k - m
+ 1,
( - 1)k\lambda y2k - m
(2k - m)2k
\Biggr]
, i = 0, 1, . . . , 2k - 1,
where
pFq
\Biggl[
a1, . . . , ap, x
b1, . . . , bq
\Biggr]
=
\infty \sum
k=0
(a1)k . . . (ap)k
(b1)k . . . (bq)k
xk
k!
is the generalized hypergeometric function
(a)k = a(a+ 1) . . . (a+ k - 1)
is the Pochhammer symbol.
In particular, for k = 1 we have (c0, . . . , c3 are constants)
Y0(t) = c0
\Biggl( \surd
\lambda y
2 - m
2
2 - m
\Biggr) 1
2 - m \infty \sum
j=0
( - 1)j
\Biggl(
2
\surd
\lambda y
2 - m
2
2(2 - m)
\Biggr) 2j - 1
2 - m
j!\Gamma
\biggl(
j - 1
2 - m
+ 1
\biggr) = c1
\surd
yJ - 1
2 - m
\Biggl(
2
\surd
\lambda y
2 - m
2
2 - m
\Biggr)
,
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1330 B. YU. IRGASHEV
Y1(y) = c2y
\infty \sum
j=0
( - 1)j
\biggl(
\lambda y2 - m
(2 - m)2
\biggr) j
j!\Gamma
\biggl(
j +
1
2 - m
+ 1
\biggr) = c3
\surd
y
\infty \sum
j=0
( - 1)j
\Biggl(
2
\surd
\lambda y
2 - m
2
2(2 - m)
\Biggr) 2j+ 1
2 - m
j!\Gamma
\biggl(
j +
1
2 - m
+ 1
\biggr) =
= c3
\surd
yJ 1
2 - m
\Biggl(
2
\surd
\lambda y
2 - m
2
2 - m
\Biggr)
,
where
J\nu (z) =
\infty \sum
j=0
( - 1)j
\Bigl( z
2
\Bigr) 2j+\nu
j!\Gamma (j + \nu + 1)
are Bessel functions [15].
Satisfying the boundary conditions, we obtain the condition for the existence of eigenvalues
J 1
2 - m
\Biggl(
2
\surd
\lambda
2 - m
\Biggr)
= 0.
Let us get back to the general case. Because (2k - m) /\in N, then the system of functions\bigl\{
Yi(y)
\bigr\} i=2k - 1
i=0
is the forms a fundamental system of solutions. Hence, the general solution of
equation (6) has the form
Y (y) = c0Y0(y) + c1Y1(y) + . . .+ c2k - 1Y2k - 1(y),
and from the boundary conditions at the point x = 0, we have
Y (y) = ckYk(y) + ck+1Yk+1(y) + . . .+ c2k - 1Y2k - 1(y).
It mean that
Y (y) = O(yk), y \rightarrow +0.
From the conditions at the point x = 1, we obtain the system
ckYk(1) + ck+1Yk+1(1) + . . .+ c2k - 2Y2k - 2(1) + c2k - 1Y2k - 1(1) = 0,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\bigl(
ckYk(y) + ck+1Yk+1(y) + . . .+ c2k - 2Y2k - 2(y) + c2k - 1Y2k - 1(y)
\bigr) (k - 1)
y=1
= 0.
Equating to zero the main determinant of the system, one can find the eigenvalues of problem (6).
But in view of the complexity of this process, we will proceed in a different way, namely: we reduce
problem (6) to the integral equation using the Green function and obtain the necessary estimates for
the eigenfunctions. But first, we show that \lambda > 0. Indeed, we have
1\int
0
Y (y)Y (2k)(y)dy = ( - 1)k\lambda
1\int
0
y - mY 2(y)dy,
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
MIXED PROBLEM FOR HIGHER-ORDER EQUATIONS WITH FRACTIONAL DERIVATIVE . . . 1331
1\int
0
(Y (k))
2
dy = \lambda
1\int
0
y - mY 2(y)dy,
because \lambda = 0 is not an eigenvalue, it follows that \lambda > 0. It remains to show the existence of
eigenvalues and eigenfunctions of problem (6). The integral equation equivalent to problem (6) has
the form
Y (y) = ( - 1)k\lambda
1\int
0
\xi - mG(y, \xi )Y (\xi )d\xi , (7)
where
G(y, \xi ) = - 1
(2k - 1)!
