Next Article in Journal
ICT Use and Successful Learning: The Role of the Stock of Human Capital
Next Article in Special Issue
Important Criteria for Asymptotic Properties of Nonlinear Differential Equations
Previous Article in Journal
Correction: Pfeifer, D.; Ragulina, O. Adaptive Bernstein Copulas and Risk Management. Mathematics 2020, 8, 2221
Previous Article in Special Issue
On-Off Intermittency in a Three-Species Food Chain
 
 
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

Inverse Problem for the Sobolev Type Equation of Higher Order

Department of Applied Mathematics and Programming, South Ural State University, 454080 Chelyabinsk, Russia
*
Author to whom correspondence should be addressed.
Mathematics 2021, 9(14), 1647; https://doi.org/10.3390/math9141647
Submission received: 8 June 2021 / Revised: 7 July 2021 / Accepted: 9 July 2021 / Published: 13 July 2021

Abstract

:
The article investigates the inverse problem for a complete, inhomogeneous, higher-order Sobolev type equation, together with the Cauchy and overdetermination conditions. This problem was reduced to two equivalent problems in the aggregate: regular and singular. For these problems, the theory of polynomially bounded operator pencils is used. The unknown coefficient of the original equation is restored using the method of successive approximations. The main result of this work is a theorem on the unique solvability of the original problem. This study continues and generalizes the authors’ previous research in this area. All the obtained results can be applied to the mathematical modeling of various processes and phenomena that fit the problem under study.

1. Introduction

Let U , F , Y be Banach spaces, operators A , B 0 , B 1 , , B n 1 L ( U ; F ) , i.e., linear and continuous operators defined on U and acting to F , ker A { 0 } , C L ( U ; Y ) , given functions χ : [ 0 , T ] L ( Y ; F ) , f : [ 0 , T ] F , Ψ : [ 0 , T ] Y . Consider the following problem with t [ 0 , T ]
A v ( n ) ( t ) = B n 1 v ( n 1 ) ( t ) + + B 1 v ( t ) + B 0 v ( t ) + q ( t ) χ ( t ) + f ( t ) ,
v ( 0 ) = v 0 , v ( 0 ) = v 1 , , v ( n 1 ) ( 0 ) = v n 1 ,
C v ( t ) = Ψ ( t ) .
The problem of finding a pair of functions v ( t ) C n ( [ 0 , T ] ; U ) (n times continuously diferentiable) and q ( t ) C 1 ( [ 0 , T ] ; Y ) (continuously diferentiable) from relations (1)–(3) is called an inverse problem. At present, the authors have obtained the result when studying the inverse problem, but only in the case of the second-order, Sobolev type equation [1].
The degeneracy of the operator A allows us to classify Equation (1) as a Sobolev type equation. Additionally, one can see that this equation is complete, since all the components v ( t ) , v ( t ) , , v ( n ) ( t ) are present. In addition, the Cauchy condition (2) is posed. The overdetermination condition (3) arises due to the need to restore the parameter q ( t ) of the equation.
The study of Sobolev type equations was carried out repeatedly [1,2,3,4,5,6,7,8,9,10,11,12]. There are articles devoted to both the first [2,3,4], the second [1,5,6], the third [7], and the higher [8,9,10] order. In [2], sufficient conditions for the existence of positive solutions to the Showalter–Sidorov and the Cauchy problem for an abstract linear equation of this type were presented. The linear representatives of Sobolev type equations, such as the Barenblatt–Zheltov–Kochina equation and the Hoff equation are studied in [3]. The paper [7] contains a condition for the existence of a weak, local, timely solution to the Cauchy problem for a model Sobolev type equation. In the study of the direct problem for a higher-order, Sobolev type equation, the phase space method was used [10]. Papers [11,12] are among the first investigations of Sobolev type equations, and the recent works devoted to applications of Sobolev type equations to real-life models are as follows: [13,14].
The works [1,5,15,16,17,18,19,20,21,22,23,24] were devoted to the consideration of inverse problems. In [15], the process of unsteady flow of a viscous incompressible fluid in a pipe with a permeable wall was considered. The dependence on the choice of the boundary of the rectangular region and the unique solvability of the inverse problem were investigated in [16]. The uniqueness criterion for the Lavrent’ev–Bitsadze equation is established in [17]. The correctness in Sobolev spaces of the problem of determining the function of sources in the heat and mass transfer Navier–Stokes system was proved [18]. The problem was finding the area where the vector of boundary displacements and forces is given in parametric form [19]. In [20], the inverse boundary value problem for the heat equation was studied, and the error of the obtained approximate solution was estimated.
The article consists of four sections. The second section combines the necessary, previously obtained, results of the theory of polynomially A-bounded of operator pencils formulated in the form of definitions, theorems and lemmas. Section «Results» has three subsections. The first one presents the result of applying the splitting theorem; thus, the original problem is divided into two equivalent problems in the aggregate: regular and singular. In the second subsection, we study the unique solvability of the regular problem by reducing it to an equivalent problem of the first order and achieving the necessary smoothness for the required function q using the method of successive approximations. The third subsection generalizes the result of studying the singular problem obtained earlier in the work [9], thus obtaining the theorem on the existence and uniqueness of the solution to the problem (1)–(3). In the last section, the significance of the obtained results is given in both the development of the studied theory and their practical application.

