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Fracti onal
Differential
Calculus
Volume 10, Number 2 (2020), 291–306 doi:10.7153/fdc-2020-10-18
INITIAL–BOUNDARY VALUE AND INVERSE
PROBLEMS FOR SUBDIFFUSION EQUATIONS IN RN
RAVSHAN ASHUROV AND RAKHIM ZUNNUNOV
(Communicated by A. Pskhu)
Abstract. An initial-boundary value problem for a subdiffusion equation with an elliptic oper-
ator A(D)in RNis considered. The existence and uniqueness theorems for a solution of this
problem are proved by the Fourier method. Considering the order of the Caputo time-fractional
derivative as an unknown parameter, the corresponding inverse problem of determining this order
is studied. It is proved, that the Fourier transform of the solution ˆu(
ξ
,t)at a fixed time instance
recovers uniquely the unknown parameter. Further, a similar initial-boundary value problem is
investigated in the case when operator A(D)is replaced by its power A
σ
. Finally, the existence
and uniqueness theorems for a solution of the inverse problem of determining both the orders of
fractional derivatives with respect to time and the degree
σ
are proved. We also note that when
solving the inverse problems, a decrease in the parameter
ρ
of the Mittag-Leffler functions E
ρ
has been proved.
1. Introduction and main results
The theory of differential equations with fractional derivatives has gained signifi-
cant popularity and importance in the last few decades, mainly due to its applications
in many seemingly distant fields of science and technology (see, for example, [1]–[6]).
One of the most important time-fractional equations is the subdiffusion equation,
which models anomalous or slow diffusion processes. This equation is a partial integro-
differential equation obtained from the classical heat equation by replacing the first-
order derivative with a time-fractional derivative of order
ρ
∈(0,1).
When considering the subdiffusion equation as a model equation in the analysis of
anomalous diffusion processes, the order of the fractional derivative is often unknown
and difficult to measure directly. To determine this parameter, it is necessary to in-
vestigate the inverse problems of identifying these physical quantities based on some
indirectly observable information about solutions (see a survey paper Li, Liu and Ya-
mamoto [7]).
In this paper, we investigate the existence and uniqueness of solutions to initial-
boundary value problems for subdiffusion equations with the Caputo derivative and
an elliptic operator A(D)in RN, having constant coefficients. Inverse problems of
Mathematics subject classification (2010): Primary 35R11; Secondary 74S25.
Keywords and phrases: Subdiffusion equation, Caputo derivatives, inverse and initial-boundary value
problem, determination of order of derivatives, Fourier method.
c
,Zagreb
Paper FDC-10-18 291
292 R. ASHUROV AND R. ZUNNUNOV
determining the order of the fractional derivative with respect to time and with respect
to the spatial variable will also be investigated.
Let us proceed to a rigorous formulation of the main results of this article.
1. Let A(D)= ∑
|
α
|=m
a
α
D
α
be a homogeneous symmetric elliptic differential ex-
pression of even order m=2l, with constant coefficients, i.e. A(
ξ
)>0, for all
ξ
=0,
where
α
=(
α
1,
α
2,...,
α
N)- multi-index and D=(D1,D2,...,DN),Dj=1
i
∂
∂
xj,i=
√−1.
The fractional integration in the Riemann-Liouville sense of order
ρ
<0ofa
function hdefined on [0,∞)has the form
∂ρ
th(t)= 1
Γ(−
ρ
)
t
0
h(
ξ
)
(t−
ξ
)
ρ
+1d
ξ
,t>0,
provided the right-hand side exists. Here Γ(
ρ
)is Euler’s gamma function. Using this
definition one can define the Caputo fractional derivative of order
ρ
,0<
ρ
<1, as
D
ρ
th(t)=
∂ρ
−1
t
d
dt h(t).
Note that if
ρ
=1, then fractional derivative coincides with the ordinary classical
derivative of the first order: Dth(t)= d
dt h(t).
Let
ρ
∈(0,1]be a given number and L
τ
2(RN)stand for the Sobolev classes (see
the definition in the next section). Consider the initial-boundary value problem: find a
function u(x,t)∈Lm
2(RN),t∈[0,T), such that (note that this inclusion is considered
as a boundary condition at infinity)
D
ρ
tu(x,t)+A(D)u(x,t)=0,x∈RN,0<t<T,(1)
u(x,0)=
ϕ
(x),x∈RN,(2)
where
ϕ
(x)is a given continuous function.
We call problem (1)–(2)the forward problem.
We draw attention to the fact, that in the statement of the forward problem the
requirement u(x,t)∈Lm
2(RN)is not caused by the merits. However, on the one hand,
the uniqueness of just such a solution is proved quite simply, and on the other, the
solution found by the Fourier method satisfies the above condition.
DEFINITION 1. A function u(x,t)∈C(RN×[0,T)) with the properties
D
ρ
tu(x,t)and A(D)u(x,t)∈C(RN×(0,T))
and satisfying conditions (1)–(2) is called the classical solution (or simply, the solution)
of the forward problem.