\left\{ G1(y, \xi ), 0 \leq y \leq \xi ,
G2(y, \xi ), \xi \leq y \leq 1,
is the Green function of problem (6) (see [16]). Here,
G1(y, \xi ) = (1 - \xi )kyk
k - 1\sum
i=0
k - i - 1\sum
j=0
( - 1)iCi
2k - 1C
j
k - 1+jy
k - i - 1\xi j+i,
G2(y, \xi ) = (1 - y)k\xi k
k - 1\sum
i=0
k - i - 1\sum
j=0
( - 1)iCi
2k - 1C
j
k - 1+j\xi
k - i - 1yj+i,
Ck
n =
n!
k!(n - k)!
.
Rewrite (7) as
y -
m
2 Y (y) = \lambda
1\int
0
\xi -
m
2
\Bigl[
( - 1)kG(y, \xi )
\Bigr]
y -
m
2
\Bigl(
\xi -
m
2 Y (\xi )
\Bigr)
d\xi ,
we introduce the notation
Y (y) = y -
m
2 Y (y),
G(y, \xi ) = \xi -
m
2
\Bigl[
( - 1)kG(y, \xi )
\Bigr]
y -
m
2 .
Then we have
Y (y) = \lambda
1\int
0
G(y, \xi )Y (\xi )d\xi . (8)
Equation (8) is an integral equation with a continuous, in both variables, and a symmetric kernel.
According to the theory of equations with symmetric kernels, equation (8) has no more than a
countable number of eigenvalues and eigenfunctions. So, problem (6) has eigenvalues \lambda n > 0,
n = 1, 2, . . . , and the corresponding eigenfunctions are Yn(y). Further, we assume that
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1332 B. YU. IRGASHEV
\bigm\| \bigm\| Yn(y)\bigm\| \bigm\| 2 = 1\int
0
y - mY 2
n (y)dy = 1.
Then, taking into account (8), we have the Bessel inequality
\infty \sum
n=0
\biggl(
Yn(y)
\lambda n
\biggr) 2
\leq
1\int
0
y - mG2(y, \xi )dy < \infty . (9)
Now we find the conditions under which the given function \varphi (y) is expanded in a series according
to the eigenfunctions Yn(y). For this we use the Hilbert – Schmidt theorem.
Theorem 1. Let the function \varphi (y) satisfies the following conditions:
\varphi (y) \in C2k[0, 1],
\varphi (i)(0) = \varphi (i)(1) = 0, i = 0, 1, . . . , k - 1.
Then it can be expanded in a uniformly and absolutely converging series of the form
\varphi (y) =
\infty \sum
n=1
\varphi nYn(y),
where
\varphi n =
1\int
0
y - m\varphi (y)Yn(y)dy.
Proof. We show the equality
y -
m
2 \varphi (y) =
1\int
0
G(y, \xi )
\biggl(
( - 1)k\xi
m
2
d2k\varphi (\xi )
d\xi 2k
\biggr)
d\xi ,
really
1\int
0
\xi -
m
2
\Bigl[
( - 1)kG(y, \xi )
\Bigr]
y -
m
2
\biggl(
( - 1)k\xi
m
2
d2k\varphi (\xi )
d\xi 2k
\biggr)
d\xi =
= y -
m
2
1\int
0
G(y, \xi )
d2k\varphi (\xi )
d\xi 2k
d\xi = y -
m
2 \varphi (y).