2. Preliminary Information

To find a pair of functions v ( t ) and q ( t ) , we use the results obtained in the research into higher-order, Sobolev type equations [8]. Thus, we will apply the theory of polynomially A-bounded operator pencils. Denote by B the pencil of operators B 0 , B 1 , , B n 1 .
Definition 1.
The sets
ρ A ( B ) = { μ C : ( μ n A μ n 1 B n 1 μ B 1 B 0 ) 1 L ( F ; U ) }
and σ A ( B ) = C ¯ \ ρ A ( B ) will be called the A-resolvent set and the A-spectrum of the pencil B , respectively.
Definition 2.
The operator-function complex variable
R μ A ( B ) = ( μ n A μ n 1 B n 1 μ B 1 B 0 ) 1
with domain ρ A ( B ) will be called the A-resolvent of the pencil B .
Definition 3.
Let the pencil B be polynomially A-bounded if
a R + μ C ( | μ | > a ) R μ A ( B ) L ( F ; U ) .
Let the pencil B be polynomially A-bounded. Introduce an important condition
γ μ k R μ A ( B ) d μ O , k = 0 , 1 , , n 2 ,
where γ = { μ C : | μ | = r > a } .
Lemma 1.
Let the pencil B be polynomially A-bounded and condition (4) be fulfilled. Then, the operators
P = 1 2 π i γ R μ A ( B ) μ n 1 A d μ L ( U ) ,
Q = 1 2 π i γ μ n 1 A R μ A ( B ) d μ L ( F )
are projectors.
Put U 0 = ker P ,   F 0 = ker Q ,   U 1 = im P ,   F 1 = im Q . From the previous Lemma it follows that U = U 0 U 1 ,   F = F 0 F 1 . Let A k ( B l k ) denote the restriction of the operator A ( B l ) onto U k ,   k = 0 , 1 ;   l = 0 , 1 , , n 1 .
Theorem 1.
Let the pencil B be polynomially A-bounded and condition (4) be fulfilled. Then, the actions of the operators split:
1. 
A k L ( U k ; F k ) , k = 0 , 1 ;
2. 
B l k L ( U k ; F k ) , k = 0 , 1 ; l = 0 , 1 , , n 1 ;
3. 
There exists an operator ( A 1 ) 1 L ( F 1 ; U 1 ) ;
4. 
There exists an operator ( B 0 0 ) 1 L ( F 0 ; U 0 ) .
Definition 4.
Define the family of operators { K q 1 , K q 2 , , K q n } as follows
K 1 1 = H 0 , K 1 2 = H 1 , , K 1 n = H n 1 ,
K q + 1 1 = K q n H 0 , K q + 1 2 = K q 1 K q n H 1 , , K q + 1 n = K q n 1 K q n H n 1 ; q = 1 , 2 , ,
where H 0 = ( B 0 0 ) 1 A 0 , H 1 = ( B 0 0 ) 1 B 1 0 , , H n 1 = ( B 0 0 ) 1 B n 1 0 .
Definition 5.
The point ∞ is called
1. 
Removable singular point of the A-resolvent of pencil B , if K 1 1 O , K 1 2 O , , K 1 n O ;
2. 
A pole of order p N of the A-resolvent of pencil B , if p such, that K p 1 O , K p 2 O , , K p n O , but K p + 1 1 O , K p + 1 2 O , , K p + 1 n O ;
3. 
An essentially singular point of the A-resolvent of the pencil B , if K p n O for any p N .

3. Results

3.1. Reduction of the Initial Inverse Problem

Let the pencil B be polynomially A-bounded and condition (4) be fulfilled, then v ( t ) can be represented as v ( t ) = P v ( t ) + ( I P ) v ( t ) = u ( t ) + ω ( t ) . Suppose that U 0 ker C . Then, by virtue of Theorem 1 and Lemma 1 problem (1)–(3) is equivalent to the problem of finding the functions u C n ( [ 0 , T ] ; U 1 ) ,   ω C n ( [ 0 , T ] ; U 0 ) , q C 1 ( [ 0 , T ] ; Y ) from the relations
u ( n ) ( t ) = S n 1 u ( n 1 ) ( t ) + + S 1 u ( t ) + S 0 u ( t ) + q ( t ) ( A 1 ) 1 Q χ ( t ) + ( A 1 ) 1 Q f ( t ) ,
u ( 0 ) = u 0 , u ( 0 ) = u 1 , , u ( n 1 ) ( 0 ) = u n 1 ,
C u ( t ) = Ψ ( t ) C v ( t ) ,
H 0 ω ( n ) ( t ) = H n 1 ω ( n 1 ) ( t ) + + H 2 ω ( t ) + H 1 ω ( t ) + ω ( t ) + + q ( t ) ( B 0 0 ) 1 ( I Q ) χ ( t ) + ( B 0 0 ) 1 ( I Q ) f ( t ) ,
ω ( 0 ) = ω 0 , ω ( 0 ) = ω 1 , , ω ( n 1 ) ( 0 ) = ω n 1 ,
where
S 0 = ( A 1 ) 1 B 0 1 , S 1 = ( A 1 ) 1 B 1 1 , , S n 1 = ( A 1 ) 1 B n 1 1 ,
u 0 = P v 0 , u 1 = P v 1 , , u n 1 = P v n 1 ,
ω 0 = ( I P ) v 0 , ω 1 = ( I P ) v 1 , , ω n 1 = ( I P ) v n 1 , t [ 0 , T ] .
The inverse problem (5)–(7) is called regular, and problem (8), (9) is called singular.

