Dirichlet-Neumann and Neumann-Neumann Methods

Dirichlet-Neumann and Neumann-Neumann Methods Felix Kwok Hong Kong Baptist University Introductory Domain Decomposition Short Course DD25, Memorial University of Newfoundland July 22, 2018

Outline Methods Without Overlap Dirichlet Neumann Method Neumann Neumann Method Discrete Formulation

DD methods without overlap Methods that do not require overlap between subdomains: 1. Optimized Schwarz methods (lecture by Laurence Halpern) 2. Dirichlet-Neumann and Neumann-Neumann methods (this lecture)

DD methods without overlap Method converges to monodomain solution if the subdomain solutions have the same function values and derivatives across the interface: u 1 = u 2 u 1 n 1 = u 2 n 2 on Γ on Γ

Outline Methods Without Overlap Dirichlet Neumann Method Neumann Neumann Method Discrete Formulation

Dirichlet Neumann Method (Bjørstad & Widlund 1986) Geometry: assume two non-overlapping subdomains Iterates: solution on the interface u 1 Γ, u2 Γ,... Once interface values converge, solve subdomain problems to get solution everywhere

Dirichlet Neumann Method (Bjørstad & Widlund 1986) Iteration n consists of two substeps: 1. Solve Dirichlet problem on Ω 1, using uγ n as Dirichlet value on Γ u n+1/2 1 = f in Ω 1 u n+1/2 1 = 0 on Ω 1 \ Γ u n+1/2 1 = uγ n on Γ

Dirichlet Neumann Method (Bjørstad & Widlund 1986) Iteration n consists of two substeps: 2. Solve Neumann problem on Ω 2 by matching normal derivatives with u n+1/2 1 : u n+1 2 = f in Ω 2 u n+1 2 = 0 on Ω 2 \ Γ u n+1 2 n 2 = un+1/2 1 n 1 on Γ

Dirichlet Neumann Method (Bjørstad & Widlund 1986) Iteration n consists of two substeps: 2. Solve Neumann problem on Ω 2 by matching normal derivatives with u n+1/2 1 : Update interface trace: u n+1 2 = f in Ω 2 u n+1 2 = 0 on Ω 2 \ Γ u n+1 2 n 2 = un+1/2 1 n 1 on Γ u n+1 Γ = (1 θ)u n Γ + θu n+1 2 Γ, 0 < θ 1

Why does this work? For i = 1, 2, define the Dirichlet-to-Neumann map DtN i (, ) so that DtN i (f, g) = u i n i Γ, where u i satisfies u i = f on Ω i and u i = g on Γ; the Neumann-to-Dirichlet map NtD i (, ) so that NtD i (f, g) = u i Γ, where u i satisfies u i = f on Ω i and u i / n i = g on Γ. Note that for fixed f and i, DtN and NtD are inverses of each other: DtN i (f, NtD i (f, g)) = g, NtD i (f, DtN i (f, g)) = g.

Why does it work? Then the Dirichlet-Neumann (DN) method can be written as u n+1 Γ = (1 θ)u n Γ + θntd 2 (f, DtN 1 (f, u n Γ)). If the method converges, then at the fixed point u Γ = lim n u n Γ, we have u Γ = NtD 2 (f, DtN 1 (f, u Γ )),

Why does it work? Then the Dirichlet-Neumann (DN) method can be written as u n+1 Γ = (1 θ)u n Γ + θntd 2 (f, DtN 1 (f, u n Γ)). If the method converges, then at the fixed point u Γ = lim n uγ n, we have DtN 1 (f, u Γ ) + DtN 2 (f, u Γ ) = 0 ( ) In other words, we have u 1 and u 2 such that u i = f u 1 = u 2 = u Γ u 1 n 1 + u 2 n 2 = 0 on Ω i on Γ on Γ = continuity and smoothness conditions satisfied.

Multiple subdomains Start with one subdomain and solve Dirichlet problem

Multiple subdomains Start with one subdomain and solve Dirichlet problem Solve mixed Dirichlet-Neumann problems on neighbours

Multiple subdomains Start with one subdomain and solve Dirichlet problem Solve mixed Dirichlet-Neumann problems on neighbours Update interface traces

Multiple subdomains Start with one subdomain and solve Dirichlet problem Solve mixed Dirichlet-Neumann problems on neighbours Update interface traces Move on to next set of neighbours, etc.

Multiple subdomains Start with one subdomain and solve Dirichlet problem Solve mixed Dirichlet-Neumann problems on neighbours Update interface traces Move on to next set of neighbours, etc. Many configurations possible!

