Numerical simulation of a time dependent lubrication problem arising in magnetic reading processes

Iñigo Arregui; José Jesús Cendán; María González

doi:10.3934/dcdss.2023083

Article Contents

2024, Volume 17, Issue 7: 2436-2449. Doi: 10.3934/dcdss.2023083

This issue Previous Article A comparison of force computations for modeling fluid - structure interaction problems with the lattice Boltzmann volume penalisation method Next Article Group theoretic approach and analytical solutions of the compressible Navier–Stokes equations

Numerical simulation of a time dependent lubrication problem arising in magnetic reading processes

1.
Dept. of Mathematics, Universidade da Coruña, 15071 A Coruña, Spain
2.
CITIC, 15071 A Coruña, Spain

^*Corresponding author: Iñigo Arregui
^*Corresponding author: Iñigo Arregui

Dedicated to Philippe Destuynder

Received: March 1, 2023

Revised: March 1, 2023

Early access: April 17, 2023

Published online: July 1, 2024

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Abstract

We present in this article a numerical technique to simulate the hydrodynamic behaviour of a magnetic reading device. In the transient regime, the evolution of the air pressure is computed as the solution of a time dependent compressible Reynolds equation. The proposed numerical method is based on a local discontinuous Galerkin method combined with a $ \theta- $method for the time derivative. We present some numerical tests to show the good behaviour of the numerical scheme.

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Mathematics Subject Classification: Primary: 65M60.

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Figure 1. Flexible tape device [12,8]

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Figure 2. Rigid device [12]

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Figure 3. Computed pressure at $ t = 0.25 $, $ t = 0.50 $, $ t = 0.75 $ and $ t = 1.0 $ (from top to bottom and from left to right). Test 2

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Figure 4. Evolution of the infinity norm. Fully implicit Euler method ($ \theta = 1 $). Test 2

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Figure 5. Gap function (at $ t = 0 $) in Test 3

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Figure 6. Computed pressure at $ t = 0.25 $, $ t = 0.50 $, $ t = 0.75 $ and $ t = 1.0 $ (from top to bottom and from left to right). Test 3

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Figure 7. Computed pressure at instants $ 0.25 $, $ 0.50 $, $ 0.75 $ and $ 1.0 $, without slots (left) and with slots (right). Test 4

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Table 1. Errors and rates of convergence for $ k_2 = 1 $ and $ k_1 = k_3 = 0 $. Fully implicit Euler method ($ \theta = 1 $). Test 1

$ \Delta x_1 = \Delta x_2 $	$ \Delta t $	$ \\|p-p_h\\|_{\infty} $	$ r(p) $	$ {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert {p-p_h} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_h $	$ r_h(p) $	$ \\|\boldsymbol \sigma-\boldsymbol\sigma_h\\|_{\infty} $	$ r(\boldsymbol\sigma) $
$ 1/5 $	0.128	5.1993E-06	–	2.5705E-05	–	4.1925E-07	–
$ 1/10 $	0.064	2.1445E-06	1.2777	1.2891E-05	0.9957	2.4851E-07	0.7545
$ 1/20 $	0.032	9.3242E-07	1.2016	6.6371E-06	0.9577	1.3326E-07	0.8991
$ 1/40 $	0.016	3.9289E-07	1.2469	3.3664E-06	0.9793	6.8870E-08	0.9523
$ 1/80 $	0.008	1.8374E-07	1.0965	1.7373E-06	0.9544	5.0225E-08	0.4555
$ 1/160 $	0.004	8.9604E-08	1.0360	8.8140E-07	0.9790	3.7023E-08	0.4400

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Table 2. Errors and rates of convergence for $ k_2 = 1 $ and $ k_1 = k_3 = 0 $. Crank-Nicolson method ($ \theta = 0.5 $). Test 1

$ \Delta x_1 = \Delta x_2 $	$ \Delta t $	$ \\|p-p_h\\|_{\infty} $	$ r(p) $	$ {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert {p-p_h} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_h $	$ r_h(p) $	$ \\|\boldsymbol \sigma-\boldsymbol\sigma_h\\|_{\infty} $	$ r(\boldsymbol\sigma) $
$ 1/5 $	0.128	5.4997E-06	–	2.5597E-05	–	4.5497E-07	–
$ 1/10 $	0.064	1.3872E-06	1.9872	1.1968E-05	1.0968	2.4934E-07	0.8677
$ 1/20 $	0.032	5.3585E-07	1.3723	5.8561E-06	1.0312	1.3380E-07	0.8980
$ 1/40 $	0.016	2.1643E-07	1.3079	2.8962E-06	1.0158	6.8838E-08	0.9588
$ 1/80 $	0.008	1.2519E-07	0.7898	1.5042E-06	0.9452	3.7036E-08	0.8943
$ 1/160 $	0.004	6.0052E-08	1.0598	5.8254E-07	1.3686	2.3320E-08	0.6674

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Table 3. Errors and rates of convergence for $ k_2 = 1 $ and $ k_1 = k_3 = 1 $. Fully implicit Euler method ($ \theta = 1 $). Test 1

$ \Delta x_1 = \Delta x_2 $	$ \Delta t $	$ \\|p-p_h\\|_{\infty} $	$ r(p) $	$ {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert {p-p_h} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_h $	$ r_h(p) $	$ \\|\boldsymbol \sigma-\boldsymbol\sigma_h\\|_{\infty} $	$ r(\boldsymbol\sigma) $
$ 1/5 $	0.128	5.2283E-06	–	3.3677E-05	–	1.3622E-06	–
$ 1/10 $	0.064	2.1092E-06	1.3096	1.6568E-05	1.0234	8.0690E-07	0.7555
$ 1/20 $	0.032	9.3481E-07	1.1740	8.3742E-06	0.9844	4.3761E-07	0.8827
$ 1/40 $	0.016	3.9822E-07	1.2311	4.2101E-06	0.9921	2.3721E-07	0.8835
$ 1/80 $	0.008	1.7406E-07	1.1940	2.1583E-06	0.9640	1.2435E-07	0.9318

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Table 4. Errors and rates of convergence for $ k_2 = 1 $ and $ k_1 = k_3 = 1 $. Crank-Nicolson method ($ \theta = 0.5 $). Test 1

$ \Delta x_1 = \Delta x_2 $	$ \Delta t $	$ \\|p-p_h\\|_{\infty} $	$ r(p) $	$ {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert {p-p_h} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_h $	$ r_h(p) $	$ \\|\boldsymbol \sigma-\boldsymbol\sigma_h\\|_{\infty} $	$ r(\boldsymbol\sigma) $
$ 1/5 $	0.128	5.5222E-06	–	3.3553E-05	–	1.4569E-06	–
$ 1/10 $	0.064	1.3912E-06	1.9889	1.5874E-05	1.0798	8.3159E-07	0.8090
$ 1/20 $	0.032	5.4606E-07	1.3492	7.7746E-06	1.0298	4.3707E-07	0.9280
$ 1/40 $	0.016	2.2329E-07	1.2901	3.8408E-06	1.0174	2.3544E-07	0.8925
$ 1/80 $	0.008	1.1259E-07	0.9878	1.9405E-06	0.9850	1.2540E-07	0.9088

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