3.2. Solution of the Regular Inverse Problem

Rewrite problem (5)–(7) in the notation [25]. Let X = U 1 , operators S 0 , S 1 , , S n 1 C l ( X ) ,   C L ( X , Y ) , operator-function Φ : [ 0 , T ] L ( Y ; X ) , functions h : [ 0 , T ] X ,   Ψ : [ 0 , T ] Y
u ( n ) ( t ) = S n 1 u ( n 1 ) ( t ) + + S 1 u ( t ) + S 0 u ( t ) + q ( t ) Φ ( t ) + h ( t ) , t [ 0 , T ] ,
u ( 0 ) = u 0 , u ( 0 ) = u 1 , , u ( n 1 ) ( 0 ) = u n 1 ,
C u ( t ) = Ψ ( t ) .
Theorem 2.
Let the pencil B be polynomially A-bounded and condition (4) be fulfilled; moreover, C L ( X ; Y ) ,   Φ C 1 ( [ 0 , T ] ; L ( Y ; X ) ) ,   h C 1 ( [ 0 , T ] ; X ) ,   Ψ C n + 1 ( [ 0 , T ] ; Y ) , for any t [ 0 , T ] the operator C Φ ( t ) be invertible and ( C Φ ) 1 C 1 ( [ 0 , T ] ; L ( Y ) ) . If the compatibility condition C u n 1 = Ψ n 1 ( 0 ) is satisfied, then the solution to the inverse problem (10)–(12) exists and is unique in the class of functions q C 1 ( [ 0 , T ] ; Y ) , u C n ( [ 0 , T ] ; X ) .
Proof of Theorem 2.
Reduce problem (10)–(12) to the problem for the first-order equation
z ( t ) = A z ( t ) + q ( t ) Q ( t ) + H ( t ) , t [ 0 , T ] ,
z ( 0 ) = z 0 ,
B z ( t ) = Ψ ¯ ( t ) ,
where z ( t ) = u ( t ) u ( n 2 ) ( t ) u ( n 1 ) ( t ) ,   A = 0 I 0 0 0 I S 0 S 1 S n 1 ,   Q ( t ) = 0 0 Φ ( t ) , H ( t ) = 0 0 h ( t ) ,   z ( 0 ) = u ( 0 ) u ( n 2 ) ( 0 ) u ( n 1 ) ( 0 ) ,   z 0 = u 0 u n 2 u n 1 ,   B = 0 0 C ,   Ψ ¯ ( t ) = 0 0 Ψ ( n 1 ) ( t ) .
Put R ( t ) = ( C Φ ( t ) ) 1 . Therefore, all the conditions of Theorem 6.2.3 from [25], are fulfilled, and the function q ( t ) satisfies the integral equation
q ( t ) = q 0 ( t ) + R ( t ) ( C S 0 0 t V 1 , n ( t s ) q ( s ) Φ ( s ) d s + + C S 1 0 t V 2 , n ( t s ) q ( s ) Φ ( s ) d s + + C S n 1 0 t V n , n ( t s ) q ( s ) Φ ( s ) d s ) ,
where
q 0 ( t ) = R ( t ) ( Ψ ( n ) ( t ) C S 0 V 1 , 1 ( t ) u 0 C S 1 V 2 , 1 ( t ) u 0 C S n 1 V n , 1 ( t ) u 0
C S 0 V 1 , 2 ( t ) u 1 C S 1 V 2 , 2 ( t ) u 1 C S n 1 V n , 2 ( t ) u 1
C S 0 V 1 , n ( t ) u n 1 C S 1 V 2 , n ( t ) u n 1 C S n 1 V n , n ( t ) u n 1
C S 0 0 t V 1 , n ( t s ) h ( s ) d s C S 1 0 t V 2 , n ( t s ) h ( s ) d s
C S n 1 0 t V n , n ( t s ) h ( s ) d s C h ( t ) ) .
Thus, there exists a unique solution q C 1 ( [ 0 , T ] ; Y ) , z C 1 ( [ 0 , T ] ; X n ) to the inverse problem (13)–(15). And we obtain that the solution to the regular inverse problem (10)–(12) exists and is unique, with q C 1 ( [ 0 , T ] ; Y ) ,   u C n ( [ 0 , T ] ; X ) .  □
In order to obtain a solution to a singular problem, we need a greater smoothness of the function q from the solution of a regular problem than class C 1 ( [ 0 , T ] ; Y ) . Next, we need the following Lemma from [1].
Lemma 2.
Let l N ,   V C l 1 ( [ 0 , T ] ; L ( X ) ) ,   g C l ( [ 0 , T ] ; X ) . Then
0 t V ( t s ) g ( s ) d s ( l ) = k = 0 l 1 V ( l k 1 ) ( t ) g ( k ) ( 0 ) + 0 t V ( t s ) g ( l ) ( s ) d s .
The following theorem provides sufficient conditions for the existence of a more smooth (as p N ) solution q C p + n ( [ 0 , T ] , Y ) of a regular problem.
Theorem 3.
Let the pencil B be polynomially A-bounded and condition (4) be fulfilled, p N 0 ; moreover, C L ( X ; Y ) ,   Φ C p + n ( [ 0 , T ] ; L ( Y ; X ) ) ,   h C p + n ( [ 0 , T ] ; X ) , Ψ C p + 2 n ( [ 0 , T ] ; Y ) , for any t [ 0 , T ] operator C Φ ( t ) be invertible, with ( C Φ ) 1 C p + n ( [ 0 , T ] ; L ( Y ) ) and the compatibility condition C u n 1 = Ψ ( n 1 ) ( 0 ) be satisfied for some u n 1 U 1 . Then there exists and a unique solution of (10)–(12) and q C p + n ( [ 0 , T ] ; Y ) .
Proof of Theorem 3.
Write the propagators of the homogeneous Equation (10) in a matrix, denoting the resolving group of homogeneous Equation (13)