INITIAL-BOUNDARYVALUE AND INVERSE PROBLEMS 293
Let us denote by E
ρ
(t)the Mittag-Leffler function of the form
E
ρ
(t)=
∞
∑
k=0
tk
Γ(
ρ
k+1),
and denote by ˆ
f(
ξ
)the Fourier transform of a function f(x)∈L2(RN):
ˆ
f(
ξ
)=(2
π
)−N
RN
f(x)e−ix
ξ
dx.
Now we can formulate the existence and uniqueness theorem for the forward prob-
lem.
THEOREM 1. Let
τ
>N
2and
ϕ
∈L
τ
2(RN). Then the forward problem has a
unique solution and this solution has the form
u(x,t)=
RN
E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)eix
ξ
d
ξ
.(3)
The integral uniformly and absolutely converges with respect to x ∈RNand for each
t∈[0,T). Moreover, solution (3) has the property
lim
|x|→∞D
α
u(x,t)=0,|
α
|m,0<t<T,(4)
In recent years, many works by specialists have appeared in which various initial-
boundary value problems for various subdiffusion equations are investigated. Let us
mention only some of these works. Basically, the case of one spatial variable x∈Rand
subdiffusion equation with “the elliptical part” uxx were considered (see, for example,
handbook Machado, editor [1], book of A. A. Kilbas et al. [3] and monograph of A. V.
Pskhu [8], and references in these works). The paper Gorenflo, Luchko and Yamamoto
[9] is devoted to the study of subdiffusion equations in Sobelev spaces. In the paper by
Kubica and Yamamoto [10], initial-boundary value problems for equations with time-
dependent coefficients are considered. In the multidimensional case (x∈RN), instead
of the differential expression uxx , authors considered either the Laplace operator ([3],
[11]–[13]) or pseudodifferential operators with constant coefficients in the whole space
RN(Umarov [14]). In the last work the initial function
ϕ
∈Lp(RN)is such, that the
Fourier transform ˆ
ϕ
is compactly supported. The authors of the recent paper [15]
considered initial-boundary value problems for subdiffusion equations with arbitrary
elliptic differential operators in bounded domains.
2. Determining the correct order of an equation in applied fractional modeling
plays an important role. The corresponding inverse problem for subdiffusion equations
has been considered by a number of authors (see a survey paper Li, Liu and Yamamoto
[7] and references therein, [16]–[22]). Note that in all known works the subdiffusion
equation was considered in a bounded domain Ω⊂RN. In addition, it should be noted
294 R. ASHUROV AND R. ZUNNUNOV
that in publications [16]–[19] the following relation was taken as an additional condi-
tion
u(x0,t)=h(t),0<t<T,(5)
at a monitoring point x0∈Ω. But this condition, as a rule (an exception is the work
[19] by J. Janno, where both the uniqueness and existence are proved), can ensure
only the uniqueness of the solution of the inverse problem [16]–[18]. The authors
of the article Ashurov and Umarov [20] considered the value of the projection of the
solution onto the first eigenfunction of the elliptic part of the subdiffusion equation
as additional information. Note that the results from [20] are only applicable when
the first eigenvalue is zero. The uniqueness and existence of an unknown order of
the fractional derivative in the subdiffusion equation were proved in the recent work of
Alimov and Ashurov [21]. In this case, the additional condition is ||u(x,t0)||2=d0,and
the boundary condition is not necessarily homogeneous. The authors of the article [22]
investigated the inverse problem for the simultaneous determination of the order of the
Riemann-Liouville time fractional derivative andthe source function in the subdiffusion
equations.
In what follows, we will assume that the initial function
ϕ
belongs to the class
L
τ
2(RN)with
τ
>N
2. Then, by Theorem 1, the forward problem has a unique solution
of the form (3)forany
ρ
∈(0,1].
Let us consider the order of fractional derivative
ρ
in equation (1) as an unknown
parameter. To formulate our inverse problem we will additionally assume that
ϕ
(x)∈
L1(RN). This implies that both functions ˆ
ϕ
(
ξ
)and ˆu(
ξ
,t)=E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
),t∈
[0,T), are continuous in the variable
ξ
∈RN.Letusfix a vector
ξ
0=0, such that
ˆ
ϕ
(
ξ
0)=0 and put
λ
0=A(
ξ
0)>0. To determine the order
ρ
we use the following
extra data:
U(t0,
ρ
)≡|ˆu(
ξ
0,t0)|=d0,(6)
where t0,0<t0<T,isafixed time instant.
The problem (1)–(2) together with extra condition (6) is called the inverse problem.
To solve the inverse problem fix the number
ρ
0∈(0,1)and consider the problem
for
ρ
∈[
ρ
0,1].
DEFINITION 2. The pair {u(x,t),
ρ
}of the solution u(x,t)to the forward prob-
lem and the parameter
ρ
∈[
ρ
0,1]is called the classical solution (or simply, the solution)
of the inverse problem.
The following property of the Fourier transform ˆu(
ξ
,t)of the forward problem’s
solution plays an important role in the solution of the inverse problem and, in our opin-
ion, is of independent interest.
