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December  2020, 28(4): 1503-1528. doi: 10.3934/era.2020079

## Efficient finite difference methods for the nonlinear Helmholtz equation in Kerr medium

 1 College of Mathematics and Statistics, Chongqing University, Chongqing 401331, China 2 School of Mathematical Sciences, University of Electronic Science and Technology of China, Sichuan 611731, China

Received  April 2020 Revised  June 2020 Published  December 2020 Early access  July 2020

Fund Project: This work is supported by the Natural Science Foundation of China (No. 91630205)

In this paper, we consider a kind of efficient finite difference methods (FDMs) for solving the nonlinear Helmholtz equation in the Kerr medium. Firstly, by applying several iteration methods, we linearize the nonlinear Helmholtz equation in several different ways. Then, based on the resulted linearized problem at each iterative step, by rearranging the Taylor expansion and using the ADI method, we deduce a kind of new FDMs, which also provide a route to deal with the problem with discontinuous coefficients.Finally, some numerical results are presented to validate the efficiency of the proposed schemes, and to show that our schemes perform with much higher accuracy and better convergence compared with the classical ones.

Citation: Xuefei He, Kun Wang, Liwei Xu. Efficient finite difference methods for the nonlinear Helmholtz equation in Kerr medium. Electronic Research Archive, 2020, 28 (4) : 1503-1528. doi: 10.3934/era.2020079
##### References:

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##### References:
Kerr medium
Computational domain in 2D
Numerical solutions with $\varepsilon = 0.01$ in 1D problem (Red: new scheme with $k_0h = 1$; Blue: reference solution)
Numerical solutions with $\varepsilon = 0.1$ for the 1D problem (Red: new scheme with $k_0h = 1$; Blue: reference solution)
$|T|^2$ with respect to $\varepsilon$ for the 1D problem
Switchback-type non-uniqueness of $|T|^2$ near $\varepsilon = 0.724$ for the 1D problem
Solutions for the 2D problem with $k_0 = 100,h = 1/400$ (Left: numerical solution; Right: exact solution)
Transmission of a single soliton
Collision of two solitons
Errors in $l^\infty$-norm for the 1D problem with $\varepsilon = 0.01$
 $N$ 100 200 400 800 1600 $k_0=10$ SFD 2.14 1.05 2.69e-1 6.71e-2 1.67e-3 FV[2] 1.59 5.03e-1 1.29e-1 3.23e-2 8.09e-3 CFD 5.38e-1 3.70e-2 3.27e-3 6.67e-4 2.72e-4 Scheme (29) 1.26e-3 2.99e-4 7.43e-5 1.89e-5 5.16e-6 Scheme (30) 2.16e-4 5.68e-5 1.43e-5 3.68e-6 1.16e-6 $k_0=20$ SFD 2.31 2.13 1.80 5.33e-1 1.34e-1 FV[2] 2.00 2.00 9.76e-1 2.60e-1 6.52e-2 CFD 2.17 1.02 7.16e-2 5.46e-3 8.00e-4 Scheme (29) 8.56e-3 1.57e-3 3.75e-4 9.57e-5 2.70e-5 Scheme (30) 1.29e-3 3.16e-4 7.56e-5 1.87e-5 4.95e-6 $k_0=40$ SFD 1.24 2.35 2.13 2.03 1.03 FV[2] - 2.00 1.99 1.70 5.16e-1 CFD 1.22 2.36 1.76 1.40e-1 9.86e-3 Scheme (29) 1.12e-2 9.70e-3 1.80e-3 4.49e-4 1.31e-4 Scheme (30) 5.64e-3 2.68e-3 6.86e-4 1.05e-4 2.61e-5 $k_0=80$ SFD 1.07 1.05 2.32 2.29 2.02 FV[2] - - 2.00 1.98 1.97 CFD 1.04 1.21 2.31 1.99 0.29 Scheme (29) 7.38e-3 8.68e-3 4.92e-3 1.99e-3 1.52e-3 Scheme (30) 1.47e-3 1.05e-3 2.48e-4 2.56e-4 2.04e-4
 $N$ 100 200 400 800 1600 $k_0=10$ SFD 2.14 1.05 2.69e-1 6.71e-2 1.67e-3 FV[2] 1.59 5.03e-1 1.29e-1 3.23e-2 8.09e-3 CFD 5.38e-1 3.70e-2 3.27e-3 6.67e-4 2.72e-4 Scheme (29) 1.26e-3 2.99e-4 7.43e-5 1.89e-5 5.16e-6 Scheme (30) 2.16e-4 5.68e-5 1.43e-5 3.68e-6 1.16e-6 $k_0=20$ SFD 2.31 2.13 1.80 5.33e-1 1.34e-1 FV[2] 2.00 2.00 9.76e-1 2.60e-1 6.52e-2 CFD 2.17 1.02 7.16e-2 5.46e-3 8.00e-4 Scheme (29) 8.56e-3 1.57e-3 3.75e-4 9.57e-5 2.70e-5 Scheme (30) 1.29e-3 3.16e-4 7.56e-5 1.87e-5 4.95e-6 $k_0=40$ SFD 1.24 2.35 2.13 2.03 1.03 FV[2] - 2.00 1.99 1.70 5.16e-1 CFD 1.22 2.36 1.76 1.40e-1 9.86e-3 Scheme (29) 1.12e-2 9.70e-3 1.80e-3 4.49e-4 1.31e-4 Scheme (30) 5.64e-3 2.68e-3 6.86e-4 1.05e-4 2.61e-5 $k_0=80$ SFD 1.07 1.05 2.32 2.29 2.02 FV[2] - - 2.00 1.98 1.97 CFD 1.04 1.21 2.31 1.99 0.29 Scheme (29) 7.38e-3 8.68e-3 4.92e-3 1.99e-3 1.52e-3 Scheme (30) 1.47e-3 1.05e-3 2.48e-4 2.56e-4 2.04e-4
Iteration numbers of different iteration methods for the 1D problem
 $k_0$ 10 20 40 80 160 320 640 1280 $\varepsilon=0.01$ Frozen-nonlinearity 5 5 6 7 9 12 17 38 Error Correction 3 4 4 4 4 4 5 6 Modified Newton 5 6 7 8 10 14 22 - Newton's method 4 4 5 5 6 8 11 - $\varepsilon=0.02$ Frozen-nonlinearity 5 6 7 9 12 19 45 - Error Correction 4 4 4 4 5 5 7 9 Modified Newton 6 7 8 10 14 23 - - Newton's method 4 5 5 6 8 11 - - $\varepsilon=0.04$ Frozen-nonlinearity 6 8 9 13 22 55 - - Error Correction 4 4 5 5 6 8 13 - Modified Newton 7 9 10 16 25 - - - Newton's method 5 5 6 8 10 - - - $\varepsilon=0.06$ Frozen-nonlinearity 7 9 12 18 35 - - - Error Correction 4 5 5 6 7 10 - - Modified Newton 8 9 14 20 39 - - - Newton's method 5 6 7 10 - - - - $\varepsilon=0.08$ Frozen-nonlinearity 8 10 14 20 89 - - - Error Correction 5 5 6 7 9 - - - Modified Newton 9 11 17 25 - - - - Newton's method 5 6 8 11 - - - - $\varepsilon=0.1$ Frozen-nonlinearity 9 10 16 35 - - - - Error Correction 5 6 6 8 12 - - - Modified Newton 10 13 18 36 - - - - Newton's method 6 7 9 - - - - -
 $k_0$ 10 20 40 80 160 320 640 1280 $\varepsilon=0.01$ Frozen-nonlinearity 5 5 6 7 9 12 17 38 Error Correction 3 4 4 4 4 4 5 6 Modified Newton 5 6 7 8 10 14 22 - Newton's method 4 4 5 5 6 8 11 - $\varepsilon=0.02$ Frozen-nonlinearity 5 6 7 9 12 19 45 - Error Correction 4 4 4 4 5 5 7 9 Modified Newton 6 7 8 10 14 23 - - Newton's method 4 5 5 6 8 11 - - $\varepsilon=0.04$ Frozen-nonlinearity 6 8 9 13 22 55 - - Error Correction 4 4 5 5 6 8 13 - Modified Newton 7 9 10 16 25 - - - Newton's method 5 5 6 8 10 - - - $\varepsilon=0.06$ Frozen-nonlinearity 7 9 12 18 35 - - - Error Correction 4 5 5 6 7 10 - - Modified Newton 8 9 14 20 39 - - - Newton's method 5 6 7 10 - - - - $\varepsilon=0.08$ Frozen-nonlinearity 8 10 14 20 89 - - - Error Correction 5 5 6 7 9 - - - Modified Newton 9 11 17 25 - - - - Newton's method 5 6 8 11 - - - - $\varepsilon=0.1$ Frozen-nonlinearity 9 10 16 35 - - - - Error Correction 5 6 6 8 12 - - - Modified Newton 10 13 18 36 - - - - Newton's method 6 7 9 - - - - -
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