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Article

Regularized Solution of the Cauchy Problem in an Unbounded Domain

by
Davron Aslonqulovich Juraev
1,2,
Ali Shokri
3 and
Daniela Marian
4,*
1
Department of Natural Science Disciplines, Higher Military Aviation School of the Republic of Uzbekistan, Karshi 180100, Uzbekistan
2
Department of Mathematics, Anand International College of Engineering, Jaipur 303012, India
3
Department of Mathematics, Faculty of Basic Sciences, University of Maragheh, Maragheh 83111-55181, Iran
4
Department of Mathematics, Technical University of Cluj-Napoca, 28 Memorandumului Street, 400114 Cluj-Napoca, Romania
*
Author to whom correspondence should be addressed.
Symmetry 2022, 14(8), 1682; https://doi.org/10.3390/sym14081682
Submission received: 15 July 2022 / Revised: 8 August 2022 / Accepted: 10 August 2022 / Published: 12 August 2022
(This article belongs to the Special Issue Symmetry and Approximation Methods II)

Abstract

:
In this paper, using the construction of the Carleman matrix, we explicitly find a regularized solution of the Cauchy problem for matrix factorizations of the Helmholtz equation in a three-dimensional unbounded domain.

1. Introduction

A fundamental problem in the theory of differential equations (ordinary and partial) is the determination of a solution that verifies certain initial conditions.
Regarding Cauchy problems, certain questions arise: Does a solution exist (even locally only)? Is this unique? In this case, the solution continuously depends on the initial data, that is, is the problem well posed? The concept of a well-posed problem is connected with investigations by the famous French mathematician Hadamard [1]. The problems that are not well-posed are called ill-posed problems. The theory of ill-posed problems has been the subject of research by many mathematicians in the last years, with applicability in various fields: theoretical physics, optimization of control, astronomy, management and planning, automatic systems, etc., all of which have been influenced by the rapid development of computing technology.
Tikhonov [2] answered certain questions that are posed in the class of ill-posed problems, such as: what does an approximate solution mean, and what algorithm can be used to find such an approximate solution? This involves including additional assumptions. This process is known as regularization. Tikhonov regularization is one of the most commonly used for the regularization of linear ill-posed problems. Lavrent’ev [3,4] also established a regularization method. Based on this method, Yarmukhamedov [5,6] constructed the Carleman functions for the Laplace and Helmholtz, when the data is unknown on a conical surface or a hyper surface. Carleman-type formulas allow a solution to an elliptic equation to be found when the Cauchy data are known only on a part of the boundary of the domain. Carleman [7] obtained a formula for a solution to Cauchy–Riemann equations, on domains of certain forms. Based on [7], Goluzin and Krylov [8] gave a formula for establishing the values of analytic functions on arbitrary domains. The multidimensional case was treated in [9]. The Cauchy problem for elliptic equations was considered by Tarkhanov [10,11]. In [12], the Cauchy problem for the Helmholtz equation in an arbitrary bounded plane domain was considered. Certain boundary value problems and the determination of numerical solutions was investigated in [13,14,15,16,17,18,19,20,21,22,23,24,25]. In [21] is studied the Cauchy problem of a modified Helmholtz equation. An efficient D-N alternating algorithm for solving an inverse problem for Helmholtz equation was investigated in [18]. The Cauchy problem for elliptic equations, was studied in [2,3,4,5,6,7,8,9,10,11] and then it was investigated in [12,26,27,28,29,30,31,32,33,34,35,36,37].
In this article, based on previous works [30,31,32,37], we find an explicit formula for an approximate solution of the Cauchy problem for matrix factorizations of the Helmholtz equation in a three-dimensional unbounded domain of R 3 . The approximate solution formula requires construction of a family of fundamental solutions of the Helmholtz operator in space. This family is parametrized by some entire function K ( z ) . Relying on the works [30,31,32,33,34,35,36,37], we obtain better results, due to a special selection of the function K ( z ) . This helped us to obtain good results when finding an approximate solution based on the Carleman matrix.
Let
ζ = ( ζ 1 , ζ 2 , ζ 3 ) R 3 , η = ( η 1 , η 2 , η 3 ) R 3 ,
ζ = ( ζ 1 , ζ 2 ) R 2 , η = ( η 1 , η 2 ) R 2 ,
and Ω R 3 an unbounded, simply connected domain, having the boundary Ω piecewise smooth, such that Ω = Σ D , where D is the plane η 3 = 0 and Σ is a smooth surface lying in the half-space η 3 > 0 .
The following notations are used in the paper:
r = η ζ , α = η ζ , z = i a 2 + α 2 + η 3 , a 0 ,
ζ = ζ 1 , ζ 2 , ζ 3 T , ζ χ T , χ T = χ 1 χ 2 χ 3 transposed vector χ ,
W ( ζ ) = ( W 1 ( ζ ) , , W n ( ζ ) ) T , v 0 = ( 1 , , 1 ) R n , n = 2 m , m = 3 ,
E ( w ) = w 1 0 0 0 w 2 0 0 0 0 w n diagonal matrix , w = ( w 1 , , w n ) R n .
P ( χ T ) is an n × n matrix, having the elements linear functions with constant coefficients from C , such that
P * ( χ T ) P ( χ T ) = E ( ( χ 2 + λ 2 ) v 0 ) ,
where P * ( χ T ) is the Hermitian conjugate matrix of P ( χ T ) and λ R .
Next, we consider the system
P ζ W ( ζ ) = 0 , η Ω ,
where P ζ is the matrix differential operator of order one.
Additionally, consider the set
S Ω = W : Ω ¯ R n ,
where W is continuous on Ω ¯ = Ω Ω and W satisfies (1).