Convergence Analysis by Fourier Assume 2D rectangular domain ( a, b) (0, π) Two subdomains: Ω 1 = (0, b) (0, π), Ω 2 = ( a, 0) (0, π) Solve u = f with Dirichlet boundary conditions Since problem is linear, assume f = 0 and study how the error e n := u n u tends to zero as n

Convergence Analysis by Fourier Expand eγ n into a Fourier sine series: e n Γ = k 1 ê n Γ(k) sin(ky) Solve e n 1 = 0 on Ω 1 using separation of variables: e n+1/2 1 (x, y) = k 1 A k sin(ky) sinh(k(b x)) e n+1/2 1 (0, y) = A k sin(ky) sinh(kb) = êγ(k) n sin(ky) k 1 k 1 = e n+1/2 1 (x, y) = k 1 sinh(k(b x)) ê n sinh(kb) Γ(k) sin(ky)

Convergence Analysis by Fourier Now solve e n+1 2 = 0 on Ω 2 using separation of variables: e n+1 2 (x, y) = k 1 B k sin(ky) sinh(k(x + a)) e n+1 2 x (0, y) = kb k sin(ky) cosh(ka) k 1 But e n+1/2 1 (0, y) = k cosh(kb) x sinh(kb) ên Γ(k) sin(ky) k 1 = e n+1 2 (x, y) = k 1 cosh(kb) sinh(k(x + a)) ê n cosh(ka) sinh(kb) Γ(k) sin(ky)

Convergence Analysis by Fourier Therefore, e n+1 Γ [ = ê n+1 Γ (k) = = (1 θ)eγ n + θe n+! 2 (0, y) = [ ( cosh(kb) sinh(ka) 1 θ 1 + cosh(ka) sinh(kb) k 1 ( 1 θ 1 + tanh(ka) )] ê n tanh(kb) Γ(k) } {{ } ρ(k,a,b,θ) Contraction factor depends on 1. Frequency k, 2. Geometry, i.e., relative sizes b and a of Ω 1 and Ω 2, 3. Relaxation parameter θ. If ρ(k, a, b, θ) < 1 for all k, then e n Γ 0 as n If ρ(k, a, b, θ) > 1 for some k, then iteration diverges )] ê n Γ(k) sin(ky)

Symmetric Domains (k) = ρ(k, a, b, θ)êγ(k) n ( ρ(k, a, b, θ) = 1 θ 1 + tanh(ka) tanh(kb) ê n+1 Γ ) If Ω 1 and Ω 2 have the same size, i.e., a = b, then ρ(k, a, b, θ) = 1 2θ Convergence independent of frequency k If θ = 1/2, then e n+1 Γ = 0 for all n 0 Exact solution obtained on Γ after one iteration One more subdomain solve to get solution everywhere

; Unsymmetric Domains Example for a = 3, b = 1: 3 2 1 0-1 -2 a = 3, b = 1 3=0.1 3=0.2 3=0.3 3=0.4 3=0.5 3=0.6 3=0.7 3=0.8 3=0.9 3=1-3 0 2 4 6 8 10 k

; Unsymmetric Domains Example for a = 1, b = 3: 3 2 1 0-1 -2 a = 1, b = 3 3=0.1 3=0.2 3=0.3 3=0.4 3=0.5 3=0.6 3=0.7 3=0.8 3=0.9 3=1-3 0 2 4 6 8 10 k

; Unsymmetric Domains Example for a = 1, b = 3: 3 2 1 0-1 -2 a = 1, b = 3 3=0.1 3=0.2 3=0.3 3=0.4 3=0.5 3=0.6 3=0.7 3=0.8 3=0.9 3=1-3 0 2 4 6 8 10 k Want to choose θ to minimize ρ(k, a, b, θ) over all frequencies k, i.e., solve min max ρ(k, a, b, θ), θ k 0 where ( ρ(k, a, b, θ) = 1 θ 1 + tanh(ka) ) tanh(kb)

Unsymmetric Domains ( ρ k = θ sinh(2ka) sinh(2kb) ) { > 0, a > b, = pos.terms 2a 2b < 0, a < b. So ρ(k, a, b, θ) is monotonic in k with lim k 0 ρ = (1 θ) θa b, lim ρ = 1 2θ. k Therefore, the value of θ that minimizes ρ over all k is given by the equioscillation condition lim ρ k 0 = lim ρ k = θ opt = 2b a + 3b, ρ a b opt = a + 3b < 1.