LEMMA 1. Fo r
ρ
0from the interval 0<
ρ
0<1, there is a number T0=T0(
λ
0,
ρ
0)
such that for all t0,T
0t0<T , the function U (t0,
ρ
)decreases monotonically with
respect to
ρ
∈[
ρ
0,1].
The result related to the inverse problem has the form.
INITIAL-BOUNDARYVALUE AND INVERSE PROBLEMS 295
THEOREM 2. Let T0t0<T . Then the inverse problem has a unique solution
{u(x,t),
ρ
}if and only if
e−
λ
0t0d0
|ˆ
ϕ
(
ξ
0)|E
ρ
0(−
λ
0t
ρ
0
0).(7)
3. Finally, we will consider another inverse problem of determining both the orders
of fractional derivatives with respect to time and the spatial derivatives in the subdiffu-
sion equations.
For the best of our knowledge, only in the following two papers [23]and[24]such
inverse problems were studied and only the uniqueness theorems ware proved (note that
the uniqueness is a very important property of a solution from an application point of
view). In the paper [23] by Tatar and Ulusoy it is considered the initial-boundary value
problem for the differential equation
∂ρ
tu(t,x)=−(−)
σ
u(t,x),t>0,x∈(0,1),
where
σ
is the one-dimensional fractional Laplace operator,
ρ
∈(0,1)and
σ
∈
(1/4,1). The authors have proved that if the initial function
ϕ
(x)is sufficiently smooth
and all its Fourier coefficients are positive, then the two-parameter inverse problemwith
additional information (5) may have only one solution. As for physical backgrounds for
two-parameter differential equations, see, for example, [25].
In [24], M. Yamamoto provedthe uniqueness theorem for the above two-parameter
inverse problem in an N-dimensional bounded domain Ωwith a smooth boundary
∂
Ω. The conditions for the initial function found in this work are less restrictive, for
example, if
ϕ
is zero on
∂
Ω,
ϕ
∈L
τ
2(Ω),
τ
>N/2,
ϕ
0inΩand
ϕ
(x0)=0, then
the uniqueness theorem is true.
Let us denote by Aan operator in L2(RN)with the domain of definition D(A)=
C∞
0(RN), acting as Af(x)=A(D)f(x). It is easy to verify that the closure ˆ
Aof operator
Ais nonnegative and selfadjoint. Therefore, by virtue of the von Neumann theorem,
for any
σ
>0, we can introduce the degree of the operator ˆ
Aas
ˆ
A
σ
f(x)=
∞
0
λσ
dP
λ
f(x)=
RN
A
σ
(
ξ
)ˆ
f(
ξ
)eix
ξ
d
ξ
,
where projectors P
λ
defined as
P
λ
f(x)=
A(
ξ
)<
λ
ˆ
f(
ξ
)eix
ξ
d
ξ
.
The domain of definition of this operator is determined from the condition ˆ
A
σ
f(x)∈
L2(RN)and has the form
D(ˆ
A
σ
)={f∈L2(RN):
RN
A2
σ
(
ξ
)|ˆ
f(
ξ
)|2d
ξ
<∞}.
296 R. ASHUROV AND R. ZUNNUNOV
Suppose that
ρ
∈(0,1]and
σ
∈(0,1]are given numbers and consider the initial-
boundary value (the second forward) problem: find a function v(x,t)∈D(ˆ
A
σ
)such
that (note that this inclusion is also considered as a boundary condition)
D
ρ
tv(x,t)+ ˆ
A
σ
v(x,t)=0,x∈RN,0<t<T,(8)
v(x,0)=
ϕ
(x),x∈RN,(9)
where
ϕ
(x)is a given function and as mentioned above, we assume
ϕ
∈L
τ
2(RN)for
some
τ
>N
2.
The solution to this problem is defined similarly to the solution to problem (1)–(2)
(see Definition 1). In exactly the same way as Theorem 1, it is proved that the unique
solution of the second forward problem has the form
v(x,t)=
RN
E
ρ
(−A
σ
(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)eix
ξ
d
ξ
,(10)
where the integral uniformly and absolutely converges in x∈RNand for each t∈
[0,T).
Now let
ρ
0>0and
σ
0>0befixed numbers and assume, that in the second for-
ward problem, the parameters
ρ
∈[
ρ
0,1]and
σ
∈[
σ
0,1]are unknown. Since there are
two unknown numbers, then one obviously needs two extra conditions. To formulate
these conditions, we again assume that
ϕ
∈L1(RN). Then both functions ˆ
ϕ
(
ξ
)and
ˆv(
ξ
,t)are continuous in
ξ
. It should be noted, that the proposed in this paper method,
for simultaneously finding both the order of fractional differentiation
ρ
and the power
σ
is applicable if there exists
ξ
0∈
∂
ΩA≡{
ξ
∈RN;A(
ξ
)=1}, such that ˆ
ϕ
(
ξ
0)=0.
Note, that if A(D)is the Laplace operator, then
∂
ΩAis the N-dimensional unit sphere.
