On the topological gradient method for an inverse problem resolution

Mohamed Abdelwahed; Nejmeddine Chorfi

doi:10.1515/anona-2023-0109

Article Open Access

On the topological gradient method for an inverse problem resolution

Mohamed Abdelwahed and Nejmeddine Chorfi

Published/Copyright: October 13, 2023

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From the journal Advances in Nonlinear Analysis Volume 12 Issue 1

Abstract

In this work, we consider the topological gradient method to deal with an inverse problem associated with three-dimensional Stokes equations. The problem consists in detecting unknown point forces acting on fluid from measurements on the boundary of the domain. We present an asymptotic expansion of the considered cost function using the topological sensitivity analysis method. A detection algorithm is then presented using the developed formula. Some numerical tests are presented to show the efficiency of the presented algorithm.

Keywords: inverse problem; topological gradient; sensitivity analysis

MSC 2010: 35M12; 49Q12; 74P15

1 Introduction

Inverse problems are used in many scientific fields. We cite, as real applications, the detection of gas bubbles in a fluid during mold fillings [8], the detection of an obstacle immersed fluid flow [9], and the optimization of injector location in reservoir aeration process [3]. In some other cases, we use inverse problems in the study of complex systems for which we only have access to a small number of measurements, for example, the Earth in geophysics, organic tissues in medical imaging, the Universe in cosmology, and a concert hall in architectural acoustics.

One of the relevant subjects in inverse problems is the detection of the position and shape of objects immersed in a fluid from measured data at its boundary [5,6,13,21]. In this work, we propose a reconstruction method combining the Kohn-Vogelius formulation and the topological sensitivity analysis method. The Kohn-Vogelius formulation is a self-regularization technique that rephrases the geometrical inverse problem into a shape optimization one [4]. Besides, the topological gradient method is an optimization method based on studying the variation of a given shape function with respect to the domain perturbation [1,2,15,16,18,19,22].

In this work, we study the problem of determining the unknown point forces acting on the Stokes fluid from measurements on the boundary of the domain.

Consider a fluid flow occupying a bounded open three-dimensional domain D . We assume that a finite number of small particles exert point forces on this fluid flow. Each particle is considered to be not larger than a single point, which is mathematically expressed in terms of the distribution of the Dirac function. The total acting force is then considered equal to P = ∑ i = 1 N p η i δ z i , where z i ∈ D ⊂ R 3 is the position of the particle, N P is the total number of point forces, η i ∈ R 3 is a constant force, and δ z i represents the Dirac function.

We study the following inverse problem: from the given velocity u ∂ D on ∂ D and fluid stress S on Σ prescribed on the boundary ∂ D = Γ ∪ Σ , such that Γ ∩ Σ = ∅ , identify the unknown point forces such that the fluid velocity u and the pressure p are the solutions of the following Stokes problem:

(1) − div ( σ ( u , p ) ) = P in D , div u = 0 in D , u = u d on Σ , u = 0 on Γ , σ ( u , p ) n = S on Σ ,

where n is the unit outward normal vector on the boundary ∂ D , and σ ( u , p ) = − p I + 2 ν D ( u ) , where I denotes the 3 × 3 identity matrix, ν > 0 is the viscosity coefficient, and D ( u ) = 1 2 ( ∇ u + ∇ u T ) .

The considered problem can be viewed as a prototype of a geometric inverse problem arising in many applications. The presented approach is general and can be adapted for various partial differential equations. This article is organized as follows:

Section 2 is devoted to the studied optimization problem presentation.
In Section 3, we study the asymptotic development analysis based on the topological sensitivity method.
The numerical algorithm and some numerical investigations are finally illustrated in Section 4.

2 Optimization problem

The optimization problem that we consider can be formulated as follows: given a measured velocity on the boundary ∂ D , find the set of point forces P = ∑ i = 1 N P η i δ z i such that ( u P , p P ) is solution to

− div ( σ ( u P , p P ) ) = P in D , div u P = 0 in D , u P = u d on Σ , u P = 0 on Γ ,

which satisfies σ ( u P , p P ) n = S on Σ .

