Post on 13-Sep-2019
Boundary element solutionof a parabolic equation
based on the Laplace transform
Thorsten Hohage, University of Göttingen, Germany
and
Javier Sayas, University of Zaragoza, Spain
Third International Conference on Boundary Integral Methods:Theory and Applications
Reading, UK, September 14-18, 2004
. – p.1/21
overview
1. problem formulation
2. Laplace transform method
3. boundary integral equations
4. convergence analysis and numerical results
. – p.2/21
problem description
Ω−
Ω+
measurements / heat sources
F. Garrido and A. Salazar. Thermal wave scattering by spheres. J. Appl. Phys,95:140–149, 2004.
A. Ocariz, A. Sánchez-Lavega, and A. Salazar. Photothermal study of subsurfacecylindrical structures. 2. experimental results. J. Appl. Phys, 81:7561–7566, 1997.
J. M. Terrón, A. Salazar, and A. Sánchez-Lavega. General solution for the thermalwave scattering in fiber composites. J. Appl. Phys, 91:1087–1098, 2002.
. – p.3/21
experimental setup
Grupo de Fototermia, Univ. Pais Vasco (Spain)
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mathematical formulation
differential equation: Ω+ = exterior domain, Ω− = interior domainκ± = conductivity/(density · specific heat) in Ω±
∂tu+ = κ+∆u+, in Ω+ × (0,∞),
∂tu− = κ−∆u−, in Ω− × (0,∞)
transmission conditions: uhom = solution without inclusionf proportional to corrosion, unknown of inverse problem!α = interior conductivity/ exterior conductivity
u+ + uhom = u− + f∂νu− on Γ × (0,∞)
∂νu+ + ∂νuhom = α∂νu− on Γ × (0,∞)
boundary condition at top barrier H : ∂νu+|H = 0.
initial condition: u+( · , 0) = 0, u−( · , 0) = 0.
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2. Laplace transform method
. – p.6/21
stable inversion of Laplace transform
The Laplace transform of a continuous, bounded mappingu : [0,∞) → X taking values in a Banach space X is
U(s) =
∫ ∞
0e−stu(t) dt, Re s > 0.
assumption: U has a holomorphic extension U : C \ Σδ,r → X tothe complement of Σδ,r := s ∈ C : | arg(−s)| ≤ δ or |s| ≤ r,δ ∈ (0, π
2 ), r > 0, and there exists M > 0 such that
‖U(s)‖ ≤ M
|s| for all s ∈ C \ Σδ,r.
stable inversion formula: Γ = path connecting −i∞ to i∞
u(t) =1
2πi
∫
ΓetsU(s) ds, t > 0.
. – p.7/21
inversion contours
−1.5 −1 −0.5 0 0.5−3
−2
−1
0
1
2
3
Re(s)
Im(s
)Lopez−F./PalenciaTalbot
. – p.8/21
numerical inversion of Laplace transform
quadrature: For t in a compact interval [t0, T ] ⊂ (0,∞)
1
2πi
∫
ΓetsU(s) ds ≈
N∑
j=−N
γ(N)j es
(N)j tU(x, s
(N)j ) =: uN (t)
Talbot, 1979:
‖uN (t) − u(t)‖ ≤ CM exp(
−c√N
)
, t ∈ [t0, T ] ???
Sheen, Sloan & Thomée, 1999/2003: (applications to FEM)If Ml is a bound on U (l)(s), l ∈ N, then
‖uN (t) − u(t)‖ ≤ ClMlN−l, t ∈ [t0, T ].
López-Fernández & Palencia, 2004:
‖uN (t) − u(t)‖ ≤ CM exp (−cN/ lnN) , t ∈ [t0, T ]. – p.9/21
other boundary integral methods for heat equation
boundary integral formulation using fundamental solution tothe heat equation as kernel.see Costabel ’90, Hsiao ’94, Kress ’89, ...
Lubich & Schneider ’92: operational quadrature methods tosolve the Volterrà–Fredholm integral equations ofconvolution type sequence of boundary integralequations involving Laplace transformed kernels.see also Chapko & Kress ’97
advantages of Laplace transform method:
exponential instead of polynomial order of convergence.
Laplace transformed problems can be solved in parallel.
no need to store previous time-steps.
disadvantage:
convergence only on compact intervals [t0, T ] ⊂ (0,∞).Estimates deteriorate as T/t0 → ∞.
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3. boundary integral equations
. – p.11/21
Laplace transformed transmission problem
differential equation:
κ+∆U+ − sU+ = 0 in Ω+,
κ−∆U− − sU− = 0 in Ω−
“radiation condition”:
U+(s, ·) ∈ H2(Ω+)
transmission/boundary conditions:
U+ + Uhom = U− + f∂νU− on Γ,
∂νU+ + ∂νUhom = α∂νU− on Γ,
∂νU+ = 0 on H.
