New iterative regularization methods for solving split variational inclusion problems

Dang Van Hieu; Le Dung Muu; Pham Kim Quy

doi:10.3934/jimo.2021185

Article Contents

2023, Volume 19, Issue 1: 300-320. Doi: 10.3934/jimo.2021185

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New iterative regularization methods for solving split variational inclusion problems

1.
TIMAS - Thang Long University, Ha Noi, Vietnam
2.
Institute of Mathematics, VAST, Hanoi, 18 Hoang Quoc Viet, Hanoi, Vietnam
3.
Department of Basic Sciences, College of Air Force, Nha Trang City, Vietnam

^* Corresponding author: Dang Van Hieu (dangvanhieu@tdtu.edu.vn)
^* Corresponding author: Dang Van Hieu (dangvanhieu@tdtu.edu.vn)

Received: September 2020

Revised: March 2021

Early access: November 2021

Published: January 2023

Abstract Full Text(HTML) Figure(6) / Table(2) Related Papers Cited by

Abstract

The paper proposes some new iterative algorithms for solving a split variational inclusion problem involving maximally monotone multi-valued operators in a Hilbert space. The algorithms are constructed around the resolvent of operator and the regularization technique to get the strong convergence. Some stepsize rules are incorporated to allow the algorithms to work easily. An application of the proposed algorithms to split feasibility problems is also studied. The computational performance of the new algorithms in comparison with others is shown by some numerical experiments.

Keywords:

Mathematics Subject Classification: Primary: 65Y05, 65K15; Secondary: 90C33.

Citation:

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Figure 1. Example 1 for $ M = 100, N = 200 $. Number of iteration is 1393, 1395, 1365, 1414, 1402, respectively

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Figure 2. Example 1 for M = 200, N = 500. Number of iteration is 202, 204, 198, 209, 210, respectively

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Figure 3. Example 2 for M = 1024, N = 256. Number of iteration is 763, 759, 228, 748, 756, respectively

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Figure 4. Example 2 for M = 2048, N = 512. Number of iteration is 357,358,171,359,353, respectively

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Figure 5. Example 3 for x₀(t) = 1. Number of iteration is 350,361,189,414,340, respectively

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Figure 6. Example 3 for x₀(t) = exp(−t). Number of iteration is 238,217,119,335,283, respectively

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Algorithm 1:

Initialization: Take $ x_0 \in \mathcal{H}_1 $ and two sequences $ \left\{\lambda_n\right\},\,\left\{\alpha_n\right\} \subset \left(0,+\infty\right) $.
Iterative Steps:
Step 1. Compute $ x_{n+1}=J_{\lambda_n\mathcal{B}_1}\left(x_n-\lambda_n\mathcal{A}^*(I-J_{\mathcal{B}_2})\mathcal{A}x_n-\lambda_n\alpha_n\mathcal{F}x_n\right). $
Step 2. Set $ n:=n+1 $ and return to Step 1.

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Algorithm 2:

Initialization: Take $ x_0 \in \mathcal{H}_1 $ and $ \lambda_0>0 $, $ \mu \in (0,1) $. Choose a sequence $ \left\{\alpha_n\right\} \subset \left(0,+\infty\right) $ such that conditions $ \rm (C2) - (C4) $ hold.
Iterative Steps:
Step 1. Compute
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left\{ \begin{array}{ll} y_n = J_{\lambda_n\mathcal{B}_1}\left(x_n-\lambda_n\mathcal{A}^*(I-J_{\mathcal{B}_2})\mathcal{A}x_n-\lambda_n\alpha_n\mathcal{F}x_n\right).\\ x_{n+1} = y_n+\lambda_n\left(\mathcal{A}^*(I-J_{\mathcal{B}_2})\mathcal{A}x_n-\mathcal{A}^*(I-J_{\mathcal{B}_2})\mathcal{A}y_n\right). \end{array} \right. $
Step 2. Update $ \lambda_n $:
$ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \lambda_{n+1} = \min \left\{\lambda_n,\frac{\mu ||x_n-y_n||}{||\mathcal{A}^*(I-J_{\mathcal{B}_2})\mathcal{A}x_n-\mathcal{A}^*(I-J_{\mathcal{B}_2})\mathcal{A}y_n||}\right\}.$

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