Open Access

Two regularization methods for a class of inverse boundary value problems of elliptic type

  • Abdallah Bouzitouna1,
  • Nadjib Boussetila2Email author and
  • Faouzia Rebbani1
Boundary Value Problems20132013:178

DOI: 10.1186/1687-2770-2013-178

Received: 13 March 2013

Accepted: 18 July 2013

Published: 2 August 2013

Abstract

This paper deals with the problem of determining an unknown boundary condition u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq1_HTML.gif in the boundary value problem u y y ( y ) A u ( y ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq2_HTML.gif, u ( 0 ) = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq3_HTML.gif, u ( + ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq4_HTML.gif, with the aid of an extra measurement at an internal point. It is well known that such a problem is severely ill-posed, i.e., the solution does not depend continuously on the data. In order to overcome the instability of the ill-posed problem, we propose two regularization procedures: the first method is based on the spectral truncation, and the second is a version of the Kozlov-Maz’ya iteration method. Finally, some other convergence results including some explicit convergence rates are also established under a priori bound assumptions on the exact solution.

MSC: 35R25, 65J20, 35J25.

Keywords

ill-posed problems elliptic problems cut-off spectral regularization iterative regularization

1 Formulation of the problem

Throughout this paper, H denotes a complex separable Hilbert space endowed with the inner product ( , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq5_HTML.gif and the norm https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq6_HTML.gif, L ( H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq7_HTML.gif stands for the Banach algebra of bounded linear operators on H.

Let A : D ( A ) H H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq8_HTML.gif be a positive, self-adjoint operator with compact resolvent, so that A has an orthonormal basis of eigenvectors ( ϕ n ) H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq9_HTML.gif with real eigenvalues ( λ n ) R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq10_HTML.gif, i.e.,
A ϕ n = λ n ϕ n , n N , ( ϕ i , ϕ j ) = δ i j = { 1 if  i = j , 0 if  i j , 0 < ν λ 1 λ 2 λ 3 , lim n λ n = , h H , h = n = 1 h n ϕ n , h n = ( h , ϕ n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equa_HTML.gif
In this paper, we are interested in the following inverse boundary value problem: find ( u ( y ) , u ( 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq11_HTML.gif satisfying
{ u y y A u = 0 , 0 < y < , u ( 0 ) = f , u ( + ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ1_HTML.gif
(1.1)
where f is the unknown boundary condition to be determined from the interior data
u ( b ) = g H , 0 < b < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ2_HTML.gif
(1.2)

This problem is an abstract version of an inverse boundary value problem, which generalizes inverse problems for second-order elliptic partial differential equations in a cylindrical domain, for example we mention the following problem.

Example 1.1 An example of (1.1) is the boundary value problem for the Laplace equation in the strip ( 0 , π ) × ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq12_HTML.gif, where the operator A is given by
A = 2 x 2 , D ( A ) = H 0 1 ( 0 , π ) H 2 ( 0 , π ) H = L 2 ( 0 , π ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equb_HTML.gif
which takes the form
{ u y y ( x , y ) + u x x ( x , y ) = 0 , x ( 0 , π ) , y ( 0 , + ) , u ( 0 , y ) = u ( π , y ) = 0 , y ( 0 , + ) , u ( x , 0 ) = f ( x ) , u ( x , + ) = 0 , x [ 0 , π ] , u ( x , y = b ) = g ( x ) , x [ 0 , π ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equc_HTML.gif

To our knowledge, there are few papers devoted to this class of problems in the abstract setting, except for [1, 2]. In [3], the author studied a similar problem posed on a bounded interval. In this study, the algebraic invertibility of the inverse problem was established. However, the regularization aspect was not investigated.

We note here that this inverse problem was studied by Levine and Vessella [2], where the authors considered the problem of recovering u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq1_HTML.gif from the experimental data g 1 , , g n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq13_HTML.gif associated to the internal measurements u ( b 1 ) , , u ( b n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq14_HTML.gif, in which the temperature is measured at various depths 0 < b 1 < < b n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq15_HTML.gif as approximate functions g 1 , , g n H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq16_HTML.gif such that
i = 1 n p i u ( b i ) g i 2 ε 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equd_HTML.gif

where p 1 , , p n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq17_HTML.gif are positive weights with i = 1 n p i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq18_HTML.gif and ε denotes the level noisy.

The regularizing strategy employed in [2] is essentially based on the Tikhonov regularization and the conditional stability estimate u y ( 0 ) E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq19_HTML.gif for some a priori constant E.

In practice, the use of N-measurements or the average of a series of measurements is an expensive operation, and sometimes unrealizable. Moreover, the numerical implementation of the stabilized solutions by the Tikhonov regularization method for this class of problems will be a very complex task.

For these reasons, we propose in our study a practical regularizing strategy. We show that we can recover u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq1_HTML.gif from the internal measurement u ( b ) = g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq20_HTML.gif under the conditional stability estimate u ( 0 ) E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq21_HTML.gif for some a priori constant E. Moreover, our investigation is supplemented by numerical simulations justifying the feasibility of our approach.

2 Preliminaries and basic results

In this section we present the notation and the functional setting which will be used in this paper and prepare some material which will be used in our analysis.

2.1 Notation

We denote by C ( H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq22_HTML.gif the set of all closed linear operators densely defined in H. The domain, range and kernel of a linear operator B C ( H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq23_HTML.gif are denoted as D ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq24_HTML.gif, R ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq25_HTML.gif and N ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq26_HTML.gif; the symbols ρ ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq27_HTML.gif, σ ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq28_HTML.gif and σ p ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq29_HTML.gif are used for the resolvent set, spectrum and point spectrum of B, respectively. If V is a closed subspace of H, we denote by Π V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq30_HTML.gif the orthogonal projection from H to V.

For the ease of reading, we summarize some well-known facts in spectral theory.