; ; Unsymmetric Domains Example for a = 1, b = 3: θ opt = 0.6, ρ opt = 0.2 1 a = 1, b = 3,3 = 0.6 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 k Example for a = 3, b = 1: θ opt = ρ opt = 1/3 1 a = 3, b = 1,3 = 0.33333 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 k

Outline Methods Without Overlap Dirichlet Neumann Method Neumann Neumann Method Discrete Formulation

Neumann Neumann Method (Bourgat, Glowinski, Le Tallec & Vidrascu 1989) Consider the elliptic problem u = f

Neumann Neumann (NN) Method Given Dirichlet traces g n 1,..., gn j 1 :

Neumann Neumann (NN) Method Given Dirichlet traces g n 1,..., gn j 1 : 1. Solve Dirichlet problems on Ω 1,..., Ω J : u n+1/2 i = f on Ω i u n+1/2 i = g i 1 on Γ i 1,i u n+1/2 i = g i on Γ i,i+1

Neumann Neumann (NN) Method Given Dirichlet traces g n 1,..., gn j 1 : 1. Solve Dirichlet problems on Ω 1,..., Ω J : 2. Calculate jumps in Neumann traces along Γ i 1,i : r n+1/2 i = un+1/2 i n i + un+1/2 i+1 n i+1

Neumann Neumann (NN) Method Given Dirichlet traces g n 1,..., gn j 1 : 3. Solve Neumann problems for corrections ψ n+1 i on Ω 1,..., Ω J : ψ n+1/2 i n i 1 ψ n+1/2 i = 0 on Ω i = r i 1, Γi 1,i ψ n+1/2 i = r n i i. Γi,i+1

Neumann Neumann (NN) Method Given Dirichlet traces g n 1,..., gn j 1 : 3. Solve Neumann problems for corrections ψ n+1 i on Ω 1,..., Ω J : 4. Update Dirichlet traces: g n+1 i = gi n θ(ψ n+1 i Γi + ψ n+1 i+1 Γ i ).

Convergence Analysis Assume two rectangular subdomains again: Analyze how the error goes to zero for the homogeneous problem

Convergence Analysis Fourier analysis gives for the Dirichlet solve e n+1/2 1 (x, y) = k 1 e n+1/2 2 (x, y) = k 1 sinh(k(b x)) ê n sinh(kb) Γ(k) sin(ky) sinh(k(x + a)) ê n sinh(ka) Γ(k) sin(ky)

Convergence Analysis The normal derivative thus becomes e n+1/2 1 (0, y) = k cosh(kb) x sinh(kb) ên Γ(k) sin(ky) k 1 n+1/2 e 2 (0, y) = k cosh(ka) x k 1 = r n+1/2 = k 1 k sinh(ka) ên Γ(k) sin(ky) ( cosh(ka) sinh(ka) + cosh(kb) sinh(kb) ) ê n Γ(k) sin(ky)

Convergence Analysis r n+1/2 = k 1 k The corrections ψ n+1 1, ψ n+1 2 satisfy ( cosh(ka) sinh(ka) + cosh(kb) ) ê n sinh(kb) Γ(k) sin(ky) ψ n+1 1 (x, y) = k 1 A k sinh(k(b x)) sin(ky) ψ n+1 2 (x, y) = k 1 B k sinh(k(x + a)) sin(ky)

Convergence Analysis r n+1/2 = k 1 k The corrections ψ n+1 1, ψ n+1 2 satisfy ( cosh(ka) sinh(ka) + cosh(kb) ) ê n sinh(kb) Γ(k) sin(ky) ψn+1 1 x (0, y) = ka k cosh(kb) sin(ky) = r n+1/2 k 1 ψ n+1 2 x (0, y) = kb k cosh(ka) sin(ky) = r n+1/2 k 1

Convergence Analysis r n+1/2 = k 1 k The corrections ψ n+1 1, ψ n+1 ( cosh(ka) sinh(ka) + cosh(kb) ) ê n sinh(kb) Γ(k) sin(ky) 2 satisfy ψ n+1 1 (x, y) = ( sinh(k(b x)) cosh(ka) cosh(kb) k 1 sinh(ka) + cosh(kb) sinh(kb) ψ n+1 2 (x, y) = ( sinh(k(x + a)) cosh(ka) cosh(ka) sinh(ka) + cosh(kb) sinh(kb) k 1 ) ê n Γ(k) sin(ky) ) ê n Γ(k) sin(ky)