V ( t ) = V 1 , 1 ( t ) V 1 , 2 ( t ) V 1 , n 1 ( t ) V 1 , n ( t ) V 2 , 1 ( t ) V 2 , 2 ( t ) V 2 , n 1 ( t ) V 2 , n ( t ) V n 1 , 1 ( t ) V n 1 , 2 ( t ) V n 1 , n 1 ( t ) V n 1 , n ( t ) V n , 1 ( t ) V n , 2 ( t ) V n , n 1 ( t ) V n , n ( t ) = 1 2 π i γ R μ A ( B ) ×
× μ n 1 A μ n 2 B n 1 B 1 μ n 2 A μ n 3 B n 1 B 2 B 0 μ n 1 A μ n 2 B n 1 μ B 2 μ n 3 B 0 μ n 3 B 1 + μ n 4 B 0 μ n 2 B 0 μ n 2 B 1 + μ n 3 B 0
μ A B n 1 I μ 2 A μ B n 1 μ I μ n 1 A μ n 2 B n 1 μ n 2 I μ n 2 B n 2 + μ n 3 B n 3 + + B 0 μ n 1 I e μ t d μ ,
where I is the identity operator. Earlier, in the proof of Theorem 2, it was established that the function q ( t ) satisfies the integral Equation (16). Take the natural number l p + n . Assuming that q C l ( [ 0 , T ] ; Y ) by Lemma 2, we obtain the equality
q ( l ) ( t ) = q 0 ( l ) ( t ) + k = 0 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m 1 ) ( t ) ( q Φ ) ( m ) ( 0 ) +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ( l k m ) ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l 1 C l k R ( k ) ( t ) C S 1 m = 0 l k 1 V 2 , n ( l k m 1 ) ( t ) ( q Φ ) ( m ) ( 0 ) +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 1 0 t V 2 , n ( t s ) q ( l k m ) ( s ) Φ ( m ) ( s ) d s + +
+ k = 0 l 1 C l k R ( k ) ( t ) C S n 1 m = 0 l k 1 V n , n ( l k m 1 ) ( t ) ( q Φ ) ( m ) ( 0 ) +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S n 1 0 t V n , n ( t s ) q ( l k m ) ( s ) Φ ( m ) ( s ) d s ,
where C l k = l ! k ! ( l k ) ! , C l k , m = l ! k ! m ! ( l k m ) ! and
q 0 ( l ) ( t ) = k = 0 l C l k R ( k ) ( t ) ( Ψ ( l k + n ) ( t )
C S 0 V 1 , 1 ( l k ) ( t ) u 0 C S 1 V 2 , 1 ( l k ) ( t ) u 0 C S n 1 V n , 1 ( l k ) ( t ) u 0
C S 0 V 1 , 2 ( l k ) ( t ) u 1 C S 1 V 2 , 2 ( l k ) ( t ) u 1 C S n 1 V n , 2 ( l k ) ( t ) u 1
C S 0 V 1 , n ( l k ) ( t ) u n 1 C S 1 V 2 , n ( l k ) ( t ) u n 1 C S n 1 V n , n ( l k ) ( t ) u n 1
C S 0 0 t V 1 , n ( t s ) h ( l k ) ( s ) d s C S 1 0 t V 2 , n ( t s ) h ( l k ) ( s ) d s
C S n 1 0 t V n , n ( t s ) h ( l k ) ( s ) d s C h ( l k ) ( t ) ) +
+ k = 0 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m 1 ) ( t ) h ( m ) ( 0 ) +
+ k = 0 l 1 C l k R ( k ) ( t ) C S 1 m = 0 l k 1 V 2 , n ( l k m 1 ) ( t ) h ( m ) ( 0 ) + +
+ k = 0 l 1 C l k R ( k ) ( t ) C S n 1 m = 0 l k 1 V n , n ( l k m 1 ) ( t ) h ( m ) ( 0 )
exists from the conditions of this theorem for l = 0 , 1 , , p + n .
Show that q C p + n ( [ 0 , T ] , Y ) ; for this purpose, denote r 0 = q 0 ( 0 ) , and for l = 1 , 2 , , p + n , determine the following values
r l = q 0 ( l ) ( 0 ) + k = 0 l 1 C l k R ( k ) ( 0 ) C S 0 m = 0 l k 1 V 1 , n ( l k m 1 ) ( 0 ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 0 l 1 C l k R ( k ) ( 0 ) C S 1 m = 0 l k 1 V 2 , n ( l k m 1 ) ( 0 ) j = 0 m C m j r m j Φ ( j ) ( 0 ) + +
+ k = 0 l 1 C l k R ( k ) ( 0 ) C S n 1 m = 0 l k 1 V n , n ( l k m 1 ) ( 0 ) j = 0 m C m j r m j Φ ( j ) ( 0 ) .
Consider the system of integral equations
q ˜ 0 ( t ) = q 0 ( t ) + R ( t ) ( C S 0 0 t V 1 , n ( t s ) q ˜ 0 ( s ) Φ ( s ) d s +
+ C S 1 0 t V 2 , n ( t s ) q ˜ 0 ( s ) Φ ( s ) d s + + C S n 1 0 t V n , n ( t s ) q ˜ 0 ( s ) Φ ( s ) d s ) ,
q ˜ l ( t ) = q 0 ( l ) ( t ) + k = 0 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 0 l 1 C l k R ( k ) ( t ) C S 1 m = 0 l k 1 V 2 , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) + +
+ k = 0 l 1 C l k R ( k ) ( t ) C S n 1 m = 0 l k 1 V n , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 1 0 t V 2 , n ( t s ) q ˜ l k m ( s ) Φ ( m ) ( s ) d s + +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S n 1 0 t V n , n ( t s ) q ˜ l k m ( s ) Φ ( m ) ( s ) d s ,