2.2 Spectral theorem and properties

By the spectral theorem, for each strictly positive self-adjoint operator B,
B : D ( B ) H H , D ( B ) ¯ = H , B = B and ( B u , u ) γ u 2 , u D ( B ) ( γ > 0 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Eque_HTML.gif
there is a unique right continuous family { E λ , λ [ γ , [ } L ( H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq31_HTML.gif of orthogonal projection operators such that B = γ λ d E λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq32_HTML.gif with
D ( B ) = { v H : γ λ 2 d ( E λ v , v ) < } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equf_HTML.gif

Theorem 2.1 [[4], Theorem 6, XII.2.5, pp.1196-1198]

Let { E λ , λ γ > 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq33_HTML.gif be the spectral resolution of the identity associated to B, and let Φ be a complex Borel function defined E-almost everywhere on the real axis. Then Φ ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq34_HTML.gif is a closed operator with dense domain. Moreover,
  1. (i)

    D ( Φ ( B ) ) : = { h H : γ | Φ ( λ ) | 2 d ( E λ v , v ) < } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq35_HTML.gif,

     
  2. (ii)

    ( Φ ( B ) h , y ) = γ Φ ( λ ) d ( E λ h , y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq36_HTML.gif, h D ( Φ ( B ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq37_HTML.gif, y H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq38_HTML.gif,

     
  3. (iii)

    Φ ( B ) h 2 = γ | Φ ( λ ) | 2 d ( E λ h , h ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq39_HTML.gif, h D ( Φ ( B ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq37_HTML.gif,

     
  4. (iv)

    Φ ( B ) = Φ ¯ ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq40_HTML.gif. In particular, if Φ is a real Borel function, then Φ ( B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq34_HTML.gif is self-adjoint.

     

We denote by S ( y ) = e y A = n = 1 + e y λ n ( , ϕ n ) ϕ n L ( H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq41_HTML.gif, y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq42_HTML.gif, the C 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq43_HTML.gif-semigroup generated by A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq44_HTML.gif. Some basic properties of S ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq45_HTML.gif are listed in the following theorem.

Theorem 2.2 (see [5], Chapter 2, Theorem 6.13, p.74)

For this family of operators, we have:
  1. 1.

    S ( y ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq46_HTML.gif, y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq47_HTML.gif;

     
  2. 2.

    the function y S ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq48_HTML.gif, y > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq49_HTML.gif, is analytic;

     
  3. 3.

    for every real r 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq50_HTML.gif and y > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq51_HTML.gif, the operator S ( y ) L ( H , D ( A r / 2 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq52_HTML.gif;

     
  4. 4.

    for every integer k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq53_HTML.gif and y > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq54_HTML.gif, S ( k ) ( y ) = A k / 2 S ( y ) c ( k ) y k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq55_HTML.gif;

     
  5. 5.

    for every h D ( A r / 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq56_HTML.gif, r 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq50_HTML.gif, we have S ( t ) A r / 2 h = A r / 2 S ( y ) h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq57_HTML.gif.

     

Theorem 2.3 For y > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq51_HTML.gif, S ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq45_HTML.gif is self-adjoint and one-to-one operator with dense range ( S ( y ) = S ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq58_HTML.gif, R ( S ( y ) ) ¯ = H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq59_HTML.gif).

Proof Let ϕ y : [ 0 , + [ R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq60_HTML.gif, s ϕ y ( s ) = e y s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq61_HTML.gif. Then, by virtue of (iv) of Theorem 2.1, we can write S ( y ) = ϕ y ¯ ( A ) = ϕ y ( A ) = e y A = S ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq62_HTML.gif.

Let h N ( S ( y 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq63_HTML.gif, y 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq64_HTML.gif, then S ( y 0 ) h = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq65_HTML.gif, which implies that S ( y ) S ( y 0 ) h = S ( t + t 0 ) h = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq66_HTML.gif, y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq67_HTML.gif. Using analyticity, we obtain that S ( y ) h = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq68_HTML.gif, y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq42_HTML.gif. Strong continuity at 0 now gives h = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq69_HTML.gif. This shows that N ( S ( y 0 ) ) = { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq70_HTML.gif.

Thanks to
R ( S ( y 0 ) ) ¯ = N ( S ( y 0 ) ) = { 0 } = H , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equg_HTML.gif

we conclude that R ( S ( y 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq71_HTML.gif is dense in H. □

Remark 2.1 For y = b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq72_HTML.gif, this theorem ensures that S ( b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq73_HTML.gif is self-adjoint and one-to-one operator with dense range R ( S ( b ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq74_HTML.gif. Then we can define its inverse S ( b ) 1 = e b A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq75_HTML.gif, which is an unbounded self-adjoint strictly positive definite operator in H with dense domain
D ( S ( b ) 1 ) = R ( S ( b ) ) = { h H : e b A h 2 = n = 1 + e b λ n | ( h , ϕ n ) | 2 < + } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equh_HTML.gif
Let us consider the following problem: for ξ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq76_HTML.gif find v C 1 ( ] 0 , + [ ; H ) C ( [ 0 , + [ ; H ) C ( ] 0 , + [ ; D ( A ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq77_HTML.gif such that
v ( y ) + A u ( y ) = 0 , 0 < y < + , v ( 0 ) = ξ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ3_HTML.gif
(2.1)

Theorem 2.4 [[6], Theorem 7.5, p.191]

For any ξ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq76_HTML.gif, problem (2.1) has a unique solution, given by
v ( y ) = S ( y ) ξ = n = 1 e y λ n ( ξ , ϕ n ) ϕ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ4_HTML.gif
(2.2)
Moreover, for all integer k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq78_HTML.gif, v C ( ] 0 , + [ ; D ( A k / 2 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq79_HTML.gif. If, in addition, ξ D ( A j / 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq80_HTML.gif, then v C ( [ 0 , + [ ; D ( A j / 2 ) ) C j ( [ 0 , + [ ; H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq81_HTML.gif and
k , j N , d ( k + j ) d y v ( y ) = A k / 2 u ( y ) ( j ) c ( k ) y k A j / 2 ξ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equi_HTML.gif

On the other hand, Theorem 2.4 provides smoothness results with respect to y: v C ( ] 0 , + [ ; H ) C j ( [ 0 , + [ ; H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq82_HTML.gif whenever ξ D ( A j / 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq83_HTML.gif, j N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq84_HTML.gif. Under this same hypothesis, we also have smoothness in space: v C ( [ 0 , + [ ; D ( A j / 2 ) ) C j k ( [ 0 , + [ ; D ( A k / 2 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq85_HTML.gif, k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq86_HTML.gif.