As the boundary condition is overspecified, we use the Kohn-Vogelius technique [7,14,20] to construct the criteria. The idea consists in replacing the original over-determined boundary value problem with two auxiliary well-posed problems:

− div ( σ ( u 1 P , p 1 P ) ) = P in D , div u 1 P = 0 in D , u 1 P = u d on Γ , u 1 P = 0 on Σ ,

and

− div ( σ ( u 2 P , p 2 P ) ) = P in D , div u 2 P = 0 in D , σ ( u 2 P , p 2 P ) n = S on Σ , u 2 P = 0 on Γ .

One can remark here that the misfit between the solutions u 1 and u 2 vanishes at the points where the force is imposed. Starting from this observation, we propose an identification process based on the minimization of the following Kohn-Vogelius functional:

k ( P ) = K ( u 1 P , u 2 P ) = ∫ D ∣ u 1 P − u 2 P ∣ 2 d z .

To solve the inverse studied problem, we need to minimize k . To reach this aim, we derive an asymptotic expansion of the cost function k .

3 Asymptotic expansion

Throughout the two past decades, topological gradient-based methods provide important advances in the field of shape and topology optimization techniques. Such kind of optimization method is based on a topological sensitivity analysis of a given design function with respect to a small perturbation of the computational domain. In the case of structural topology optimization, a perturbation means simply removing some material. In the case of fluid mechanics where the domain represents the fluid, creating a perturbation means inserting an obstacle O ε = z + ε θ of small size ε > 0 around a point z ∈ D , where θ ∈ R 3 is a smooth domain representing the obstacle shape.

3.1 General case

Consider K a differentiable functional on L p ( D ) 3 , 1 ≤ p < 2 , defined by:

(2) K : L p ( D ) 3 → R , g ↦ K ( g ) .

We want to study the variation of the functional K with respect to a finite topological perturbation of g on the form:

(3) δ g ε ( z ) = η , if z ∈ O ε , 0 , if z ∈ D \ O ε ¯ ,

where η ∈ R 3 is a constant vector and O ε is a small geometry perturbation of size ε > 0 . We have the following general asymptotic expansion result.

Theorem 3.1

Assume that:

The function K is differentiable and there exists constant c 1 > 0 such that for all g ∈ L p ( D ) 3 ,
(4) ∣ K ( g + δ g ) − K ( g ) − ∇ K ( g ) δ g ∣ ≤ c 1 ∥ δ g ∥ L p ( D ) 3 2 , ∀ δ g ∈ L p ( D ) 3 ,
where R denotes the Riez representative of the differential ∇ K ( g ) verifying
(5) ∇ K ( g ) δ g = ∫ D R ( z ) δ g ( z ) d z , ∀ δ g ∈ L p ( D ) 3 .
The function R is Lipschitz continuous, i.e., there exists a constant c 2 > 0 such that:
(6) ∥ R ( z 1 ) − R ( z 2 ) ∥ ≤ c 2 ∥ z 1 − z 2 ∥ , ∀ z 1 , z 1 ∈ D .

Then, the variation of K is given by:

(7) K ( g + δ g ε ) − K ( g ) = ρ ( ε ) { η ⋅ R ( z 0 ) } + o ( ρ ( ε ) ) ,

where ρ ( ε ) = meas ( O ε ) .

Proof 1

By triangular inequality, we have

(8) ∣ K ( g + δ g ε ) − K ( g ) − ρ ( ε ) η . R ( z 0 ) ∣ ≤ ∣ K ( g + δ g ε ) − K ( g ) − ∇ K ( g ) δ g ε ∣ + ∣ ∇ K ( g ) δ g ε − ρ ( ε ) η . R ( z 0 ) ∣ .

Using (4), we obtain

∣ K ( g + δ g ε ) − K ( g ) − ∇ K ( g ) δ g ε ∣ ≤ c 1 ∥ δ g ε ∥ L p ( D ) 3 2 .

By (3) and the fact that p < 2 , we obtain

∥ δ g ε ∥ L p ( D ) 3 2 ≤ c ( ρ ( ε ) 1 ⁄ p ) 2 = c ρ ( ε ) 2 ⁄ p = o ( ρ ( ε ) ) .

Then,

(9) ∣ K ( g + δ g ε ) − K ( g ) − ∇ K ( g ) δ g ε ∣ = o ( ρ ( ε ) ) .

For the second term in (8), we have

∫ D R ( z ) . δ g ε ( z ) d z − ρ ( ε ) η . R ( z 0 ) = ∫ D ( R ( z ) − R ( z 0 ) ) . δ g ε ( z ) d z .