l = 1 , 2 , , p + n .
Reduce (17) to the Volterra equation of the second kind
g ( t ) = g 0 ( t ) + 0 t K ( t , s ) g ( s ) d s
on the space ( C ( [ 0 , T ] ; Y ) ) p + n + 1 with a matrix operator function K ( t , s ) , given on the triangle Δ = { ( t , s ) R 2 : 0 t T , 0 s t } . By virtue of the continuity of all data of system (17), this has a unique solution
( q ˜ 0 , q ˜ 1 , , q ˜ p + n ) ( C ( [ 0 , T ] ; Y ) ) p + n + 1 .
This solution will be the limit of the sequence of approximations
q ˜ 0 , i ( t ) = q 0 ( t ) + R ( t ) ( C S 0 0 t V 1 , n ( t s ) q ˜ 0 , i 1 ( s ) Φ ( s ) d s +
+ C S 1 0 t V 2 , n ( t s ) q ˜ 0 , i 1 ( s ) Φ ( s ) d s + + C S n 1 0 t V n , n ( t s ) q ˜ 0 , i 1 ( s ) Φ ( s ) d s ) ,
q ˜ l , i ( t ) = q 0 ( l ) ( t ) + k = 0 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 0 l 1 C l k R ( k ) ( t ) C S 1 m = 0 l k 1 V 2 , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) + +
+ k = 0 l 1 C l k R ( k ) ( t ) C S n 1 m = 0 l k 1 V n , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m , i 1 ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 1 0 t V 2 , n ( t s ) q ˜ l k m , i 1 ( s ) Φ ( m ) ( s ) d s + +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S n 1 0 t V n , n ( t s ) q ˜ l k m , i 1 ( s ) Φ ( m ) ( s ) d s ,
l = 1 , 2 , , p + n ; i N ,
which for i on the interval [ 0 , T ] converge uniformly to the functions q ˜ l , l = 0 , 1 , , p + n . Set the initial approximation q ˜ l , 0 0 ;   l = 0 , 1 , , p + n , then q ˜ l + 1 , 0 = q ˜ l , 0 ;   l = 0 , 1 , , p + n 1 . In addition, from (18), it follows that
q ˜ l , i ( 0 ) = r l ; l = 0 , 1 , , p + n ; i N .
Assume that for all τ = 1 , 2 , , i the equalities q ˜ l + 1 , τ ( t ) = q ˜ l , τ ( t ) ,   l = 0 , 1 , , p + n 1 are true. Then, using Lemma 2 and equalities (18), we obtain
d d t k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m , i ( s ) Φ ( m ) ( s ) d s =
= k = 0 l m = 0 l k C l k , m R ( k + 1 ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m , i ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 V 1 , n ( t ) q ˜ l k m , i ( 0 ) Φ ( m ) ( 0 ) +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m , i ( s ) Φ ( m + 1 ) ( s ) d s +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s =
= k = 1 l + 1 m = 0 l k + 1 C l k 1 , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l C l k R ( k ) ( t ) C S 0 V 1 , n ( t ) m = 0 l k C l k m r l k m Φ ( m ) ( 0 ) +
+ k = 0 l m = 1 l k + 1 C l k , m 1 R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s .
Denote by
a k , m = R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s , l = 2 , 3 , , p + n .
Taking into account the equalities
C l k + C l k 1 = C l + 1 k , C l k , m + C l k 1 , m + C l k , m 1 = C l + 1 k , m
we obtain
k = 0 l m = 0 l k C l k , m a k , m + k = 1 l + 1 m = 0 l k + 1 C l k 1 , m a k , m + k = 0 l m = 1 l k + 1 C l k , m 1 a k , m =
= k = 1 l m = 1 l k C l k , m a k , m + k = 1 l C l k , 0 a k , 0 + m = 0 l C l 0 , m a 0 , m +
+ k = 1 l m = 1 l k C l k 1 , m a k , m + k = 1 l C l k 1 , 0 a k , 0 + k = 1 l C l k 1 , l k + 1 a k , l k + 1 + C l l , 0 a l + 1 , 0 +
+ k = 1 l m = 1 l k C l k , m 1 a k , m + m = 1 l + 1 C l 0 , m 1 a 0 , m + k = 1 l C l k , l k a k , l k + 1 =
= k = 1 l m = 1 l k C l + 1 k , m a k , m + k = 1 l C l + 1 k , 0 a k , 0 + m = 1 l C l + 1 0 , m a 0 , m + k = 1 l C l + 1 k , 0 a k , l k + 1 +
+ C l 0 , 0 a 0 , 0 + C l 0 , l a 0 , l + 1 + C l l , 0 a l + 1 , 0 = k = 0 l + 1 m = 0 l k + 1 C l + 1 k , m a k , m .
For l = 0 , 1 fullment of (21) can be checked directly.
From (20) and (21), it follows that
d d t k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m , i ( s ) Φ ( m ) ( s ) d s =