2. Statement of the Cauchy Problem

We formulate now the following Cauchy problem for the system (1):
Let f : Σ R n be a continuous given function on Σ .
Suppose W ( η ) S ( Ω ) and
W ( η ) Σ = f ( η ) , η Σ .
Our purpose is to determine the function W ( η ) in the domain Ω when its values are known on Σ .
If W ( η ) S ( Ω ) , then
W ( ζ ) = Ω L ( η , ζ ; λ ) W ( η ) d s η , ζ Ω ,
where
L ( η , ζ ; λ ) = E Γ 3 ( λ r ) v 0 P * ζ P ( t T ) ,
t = ( t 1 , t 2 , t 3 ) is the unit exterior normal at a point η on the surface Ω and Γ 3 ( λ r ) denotes the fundamental solution of the Helmholtz equation in R 3 (see, [38]), that is
Γ 3 ( λ r ) = e i λ r 4 π r .
Let K ( z ) be an entire function taking real values for real z ( z = a + i b , a , b R ), satisfying
K ( a ) 0 , sup b 1 b p K ( p ) ( z ) = B ( a , p ) < , < a < , p = 0 , 3 ¯ .
Define
Ψ ( η , ζ ; λ ) = 1 2 π 2 K ( ζ 3 ) 0 Im K ( z ) z ζ 3 cos ( λ a ) a 2 + α 2 d a , for η ζ .
Consider Ψ ( η , ζ ; λ ) in (3) instead Γ 3 ( λ r ) , where
Ψ ( η , ζ ; λ ) = Γ 3 ( λ r ) + G ( η , ζ ; λ ) ,
G ( η , ζ ; λ ) being the regular solution of Helmholtz’s equation with respect to η , including the case η = ζ .
We obtain
W ( ζ ) = Ω L ( η , ζ ; λ ) W ( η ) d s η , ζ Ω ,
where
L ( η , ζ ; λ ) = E Ψ ( η , ζ ; λ ) v 0 P * ζ P ( t T ) .
We generalize (8) for the case when the domain Ω is unbounded.
Hence, in what follows, we consider the domain Ω R 3 be unbounded.
Suppose that Ω is situated inside the layer of smallest width defined by the inequality
0 < η 3 < h , h = π ρ , ρ > 0 ,
and Ω extends to infinity.
Let
Ω R = η : η Ω , η < R , Ω R = Ω Ω R , R > 0 .
Theorem 1.
Let W ( η ) S ( Ω ) . If for each fixed ζ Ω we have the equality
lim R Ω R L ( η , ζ ; λ ) W ( η ) d s η = 0 ,
then (8) is satisfied.
Proof. 
Fix ζ Ω , ζ < R . Using (8) we obtain
Ω L ( η , ζ ; λ ) W ( η ) d s η = Ω R L ( η , ζ ; λ ) W ( η ) d s y + Ω R L ( η , ζ ; λ ) W ( η ) d s η = W ( ζ ) + Ω R L ( η , ζ ; λ ) W ( η ) d s η , ζ Ω R .
Using (9), we obtain (8).
Also assume that the length Ω satisfies the following growth condition
Ω exp d 0 ρ 0 η d s η < , 0 < ρ 0 < ρ ,
for some d 0 > 0 . Suppose W ( η ) S ( Ω ) satisfies
W ( η ) exp exp ρ 2 η , ρ 2 < ρ , η Ω .
We consider in (6):
K ( z ) = exp d i ρ 1 z h 2 d 1 i ρ 0 z h 2 , K ( ζ 3 ) = exp d cos ρ 1 ζ 3 h 2 + d 1 cos i ρ 0 ζ 3 h 2 , 0 < ρ 1 < ρ , 0 < ζ 3 < h ,
where
d = 2 c exp ρ 1 ζ , d 1 > d 0 cos ρ 0 h 2 , c 0 , d > 0 .
Then (8) is valid.
Let ζ Ω be fixed and η . We estimate Ψ ( η , ζ ; λ ) , Ψ ( η , ζ ; λ ) η j , j = 1 , 2 ¯ and Ψ ( η , ζ ; λ ) η 3 . To estimate Ψ ( η , ζ ; λ ) η j , we use the equality
Ψ ( η , ζ ; λ ) η j = Ψ ( η , ζ ; λ ) s s η j = 2 ( η j ζ j ) Ψ ( η , ζ ; λ ) s , j = 1 , 2 ¯ .
Really,
exp d i ρ 1 z h 2 d 1 i ρ 0 z h 2 = exp Re d i ρ 1 z h 2 d 1 i ρ 0 z h 2 = exp d ρ 1 a 2 + α 2 cos ρ 1 η 3 h 2 d 1 ρ 0 a 2 + α 2 cos ρ 0 η 3 h 2 .
As
π 2 ρ 1 ρ · π 2 ρ 1 ρ · π 2 < π 2 , π 2 ρ 1 ρ · π 2 ρ 0 y 3 h 2 ρ 1 ρ · π 2 < π 2 .
Consequently,
cos ρ η 3 h 2 > 0 , cos ρ 0 η 3 h 2 cos h ρ 0 2 > δ 0 > 0 .
It does not vanish in the region Ω and