Therefore, using (6),

(10) ∫ D R ( z ) δ g ε ( z ) d z − ρ ( ε ) η ⋅ R ( z 0 ) ≤ c 2 ∫ D ∥ z − z 0 ∥ ∥ δ g ε ( z ) ∥ d z = c 2 ∥ η ∥ ∫ O ε ∥ z − z 0 ∥ d z ≤ c 2 ∥ η ∥ ε ρ ( ε ) .

We deduce the desired result using (9) and (10).□

3.2 Studied problem case

Consider the studied Problem (1). For all G ∈ L 2 ( D ) and u Γ ∈ H 1 ⁄ 2 ( Γ ) , Problem (1) has one solution ( u , p ) ∈ H 1 ( D ) × L 0 2 ( D ) (see [17,23]).

We define V = { v ∈ H 1 ( D ) , div v = 0 } , and V 0 = { v ∈ V , v ∣ Γ = 0 } .

The weak formulation of Problem (1) is: find u ∈ V solution to the variational problem:

(11) a ( u , v ) = l ( v ) , ∀ v ∈ V 0 ,

where, ∀ u , v ∈ V ,

a ( u , v ) = ν ∫ D ∇ u ∇ v d z is a continuous bilinear form on V × V

and l ( v ) = ∫ D G v d z is a continuous linear form on V .

We have the following lemma.

Lemma 3.1

Let l g the linear form be defined by:

l g : V ⟶ R v ⟼ l g ( v ) = l ( v ) + ∫ D g ( t ) v ( t ) d t .

Then, the map

(12) L p ( D ) 3 ⟶ ℒ ( V ) g ⟼ l g

is continuous for p > 6 5 .

Proof 2

We have H 1 ( D ) ⊂ L q ( D ) for 1 ≤ q < 6 (see [12]).

It is well known that if v ∈ L p 1 ( D ) and w ∈ L p 2 ( D ) , the product v ⋅ w ∈ L p ( D ) with 1 p = 1 p 1 + 1 p 2 . Consequently, the map g ↦ l g is continuous as soon as

(13) 1 p + 1 q = 1 .

Then, combining (13) and the conditions on q for which we have H 1 ( D ) ⊂ L q ( D ) , we deduce that the map is continuous for p > 6 5 .□

We define the following:

u g ∈ V , the unique solution of the variational problem:
(14) a ( u g , v ) = l g ( v ) , ∀ v ∈ V 0 ,
K , the cost function defined by:
(15) K : L p ( D ) 3 ⟶ R g ⟼ K ( g ) = K ( u g ) ,
where K is a given functional defined on D .

Using the Lagrangian method, one can prove that if K is differentiable, the function K is differentiable and we have

(16) ∇ K ( g ) δ g = ∫ D δ g ( z ) . w g ( z ) d z , ∀ δ g ∈ L p ( D ) 3 ,

where w g ∈ V 0 is the solution to the associated adjoint Problem [11]:

(17) a ( v , w g ) = − D K ( u g ) v , ∀ v ∈ V 0 .

From (5) and (16), we have

(18) R = w g .

Using Theorem 3.1 in the particular case when g ≡ 0 and δ g = δ g ε , by denoting k ( ε ) = K ( u ε ) and u ε = u 0 + δ g ε , we deduce the following result.

Corollary 3.1

The cost function j has the following asymptotic expansion:

(19) k ( ε ) − k ( 0 ) = ρ ( ε ) η . w 0 ( z ) + o ( ρ ( ε ) ) ,

where w 0 is the solution of (17), where g = 0 .

3.3 Standard cost function case

The asymptotic expansion obtained in Theorem 3.1 is valid for all cost functions verifying the hypotheses ( H 1 ) and ( H 2 ) . In the following, we study the feasibility of these conditions on the following two standard examples of cost functions defined for all Ů ∈ H 1 ( D ) by:

(20) K ( u ) = ∫ D ∣ u − Ů ∣ 2 d z

and

(21) K ( u ) = ν ∫ D ∣ ∇ u − ∇ Ů ∣ 2 d z .

