IFE for Stokes interface problem Nabil Chaabane Slimane Adjerid, Tao Lin Virginia Tech SIAM chapter February 4, 24 Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 / 2
Problem statement Consider the Stokes problem where the Stress tensor is S(u, p) = f, in Ω, (a) u =, in Ω, (b) u = g, on Ω, (c) S lj (u, p) = ν(d l u j + D j u l ) pδ lj, l, j =, 2. Ω is occupied by two fluids ν + and ν separated by an interface Γ. The jump conditions are [S(u, p) n] Γ =, (2) [u] Γ =. (3) In this work, we assume that the pressure is continuous. Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 2 / 2
Ω Γ Ω + Ω Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 3 / 2
Ω Γ Ω + Ω Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 4 / 2
IFE basis functions We approximate the interface with a straight line. We map the physical element to the reference element using appropriate mapping. Then, we construct a piecewise vector-valued function ˆΦ s.t: { ˆΦ ˆΦ + if (ˆx, ŷ) ˆT + = ˆΦ if (ˆx, ŷ) ˆT, Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 5 / 2
IFE basis functions ˆΦ (ˆx, ŷ) = ( a + b ˆx + c ŷ + d ˆxŷ a 2 + b 2 ˆx + c 2 ŷ + d 2 ˆxŷ ˆΦ + (ˆx, ŷ) = ( a + + b + ˆx + c+ ŷ + d + ˆxŷ a + 2 + b+ 2 ˆx + c+ 2 ŷ + d + 2 ˆxŷ The IFE basis functions satisfy the following jump constraints ), (4a) ), (4b) ˆΦ (Ê) = ˆΦ + (Ê), ˆΦ j (Â i ) = δ ij, i, j =, 2,..., 8, (5) ˆΦ ( ˆD) = ˆΦ + ( ˆD), 2 ˆΦ ˆx ŷ = 2 ˆΦ + ˆx ŷ, (6) [S( ˆΦ, p) n ]ds =. (7) ˆDÊ ˆDÊ The jump conditions yield 6 constraints that are used to solve for the unknowns. Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 6 / 2
Properties of the bilinear IFE space On every interface element T the IFE basis: are well defined and are unique. are continuous. form a partition of unity i.e 4 Φ i (x, y) = i= ( ), 8 Φ i (x, y) = i=4 ( ), (x, y) t T. Furthermore, If the viscosity parameters have no discontinuity i.e ν + = ν, then Φ i becomes the standard vector-valued bilinear nodal basis Ψ i. If min{ T +, T } shrinks to, then Φ i becomes the standard vector bilinear nodal basis Ψ i. The interpolation error converges with optimal rate. Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 7 / 2
IFE basis functions.8.6.8.4.6.2.4.2.8.6.8.4.6.8.6.4.2.4..2.4.3.8.7.6.5.9.2.2.6.5.5.4.3.2.5...8.8.6.6.4.4.2.2.8.6.8.6.4.4.2.2 Figure: First and second components of the IFE basis functions with ν =, ν + = 5.3 on the left. The first and second components of the standard nodal finite element basis on the right. Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 8 / 2
Interpolation error h u I h u,ω Order u 2 I h u 2,Ω Order.6844e-.6844e- 2 7.926e-2.2478 7.926e-2.2478 4 2.96e-2.267 2.96e-2.267 8 7.4436e-3.995 7.4436e-3.995 6 3.23e-3.245 3.23e-3.245 32.2964e-3.37.2964e-3.37 Table: L 2 error of the interpolation error with ν = and ν + = 5.3 Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 9 / 2
Interpolation error h u (I h u),ω Order u 2 (I h u) 2,Ω Order 4.623525e-3.99738 2.93688e-3.994272 8 2.963685e-4.994828 5.4952759e-4.99794 6 7.38597e-5.9978372.3755382e-4.998957 32.82763e-5.9987745 3.44729e-5.9992 64 4.5693233e-6.999437 8.6426e-6.999637 Table: L 2 error of the interpolation error with ν = and ν + = 5.3 h u (I h u),ω Order u 2 (I h u) 2,Ω Order 4.77835e-.9642986 2.235568e-.97399434 8 5.96752e-2.9893395.27742e-.98693996 6 3.2227e-2.994885 5.6645e-2.9935397 32.558924e-2.99536427 2.83892e-2.9969376 64 7.543788e-3.9977844.42629e-2.9984784 Table: H error of the interpolation error with ν = and ν + = 5.3 Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 / 2