Convergence Analysis ψ n+1 1 (x, y) = ( sinh(k(b x)) cosh(ka) cosh(kb) k 1 sinh(ka) + cosh(kb) sinh(kb) ψ n+1 2 (x, y) = ( sinh(k(x + a)) cosh(ka) cosh(ka) sinh(ka) + cosh(kb) sinh(kb) k 1 ) ê n Γ(k) sin(ky) ) ê n Γ(k) sin(ky) New interface trace satisfies where ρ(k, a, b, θ) = 1 θ e n+1 Γ = e n Γ θ(ψ n+1 1 (0, y) + ψ n+1 2 (0, y)) = k 1 ρ(k, a, b, θ)ê n Γ sin(ky) ( sinh(ka) cosh(ka) + sinh(kb) ) ( cosh(ka) cosh(kb) sinh(ka) + cosh(kb) ) sinh(kb)

Convergence Analysis ( sinh(ka) ρ(k, a, b, θ) = 1 θ cosh(ka) + sinh(kb) ) ( cosh(ka) cosh(kb) sinh(ka) + cosh(kb) ) sinh(kb) Expression symmetric in a and b If a = b, then ρ(k, a, b, θ) = 1 4θ = convergence independent of k = exact convergence after 1 iteration if θ = 1/4

; Unsymmetric Domains Example for a = 3, b = 1: 3 2 1 0-1 -2 a = 3, b = 1 3=0.1 3=0.2 3=0.3 3=0.4 3=0.5-3 0 2 4 6 8 10 k ρ is increasing in k with ( a lim ρ = (1 2θ) θ k 0 b + b ), lim ρ = 1 4θ. a k

; Unsymmetric Domains Equioscillation gives θ opt = 2ab (a + b) 2 + 4ab, ρ (a b) 2 opt = (a + b) 2 + 4ab Example for a = 3, b = 1: θ opt = 3/14, ρ opt = 1/7 1 a = 3, b = 1,3 = 0.21429 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 k

Outline Methods Without Overlap Dirichlet Neumann Method Neumann Neumann Method Discrete Formulation

Discrete subdomain problems Denote the (unknown) Neumann traces by u 1 λ(v) := v = n 1 Γ Γ u 2 n 2 v, for all v V h. Then integration by parts on Ω 1 gives u 1 u 1 v v = fv Ω i Γ n 1 Ω i Similarly, for Ω 2, we have u 1 = f 1 + λ A (2) u 2 = f 2 λ.

Matrix Formulation Partition the problem on Ω 1 into interior and interface unknowns: [ ] (u ) ( ) k IΓ 1I f1i ΓI u k = ΓΓ 1Γ f 1Γ + λ k ;

Matrix Formulation Partition the problem on Ω 1 into interior and interface unknowns: [ ] (u ) ( ) k IΓ 1I f1i ΓI u k = ΓΓ 1Γ f 1Γ + λ k ; Do the same for Ω 2 ; [ ] (u A (2) A (2) ) k IΓ 2I A (2) ΓI A (2) u k = ΓΓ 2Γ ( ) f2i f 2Γ λ k ; This is consistent with the assembled matrix 11 0 IΓ u 1I f 1I 0 A (2) A (2) IΓ u 2I = f 2I. ΓI A (2) ΓI ΓΓ + A(2) u Γ f 1Γ + f 2Γ ΓΓ

Dirichlet-Neumann in Matrix Form Recall subdomain problem on Ω 1 : [ ] (u ) k IΓ 1I ΓI u k = ΓΓ 1Γ ( f1i ) f 1Γ + λ k 1. Solve Dirichlet problem on Ω 1 with u k 1 = gk on Γ: u k 1I + IΓ gk = f 1I u k 1I = ( ) 1 (f 1I IΓ gk )

Dirichlet-Neumann in Matrix Form Recall subdomain problem on Ω 1 : [ ] (u ) k IΓ 1I ΓI u k = ΓΓ 1Γ ( f1i ) f 1Γ + λ k 1. Solve Dirichlet problem on Ω 1 with u k 1 = gk on Γ: 2. Calculate interface condition λ k : u k 1I + IΓ gk = f 1I u k 1I = ( ) 1 (f 1I IΓ gk ) λ k = ΓI u k 1I + ΓΓ gk f 1Γ

Dirichlet-Neumann in Matrix Form 3. Solve Neumann problem on Ω 2 : [ ] (u A (2) A (2) ) k IΓ 2I A (2) ΓI A (2) u k = ΓΓ 2Γ A (2) u k 2I + A (2) IΓ uk 2Γ = f 2I ( f2i ) f 2Γ λ k A (2) ΓI u k 2I + A (2) ΓΓ uk 2Γ = f 2Γ λ k

Dirichlet-Neumann in Matrix Form 3. Solve Neumann problem on Ω 2 : [ ] (u A (2) A (2) ) k IΓ 2I A (2) ΓI A (2) u k = ΓΓ 2Γ ( f2i ) f 2Γ λ k u k 2I = (A (2) ) 1 (f 2I A (2) IΓ uk 2Γ) A (2) ΓI u k 2I + A (2) ΓΓ uk 2Γ = f 2Γ λ k

