# American Institute of Mathematical Sciences

doi: 10.3934/ipi.2021056
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## Small defects reconstruction in waveguides from multifrequency one-side scattering data

 1 Institut Fourier, Université Grenoble Alpes, France 2 Institut Camille Jordan, École Centrale Lyon, France

* Corresponding author: Angèle Niclas

Received  January 2021 Revised  June 2021 Early access September 2021

Localization and reconstruction of small defects in acoustic or electromagnetic waveguides is of crucial interest in nondestructive evaluation of structures. The aim of this work is to present a new multi-frequency inversion method to reconstruct small defects in a 2D waveguide. Given one-side multi-frequency wave field measurements of propagating modes, we use a Born approximation to provide a $\text{L}^2$-stable reconstruction of three types of defects: a local perturbation inside the waveguide, a bending of the waveguide, and a localized defect in the geometry of the waveguide. This method is based on a mode-by-mode spacial Fourier inversion from the available partial data in the Fourier domain. Indeed, in the available data, some high and low spatial frequency information on the defect are missing. We overcome this issue using both a compact support hypothesis and a minimal smoothness hypothesis on the defects. We also provide a suitable numerical method for efficient reconstruction of such defects and we discuss its applications and limits.

Citation: Éric Bonnetier, Angèle Niclas, Laurent Seppecher, Grégory Vial. Small defects reconstruction in waveguides from multifrequency one-side scattering data. Inverse Problems & Imaging, doi: 10.3934/ipi.2021056
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##### References:
Representation of the three types of defects: in $(1)$ a local perturbation $q$, in $(2)$ a bending of the waveguide, in $(3)$ a localized defect in the geometry of $\Omega$. A controlled source $s$ generates a wave field $u^\text{inc}_k$. When it crosses the defect, it generates a scattered wave field $u^s_k$. Both $u^\text{inc}_k$ and $u^s_k$ are measured on the section $\Sigma$
Condition number of $M_t^TM_t$ for different sizes of support and values of $\omega_0$. Here, $X$ is the discretization of $[1-r, 1+r]$ with $500r+1$ points. The $x$-axis represents the evolution of $r$, and the $y$-axis $\text{cond}_2(M_t^TM_t)$. Each curve corresponds to value of $\omega_0$ as indicated in the left rectangle
Representation of a bend in a waveguide
Representation of a shape defect in a waveguide
Reconstruction of $f(x) = (x-0.8)(1.2-x)\textbf{1}_{0.8\leq x\leq 1.2}$ for different values of $\omega_1$ using the discrete operator $\gamma$ and the algorithm (91) with $\lambda = 0.001$. Here, $X$ is the discretization of $[0.5, 1.5]$ with $10\omega_1$ points, and $K$ is the discretization of $[0.01, \omega_1]$ with $1000$ points
$\text{L}^2$-error between $f(x) = (x-0.8)(1.2-x)\textbf{1}_{0.8\leq x\leq 1.2}$ and its reconstruction $f_{app}$ for different values of $\omega_1$ using the discrete operator $\gamma$ and the algorithm (91) with $\lambda = 0.001$. Here, $X$ is the discretization of $[0.5, 1.5]$ with $10\omega_1$ points, and $K$ is the discretization of $[0.01, \omega_1]$ with $1000$ points
Reconstruction of $f(x) = (x-0.8)(1.2-x)\textbf{1}_{0.8\leq x\leq 1.2}$ for different values of $\omega_0$ and $r = 0.5$ using the discrete operator $\gamma$ and the algorithm (91) with $\lambda = 0.001$. Here, $X$ is the discretization of $[0.5, 1.5]$ with $251$ points, and $K$ is the discretization of $[\omega_0, 50]$ with $1000$ points
Reconstruction of $f(x) = (x-0.8)(1.2-x)\textbf{1}_{0.8\leq x\leq 1.2}$ for different sizes of support $r$ and $\omega_0 = 3\pi$ using the discrete operator $\gamma$ and the algorithm (91) with $\lambda = 0.001$. Here, $X$ is the discretization of $[1-r, 1+r]$ with $500r+1$ points, and $K$ is the discretization of $[3\pi, 50]$ with $1000$ points
Reconstruction of two different bends. The black lines represent the initial shape of $\Omega$, and the red the reconstruction of $\Omega$. In both cases, $K$ is the discretization of $[0.01, 40]$ with $100$ points, and the reconstruction is obtain by (94). On the left, the initial parameters of the bend are $(x_c, r, \theta) = (4, 10, \pi/12)$ and on the right, $(x_c, r, \theta) = (2, 5, \pi/6)$