Those for the function y -
m
2 \varphi (y) the conditions of the Hilbert – Schmidt theorem are satisfied and,
therefore,
y -
m
2 \varphi (y) =
\infty \sum
n=1
y -
m
2 \varphi nYn(y),
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
MIXED PROBLEM FOR HIGHER-ORDER EQUATIONS WITH FRACTIONAL DERIVATIVE . . . 1333
dividing by y -
m
2 , we have
\varphi (y) =
\infty \sum
n=1
\varphi nYn(y).
Theorem 1 is proved.
In what follows we will assume that the function \varphi (x) satisfies the conditions of Theorem 1. We
proceed to solve the equation in the variable x. Taking into account condition (5), we obtain the
following initial problem:
D\alpha
0xXn(x) = - \lambda nx
sXn(x),
(10)
\mathrm{l}\mathrm{i}\mathrm{m}
x\rightarrow 0
x1 - \alpha Xn(x) =
\varphi n
\Gamma (\alpha )
,
where
\varphi n =
1\int
0
\varphi (y)y - mYn(y)dy.
Using the results of [17], the solution to problem (10) can be written in the form
Xn(x) =
\varphi nx
\alpha - 1
\Gamma
\biggl(
\alpha
\alpha + s
\biggr)
. . .\Gamma
\biggl(
\alpha + s - 1
\alpha + s
\biggr) \times
\times
\infty \sum
m=0
\Gamma
\biggl(
m+
\alpha
\alpha + s
\biggr)
. . .\Gamma
\biggl(
m+
\alpha + s - 1
\alpha + s
\biggr)
\Gamma (m+ 1)
\Gamma ((\alpha + s)m+ \alpha )
( - \lambda n(\alpha + s)x\alpha +s)
m
m!
.
This representation implies the uniqueness of the solution to problem (10).
Because \alpha + s + 1 > s + 1, then the last series converges absolutely and uniformly for fixed
values of \lambda n and for bounded values of x (see [18]). This means that the permutation of the series
and the integral in above was legal.
In terms of special functions, the solution to problem (10) can be written in the form
Xn(x) =
\varphi nx
\alpha - 1
\Gamma
\biggl(
\alpha
\alpha + s
\biggr)
. . .\Gamma
\biggl(
\alpha + s - 1
\alpha + s
\biggr) \times
\times s+1\Psi 1
\left( \left( \biggl( 1, \alpha
\alpha + s
\biggr)
, . . . ,
\biggl(
1,
\alpha + s - 1
\alpha + s
\biggr)
, (1, 1), - \lambda n(\alpha + s)sx\alpha +s
(\alpha + s, \alpha )
\right) \right) ,
where
p\Psi q
\Biggl( \Biggl(
(\alpha 1, a1), . . . , (\alpha p, ap), z
(\beta 1, b1), . . . , (\beta q, bp)
\Biggr) \Biggr)
=
\infty \sum
m=0
\prod p
i=1
\Gamma (\alpha im+ ai)\prod q
i=1
\Gamma (\beta im+ bi)
zm
m!
is the generalized Wright function (see [18]).
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1334 B. YU. IRGASHEV
Using the results obtained in [18, 19] (Theorem 4), we obtain an asymptotic expansion of the
generalized Wright function for large values \lambda n and x > \delta > 0:
s+1\Psi 1
\left( \left( \biggl( 1, \alpha
\alpha + s
\biggr)
, . . . ,
\biggl(
1,
\alpha + s - 1
\alpha + s
\biggr)
, (1, 1), - \lambda n(\alpha + s)x\alpha +s
(\alpha + s, \alpha )
\right) \right) \sim
\sim Hs+1,1
\bigl(
\lambda n(\alpha + s)x\alpha +s
\bigr)
= Hs+1,1(t) =
=
s\sum
m=1
t -
\alpha +m - 1
\alpha +s Ss+1,1(t;m) + t - 1Ss+1,1(t; s+ 1),
where
Ss+1,1(t;m) =
=
\infty \sum
k=0
( - 1)k
k!