Dirichlet-Neumann in Matrix Form 3. Solve Neumann problem on Ω 2 : [ ] (u A (2) A (2) ) k IΓ 2I A (2) ΓI A (2) u k = ΓΓ 2Γ ( f2i ) f 2Γ λ k u k 2I = (A (2) ) 1 (f 2I A (2) IΓ uk 2Γ) (A (2) ΓΓ A(2) ΓI (A (2) ) 1 A (2) IΓ }{{} ) S 2 k u 2Γ = f 2Γ A (2) ΓI (A (2) ) 1 f 2I }{{} f2 + f 1 S 1 g k }{{} λ k

Dirichlet-Neumann in Matrix Form g k+1 = [I θs 1 2 (S 1 + S 2 )]g k + θs 1 2 ( f 1 + f 2 ) At convergence, the fixed point g satisfies θs 1 2 (S 1 + S 2 )g = S 1 2 ( f 1 + f 2 ) Thus, DN is an iteration for solving the primal Schur complment problem (S 1 + S 2 )g = f 1 + f 2 with preconditioner M = θ 1 S 2.

Neumann-Neumann in Matrix form By a similar argument, one can show that the Neumann-Neumann method is equivalent to the stationary iteration g k+1 = [I θ(s 1 1 + S 1 2 )(S 1 + S 2 )]g k + θ(s 1 1 + S 1 2 )( f 1 + f 2 ) So we are still solving the primal Schur system (S 1 + S 2 )g = f 1 + f 2, but with the preconditioner M = θ(s 1 1 + S 1 2 ) instead.

Why preconditioning? A fixed point iteration can be accelerated using Krylov methods such as Conjugate Gradient or GMRES (see talk by M.J. Gander) For CG, convergence is determined by the ratio of extreme eigenvalues λ max /λ min For our two-subdomain example with rectangular geometry, it was possible use Fourier to diagonalize A or M 1 A λ min corresponds to low frequency k 0, and λ max to high frequency k = k max = π/h

Why preconditioning? No precond: DN: NN: ( ) 1 λ k (A) = k tanh(ka) + 1 tanh(kb) λ min = 1 a + 1 b, λ max 2k max λ k (M 1 A) = 1 + tanh(ka) tanh(kb) λ min = 2, λ max = 1 + a (a > b) b λ k (M 1 A) = 2 + tanh(ka) tanh(kb) + tanh(kb) tanh(ka) λ min = 4, λ max = 2 + a b + b a Preconditioned methods converge independently of mesh size!

DN and NN as Preconditioners Example: a = 3, b = 1 10 0 DN DN+CG NN 10-2 NN+CG 10-4 10-6 10-8 10-10 2 4 6 8 10 12 14 16 18 20

Dual Schur Problem Instead of using u Γ as primary variable, one could use the Neumann trace λ instead, where [ ] (u1i IΓ ΓI ΓΓ ) = u 1Γ ( ) f1i f 1Γ + λ

Dual Schur Problem Instead of using u Γ as primary variable, one could use the Neumann trace λ instead, where [ ] (u1i ) IΓ ΓI ΓΓ u 1Γ = S 1 u 1Γ = f 1 + λ ( ) f1i f 1Γ + λ

Dual Schur Problem Instead of using u Γ as primary variable, one could use the Neumann trace λ instead, where [ ] (u1i ) IΓ ΓI ΓΓ u 1Γ = u 1Γ = S 1 1 ( f 1 + λ) ( ) f1i f 1Γ + λ

Dual Schur Problem Instead of using u Γ as primary variable, one could use the Neumann trace λ instead, where [ ] (u1i ) IΓ ΓI ΓΓ u 1Γ = u 1Γ = S 1 1 ( f 1 + λ) ( ) f1i f 1Γ + λ u 2Γ = S 1 2 ( f 2 λ)

Dual Schur Problem Instead of using u Γ as primary variable, one could use the Neumann trace λ instead, where [ ] (u1i IΓ ΓI ΓΓ ) = u 1Γ u 1Γ = S 1 1 ( f 1 + λ) ( ) f1i f 1Γ + λ Impose continuity: u 2Γ = S 1 2 ( f 2 λ) S 1 1 ( f 1 + λ) = S 1 2 ( f 2 λ)

Summary 1. DN and NN methods: explicitly match derivatives, work for non-overlapping subdomains 2. Interpretation in terms of DtN operators 3. Analysis for two subdomains using Fourier 4. DN and NN as preconditioners = near grid-independent convergence