Here we recall a crucial theorem in the analysis of the inverse problems.

Theorem 2.5 [[7], Generalized Picard theorem, p.502]

Let B : D ( B ) H H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq87_HTML.gif be a self-adjoint operator and the Hilbert space H, and let E μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq88_HTML.gif be its spectral resolution of unity. Let θ C ( R , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq89_HTML.gif and Z ( θ ) : = { t R : θ ( t ) = 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq90_HTML.gif. We suppose that the set Z ( θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq91_HTML.gif either is empty or contains isolated point only. Then the vectorial equation
θ ( B ) φ = ψ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equj_HTML.gif
is solvable if and only if
R | θ ( λ ) | 2 d | E λ ψ | 2 < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equk_HTML.gif
Moreover,
N ( θ ( B ) ) = { 0 } σ p ( B ) Z ( θ ) = . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equl_HTML.gif
On the basis { ϕ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq92_HTML.gif, we introduce the Hilbert scale ( H s ) s R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq93_HTML.gif (resp. ( E s ) s R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq94_HTML.gif) induced by https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq95_HTML.gif as follows:
H s = { h H : n = 1 λ n 2 s | ( h , φ n ) | 2 < + } , E s = { h H : n = 1 e 2 b s λ n | ( h , φ n ) | 2 < + } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equm_HTML.gif

2.3 Non-expansive operators

Definition 2.1 A linear operator M L ( H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq96_HTML.gif is called non-expansive if
M 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equn_HTML.gif

Theorem 2.6 [[8], Theorem 2.2]

Let M L ( H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq96_HTML.gif be a positive, self-adjoint operator with M 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq97_HTML.gif. Putting V 0 = N ( M ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq98_HTML.gif and V 1 = N ( I M ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq99_HTML.gif, we have
s lim n + M n = Π V 1 , s lim n + ( I M ) n = Π V 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equo_HTML.gif
i.e.,
h H , lim n + M n h = Π V 1 h , lim n + ( I M ) n h = Π V 0 h . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equp_HTML.gif

For more details concerning the theory of non-expansive operators, we refer to Krasnosel’skii et al. [[9], p.66].

Let use consider the operator equation
S φ = ( I M ) φ = ψ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ5_HTML.gif
(2.3)

for non-expansive operators M.

Theorem 2.7 Let M be a linear self-adjoint, positive and non-expansive operator on H. Let ψ ˆ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq100_HTML.gif be such that equation (2.3) has a solution φ ˆ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq101_HTML.gif. If 1 is not an eigenvalue of M, i.e., ( I M ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq102_HTML.gif is injective ( V 1 = N ( I M ) = { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq103_HTML.gif), then the successive approximations
φ n + 1 = M φ n + ψ ˆ , n = 0 , 1 , 2 , , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equq_HTML.gif

converge to φ ˆ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq101_HTML.gif for any initial data φ 0 H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq104_HTML.gif.

Proof From the hypothesis and by virtue of Theorem 2.6, we have
φ 0 H , M n φ 0 Π V 1 φ 0 = Π { 0 } φ 0 = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ6_HTML.gif
(2.4)
By induction with respect to n, it is easily seen that φ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq105_HTML.gif has the explicit form
φ n = M n φ 0 + j = 0 n 1 M j ψ ˆ = M n φ 0 + ( I M n ) ( I M ) 1 ψ ˆ = M n φ 0 + ( I M n ) φ ˆ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equr_HTML.gif
and (2.4) allows us to conclude that
φ ˆ φ n = M n ( φ 0 φ ˆ ) 0 , n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ7_HTML.gif
(2.5)

 □

Remark 2.2 In many situations, some boundary value problems for partial differential equations which are ill-posed can be reduced to Fredholm operator equations of the first kind of the form B φ = ψ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq106_HTML.gif, where B is compact, positive, and self-adjoint operator in a Hilbert space H. This equation can be rewritten in the following way:
φ = ( I ω B ) φ + ω ψ = L φ + ω ψ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equs_HTML.gif

where L = ( I ω B ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq107_HTML.gif, and ω is a positive parameter satisfying ω < 1 B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq108_HTML.gif. It is easily seen that the operator L is non-expansive and 1 is not an eigenvalue of L. It follows from Theorem 2.7 that the sequence { φ n } n = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq109_HTML.gif converges and ( I ω B ) n ζ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq110_HTML.gif for every ζ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq111_HTML.gif as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq112_HTML.gif.

3 Ill-posedness and stabilization of the inverse boundary value problem

3.1 Cauchy problem with Dirichlet conditions

Consider the following well-posed boundary value problem:
{ v y y A v = 0 , 0 < y < , v ( 0 ) = ξ , v ( + ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ8_HTML.gif
(3.1)

where ξ is an H-valued function.

Definition 3.1 [[10], p.250]

  • A function v : [ 0 , + [ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq113_HTML.gif is called a generalized solution to equation (3.1) if v E g = C ( [ 0 , + [ ; H ) C 2 ( ] 0 , + [ ; H ) C 1 ( [ 0 , + [ ; H 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq114_HTML.gif, and for all y ] 0 , + [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq115_HTML.gif, u ( y ) D ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq116_HTML.gif and obeys equation (3.1) on the same interval ] 0 , + [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq117_HTML.gif.