Ψ ( η , ζ ; λ ) = O exp ε ρ 1 η , ε > 0 , η , η Ω Ω , Ψ ( η , ζ ; λ ) η j = O exp ε ρ 1 η , ε > 0 , η , η Ω Ω , j = 1 , 2 ¯ . Ψ ( η , ζ ; λ ) η 3 = O exp ε ρ 1 η , ε > 0 , η , η Ω Ω .
We now choose ρ 1 with the condition ρ 2 < ρ 1 < ρ . Hence, (10) is satisfied and (8) is true. □
Condition (12) can be weakened.
Denote
S ρ ( Ω ) = W ( η ) : W ( η ) S ( Ω ) , W ( η ) exp O exp ρ η 1 , η , η Ω .
Theorem 2.
If W ( η ) S ρ ( Ω ) satisfies
W ( η ) C exp c cos ρ 1 η 3 h 2 exp ρ 1 η , C constant , c 0 , 0 < ρ 1 < ρ , η Ω ,
then (8) is true.
Proof. 
Divide Ω by a line η 3 = h 2 into the domains
Ω 1 = η : 0 < η 3 < h 2 and Ω 2 = η : h 2 < η 3 < h .
Consider the domain Ω 1 . We put
K 1 ( z ) = K ( z ) exp δ i τ z h 2 δ 1 i ρ z h 2 , ρ < τ < 2 ρ , δ > 0 , δ 1 > 0 ,
in (6), K ( z ) being defined in (12) and we obtain that (10) is valid.
Really,
exp i τ z h 4 δ 1 i ρ z h 4 = exp δ τ a 2 + α 2 cos τ η 3 h 4 = exp δ τ a 2 + α 2 exp δ exp τ η ,
π 2 τ π 4 τ η 3 h 4 τ π 2 < h 2 and cos τ η 3 h 4 cos τ h 4 δ 0 > 0 .
We denote the corresponding Ψ ( η , ζ ; λ ) by Ψ + ( η , ζ ; λ ) .
Since
cos τ η 3 h 4 δ 0 , η Ω 1 Ω 1 ,
then for fixed ζ Ω 1 , η Ω 1 Ω 1 we have
Ψ + ( η , ζ ; λ ) = O exp ( δ 0 exp τ η , η , ρ < τ < 2 ρ , Ψ + ( η , ζ ; λ ) η j = O exp ( δ 0 exp τ η , η , ρ < τ < 2 ρ , j = 1 , 2 ¯ . Ψ + ( η , ζ ; λ ) η 3 = O exp ( δ 0 exp τ η , η , ρ < τ < 2 ρ .
Suppose W ( η ) S ρ ( Ω 1 ) satisfies
W ( η ) C exp exp 2 ρ ε η , ε > 0 , η Ω 1 .
Consider τ in (16) satisfying 2 ρ ε < τ < 2 ρ .
We obtain that (16) is valid in Ω 1 , and we have
W ( ζ ) = Ω 1 L ( η , ζ ; λ ) W ( η ) d s η , ζ Ω 1 .
where
L ( η , ζ ; λ ) = E Ψ + ( η , ζ ; λ ) v 0 P * ζ P ( t T ) .
If W ( η ) S ρ ( Ω 2 ) satisfies (15) in Ω 2 , then for 2 ρ ε < τ < 2 ρ analog we have
W ( ζ ) = Ω 2 L ( η , ζ ; λ ) W ( η ) d s η , ζ Ω 2 ,
where
L ( η , ζ ; λ ) = E Ψ ( η , ζ ; λ ) v 0 P * ζ P ( t T ) ,
and Ψ ( η , ζ ; λ ) it is given by (6), in which K ( z ) it is replaced by the function K 2 ( z ) :
K 2 ( z ) = K ( z ) exp δ i τ z h 1 δ 1 i ρ z h 2 ,
where
h 1 = h 2 + h 4 , h 2 < η 3 < h , h 2 < ζ 3 < h 1 , δ > 0 , δ 1 > 0 .
The integrals converge uniformly for δ 0 , and W ( η ) S ρ ( Ω ) . We consider δ = 0 and we find
W ( ζ ) = Ω L ( η , ζ ; λ ) W ( η ) d s η , ζ Ω , ζ 3 h 2 ,
where
L ( η , ζ ; λ ) = E Ψ ˜ ( η , ζ ; λ ) v 0 P * ζ P ( t T ) ,
Ψ ˜ ( η , ζ ; λ ) = ( Ψ + ( η , ζ ; λ ) ) δ = 0 = ( Ψ ( η , ζ ; λ ) ) δ = 0 .
Here, Ψ ˜ ( η , ζ ; λ ) is given by (6), and K ( z ) by (16), for δ = 0 . According to the continuation principle, Formula (21) is valid for every ζ Ω . Using (18) and (21) holds for every δ 1 0 . Supposing δ 1 = 0 , Theorem 2 is proved. □
We choose
K ( z ) = 1 ( z ζ 3 + 2 h ) 2 exp ( σ z 2 ) , K ( ζ 3 ) = 1 ( 2 h ) 2 exp ( σ ζ 3 2 ) , 0 < ζ 3 < h , h = π ρ ,
in (6) and we obtain
Ψ σ ( η , ζ ; λ ) = e σ ζ 3 2 π 2 ( 2 h ) 1 0 Im exp ( σ z 2 ) ( z ζ 3 + 2 h ) 2 ( z ζ 3 ) cos ( λ a ) a 2 + α 2 d a .
The Formula (8) becomes:
W ( ζ ) = Ω L σ ( η , ζ ; λ ) W ( η ) d s η , ζ Ω ,
where
L σ ( η , ζ ; λ ) = E Ψ σ ( η , ζ ; λ ) v 0 P * ζ P ( t T ) .