Proposition 3.1

The cost function (20) satisfies the conditions of Theorem 3.1 with

∇ K ( g ) ( δ g ) = ∫ D w g ⋅ δ g d z , ∀ δ g ∈ L p ( D ) 3 ,

and w g ∈ V 0 is solution to the adjoint problem

a ( v , w g ) = − 2 ∫ D ( w g − Ů ) v d z , ∀ v ∈ V 0 .

Proof 3

We can show that K is differentiable on V and

D K ( u ) v = 2 ∫ D ( u − Ů ) v d z , ∀ w ∈ V .

Then, the adjoint state w g ∈ V 0 is solution to

(22) a ( v , w g ) = − 2 ∫ D ( u − Ů ) v d z , ∀ v ∈ V 0 .

By (15), (16), and (20), we obtain

(23) K ( g + δ g ) − K ( g ) − ∇ K ( g ) δ g = ∫ D ∣ u g + δ g − Ů ∣ 2 d z − ∫ D ∣ u g − Ů ∣ 2 d z − ∫ D w g ⋅ δ g d z = 2 ∫ D ( u g − Ů ) . ( u g + δ g − u g ) d z + ∥ u g + δ g − u g ∥ 0 , D 2 − ∫ D w g ⋅ δ g d z .

Since u g ∈ V is solution to

(24) a ( u g , v ) = l g ( v ) , ∀ v ∈ V ,

then ( u g + δ g − u g ) ∈ V 0 satisfies

(25) a ( u g + δ g − u g , v ) = ∫ D δ g . v d z , ∀ v ∈ V 0 .

Choosing v = u g + δ g − u g in (22) and v = w g in (25), we obtain

2 ∫ D ( u g − Ů ) . ( u g + δ g − u g ) d z = ∫ D δ g . w g d z .

Thus,

(26) K ( g + δ g ) − K ( g ) − ∇ K ( g ) δ g = ∥ u g + δ g − u g ∥ 0 , D 2 .

To estimate the last term in (26), we suppose that the perturbation δ g has the form:

δ g ( x ) = η if x ∈ O ε , 0 if x ∈ D \ O ε ¯ ,

where η ∈ R 3 is a constant vector and ε > 0 is small enough.

Choosing v g = u g + δ g − u g and s g = p g + δ g − p g , then ( v g , s g ) ∈ H 0 1 ( D ) × L 0 2 ( D ) is solution to

− ν Δ v g + ∇ s g = δ g in D , div v g = 0 in D , v g = 0 on Γ .

Then, there exists a positive constant c , independent of g , such that:

∥ v g ∥ 1 , D 2 ≤ c ∫ D δ g . v g d z .

We consider p = 13 ⁄ 10 and q = 13 ⁄ 3 ; using the Holder inequality and the Sobolev embedding theorem, we derive

∫ D δ g ⋅ v g d z ≤ ∥ δ g ∥ L 13 ⁄ 10 ( ω ε ) ∥ v g ∥ L 13 ⁄ 3 ( D ) = c ρ ( ε ) 10 ⁄ 13 ∥ v g ∥ 1 , D .

Therefore,

(27) ∥ u g + δ g − u g ∥ 0 , D 2 = ∥ v g ∥ 1 , D 2 = o ( ρ ( ε ) ) .

Using (26) and (27), we deduce

□ K ( g + δ g ) − K ( g ) − ∇ K ( g ) δ g = o ( ρ ( ε ) ) .

Proposition 3.2

The cost function (21) satisfies the hypothesis of Theorem 3.1 with

∇ K ( g ) δ g = ∫ D w g ⋅ δ g d z ∀ δ g ∈ L p ( D ) 3 ,

and w g ∈ V 0 is solution to the adjoint problem

a ( v , w g ) = − 2 ∫ D ( ∇ u g − ∇ Ů ) v d z , ∀ v ∈ V 0 .

Proof 4

It is easy to show that K is differentiable on V and

D K ( u ) v = 2 ∫ D ( ∇ u − ∇ Ů ) . v d z ∀ v ∈ V .

The adjoint solution w g belongs to V 0 and satisfies

(28) a ( v , w g ) = − 2 ∫ D ( ∇ u g − ∇ Ů ) v d z , ∀ v ∈ V 0 .

Using (24), (21), and (28), we obtain

K ( g + δ g ) − K ( g ) − ∇ K ( g ) δ g = ∣ u g + δ g − u g ∣ 1 , D 2 .