  • A function v : [ 0 , + [ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq113_HTML.gif is called a classical solution to equation (3.1) if v E c = C 1 ( [ 0 , + [ ; H ) C 2 ( ] 0 , + [ ; H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq118_HTML.gif, and for all y ] 0 , + [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq115_HTML.gif, u ( y ) D ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq116_HTML.gif and obeys equation (3.1) on the same interval ] 0 , + [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq117_HTML.gif.

Theorem 3.1 Problem (3.1) admits a unique generalized (resp. classical) solution if and only if ξ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq76_HTML.gif (resp. ξ H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq119_HTML.gif).

Proof By using the Fourier expansion and the given Dirichlet boundary conditions
v ( y ) = n = 1 + v n ( y ) ϕ n , v ( 0 ) = n = 1 + v n ( 0 ) ϕ n = ξ = n = 1 + ξ n ϕ n , v ( + ) = n = 1 + v n ( + ) ϕ n = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equt_HTML.gif
we obtain
{ v n λ n v n ( y ) = 0 , 0 < y < , v n ( 0 ) = ξ n , v n ( + ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ9_HTML.gif
(3.2)
This differential equation admits two linearly independent fundamental solutions
φ n + ( y ) = e + y λ n , φ n ( y ) = e y λ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equu_HTML.gif
Thus, its general solution can be written as
v n ( y ) = c n + e + y λ n + c n e y λ n , c n + , c n R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equv_HTML.gif
Applying v n ( + ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq120_HTML.gif and v n ( 0 ) = ξ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq121_HTML.gif yields c n + = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq122_HTML.gif and c n = ξ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq123_HTML.gif. Finally, the solution of (3.2) is
v ( y ) = S ( y ) ξ = e y A ξ = n = 1 + e y λ n ξ n ϕ n , ξ n = ( ξ , ϕ n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ10_HTML.gif
(3.3)
Remark 3.1 It is easy to check that the expression (3.3) solves the problem
u ( y ) + A u ( y ) = 0 , y ] 0 , + [ , u ( 0 ) = ξ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equw_HTML.gif

If ξ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq76_HTML.gif (resp. ξ H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq119_HTML.gif), by virtue of Theorem 2.4 and Remark 3.1, we easily check the inclusion v E g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq124_HTML.gif (resp. v E c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq125_HTML.gif) and v ( y ) D ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq126_HTML.gif for y ] 0 , + [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq115_HTML.gif. □

3.2 Inverse boundary value problem

Our inverse problem is to determine v ( 0 ) = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq127_HTML.gif from the supplementary condition v ( b ) = g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq128_HTML.gif, then we get
v ( b ) = n = 1 + e b λ n f n ϕ n = g = n = 1 + g n ϕ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ11_HTML.gif
(3.4)
We define
K = S ( b ) : H H , h K h = n = 1 + e b λ n h n ϕ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ12_HTML.gif
(3.5)
The operator equation (3.5) is the main instrument in investigating problem (3.4). More precisely, we want to study the following properties:
  1. 1.

    Injectivity of K (identifiability);

     
  2. 2.

    Continuity of K and the existence of its inverse (stability);

     
  3. 3.

    The range of K.

     
It is easy to see that K is a linear compact self-adjoint operator with the singular values ( σ k = e b λ k ) k = 1 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq129_HTML.gif, and by virtue of Remark 2.1, we have
1 . N ( K ) = { 0 } , 2 . R ( K ) = { h H : e b A h 2 = n = 1 + e 2 b λ n | ( h , ϕ n ) | 2 < + } , 3 . R ( K ) ¯ = H . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equx_HTML.gif

Now, to conclude the solvability of problem (3.4) it is enough to apply Theorem 2.5.

Corollary 3.1 The inverse problem (3.4) is uniquely solvable if and only if
u ( b ) = g R ( K ) = { h H : n = 1 + e 2 b λ n | ( h , ϕ n ) | 2 < + } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ13_HTML.gif
(3.6)
In this case, we have
f = u ( 0 ) = K 1 g = n = 1 + e b λ n g n ϕ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ14_HTML.gif
(3.7)
In other words, the solution f of the inverse problem is obtained from the data g via the unbounded operator L = K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq130_HTML.gif defined on functions g in the subspace
D ( L ) = { g H : n = 1 + e 2 b λ n | ( g , ϕ n ) | 2 < + , g n = ( g , ϕ n ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equy_HTML.gif
Corollary 3.2 Problem (1.1)-(1.2) admits a unique solution u C ( [ 0 , + [ ; H ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq131_HTML.gif if and only if
u ( 0 ) H g R ( K ) = { h H : n = 1 + e b λ n | ( h , ϕ n ) | 2 < + } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equz_HTML.gif
In this case, we have
u ( y ) = e ( b y ) A g = n = 1 + e ( b y ) λ n g n ϕ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ15_HTML.gif
(3.8)

From this representation, we see that:

  • u ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq132_HTML.gif is stable in the interval [ b , + [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq133_HTML.gif ( sup y [ b , + [ u ( y ) g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq134_HTML.gif);

  • u is unstable in [ 0 , b [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq135_HTML.gif. This follows from the high-frequency ω n = e ( b y ) λ n + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq136_HTML.gif, n + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq137_HTML.gif.

3.3 Regularization by truncation method and error estimates

A natural way to stabilize the problem is to eliminate all the components of large n from the solution and instead consider (3.7) only for n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq138_HTML.gif.

Definition 3.2 For N > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq139_HTML.gif, the regularized solution of problem (1.1)-(1.2) is given by
f N = n N e b λ n g n ϕ n , g n = ( g , ϕ n ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ16_HTML.gif
(3.9)
u N ( y ) = n N e ( b y ) λ n g n ϕ n , g n = ( g , ϕ n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ17_HTML.gif
(3.10)

Remark 3.2 If the parameter N is large, f N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq140_HTML.gif is close to the exact solution f. On the other hand, if the parameter N is fixed, f N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq140_HTML.gif is bounded. So, the positive integer N plays the role of regularization parameter.