\Gamma
\biggl(
k +
\alpha +m - 1
\alpha + s
\biggr)
\Gamma
\biggl(
1 -
\biggl(
k +
\alpha +m - 1
\alpha + s
\biggr) \biggr) \prod s
r=1,r \not =m
\Gamma
\biggl(
r - m
\alpha + s
- k
\biggr)
\Gamma
\biggl(
\alpha - (\alpha + s)
\biggl(
k +
\alpha +m - 1
\alpha + s
\biggr) \biggr) t - k =
= ( - 1)m
\infty \sum
k=0
1
k!
\prod s
r=1,r \not =m
- \pi
( - 1)k+1 \mathrm{s}\mathrm{i}\mathrm{n}\pi
r - m
\alpha + s
\Gamma
\biggl(
k - r - m
\alpha + s
+ 1
\biggr)
\mathrm{s}\mathrm{i}\mathrm{n}\pi
\alpha +m - 1
\alpha + s
\mathrm{s}\mathrm{i}\mathrm{n}\pi ((\alpha + s)k)\Gamma ((\alpha + s)k +m)
t - k =
= ( - 1)m+s\pi s - 1
\infty \sum
k=0
( - 1)s(k+1)
k!
\mathrm{s}\mathrm{i}\mathrm{n}\pi ((\alpha + s)k)\Gamma ((\alpha + s)k +m)
\mathrm{s}\mathrm{i}\mathrm{n}\pi
\alpha +m - 1
\alpha + s
\prod s
r=1,r \not =m
\mathrm{s}\mathrm{i}\mathrm{n}\pi
r - m
\alpha + s
\Gamma
\biggl(
k - r - m
\alpha + s
+ 1
\biggr) t - k,
Ss+1,1(t; s+ 1) =
\infty \sum
k=0
( - 1)k
k!
\Gamma (k + 1)
\prod s
r=1
\Gamma
\biggl(
\alpha + r - 1
\alpha + s
- k - 1
\biggr)
\Gamma (\alpha - (\alpha + s)(k + 1))
t - k =
= ( - 1)s+1\pi s - 1
\infty \sum
k=0
( - 1)k+s(k+1)\Gamma ((\alpha + s)k + s+ 1) \mathrm{s}\mathrm{i}\mathrm{n}\pi
\bigl(
(\alpha + s)k
\bigr)
\prod s
r=1
\Gamma
\biggl(
k + 2 - \alpha + r - 1
\alpha + s
\biggr)
\mathrm{s}\mathrm{i}\mathrm{n}\pi
\alpha + r - 1
\alpha + s
t - k.
Taking into account the latter, we have that there exists a number K such that, for all n > K and
for x > \delta > 0, the estimate\bigm| \bigm| Xn(x)
\bigm| \bigm| \leq M
\lambda n
| \varphi n| x\alpha - 1, 0 < M is constant.
So, the formal solution of the posed problem A has the form
u(x, y) =
\infty \sum
n=0
Xn(x)Yn(y). (11)
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
MIXED PROBLEM FOR HIGHER-ORDER EQUATIONS WITH FRACTIONAL DERIVATIVE . . . 1335
Let us show that (11) is a classical solution to equation (1). We have
\bigm| \bigm| u(x, y)\bigm| \bigm| \leq \infty \sum
n=0
\bigm| \bigm| Xn(x)
\bigm| \bigm| \bigm| \bigm| Yn(y)\bigm| \bigm| \leq Mx\alpha - 1
\infty \sum
n=0
| \varphi n|
\bigm| \bigm| Yn(y)\bigm| \bigm|
\lambda n
.