By a similar technique as in Proof 3, one can show that:

∣ u g + δ g − u g ∣ 1 , D 2 = o ( ρ ( ε ) ) ,

which implies

□ K ( g + δ g ) − K ( g ) − ∇ K ( g ) δ g = o ( ρ ( ε ) ) .

4 Numerical investigations

Consider D a three-dimensional domain occupied by the fluid containing some point forces:

P = { P i = ( z i , η i ) , 1 ≤ i ≤ N P } ,

where z i and η i denote, respectively, the position and intensity of the point force P i of number i and N P the total point force number. Our aim is to identify P using a boundary measurement S of the stress tensor.

4.1 Numerical algorithm

We use the following numerical algorithm to detect locations, intensity, and number of point forces acting on fluid.

Initialization: choose D 0 = D , and set l = 0 .
Repeat until target is reached:
1. Compute the topological sensitivity δ k l ,
2. Set D l + 1 = { z ∈ D l , δ k l ( z ) ≥ e l + 1 } , where e l + 1 = 0.9 μ , where μ is the most negative minimum of δ k l ( z ) ,
3. l ← l + 1 .

Using Corollary 3.1, the topological gradient is given by:

(29) δ k ( z ) = − η . ( w 1 ( z ) + w 2 ( z ) ) , ∀ z ∈ D .

Then, to compute it, we need to solve the following two direct problems:

(30) − div ( σ ( u 1 , p 1 ) ) = 0 in D , div u 1 = 0 in D , u 1 = u d on Γ , u 1 = 0 on Σ , and − div ( σ ( u 2 , p 2 ) ) = 0 in D , div u 2 = 0 in D , u 2 = 0 on Γ , σ ( u 2 , p 2 ) n = S on Σ ,

and the two adjoint problems:

(31) − div ( σ ( w 1 , q 1 ) ) = − 2 ( u 1 − u 2 ) in D , div w 1 = 0 in D , w 1 = 0 on ∂ D , and − div ( σ ( w 2 , q 2 ) ) = 2 ( u 1 − u 2 ) in D , div w 2 = 0 in D , w 2 = 0 on Γ , σ ( w 2 , w 2 ) n = 0 on Σ .

To stop the optimization algorithm, we used the number of obstacles to detect as stopping criteria for convergence.

4.2 Numerical tests

We propose an adaptation of the previous algorithm to our context. We consider the set { z ∈ D k ; δ k l ( z ) < e l + 1 } . Each connected component of this set is a neighborhood of a point force detected by the algorithm.

In the aforementioned algorithm, Systems (30) and (31) are discretized by a finite element method using NSP1B3 code developed in INRIA [10].

Consider D = [ 0 , 1 ] 3 a three-dimensional cube and the goal is to detect some point forces z i = ( z i 1 , z i 2 , z i 3 ) in D having forces η i = ( η i 1 , η i 2 , η i 3 ) , 1 ≤ i ≤ N . We consider as example the case N = 6 and different forces η i described in Table 1.

Table 1

Data of detected point forces

i	z i = ( z i 1 , z i 2 , z i 3 )	η i = ( η i 1 , η i 2 , η i 3 )
1	z 1 = 0.8 , z 2 = 0.2 , z 3 = 0.7	η 1 = 1 , η 2 = 1 , η 3 = 1
2	z 1 = 0.3 , z 2 = ‒ 0.266 , z 3 = 0.2	η 1 = ‒ 1 , η 2 = ‒ 1 , η 3 = ‒ 1
3	z 1 = 0.733 , z 2 = 0.5 , z 3 = 0.4	η 1 = 0 , η 2 = ‒ 1 , η 3 = 1
4	z 1 = 0.266 , z 2 = 0.633 , z 3 = 0.5	η 1 = ‒ 1 , η 2 = 2 , η 3 = ‒ 1
5	z 1 = 0.7 , z 2 = 0.766 , z 3 = 0.266	η 1 = 2 , η 2 = ‒ 1 , η 3 = ‒ 1
6	z 1 = 0.2 , z 2 = 0.466 , z 3 = 0.8	η 1 = ‒ 2 , η 2 = 1 , η 3 = 2

At each iteration, a new force is localized on the detected point and its support is represented by a sphere of center X i and radius 0.01. We impose the velocity boundary conditions u d = 0 and S = ( 0 , 0 , − 1 ) on the boundary of the domain.