Let
ξ
0be one of such a vector. We consider the following information as additional
conditions:
V(
ξ
0,t0,
ρ
,
σ
)=|ˆv(
ξ
0,t0)|=d0,t0T0(1,
ρ
0),(11)
V(
ξ
1,t1,
ρ
,
σ
)=|ˆv(
ξ
1,t1)|=d1,A(
ξ
1)=
λ
1(=1)Λ1,t11,(12)
where T0is defined in Lemma 1,
ξ
1is such that ˆ
ϕ
(
ξ
1)=0andΛ1is definedin(27).
We call the problem (8)–(9) together with extra conditions (11)and(12)the second
inverse problem.
Note that since
ξ
0∈
∂
ΩA,then V(
ξ
0,t0,
ρ
,
σ
)is actually independent of
σ
:
V(
ξ
0,t0,
ρ
,
σ
)=|E
ρ
(−A
σ
(
ξ
0)t
ρ
0)ˆ
ϕ
(
ξ
0)|=|E
ρ
(−t
ρ
)ˆ
ϕ
(
ξ
0)|.
Therefore, to solve the second inverse problem, we first find the unique
ρ
that satisfies
the relation (11). Then, assuming that
ρ
is already known and using the relation (12),
we find the second unknown parameter
σ
. It should be noted that the number Λ1
from condition (12) depends on
σ
0and
ρ
.
INITIAL-BOUNDARYVALUE AND INVERSE PROBLEMS 297
THEOREM 3. There is a unique
ρ
∈[
ρ
0,1], satisfying (11), if and only if d0
satisfies the inequalities (7) with
λ
0=1.For
σ
∈[
σ
0,1]to exist, it is necessary and
sufficient that d1satisfy the inequalities
E
ρ
(−
λ
1t
ρ
1)d1
|ˆ
ϕ
(
ξ
1)|E
ρ
(−
λσ
0
1t
ρ
1).(13)
REMARK 1. As Theorems 2and 3show, in order for the inverse problems to have
solutions, the domain (0,T), where the equations are satisfied, must be large enough.
In conclusion, note that the theory and applications of various inverse problems,
on determining the coefficients of the equation, the right-hand side, and also on deter-
mining the initial or boundary functions for differential equations of integer order are
discussed in Kabanikhin [26] (see also references therein) Similar inverse problems for
fractional-order equations were considered, for example, in the works [27]–[31].
2. Forward problems
In the present section we prove Theorems 1and the equation (10).
The class of functions L2(RN)which for a given fixed number a>0makethe
norm
||f||2
La
2(RN)=
RN
(1+|
ξ
|2)a
2ˆ
f(
ξ
)eix
ξ
d
ξ
2
L2(RN)=
RN
(1+|
ξ
|2)a|ˆ
f(
ξ
)|2d
ξ
finite is termed the Sobolev class La
2(RN).Sincefor
τ
>0 and some constants c1and
c2one has the inequality
c1(1+|
ξ
|2)
τ
m1+A2
τ
(
ξ
)c2(1+|
ξ
|2)
τ
m,(14)
then D(ˆ
A
τ
)=L
τ
m
2(RN).
Let Ibe the identity operator in L2(RN).Operator (ˆ
A+I)
ν
is definedinthesame
way as operator ˆ
A
σ
.
Proof of Theorem 1.The existence of a solution to the forward problem is based
on the following lemma (see M. A. Krasnoselski et al. [32], p. 453); for the operator ˆ
A
this lemma is a simple consequence of the Sobolev embedding theorem.
LEMMA 2. Let a multi-index
α
be such that |
α
|m and
ν
>|
α
|
m+N
2m. Then the
operator D
α
(ˆ
A+I)−
ν
continuously maps from L2(RN)into C(RN)and moreover the
following estimate holds true
||D
α
(ˆ
A+I)−
ν
f||C(RN)C||f||L2(RN).(15)
Proof. For any a>N/2 one has the Sobolev embedding theorem: La
2(RN)→
C(RN),thatis
||D
α
(ˆ
A+I)−
ν
f||C(RN)C||D
α
(ˆ
A+I)−
ν
f||La
2(RN).
298 R. ASHUROV AND R. ZUNNUNOV
Therefore, it is sufficient to prove the inequality
||D
α
(ˆ
A+I)−
ν
f||La
2(RN)C||f||L2(RN).
But this is a consequence of the estimate
RN|ˆ
f(
ξ
)|2|
ξ
|2|
α
|(1+A(
ξ
))−2
ν
(1+|
ξ
|2)ad
ξ
C
RN|ˆ
f(
ξ
)|2d
ξ
,
that is valid for N
2<a
ν
m−|
α
|.
To prove the existence of the forward problem’s solution we remind the following
estimate of the Mittag-Leffler function with a negative argument (see, for example, [6],
p. 29)
|E
ρ
(−t)|C
1+t,t>0.(16)
Let a sequence {Ωk}∞
k=1of domains Ωk⊂RNhave the following two properties:
1) closure Ωk=Ωk∪
∂
Ωkof Ωkis contained in Ωk+1:
Ωk⊂Ωk+1;
2) the union of all Ωkfills the entire space RN:
∞
k=1
Ωk=RN.