3. Regularized Solution of the Problem

Theorem 3.
Let W ( η ) S ρ ( Ω ) satisfying
W ( η ) M , η D .
If
W σ ( ζ ) = Σ L σ ( η , ζ ; λ ) W ( η ) d s η , η Ω ,
then
W ( ζ ) W σ ( ζ ) K ρ ( λ , ζ ) σ 2 M e σ ζ 3 2 , ζ Ω ,
W ( ζ ) ζ j W σ ( ζ ) ζ j K ρ ( λ , ζ ) σ 2 M e σ ζ 3 2 , σ > 1 , ζ Ω , j = 1 , 3 ¯ ,
where K ρ ( λ , ζ ) are bounded on compact subsets of Ω.
Proof. 
From (24) and (26), we obtain
W ( ζ ) = Σ L σ ( η , ζ ; λ ) W ( η ) d s η + D L σ ( η , ζ ; λ ) W ( η ) d s η = W σ ( ζ ) + D L σ ( η , ζ ; λ ) W ( η ) d s η , ζ Ω .
Now using (25), we obtain
W ( ζ ) W σ ( ζ ) D L σ ( η , ζ ; λ ) W ( η ) d s η D L σ ( η , ζ ; λ ) W ( η ) d s η M D L σ ( η , ζ ; λ ) d s η , ζ Ω .
Next, we estimate the integrals D Ψ σ ( η , ζ ; λ ) d s η , D Ψ σ ( η , ζ ; λ ) η j d s η , j = 1 , 2 ¯ and D Ψ σ ( η , ζ ; λ ) ζ 3 d s η on the part D of the plane η 3 = 0 .
Separating the imaginary part of (23), we obtain
Ψ σ ( η , ζ ; λ ) = e σ ( η 3 2 ζ 3 2 ) π 2 ( 2 h 2 ) 1 0 e σ ( u 2 + α 2 ) ( α 1 2 + β 1 2 + 2 β 1 ) cos γ α 1 α 1 2 + β 1 2 2 α 1 2 + β 2 e σ ( u 2 + α 2 ) 2 α 1 2 β 1 + α 1 2 β β 1 2 β α 1 2 + β 1 2 2 α 1 2 + β 2 sin γ α 1 α 1 cos ( λ a ) d a ,
where
γ = 2 σ η 3 , α 1 2 = a 2 + α 2 , β = η 3 ζ 3 , β 1 = η 3 ζ 3 + 2 h .
Given equality (30), we have
D Ψ σ ( η , ζ ; λ ) d s η K ρ ( λ , ζ ) σ 2 e σ ζ 3 2 , σ > 1 , ζ Ω .
Now using the equality
Ψ σ ( η , ζ ; λ ) η j = Ψ σ ( η , ζ ; λ ) s s η j = 2 ( η j ζ j ) Ψ σ ( η , ζ ; λ ) s , s = α 2 , j = 1 , 2 ¯ ,
the equality (30) and (32), we have
D Ψ σ ( η , ζ ; λ ) η j d s η K ρ ( λ , ζ ) σ 2 e σ ζ 3 2 , σ > 1 , ζ Ω , j = 1 , 2 ¯ .
Now, we estimate the integral D Ψ σ ( η , ζ ; λ ) η 3 d s η .
Taking into account equality (30), we obtain
D Ψ σ ( η , ζ ; λ ) η 3 d s η K ρ ( λ , ζ ) σ 2 e σ ζ 3 2 , σ > 1 , ζ Ω ,
From inequalities (29), (31), (33), and (34), we obtain (27).
Now we prove the inequality (28). Taking the derivatives from equalities (24) and (26) with respect to ζ j , j = 1 , 3 ¯ , we obtain:
W ( ζ ) ζ j = Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η , W σ ( ζ ) ζ j = Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η , ζ Ω , j = 1 , 3 ¯
From (25) and (35), we have
W ( ζ ) ζ j σ W ( ζ ) ζ j D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η M D L σ ( η , ζ ; λ ) ζ j d s η , ζ Ω , j = 1 , 3 ¯ .
