The boundary element method discretised with the space-time method for the heat equation in 2D
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The boundary element method discretised with the space-time method - - PowerPoint PPT Presentation
The boundary element method discretised with the space-time method for the heat equation in 2D 1 Kazuki Niino, Olaf Steinbach TU Graz 2 Background Space-time method for a BEM Flexible mesh Stability Large coefficient matrix
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∂u ∂t (x, t) − ∆u(x, t) = 0
u(x, 0) = f(x) u(x, t) = g(x, t)
Ω Γ in
Q := Ω × (0, T)
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+ Z
Ω
K(x − x0, t)f(x0)dx0 where H(t) : Heaviside function
u(x, t) = Z t Z
Γ
⇢ K(x − y, t − τ) ∂u ∂ny (y, τ) − ∂K ∂ny (x − y, t − τ)g(y, τ)
K(x, t) = exp ⇣ − |x|2
4t
⌘ 4πt H(t) x ∈ Ω × (0, T) for
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Ω
K(x − x0, t)f(x0) dx0 Sϕ = Z t Z
Γ
K(x − y, t − τ)ϕ(y, τ)dydτ Dϕ = Z t Z
Γ
∂K ∂ny (x − y, t − τ)ϕ(y, τ)dydτ
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X Y Z
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(S)ij = Z T Z
Γ
dxdt ⇢ φi(x, t) Z t Z
Γ
K(x − y, t − τ)φj(y, τ)dydτ
Z T dxdtφi(x, t) ( 1 2g(x, t) + Z T Z
Γ
∂K ∂ny (x − y, t − τ)g(y, τ)dydτ − Z
Ω
K(x − x0, t)f(x0)dx0 )
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φi : piecewise linear element associated with the triangular mesh
q = X
j
qjφj
(S)ij = Z T Z
Γ
dxdt ⇢ φi(x, t) Z t Z
Γ
K(x − y, t − τ)φj(y, τ)dydτ
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AM −1y = b, x = M −1y M −1Ax = M −1b (Right preconditioning) (Left preconditioning) : Preconditioner
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Dϕ = Z
e Γ
∂K ∂ny (x − y, t − τ)ϕ(y, τ)dS
D∗ϕ = Z
e Γ
∂K ∂nx (x − y, t − τ)ϕ(y, τ)dS
Nϕ = = Z
e Γ
∂2K ∂nx∂ny (x − y, t − τ)ϕ(y, τ)dS
S0ϕ = Z
Ω
K(x − x0, t)ϕ(x0)dx0
D0ϕ = Z
Ω
∂K ∂nx (x − x0, t)ϕ(x0) dx0
X Y Ze Γ = Γ × (0, T)
Sϕ = Z
e Γ
K(x − y, t − τ)ϕ(y, τ)dS
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ψ where
Dϕ = Z
e Γ
∂K ∂ny (x − y, t − τ)ϕ(y, τ)dS
D∗ϕ = Z
e Γ
∂K ∂nx (x − y, t − τ)ϕ(y, τ)dS
Nϕ = = Z
e Γ
∂2K ∂nx∂ny (x − y, t − τ)ϕ(y, τ)dS
S0ϕ = Z
Ω
K(x − x0, t)ϕ(x0)dx0
D0ϕ = Z
Ω
∂K ∂nx (x − x0, t)ϕ(x0) dx0 Sϕ = Z
e Γ
K(x − y, t − τ)ϕ(y, τ)dS
q = 1 2φ + D∗φ − Nψ + D0f
X Y Ze Γ = Γ × (0, T)
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1 2 ✓ Sφ + 1 2ψ − Dψ + S0f ◆ =S ✓1 2φ + D∗φ − Nψ + D0f ◆ − D ✓ Sφ + 1 2ψ − Dψ + S0f ◆ + S0f
0 = (SD∗ − DS)φ + ✓ DD − SN − 1 2I ◆ ψ + ✓1 2S0f + SD0f − DS0f ◆
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(since is a solution of heat eq. with initial function )
S0f
f
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Dϕ = Z
e Γ
∂K ∂ny (x − y, t − τ)ϕ(y, τ)dS
D∗ϕ = Z
e Γ
∂K ∂nx (x − y, t − τ)ϕ(y, τ)dS
Nϕ = = Z
e Γ
∂2K ∂nx∂ny (x − y, t − τ)ϕ(y, τ)dS
Sϕ = Z
e Γ
K(x − y, t − τ)ϕ(y, τ)dS
DD − SN = 1 4I SD∗ − DS = 0 −D∗N + ND = 0 D∗D∗ − NS = 1 4I
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Sq = 1 2u(x, t) + Du − Z
Ω
K(x − x0, t)f(x0) dx0
m=1 φntn(x),
n=1 ψntn(x)
T := Z
e Γ
tm · tndS
Nh
X
n=1
Smnφn ≈ Z
e Γ
tm(x)(Sφ)(x)dSx = Z
e Γ
tm(x)ψ(x)dSx ≈ Z
e Γ
tm(x)
Nh
X
n=1
ψntn(x)dSx =
Nh
X
n=1
Z
e Γ
tm(x)tn(x)dSxψn
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SN = −1 4I + K T 1ST 1N = −1 4I + K0 ST −1NT −1 = −1 4I + K
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(N)ij = Z T Z
Γ
dxdt ⇢ φi(x, t) Z t Z
Γ
∂2K ∂nxny (x − y, t − τ)φj(y, τ)dydτ
Z T Z
Γ
φi(x, t)φj(x, t)dxdt
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Γ
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10−5
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DOF DOF 35 118 1237
114 204 2362 L2 error Calderon 0.0385 0.0117 0.00359 L2 error Jaboci 0.0385 0.0117 0.00360 L2 error No precond. 0.0385 0.0117 0.00359
X Y Z
u(x, 0) = x2
1 − x2 2
u(x, 0) = x2
1 − x2 2
Cylindrical domain
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X Y Z
5 10 15 20 25 30 35 40 45 500 1000 1500 2000 2500 3000
Iteration Numbers Number of nodes
precon3/ precon1/ precon0/
u(x, 0) = x2
1 − x2 2
u(x, 0) = x2
1 − x2 2
Cylindrical domain
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X Z Y Y 1.51
neumann
X Z
u(x, 0) = x2
1 − x2 2
u(x, 0) = x2
1 − x2 2
10 20 30 40 50 60 1000 2000 3000 4000 5000 6000
Iteration Numbers Number of nodes
precon3/ precon1/ precon0/
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