Consider the truncated integral
Sk(x,t)=
Ωk
E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)eix
ξ
d
ξ
.(17)
Step 1.It is not difficult to verify that for any kfunction Sk(x,t)satisfies equation
(1) and the initial condition (2) (see, for example, [6], page 173 and [33]). From the
Sobolev embedding theorem and the condition
ϕ
∈L
τ
2(RN),
τ
>N
2, it follows that
ϕ
∈C(RN).
Step 2.In accordance with Definition 1, we will show that for the function (3) one
has A(D)u(x,t)∈C(RN×(0,T)).
Let |
α
|m,
τ
>N
2and
ν
=1+
τ
m>|
α
|
m+N
2m.Then
Sk(x,t)=(ˆ
A+I)−
τ
/m−1
Ωk
(A(
ξ
)+1)
τ
/m+1E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)eix
ξ
d
ξ
.
Therefore by virtue of Lemma 2one has
||D
α
Sk(x,t)||2
C(RN)
=
D
α
(ˆ
A+I)−
τ
/m−1
Ωk
(A(
ξ
)+1)
τ
/m+1E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)eix
ξ
d
ξ
2
C(RN)
C
Ωk
(A(
ξ
)+1)
τ
/m+1E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)eix
ξ
d
ξ
L2(RN)
.
INITIAL-BOUNDARYVALUE AND INVERSE PROBLEMS 299
Using the Parseval equality, we will have
||D
α
Sk(x,t)||2
C(RN)C
Ωk
(A(
ξ
)+1)
τ
/m+1E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)
2d
ξ
.
Applying the inequality (16)gives |(A(
ξ
)+1)E
ρ
(−A(
ξ
)t
ρ
)|C(1+t−
ρ
). Therefore,
||D
α
Sk(x,t)||2
C(RN)C(1+t−
ρ
)2
Ωk
(A(
ξ
)+1)
τ
/mˆ
ϕ
(
ξ
)
2d
ξ
C(1+t−
ρ
)2||
ϕ
||2
L
τ
2(RN).
This implies uniform (and absolute) in x∈RNconvergence of the differentiated integral
(3)inthevariables xjfor each t∈(0,T).
Step 3.If
α
=0, then taking
ν
=
τ
mand applying the inequality (16), we establish
uniform (and absolute) convergence of the integral (3) (hence, the continuity of the
solution) in the domain t∈[0,T):
||Sk(x,t)||2
C(RN)C
Ωk
(A(
ξ
)+1)
τ
/mE
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)
2d
ξ
C
Ωk
(A(
ξ
)+1)
τ
/mˆ
ϕ
(
ξ
)
2d
ξ
C||
ϕ
||2
L
τ
2(RN).
Step 4.Further, from equation (1)wegetD
ρ
tSk(x,t)=−A(D)Sk(x,t). Therefore,
proceeding the above reasoning, we arrive at D
ρ
tu(x,t)∈C(RN×(0.T)).
Step 5.The inclusion u(x,t)∈Lm
2(RN)for all t∈(0,T), is a consequence of the
condition
ϕ
∈L2(RN). Indeed, using inequalities (14)and(16) we arrive at
||D
α
Sk(x,t)||2
L2(RN)=
Ωk
ξα
E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)
2d
ξ
C
Ωk
A(
ξ
)E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)
2d
ξ
CTt−2
ρ
||
ϕ
||2
L2(RN).
Step 6.Let us show the property (4) of the solution (3). To do this, note first that
the inclusion
ϕ
∈L
τ
2(RN),
τ
>N/2, implies ˆ
ϕ
∈L1(RN). Indeed, application of the
H¨older inequality gives
RN|ˆ
ϕ
(
ξ
)|d
ξ
=
RN|ˆ
ϕ
(
ξ
)|(1+|
ξ
|2)
τ
/2(1+|
ξ
|2)−
τ
/2d
ξ
C
τ
||
ϕ
||L
τ
2(RN).
Therefore, by virtue of inequality (16), one has E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)∈L1(RN). Simi-
larly, inequalities (14)and(16)imply
|
ξα
E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)|C|A(
ξ
)E
ρ
(−A(
ξ
)t
ρ
)ˆ
ϕ
(
ξ
)|∈L1(RN)
for all |
α
|m. Hence, D
α
u(x,t), as a function of x, is the Fourier transform of a L1-
function. Obviously, this implies the property (4).
300 R. ASHUROV AND R. ZUNNUNOV
Step 7.Let us prove the uniqueness of the forward problem’s solution.