= k = 0 l + 1 m = 0 l k + 1 C l + 1 k , m R ( k ) ( t ) C S 0 0 t V 1 , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l C l k R ( k ) ( t ) C S 0 V 1 , n ( t ) m = 0 l k C l k m r l k m Φ ( m ) ( 0 ) .
Similarly, we obtain the result for the subsequent integral element from (18)
d d t k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S 1 0 t V 2 , n ( t s ) q ˜ l k m , i ( s ) Φ ( m ) ( s ) d s =
= k = 0 l + 1 m = 0 l k + 1 C l + 1 k , m R ( k ) ( t ) C S 1 0 t V 2 , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l C l k R ( k ) ( t ) C S 1 V 2 , n ( t ) m = 0 l k C l k m r l k m Φ ( m ) ( 0 ) .
Continuing the procedure for all subsequent integral elements (18), we present the result for the last
d d t k = 0 l m = 0 l k C l k , m R ( k ) ( t ) C S n 1 0 t V n , n ( t s ) q ˜ l k m , i ( s ) Φ ( m ) ( s ) d s =
= k = 0 l + 1 m = 0 l k + 1 C l + 1 k , m R ( k ) ( t ) C S n 1 0 t V n , n ( t s ) q ˜ l k m + 1 , i ( s ) Φ ( m ) ( s ) d s +
+ k = 0 l C l k R ( k ) ( t ) C S n 1 V n , n ( t ) m = 0 l k C l k m r l k m Φ ( m ) ( 0 ) .
Changing the summation indices and re-grading the sums, we obtain
d d t k = 0 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) =
= k = 0 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 0 l 1 C l k R ( k + 1 ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) =
= k = 0 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 1 l C l k 1 R ( k ) ( t ) C S 0 m = 0 l k V 1 , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) =
= ( k = 1 l 1 C l k R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ C l 0 R ( t ) C S 0 m = 0 l 1 V 1 , n ( l m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) ) +
+ ( k = 1 l 1 C l k 1 R ( k ) ( t ) C S 0 m = 0 l k 1 V 1 , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) +
+ k = 1 l 1 C l k 1 R ( k ) ( t ) C S 0 V 1 , n ( t ) j = 0 l k C l k l r l k j Φ ( j ) ( 0 ) +
+ C l l 1 R ( l ) ( t ) C S 0 V 1 , n ( t ) C 0 0 r 0 Φ ( 0 ) ) =
= k = 0 l C l + 1 k R ( k ) ( t ) C S 0 m = 0 l k V 1 , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 )
k = 0 l C l k R ( k ) ( t ) C S 0 V 1 , n ( t ) m = 0 l k C l k m r l k m Φ ( m ) ( 0 ) .
Similarly, we obtain the result for the next non-integral element from (18)
d d t k = 0 l 1 C l k R ( k ) ( t ) C S 1 m = 0 l k 1 V 2 , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) =
= k = 0 l C l + 1 k R ( k ) ( t ) C S 1 m = 0 l k V 2 , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 )
k = 0 l C l k R ( k ) ( t ) C S 1 V 2 , n ( t ) m = 0 l k C l k m r l k m Φ ( m ) ( 0 ) .
Continuing the procedure for all subsequent non-integral elements (18), we present the result for the last
d d t k = 0 l 1 C l k R ( k ) ( t ) C S n 1 m = 0 l k 1 V n , n ( l k m 1 ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 ) =
= k = 0 l C l + 1 k R ( k ) ( t ) C S n 1 m = 0 l k V n , n ( l k m ) ( t ) j = 0 m C m j r m j Φ ( j ) ( 0 )
k = 0 l C l k R ( k ) ( t ) C S n 1 V n , n ( t ) m = 0 l k C l k m r l k m Φ ( m ) ( 0 ) .
Differentiating (18), and also using (22)–(27), we obtain the equalities q ˜ l , i + 1 = q ˜ l + 1 , i + 1 ; l = 0 , 1 , , p + n 1 . Thus, the sequence q ˜ 0 , i converges as i to the function q ˜ 0 uniformly on the interval [ 0 , T ] , and the sequence q ˜ 0 , i = q ˜ 1 , i converges as i to the function q ˜ 1 uniformly on the segment [ 0 , T ] . Therefore, the function q ˜ 0 is continuously differentiable and q ˜ 0 = q ˜ 1 . The equalities of q ˜ l = q ˜ l + 1 ;   l = 1 , 2 , , p + n 1 , are proved in the same way, which implies that q ˜ 0 q C p + n ( [ 0 , T ] ; Y ) and, therefore, q ( l ) = q ˜ l ; l = 1 , 2 , , p + n .  □