We apply the Cauchy – Bunyakovsky inequality
\infty \sum
n=0
\bigm| \bigm| \varphi n
\bigm| \bigm| \bigm| \bigm| Yn(y)\bigm| \bigm|
\lambda n
\leq
\sqrt{} \infty \sum
n=0
\varphi 2
n
\sqrt{} \infty \sum
n=0
\biggl(
Yn(y)
\lambda n
\biggr) 2
.
Taking into account inequality (9) and the Bessel inequality
\infty \sum
n=0
\varphi 2
n \leq
1\int
0
\varphi 2(y)y - mdy < \infty ,
we obtain uniform convergence of series (11) in any closed subdomain \Omega and the condition
x1 - \alpha u(x, y) \in C(\Omega x \times \Omega y). Let us now turn to the proof of the legality of differentiation. We
will act in the same way as above
\bigm| \bigm| D\alpha
0xu(x, y)
\bigm| \bigm| \leq \infty \sum
n=0
| D\alpha
0xXn(x)|
\bigm| \bigm| Yn(y)\bigm| \bigm| \leq
\leq xs
\infty \sum
n=0
\lambda n| Xn(x)|
\bigm| \bigm| Yn(y)\bigm| \bigm| \leq Mxs+\alpha - 1
\infty \sum
n=0
\bigm| \bigm| \varphi n
\bigm| \bigm| \bigm| \bigm| Yn(y)\bigm| \bigm| .
Next, we apply the Cauchy – Bunyakovsky inequality
\infty \sum
n=0
\bigm| \bigm| \varphi n
\bigm| \bigm| \bigm| \bigm| Yn(y)\bigm| \bigm| = \infty \sum
n=0
| \lambda n\varphi n|
\bigm| \bigm| \bigm| \bigm| Yn(y)\lambda n
\bigm| \bigm| \bigm| \bigm| \leq
\sqrt{} \infty \sum
n=1
\bigm| \bigm| \lambda 2
n\varphi
2
n
\bigm| \bigm| \sqrt{} \infty \sum
n=0
\bigm| \bigm| \bigm| \bigm| Y 2
n (y)
\lambda 2
n
\bigm| \bigm| \bigm| \bigm| ,
we have
\varphi n =
1\int
0
y - m\varphi (y)Yn(y)dy =
( - 1)k
\lambda n
1\int
0
\varphi (y)Y (2k)
n (y)dy =
=
( - 1)k
\lambda n
\left[ \varphi (y)Y (2k - 1)
n (y)
\bigm| \bigm| \bigm| 1
0
-
1\int
0
\varphi \prime (y)Y (2k - 1)
n (y)dy
\right] .
If
\mathrm{l}\mathrm{i}\mathrm{m}
y\rightarrow +0
Y (2k - 1)
n (y) \not = \infty ,
then
\varphi (0)Y (2k - 1)
n (0) = 0,
and if
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1336 B. YU. IRGASHEV
\mathrm{l}\mathrm{i}\mathrm{m}
y\rightarrow +0
Y (2k - 1)
n (y) = \infty ,
then, applying L’Hopital’s rule, we get
\mathrm{l}\mathrm{i}\mathrm{m}
y\rightarrow +0
Y
(2k - 1)
n (y)
(\varphi (y)) - 1 = \mathrm{l}\mathrm{i}\mathrm{m}
y\rightarrow +0
Y
(2k)
n (y)
- (\varphi (y)) - 2\varphi \prime (y)
= ( - 1)k+1\lambda n \mathrm{l}\mathrm{i}\mathrm{m}
y\rightarrow +0
\varphi 2(y)y - mYn(y)
\varphi \prime (y)
=
= ( - 1)k+1\lambda n \mathrm{l}\mathrm{i}\mathrm{m}
y\rightarrow +0
O(y2k)O(yk)y - m
O(yk - 1)
= 0,
whence
\varphi n =
( - 1)k+1
\lambda n
1\int
0
\varphi \prime (y)Y (2k - 1)
n (y)dy =
( - 1)k
\lambda n
1\int
0
\varphi (2k)(y)Yn(y)dy \Rightarrow
\Rightarrow \lambda n\varphi n =
1\int
0
\Bigl[
( - 1)kym\varphi (2k)(y)
\Bigr]
Yn(y)y
- mdy.
