Startseite Direct and inverse scalar scattering problems for the Helmholtz equation in ℝ m
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Direct and inverse scalar scattering problems for the Helmholtz equation in ℝ m

  • Yury G. Smirnov ORCID logo und Aleksei A. Tsupak ORCID logo EMAIL logo
Veröffentlicht/Copyright: 19. November 2020

Abstract

The boundary value problem for the Helmholtz equation in the m-dimensional free space is considered. The problem is reduced to the Lippmann–Schwinger integral equation over the inhomogeneity domain. The operator of the integral equation is shown to be an invertible Fredholm operator. The inverse coefficient problem is considered. An application of the two-step method reduces the inverse problem to the source-type integral equation with a smooth kernel. Special classes of solutions to this equation are introduced. The uniqueness of a solution to the integral equation of the first kind is proved in the defined function classes.

Award Identifier / Grant number: 18-01-00219 A

Funding statement: This work was supported by the Russian Foundation for Basic Research [grant No. 18-01-00219 A].

A On the fundamental solution to the Helmholtz equation in m

Let r = | x | , where x m and m 2 . The function

(A.1) G m ( r ) = i 4 ( k 0 2 π r ) m / 2 - 1 H m / 2 - 1 ( 1 ) ( k 0 r ) = c m ( 1 / r ) m / 2 - 1 H m / 2 - 1 ( 1 ) ( k 0 r )

is a solution to the equation

( + k 0 2 ) G m ( r ) = - δ ( r ) .

Since

H ν ( 1 ) ( x ) = J ν ( x ) + i N ν ( x ) ,

where J ν and N ν are Bessel and Neumann functions, respectively, then from the relations

J ν ( x ) x ν 2 ν Γ ( ν + 1 ) , x 0 , ν ( - ) ,

and

N ν ( x ) { - 2 π ln 1 x , ν = 0 , - Γ ( ν ) π ( 2 x ) ν , ν > 0 , x + 0 ,

it follows that

(A.2) G m ( r ) G m 0 ( r ) = { 1 2 π ln 1 r , m = 2 , Γ ( m 2 - 1 ) 4 π m / 2 ( 1 r ) m - 2 , m > 2 ,

as r + 0 .

Let us show that the functions G m ( r ) satisfy the radiation conditions (2.4).

Statement A.1.

The function G m ( r ) represented by (A.1) satisfies the conditions

G m ( r ) = O ( ( 1 / r ) m / 2 - 1 / 2 ) ,
r G m ( r ) - i k 0 G m ( r ) = o ( ( 1 / r ) m / 2 - 1 / 2 ) , r .

Proof.

We prove this statement in two steps.

1. Taking into account the asymptotic behavior [21] of Hankel functions at infinity, i.e.,

H ν ( 1 ) ( x ) = 2 π x e i ( x - π 2 ν - π 4 ) + O ( x - 3 / 2 ) , x + ,

we obtain

(A.3) H m / 2 - 1 ( 1 ) ( k 0 r ) = i 2 π k 0 r e i ( k 0 r - π m 4 - π 4 ) + O ( r - 3 / 2 ) , r .

From (A.1) and (A.3) it follows that G m ( r ) = O ( ( 1 / r ) m / 2 - 1 / 2 ) , r .

2. Since

d d x H ν ( 1 ) ( x ) = - H ν + 1 ( 1 ) ( x ) + ν x H ν ( 1 ) ( x ) ,

we obtain

r H ν ( 1 ) ( k 0 r ) = - H ν + 1 ( 1 ) ( k 0 r ) k 0 + ν r H ν ( 1 ) ( k 0 r ) .

Further,

r G m ( r ) = c m [ - ( m / 2 - 1 ) ( 1 / r ) m / 2 H m / 2 - 1 ( 1 ) ( k 0 r ) - ( 1 / r ) m / 2 - 1 H m / 2 ( 1 ) ( k 0 r ) k 0 + ( m / 2 - 1 ) ( 1 / r ) m / 2 H m / 2 - 1 ( 1 ) ( k 0 r ) ]
= - c m ( 1 / r ) m / 2 - 1 H m / 2 ( 1 ) ( k 0 r ) k 0 .

Finally, we arrive at

r G m ( r ) - i k 0 G m ( r ) = - c m ( 1 / r ) m / 2 - 1 H m / 2 ( 1 ) ( k 0 r ) k 0 - i k 0 c m ( 1 / r ) m / 2 - 1 H m / 2 - 1 ( 1 ) ( k 0 r )
= - c m k 0 ( 1 / r ) m / 2 - 1 [ H m / 2 ( 1 ) ( k 0 r ) + i H m / 2 - 1 ( 1 ) ( k 0 r ) ]
= - c m k 0 ( 1 / r ) m / 2 - 1 O ( r - 3 / 2 ) t
(A.4) = o ( ( 1 / r ) m / 2 - 1 / 2 ) ,

as desired. ∎

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Received: 2020-05-26
Accepted: 2020-10-16
Published Online: 2020-11-19
Published in Print: 2022-02-01

© 2020 Walter de Gruyter GmbH, Berlin/Boston

Heruntergeladen am 22.9.2025 von https://www.degruyterbrill.com/document/doi/10.1515/jiip-2020-0060/html
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