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ansatz
ansatz:
U+(s, x) =
∫
ΓΦ√
−s/κ+(x, y)ψ+(y) ds(y), x ∈ Ω+
U−(s, x) =
∫
ΓΦ√
−s/κ−
(x, y)ψ−(y) ds(y), x ∈ Ω−
fundamental solution of ∆ + ρ2 in Rd:
Φρ(x, y) :=
ı
4H
(1)0 (ρ|x− y|), when d = 2,
exp(ı ρ |x− y|)4π |x− y| , when d = 3,
fundamental solution of ∆ + ρ2 in half-space: (d = 2)
Φρ((x1, x2), y) := Φρ((x1, x2), y) + Φρ((x1,−x2), y). – p.13/21
boundary integral equations
(12ψ− + J√
−s/κ−
ψ−)f + V√−s/κ−
ψ− −V√−s/κ+
ψ+ = Uhom(s, ·)α
(
12ψ− + J√
−s/κ−
ψ−
)
12ψ+ − J√
−s/κ+ψ+ = ∂νUhom(s, ·)
potential operators:
(Vρψ)(x) :=
∫
ΓΦρ(x, y)ψ(y) ds(y), x ∈ Γ,
(Jρψ)(x) :=
∫
Γ
∂Φρ(x, y)
∂ν(x)ψ(y) ds(y), x ∈ Γ.
assumption: f(x) ≥ f0 > 0 for all x ∈ Γ
uniqueness of transmission problem:
reflection argument (Rapun & Sayas, submitted)
+ uniqueness of transmission problem in Rd (Kress & Roach, ’78)
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self-adjointness
inner product on L2(Ω− × Ω+):〈u, v〉 :=
∫
Ω−
(α/κ−)u− v−dx+∫
Ω+(1/κ+)u+ v+dx
operator:
Au :=
κ−∆u, in Ω−,
κ+∆u, in Ω+
domain of A : D(A) → L2(Ω− × Ω+):
D(A) :=
u ∈ H2(Ω− ∪ Ω+) :u|Γ,− + f ∂νu|Γ,− − u|Γ,+ = 0,α ∂νu|Γ,− − ∂νu|Γ,+ = 0,∂νu|H = 0
.
Theorem: A is self-adjoint, and σ(A) ⊂ (−∞, 0].
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4. convergence analysis and
numerical results
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bounds on boundary integral operators explicit in s
Theorem: Let W (s) : Hν(Γ) ×Hν(Γ) → Hν(Γ) ×Hν(Γ) denotethe matrix of integral operators. Then for all k ∈ −1, 0, 1, . . . there exists a constant C > 0 such that
‖W (s)−1‖k+1/2 ≤ C(1 + |s|)2⌊(k+1)/2⌋+4
dist (s, (−∞, 0])2for all s ∈ C\(−∞, 0]
and
‖W (s)‖±1/2 ≤ C(1 + |s|)2
dist (s, (−∞, 0])for all s ∈ C \ (−∞, 0].
proof. Using standard bounds on resolvent of A . . .
alternative approach by Lubich and Schneider ’92 forV : H−1/2(Γ) → H1/2(Γ) using estimates of kernel andparameter-dependent pseudodifferential calculus.
. – p.17/21
convergence result
space discretization: We consider a Galerkin method withcontinuous, piecewise polynomial functions of degree ≤ k on aquasi-uniform triangulation of Γ with mesh size h.
Theorem: Let uN,h be the numerical approximation of u meshsize h and 2N + 1 quadrature points for the Laplace inversion.Then there exists constants c, C1, C2 > 0 for each compact timeinterval [t0, T ] ⊂ (0,∞) such that
‖u(t) − uN,h(t)‖H1(Ω−∪Ω+) ≤ C1ecN/ ln N + C2h
k+3/2 for all t ∈ [t0, T ],
and on the top barrier
‖u(t) − uN,h(t)‖L∞(H) ≤ C1ecN/ ln N + C2h
2k+2 for all t ∈ [t0, T ].
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numerical tests
source
numerical experiments by Maria-Luisa Rapun
. – p.19/21
convergence as N → ∞
2 2.5 3 3.5 4
2.2
2.4
2.6
2.8
3
3.2
log(N)
log(
−lo
g(E
))
N rel. error5 3.18 e-410 1.37 e-415 1.16 e-520 1.44 e-625 2.55 e-730 1.94 e-835 5.46 e-940 2.93 e-1045 1.18 e-1050 5.32 e-1255 2.98 e-1260 2.46 e-13
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time dependence of the error
0.01 0.1 1 10 100
−15
−10
−5
0
5
10
15
20
25
t
log(
E)
Total error
λ=10λ=1λ=0.1
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