3.3. Solvability of the Original Inverse Problem

Theorem 4.
Let the pencil B be polynomially A-bounded and condition (4) be fulfilled; moreover, the ∞ be a pole of order p N 0 of the A-resolvent of the pencil B , operator C L ( U ; Y ) , U 0 ker C , χ C p + n ( [ 0 , T ] ; L ( Y ; F ) ) , f C p + n ( [ 0 , T ] ; F ) , Ψ C p + 2 n ( [ 0 , T ] ; Y ) , for any t [ 0 , T ] operator C ( A 1 ) 1 Q χ be invertible, with ( C ( A 1 ) 1 Q χ ) 1 C p + n ( [ 0 , T ] ; L ( Y ) ) , the condition C u n 1 = Ψ ( n 1 ) ( 0 ) be satisfied at some initial value u n 1 U 1 , and the initial values w k = ( I P ) v k U 0 satisfy
w k = j = 0 p K j n ( B 0 0 ) 1 d j + k d t j + k ( I Q ) ( q ( 0 ) χ ( 0 ) + f ( 0 ) ) , k = 0 , 1 , , n 1 .
Then, there exists a unique solution ( v , q ) of inverse problem (1)–(3), where q C p + n ( [ 0 , T ] ; Y ) , v = u + w , whence u C n ( [ 0 , T ] ; U 1 ) is the solution of (5)–(7), and the function w C n ( [ 0 , T ] ; U 0 ) is a solution of (8) and (9) given by
w ( t ) = j = 0 p K j n ( B 0 0 ) 1 d j d t j ( I Q ) ( q ( t ) χ ( t ) + f ( t ) ) .
Proof of Theorem 4.
The conditions of Theorems 2 and 3 are satisfied, and, therefore, there exists a unique solution ( q , u ) to problem (5)–(7), where q C p + n ( [ 0 , T ] ; Y ) , u C n ( [ 0 , T ] ; U 1 ) .
Using the result of [9] and the required smoothness of the function q, we obtain that there exists a unique solution w C n ( [ 0 , T ] ; U 0 ) to (8), (9), given by (28). □

4. Discussion

The results obtained in the article can be applied to various mathematical models, such as a model of oscillation of a rotating viscous fluid using the viscosity coefficient, a model of gravitational-gyroscopic and internal waves, and a model of sound waves in smectics, since these mathematical models can be reduced to the Sobolev type equations of higher order. One of the most typical examples of the application of the Sobolev type equations theory is the Boussinesq–Love model [5]:
( λ Δ ) v t t = α ( Δ λ ) v t + β ( Δ λ ) v + q f ,
with initial conditions
v ( x , 0 ) = v 0 ( x ) , v t ( x , 0 ) = v 1 ( x ) ,
boundary condition
v ( x , t ) | Ω = 0
and overdetermination condition
Ω v ( x , t ) K ( x ) d x = Φ ( t ) ,
where v 0 ( x ) , v 1 ( x ) , K ( x ) , Φ ( t ) are given functions, v ( x , t ) is a searched function and Ω R n is a bounded domain with a boundary Ω of class C . Equation (29) describes longitudinal vibrations in a thin elastic rod, taking into account the inertia and external load. The coefficients λ , α , λ , β , λ characterize the properties of the rod material and relate such quantities as Young’s modulus, Poisson’s ratio, material density and radius of gyration relative to the center of gravity, in addition, the function f sets a known part of the external load (if known). The integral overdetermination condition (30) arises at the moment when, in addition to finding the function v, it is necessary to restore the component of the external load q. In addition, it is planned to use the obtained results for the development of numerical methods, to find approximate solutions to some of the previously presented models.

Author Contributions

Conceptualization, A.Z. and A.L.; Methodology, A.Z.; Validation, A.Z. and A.L.; Formal Analysis, A.L.; Investigation, A.L.; Resources, A.Z.; Data Curation, A.L.; Writing—Original Draft Preparation, A.L.; Supervision, A.Z.; Project administration, A.Z.; Funding Acquisition, A.L. All authors have read and agreed to the published version of the manuscript.

Funding

The reported study was funded by RFBR, project number 19-31-90137.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