Remark 3.3 In view of
u ( y ) u N ( y ) = S ( y ) ( f f N ) ( f f N ) u u N ( f f N ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equaa_HTML.gif
and if g E 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq141_HTML.gif, i.e., n = 1 e 2 b λ n | ( g , ϕ n ) | 2 < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq142_HTML.gif, then
f f N 0 , N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equab_HTML.gif
implies
u u N = sup y [ 0 , + [ u ( y ) u N ( y ) 0 , N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equac_HTML.gif

Since the data g are based on (physical) observations and are not known with complete accuracy, we assume that g and g δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq143_HTML.gif satisfy g g δ δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq144_HTML.gif, where g δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq143_HTML.gif denotes the measured data and δ denotes the level noisy.

Let ( f N δ , u N δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq145_HTML.gif denote the regularized solution of problem (1.1), (1.2) with measured data  g δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq146_HTML.gif:
f N δ = n N e b λ n g n δ ϕ n , g n δ = ( g δ , ϕ n ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ18_HTML.gif
(3.11)
u N δ ( y ) = n N e ( b y ) λ n g n δ ϕ n , g n δ = ( g δ , ϕ n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ19_HTML.gif
(3.12)
As usual, in order to obtain convergence rate, we assume that there exists an a priori bound for problem (1.2)
A r / 2 f 2 E 2 < + n = 1 + λ n r e 2 b λ n | g n | 2 E 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ20_HTML.gif
(3.13)

where E > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq147_HTML.gif is a given constant.

Remark 3.4 For given two exact conditions g 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq148_HTML.gif and g 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq149_HTML.gif, let f 1 , N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq150_HTML.gif and f 2 , N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq151_HTML.gif be the corresponding regularized solutions, respectively. Then
f 2 , N f 1 , N 2 = n N e 2 b λ n | ( g 2 g 1 ) k | 2 e 2 b λ N g 2 g 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ21_HTML.gif
(3.14)

The main theorem of this method is as follows.

Theorem 3.2 Let f N δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq152_HTML.gif be the regularized solution given by (3.11), and let f be the exact solution given by (3.7). If A r / 2 f E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq153_HTML.gif, r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq154_HTML.gif and if we choose λ N θ b log ( 1 δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq155_HTML.gif, 0 < θ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq156_HTML.gif, then we have the error bound
f f N δ ( b θ ) r ( 1 log ( 1 / δ ) ) r E + δ 1 θ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ22_HTML.gif
(3.15)
Proof From direct computations, we have
Δ 1 = f N f N δ e b λ N g g δ e b λ N δ , Δ 2 2 = f f N 2 = n = N + 1 + e 2 b λ n | g n | 2 = n = N + 1 + 1 λ n 2 r λ n 2 r e 2 b λ n | g n | 2 1 λ N + 1 2 r n = N + 1 + λ n 2 r e 2 b λ n | g n | 2 ( 1 λ N ) 2 r E 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equad_HTML.gif
Using the triangle inequality
f f N δ f f N + f N f N δ = Δ 1 + Δ 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equae_HTML.gif
we obtain
f f N δ ( 1 λ N ) r E + e b λ N δ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ23_HTML.gif
(3.16)
By choosing λ N = θ b log ( 1 δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq157_HTML.gif, 0 < θ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq156_HTML.gif, we obtain
f f N δ ( b θ ) r ( 1 log ( 1 / δ ) ) r E + δ 1 θ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equaf_HTML.gif

 □

Finally, from (3.4) and (3.15), we deduce the following corollary.

Corollary 3.3 Let u N δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq158_HTML.gif be the regularized solution given by (3.12), and let u be the exact solution given by (3.8). If A r / 2 f E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq153_HTML.gif, r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq154_HTML.gif and if we choose λ N = θ b log ( 1 δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq157_HTML.gif, 0 < θ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq156_HTML.gif, then we have the error bound
u u N δ = sup y [ 0 , + [ u ( y ) u N δ ( y ) f f N δ ( b θ ) r ( 1 log ( 1 / δ ) ) r E + δ 1 θ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ24_HTML.gif
(3.17)

4 Regularization by the Kozlov-Maz’ya iteration method and error estimates

In [11, 12] Kozlov and Maz’ya proposed an alternating iterative method to solve boundary value problems for general strongly elliptic and formally self-adjoint systems. After that, the idea of this method has been successfully used for solving various classes of ill-posed (elliptic, parabolic and hyperbolic) problems; see, e.g., [1315].

In this section we extend this method to our ill-posed problem.

4.1 Description of the method

The iterative algorithm for solving the inverse problem (1.1)-(1.2) starts by letting f 0 H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq159_HTML.gif be arbitrary. The first approximation u 0 ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq160_HTML.gif is the solution to the direct problem
{ u y y 0 A u 0 = 0 , 0 < y < , u 0 ( 0 ) = f 0 , u ( + ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ25_HTML.gif
(4.1)
If the pair ( u k , f k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq161_HTML.gif has been constructed, let
( P ) k + 1 : f k + 1 = f k ω ( u k ( b ) f ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ26_HTML.gif
(4.2)
where ω is such that
0 < ω < 1 K = e b λ 1 , K = sup n e b λ n = e b λ 1 < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equag_HTML.gif
Finally, we get u k + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq162_HTML.gif by solving the problem
{ u y y k + 1 A u k + 1 = 0 , 0 < y < , u k + 1 ( 0 ) = f k + 1 , u k + 1 ( + ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ27_HTML.gif
(4.3)
We set G = ( I ω K ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq163_HTML.gif. If we iterate backwards in ( P ) k + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq164_HTML.gif, we obtain
f k = G k f 0 + ω i = 0 k 1 G i g = G k f 0 + ( I G k ) K 1 g = G k f 0 + f G k f . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ28_HTML.gif
(4.4)
This implies that
f k f = G k ( f 0 f ) , u k ( y ) u ( y ) = S ( y ) G k ( f 0 f ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ29_HTML.gif
(4.5)

Proposition 4.1 The operator G = ( I ω K ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq165_HTML.gif is self-adjoint and non-expansive on H. Moreover, it has not 1 as eigenvalue.

