Published by De Gruyter September 18, 2019

A Reduced Crouzeix–Raviart Immersed Finite Element Method for Elasticity Problems with Interfaces

Gwanghyun Jo and Do Young Kwak

From the journal Computational Methods in Applied Mathematics

https://doi.org/10.1515/cmam-2019-0046

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Abstract

The purpose of this paper is to develop a reduced Crouzeix–Raviart immersed finite element method (RCRIFEM) for two-dimensional elasticity problems with interface, which is based on the Kouhia–Stenberg finite element method (Kouhia et al. 1995) and Crouzeix–Raviart IFEM (CRIFEM) (Kwak et al. 2017). We use a P 1 -conforming like element for one of the components of the displacement vector, and a P 1 -nonconforming like element for the other component. The number of degrees of freedom of our scheme is reduced to two thirds of CRIFEM. Furthermore, we can choose penalty parameters independent of the Poisson ratio. One of the penalty parameters depends on Lamé’s second constant μ, and the other penalty parameter is independent of both μ and λ. We prove the optimal order error estimates in piecewise H 1 -norm, which is independent of the Poisson ratio. Numerical experiments show optimal order of convergence both in L 2 and piecewise H 1 -norms for all problems including nearly incompressible cases.

Keywords: Immersed Finite Element Method; Elasticity Equation With Interface; Kouhia–Stenberg Element; Nearly Incompressible; Locking Free; Korn’s Inequality

MSC 2010: 65N12; 65N30

Funding source: National Research Foundation of Korea

Award Identifier / Grant number: No.2017R1D1A1B03032765

Funding statement: Do Y. Kwak is supported by NRF, contract No.2017R1D1A1B03032765.

A Appendix

We prove Proposition 3.1. Suppose a typical interface element T has vertices at A ⁢ ( 0 , 0 ) , B ⁢ ( 1 , 0 ) and C ⁢ ( 0 , 1 ) . Assume that the interface meets with the edges at D = ( x 0 , 0 ) and E = ( 0 , y 0 ) (Figure 6). Other cases can be treated similarly.

Figure 6

A typical reference interface triangle.

Let 𝐜 i = ( a i + , b i + , c i + , a i - , b i - , c i - ) ( i = 1 , 2 ) be the coefficients of ϕ ^ in (3.1). Then conditions (3.2a), (3.2b) and (3.2c) give rise to

( A 𝟎 𝟎 B ) ⁢ ( 𝐜 1 𝐜 2 ) = ( 𝐠 1 𝐠 2 ) ,

where 5 × 6 matrices A and B are respectively given by

(A.1) ( 1 1 2 1 2 0 0 0 1 - y 0 0 1 2 ⁢ ( 1 - y 0 2 ) y 0 0 1 2 ⁢ y 0 2 1 - x 0 1 2 ⁢ ( 1 - x 0 2 ) 0 x 0 1 2 ⁢ x 0 2 0 - 1 - x 0 0 1 x 0 0 - 1 0 - y 0 1 0 y 0 ) , ( 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 1 x 0 0 - 1 - x 0 0 1 0 y 0 - 1 0 - y 0 ) .

Here, 𝐠 1 = [ R 1 , R 2 , R 3 , 0 , 0 ] and 𝐠 2 = [ V 1 , V 2 , V 3 , 0 , 0 ] T , where R i is the edge average of the first component of ϕ ^ and V i is the nodal value of the second component. Furthermore, condition (3.2d) is written as

(A.2) ( 𝐝 1 T 𝐝 2 T 𝐞 1 T 𝐞 2 T ) ⁢ ( 𝐜 1 𝐜 2 ) = ( 0 0 ) ,

where 𝐧 = ( n 1 , n 2 ) = ( y 0 / x 0 2 + y 0 2 , x 0 / x 0 2 + y 0 2 ) and

𝐝 1 T = ( 0 , ( 2 ⁢ μ + + λ + ) ⁢ n 1 , μ + ⁢ n 2 , 0 , - ( 2 ⁢ μ - + λ - ) ⁢ n 1 , - μ - ⁢ n 2 ) :- ( d 1 ⁢ i ) i = 1 6 ,

𝐝 2 T = ( 0 , μ + ⁢ n 2 , λ + ⁢ n 1 , 0 , - μ - ⁢ n 2 , - λ - ⁢ n 1 ) :- ( d 2 ⁢ i ) i = 1 6 ,

𝐞 1 T = ( 0 , λ + ⁢ n 2 , μ + ⁢ n 1 , 0 , - λ - ⁢ n 2 , - μ - ⁢ n 1 ) :- ( e 1 ⁢ i ) i = 1 6 ,

𝐞 2 T = ( 0 , μ + ⁢ n 1 , ( 2 ⁢ μ + + λ + ) ⁢ n 2 , 0 , - μ - ⁢ n 1 , - ( 2 ⁢ μ - + λ - ) ⁢ n 2 ) :- ( e 2 ⁢ i ) i = 1 6 .