To prove (36), we estimate D Ψ σ ( η , ζ ; λ ) ζ j d s η , j = 1 , 2 ¯ , and D Ψ σ ( η , ζ ; λ ) ζ 3 d s η , on the part D of the plane η 3 = 0 .
For the first integrals, we use:
Ψ σ ( η , ζ ; λ ) ζ j = Ψ σ ( η , ζ ; λ ) s s ζ j = 2 ( η j ζ j ) Ψ σ ( η , ζ ; λ ) s , s = α 2 , j = 1 , 2 ¯ .
Applying equality (30) and equality (37), we obtain
D Ψ σ ( η , ζ ; λ ) ζ j d s η K ρ ( λ , ζ ) σ 2 e σ ζ 3 2 , σ > 1 , ζ Ω , j = 1 , 2 ¯ .
Now, we estimate the integral D Ψ σ ( η , ζ ; λ ) ζ 3 d s η .
Taking into account equality (30), we obtain
D Ψ σ ( η , ζ ; λ ) ζ 3 d s η K ρ ( λ , ζ ) σ 2 e σ ζ 3 2 , σ > 1 , ζ Ω .
Using (36), (38) and (39), we obtain (27). □
Corollary 1.
For every ζ Ω ,
lim σ W σ ( ζ ) = W ( ζ ) , lim σ W σ ( ζ ) ζ j = W ( ζ ) ζ j , j = 1 , 3 ¯ .
We define Ω ¯ ε as
Ω ¯ ε = ( ζ 1 , ζ 2 , ζ 3 ) Ω , q > ζ 3 ε , q = max D ψ ( ζ ) , 0 < ε < q .
Here, ψ ( ζ ) is a surface. We remark that the set Ω ¯ ε Ω is compact.
Corollary 2.
If ζ Ω ¯ ε , then the families of functions W σ ( ζ ) and W σ ( ζ ) ζ j converge uniformly for σ , that is:
W σ ( ζ ) W ( ζ ) , W σ ( ζ ) ζ j W ( ζ ) ζ j , j = 1 , 3 ¯ .
Remark that E ε = Ω Ω ¯ ε is a boundary layer for this problem, as in the theory of singular perturbations, where there is no uniform convergence.
Suppose that the surface Σ is given by the equation
η m = ψ ( η ) , η R 2 ,
where ψ ( η ) satisfies the condition
ψ ( η ) C < , C = c o n s t .
Consider
q = max D ψ ( η ) , l = max D 1 + ψ 2 ( η ) .
Theorem 4.
If W ( η ) S ρ ( Ω ) satisfies (25), and the inequality
W ( η ) δ , 0 < δ < 1 , η Σ , Σ a s m o o t h s u r f a c e ,
then
W ( ζ ) K ρ ( λ , ζ ) σ 2 M 1 ζ 3 2 q 2 δ ζ 3 2 q 2 , σ > 1 , ζ Ω .
W ( ζ ) ζ j K ρ ( λ , ζ ) σ 2 M 1 ζ 3 2 q 2 δ ζ 3 2 q 2 , σ > 1 , ζ Ω , j = 1 , 3 ¯ .
Proof. 
Using (24), we obtain
W ( ζ ) = Σ L σ ( η , ζ ; λ ) W ( η ) d s η + D L σ ( η , ζ ; λ ) ) W ( η ) d s η , ζ Ω .
We estimate the following
W ( ζ ) Σ L σ ( η , ζ ; λ ) W ( η ) d s η + D L σ ( η , ζ ; λ ) W ( η ) d s η , ζ Ω .
Given inequality (40), we estimate the first integral of inequality (44).
Σ L σ ( η , ζ ; λ ) W ( η ) d s η Σ L σ ( η , ζ ; λ ) W ( η ) d s η δ Σ L σ ( η , ζ ; λ ) d s η , ζ Ω .
We estimate now the integrals Σ Ψ σ ( η , ζ ; λ ) d s η , Σ Ψ σ ( η , ζ ; λ ) η j d s η , j = 1 , 2 ¯ and Σ Ψ σ ( η , ζ ; λ ) η 3 d s η on Σ .
Using (30), we have