Suppose that problem (1)–(2) has two solutions u1(x,t)and u2(x,t).Ouraimis
to prove that u(x,t)=u1(x,t)−u2(x,t)≡0 . Since the problem is linear, then we have
the following homogenous problem for u(x,t)∈Lm
2(RN):
D
ρ
tu(x,t)+A(D)u(x,t)=0,x∈RN,0<t<T; (18)
u(x,0)=0,x∈RN.(19)
Let u(x,t)be a solution of problem (18)–(19)and
ω
(x)be an arbitrary function
with properties
ω
(x)0and
ω
∈C∞
0(RN). Obviously ˆ
ω
(
ξ
)∈L2(RN), and since
ˆu(
ξ
,t)∈L2(RN),then ˆ
ω
(
ξ
)ˆu(
ξ
,t)∈L1(RN). Therefore, by virtue of Fubini’s theo-
rem, the following function of t∈[0,T)exists for almost all
λ
:
w
λ
(t)=
A(
ξ
)=
λ
eiy
ξ
ˆ
ω
(
ξ
)ˆu(
ξ
,t)d
σλ
(
ξ
),(20)
where d
σλ
(
ξ
)is the corresponding surface element and y∈RN.
Taking into account that u(x,t)is a solution of equation (18) we have (note,
A(D)u(x,t)∈L2(RN))
D
ρ
tw
λ
(t)=−(2
π
)−N
A(
ξ
)=
λ
eiy
ξ
ˆ
ω
(
ξ
)
RN
A(D)u(x,t)e−ix
ξ
dxd
σλ
(
ξ
).
The inner integral exists as the Fourier transform of the L2-function. From the equation
A(D)u(x,t)=
RN
A(
η
)ˆu(
η
,t)eix
η
d
η
,
one has
D
ρ
tw
λ
(t)=−
A(
ξ
)=
λ
eiy
ξ
ˆ
ω
(
ξ
)A(
ξ
)ˆu(
ξ
,t)d
σλ
(
ξ
)=−
λ
w
λ
(t).
Therefore, we have the following Cauchy problem for w
λ
(t):
D
ρ
tw
λ
(t)+
λ
w
λ
(t)=0,t>0; w
λ
(0)=0.
This problem has the unique solution; hence, the function defined by (20), is identi-
cally zero (see, for example, [6], p. 173 and [33]): w
λ
(t)≡0 for almost all
λ
>0.
Integrating the equation (20) with respect to
λ
over the domain (0,+∞)we obtain,
that
RN
eiy
ξ
ˆ
ω
(
ξ
)ˆu(
ξ
,t)d
ξ
=
RN
ω
(y−x)u(x,t)dx =0,
for almost all yand since both functions
ω
(·)and u(·,t)are continuous, then for all
y∈RNand t∈[0,T). Taking into account that the function
ω
(x)is arbitrary with the
above properties, then from the last equality we have u(x,t)≡0.
Thus Theorem 1is proved.
The uniqueness of the solution to the second forward problem and the formula
(10) is established based on the above reasoning.
INITIAL-BOUNDARYVALUE AND INVERSE PROBLEMS 301
3. First inverse problem
LEMMA 3. Given
ρ
0from the interval 0<
ρ
0<1, there exists a number T0=
T0(
λ
0,
ρ
0), such that for all t0T0and
λ
λ
0the function e
λ
(
ρ
)=E
ρ
(−
λ
t
ρ
0)is
positive and monotonically decreasing with respect to
ρ
∈[
ρ
0,1]and
e
λ
(1)e
λ
(
ρ
)e
λ
(
ρ
0).
Proof. Let us denote by
δ
(1;
β
)a contour oriented by non-decreasing arg
ζ
con-
sisting of the following parts: the ray arg
ζ
=−
β
with |
ζ
|1,the arc −
β
arg
ζ
β
,
|
ζ
|=1, and the ray arg
ζ
=
β
,|
ζ
|1. If 0 <
β
<
π
, then the contour
δ
(1;
β
)di-
vides the complex
ζ
-plane into two unbounded parts, namely G(−)(1;
β
)to the left of
δ
(1;
β
)by orientation, and G(+)(1;
β
)to the right of it. The contour
δ
(1;
β
)is called
the Hankel path.
Let
β
=3
π
4
ρ
,
ρ
∈[
ρ
0,1). Then by the definition of this contour
δ
(1;
β
),we
arrive at (note, −
λ
t
ρ
0∈G(−)(1;
β
),see[6], p. 27)
E
ρ
(−
λ
t
ρ
0)= 1
λ
t
ρ
0Γ(1−
ρ
)−1
2
π
i
ρλ
t
ρ
0
δ
(1;
β
)
e
ζ
1/
ρ
ζ
ζ
+
λ
t
ρ
0
d
ζ
=f1(
ρ
)+ f2(
ρ
).(21)
To prove the lemma it suffices to show that the derivative d
d
ρ
e
λ
(
ρ
)is negative
for all
ρ
∈[
ρ
0,1), since the positivity of e
λ
(
ρ
)follows from the inequality e
λ
(1)=
e−
λ
t>0.
It is not hard to estimate the derivative f
1(
ρ
). Indeed, let Ψ(
ρ
)be the logarithmic
derivative of the gamma function Γ(
ρ
)(for the definition and properties of Ψsee [34]).
Then Γ(
ρ
)=Γ(
ρ
)Ψ(
ρ
), and therefore,
f
1(
ρ
)=−lnt0−Ψ(1−
ρ
)
λ
t
ρ
0Γ(1−
ρ
).