Proof The self-adjointness follows from the definition of G (see Theorem 2.1). Since the inequality 0 < 1 ω e b λ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq166_HTML.gif for λ σ ( A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq167_HTML.gif, we have σ p ( G ) ] 0 , 1 [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq168_HTML.gif, then 1 is not an eigenvalue of G. □

In general, the exact solution u ( 0 ) = f H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq169_HTML.gif is required to satisfy the so-called source condition; otherwise, the convergence of the regularization method approximating the problem can be arbitrarily slow. Since our problem is exponentially ill-posed (the eigenvalues s n = e b λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq170_HTML.gif of K converge exponentially to 0), it is well known in this case [16, 17] that the best choice to accelerate the convergence of the regularization method is to use logarithmic-type source conditions, i.e.,
( f 0 f ) = Ψ β ( ω K ) ξ , ξ H , ξ E , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ30_HTML.gif
(4.6)
where
Ψ β ( t ) = { ln ( e t ) β , 0 < t 1 , 0 , t = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equah_HTML.gif

with β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq171_HTML.gif.

Remark 4.1 [[16], p.34]

The logarithmic source condition ζ = ( f 0 f ) R ( Ψ β ( ω K ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq172_HTML.gif is equivalent to the inclusion ζ R ( A β / 2 ) = D ( A β / 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq173_HTML.gif.

Proof The proof is based on the following equivalence:
k = 1 ( ln ( e ω ) + λ n ) 2 β < + k = 1 ( λ n ) 2 β < + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equai_HTML.gif

 □

Lemma 4.1 [[18], Appendix, Lemma A.1]

Let β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq171_HTML.gif and k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq174_HTML.gif, k 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq175_HTML.gif. Then the real-valued function τ ( t ) = ( 1 t ) k ln ( e t ) β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq176_HTML.gif defined on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq177_HTML.gif satisfies
τ ( t ) C ln ( k ) β . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ31_HTML.gif
(4.7)
Remark 4.2 Let k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq178_HTML.gif. Then the real-valued function ϱ ( t ) = 1 ( 1 t ) k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq179_HTML.gif defined on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq177_HTML.gif satisfies
ϱ ( t ) k t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ32_HTML.gif
(4.8)
Proof Using the mean value theorem, we can write
ϱ ( t ) ϱ ( 0 ) = ( t 0 ) ϱ ( t ˆ ) , 0 < t ˆ < t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equaj_HTML.gif
then
ϱ ( t ) = t k ( 1 t ˆ ) k 1 k t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equak_HTML.gif

 □

Let us consider the following real-valued functions:
Q ( λ ) = ( 1 ω e b λ ) k ln ( e ω e b λ ) β , λ [ λ 1 , + [ , P ( λ ) = ω i = 0 k 1 ( 1 ω e b λ ) i = ω 1 ( 1 ω e b λ ) k ω e b λ , λ [ λ 1 , + [ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equal_HTML.gif
Using the change of variables t = ϑ ( λ ) = ω e b λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq180_HTML.gif, we obtain
Q ˆ ( t ) = Q ( ϑ 1 ( t ) ) = ( 1 t ) k ln ( e t ) β , t [ 0 , 1 ] , P ˆ ( t ) = P ( ϑ 1 ( t ) ) = { ω 1 ( 1 t ) k t , t ] 0 , 1 ] , ω k , t = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equam_HTML.gif

Now we are in a position to state the main result of this method.

Theorem 4.1 Let g E 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq141_HTML.gif and ω satisfy 0 < ω < e b λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq181_HTML.gif, let f 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq182_HTML.gif be an arbitrary element for the iterative procedure suggested above, and let u k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq183_HTML.gif be the kth approximate solution. Then we have
sup y [ 0 , + [ u ( y ) u k ( y ) 0 , k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ33_HTML.gif
(4.9)
Moreover, if ( f 0 f ) H β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq184_HTML.gif ( β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq171_HTML.gif), i.e., ( f 0 f ) = Ψ β ( ω K ) ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq185_HTML.gif, ξ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq76_HTML.gif, ξ E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq186_HTML.gif, then the rate of convergence of the method is given by
sup y [ 0 , + [ u ( y ) u k ( y ) C E ( 1 ln ( k ) ) β , k 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ34_HTML.gif
(4.10)
Proof By virtue of Proposition 4.1 and Theorem 2.7, it follows immediately
sup y [ 0 , + [ u ( y ) u k ( y ) G k ( f 0 f ) 0 , k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equan_HTML.gif
We have
u ( y ) u k ( y ) 2 = S ( y ) G k ( f 0 f ) 2 G k ( f 0 f ) 2 = n = 1 Q ( λ n ) 2 ( ξ , ϕ n ) 2 ( sup t [ 0 , 1 ] Q ˆ ( t ) ) 2 ξ 2 ( sup t [ 0 , 1 ] Q ˆ ( t ) ) 2 E 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equao_HTML.gif

and by virtue of Lemma 4.1 (estimate (4.7)), we conclude the desired estimate. □

Theorem 4.2 Let g E 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq141_HTML.gif and ω satisfy 0 < ω < e b λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq187_HTML.gif, let f 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq182_HTML.gif be an arbitrary element for the iterative procedure suggested above, and let u k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq183_HTML.gif (resp. u k δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq188_HTML.gif) be the kth approximate solution for the exact data g (resp. for the inexact data g δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq143_HTML.gif) such that g g δ δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq189_HTML.gif. Then, under condition (4.6), the following inequality holds:
sup y [ 0 , + [ u ( y ) u δ k ( y ) C E ( 1 ln ( k ) ) β + ε ( k ) δ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equap_HTML.gif

where ε ( k ) = ω i = 0 k 1 ( I ω K ) i k ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq190_HTML.gif.