Σ Ψ σ ( η , ζ ; λ ) d s η K ρ ( λ , ζ ) σ 2 e σ ( q 2 ζ 3 2 ) , σ > 1 , ζ Ω .
From (30) and (32), we have
Σ Ψ σ ( η , ζ ; λ ) y j d s η K ρ ( λ , ζ ) σ 2 e σ ( q 2 ζ 3 2 ) , σ > 1 , ζ Ω , j = 1 , 2 ¯ .
Using (30), we obtain
Σ Ψ σ ( η , ζ ; λ ) η 3 d s η K ρ ( λ , ζ ) σ 2 e σ ( q 2 ζ 3 2 ) , σ > 1 , ζ Ω .
From (46)–(48) and applying (45), we obtain
Σ L σ ( η , ζ ; λ ) W ( η ) d s η K ρ ( λ , ζ ) σ 2 δ e σ ( q 2 ζ 3 2 ) , σ > 1 , ζ Ω .
The following is known
D L σ ( η , ζ ; λ ) W ( η ) d s η K ρ ( λ , ζ ) σ 2 M e σ ζ 3 2 , σ > 1 , ζ Ω .
Now using (44), (49) and (50), we have
W ( ζ ) K ρ ( λ , ζ ) σ 2 2 ( δ e σ q 2 + M ) e σ ζ 3 2 , σ > 1 , ζ Ω .
Choosing
σ = 1 q 2 ln M δ ,
we obtain (41).
We compute now the partial derivative from Formula (24) with respect to ζ j , j = 1 , 3 ¯ :
W ( ζ ) ζ j = Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η = W σ ( ζ ) ζ j + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η , ζ Ω , j = 1 , 3 ¯ .
Here
W σ ( ζ ) ζ j = Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η .
Now we have
W ( ζ ) ζ j Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η W σ ( ζ ) ζ j + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η , ζ Ω , j = 1 , 3 ¯ .
Given inequality (40), we obtain:
Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η δ Σ L σ ( η , ζ ; λ ) ζ j d s η , ζ Ω , j = 1 , 3 ¯ .
To prove (56), we estimate now Σ Ψ σ ( η , ζ ; λ ) ζ j d s y , j = 1 , 2 ¯ and Σ Ψ σ ( η , ζ ; λ ) ζ 3 d s η on a smooth surface Σ .
Given equality (30) and equality (35), we obtain
Σ Ψ σ ( η , ζ ; λ ) ζ j d s η K ρ ( λ , ζ ) σ 2 e σ ( q 2 ζ 3 2 ) , σ > 1 , ζ Ω , j = 1 , 2 ¯ .
Taking into account (30), we obtain
Σ Ψ σ ( η , ζ ; λ ) ζ 3 d s y K ρ ( λ , ζ ) σ 2 e σ ( q 2 ζ 3 2 ) , σ > 1 , ζ Ω ,
From (57) and (58), bearing in mind (56), we obtain
Σ L σ ( η , ζ ; λ ) ζ j W ( η ) d s η K ρ ( λ , ζ ) σ 2 δ e σ ( q 2 ζ 3 2 ) , σ > 1 , ζ Ω , j = 1 , 3 ¯ .
We have
D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η K ρ ( λ , ζ ) σ 2 M e σ ζ 3 2 , σ > 1 , ζ Ω , j = 1 , 3 ¯ .
From (55), (59) and (60), we obtain
W ( ζ ) ζ j K ρ ( λ , ζ ) σ 2 2 ( δ e σ q 2 + M ) e σ ζ 3 2 , σ > 1 , ζ Ω , j = 1 , 3 ¯ .
Choosing σ as in (52) we get (42). □
Suppose now that W ( η ) S ρ ( Ω ) is defined on Σ and f δ ( η ) is its approximation with an error 0 < δ < 1 . Then
