Convergence to equilibrium for time and space discretizations of the Cahn-Hilliard equation

Matthieu Brachet; Philippe Parnaudeau; Morgan Pierre

doi:10.3934/dcdss.2022110

Article Contents

2022, Volume 15, Issue 8: 1987-2031. Doi: 10.3934/dcdss.2022110

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Convergence to equilibrium for time and space discretizations of the Cahn-Hilliard equation

1.
Laboratoire de Mathématiques et Applications, Université de Poitiers, CNRS, F-86073 Poitiers, France
2.
Institut Pprime, CNRS, UPR 3346, 11 Boulevard Marie et Pierre Curie, 86962 Futuroscope Chasseneuil Cedex, France

^* Corresponding author: Morgan Pierre
^* Corresponding author: Morgan Pierre

Dedicated to Maurizio Grasselli on the occasion of his 60th birthday

Received: November 30, 2021

Revised: March 31, 2022

Published: August 2022

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Abstract

We review space and time discretizations of the Cahn-Hilliard equation which are energy stable. In many cases, we prove that a solution converges to a steady state as time goes to infinity. The proof is based on Lyapunov theory and on a Lojasiewicz type inequality. In a few cases, the convergence result is only partial and this raises some interesting questions. Numerical simulations in two and three space dimensions illustrate the theoretical results. Several perspectives are discussed.

Keywords:

Mathematics Subject Classification: Primary: 65M12, 65P40; Secondary: 37M05.

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Figure 1. Mesh used for (88)

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Figure 2. Final value for (88) with $ \theta = 0 $ (left), $ \theta = \pi/4 $ (middle) and $ \theta = \pi/2 $ (right). The thickness parameter $ \varepsilon $ is equal to $ 0.2 $

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Figure 3. Final energy level vs. angle $ \theta $ for (88)

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Figure 4. Time step vs. time for the 2D secant scheme

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Figure 5. Energy vs. time for the 2D secant scheme

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Figure 6. Time $ t = 0.03\cdot 10^{-5} $ (left) and $ t = 2.3\cdot 10^{-5} $ (right)

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Figure 7. Time $ t = 40\cdot 10^{-5} $ (left) and $ t = 357\cdot 10^{-5} $ (right)

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Figure 8. Time $ t = 532\cdot 10^{-5} $ (left) and steady state (right). The thickness parameter $ \varepsilon $ is equal to $ 0.04 $

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Figure 9. Solution at time $ t = 5e-6 $ (left), $ t = 6e-4 $ (middle) and $ t = 0.1 $ (right) for $ \varepsilon = 0.04 $. The time step is $ \tau = 10^{-6} $

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Figure 10. Solution at time $ t = 5e-6 $ (left), $ t = 6e-4 $ (middle) and $ t = 0.1 $ (right) for $ \varepsilon = 0.04 $. The time step is $ \tau = 5\times 10^{-7} $

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Figure 11. Time $ t = 0 $ (left) and $ t = 2\cdot 10^{-5} $ (right)

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Figure 12. Time $ 8\cdot 10^{-5} $ (left) and $ t = 36\cdot 10^{-5} $ (right)

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Figure 13. Time $ 875\cdot 10^{-5} $ (left) and steady state (right). The thickness parameter $ \varepsilon $ is equal to $ 0.04 $

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Figure 14. Energy vs. time for 3D linear IMEX scheme

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