Proof Using (4.4) and the triangle inequality, we can write
f k = G k f 0 + ω i = 0 k 1 G i g , u k ( y ) = S ( y ) f k , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ35_HTML.gif
(4.11)
f δ k = G k f 0 + ω i = 0 k 1 G i g δ , u k δ ( y ) = S ( y ) f δ k , u ( y ) u δ k ( y ) = ( u ( y ) u k ( y ) ) + ( u k ( y ) u δ k ( y ) ) Δ 1 + Δ 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ36_HTML.gif
(4.12)
where
Δ 1 = u ( y ) u k ( y ) u ( y ) u k ( y ) C E ( 1 ln ( k ) ) β , k 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ37_HTML.gif
(4.13)
and
Δ 2 = u k ( y ) u δ k ( y ) = S ( y ) ( f k f δ k ) = ω S ( y ) i = 0 k 1 G i ( g g δ ) ω i = 0 k 1 G i ( g g δ ) ω i = 0 k 1 G i δ = Δ ˆ 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equaq_HTML.gif
By using inequality (4.8), the quantity Δ ˆ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq191_HTML.gif can be estimated as follows:
Δ ˆ 2 ω k δ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ38_HTML.gif
(4.14)

Combining (4.13) and (4.14) and taking the supremum with respect to y [ 0 , + [ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq192_HTML.gif of u ( y ) u δ k ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq193_HTML.gif, we obtain the desired bound.

Remark 4.3 Choosing k = k ( δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq194_HTML.gif such that ω k δ + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq195_HTML.gif as δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq196_HTML.gif, we obtain
sup y [ 0 , + [ u k ( y ) u δ k ( y ) 0 as  k + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equar_HTML.gif

 □

5 Numerical results

In this section we give a two-dimensional numerical test to show the feasibility and efficiency of the proposed methods. Numerical experiments were carried out using MATLAB.

We consider the following inverse problem:
{ u y y ( x , y ) + u x x ( x , y ) = 0 , x ( 0 , π ) , y ( 0 , + ) , u ( 0 , y ) = u ( π , y ) = 0 , y ( 0 , + ) , u ( x , 0 ) = f ( x ) , u ( x , + ) = 0 , x [ 0 , π ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ39_HTML.gif
(5.1)

where f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq197_HTML.gif is the unknown source and u ( x , 1 ) = g ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq198_HTML.gif is the supplementary condition.

It is easy to check that the operator
A = 2 x 2 , D ( A ) = H 0 1 ( 0 , π ) H 2 ( 0 , π ) H = L 2 ( 0 , π ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equas_HTML.gif

is positive, self-adjoint with compact resolvent (A is diagonalizable).

The eigenpairs ( λ n , ϕ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq199_HTML.gif of A are
λ n = n 2 , ϕ n ( x ) = 2 π sin ( n x ) , n N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equat_HTML.gif
In this case, formula (3.7) takes the form
f ( x ) = u ( x , 0 ) = K 1 g ( x ) = 2 π n = 1 + e n ( 0 π g ( x ) sin ( n x ) d x ) sin ( n x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ40_HTML.gif
(5.2)

Truncation method

We use trapezoid’s rule to approach the integral and do an approximate truncation for the series by choosing the sum of the front M + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq200_HTML.gif terms. After considering an equidistant grid 0 = x 1 < < x M + 1 = π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq201_HTML.gif, x j = ( j 1 ) h = ( j 1 ) π M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq202_HTML.gif, j = 1 ( M + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq203_HTML.gif, we get
f ( x j ) = 2 π i = 1 M + 1 n = 1 + e n ( h g ( x i ) sin ( n x i ) ) sin ( n x j ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ41_HTML.gif
(5.3)
f N ( x j ) = 2 π i = 1 M + 1 n = 1 N e n ( h g ( x i ) sin ( n x i ) ) sin ( n x j ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ42_HTML.gif
(5.4)
f N δ ( x j ) = 2 π i = 1 M + 1 n = 1 N e n ( h g δ ( x i ) sin ( n x i ) ) sin ( n x j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ43_HTML.gif
(5.5)

In the following, we consider an example which has an exact expression of solutions ( u ( x , y ) , f ( x ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq204_HTML.gif.

Example

If u ( x , 0 ) = 2 π e sin ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq205_HTML.gif, then the function u ( x , y ) = 2 π e 1 y sin ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq206_HTML.gif is the exact solution of problem (5.1). Consequently, the data function is g ( x ) = u ( x , 1 ) = 2 π sin ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq207_HTML.gif.

Adding a random distributed perturbation (obtained by the Matlab command randn) to each data function, we obtain the vector g δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq143_HTML.gif:
g δ = g + ε randn ( size ( g ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equau_HTML.gif
where ε indicates the noise level of the measurement data and the function ‘ randn ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq208_HTML.gif’ generates arrays of random numbers whose elements are normally distributed with mean 0, variance σ 2 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq209_HTML.gif, and standard deviation σ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq210_HTML.gif. ‘ randn ( size ( g ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq211_HTML.gif’ returns an array of random entries that is the same size as g. The bound on the measurement error δ can be measured in the sense of Root Mean Square Error (RMSE) according to
δ = g δ g = ( 1 M + 1 i = 1 M + 1 ( g ( x i ) g δ ( x i ) ) 2 ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equav_HTML.gif
Using g δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq143_HTML.gif as a data function, we obtain the computed approximation f N δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq152_HTML.gif by (5.5). The relative error E r ( f ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq212_HTML.gif is given by
E r ( f ) = f N δ f f . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ44_HTML.gif
(5.6)