max Σ W ( η ) f δ ( η ) δ .
We put
W σ ( δ ) ( ζ ) = Σ L σ ( η , ζ ; λ ) f δ ( η ) d s η , ζ Ω .
Theorem 5.
Let W ( η ) S ρ ( Ω ) satisfying the condition (25) on the part of the plane η 3 = 0 .
Then
W ( ζ ) W σ ( δ ) ( ζ ) K ρ ( λ , ζ ) σ 2 M 1 ζ 3 2 q 2 δ ζ 3 2 q 2 , σ > 1 , ζ Ω .
W ( ζ ) ζ j W σ ( δ ) ( ζ ) ζ j K ρ ( λ , ζ ) σ 2 M 1 ζ 3 2 q 2 δ ζ 3 2 q 2 , σ > 1 , ζ Ω , j = 1 , 3 ¯ .
Proof. 
From (24) and (63), we obtain
W ( ζ ) W σ ( δ ) ( ζ ) = Ω L σ ( η , ζ ; λ ) W ( η ) d s η Σ L σ ( η , ζ ; λ ) f δ ( η ) d s η = Σ L σ ( η , ζ ; λ ) W ( η ) d s η + D L σ ( η , ζ ; λ ) W ( η ) d s η Σ L σ ( η , ζ ; λ ) f δ ( η ) d s η = Σ L σ ( η , ζ ; λ ) W ( η ) f δ ( η ) d s η + D L σ ( η , ζ ; λ ) W ( η ) d s η .
and
W ( ζ ) ζ j W σ ( δ ) ( ζ ) ζ j = Ω L σ ( η , ζ ; λ ) ζ j W ( η ) d s η Σ L σ ( η , ζ ; λ ) ζ j f δ ( y ) d s y = S N σ ( y , x ; λ ) x j U ( y ) d s y + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η Σ L σ ( η , ζ ; λ ) ζ j f δ ( η ) d s η = Σ L σ ( η , ζ ; λ ) ζ j W ( η ) f δ ( η ) d s η + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η , j = 1 , 3 ¯ .
Using (25) and (62), we obtain:
W ( ζ ) W σ ( δ ) ( ζ ) = Σ L σ ( η , ζ ; λ ) W ( η ) f δ ( η ) d s η + D L σ ( η , ζ ; λ ) W ( η ) d s η Σ L σ ( η , ζ ; λ ) W ( η ) f δ ( η ) d s η + D L σ ( η , ζ ; λ ) W ( η ) d s η δ Σ L σ ( η , ζ ; λ ) d s η + M D L σ ( η , ζ ; λ ) d s η .
and
W ( ζ ) ζ j W σ ( δ ) ( ζ ) ζ j = σ L σ ( η , ζ ; λ ) ζ j W ( η ) f δ ( η ) d s η + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η Σ L σ ( η , ζ ; λ ) ζ j W ( η ) f δ ( η ) d s η + D L σ ( η , ζ ; λ ) ζ j W ( η ) d s η δ Σ L σ ( η , ζ ; λ ) ζ j d s η + M D L σ ( η , ζ ; λ ) ζ j d s η , j = 1 , 3 ¯ .
We obtain, similarly repeating the proof of Theorems 3 and 4, that
W ( ζ ) W σ ( δ ) ( ζ ) K ρ ( λ , ζ ) σ 2 2 ( δ e σ q 2 + M ) e σ ζ 3 2 .
W ( ζ ) ζ j W σ ( δ ) ( ζ ) ζ j K ρ ( λ , ζ ) σ 2 2 ( δ e σ q 2 + M ) e σ ζ 3 2 , j = 1 , 3 ¯ .
Considering σ from (52), we obtain (64) and (65). □
Corollary 3.
For every ζ Ω ,
lim δ 0 W σ ( δ ) ( ζ ) = W ( ζ ) , lim δ 0 W σ ( δ ) ( ζ ) ζ j = W ( ζ ) ζ j , j = 1 , 3 ¯ .
Corollary 4.
If ζ Ω ¯ ε , then the families of functions W σ ( δ ) ( ζ ) and W σ ( δ ) ( ζ ) ζ j are convergent uniformly for δ 0 , that is:
W σ ( δ ) ( ζ ) W ( ζ ) , W σ ( δ ) ( ζ ) ζ j W ( ζ ) ζ j , j = 1 , 3 ¯ .