Since 1
Γ(1−
ρ
)=1−
ρ
Γ(2−
ρ
),Ψ(1−
ρ
)=Ψ(2−
ρ
)−1
1−
ρ
,
the function f
1(
ρ
)can be represented as follows
f
1(
ρ
)=−1
λ
t
ρ
0
(1−
ρ
)[lnt0−Ψ(2−
ρ
)] + 1
Γ(2−
ρ
).
If
γ
≈0,57722 is the Euler-Mascheroni constant, then −
γ
<Ψ(2−
ρ
)<1−
γ
.By
virtue of this estimate we may write
−f
1(
ρ
)(1−
ρ
)[lnt0−(1−
γ
)] + 1
Γ(2−
ρ
)
λ
t
ρ
0
1
λ
t
ρ
0
,(22)
provided lnt0>1−
γ
or t02.
302 R. ASHUROV AND R. ZUNNUNOV
To estimate the derivative f
2(
ρ
), we denote the integrand in (21)byF(
ζ
,
ρ
):
F(
ζ
,
ρ
)= 1
2
π
i
ρλ
t
ρ
0·e
ζ
1/
ρ
ζ
ζ
+
λ
t
ρ
0
.
Note, that the domain of integration
δ
(1;
β
)also depends on
ρ
. To take this circum-
stance into account when differentiating the function f
2(
ρ
), we rewrite the integral (21)
in the form:
f2(
ρ
)= f2+(
ρ
)+ f2−(
ρ
)+ f21(
ρ
),
where
f2±(
ρ
)=e±i
β
∞
1
F(se±i
β
,
ρ
)ds,
f21(
ρ
)=i
β
−
β
F(eiy,
ρ
)eiydy =i
β
1
−1
F(ei
β
s,
ρ
)ei
β
sds.
Let us consider the function f2+(
ρ
).Since
β
=3
π
4
ρ
and
ζ
=sei
β
,then
e
ζ
1/
ρ
=e1
2(i−1)s
1
ρ
.
The derivative of the function f2+(
ρ
)has the form
f
2+(
ρ
)= 1
2
π
i
ρλ
t
ρ
0
∞
1
e1
2(i−1)s1/
ρ
se2ia
ρ
−i−1
2
ρ
2s1/
ρ
ln s+2ia−1
ρ
−lnt0−ias eia
ρ
+
λ
t
ρ
0lnt0
se
ia
ρ
+
λ
t
ρ
0
seia
ρ
+
λ
t
ρ
0
ds,
where a=3
π
4. By virtue of the inequality |seia
ρ
+
λ
t
ρ
0|
λ
t
ρ
0we arrive at
|f
2+(
ρ
)|C
ρ
(
λ
t
ρ
0)2
∞
1
e−1
2s1/
ρ
s1
ρ
2s1/
ρ
ln s+lnt0ds.
LEMMA 4. Let 0<
ρ
1and m ∈N.Then
K(
ρ
)= 1
ρ
∞
1
e−1
2s
1
ρ
sm
ρ
+1ds Cm.
Proof. Set r=s1
ρ
.Then
s=r
ρ
,ds =
ρ
r
ρ
−1dr.
INITIAL-BOUNDARYVALUE AND INVERSE PROBLEMS 303
Therefore,
K(
ρ
)=
∞
1
e−1
2rrm−1+2
ρ
dr
∞
1
e−1
2rrm+1dr =Cm.
Since ln s1
ρ
<s1
ρ
,thenbyvirtueofLemma4,
|f
2+(
ρ
)|C
(
λ
t
ρ
0)2C2
ρ
+C0lnt0C
(
λ
t
ρ
0)21
ρ
+lnt0.
Function f
2−(
ρ
)has exactly the same estimate.
Now consider the function f21 (
ρ
). It is not hard to verify that
f
21(
ρ
)= a
2
πλ
t
ρ
0
1
−1
eeias e2ia
ρ
s2ias −lnt0−iaseia
ρ
s+
λ
t
ρ
0lnt0
eia
ρ
s+
λ
t
ρ
0
eia
ρ
s+
λ
t
ρ
0
ds.
Therefore,
|f
21(
ρ
)|Clnt0
(
λ
t
ρ
0)2.
Taking into account estimate (22) and the estimates of f
2±and f
21 ,wehave
d
d
ρ
e
λ
(
ρ
)<−1
λ
t
ρ
0
+C1/
ρ
+lnt0
(
λ
t
ρ
0)2.(23)
In other words, this derivative is negative if
t
ρ
0>C1/
ρ
+lnt0
λ
for all
ρ
∈[
ρ
0,1)or
t
ρ
0
0>C1/
ρ
0+lnt0
λ
.(24)
Thus, there exists a number T0=T0(
λ
0,
ρ
0)such, that for all t0T0we have the
estimate d
d
ρ
e
λ
(
ρ
)<0,
λ
λ
0,
ρ
∈[
ρ
0,1].