Hence, \lambda n\varphi n are the Fourier coefficients of the function ( - 1)kym\varphi (2k)(y). Then, by Bessel’s
inequality, we obtain
\infty \sum
n=0
\lambda 2
n
\bigm| \bigm| \varphi n(y)
\bigm| \bigm| 2 \leq 1\int
0
ym
\bigl(
\varphi (2k)(y)
\bigr) 2
dy. (12)
Now, in order for the calculations made above to be legal, we impose the following restrictions on
the function \varphi (y):
\varphi (j)(0) = \varphi (j)(1) = 0, \varphi (y) \in C2k[0, 1], j = 0, 1, . . . , k - 1.
Taking into account (9) and (12), we have that the series
D\alpha
0xu(x, y) =
\infty \sum
n=0
D\alpha
0xXn(x)Yn(y)
converges uniformly in any closed subdomain \Omega for s = 0 and converges uniformly in \Omega x \times \Omega y for
s = 1, 2, 3, . . .. The uniform convergence of the series
\partial 2ku(x, y)
\partial y2k
=
\infty \sum
n=0
Xn(x)
\partial 2kYn(y)
\partial y2k
= ( - 1)ky - m
\infty \sum
n=0
\lambda nXn(x)Yn(y). (13)
Theorem 1 is proved.
Theorem \bfone \prime . Let the function \varphi (y) satisfies the following conditions:
\varphi (y) \in C2k[0, 1], \varphi (j)(0) = \varphi (j)(1) = 0, j = 0, 1, . . . , k - 1.
Then a solution to Problem A exists.
Remark. It can be seen from the construction of the solution that (see (13))
\partial 2ku(x, y)
\partial y2k
\in C
\bigl(
\Omega x \times \Omega y
\bigr)
. (14)
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
MIXED PROBLEM FOR HIGHER-ORDER EQUATIONS WITH FRACTIONAL DERIVATIVE . . . 1337
3. Uniqueness.
Theorem 2. If there is a solution to problem A from class (2), (14), then it is unique.
Proof. Let the function u(x, y) be a solution to Problem A with zero initial and boundary con-
ditions. Consider its Fourier coefficients with respect to the system of eigenfunctions of problem (6)
un(x) =
1\int
0
y - mu(x, y)Yn(y)dy,
it is easy to show that un(x) is a solution to the problem
D\alpha
0xun(x) = - \lambda nx
sun(x), \mathrm{l}\mathrm{i}\mathrm{m}
x\rightarrow 0
(x1 - \alpha un(x)) = 0.
This problem has only a zero solution, i.e.,
1\int
0
y - mu(x, y)Yn(y)dy = 0 for all n.
Because G(y, \xi ) symmetric, continuous,
1\int
0
G
2
(y, \xi )d\xi < \infty ,
1\int
0
G
2
(y, \xi )dy < \infty ,
1\int
0
1\int
0
G
2
(y, \xi )dyd\xi < \infty , \lambda n > 0 for all n,
then the conditions of Mercer’s theorem are fulfilled and
G(y, \xi ) =
\infty \sum
n=0
Yn(y)Yn(\xi )
\lambda n
.