IFE spaces using Q 2 iso Q /Q elements Consider an arbitrary uniform cartesian mesh T h and its refinement T h/2 and define the following on T h/2 { span{φ i (x, y), i =, 2,..., 8}, T is an interface element S h/2 (T ) = span{ψ i (x, y), i =, 2,..., 8}, otherwise, and define S h/2 (Ω) = T Th/2 S h/2 (T ), and S h/2, (Ω) = {v S h/2 (Ω) : v e =, e Eh/2 B \Ei h/2 }, where Eh/2 i is the set of interface edges. On the mesh T h, define M h (Ω) = {q C L 2 : q T Q }. Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 / 2
IFE formulation The finite element formulation is: Find (u h, p h ) S h (Ω) M h (Ω) s.t 2 S(u h, p h ) : v h dx = f v h dx, v h S h,(ω) (8a) 2 Ω Ω u h q h dx =, q h M h (Ω), (8b) Ω Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 2 / 2
Numerical results The domain Ω is the square [, ] 2, the interface is a half-circle with radius r =.3, and center x =.4327, y =.329 that separates Ω into two regions Ω + = {(x, y) t : (x x ) 2 + (y y ) 2 > r 2 } and Ω = {(x, y) t : (x x ) 2 + (y y ) 2 < r 2 }. The function u is defined as { u + u + (x, y) = = ((x x ) 2 + (y y ) 2 r 2 )(y y ), u 2 + = ((x x ) 2 + (y y ) 2 r 2, (9a) )(x x ), u = ν+ ν u+. (9b) Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 3 / 2
Numerical results h u u,h,ω u 2 u 2,h,Ω p p h,ω 4.5992748e-3 6.9526529e-3.7285727e- 2.345294e-3.746736e-3 8.2775e-2 4 4.22835e-4 4.73392e-4 4.89925e-2 8.45795e-4.559823e-4 3.387833e-2 6 5.5487582e-5 6.737782e-5 2.97387e-2 32 2.349e-5 3.959e-5.577559e-2 Table: L 2 errors with ν = and ν + = 5.3 for the non-penalized method. u u,h,ω Ch.587, u 2 u 2,h,Ω Ch.6484, p p h,ω Ch.7582. Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 4 / 2
.2.8.6.4.2.2.8.6.4.2.8.6.4.2.2.4.6.8 Figure: Standard Lagrange finite element basis Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 5 / 2
.2.8.6.4.2.2.8.6.4.2.8.6.4.2.2.4.6.8 Figure: Immersed finite element basis Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 6 / 2
IFE formulation The finite element formulation is: Find (u h, p h ) S h (Ω) M h (Ω) s.t 2 S(u h, p h ) : v h dx Ω e Eh i + e E i h \EB h e e E i h \EB h [u h ] [v h ]ds = Ω Ω e {(S(u h, p h ) n)} [v h ]ds e {(S(v h, ) n)} [u h ]ds f v h dx, (a) v h S h,(ω) (b) 2 u h q h dx =, q h M h (Ω), (c) Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 7 / 2
Numerical results The domain Ω is the square [, ] 2, the interface is a half-circle with radius r =.3, and center x =.4327, y =.329 that separates Ω into two regions Ω + = {(x, y) t : (x x ) 2 + (y y ) 2 > r 2 } and Ω = {(x, y) t : (x x ) 2 + (y y ) 2 < r 2 }. The function u is defined as { u + u + (x, y) = = ((x x ) 2 + (y y ) 2 r 2 )(y y ), u 2 + = ((x x ) 2 + (y y ) 2 r 2, (a) )(x x ), u = ν+ ν u+. (b) Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 8 / 2
Numerical results h u u,h,ω Order u 2 u 2,h,Ω Order.592e-2 2.726e-2 2 9.6973e-4.98.7357e-3.99 4 2.4288e-4.99 4.346e-4.99 8 6.722e-5.99.9e-4.99 6.577e-5 2. 2.7225e-5 2. 32 3.7986e-6.99 6.894e-6.99 Table: L 2 error and order of convergence with ν = and ν + = 5.3 Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 9 / 2
Numerical results h p p h,ω Order 5.363e- 2 4.53e-2.76 4.4236e-2.49 8 9.7344e-3.54 6 2.9e-3 2.2 32 7.644e-4.45 Table: L 2 error and order of convergence with ν = and ν + = 5.3 Nabil Chaabane (VT) IFE for Stokes interface problem February 4, 24 2 / 2
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