Since
U(t,
ρ
)=|ˆu(
ξ
0,t)|=E
ρ
(−A(
ξ
0)t
ρ
)|ˆ
ϕ
(
ξ
0)|=E
ρ
(−
λ
0t
ρ
)|ˆ
ϕ
(
ξ
0)|,
Lemma 1follows immediately from Lemma 3. Theorem 2is an easy consequence of
these two lemmas.
In conclusion, we make the following remark. If the elliptic polynomial A(
ξ
)is
nonhomogeneous, that is A(
ξ
)= ∑
|
α
|m
a
αξα
and moreover, A(
ξ
)
λ
0>0, then from
Lemma 3it follows:
If t0T0and T0is as above, then E
ρ
(−A(
ξ
)t
ρ
), as a function of
ρ
, is positive
and decreases monotonically in
ρ
∈[
ρ
0,1]for any
ξ
∈RN.
Therefore, in this case you can also consider various options for the function
U(t,
ρ
).ExamplesU(t,
ρ
)=||Au(x,t)||2and U(t,
ρ
)=||u(x,t)||2.
304 R. ASHUROV AND R. ZUNNUNOV
4. Second inverse problem
To prove Theorem 3,wefirst find the unknown parameter
ρ
. Suppose, as required
by Theorem 3,that d0satisfies condition (7) with
λ
0=A(
ξ
0)=1. Then, as it follows
from Lemma 3,forallt0T0(1,
ρ
0)the equation
V(
ξ
0,t0,
ρ
,
σ
)=|ˆv(
ξ
0,t0)|=E
ρ
(−t
ρ
)|ˆ
ϕ
(
ξ
0)|=d0
has the unique solution
ρ
∈[
ρ
0,1].
Now let us define
σ
∈[
σ
0,1], which corresponds to the already found
ρ
and
satisfies condition (12).
We first assume, that
ρ
<1andlet
β
=3
π
4
ρ
. Then formula (21) will have the
form
E
ρ
(−
λσ
t
ρ
1)= 1
λσ
t
ρ
1Γ(1−
ρ
)−1
2
π
i
ρ
λσ
t
ρ
1
δ
(1;
β
)
e
ζ
1/
ρ
ζ
ζ
+
λσ
t
ρ
1
d
ζ
=g1(
σ
)+g2(
σ
).
(25)
One has
g
1(
σ
)=−ln
λ
λσ
t
ρ
1Γ(1−
ρ
)
and
g
2(
σ
)=(1+t
ρ
1)ln
λ
2
π
i
ρ
λσ
t
ρ
1
δ
(1;
β
)
e
ζ
1/
ρ
ζ
ζ
+
λσ
t
ρ
1
d
ζ
.
It is easy to check that g
2(
σ
)has an estimate (it is proved similarly to the estimate for
f
2±)
|g
2(
σ
)|(1+t
ρ
1)ln
λ
π
(
λσ
t
ρ
1)2C0+4
3
π
<5ln
λ
λ
2
σ
t
ρ
1
.
Therefore, for all t11wehave
d
d
σ
E
ρ
(−
λσ
t
ρ
1)<−ln
λ
λσ
t
ρ
1Γ(1−
ρ
)+5ln
λ
λ
2
σ
t
ρ
1
.(26)
Hence this derivative is negative if
λσ
λσ
05Γ(1−
ρ
).
Thus, if
λ
1Λ1=Λ1(
ρ
,
σ
0),and(see(12))
Λ1=en,nln(5Γ(1−
ρ
))
σ
0
,(27)
then E
ρ
(−
λσ
t
ρ
1), as a function of
σ
∈[
σ
0,1], strictly decreases for all t11.
Now let
ρ
=1. Then E
ρ
(−
λσ
t
ρ
1)=e−
λ
σ
t1and the derivative (26)isnegative
for all
λ
>1andt11.
INITIAL-BOUNDARYVALUE AND INVERSE PROBLEMS 305
Since the function E
ρ
(−
λσ
t
ρ
1)is decreasing, then the following estimates hold
E
ρ
(−
λ
1t
ρ
1)E
ρ
(−
λσ
1t
ρ
1)E
ρ
(−
λσ
0
1t
ρ
1),
λ
1Λ1,
σ
∈[
σ
0,1].
The last estimate shows that if d1satisfies condition (13), then, assuming
ρ
has al-
ready been found, we can uniquely determine the parameter
σ
from equality (12), that
is, from
E
ρ
(−
λσ
1t
ρ
1)|ˆ
ϕ
(
ξ
1)|=d1.
Acknowledgement. The authors convey thanks to Sh. A. Alimov for discussions
of these results.
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(Received October 24, 2020) Ravshan Ashurov
Institute of Mathematics
Uzbekistan Academy of Science
Tashkent, 81 Mirzo Ulugbek str. 100170
e-mail: ashurovr@gmail.com
Rakhim Zunnunov
Institute of Mathematics
Uzbekistan Academy of Science
Tashkent, 81 Mirzo Ulugbek str. 100170
e-mail: zunnunov@mail.ru
Fractional Differential Calculus
www.ele-math.com
fdc@ele-math.com