Arranging the equations (A.1) and (A.2), we get the 12-by-12 systems

M :- ( A 𝟎 𝐝 1 T 𝐝 2 T 𝟎 B 𝐞 1 T 𝐞 2 T ) ⁢ ( 𝐜 1 𝐜 2 ) = ( 𝐠 1 0 𝐠 2 0 ) .

It suffices to show that the determinant of M is nonzero. By adding columns 6, 5 and 4 to 3, 2 and 1, respectively, and using the row eliminations, we see that the determinant of M is the same as the determinant of the matrix

M ′ :- ( U 0 𝐎 0 d ¯ 66 𝐝 2 T 𝐎 0 B 0 e ¯ 16 𝐞 2 T ) , where U :- ( 1 1 2 1 2 0 0 0 - 1 2 0 y 0 0 0 0 - 1 2 x 0 1 2 ⁢ x 0 2 0 0 0 1 x 0 0 0 0 0 - x 0 ) .

Here, d ¯ 66 and e ¯ 66 satisfy

x 0 ⁢ d ¯ 66 = - n 1 ⁢ y 0 ⁢ { ( 2 ⁢ μ + + λ + ) ⁢ x 0 ⁢ y 0 + ( 2 ⁢ μ - + λ - ) ⁢ ( 1 - x 0 ⁢ y 0 ) } - x 0 ⁢ n 2 ⁢ { μ + ⁢ x 0 ⁢ y 0 + μ - ⁢ ( 1 - x 0 ⁢ y 0 ) } ,

x 0 ⁢ e ¯ 16 = - n 2 ⁢ y 0 ⁢ { λ + ⁢ x 0 ⁢ y 0 + λ - ⁢ ( 1 - x 0 ⁢ y 0 ) } - n 2 ⁢ x 0 ⁢ { μ + ⁢ x 0 ⁢ y 0 + μ - ⁢ ( 1 - x 0 ⁢ y 0 ) } .

By applying row operations to rows 7–12 of M ′ , we have

M ′′ :- ( U 0 𝐎 0 d ¯ 66 𝐝 2 T 𝐎 0 C 0 e ¯ 16 ) , where C :- ( 1 1 0 0 0 0 0 - 1 1 0 0 0 0 0 0 1 0 0 0 0 x 0 - 1 0 - x 0 0 0 0 y 0 - 1 0 0 - y 0 0 0 e 22 + e 23 0 e 25 e 26 ) .

Here, it is easy to see that

x 0 2 + y 0 2 ⁢ det ⁡ ( C ) = y 0 2 ⁢ ( x 0 ⁢ μ + + ( 1 - x 0 ) ⁢ μ - ) + 2 ⁢ x 0 2 ⁢ ( y 0 ⁢ μ + + ( 1 - y 0 ) ⁢ μ - ) + x 0 2 ⁢ ( y 0 ⁢ λ + + ( 1 - y 0 ) ⁢ λ - ) .

Lemma A.1.

The determinant of M is given by

det ⁡ ( M ) = det ⁡ ( U ) ⁢ { d ¯ 66 ⁢ det ⁡ ( C ) + e ¯ 16 ⁢ cofac } ,

where

cofac = det ⁡ ( 0 μ + ⁢ n 2 λ + ⁢ n 1 0 - μ - ⁢ n 2 - λ - ⁢ n 1 1 1 0 0 0 0 0 - 1 1 0 0 0 0 0 0 1 0 0 0 0 x 0 - 1 0 - x 0 0 0 0 y 0 - 1 0 0 - y 0 ) = - det ⁡ ( μ + ⁢ n 2 λ + ⁢ n 1 - μ - ⁢ n 2 - λ - ⁢ n 1 - 1 1 0 0 0 x 0 - 1 - x 0 0 0 y 0 - 1 0 - y 0 ) = x 0 ⁢ y 0 ⁢ ( μ + ⁢ n 2 + λ + ⁢ n 1 ) + ( 1 - x 0 ) ⁢ y 0 ⁢ μ - ⁢ n 2 + λ - ⁢ n 1 ⁢ ( 1 - y 0 ) ⁢ x 0 .

Proof.

This can be obtained by expanding the determinants with respect to column 7 of M ′′ . ∎

Proposition A.2.

The determinant of M is always positive.

Proof.

This can be proven directly by Lemma A.1 and the fact that

det ⁡ ( U ) < 0 , det ⁡ ( C ) > 0 , d ¯ 66 < 0 , e ¯ 16 < 0 , cofac > 0 . ∎

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Received: 2019-03-09

Revised: 2019-06-15

Accepted: 2019-09-04

Published Online: 2019-09-18

Published in Print: 2020-07-01