Using Algorithm 4.1, we show in Figures 1, 2, 3, 4, 5, 6 a two-dimensional (2D) cut on x , y , and z of the obtained isovalues of δ k at each iteration. The obtained global minimum at each iteration corresponds to one of the exact point force location.

$Figure 1 δ k \delta k showing the exact location corresponding to the point force 1: (a) 2D cut on x = 0.8 x=0.8 , (b) 2D cut on y = 0.2 y=0.2 , and (c) 2D cut on z = 0.7 z=0.7 .$

Figure 1

δ k showing the exact location corresponding to the point force 1: (a) 2D cut on x = 0.8 , (b) 2D cut on y = 0.2 , and (c) 2D cut on z = 0.7 .

$Figure 2 δ k \delta k showing the exact location corresponding to the point force 2: (a) 2D cut on x = 0.3 x=0.3 , (b) 2D cut on y = 0.266 y=0.266 , and (c) 2D cut on z = 0.2 z=0.2 .$

Figure 2

δ k showing the exact location corresponding to the point force 2: (a) 2D cut on x = 0.3 , (b) 2D cut on y = 0.266 , and (c) 2D cut on z = 0.2 .

$Figure 3 δ k \delta k showing the exact location corresponding to the point force 3: (a) 2D cut on x = 0.733 x=0.733 , (b) 2D cut on y = 0.5 y=0.5 , and (c) 2D cut on z = 0.4 z=0.4 .$

Figure 3

δ k showing the exact location corresponding to the point force 3: (a) 2D cut on x = 0.733 , (b) 2D cut on y = 0.5 , and (c) 2D cut on z = 0.4 .

$Figure 4 δ k \delta k showing the exact location corresponding to the point force 4: (a) 2D cut on x = 0.266 x=0.266 , (b) 2D cut on y = 0.633 y=0.633 , and (c) 2D cut on z = 0.5 z=0.5 .$

Figure 4

δ k showing the exact location corresponding to the point force 4: (a) 2D cut on x = 0.266 , (b) 2D cut on y = 0.633 , and (c) 2D cut on z = 0.5 .

$Figure 5 δ k \delta k showing the exact location corresponding to the point force 5: (a) 2D cut on x = 0.7 x=0.7 , (b) 2D cut on y = 0.766 y=0.766 , and (c) 2D cut on z = 0.266 z=0.266 .$

Figure 5

δ k showing the exact location corresponding to the point force 5: (a) 2D cut on x = 0.7 , (b) 2D cut on y = 0.766 , and (c) 2D cut on z = 0.266 .

$Figure 6 δ k \delta k showing the exact location corresponding to the point force 6: (a) 2D cut on x = 0.2 x=0.2 , (b) 2D cut on y = 0.466 y=0.466 , and (c) 2D cut on z = 0.8 z=0.8 .$

Figure 6

δ k showing the exact location corresponding to the point force 6: (a) 2D cut on x = 0.2 , (b) 2D cut on y = 0.466 , and (c) 2D cut on z = 0.8 .

To study the effects of the distance d , separating two point forces, on the identification result, we present in Figure 7 the obtained isovalues of δ k for different values of d . We notice that the exact position was detected until ( d ≥ 0.15 ). However, for small distance, only the zone containing the location can be detected.

$Figure 7 Isovalues of δ k \delta k for different values of d d .$

Figure 7

Isovalues of δ k for different values of d .

5 Conclusion

In this work, we studied the problem of point force detection from prescribed boundary data in three dimension domain. The used method is based on the topological gradient method. The presented technique can be generalized to other applications using more types of partial differential equations. The numerical algorithm based on the developed asymptotic analysis is of higher interest since we obtain fast accurate results. We will focus in future work on the generalization of this technique to the nonstationary case and when the boundary data are missing on a part of the domain.

Acknowledgments

The authors extend their sincere appreciation to deputyship for research and innovation, “Ministry of Education” in Saudi Arabia for funding this research (IFKSUOR3-450-1).

Conflict of interest: Authors state no conflict of interest.

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Received: 2023-06-17

Revised: 2023-09-08

Accepted: 2023-09-12

Published Online: 2023-10-13

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/anona-2023-0109

Keywords for this article

inverse problem; topological gradient; sensitivity analysis

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