Kozlov-Maz’ya iteration method

By using the central difference with step length h = π N + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq213_HTML.gif to approximate the first derivative u x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq214_HTML.gif and the second derivative u x x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq215_HTML.gif, we can get the following semi-discrete problem (ordinary differential equation):
{ u y y ( x i , y ) A h ( x i , y ) = 0 , x i = i h , i = 1 , , N , y ( 0 , + ) , u ( x 0 = 0 , y ) = u ( x N + 1 = π , y ) = 0 , y ( 0 , + ) , u ( x i , 0 ) = f ( x i ) , u ( x i , + ) = 0 , x i = i h , i = 1 , , N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ45_HTML.gif
(5.7)
where A h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq216_HTML.gif is the discretization matrix stemming from the operator A = d 2 d x 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq217_HTML.gif:
A h = 1 h 2 Tridiag ( 1 , 2 , 1 ) M N ( R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equaw_HTML.gif
is a symmetric, positive definite matrix. We assume that it is fine enough so that the discretization errors are small compared to the uncertainty δ of the data; this means that A h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq216_HTML.gif is a good approximation of the differential operator A = d 2 d x 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq218_HTML.gif, whose unboundedness is reflected in a large norm of A h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq216_HTML.gif (see [[19], p.5]). The eigenpairs ( μ k , e k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq219_HTML.gif of A h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq216_HTML.gif are given by
μ k = 4 ( N + 1 π ) 2 sin 2 ( k π 2 ( N + 1 ) ) , e k = ( sin ( j k π N + 1 ) ) j = 1 N , k = 1 , , N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equax_HTML.gif
The discrete iterative approximation of (4.12) takes the form
f k δ ( x j ) = ( I ω K h ) k f 0 ( x j ) + ω i = 0 k 1 ( I ω K h ) i g δ ( x j ) , j = 1 , , N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Equ46_HTML.gif
(5.8)

where K h = e A h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq220_HTML.gif and ω < 1 K h = e μ 1 = 2.7881 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq221_HTML.gif.

Figures 1-4, Table 1 show the comparisons between the exact solution and its computed approximations for different values N, M and ε.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig1_HTML.jpg
Figure 1

TM with ( noise level = 0.01 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq222_HTML.gif , truncation term = 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq223_HTML.gif , grid points of TR = 21 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq224_HTML.gif ).

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig2_HTML.jpg
Figure 2

TM with ( noise level = 0.01 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq222_HTML.gif , truncation term = 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq223_HTML.gif , grid points of TR = 41 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq225_HTML.gif ).

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig3_HTML.jpg
Figure 3

TM with ( noise level = 0.001 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq226_HTML.gif , truncation term = 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq223_HTML.gif , grid points of TR = 21 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq224_HTML.gif ).

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig4_HTML.jpg
Figure 4

TM with ( noise level = 0.001 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq226_HTML.gif , truncation term = 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq223_HTML.gif , grid points of TR = 41 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq225_HTML.gif ).

Table 1

Truncation method: Relative error E r ( f ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq212_HTML.gif

N

M

ϵ

E r ( f ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq227_HTML.gif

20

4

0.01

0.0361

40

4

0.01

0.0155

20

4

0.001

0.0022

40

4

0.001

0.0019

Figures 5-12, Table 2 show the comparisons between the exact solution and its computed approximations for different values N, k, ω and ε.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig5_HTML.jpg
Figure 5

ω = 1.3941 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq228_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig6_HTML.jpg
Figure 6

ω = 1.8587 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq229_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig7_HTML.jpg
Figure 7

ω = 1.9517 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq230_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig8_HTML.jpg
Figure 8

ω = 2.2305 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq231_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig9_HTML.jpg
Figure 9

ω = 1.3941 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq228_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig10_HTML.jpg
Figure 10

ω = 1.8587 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq229_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig11_HTML.jpg
Figure 11

ω = 1.9517 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq230_HTML.gif .

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_Fig12_HTML.jpg
Figure 12

ω = 2.2305 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq231_HTML.gif .

Table 2

Kozlov-Maz’ya method: Relative error E r ( f ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq212_HTML.gif

N

k

ϵ

ω

E r ( f ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq227_HTML.gif

40

4

0.01

0.5 × 2.7881 = 1.3941

0.0790

40

4

0.01

2/3 × 2.7881 = 1.8587

0.0205

40

4

0.01

0.7 × 2.7881 = 1.9517

0.0223

40

4

0.01

0.8 × 2.7881 = 2.2305

0.0214

40

4

0.001

0.5 × 2.7881 = 1.3941

0.0792

40

4

0.001

2/3 × 2.7881 = 1.8587

0.0082

40

4

0.001

0.7 × 2.7881 = 1.9517

0.0026

Conclusion

The numerical results (Figures 1-4) are quite satisfactory. Even with the noise level ε = 0.01 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq232_HTML.gif, the numerical solutions are still in good agreement with the exact solution. In addition, the numerical results (Figures 5-12) are better for ( ω = 2.2305 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq233_HTML.gif, ε = 0.01 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq232_HTML.gif) and ( ω = 1.9517 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq234_HTML.gif, ε = 0.001 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-178/MediaObjects/13661_2013_Article_439_IEq235_HTML.gif) and the other values are also acceptable.

In this study, a convergent and stable reconstruction of an unknown boundary condition has been obtained using two regularizing methods: truncation method and Kozlov-Maz’ya iteration method. Both theoretical and numerical studies have been provided.

Future work will involve the error effect arising in computing eigenfunctions and eigenvalues of the operator A on the truncation method. The question is how to obtain some optimal balance between the accuracy of eigensystem and the noise level of input data.

Declarations

Acknowledgements

The authors would like to thank the editor and the anonymous referees for their valuable comments and helpful suggestions that improved the quality of our paper. This work is supported by the DGRST of Algeria (PNR Project 2011-code: 8\92 u23\92 997).

Authors’ Affiliations

(1)
Applied Mathematics Laboratory, University Badji Mokhtar Annaba
(2)
Department of Mathematics, 8 Mai 1945 Guelma University

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