References

  1. Zamyshlyaeva, A.A.; Lut, A.V. Inverse Problem for Sobolev Type Mathematical Models. Bull. South Ural State Univ. Ser. Math. Model. Program. 2019, 12, 25–36. [Google Scholar] [CrossRef]
  2. Banasiak, J.; Manakova, N.A.; Sviridyuk, G.A. Positive solutions to Sobolev type equations with relatively p-sectorial operators. Bull. South Ural. State Univ. Ser. Math. Model. Program. 2020, 13, 17–32. [Google Scholar] [CrossRef]
  3. Shafranov, D.E. On numerical solution in the space of differential forms for one stochastic sobolev-type equation with a relatively radial operator. J. Comput. Eng. Math. 2020, 7, 48–55. [Google Scholar] [CrossRef]
  4. Shafranov, D.E.; Adukova, N.V. Solvability of the Showalter–Sidorov problem for Sobolev type equations with operators in the form of first-order polynomials from the Laplace–Beltrami operator on differential forms. J. Comput. Eng. Math. 2017, 4, 27–34. [Google Scholar] [CrossRef]
  5. Lut, A.V. Numerical Investigation of The Inverse Problem for The Boussinesq–Love Mathematical Model. J. Comput. Eng. Math. 2020, 7, 45–59. [Google Scholar] [CrossRef]
  6. Zamyshlyaeva, A.A.; Lut, A.V. Numerical investigation of the Boussinesq–Love mathematical models on geometrical graphs. Bull. South Ural State Univ. Ser. Math. Model. Program. 2017, 10, 137–143. [Google Scholar] [CrossRef]
  7. Korpusov, M.O.; Panin, A.A.; Shishkov, A.E. On the Critical Exponent «Instantaneous Blow-up» Versus «Local Solubility» in the Cauchy Problem for a Model Equation of Sobolev Type. Izv. Math. 2021, 85, 111–144. [Google Scholar]
  8. Zamyshlyaeva, A.A.; Bychkov, E.V. The Cauchy Problem for the Sobolev Type Equation of Higher Order. Bull. South Ural State Univ. Ser. Math. Model. Program. 2018, 11, 5–14. [Google Scholar] [CrossRef]
  9. Zamyshlyaeva, A.A.; Tsyplenkova, O.N.; Bychkov, E.V. Optimal Control of Solutions to the Initial-final Problem for the Sobolev Type Equation of Higher Order. J. Comput. Eng. Math. 2016, 3, 57–67. [Google Scholar] [CrossRef] [Green Version]
  10. Zamyshlyaeva, A.A.; Manakova, N.A.; Tsyplenkova, O.N. Optimal Control in Linear Sobolev Type Mathematical Models. Bull. South Ural State Univ. Ser. Math. Model. Program. 2020, 13, 5–27. [Google Scholar]
  11. Sobolev, S.L. On a new problem of mathematical physics. Izv. Akad. Nauk SSSR Ser. Mat. 1954, 18, 3–50. [Google Scholar]
  12. Showalter, R.E. The Sobolev type equations. I. Appl. Anal. 1975, 5, 15–22. [Google Scholar] [CrossRef]
  13. Mohan, M.T. On the three dimensional Kelvin-Voigt fluids: Global solvability, exponential stability and exact controllability of Galerkin approximations. Evol. Equ. Control Theory 2020, 9, 301–339. [Google Scholar] [CrossRef] [Green Version]
  14. Baranovskii, E.S. Strong solutions of the incompressible Navier–Stokes–Voigt model. Mathematics 2020, 8, 181. [Google Scholar] [CrossRef] [Green Version]
  15. Gamzaev, K.M. Inverse Problem of Unsteady Incompressible Fluid Flow in a Pipe with a Permeable Wall. Bull. South Ural State Univ. Ser. Math. Mech. Phys. 2020, 12, 24–30. [Google Scholar]
  16. Islomov, B.I.; Ubaydullayev, U.S. The Inverse Problem for a Mixed Type Equation with a Fractional Order Operator in a Rectangular Domain. Russ. Math. 2021, 65, 25–42. [Google Scholar] [CrossRef]
  17. Martemyanova, N.V. Nonlocal Inverse Problem of Finding Unknown Multipliers in the Right-Hand Part of Lavrentiev–Bitsadze Equation. Russ. Math. 2020, 64, 40–57. [Google Scholar] [CrossRef]
  18. Pyatkov, S.G. On Some Classes of Inverse Problems on Determining Source Functions for Heat and Mass Transfer Systems. J. Math. Sci. 2020, 188, 23–42. [Google Scholar]
  19. Shirokova, E.A. Inverse Boundary Value Problem of the Plane Theory of Elasticity. Russ. Math. 2020, 64, 66–73. [Google Scholar] [CrossRef]
  20. Tanana, V.P. On Reducing an Inverse Boundary-Value Problem to the Synthesis of Two Ill-Posed Problems and Their Solution. Numer. Anal. Appl. 2020, 13, 180–192. [Google Scholar] [CrossRef]
  21. Baev, A.V. On the Solution of an Inverse Problem for Equations of Shallow Water in a Pool with Variable Depth. Math. Model. Comput. Simulations 2020, 32, 3–15. [Google Scholar]
  22. Glasko, Y.V. The Inverse Problem of Interpretation of Gravitational and Magnetic Anomalies of Hydrocarbon Deposits. J. Appl. Ind. Math. 2020, 14, 46–55. [Google Scholar] [CrossRef]
  23. Shimelevich, M.I. On the Method of Calculating the Modulus of Continuity of The Inverse Operator and its Modifications with Application to Non-linear Problems of Geoelectrics. Numer. Methods Program. 2020, 21, 350–372. [Google Scholar]
  24. Timonov, A. A novel method for the numerical solution of a hybrid inverse problem of electrical conductivity imaging. Investig. Appl. Math. Inform. Part I Zap. Nauchnykh Semin. POMI 2021, 499, 105–128. [Google Scholar]
  25. Prilepko, A.I.; Orlovsky, D.G.; Vasin, I.A. Methods for Solving Inverse Problems in Mathematical Physics; Marcel Dekker: New York, NY, USA, 2000. [Google Scholar]
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Share and Cite

MDPI and ACS Style

Zamyshlyaeva, A.; Lut, A. Inverse Problem for the Sobolev Type Equation of Higher Order. Mathematics 2021, 9, 1647. https://doi.org/10.3390/math9141647

AMA Style

Zamyshlyaeva A, Lut A. Inverse Problem for the Sobolev Type Equation of Higher Order. Mathematics. 2021; 9(14):1647. https://doi.org/10.3390/math9141647

Chicago/Turabian Style

Zamyshlyaeva, Alyona, and Aleksandr Lut. 2021. "Inverse Problem for the Sobolev Type Equation of Higher Order" Mathematics 9, no. 14: 1647. https://doi.org/10.3390/math9141647

Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.

Article Metrics

Back to TopTop