Reconstruction of a waveguide with two successive bends. The black lines represent the initial shape of $\Omega$, and the red the reconstruction of $\Omega$, slightly shifted for comparison purposes. In both cases, $K$ is the discretization of $[0.01, 40]$ with $100$ points. The parameters of the two bends are $(x_c^{(1)}, r^{(1)}, \theta^{(1)}) = (2, 10, \pi/30))$ and $(x_c^{(2)}, r^{(2)}, \theta^{(2)}) = (3.8, 8, -\pi/20))$
Reconstruction of two shape defects. In black, the initial shape of $\Omega$, and in red the reconstruction, slightly shifted for comparison purposes. In both cases, $K$ is the discretization of $[0.01, 70]\setminus \{[n\pi-0.2, n\pi+0.2], n\in \mathbb{N}\}$ with $300$ points, $X$ is the discretization of $[3, 4.5]$ with $151$ points and we use the algorithm (91) with $\lambda = 0.08$ to reconstruct $s_0$ and $s_1$. On the left, $h(x) = \frac{5}{16}\textbf{1}_{3.2\leq x\leq 4.2}(x-3.2)^2(4.2-x)^2$ and $g(x) = -\frac{35}{16}\textbf{1}_{3.4\leq x\leq 4}(x-3.4)^2(4-x)^2$. On the right, $h(x) = \frac{125}{16}\textbf{1}_{3.7\leq x\leq 4.2}(x-3.7)^2(4.2-x)^2$ and $g(x) = \frac{125}{16}\textbf{1}_{3.4\leq x\leq 4}(x-3.4)^2(4-x)^2$
Recontruction of $h_n$ for $0\leq n\leq 9$, where $h(x) = 0.05\textbf{1}_{\left|\left(\frac{x-4}{0.05}, \frac{y-0.6}{0.15}\right)\right|\leq 1}\left|\left(\frac{x-4}{0.05}, \frac{y-0.6}{0.15}\right)\right|^2$. In blue, we represent $h_n$ and in red the reconstruction of $h_{n_{\text{app}}}$. In every reconstruction, $K$ is the discretization of $[0.01, 150]$ with $200$ points, $X$ is the discretization of $[3.8, 4.2]$ with $101$ points and we use the algorithm (91) with $\lambda = 0.002$ to reconstruct every $h_n$
Recontruction of an inhomogeneity $h$, where $h(x) = 0.05\textbf{1}_{\left|\left(\frac{x-4}{0.05}, \frac{y-0.6}{0.15}\right)\right|\leq 1}\left|\left(\frac{x-4}{0.05}, \frac{y-0.6}{0.15}\right)\right|^2$. On the left, we represent the initial shape of $h$, and on the right the reconstruction $h_{\text{app}}$. Here, $K$ is the discretization of $[0.01, 150]$ with $200$ points, $X$ is the discretization of $[3.8, 4.2]$ with $101$ points and we use the algorithm (91) with $\lambda = 0.002$ to reconstruct every $h_n$. We used $N = 20$ modes to reconstruct $h$
Recontruction of an inhomogeneity $h$. From top to bottom, the initial representation of $h$, the reconstruction $h_{\text{app}}$ and the reconstruction $h_{\text{app}}$ with the knowledge of the positivity of $h$. Here, $K$ is the discretization of $[0.01, 150]$ with $200$ points, $X$ is the discretization of $[3, 6]$ with $3001$ points and we use the algorithm (91) with $\lambda = 0.01$ to reconstruct every $h_n$. We choose used $N = 20$ modes to reconstruct $h$
Relative errors on the reconstruction of $(x_c, r, \theta)$ for different bends. In each case, $K$ is the discretization of $[0.01, 40]$ with $100$ points, and the reconstruction is obtain by (94)
 $(x_c, r, \theta)$ $(2.5, 40, \pi/80)$ $(4, 10, \pi/12)$ $(2, 5, \pi/6)$ relative error on $x_c$ $1.8\%$ $0\%$ $7.6\%$ relative error on $r$ $3.0\%$ $7.5\%$ $23.8\%$ relative error on $\theta$ $1.6\%$ $10.7\%$ $16.9\%$
 $(x_c, r, \theta)$ $(2.5, 40, \pi/80)$ $(4, 10, \pi/12)$ $(2, 5, \pi/6)$ relative error on $x_c$ $1.8\%$ $0\%$ $7.6\%$ relative error on $r$ $3.0\%$ $7.5\%$ $23.8\%$ relative error on $\theta$ $1.6\%$ $10.7\%$ $16.9\%$
Relative errors on the reconstruction of $h$ for different amplitudes $A$. We choose $h(x) = A\textbf{1}_{3\leq x\leq 5}(x-3)^2(5-x)^2$ and $g(x) = 0$. In every reconstruction, $K$ is the discretization of $[0.01, 40]\setminus \{[n\pi-0.2, n\pi+0.2], n\in \mathbb{N}\}$ with $100$ points, $X$ is the discretization of $[1, 7]$ with $601$ points and we use the algorithm (91) with $\lambda = 0.08$ to reconstruct $h'$
 $A$ $0.1$ $0.2$ $0.3$ $0.5$ $\Vert h-h_{\text{app}}\Vert_{\text{L}^2( \mathbb{R})}/\Vert h\Vert_{\text{L}^2( \mathbb{R})}$ $8.82\%$ $10.41\%$ $15.12\%$ $54.99\%$
 $A$ $0.1$ $0.2$ $0.3$ $0.5$ $\Vert h-h_{\text{app}}\Vert_{\text{L}^2( \mathbb{R})}/\Vert h\Vert_{\text{L}^2( \mathbb{R})}$ $8.82\%$ $10.41\%$ $15.12\%$ $54.99\%$
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