Inverse obstacle scattering for acoustic waves in the time domain

Lu Zhao; Heping Dong; Fuming Ma

doi:10.3934/ipi.2021037

Article Contents

2021, Volume 15, Issue 5: 1269-1286. Doi: 10.3934/ipi.2021037

This issue Previous Article A linear sampling method for inverse acoustic scattering by a locally rough interface Next Article

Inverse obstacle scattering for acoustic waves in the time domain

School of Mathematics, Jilin University, Changchun 130012, China

^* Corresponding author: Heping Dong
^* Corresponding author: Heping Dong

Received: December 2020

Revised: February 2021

Early access: April 2021

Published: October 2021

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Abstract

This paper concerns an inverse acoustic scattering problem which is to determine the location and shape of a rigid obstacle from time domain scattered field data. An efficient convolution quadrature method combined with nonlinear integral equation method is proposed to solve the inverse problem. In particular, replacing the classic Fourier transform with the convolution quadrature method for time discretization, the boundary integral equations for the Helmholtz equation with complex wave numbers can be obtained to guarantee the numerically approximate causality property of the scattered field under some condition. Numerical experiments are presented to demonstrate the effectiveness and robustness of the proposed method.

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Mathematics Subject Classification: 78A46.

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Figure 1. Reconstructions of an apple-shaped obstacle at different levels of noise, the radius of the initial guess is $ r_0 = 0.3 $, the position of the initial guess is $ (c_1^{(0)},c_2^{(0)}) = (-0.7,0.3) $ and the launch position of the incident wave is $ (1.5\cos\theta,1.5\sin\theta ) $, $ \theta = 7\pi/10 $

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Figure 2. Reconstructions of a peanut-shaped obstacle at different levels of noise, the radius of the initial guess is $ r_0 = 0.3 $, the position of the initial guess is $ (c_1^{(0)},c_2^{(0)}) = (0.75,-0.55) $ and the launch position of the incident wave is $ (1.5\cos\theta,1.5\sin\theta ) $, $ \theta = 7\pi/20 $

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Figure 3. Reconstructions of an apple-shaped obstacle at different positions of the initial guess $ (c_1^{(0)},c_2^{(0)}) = (-0.75.-0.4) $ and $ (c_1^{(0)},c_2^{(0)}) = (1,0.2) $, the radius of the initial guess is $ r_0 = 0.3 $, the noise level is 1$ \% $ and the launch position of the incident wave is $ (1.5\cos\theta,1.5\sin\theta ) $, $ \theta = \pi/2 $

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Figure 4. Reconstructions of a peanut-shaped obstacle at different positions of the initial guess $ (c_1^{(0)},c_2^{(0)}) = (-0.5,-0.65) $ and $ (c_1^{(0)},c_2^{(0)}) = (0.46,0.7) $, the radius of the initial guess is $ r_0 = 0.3 $, the noise level is 1$ \% $ and the launch position of the incident wave is $ (1.5\cos\theta,1.5\sin\theta ) $, $ \theta = 13\pi/20 $

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Figure 5. Reconstructions of an apple-shaped obstacle at different launch positions of the incident wave $ (1.5\cos\theta,1.5\sin\theta) $, the values of $ \theta $ are given in figure. The radius of the initial guess is $ r_0 = 0.3 $, the position of the initial guess is $ (c_1^{(0)},c_2^{(0)}) = (-0.9,-0.2) $ and the noise level is 1$ \% $

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Figure 6. Reconstructions of a peanut-shaped obstacle at different launch positions of the incident wave $ (1.5\cos\theta,1.5\sin\theta) $, the values of $ \theta $ are given in figure. The radius of the initial guess is $ r_0 = 0.3 $, the position of the initial guess is $ (c_1^{(0)},c_2^{(0)}) = (0.7,0.3) $ and the noise level is 1$ \% $

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Figure 7. Reconstructions of an apple-shaped obstacle at different numbers of launch position or different levels of noise, the radius of the initial guess is $ r_0 = 0.3 $, the position of the initial guess is $ (c_1^{(0)},c_2^{(0)}) = (0.9,-0.2) $ and the launch position of the incident wave is $ (1.5\cos\theta,1.5\sin\theta ) $, $ \theta_1 = 17\pi/20 $ and $ \theta_2 = 37\pi/20 $

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Figure 8. Reconstructions of a peanut-shaped obstacle at different numbers of launch position or different levles of noise, the radius of the initial guess is $ r_0 = 0.3 $, the position of the initial guess is $ (c_1^{(0)},c_2^{(0)}) = (-0.7,-0.4) $ and the launch position of the incident wave is $ (1.5\cos\theta,1.5\sin\theta ) $, $ \theta_1 = 11\pi/20 $ and $ \theta_2 = 31\pi/20 $

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Table 1. Parametrization of the exact boundary curves

Type	Parametrization
apple-shaped	$ p_D(\theta)= \frac{0.55(1+0.9\cos{\theta}+0.1\sin{2\theta})}{1+0.75\cos{\theta}}(\cos{\theta},\sin{\theta}), \quad \theta\in [0,2\pi] $
peanut-shaped	$ p_D(t)=0.5\sqrt{0.25\cos^2{\theta}+\sin^2{\theta}}(\cos{\theta},\sin{\theta}), \quad \theta\in[0,2\pi] $

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