Hence, we have
y -
m
2 u(x, y) =
1\int
0
G(y, \xi )
\biggl(
( - 1)k\xi
m
2
\partial 2ku(x, \xi )
\partial \xi 2k
\biggr)
d\xi =
= ( - 1)k
1\int
0
\infty \sum
n=0
Yn(y)Yn(\xi )
\lambda n
\biggl(
\xi
m
2
\partial 2ku(x, \xi )
\partial \xi 2k
\biggr)
d\xi =
= ( - 1)k
\infty \sum
n=0
y -
m
2 Yn(y)
\lambda n
1\int
0
\xi -
m
2 Yn(\xi )\xi
m
2
\partial 2ku(x, \xi )
\partial \xi 2k
d\xi =
= ( - 1)k
\infty \sum
n=0
y -
m
2 Yn(y)
\lambda n
1\int
0
Yn(\xi )
\partial 2ku(x, \xi )
\partial \xi 2k
d\xi =
= ( - 1)k
\infty \sum
n=0
y -
m
2 Yn(y)
\lambda n
1\int
0
Y (2k)
n (\xi )u(x, \xi )d\xi =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1338 B. YU. IRGASHEV
= ( - 1)k
\infty \sum
n=0
y -
m
2 Yn(y)
\lambda n
1\int
0
\lambda n( - 1)k\xi - mYn(\xi )u(x, \xi )d\xi =
= y -
m
2
\infty \sum
n=0
Yn(y)
1\int
0
\xi - mYn(\xi )u(x, \xi )d\xi = 0 \Rightarrow u(x, y) \equiv 0.
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Received 02.09.20
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
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| id | umjimathkievua-article-6298 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:26:59Z |
| publishDate | 2022 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/f9/d47483fec101bfb78f7294a0d9408df9.pdf |
| spelling | umjimathkievua-article-62982023-01-07T13:45:39Z Mixed problem for higher-order equations with fractional derivative and degeneration in both variables Mixed problem for higher-order equations with fractional derivative and degeneration in both variables Irgashev, B. Yu. irgashev, bahrom Irgashev, B. Yu. Differential equation, high order, degeneration, fractional Riemann-Liouville derivative, existence, uniqueness, series, uniform convergence. Mathematics Subject Classification 2010. 35G16 UDC 517.9 We consider an initial-boundary-value problem for a higher-order equation with fractional Riemann – Liouville derivative in a rectangular domain degenerating in both variables.&nbsp;The solution to the problem is constructed in the explicit form by the method of separation of variables.&nbsp;Uniqueness is proved by the spectral method. УДК 517.9 Мішана задача для рівнянь вищого порядку з дробовою похідною, що має виродження за обома змінними&nbsp; Pозглянуто початково-крайову задачу для рівняння вищого порядку з дробовою похідною Рімана – Ліувілля в прямокутній області, що вироджується за обома змінними.&nbsp;Розв'язок задачі отримано в явному вигляді методом&nbsp; відокремлення змінних.&nbsp;Єдиність доводиться за допомогою спектрального методу. Institute of Mathematics, NAS of Ukraine 2022-11-27 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6298 10.37863/umzh.v74i10.6298 Ukrains’kyi Matematychnyi Zhurnal; Vol. 74 No. 10 (2022); 1328 -1338 Український математичний журнал; Том 74 № 10 (2022); 1328 -1338 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6298/9308 Copyright (c) 2022 bahrom irgashev |
| spellingShingle | Irgashev, B. Yu. irgashev, bahrom Irgashev, B. Yu. Mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| title | Mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| title_alt | Mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| title_full | Mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| title_fullStr | Mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| title_full_unstemmed | Mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| title_short | Mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| title_sort | mixed problem for higher-order equations with fractional derivative and degeneration in both variables |
| topic_facet | Differential equation high order degeneration fractional Riemann-Liouville derivative existence uniqueness series uniform convergence. Mathematics Subject Classification 2010. 35G16 |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6298 |
| work_keys_str_mv | AT irgashevbyu mixedproblemforhigherorderequationswithfractionalderivativeanddegenerationinbothvariables AT irgashevbahrom mixedproblemforhigherorderequationswithfractionalderivativeanddegenerationinbothvariables AT irgashevbyu mixedproblemforhigherorderequationswithfractionalderivativeanddegenerationinbothvariables |