4. Conclusions

In this paper, as a continuation of some previous papers, we explicitly found a regularized solution of the Cauchy problem for the matrix factorization of the Helmholtz equation in an unbounded domain from R 3 . When applied problems are solved, the approximate values of W ( ζ ) and W ( ζ ) ζ j , ζ Ω , j = 1 , 3 ¯ must be found.
We have built, in this paper, a family of vector-functions W ( ζ , f δ ) = W σ ( δ ) ( ζ ) and W ( ζ , f δ ) ζ j = W σ ( δ ) ( ζ ) ζ j , j = 1 , 3 ¯ , depending on σ . Moreover, we have proved that for σ = σ ( δ ) , at δ 0 , specially chosen, W σ ( δ ) ( ζ ) and W σ ( δ ) ( ζ ) ζ j are convergent to a solution W ( ζ ) and its derivative W ( ζ ) ζ j , ζ Ω . Such a family of vector functions W σ ( δ ) ( ζ ) and W σ ( δ ) ( ζ ) ζ j are called a regularized solution of the problem. A regularized solution determines a stable method to find the approximate solution of the problem.

Author Contributions

Conceptualisation, D.A.J.; methodology, A.S. and D.M.; formal analysis, D.A.J., A.S. and D.M.; writing—original draft preparation, D.A.J., A.S. and D.M. All authors 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

The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University, Abha, Kingdom of Saudi Arabia for funding this work through Large Groups RGP.2/43/43. This work also has been supported by Walailak University Master Degree Excellence Scholarships (Contract No. ME03/2021).

Conflicts of Interest

The authors declare no conflict of interest.

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Juraev, D.A.; Shokri, A.; Marian, D. Regularized Solution of the Cauchy Problem in an Unbounded Domain. Symmetry 2022, 14, 1682. https://doi.org/10.3390/sym14081682

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Juraev DA, Shokri A, Marian D. Regularized Solution of the Cauchy Problem in an Unbounded Domain. Symmetry. 2022; 14(8):1682. https://doi.org/10.3390/sym14081682

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Juraev, Davron Aslonqulovich, Ali Shokri, and Daniela Marian. 2022. "Regularized Solution of the Cauchy Problem in an Unbounded Domain" Symmetry 14, no. 8: 1682. https://doi.org/10.3390/sym14081682

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